SOA exam FM CAS

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Chapter 1. Basic Interest Theory.

Manual for SOA Exam FM/CAS Exam 2.Chapter 1. Basic Interest Theory.

Section 1.1. Amount and accumulation functions.

c©2009. Miguel A. Arcones. All rights reserved.

Extract from:”Arcones’ Manual for the SOA Exam FM/CAS Exam 2,

Financial Mathematics. Fall 2009 Edition”,available at http://www.actexmadriver.com/

c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2.

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Chapter 1. Basic Interest Theory. Section 1.1. Amount and accumulation functions.

Interest

I When money is invested or loaned the amount of moneyreturned is different from the initial one.

I The amount of money invested (or loaned) is called theprincipal or principle.

I The amount of interest earned during a period of time is

I = final balance− invested amount.

I The effective rate of interest earned in the period [s, t] is

final balance− invested amountinvested amount

.


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Interest







.


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Interest







.


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Interest







.


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Example 1

Simon invests $1000 in a bank account. Six months later, theamount in his bank account is $1049.23.(i) Find the amount of interest earned by Simon in those 6 months.(ii) Find the (semiannual) effective rate of interest earned in those6 months.

Solution: (i) The amount of interest earned by Simon in those 6months is I = 1049.23− 1000 = 49.23.(ii) The (semiannual) effective rate of interest earned is

1049.23− 1000

1000= 0.004923 = 0.4923%.


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Example 1


Solution: (i) The amount of interest earned by Simon in those 6months is I = 1049.23− 1000 = 49.23.

(ii) The (semiannual) effective rate of interest earned is

1049.23− 1000

1000= 0.004923 = 0.4923%.


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Example 1


Solution: (i) The amount of interest earned by Simon in those 6months is I = 1049.23− 1000 = 49.23.(ii) The (semiannual) effective rate of interest earned is

1049.23− 1000

1000= 0.004923 = 0.4923%.


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Amount function

Suppose that an amount A(0) of money is invested at time 0.A(0) is the principal. Let A(t) denote the value at time t of theinitial investment A(0). The function A(t), t ≥ 0, is called theamount function. Usually, we assume that the amount functionsatisfies the following properties:(i) For each t ≥ 0, A(t) > 0.(ii) A is nondecreasing.

In this situation,I The amount of interest earned over the period [s, t] is

A(t)− A(s).


A(t)− A(s)

A(s).


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Amount function

Suppose that an amount A(0) of money is invested at time 0.A(0) is the principal. Let A(t) denote the value at time t of theinitial investment A(0). The function A(t), t ≥ 0, is called theamount function. Usually, we assume that the amount functionsatisfies the following properties:(i) For each t ≥ 0, A(t) > 0.(ii) A is nondecreasing.In this situation,

I The amount of interest earned over the period [s, t] is

A(t)− A(s).


A(t)− A(s)

A(s).


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Amount function

Suppose that an amount A(0) of money is invested at time 0.A(0) is the principal. Let A(t) denote the value at time t of theinitial investment A(0). The function A(t), t ≥ 0, is called theamount function. Usually, we assume that the amount functionsatisfies the following properties:(i) For each t ≥ 0, A(t) > 0.(ii) A is nondecreasing.In this situation,

I The amount of interest earned over the period [s, t] is

A(t)− A(s).


A(t)− A(s)

A(s).


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Example 2

Jessica invests $5000 on March 1, 2008, in a fund which followsthe accumulation function A(t) = (5000)

(1 + t

40

), where t is the

number of years after March 1, 2008.(i) Find the balance in Jessica’s account on October 1, 2008.(ii) Find the amount of interest earned in those 7 months.(iii) Find the effective rate of interest earned in that period.

Solution: (i) The balance of Jessica’s account on 10–1–2008 is

A(7/12) = (5000)

(1 +

7/12

40

)= 5072.917.

(ii) The amount of interest earned in those 7 months is

A(7/12)− A(0) = 5072.917− 5000 = 72.917.

(iii) The effective rate of interest earned in that period is

A(7/12)− A(0)

A(0)=

72.917

5000= 0.0145834 = 1.45834%.


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Example 2


(1 + t

40

), where t is the



A(7/12) = (5000)

(1 +

7/12

40

)= 5072.917.


A(7/12)− A(0) = 5072.917− 5000 = 72.917.


A(7/12)− A(0)

A(0)=

72.917

5000= 0.0145834 = 1.45834%.


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Example 2


(1 + t

40

), where t is the



A(7/12) = (5000)

(1 +

7/12

40

)= 5072.917.


A(7/12)− A(0) = 5072.917− 5000 = 72.917.


A(7/12)− A(0)

A(0)=

72.917

5000= 0.0145834 = 1.45834%.


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Example 2


(1 + t

40

), where t is the



A(7/12) = (5000)

(1 +

7/12

40

)= 5072.917.


A(7/12)− A(0) = 5072.917− 5000 = 72.917.


A(7/12)− A(0)

A(0)=

72.917

5000= 0.0145834 = 1.45834%.


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Cashflows

Often, we consider the case when several deposits/withdrawals aremade into an account following certain amount function. A seriesof (deposits/withdrawals) payments made at different times iscalled a cashflow. The payments can be either made by theindividual or to the individual. An inflow is payment to theindividual. An outflow is a payment by the individual. Werepresent inflows by positive numbers and outflows by negativenumbers. In a cashflow, we have a contribution of Cj at time tj ,for each j = 1, . . . , n. Cj can be either positive or negative. Wecan represent a cashflow in a table:

Investments C1 C2 · · · Cn

Time (in years) t1 t2 · · · tn


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Cashflow rules

Rule 1: Proportionality. If an investment strategy follows theamount function A(t), t > 0, an investment of $k made at time 0

with the previous investment strategy, has a value of $kA(t)A(0) at time

t.

Using the amount function A(·) and proportionality:

I Investing A(0) at time zero, we get A(t) at time t.

I Investing 1 at time zero, we get A(t)A(0) at time t.

I Investing k at time zero, we get kA(t)A(0) at time t.


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Cashflow rules



t.Using the amount function A(·) and proportionality:





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Cashflow rules








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Cashflow rules








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Cashflow rules








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Present value

Definition 1The present value at certain time of a cashflow is the amount ofthe money which need to invest at certain time in other to get thesame balance as that obtained from a cashflow.

Since investing k at time zero, we get kA(t)A(0) at time t, we have

that: the present value at time t of a deposit of k made at timezero is kA(t)

A(0) .

Let x be the amount which need to invest at time zero to get abalance of k at time t. We have that k = xA(t)

A(0) . So, x = kA(0)A(t) .

Hence, the present value at time 0 of a balance of k had at time tis kA(0)

A(t) .


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Present value




A(0) .


A(0) . So, x = kA(0)A(t) .


A(t) .


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Present value




A(0) .


A(0) . So, x = kA(0)A(t) .


A(t) .


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To know how the value of money changes over time we need to seehow the value of $1 varies over time. The accumulation functiona(t), t ≥ 0, is defined as the value at time t of $1 invested at time0.

By proportionality, a(t) = A(t)A(0) . Observe that a(0) = 1.

Knowing the value function a(t) and the principal A(0), we canfind the amount function A(t) using the formula A(t) = A(0)a(t).Using the accumulation function a(t), t ≥ 0, we have:

I The present value at time t of a deposit of k made at timezero is ka(t) (= kA(t)

A(0) ).

I The present value at time 0 of a balance of k had at time t isk

a(t) (= kA(0)A(t) ).


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To know how the value of money changes over time we need to seehow the value of $1 varies over time. The accumulation functiona(t), t ≥ 0, is defined as the value at time t of $1 invested at time0.By proportionality, a(t) = A(t)

A(0) . Observe that a(0) = 1.



A(0) ).


a(t) (= kA(0)A(t) ).


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Knowing the value function a(t) and the principal A(0), we canfind the amount function A(t) using the formula A(t) = A(0)a(t).

Using the accumulation function a(t), t ≥ 0, we have:


A(0) ).


a(t) (= kA(0)A(t) ).


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A(0) ).


a(t) (= kA(0)A(t) ).


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A(0) ).


a(t) (= kA(0)A(t) ).


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Example 1

The accumulation function of a fund is a(t) = (1.03)2t , t ≥ 0.(i) Amanda invests $5000 at time zero in this fund. Find thebalance into Amanda’s fund at time 2.5 years.(ii) How much money does Kevin need to invest into the fund attime 0 to accumulate $10000 at time 3?

Solution: (i) The balance into Amanda’s fund at time 2.5 years is

ka(2.5) = (5000)(1.03)2(2.5) = 5796.370371.

(ii) The amount which Kevin needs to invest at time 0 toaccumulate $10000 at time 3 is

10000

a(3)=

10000

(1.03)2(3)= 8374.842567.


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Example 1



ka(2.5) = (5000)(1.03)2(2.5) = 5796.370371.


10000

a(3)=

10000

(1.03)2(3)= 8374.842567.


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Example 1



ka(2.5) = (5000)(1.03)2(2.5) = 5796.370371.


10000

a(3)=

10000

(1.03)2(3)= 8374.842567.


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Cashflow rules

Rule 2. Grows–depends–on–balance rule. If an investmentfollows the amount function A(t), t ≥ 0, the growth during certainperiod where no deposits/withdrawals are made depends on thebalance on the account at the beginning of the period.

If an account has a balance of k at time t and nodeposits/withdrawals are made in the future, then the futurebalance in this account does not depend on how the balance of kat time t was attained.In particular, the following two accounts have the same balance fortimes bigger than t:1. An account where a unique deposit of k is made at time t.2. An account where a unique deposit of k

A(t) is made at time zero.


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Cashflow rules

Rule 2. Grows–depends–on–balance rule. If an investmentfollows the amount function A(t), t ≥ 0, the growth during certainperiod where no deposits/withdrawals are made depends on thebalance on the account at the beginning of the period.If an account has a balance of k at time t and nodeposits/withdrawals are made in the future, then the futurebalance in this account does not depend on how the balance of kat time t was attained.

In particular, the following two accounts have the same balance fortimes bigger than t:1. An account where a unique deposit of k is made at time t.2. An account where a unique deposit of k



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Cashflow rules

Rule 2. Grows–depends–on–balance rule. If an investmentfollows the amount function A(t), t ≥ 0, the growth during certainperiod where no deposits/withdrawals are made depends on thebalance on the account at the beginning of the period.If an account has a balance of k at time t and nodeposits/withdrawals are made in the future, then the futurebalance in this account does not depend on how the balance of kat time t was attained.In particular, the following two accounts have the same balance fortimes bigger than t:1. An account where a unique deposit of k is made at time t.2. An account where a unique deposit of k



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Theorem 1If an investment follows the amount function A(·), the present

value at time t of a deposit of $k made at time s is $kA(t)A(s) = ka(t)

a(s) .

Proof. We need to invest kA(s) at time 0 to get a balance of k at

time s. So, investing k at time s is equivalent to investing kA(s) at

time 0. The future value at time t of an investment of kA(s) at time

0 is kA(t)A(s) . Hence, investing k at time s is equivalent to investing

kA(t)A(s) at time t.


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Theorem 1If an investment follows the amount function A(·), the present

value at time t of a deposit of $k made at time s is $kA(t)A(s) = ka(t)

a(s) .

Proof. We need to invest kA(s) at time 0 to get a balance of k at

time s. So, investing k at time s is equivalent to investing kA(s) at

time 0. The future value at time t of an investment of kA(s) at time

0 is kA(t)A(s) . Hence, investing k at time s is equivalent to investing

kA(t)A(s) at time t.


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Another way to see the previous theorem is as follows. Thefollowing three accounts have the same balance at any time biggerthan t:1. An account where a unique deposit of A(0) is made at timezero.2. An account where a unique deposit of A(s) is made at time s.3. An account where a unique deposit of A(t) is made at time t.

Hence,

The present value at time t of an investment of A(s) made at times is A(t).

This means that:

I If t > s, an investment of A(s) made at time s has anaccumulation value of A(t) at time t.

I If t < s, to get an accumulation of A(s) at time s, we need toinvest A(t) at time t.


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Another way to see the previous theorem is as follows. Thefollowing three accounts have the same balance at any time biggerthan t:1. An account where a unique deposit of A(0) is made at timezero.2. An account where a unique deposit of A(s) is made at time s.3. An account where a unique deposit of A(t) is made at time t.Hence,


This means that:




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Another way to see the previous theorem is as follows. Thefollowing three accounts have the same balance at any time biggerthan t:1. An account where a unique deposit of A(0) is made at timezero.2. An account where a unique deposit of A(s) is made at time s.3. An account where a unique deposit of A(t) is made at time t.Hence,


This means that:




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We know that:The present value at time t of an investment of A(s) made at times is A(t), i.e.

A(s) at time s is equivalent to A(t) at time t.

By proportionality,

I The present value at time t of an investment of 1 made attime s is A(t)

A(s) , i.e.

1 at time s is equivalent to A(t)A(s) at time t.

I The present value at time t of an investment of k made attime s is kA(t)

A(s) , i.e.

k at time s is equivalent to kA(t)A(s) at time t.


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By proportionality,


A(s) , i.e.



A(s) , i.e.



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By proportionality,


A(s) , i.e.



A(s) , i.e.



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Example 2

The accumulation function of a fund follows the functiona(t) = 1 + t

20 , t > 0.(i) Michael invests $3500 into the fund at time 1. Find the valueof Michael’s fund account at time 4.(ii) How much money needs Jason to invest at time 2 toaccumulate $700 at time 4.

Solution: (i) The value of Michael’s account at time 4 is

3500a(4)a(1) = (3500)

1+ 420

1+ 120

= (3500)1.201.05 = 4000.

(ii) To accumulate $700 at time 4, Jason needs to invest at time 2,

700a(2)a(4) = 7001.1

1.2 = 641.67.


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Example 2




3500a(4)a(1) = (3500)

1+ 420

1+ 120

= (3500)1.201.05 = 4000.


700a(2)a(4) = 7001.1

1.2 = 641.67.


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Example 2




3500a(4)a(1) = (3500)

1+ 420

1+ 120

= (3500)1.201.05 = 4000.


700a(2)a(4) = 7001.1

1.2 = 641.67.


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Theorem 3Present value of a cashflow. If an investment account followsthe amount function A(t), t > 0, the (equation of value) presentvalue at time t of the cashflow

Deposits C1 C2 · · · Cn

Time t1 t2 · · · tn

where 0 ≤ t1 < t2 < · · · < tn, is

V (t) =n∑

j=1

CjA(t)

A(tj).


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Proof. Let s > tn.

Time Balance before deposit Balance after deposit

t1 0 C1

t2 C1a(t2)a(t1)

=∑1

j=1 Cja(t2)a(tj )

C1a(t2)a(t1)

+ C2 =∑2

j=1 Cja(t2)a(tj )

t3∑2

j=1 Cja(t3)a(tj )

∑3j=1 Cj

a(t3)a(tj )

t4∑3

j=1 Cja(t4)a(tj )

∑4j=1 Cj

a(t4)a(tj )

· · · · · · · · ·tn

∑n−1j=1 Cj

a(tn)a(tj )

∑nj=1 Cj

a(tn)a(tj )

Since the balance after deposit at time t1 is C1, the balance beforedeposit at time t2 is a(t2)

a(t1)C1.

Since the balance after deposit at time t2 is∑2

j=1 Cja(t2)a(tj )

, the

balance before deposit at time t3 isa(t3)a(t2)

∑2j=1 Cj

a(t2)a(tj )

=∑2

j=1 Cja(t3)a(tj )

.


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Hence, the balance at time s is

a(s)

a(tn)

n∑j=1

Cja(tn)

a(tj)=

n∑j=1

Cja(s)

a(tj).

The future value of the cashflow at time t is

a(t)

a(s)

n∑j=1

Cja(s)

a(tj)=

n∑j=1

Cja(t)

a(tj).

end of proof.


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Notice that the present value at time t of the cashflow



is the same as the sum of the present values at time t of nseparated investment accounts each following the amount functionA, with the j–the account having a unique deposit of Cj at time tj .


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Example 4


20 , t > 0. Jared invests $1000 into the fund at time 1and he withdraws $500 at time 3. Find the value of Jared’s fundaccount at time 5.

Solution: The cashflow is

deposit/withdrawal 1000 −500

Time (in years) 1 3.

The value of Jared’s account at time 5 is

1000a(5)

a(1)− 500

a(5)

a(3)= 1000

1 + 520

1 + 120

− 5001 + 5

20

1 + 320

=(1000)1.25

1.05− (500)

1.25

1.15= 1190.48− 543.48 = 647.00.


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Example 4


20 , t > 0. Jared invests $1000 into the fund at time 1and he withdraws $500 at time 3. Find the value of Jared’s fundaccount at time 5.



Time (in years) 1 3.

The value of Jared’s account at time 5 is

1000a(5)

a(1)− 500

a(5)

a(3)= 1000

1 + 520

1 + 120

− 5001 + 5

20

1 + 320

=(1000)1.25

1.05− (500)

1.25

1.15= 1190.48− 543.48 = 647.00.


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Section 1.2. Simple interest.





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Chapter 1. Basic Interest Theory. Section 1.2. Simple interest.

Simple interest

Under simple interest:

I the interest paid over certain period of time is proportional tothe length of this period of time and the principal.

I if i is the effective annual rate of simple interest, the amountof interest earned by a deposit of k held for t years is kit. Thebalance at time t years is k + kit = k(1 + it).

I interest is found using the principal not the earned interest.To find the earned interest, we need to know the amount ofprincipal, not the balance.

I balances under simple interest follow the proportionality ruleand rule about the addition of several deposits/withdrawals.However, the rule ”grows–depends–on–balance” does nothold.


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Simple interest







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Simple interest







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Simple interest







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Suppose that an account earns simple interest with annualeffective rate of i .

I If an investment of 1 is made at time zero, then the balancein this account at time t years is a(t) = 1 + it .

I If an investment of k is made at time zero, then the balancein this account at time t years is k(1 + it).

I If an investment of k is made at time s years, then thebalance in this account at time t years, t > s, isk(1 + i(t − s)). Notice that the investment is held for t − syears, and the earned interest is ki(t − s).


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Notice that the amount k(1 + i(t − s)) is not ka(t)a(s) = k(1+it)

(1+is) .

Making an investment of k(1+is) at time zero, we have a balance of

k(1+is)(1 + is) = k at time s. Making an investment of k

(1+is) at

time zero, we have a balance of k(1+is)(1 + it) at time t. This is

not the balance at time t years in an account with an investmentof k made at time s years.

I Making an investment of k(1+is) at time zero, we have a

balance of k(1+is)(1 + is) = k at time s. But since interest

does not earn interest, the amount of interest earned in theperiod [s, t] is k

(1+is) i(t − s). Hence, the balance at time t is

k +k

(1 + is)i(t − s) =

k(1 + is) + ki(t − s)

(1 + is)=

k(1 + it)

(1 + is).

I Making an investment of k at time s years, we have a balanceof k(1 + i(t − s)) at time t.


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Present value for simple interest

I A deposit of k made at time s has a future value ofk(1 + i(t − s)) at time t, if t > s.

I To get a balance of k time s, we need to make a deposit ofk 1

1+i(s−t) at time t, if t < s.


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I A deposit of k made at time s has a future value ofk(1 + i(t − s)) at time t, if t > s.

I To get a balance of k time s, we need to make a deposit ofk 1

1+i(s−t) at time t, if t < s.


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Theorem 1If deposits/withdrawals are make according with the table,



where 0 ≤ t1 < t2 < · · · < tn to an account earning simple interestwith annual effective rate of i , then the balance at time t years,where t > tn, is given by

B =n∑

j=1

Cj(1 + i(t − tj)) =n∑

j=1

Cj +n∑

j=1

Cj i(t − tj).


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Proof.

Time Deposit/withdr. Principal Amount of interestat that time after the deposit earned up

to that time

t1 C1 C1 0t2 C2 C1 + C2 C1i(t2 − t1)

t3 C3∑3

j=1 Cj∑2

j=1 Cj i(t3 − tj)

· · · · · · · · · · · ·tk Ck

∑kj=1 Cj

∑k−1j=1 Cj i(tk − tj)

· · · · · · · · · · · ·tn Cn

∑nj=1 Cj

∑n−1j=1 Cj i(tn − tj)

t 0∑n

j=1 Cj∑n

j=1 Cj i(t − tj)


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The amount of interest earned up to time t3 is

C1i(t2 − t1) + (C1 + C2)i(t3 − t2) = C1i(t3 − t1) + C2i(t3 − t2)

=2∑

j=1

Cj i(t3 − tj).

The amount of interest earned up to time tk is the amount ofinterest earned up to time tk−1 plus the amount of interest earnedin the period [tk−1, tk ], which is

k−2∑j=1

Cj i(tk−1 − tj) +k−1∑j=1

Cj i(tk − tk−1)

=k−1∑j=1

Cj i(tk−1 − tj) +k−1∑j=1

Cj i(tk − tk−1)

=k−1∑j=1

Cj i(tk − tj).


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Theorem 2If deposits/withdrawals are make according with the table,



where 0 ≤ t1 < t2 < · · · < tn to an account earning simple interestand the balance at time t years, where t > tn, is B, then theannual effective rate of i

i =B −

∑nj=1 Cj∑n

j=1 Cj(t − tj).

Proof. Solving for i in B =∑n

j=1 Cj +∑n

j=1 Cj i(t − tj), we get thevalue of i .


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In the formula,

i =B −

∑nj=1 Cj∑n

j=1 Cj(t − tj),

B −∑n

j=1 Cj is the total amount of interest earned,∑nj=1 Cj(t − tj) is the sum of the balances times the amount

balances are in the account.


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Example 3

Jeremy invests $1000 into a bank account which pays simpleinterest with an annual rate of 7%. Nine months later, Jeremywithdraws $600 from the account. Find the balance in Jeremy’saccount one year after the first deposit was made.

Solution: The cashflow of deposits is


Time (in years) 0 0.75.

The balance one year after the first deposit was made is

n∑j=1

Cj(1 + i(t − tj))

=(1000)(1 + (1− 0)(0.07)) + (−600)(1 + (1− 0.75)(0.07))

=459.5.


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Example 3

Jeremy invests $1000 into a bank account which pays simpleinterest with an annual rate of 7%. Nine months later, Jeremywithdraws $600 from the account. Find the balance in Jeremy’saccount one year after the first deposit was made.

Solution: The cashflow of deposits is


Time (in years) 0 0.75.


n∑j=1

Cj(1 + i(t − tj))

=(1000)(1 + (1− 0)(0.07)) + (−600)(1 + (1− 0.75)(0.07))

=459.5.


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Time Deposit Principal Amountmade after deposit of interest

at this time earned inthe last period

0 1000 1000 00.75 −600 400 (1000)(0.07)(0.75) = 52.5

1 0 400 (400)(0.07)(1− 0.75) = 7


400 + 52.5 + 7 = 459.5.


20/21


Example 4

On September 1, 2006, John invested $25000 into a bank accountwhich pays simple interest. On March 1, 2007, John’s wife made awithdrawal of 5000. The accumulated value of the bank accounton July 1, 2007 was $20575. Calculate the annual effective rate ofinterest earned by this account.

Solution: Let September 1, 2006 be time 0. Then, March 1, 2007is time 6

12 years; and July 1, 2007 is time 1012 years. The annual

effective rate of interest earned by this account is

i =B −

∑nj=1 Cj∑n

j=1 Cj(t − tj)=

20575− 25000 + 5000

25000(

1012

)− 5000

(1012 −

612

)=

575

19166.66667= 3%.


21/21


Example 4

On September 1, 2006, John invested $25000 into a bank accountwhich pays simple interest. On March 1, 2007, John’s wife made awithdrawal of 5000. The accumulated value of the bank accounton July 1, 2007 was $20575. Calculate the annual effective rate ofinterest earned by this account.

Solution: Let September 1, 2006 be time 0. Then, March 1, 2007is time 6

12 years; and July 1, 2007 is time 1012 years. The annual

effective rate of interest earned by this account is

i =B −

∑nj=1 Cj∑n

j=1 Cj(t − tj)=

20575− 25000 + 5000

25000(

1012

)− 5000

(1012 −

612

)=

575

19166.66667= 3%.


1/27


Manual for SOA Exam FM/CAS Exam 2.Chapter 1. Basic Interest Theory.Section 1.3. Compounded interest.





2/27

Chapter 1. Basic Interest Theory. Section 1.3. Compounded interest.

Compound interest

Under compound interest the amount function is

A(t) = A(0)(1 + i)t , t ≥ 0,

where i is the effective annual rate of interest.

Under compound interest, the effective rate of interest over acertain period of time depends only on the length of this period, i.e.

for each 0 ≤ s < t,A(t)− A(s)

A(s)=

A(t − s)− A(0)

A(0).

Notice that

A(t)− A(s)

A(s)=

A(0)(1 + i)t − A(0)(1 + i)s

A(0)(1 + i)s= (1 + i)t−s − 1.

The effective rate of interest earned in the n–th year is

in =A(n)− A(n − 1)

A(n − 1)=

A(0)(1 + i)n − A(0)(1 + i)n−1

A(0)(1 + i)n−1= i .


3/27


Compound interest

Under compound interest the amount function is

A(t) = A(0)(1 + i)t , t ≥ 0,

where i is the effective annual rate of interest.Under compound interest, the effective rate of interest over acertain period of time depends only on the length of this period, i.e.

for each 0 ≤ s < t,A(t)− A(s)

A(s)=

A(t − s)− A(0)

A(0).

Notice that

A(t)− A(s)

A(s)=

A(0)(1 + i)t − A(0)(1 + i)s

A(0)(1 + i)s= (1 + i)t−s − 1.

The effective rate of interest earned in the n–th year is

in =A(n)− A(n − 1)

A(n − 1)=

A(0)(1 + i)n − A(0)(1 + i)n−1

A(0)(1 + i)n−1= i .


4/27


Under compound interest, the present value at time t of a depositof k made at time s is

kA(t)

A(s)=

kA(0)(1 + i)t

A(0)(1 + i)s= k(1 + i)t−s .

If deposits/withdrawals are made according with the table



where 0 ≤ t1 < t2 < · · · < tn, into an account earning compoundinterest with an annual effective rate of interest of i , then thepresent value at time t of the cashflow is

V (t) =n∑

j=1

Cj(1 + i)t−tj .

In particular, the present value of the considered cashflow at timezero is

∑nj=1 Cj(1 + i)−tj .


5/27


Under compound interest, the present value at time t of a depositof k made at time s is

kA(t)

A(s)=

kA(0)(1 + i)t

A(0)(1 + i)s= k(1 + i)t−s .

If deposits/withdrawals are made according with the table



where 0 ≤ t1 < t2 < · · · < tn, into an account earning compoundinterest with an annual effective rate of interest of i , then thepresent value at time t of the cashflow is

V (t) =n∑

j=1

Cj(1 + i)t−tj .

In particular, the present value of the considered cashflow at timezero is

∑nj=1 Cj(1 + i)−tj .


6/27


Example 1

A loan with an effective annual interest rate of 5.5% is to berepaid with the following payments:(i) 1000 at the end of the first year.(ii) 2000 at the end of the second year.(iii) 5000 at the end of the third year.Calculate the loaned amount at time 0.

Solution: The cashflow of payments to the loan is

Payments 1000 2000 5000

Time 1 2 3

The loaned amount at time zero is the present value at time zeroof the cashflow of payments, which is

(1000)(1.055)−1 + (2000)(1.055)−2 + (5000)(1.055)−3

=947.8672986 + 1796.904831 + 4258.068321 = 7002.840451.


7/27


Example 1



Payments 1000 2000 5000

Time 1 2 3


(1000)(1.055)−1 + (2000)(1.055)−2 + (5000)(1.055)−3

=947.8672986 + 1796.904831 + 4258.068321 = 7002.840451.


8/27


The accumulation function for simple interest is a(t) = 1 + it,which is a linear function.The accumulation function for compound interest isa(t) = (1 + i)t , which is an increasing convex function.We have that(i) If 0 < t < 1, then (1 + i)t < 1 + it.(ii) If 1 < t, then 1 + it < (1 + i)t .


9/27


Figure 1: comparison of simple and compound accumulation functions


10/27


Usually, we solve for variables in the formula, A(t) = A(0)(1 + i)t ,using the TI–BA–II-Plus calculator.

To turn on the calculator press ON/OFF .

To clear errors press CE/C . It clears the current displays

(including error messages) and tentative operations.When entering a number, you realized that you make a mistake

you can clear the whole display by pressing CE/C .

When entering numbers, if you would like to save some of theentered digits, you can press → as many times as digits youwould like to remove. Digits are deleted starting from the lastentered digit.It is recommended to set–up the TI-BA–II–Plus calculator to 9decimals. You can do that doing2nd , FORMAT , 9 , ENTER , 2nd , QUIT .


11/27


We often will use the time value of the money worksheet of thecalculator. There are 5 main financial variables in this worksheet:

I The number of periods N .

I The nominal interest for year I/Y .

I The present value PV .

I The payment per period PMT .

I The future value FV .

You can use the calculator to find one of these financial variables,by entering the rest of the variables in the memory of thecalculator and then pressing CPT financial key , where financial

key is either N , % i , PV , PMT or FV .


12/27


Here, financial key is either N , % i , PV , PMT or FV .

I You can recall the entries in the time value of the moneyworksheet, by pressing RCL financial key .

I To enter a variable in the entry financial key , type the entry

and press financial key . The entry of variables can be donein any order.

I To find the value of any of the five variables (after entering

the rest of the variables in the memory) press CPT

financial key .

I When computing a variable, a formula using all five variablesand two auxiliary variables is used


13/27


To set–up C/Y =1 and P/Y =1, do

2nd , P/Y , 1 , ENTER , ↓ , 1 , ENTER , 2nd , QUIT .

To check that this is so, do

2nd P/Y ↓ 2nd QUIT .

If PMT equals zero, C/Y =1 and P/Y =1, you have the

formula,

PV + FV

1 +I/Y

100

− N

= 0. (1)

You can use this to solve for any element of the four elements inthe formula A(t) = A(0)(1 + i)t . Unless it is said otherwise, we

will assume that the entries for C/Y and P/Y are both 1

and PMT is 0.


14/27


Example 2

Mary invested $12000 on January 1, 1995. Assuming compositeinterest at 5 % per year, find the accumulated value on January 1,2002.

Solution: A(t) = 12000(1 + 0.05)7 = 16885.21. You can do thisin the calculator by entering:

0 PMT 7 N 5 I/Y 12000 PV CPT FV .

Note that since the calculator, uses the formula

PV + FV

1 +I/Y

100

− N

= 0.

the display in your calculator is negative.


15/27


Example 2

Mary invested $12000 on January 1, 1995. Assuming compositeinterest at 5 % per year, find the accumulated value on January 1,2002.

Solution: A(t) = 12000(1 + 0.05)7 = 16885.21. You can do thisin the calculator by entering:

0 PMT 7 N 5 I/Y 12000 PV CPT FV .

Note that since the calculator, uses the formula

PV + FV

1 +I/Y

100

− N

= 0.

the display in your calculator is negative.


16/27


Example 3

At what annual rate of compound interest will $200 grow to $275in 5 years?

Solution: We solve for i in 275 = 200(1 + i)5 and geti = 6.5763%. In the calculator, you do

−275 FV 5 N 200 PV CPT I/Y .

Since the calculator, uses the formula (1), either the present valueor the future value has to be entered as negative number (and theother one as a positive number). If you enter both the presentvalue and the future value as positive values, you get the error

message Error 5 . To clear this error message press CE/C .


17/27


Example 3

At what annual rate of compound interest will $200 grow to $275in 5 years?

Solution: We solve for i in 275 = 200(1 + i)5 and geti = 6.5763%. In the calculator, you do

−275 FV 5 N 200 PV CPT I/Y .

Since the calculator, uses the formula (1), either the present valueor the future value has to be entered as negative number (and theother one as a positive number). If you enter both the presentvalue and the future value as positive values, you get the error

message Error 5 . To clear this error message press CE/C .


18/27


Example 4

How many years does it take $200 grow to $275 at an effectiveannual rate of 5%?

Solution: We solve for t in 275 = 200(1 + 0.05)t and get thatt = 6.5270 years. In the calculator, you do

−275 FV 5 I/Y 200 PV CPT N .


19/27


Example 4

How many years does it take $200 grow to $275 at an effectiveannual rate of 5%?

Solution: We solve for t in 275 = 200(1 + 0.05)t and get thatt = 6.5270 years. In the calculator, you do

−275 FV 5 I/Y 200 PV CPT N .


20/27


Example 5

At an annual effective rate of interest of 8% how long would ittake to triple your money?

Solution: We solve for t in 3 = (1 + 0.08)t and get t = 14.2749years. In the calculator, you do

−3 FV 8 I/Y 1 PV CPT N .


21/27


Example 5

At an annual effective rate of interest of 8% how long would ittake to triple your money?

Solution: We solve for t in 3 = (1 + 0.08)t and get t = 14.2749years. In the calculator, you do

−3 FV 8 I/Y 1 PV CPT N .


22/27


Example 6

How much money was needed to invest 10 years in the past toaccumulate $ 10000 at an effective annual rate of 5%?

Solution: We solve for A(0) in 10000 = A(0)(1 + 0.05)10 and getthat A(0) = 6139.13. In the calculator, you do

10000 FV 5 I/Y 10 N CPT PV .


23/27


Example 6

How much money was needed to invest 10 years in the past toaccumulate $ 10000 at an effective annual rate of 5%?

Solution: We solve for A(0) in 10000 = A(0)(1 + 0.05)10 and getthat A(0) = 6139.13. In the calculator, you do

10000 FV 5 I/Y 10 N CPT PV .


24/27


The calculator has a memory worksheet with values in the memory,which stores ten numbers. These ten numbers are called: M0 ,

· · · , M9 .To enter the number in the display into the i–th entry of

the memory, press STO i , where i is an integer from 0 to 9. To

recall the number in the memory entry i , press RCL i , where i is

an integer from 0 to 9. The command STO + i adds thevalue in display to the entry i in the memory. You can see all thenumbers in the memory by accessing the memory worksheet. Toenter this worksheet press 2nd MEM . Use the arrows ↑ , ↓ tomove from entry to another. To entry a new value in one entry,type the number and press ENTER .


25/27


Example 7



Payments 1000 2000 5000

Time 1 2 3


(1000)(1.055)−1 + (2000)(1.055)−2 + (5000)(1.055)−3

=947.8672986 + 1796.904831 + 4258.068321 = 7002.840451.


26/27


Example 7



Payments 1000 2000 5000

Time 1 2 3


(1000)(1.055)−1 + (2000)(1.055)−2 + (5000)(1.055)−3

=947.8672986 + 1796.904831 + 4258.068321 = 7002.840451.


27/27


Using the calculator, you do

−1000 FV 1 N 5.5 I/Y CPT PV

and get (1000)(1.055)−1 = 947.8672986. You enter this number in

the memory of the calculator doing STO 1Next doing−2000 FV 2 N CPT PVyou find (2000)(1.055)−2 = 1796.904831. Notice that you do nothave to reenter the percentage interest rate. You enter thisnumber in the memory of the calculator doing STO 2Next doing−5000 FV 3 N CPT PVyou get (5000)(1.055)−3 = 4258.068321. You enter this number in

the memory of the calculator doing STO 3 .You can recall and add the three numbers doingCRCL 1 + CRCL 2 + CRCL 3 =and get 7002.840451.


1/19



Section 1.4. Present value and discount.



Financial Mathematics. Spring 2009 Edition”,available at http://www.actexmadriver.com/


2/19

Chapter 1. Basic Interest Theory. Section 1.4. Present value and discount.

Present value and discount

Suppose that we make an investment of $k in an account earningcompound interest with effective annual rate of interest i . t yearslater the balance in this account is k(1 + i)t . Here, k(1 + i)t is thefuture value of the investment t years in the future. Undercompound interest, balances multiply by (1 + i)t every t years. $kat time s is worth $k(1 + i)t at time s + t.

The quantity (1 + i)t is called the t–year interest factor.The quantity (1 + i) is called the interest factor. $k at time s isworth $k(1 + i) at time s + 1.


3/19



Suppose that we make an investment of $k in an account earningcompound interest with effective annual rate of interest i . t yearslater the balance in this account is k(1 + i)t . Here, k(1 + i)t is thefuture value of the investment t years in the future. Undercompound interest, balances multiply by (1 + i)t every t years. $kat time s is worth $k(1 + i)t at time s + t.The quantity (1 + i)t is called the t–year interest factor.

The quantity (1 + i) is called the interest factor. $k at time s isworth $k(1 + i) at time s + 1.


4/19



Suppose that we make an investment of $k in an account earningcompound interest with effective annual rate of interest i . t yearslater the balance in this account is k(1 + i)t . Here, k(1 + i)t is thefuture value of the investment t years in the future. Undercompound interest, balances multiply by (1 + i)t every t years. $kat time s is worth $k(1 + i)t at time s + t.The quantity (1 + i)t is called the t–year interest factor.The quantity (1 + i) is called the interest factor. $k at time s isworth $k(1 + i) at time s + 1.


5/19


Often, we need to find the amount of money t years in the pastneeded to accumulate certain principal. The present value t yearsin the past is the amount of money which will accumulate to theprincipal over t years.In the case of compound interest with effective annual rate ofinterest i , the present value of $1 t years in the past is 1

(1+i)t . If we

invested 1(1+i)t t years ago in account earning compound interest,

then the current balance is $1. The quantity 1(1+i)t is called the t

year discount. $k at time s is worth $kνt at time s − t.The quantity ν = 1

1+i is called the discount factor. In order toaccumulate $1, we need $ν one year in the past.


6/19


Under the accumulation function a(t),

I The n–th year interest factor is a(n)a(n−1) .

I The effective rate of interest in the n–th year isin = a(n)−a(n−1)

a(n−1) .

I The n year discount factor is νn = a(n−1)a(n) .

I The effective rate of discount in the n–th year isdn = a(n)−a(n−1)

a(n) .

in and dn are both proportions of interest over amount values, butin uses the amount value in the past and dn uses the amount valuein the future. Since the amount value in the future is bigger thanthe amount value in the past, dn < in.


7/19


Notice that

I the n–th year interest factor is equal to 1 + in.

I νn = 1− dn.

I 1 = (1 + in)νn = (1 + in)(1− dn)

I {1 unit at time n − 1} ≡ {1 + in units at time n}. So,dn = in

1+in

I {1− dn unit at time n − 1} ≡ {1 units at time n}. So,in = dn

1−dn.


8/19


Under compound interest, the effective rate of discount dn is

constant dn = a(n)−a(n−1)a(n) = (1+i)n−(1+i)n−1

(1+i)n = i1+i . Under

compound interest,

ν =1

1 + i, d = 1− ν, d =

i

i + 1and (1− d)(1 + i) = 1.


9/19


Example 1

Peter invests $738 in a bank account. One year later, his bankaccount is $765.(i) Find the effective annual interest rate earned by Peter in thatyear.(ii) Find the effective annual discount rate earned by Peter in thatyear.

Solution: (i)Peter earns an interest amount of 765− 738 = 27.The effective annual interest rate earned by Peter is27738 = 3.658537%.(ii) The effective annual discount rate earned by Peter is27765 = 3.529412%.


10/19


Example 1


Solution: (i)Peter earns an interest amount of 765− 738 = 27.The effective annual interest rate earned by Peter is27738 = 3.658537%.

(ii) The effective annual discount rate earned by Peter is27765 = 3.529412%.


11/19


Example 1


Solution: (i)Peter earns an interest amount of 765− 738 = 27.The effective annual interest rate earned by Peter is27738 = 3.658537%.(ii) The effective annual discount rate earned by Peter is27765 = 3.529412%.


12/19


Example 2

If i = 7%, what are d and ν?

Solution: We have that d = i1+i = 0.07

1+0.07 = 6.5421% and

ν = 11+i = 1

1+0.07 = 0.934579.


13/19


Example 2

If i = 7%, what are d and ν?

Solution: We have that d = i1+i = 0.07

1+0.07 = 6.5421% and

ν = 11+i = 1

1+0.07 = 0.934579.


14/19


Example 3

If ν = 0.95, what are d and i?

Solution: We have that d = 1− ν = 1− 0.95 = 0.05 andi = 1

ν − 1 = 1−0.950.95 = 5.2632%.


15/19


Example 3

If ν = 0.95, what are d and i?

Solution: We have that d = 1− ν = 1− 0.95 = 0.05 andi = 1

ν − 1 = 1−0.950.95 = 5.2632%.


16/19


Example 4

What is the present value of $5,000 to be received in 7 years at anannual effective rate of discount of 7%?

Solution: The value is (5000)(1− 0.07)7 = 3008.504353.


17/19


Example 4

What is the present value of $5,000 to be received in 7 years at anannual effective rate of discount of 7%?

Solution: The value is (5000)(1− 0.07)7 = 3008.504353.


18/19


Example 5

At time t = 0, Paul deposits $3500 into a fund crediting interestwith an annual discount factor of 0.96. Find the fund value at time2.5.

Solution: (3500)(0.96)−2.5 = 3876.055.


19/19


Example 5

At time t = 0, Paul deposits $3500 into a fund crediting interestwith an annual discount factor of 0.96. Find the fund value at time2.5.

Solution: (3500)(0.96)−2.5 = 3876.055.


1/24



Section 1.5. Nominal rates of interest and discount.





2/24

Chapter 1. Basic Interest Theory. Section 1.5. Nominal rates of interest and discount.

Nominal rate of interest

When dealing with compound interest, often we will rates differentfrom the annual effective interest rate. Suppose that an accountfollows compound interest with an annual nominal rate ofinterest compounded m times a year of i (m), then

I $1 at time zero accrues to $(1 + i (m)

m ) at time 1m years.

I The 1m–year interest factor is (1 + i (m)

m ).

I The ( 1m -year ) m–thly effective interest rate is i (m)

m .

I $1 at time zero grows to $(1 + i (m)

m

)min one year.


m

)mtin t years.

I The accumulation function is a(t) =(1 + i (m)

m

)mt.


3/24







m ).


m .


m

)min one year.


m

)mtin t years.


m

)mt.


4/24







m ).


m .


m

)min one year.


m

)mtin t years.


m

)mt.


5/24







m ).


m .


m

)min one year.


m

)mtin t years.


m

)mt.


6/24







m ).


m .


m

)min one year.


m

)mtin t years.


m

)mt.


7/24







m ).


m .


m

)min one year.


m

)mtin t years.


m

)mt.


8/24


Example 1

Paul takes a loan of $569. Interest in the loan is charged usingcompound interest. One month after a loan is taken the balance inthis loan is $581.(i) Find the monthly effective interest rate, which Paul is chargedin his loan.(ii) Find the annual nominal interest rate compounded monthly,which Paul is charged in his loan.

Solution: (i) The monthly effective interest rate, which Paul ischarged in his loan is

i (12)

12=

581− 569

569= 2.108963093%.

(ii) The annual nominal interest rate compounded monthly, whichPaul is charged in his loan is

i (12) = (12)(0.02108963093) = 25.30755712%.


9/24


Example 1



i (12)

12=

581− 569

569= 2.108963093%.


i (12) = (12)(0.02108963093) = 25.30755712%.


10/24


Example 1



i (12)

12=

581− 569

569= 2.108963093%.


i (12) = (12)(0.02108963093) = 25.30755712%.


11/24


Two rates of interest or discount are said to be equivalent if theygive rise to same accumulation function. Since, the accumulationfunction under an annual effective rate of interest i isa(t) = (1 + i)t , we have that a nominal annual rate of interest i (m)

compounded m times a year is equivalent to an annual effectiverate of interest i , if the rates

a(t) =

(1 +

i (m)

m

)mt

anda(t) = (1 + i)t

agree. This happens if and only if(1 +

i (m)

m

)m

= 1 + i .


12/24


Example 1

John takes a loan of 8,000 at a nominal annual rate of interest of10% per year convertible quarterly. How much does he owe after30 months?

Solution: We find

8000

(1 +

0.10

4

) 3012·4

= 8000 (1 + 0.025)10 = 10240.68.

In the calculator, we do:8000 PV 2.5 I/Y 10 N CPT FV .


13/24


Example 1

John takes a loan of 8,000 at a nominal annual rate of interest of10% per year convertible quarterly. How much does he owe after30 months?

Solution: We find

8000

(1 +

0.10

4

) 3012·4

= 8000 (1 + 0.025)10 = 10240.68.

In the calculator, we do:8000 PV 2.5 I/Y 10 N CPT FV .


14/24


The calculator TI–BA–II–Plus has a worksheet to convert nominalrates of interest into effective rates of interest and vice versa. Toenter this worksheet press 2nd ICONV . There are 3 entries in

this worksheet: NOM , EFF and C/Y . C/Y is the number of

times the nominal interest is converted in a year. The relationbetween these variables is

1 +EFF

100=

1 +NOM

100 C/Y

C/Y

.

You can enter a value in any of these entries by moving to thatentry using the arrows: ↑ and ↓ . To enter a value in one entry,

type the value and press ENTER . You can compute thecorresponding nominal (effective) rate by moving to the entry

NOM ( EFF ) and pressing the key CPT . It is possible to enter

negative values in the entries NOM and EFF . However, the

value in the entry C/Y has to be positive.


15/24


Example 2

If i (4) = 5% find the equivalent effective annual rate of interest.

Solution: We solve 1 + i =(1 + 0.05

4

)4and get i = 5.0945%. In

the calculator, you enter the worksheet ICONV and enter: NOM

equal to 5 and C/Y equal to 4. Then, go to EFF and press

CPT . To quit, press 2nd , QUIT .


16/24


Example 2

If i (4) = 5% find the equivalent effective annual rate of interest.

Solution: We solve 1 + i =(1 + 0.05

4

)4and get i = 5.0945%. In

the calculator, you enter the worksheet ICONV and enter: NOM

equal to 5 and C/Y equal to 4. Then, go to EFF and press

CPT . To quit, press 2nd , QUIT .


17/24


Example 3

If i = 5%, what is the equivalent i (4)?

Solution: We solve(1 + i (4)

4

)4= 1 + 0.05 we get that

i (4) = 4((1 + 0.05)1/4 − 1

)= 4.9089%. In the calculator, you

enter the worksheet ICONV and enter: EFF equal to 5 and

C/Y equal to 4. Then, go to NOM and press CPT . To quit,

press 2nd , QUIT .


18/24


Example 3

If i = 5%, what is the equivalent i (4)?

Solution: We solve(1 + i (4)

4

)4= 1 + 0.05 we get that

i (4) = 4((1 + 0.05)1/4 − 1

)= 4.9089%. In the calculator, you

enter the worksheet ICONV and enter: EFF equal to 5 and

C/Y equal to 4. Then, go to NOM and press CPT . To quit,

press 2nd , QUIT .


19/24


The nominal rate of discount d (m) is defined as the value suchthat 1 unit at the present is equivalent to 1− d (m)

m units invested 1m

years ago, i.e.

{1− d (m)

munits at time 0} ≡

{1 unit at time

1

m

}.

This implies that

{1 unit at time 0} ≡

{1

1− d (m)

m

units at time1

m

}.

The accumulation function for compound interest under a thenominal rate of discount d (m) convertible m times a year is

a(t) =(1− d (m)

m

)−mt. We have that

1 + i =

(1 +

i (m)

m

)m

= (1− d)−1 =

(1− d (m)

m

)−m

.


20/24


In the calculator TI–BA–II–Plus, you may:

I given i (m), find i , by entering

i (m) → NOM and m → C/Y , then in EFF press CPT .

I given i , find i (m), by entering

i → EFF and m → C/Y , then in NOM press CPT .

I given d (m), find d , by entering

−d (m) → NOM and m → C/Y , then in EFF press CPT .

d appears with a negative sign.

I given d , find d (m), by entering

−d → EFF and m → C/Y , then in NOM press CPT .

d (m) appears with a negative sign.

I given i , find d , by using the formula i = 11−d − 1.

I given d , find i , by using the formula d = 1− 11+i .


21/24


Example 4

If d (4) = 5% find i .

Solution: We solve(1− d (4)

4

)−4= 1 + i to get d = 4.9070% and

i = 5.1602%. In the calculator, in the ICONV worksheet, we

enter −5 in NOM , 4 in C/Y and we find that EFF is

−4.9070%, then we do−4.9070 % + 1 = 1/x − 1 =

to get i = 5.1602%.


22/24


Example 4

If d (4) = 5% find i .

Solution: We solve(1− d (4)

4

)−4= 1 + i to get d = 4.9070% and

i = 5.1602%. In the calculator, in the ICONV worksheet, we

enter −5 in NOM , 4 in C/Y and we find that EFF is

−4.9070%, then we do−4.9070 % + 1 = 1/x − 1 =

to get i = 5.1602%.


23/24


Example 5

If i = 3% find d (2).

Solution: We solve for d (2) in(1− d (2)

2

)−2= 1 + i . First we find

that d = 2.9126% doing

3 % + 1 = 1/x − 1 =

Then, using the ICONV worksheet, we get that d (2) = 2.9341%.


24/24


Example 5

If i = 3% find d (2).

Solution: We solve for d (2) in(1− d (2)

2

)−2= 1 + i . First we find

that d = 2.9126% doing

3 % + 1 = 1/x − 1 =

Then, using the ICONV worksheet, we get that d (2) = 2.9341%.


1/12



Section 1.6. Force of interest.





2/12

Chapter 1. Basic Interest Theory. Section 1.6. Force of interest.

Force of interest

The force of interest δt of an amount function A(t) is defined by

δt = ddt lnA(t) = A′(t)

A(t) .

The force of interest is the fraction of the instantaneous rate ofchange of the accumulation function and the accumulationfunction.To find the force of interest, we may use the accumulationfunction,

d

dtlnA(t) =

d

dtln(A(0)a(t)) =

d

dtln(A(0)) +

d

dtln(a(t))

=d

dtln(a(t)).


3/12


Force of interest



A(t) .The force of interest is the fraction of the instantaneous rate ofchange of the accumulation function and the accumulationfunction.

To find the force of interest, we may use the accumulationfunction,

d

dtlnA(t) =

d

dtln(A(0)a(t)) =

d

dtln(A(0)) +

d

dtln(a(t))

=d

dtln(a(t)).


4/12


Force of interest



A(t) .The force of interest is the fraction of the instantaneous rate ofchange of the accumulation function and the accumulationfunction.To find the force of interest, we may use the accumulationfunction,

d

dtlnA(t) =

d

dtln(A(0)a(t)) =

d

dtln(A(0)) +

d

dtln(a(t))

=d

dtln(a(t)).


5/12


Example 1

Consider the amount function A(t) = 25(1 + t

4

)3. At what time is

the force of interest equal of 0.5.

Solution: We have that

ln(A(t)) = ln

(25

(1 +

t

4

)3)

= ln 25 + 3 ln(1 +

t

4

).

The force of interest is

δt =d

dtln(A(t)) =

d

dt

(ln 25 + 3 ln

(1 +

t

4

))= 3

14

1 + t4

=3

4 + t.

From the equation, 34+t = 1

2 , we get that t = 2.


6/12


Example 1

Consider the amount function A(t) = 25(1 + t

4

)3. At what time is

the force of interest equal of 0.5.


ln(A(t)) = ln

(25

(1 +

t

4

)3)

= ln 25 + 3 ln(1 +

t

4

).

The force of interest is

δt =d

dtln(A(t)) =

d

dt

(ln 25 + 3 ln

(1 +

t

4

))= 3

14

1 + t4

=3

4 + t.

From the equation, 34+t = 1

2 , we get that t = 2.


7/12


The force of interest is also called the rate of interestcontinuously compounded and the continuous interest rate.We have that

δt = limh→0

A(t + h)− A(t)

A(t) · h

= limh→0

interest earned over the next h yearsinvestment at time t · h

.

The nominal annual rate earned in the next 1m years compounded

m times a year at time t is

m(a(t + 1

m

)− a(t))

a(t)=

a(t + 1

m

)− a(t)

a(t) 1m

.

We have that

limm→∞

a(t + 1

m

)− a(t)

a(t) 1m

= δt .


8/12


Under compound interest, a(t) = (1 + i)t and

δt =d

dtln a(t) =

d

dtln(1 + i)t =

d

dtt ln(1 + i) = ln(1 + i)

Under compound interest, the force of interest is a constant δ,such that δ = ln(1 + i) = − ln ν.Under compound interest,

limm→∞

i (m) = limm→∞

d (m) = δ.

In the case of simple interest, a(t) = 1 + it andδt = d

dt ln(1 + it) = i1+it . The force of interest is decreasing with t.


9/12


From the force of interest δt , we may find the accumulationfunction a(t), using

Theorem 2For each t ≥ 0, a(t) = e

R t0 δs ds .

Proof.Since δs = d

ds ln a(s) and a(0) = 1,∫ t

0δs ds =

∫ t

0

d

dsln a(s) ds = ln a(s)

∣∣∣∣t0

= ln a(t).

So, a(t) = eR t0 δs ds .


10/12


From the force of interest δt , we may find the accumulationfunction a(t), using

Theorem 2For each t ≥ 0, a(t) = e

R t0 δs ds .

Proof.Since δs = d

ds ln a(s) and a(0) = 1,∫ t

0δs ds =

∫ t

0

d

dsln a(s) ds = ln a(s)

∣∣∣∣t0

= ln a(t).

So, a(t) = eR t0 δs ds .


11/12


Example 3

A bank account credits interest using a force of interest δt = 3t2

t3+2.

A deposit of 100 is made in the account at time t = 0. Find theamount of interest earned by the account from the end of the 4–thyear until the end of the 8–th year.

Solution: First, we find a(t) = eR t0 δs ds .∫ t

0δs ds =

∫ t

0

3s2

s3 + 2ds = ln(s3 + 2)

∣∣∣∣t0

= ln(t3 + 2)− ln 2 = ln

(t3 + 2

2

)and

a(t) = eR t0 δs ds = e

ln“

t3+22

”=

t3 + 2

2= 1 +

t3

2.

The amount of interest earned in the considered period is

100(a(8)− a(4)) = (100)

(1 +

83

2−

(1 +

43

2

))= 22400.


12/12


Example 3

A bank account credits interest using a force of interest δt = 3t2

t3+2.

A deposit of 100 is made in the account at time t = 0. Find theamount of interest earned by the account from the end of the 4–thyear until the end of the 8–th year.

Solution: First, we find a(t) = eR t0 δs ds .∫ t

0δs ds =

∫ t

0

3s2

s3 + 2ds = ln(s3 + 2)

∣∣∣∣t0

= ln(t3 + 2)− ln 2 = ln

(t3 + 2

2

)and

a(t) = eR t0 δs ds = e

ln“

t3+22

”=

t3 + 2

2= 1 +

t3

2.

The amount of interest earned in the considered period is

100(a(8)− a(4)) = (100)

(1 +

83

2−

(1 +

43

2

))= 22400.


1/18

Chapter 2. Cashflows.

Manual for SOA Exam FM/CAS Exam 2.Chapter 2. Cashflows.Section 2.1. Cashflows.





2/18

Chapter 2. Cashflows. Section 2.1. Cashflows.

Net present value of cash flows

Recall that a cashflow is a series of payments made at differenttimes. We can represent a cashflow in a table:


Time (in periods) t1 t2 · · · tn

Assuming compound interest, the present value of a cashflow attime t is

V (t) =n∑

j=1

Cjνtj−t =

n∑j=1

Cj(1 + i)t−tj ,

where i is the effective interest per period and ν is the discountfactor per period. The previous equation is the equation of value.Under the accumulation function a(·), the equation of value is

V (t) =n∑

j=1

Cja(t)

a(tj).


3/18


Example 1

An investor can invest in a project which requires an investment of$37400 at time 0. The investment pays $25000 at time 1 and$15000 at time 2. The investor’s capital is currently earning aneffective annual rate of interest of 4.5%. Should the investor investin the project?

Solution: The net present value of the investment in the project is

25000(1.045)−1 + 15000(1.045)−2 − 37400

=23923.445 + 13735.9493− 37400 = 259.3943.

Yes, the investor should invest in the project.


4/18


Example 1

An investor can invest in a project which requires an investment of$37400 at time 0. The investment pays $25000 at time 1 and$15000 at time 2. The investor’s capital is currently earning aneffective annual rate of interest of 4.5%. Should the investor investin the project?

Solution: The net present value of the investment in the project is

25000(1.045)−1 + 15000(1.045)−2 − 37400

=23923.445 + 13735.9493− 37400 = 259.3943.

Yes, the investor should invest in the project.


5/18


A way to analyze the profitability of an investment is to find thepresent value at time of the cashflow derived from the investment.The net present value of an investment is the present value ofthe inflows minus the present value of the outflows.Suppose that a company can select between taking twoinvestments. Which investment has the biggest net present valuedepends on the used interest rate. To valuate investments,companies use their cost of capital. The cost of capital of acompany is an estimation of how much the company has to pay forevery dollar it borrows. This cost of capital is found using thewhole capital components of the company.


6/18


Example 2

A company has cost of capital of 7.5% as an annual effective rateof interest. Two investment projects have the following forecastedcash flows:

Project A −$20,000 0 0 $25,000 $10,000

Project B −$20,000 0 0 $10,000 $26,000

Time in years 0 1 2 3 4

(i) Find the profit made under each investment project.(ii) Which project has the highest profit?(iii) Compute the net present value for each project using thecompany’s cost of capital.(iv) Which project has the highest net present value?


7/18


Solution:(i) The profit for project A is

−20000 + (25000) + (10000) = 15000.

The profit for project B is

−20000 + (10000) + (26000) = 16000.

(ii) Project B has the highest profit.(iii) The present value of project A is

−20000 + (25000)(1.075)−3 + (10000)(1.075)−4 = 7612.02.

The present value of project B is

−20000 + (10000)(1.075)−3 + (26000)(1.075)−4 = 7518.419.

(iv) Project A has the highest net present value.


8/18


The calculator TI–BA–II–Plus has a cashflow worksheet, whichallows to work with cashflows when the deposit times arenonnegative numbers. After entering the cashflow in thecalculator, you can find either the present value of the cashflowor the internal rate of return. The internal rate of return is theeffective periodic rate of interest. There are 3 keys to enterdifferent parts of this worksheet.

I Pressing the key CF , you can enter the cash flow data.

I Pressing the key NPV opens a worksheet with two variables

NPV and I . Using this worksheet, you can compute the netpresent value.

I Pressing the IRR you compute the internal rate of return.


9/18


When you press the key CF , you can change the entries: CFo ,

C01 , F01 , C02 , F02 , . . . ., C24 , F24 . CFo is the initial

contribution. C01 is the amount of the first round of

contributions. F01 is the (number of payments) frequency of the

first round of contributions. C02 is the amount of the second

round of contributions. F02 is the frequency of the second round

of contributions. C03 , F03 , . . . , C24 , F24 are defined

similarly. To move from one entry to the next, use the arrows ↓and ↑ . To enter a number in one entry, type the number, and

then press ENTER . To clear the values in the worksheet, press

2nd CLR WORK while in the worksheet. To exit the worksheet

type 2nd QUIT . The cashflow, which we have is:

Payments CFo C01 C02 · · ·Time 0 1 to F01 F01+1 to F01+F02 · · ·


10/18


After you have entered the date you can calculate either the netpresent value or the internal rate of return.

I To calculate the net present value, press the key NPV , enter

the periodic interest rate in the entry I , then go the entry

NPV using one of the arrows ↓ and ↑ . Finally press

CPT .

I To calculate the internal rate of return, press the keys IRR

and CPT . If the equation does not have a solution, you get”Error 5”. If the equation has several solutions, you get theone with smallest absolute value.


11/18


Example 3

Joel wishes to borrow a sum of money. In return, he is prepared topay as follows: $100 after 1 year, $200 after 2 years, $300 after 3years, and $400 after 4 years. If i = 12%, how much can heborrow?


contributions 100 200 300 400

Time 1 2 3 4

He can borrow:

(100)(1 + 0.12)−1 + (200)(1 + 0.12)−2

+(300)(1 + 0.12)−3 + (400)(1 + 0.12)−4 = 716.4657955

To do this problem using the CF worksheet. Press CF , and enter

CFo =0, C01 =100, F01 =1, C02 =200, F02 =1, C03 =300,

F03 =1, C04 =400, F04 =1, 2nd QUIT . Go to NPV , enter

I =12, and compute NPV and get 716.4657955.


12/18


Example 3

Joel wishes to borrow a sum of money. In return, he is prepared topay as follows: $100 after 1 year, $200 after 2 years, $300 after 3years, and $400 after 4 years. If i = 12%, how much can heborrow?


contributions 100 200 300 400

Time 1 2 3 4

He can borrow:

(100)(1 + 0.12)−1 + (200)(1 + 0.12)−2

+(300)(1 + 0.12)−3 + (400)(1 + 0.12)−4 = 716.4657955

To do this problem using the CF worksheet. Press CF , and enter

CFo =0, C01 =100, F01 =1, C02 =200, F02 =1, C03 =300,

F03 =1, C04 =400, F04 =1, 2nd QUIT . Go to NPV , enter

I =12, and compute NPV and get 716.4657955.


13/18


Example 4



Payments 1000 2000 5000

Time 1 2 3


(1000)(1.055)−1 + (2000)(1.055)−2 + (5000)(1.055)−3 = 7002.840451.

Press CF , and enter CFo =0, C01 =1000, F01 =1,

C02 =2000, F02 =1, C03 =5000, F03 =1, 2nd QUIT . Go to

NPV , enter I =5.5, and compute NPV and get 7002.840451.


14/18


Example 4



Payments 1000 2000 5000

Time 1 2 3


(1000)(1.055)−1 + (2000)(1.055)−2 + (5000)(1.055)−3 = 7002.840451.


C02 =2000, F02 =1, C03 =5000, F03 =1, 2nd QUIT . Go to

NPV , enter I =5.5, and compute NPV and get 7002.840451.c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2.

15/18


Example 5

Helen borrows $5000 from her credit card account at a nominalannual interest rate of 20% per year convertible monthly. Twomonths later, she pays $1000 back. Four months after thepayment she borrows $2000. How much does she owe one yearafter the loan is taken out?


inflow/outflow 5000 0 −1000 0 0 0 2000

Time (in months) 0 1 2 3 4 5 6


16/18


Example 5



inflow/outflow 5000 0 −1000 0 0 0 2000



17/18


Example 5



inflow/outflow 5000 0 −1000 0 0 0 2000


The equation of value at time 1 year is

5000

(1 +

0.20

12

)12

− 1000

(1 +

0.20

12

)10

+ 2000

(1 +

0.20

12

)6

=7125.737519


18/18


Example 5



inflow/outflow 5000 0 −1000 0 0 0 2000



C02 =−1000, F02 =1, C03 =0, F03 =3, C04 =2000, F04 =1,

2nd QUIT . Go to NPV , enter I =1.66666(=20/12), and

compute NPV = 5843.69. This is the present value at time0 of the loan. The future value of the loan at time 1 year is5843.69(1.01666)12 = 7125.737519.


1/7


Manual for SOA Exam FM/CAS Exam 2.Chapter 2. Cashflows.

Section 2.2. Method of equated time.





2/7

Chapter 2. Cashflows. Section 2.2. Method of equated time.

Given the cashflow



we would like to find a time t such that a lump sum C =∑n

j=1 Cj

invested at time t is equivalent to the previous cashflow. To find t,we solve the equation,

Cν t =n∑

j=1

Cjνtj , (1)

and get

t =ln(

∑nj=1 Cjν

tj /C )

ln ν(2)


3/7


Method of equated time

The method of equated time consists on approximating t by

t =

∑nj=1 Cj tj

C=

∑nj=1 Cj tj∑nj=1 Cj

. (3)

This is the average time of all the times tj with the weightCjPn

k=1 Ck

at tj .


4/7


The first order Taylor expansion of νt = (1 + i)−t on i is 1− ti .So, Equation (1) is approximately

C (1− ti) =n∑

j=1

Cj(1− tj i),

whose solution is

t =

∑nj=1 Cj tj

C=

∑nj=1 Cj tj∑nj=1 Cj

.

If the interest were simple, the future value at time tn of theconsidered cashflow would be

C (1 + (tn − t)i) =n∑

j=1

Cj(1 + (tn − tj)i),

whose solution is t =Pn

j=1 Cj tjC .

The approximation to t using the method of equating time is thesolution to the considered problem when the accumulation functionfollows simple interest.


5/7


Example 1

Payments of $300, $100 and $200 are due at the ends of years 1,3, and 5, respectively. Assume an annual effective rate of interestof 5% per year. (i) Find the point in time at which a payment of$600 would be equivalent. (ii) Find the approximation to this pointusing the method of equated time.

Solution: (i) The time t solves the equation

600ν t = 300ν + 100ν3 + 200ν5

=285.71429 + 86.38376 + 156.70523 = 528.80328,

where ν = (1.05)−1. Hence, t = ln(600/528.80328)ln(1.05) = 2.58891.

(ii) The equated time approximation to this point is

t =(300)(1) + (100)(3) + (200)(5)

600= 2.666667.


6/7


Example 1



600ν t = 300ν + 100ν3 + 200ν5

=285.71429 + 86.38376 + 156.70523 = 528.80328,



t =(300)(1) + (100)(3) + (200)(5)

600= 2.666667.


7/7


Example 1



600ν t = 300ν + 100ν3 + 200ν5

=285.71429 + 86.38376 + 156.70523 = 528.80328,



t =(300)(1) + (100)(3) + (200)(5)

600= 2.666667.


1/16



Section 2.3. Yield rates.





2/16

Chapter 2. Cashflows. Section 2.3. Yield rates.

Yield of return

Suppose that the future value at time t of the cashflow:

Investments V0 C1 C2 · · · Cn

Time 0 t1 t2 · · · tn

is FV . Then, the rate of return i of the investment satisfies theequation,

FV = V0ν−t +

n∑j=1

Cjνtj−t = V0(1 + i)t +

n∑j=1

Cj(1 + i)t−tj .

The rate of return i , i > −1, solving this equation is called theyield rate of return or internal rate of return. This equationcan have either no solutions, or one solution, or several solutions.We are interested in values of i with i > −1. If i < −1, then(1 + i)n > 0 is n is even and (1 + i)n < 0 is n is odd. Values of iwith i ≤ −1 do not make any sense.


3/16


Example 1

Suppose that John invest $3000 in a business. One year later,John sells half of this business to a partner for $6000. Two yearsafter the beginning, the business is in red and John has to pay$4000 to close this business. What is the rate of interest John’sgot in his investment?

Solution: The cashflow is:

Inflow −3000 6000 −4000

Time 0 1 2

Since John lost money, one expect that i is negative. However,there is no solution. We have to solve−3000(1 + i)2 + 6000(1 + i)− 4000 = 0, or3(1 + i)2 − 6(1 + i) + 4 = 0. Using the quadratic formula,

1 + i =6±

√62 − 4 · 4 · 3

2=

6±√−12

2.

There is no solution.


4/16


Example 1

Suppose that John invest $3000 in a business. One year later,John sells half of this business to a partner for $6000. Two yearsafter the beginning, the business is in red and John has to pay$4000 to close this business. What is the rate of interest John’sgot in his investment?


Inflow −3000 6000 −4000

Time 0 1 2

Since John lost money, one expect that i is negative. However,there is no solution. We have to solve−3000(1 + i)2 + 6000(1 + i)− 4000 = 0, or3(1 + i)2 − 6(1 + i) + 4 = 0. Using the quadratic formula,

1 + i =6±

√62 − 4 · 4 · 3

2=

6±√−12

2.

There is no solution.c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2.

5/16


Example 2

What is the yield rate on a transaction in which a person makespayments of $100 immediately and $100 at the end of two years,in exchange for a payment of $201 at the end of one year? Find allpossible solutions.


Inflow −100 201 −100

Time 0 1 2

We have to solve −100 + 201(1 + i)−1 − 100(1 + i)−2 = 0, or100(1 + i)2 − 201(1 + i)1 + 100 = 0. Using the quadratic formula,

1 + i =201±

√2012 − 4 · 100 · 100

200=

201±√

201

200.

The two solutions are i = 10.5124922% and i = −9.512492197%.


6/16


Example 2

What is the yield rate on a transaction in which a person makespayments of $100 immediately and $100 at the end of two years,in exchange for a payment of $201 at the end of one year? Find allpossible solutions.


Inflow −100 201 −100

Time 0 1 2

We have to solve −100 + 201(1 + i)−1 − 100(1 + i)−2 = 0, or100(1 + i)2 − 201(1 + i)1 + 100 = 0. Using the quadratic formula,

1 + i =201±

√2012 − 4 · 100 · 100

200=

201±√

201

200.

The two solutions are i = 10.5124922% and i = −9.512492197%.


7/16


Since the internal rate of return could either do not exist or haveseveral solutions, it is not a good indication of the performance ofgeneral investment strategy. However there exists a unique rate ofreturn i with i > −1 if either all outflows happen before all theinflows, or all inflows happen before all the outflows.


8/16


Suppose that you an investment strategy consisting of investing(positive) payments of C1, . . . ,Cm at times t1 < · · · < tm. Attimes s1 < · · · < sn, we get respective (positive) returnsP1, . . . ,Pn, where s1 > tm. The cashflow is

Inflows −C1 −C2 · · · −Cm P1 P2 · · · Pn

Time t1 t2 · · · tm s1 s2 · · · sn

In this case, there exists a unique solution to the equation

n∑k=1

Pk(1 + i)−sk −m∑

j=1

Cj(1 + i)−tj = 0, i > −1.

Besides,

I∑n

k=1 Pk >∑m

j=1 Cj , then i > 0.

I∑n

k=1 Pk <∑m

j=1 Cj , then i < 0.

I∑n

k=1 Pk =∑m

j=1 Cj , then i = 0.


9/16


Example 3

As the budgeting officer for Road Kill Motors Inc., you areevaluating the purchase of a new car factory. The cost of thefactory is $4 million today. It will provide inflows of $1.4 million atthe end of each of the first three years. Find the effective rate ofinterest which this investment will provide your company.


Contributions −4 1.4 1.4 1.4

Time 0 1 2 3

An equation of value for the cashflow is

0 = 4− (1.4)(1 + i)−1 − (1.4)(1 + i)−2 − (1.4)(1 + i)−3.

In the TI–BA–II–Plus calculator, we can find i , by going to CF ,

and enter CFo =−4, C01 =1.4, F01 =3, 2nd , QUIT . We can

move between different entries using the arrows ↓ and ↑ . Press

IRR CPT and get IRR = i = 2.47974%.


10/16


Example 3

As the budgeting officer for Road Kill Motors Inc., you areevaluating the purchase of a new car factory. The cost of thefactory is $4 million today. It will provide inflows of $1.4 million atthe end of each of the first three years. Find the effective rate ofinterest which this investment will provide your company.


Contributions −4 1.4 1.4 1.4

Time 0 1 2 3


0 = 4− (1.4)(1 + i)−1 − (1.4)(1 + i)−2 − (1.4)(1 + i)−3.


and enter CFo =−4, C01 =1.4, F01 =3, 2nd , QUIT . We can

move between different entries using the arrows ↓ and ↑ . Press

IRR CPT and get IRR = i = 2.47974%.


11/16


Example 4

Find the internal rate of return such that a payment of 400 at thepresent, 200 at the end of one year, and 300 at the end of twoyears, accumulate to 1000 at the end of 3 years.


Contributions −400 −200 −300 1000

Time 0 1 2 3


0 = −400(1 + i)3 − 200(1 + i)2 − 300(1 + i) + 1000.


and enter CFo =−400, C01 =−200, F01 =1, C02 =−300,

F02 =1, C03 =1000, F03 =1, 2nd , QUIT . We can move

between different entries using the arrows ↓ and ↑ . Press IRR

CPT and get IRR = 5.0709%.


12/16


Example 4

Find the internal rate of return such that a payment of 400 at thepresent, 200 at the end of one year, and 300 at the end of twoyears, accumulate to 1000 at the end of 3 years.


Contributions −400 −200 −300 1000

Time 0 1 2 3


0 = −400(1 + i)3 − 200(1 + i)2 − 300(1 + i) + 1000.


and enter CFo =−400, C01 =−200, F01 =1, C02 =−300,

F02 =1, C03 =1000, F03 =1, 2nd , QUIT . We can move

between different entries using the arrows ↓ and ↑ . Press IRR

CPT and get IRR = 5.0709%.


13/16


Example 5

An investment fund is established at time 0 with a deposit of$5000. $6000 is added at the end of 6 months. The fund value,including interest, is $11500 at the end of 1 year. Find the internalrate of return as a annual nominal rate convertible semiannually.


Investments 5000 6000 11500

Time (in half years) 0 1 2


0 = (5000) + (6000)

(1 +

i (2)

2

)−1

− (11500)

(1 +

i (2)

2

)−2

.

In the TI–BA–II–Plus calculator, press CF , and enter

CFo =5000, C01 =6000, F01 =1, C02 =−11500, F02 =1.

Press IRR , CPT and get IRR = i (2)

2 = 3.095064303% and

i (2) = 6.190128606%. The six–month effective interest rate is i (2)

2 .


14/16


Example 5

An investment fund is established at time 0 with a deposit of$5000. $6000 is added at the end of 6 months. The fund value,including interest, is $11500 at the end of 1 year. Find the internalrate of return as a annual nominal rate convertible semiannually.



Time (in half years) 0 1 2


0 = (5000) + (6000)

(1 +

i (2)

2

)−1

− (11500)

(1 +

i (2)

2

)−2

.


CFo =5000, C01 =6000, F01 =1, C02 =−11500, F02 =1.

Press IRR , CPT and get IRR = i (2)

2 = 3.095064303% and

i (2) = 6.190128606%. The six–month effective interest rate is i (2)

2 .


15/16


Example 6

An investment fund is established at time 0 with a deposit of$5000. $6000 is added at the end of 6 months. The fund value,including interest, is $11500 at the end of 1 year. Find the internalrate of return as a annual nominal rate convertible monthly.



Time (in months) 0 6 12


0 = (5000) + (6000)

(1 +

i (12)

12

)−6

− (11500)

(1 +

i (12)

12

)−12

.


CFo =5000, C01 =0, F01 =5, C02 =6000, F02 =1, C03 =0,

F03 =5, C04 =−11500, F04 =1. Press IRR , CPT and get

IRR = i (12)

12 = 0.509314804% and i (12) = 6.111777648%.


16/16


Example 6

An investment fund is established at time 0 with a deposit of$5000. $6000 is added at the end of 6 months. The fund value,including interest, is $11500 at the end of 1 year. Find the internalrate of return as a annual nominal rate convertible monthly.



Time (in months) 0 6 12


0 = (5000) + (6000)

(1 +

i (12)

12

)−6

− (11500)

(1 +

i (12)

12

)−12

.


CFo =5000, C01 =0, F01 =5, C02 =6000, F02 =1, C03 =0,

F03 =5, C04 =−11500, F04 =1. Press IRR , CPT and get

IRR = i (12)

12 = 0.509314804% and i (12) = 6.111777648%.


1/9



Section 2.4. Dollar–weighted and time–weighted rates ofreturn.





2/9

Chapter 2. Cashflows. Section 2.4. Dollar–weighted and time–weighted rates of return.

Dollar–weighted and time–weighted rates of return

If the cashflow

Investments V0 C1 C2 · · · Cn

Time 0 t1 t2 · · · tn

has future value FV at time t, then its equation of value is

FV = V0(1 + i)t +n∑

j=1

Cj(1 + i)t−tj .

Using the first order Taylor expansion 1 + it of (1 + i)t , theprevious equation of value is approximately,

FV = V0(1 + it) +n∑

j=1

Cj(1 + i(t − tj)). (1)

Observe Equation (1) represents the future value of the cashflowwhen simple interest is used.


3/9


The interest rate i which solves

FFV = V0(1 + it) +n∑

j=1

Cj(1 + i(t − tj)).

is called the dollar weighted rate of return, which is

i =FV − V0 −

∑nj=1 Cj

V0t +∑n

j=1(t − tj)Cj. (2)

V0 can be interpreted as the initial balance in an account. Cj is thedeposit at time tj . FV is the final balance in the account at timet. Hence, I = FV − V0 −

∑nj=1 Cj is the interest earned in the

account. V0t +∑n

j=1(t − tj)Cj is the sum of the depositsmultiplied by the time which the deposits are in the account.


4/9


Example 1

On January 1, 2000, the balance in account is $25200. On April 1,2000, $500 are deposited in this account and on July 1, 2001, awithdraw of $1000 is made. The balance in the account onOctober 1, 2001 is $25900. What is the annual rate of interest inthis account according with the dollar–weighted method?

Solution: The cashflow

Investments 25200 500 −1000

Time in years 0 3/12 18/12

has a FV at time 21/12 of $25900. So, the annual dollar–weightedrate of interest is

25900− 25200− 500 + 1000

(25200)(21/12) + (500)(18/12)− 1000(3/12)

=1200

44100 + 750− 250=

1200

44600= 2.6906%


5/9


Example 1

On January 1, 2000, the balance in account is $25200. On April 1,2000, $500 are deposited in this account and on July 1, 2001, awithdraw of $1000 is made. The balance in the account onOctober 1, 2001 is $25900. What is the annual rate of interest inthis account according with the dollar–weighted method?

Solution: The cashflow

Investments 25200 500 −1000

Time in years 0 3/12 18/12

has a FV at time 21/12 of $25900. So, the annual dollar–weightedrate of interest is

25900− 25200− 500 + 1000

(25200)(21/12) + (500)(18/12)− 1000(3/12)

=1200

44100 + 750− 250=

1200

44600= 2.6906%


6/9


Suppose that we make investments in a fund over time and weknow the outstanding balance before each deposit or withdrawaloccurs. Let B0 be the initial balance in the fund. Let Bj be thebalance in the fund immediately before time tj , for 1 ≤ j ≤ n. LetWj be the amount of each deposit or withdrawal at time tj .Wj > 0 for deposits and Wj < 0 for withdrawal. In a table, wehave:

Time 0 t1 t2 · · · tn−1 tnBalance beforedepos./withdr.

− B1 B2 · · · Bn−1 Bn

Depos./Withdr. − W1 W2 · · · Wn−1 −Balance afterdepos./withdr.

B0 B1 + W1 B2 + W2 · · · Bn−1 + Wn−1 −

The time–weighted annual rate of return i is the solution of

(1 + i)tn =B1

B0· B2

B1 + W1· B3

B2 + W2· · · Bn

Bn−1 + Wn−1.


7/9


In the j–th period of time, the balance of the fund has changedfrom Bj−1 + Wj−1 to Bj . So, the interest factor rate in the j–th

period of time is 1 + ij =Bj

Bj−1+Wj−1, where ij is the effective rate

of return in the period [tj−1, tj ]. Observe that if the investmentfollowed an annual effective rate of interest of i , the interest factorfrom time tj−1 to time tj would be (1 + i)tj−tj−1 . Assuming that1 + ij = (1 + i)tj−tj−1 , we get that

(1+i1)(1+i2) · · · (1+in) = (1+i)t1(1+i)t2−t1 · · · (1+i)tn−tn−1 = (1+i)tn .

The time–weighted annual rate of return i is the solution of

(1 + i)tn =B1

B0· B2

B1 + W1· B3

B2 + W2· · · Bn

Bn−1 + Wn−1.

Usually, the account balance does not follow compound interestwith a fixed effective rate i . Usually, 1 + ij and (1 + i)tj−tj−1 maybe different.


8/9


Example 2

For an investment account, you are given:

Date 11/1/04 3/1/05 8/1/05 2/1/06 4/1/06

Account Balance(before depositor withdrawal)

14,516 14,547 18,351 16,969 18,542

Deposit – 3,000 – 2500 –

Withdrawal – – 2,000 – –

Calculate the annual effective yield rate by the time weightedmethod.

Solution: The annual effective yield rate i by the time weightedmethod satisfies

(1 + i)17/12 = B1B0· B2

B1+W1· B3

B2+W2· s Bn

Bn−1+Wn−1

= 1454714516

1835114547+3000

1696918351−2000

1854216969+2500 = 1.035877

and i = 2.5193371%.


9/9


Example 2

For an investment account, you are given:

Date 11/1/04 3/1/05 8/1/05 2/1/06 4/1/06

Account Balance(before depositor withdrawal)

14,516 14,547 18,351 16,969 18,542

Deposit – 3,000 – 2500 –

Withdrawal – – 2,000 – –

Calculate the annual effective yield rate by the time weightedmethod.

Solution: The annual effective yield rate i by the time weightedmethod satisfies

(1 + i)17/12 = B1B0· B2

B1+W1· B3

B2+W2· s Bn

Bn−1+Wn−1

= 1454714516

1835114547+3000

1696918351−2000

1854216969+2500 = 1.035877

and i = 2.5193371%.c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2.

1/8



Section 2.5. Investment and portfolio methods.





2/8

Chapter 2. Cashflows. Section 2.5. Investment and portfolio methods.

Investment and portfolio methods

Suppose that an investment fund pools money from severalidentities (individuals or corporations) and makes investments onbehalf of them. Then, the fund faces the question: how to allocatethe returns between the different identities? There are two mainways to allocate interest to the various accounts: the portfoliomethod and the investment year method.The portfolio method is an accounting method that credits allfunds one specified current rate of interest, regardless of when themoney was placed in the account. Usually this rate of interestchanges from year to year. Let iy denote the annual interest ratecredited in year y . If x is invested at the beginning of the year y ,then the balance in the account in the year y + t is

xt−1∏j=0

(1 + iy+j) = x(1 + iy )(1 + iy+1) · · · (1 + iy+t−1).


3/8


Example 1

Suppose that an investment account credits investors using theportfolio method with the annual rates in the following table:

Calendar year Portfolioof portfolio rate rates

y iy

1999 4.50%2000 5.50%2001 4.00%2002 6.50%

Suppose that 100 was invested on January 1, 1999.(i) Find the balance on January 1, 2000.(ii) The balance on January 1, 2001.(iii) The balance on July 1, 2001.


4/8


Solution: (i) The balance on January 1, 2000, is(100)(1.045) = 104.5.(ii) The balance on January 1, 2001, is(100)(1.045)(1.055) = 110.2475.(iii) The balance on July 1, 2001, is

(100)(1.045)(1.055)(1.04)1/2 = 112.4308.


5/8


The investment year method is an accounting method in which aninvestment fund keeps records of the interest rates it earns annuallyon funds assigned each year to accounts within the generalaccount. The investment year method is also called the newmoney method. We will assume that accounts are made accordingwith the year at which the money was invested. For example,suppose that an investment account credits investors accordingwith the investment year method using the following table:

Calendar year of Investment year ratesoriginal investment

y iy1 iy2 iy3 iy4 iy51999 4.25% 4.35% 4.47% 4.57% 4.70%2000 4.56% 4.73% 4.75% 4.98% 4.04%2001 4.05% 4.04% 4.13% 4.17% 4.24%2002 4.45% 4.15% 4.23% 4.36% 4.44%2003 4.25% 4.35% 4.55% 9.55% 5.65%


6/8


Calendar year of Investment year ratesoriginal investment

y iy1 iy2 iy3 iy4 iy51999 4.25% 4.35% 4.47% 4.57% 4.70%2000 4.56% 4.73% 4.75% 4.98% 4.04%2001 4.05% 4.04% 4.13% 4.17% 4.24%2002 4.45% 4.15% 4.23% 4.36% 4.44%2003 4.25% 4.35% 4.55% 9.55% 5.65%

This means that money invested during 2000 earns an effectiverate of interest of 4.56% during 2000, it earns an effective rate ofinterest of 4.73% during 2001, and so on. For example, if anaccount is open with an investment of x invested on January 1,2000, then:the balance on January 1, 2001 is (100)(1.0456);the balance on January 1, 2002 is (100)(1.0456)(1.0473); and soon.


7/8


Example 2

An investment fund applies the investment year method for thefirst two years, after which a portfolio rate is used. The followingtable of interest rates is used:

Calendar year Investment Portfolioof original investment year rates rates

y iy1 iy2 iy+2

2000 5.25% 5.25% 5.40%2001 5.35% 5.35% 5.65%2002 5.45% 5.45% 5.10%2003 5.45% 5.45% 5.34%2004 5.50% 5.35% 5.55%2005 5.50% 5.55% 5.65%


8/8


(i) Ashley invests $1000 into the fund on January 1, 2000. Theinvestment year method is applicable for the first two years, afterwhich a portfolio rate is used. Calculate Ashley’s accountaccumulation on January 1, 2006.(ii) Elizabeth invests $1000 into the fund on January 1, 2000. But,she redeemed her investment from the fund at the end of everyyear and reinvested the money at the new money rate. CalculateElizabeth’s accumulation account on January 1, 2006.Solution: (i) Ashley’s account accumulation on January 1, 2006 is

(1000)(1.0525)(1.0525)(1.0540)(1.0565)(1.0510)(1.0534) = 1365.684.

(ii) Elizabeth’s investment value on January 1, 2006 is

(1000)(1.0525)(1.0535)(1.0545)(1.0545)(1.055)(1.050) = 1365.814.


1/10



Section 2.6. Continuous payments.





2/10

Chapter 2. Cashflows. Section 2.6. Continuous payments.

Continuous payments

Suppose that the payments are made very often. Then byapproximation, instead of a sum, we have an integral. It is like thepayments are made continuously. Let V (t) be the outstandingfund balance at time t of the cashflow. Assume that contributionsare made continuously at an instantaneous rate C (t), then theequation of value is

V (t) = V (0)(1 + i)t +

∫ t

0C (s)(1 + i)t−s ds. (1)


3/10


(1) appears as the limit of the equation of value for the cashflow:

Inflow V (0) C (t1)(t1 − 0) C (t2)(t2 − t1) · · · C (tn)(tn − tn−1)Time 0 t1 t2 · · · tn

as max1≤j≤n(tj − tj−1) → 0, where0 = t0 < t1 < t2 < · · · < tm = t. The equation of value at time tfor this cashflow is

V (t) = V (0)(1 + i)t +n∑

j=1

C (tj)(tj − tj−1)(1 + i)t−tj ,

which tends to

V (t) = V (0)(1 + i)t +

∫ t

0C (s)(1 + i)t−s ds

as max1≤j≤n(tj − tj−1) → 0.Recall that the Riemann integral of a function f is defined as∫ t

0f (s) ds = lim

max1≤j≤n(tj−tj−1)→0

n∑j=1

f (tj)(tj − tj−1).


4/10


Example 1

A continuous–year annuity pays a constant rate 1 at time t where0 ≤ t ≤ n. Interest is compounded with an annual rate of interestof i .(i) Find the present value of the annuity at time 0.(ii) Find the future value of the annuity at time n.

Solution: (i) The present value of this continuous annuity is

PV =

∫ n

0(1 + i)−t dt =

∫ n

0e−t ln(1+i) dt =

−e−t ln(1+i)

ln(1 + i)

∣∣∣∣n0

=1

ln(1 + i)− e−n ln(1+i)

ln(1 + i)=

1− (1 + i)−n

ln(1 + i).

(ii) The future value of the continuous annuity at time n is

FV =

∫ n

0(1 + i)n−t dt = (1 + i)nPV =

(1 + i)n − 1

ln(1 + i).


5/10


Example 1

A continuous–year annuity pays a constant rate 1 at time t where0 ≤ t ≤ n. Interest is compounded with an annual rate of interestof i .(i) Find the present value of the annuity at time 0.(ii) Find the future value of the annuity at time n.

Solution: (i) The present value of this continuous annuity is

PV =

∫ n

0(1 + i)−t dt =

∫ n

0e−t ln(1+i) dt =

−e−t ln(1+i)

ln(1 + i)

∣∣∣∣n0

=1

ln(1 + i)− e−n ln(1+i)

ln(1 + i)=

1− (1 + i)−n

ln(1 + i).

(ii) The future value of the continuous annuity at time n is

FV =

∫ n

0(1 + i)n−t dt = (1 + i)nPV =

(1 + i)n − 1

ln(1 + i).


6/10


If instead of compound interest, the time value of money followsthe accumulation function a(t), then the future value at time t ofan initial outstanding balance V (0) and continuous paymentsC (s), in the interval 0 ≤ s ≤ t is

V (t) = V (0)a(t) +

∫ t

0C (s)

a(t)

a(s)ds.


7/10


Example 2

The force of interest at time t is δt = t3

10 . Find the present value ofa four–year continuous annuity which has a rate of payments attime t of 5t3.

Solution: The accumulation function is

a(t) = exp

(∫ t

0δs ds

)

= exp

(∫ t

0

s3

10ds

)= e

t4

40 .

The present value of the four-year continuous annuity is∫ 4

0

C (s)

a(s)ds =

∫ 4

05s3e−

s4

40 ds = −50e−s4

40

∣∣∣∣40

= 50− 50e−6.4

=49.91692.


8/10


Example 2




a(t) = exp

(∫ t

0δs ds

)= exp

(∫ t

0

s3

10ds

)

= et4

40 .


0

C (s)

a(s)ds =

∫ 4

05s3e−

s4

40 ds = −50e−s4

40

∣∣∣∣40

= 50− 50e−6.4

=49.91692.


9/10


Example 2




a(t) = exp

(∫ t

0δs ds

)= exp

(∫ t

0

s3

10ds

)= e

t4

40 .


0

C (s)

a(s)ds =

∫ 4

05s3e−

s4

40 ds = −50e−s4

40

∣∣∣∣40

= 50− 50e−6.4

=49.91692.


10/10


Example 2




a(t) = exp

(∫ t

0δs ds

)= exp

(∫ t

0

s3

10ds

)= e

t4

40 .


0

C (s)

a(s)ds =

∫ 4

05s3e−

s4

40 ds = −50e−s4

40

∣∣∣∣40

= 50− 50e−6.4

=49.91692.


1/??

Chapter 3. Annuities.

Manual for SOA Exam FM/CAS Exam 2.Chapter 3. Annuities.

Section 3.1. Geometric series.





2/??

Chapter 3. Annuities. Section 3.1. Geometric series.

We use the summation notation∑n

i=m xi to mean

xm + xm+1 + · · ·+ xn−1 + xn.

Usually arithmetic rules hold. In particular:

I∑n

i=m(xi + yi ) =∑n

i=m xi +∑n

i=m yi

I∑n

i=m axi = a∑n

i=m xi .

I If m ≤ k ≤ n,

n∑i=m

xi =k∑

i=m

xi +n∑

i=k+1

xi .

I∑n

i=m 1 = n −m + 1.


3/??



i=m xi to mean

xm + xm+1 + · · ·+ xn−1 + xn.


I∑n

i=m(xi + yi ) =∑n

i=m xi +∑n

i=m yi

I∑n

i=m axi = a∑n

i=m xi .

I If m ≤ k ≤ n,

n∑i=m

xi =k∑

i=m

xi +n∑

i=k+1

xi .

I∑n

i=m 1 = n −m + 1.


4/??



i=m xi to mean

xm + xm+1 + · · ·+ xn−1 + xn.


I∑n

i=m(xi + yi ) =∑n

i=m xi +∑n

i=m yi

I∑n

i=m axi = a∑n

i=m xi .

I If m ≤ k ≤ n,

n∑i=m

xi =k∑

i=m

xi +n∑

i=k+1

xi .

I∑n

i=m 1 = n −m + 1.


5/??



i=m xi to mean

xm + xm+1 + · · ·+ xn−1 + xn.


I∑n

i=m(xi + yi ) =∑n

i=m xi +∑n

i=m yi

I∑n

i=m axi = a∑n

i=m xi .

I If m ≤ k ≤ n,

n∑i=m

xi =k∑

i=m

xi +n∑

i=k+1

xi .

I∑n

i=m 1 = n −m + 1.


6/??


Definition 1The sequence of real numbers ∞n=0 = {a + nd}∞n=0 is called anarithmetic sequence.

The sequence {xn}∞n=0 = {a + nd}∞n=0 satisfies that for each n ≥ 1,xn−1 + d = xn. Notice that

xn−1 + d = a + (n − 1)d + d = a + nd = xn.

Theorem 1If a sequence {xn}∞n=0 of real numbers satisfies xn = xn−1 + d, foreach n ≥ 1, then xn = x0 + nd for each n ≥ 1.

Proof.The proof is by induction on n. The case n = 0 is obvious. Assumethat the case n holds. Then,xn+1 = xn + d = x0 + nd + d = x0 + (n + 1)d .


7/??




xn−1 + d = a + (n − 1)d + d = a + nd = xn.




8/??




xn−1 + d = a + (n − 1)d + d = a + nd = xn.




9/??




xn−1 + d = a + (n − 1)d + d = a + nd = xn.




10/??


Theorem 2∑nj=1 j = n(n+1)

2 .

Proof.

2n∑

j=1

j = (1 + 2 + · · ·+ (n − 1) + n) + (1 + 2 + · · ·+ (n − 1) + n)

=(1 + 2 + · · ·+ (n − 1) + n) + (n + (n − 1) + · · ·+ 2 + 1)

=(1 + n) + (2 + n − 1) + (3 + n − 2) · · ·+ (n + 1)

=(n + 1) + (n + 1) + (n + 1) · · ·+ (n + 1) = n(n + 1).

Previous theorem can be proved by induction.Note that in the summation

∑nj=1 j , there are n numbers and the

average of these numbers is n+12 . Hence,

∑nj=1 j = n(n+1)

2 .


11/??



2 .

Proof.

2n∑

j=1

j = (1 + 2 + · · ·+ (n − 1) + n) + (1 + 2 + · · ·+ (n − 1) + n)

=(1 + 2 + · · ·+ (n − 1) + n) + (n + (n − 1) + · · ·+ 2 + 1)

=(1 + n) + (2 + n − 1) + (3 + n − 2) · · ·+ (n + 1)

=(n + 1) + (n + 1) + (n + 1) · · ·+ (n + 1) = n(n + 1).




∑nj=1 j = n(n+1)

2 .


12/??



2 .

Proof.

2n∑

j=1

j = (1 + 2 + · · ·+ (n − 1) + n) + (1 + 2 + · · ·+ (n − 1) + n)

=(1 + 2 + · · ·+ (n − 1) + n) + (n + (n − 1) + · · ·+ 2 + 1)

=(1 + n) + (2 + n − 1) + (3 + n − 2) · · ·+ (n + 1)

=(n + 1) + (n + 1) + (n + 1) · · ·+ (n + 1) = n(n + 1).

Previous theorem can be proved by induction.

Note that in the summation∑n

j=1 j , there are n numbers and the


∑nj=1 j = n(n+1)

2 .


13/??



2 .

Proof.

2n∑

j=1

j = (1 + 2 + · · ·+ (n − 1) + n) + (1 + 2 + · · ·+ (n − 1) + n)

=(1 + 2 + · · ·+ (n − 1) + n) + (n + (n − 1) + · · ·+ 2 + 1)

=(1 + n) + (2 + n − 1) + (3 + n − 2) · · ·+ (n + 1)

=(n + 1) + (n + 1) + (n + 1) · · ·+ (n + 1) = n(n + 1).




∑nj=1 j = n(n+1)

2 .


14/??


For an arithmetic sequence,

n∑j=0

(a + jd) =n∑

j=0

a + dn∑

j=0

j = (n + 1)a + dn(n + 1)

2.


15/??


Example 1

Find∑100

k=10 k.

Solution 1:∑100k=10 k =

∑100k=1 k −

∑9k=1 k = (100)(101)

2 − (9)(10)2 = 5005.

Solution 2: By the change of variables k = j + 9,

100∑k=10

k =91∑j=1

(j + 9) =(91)(92)

2+ (9)(91) = 5005.

Notice that if k = 10, then j = 9; and if k = 100, then j = 91.


16/??


Example 1

Find∑100

k=10 k.


∑100k=1 k −

∑9k=1 k = (100)(101)

2 − (9)(10)2 = 5005.


100∑k=10

k =91∑j=1

(j + 9) =(91)(92)

2+ (9)(91) = 5005.



17/??


Example 1

Find∑100

k=10 k.


∑100k=1 k −

∑9k=1 k = (100)(101)

2 − (9)(10)2 = 5005.


100∑k=10

k =91∑j=1

(j + 9) =(91)(92)

2+ (9)(91) = 5005.



18/??


Definition 2The sequence {arn}∞n=0 is called a geometric sequence.

The geometric sequence {arn}∞n=0 satisfies that for each n ≥ 1,rxn−1 = rarn−1 = arn = xn, where xn = arn.


19/??


Definition 2The sequence {arn}∞n=0 is called a geometric sequence.

The geometric sequence {arn}∞n=0 satisfies that for each n ≥ 1,rxn−1 = rarn−1 = arn = xn, where xn = arn.


20/??


Theorem 3If a sequence satisfies xn = rxn−1, for each n ≥ 1, then xn = x0r

n

for each n ≥ 1.

Proof.The proof is by induction on n. The case n = 0 is obvious. Assumethat the case n holds. Then,xn+1 = rxn = rx0r

n = x0rn+1 = xn+1.


21/??


Theorem 3If a sequence satisfies xn = rxn−1, for each n ≥ 1, then xn = x0r

n

for each n ≥ 1.

Proof.The proof is by induction on n. The case n = 0 is obvious. Assumethat the case n holds. Then,xn+1 = rxn = rx0r

n = x0rn+1 = xn+1.


22/??


Theorem 4For any r ∈ R,

rn+1 − 1 = (r − 1)n∑

j=0

r j .

Proof.

∑nj=0 r j(r − 1) = (1 + r + r2 + · · ·+ rn)(r − 1)

= (r + r2 + · · ·+ rn+1)− (1 + r + r2 + · · ·+ rn) = rn+1 − 1.


23/??


In particular, we that

(x2 − 1) = (x − 1)(1 + x),

(x3 − 1) = (x − 1)(1 + x + x2),

(x4 − 1) = (x − 1)(1 + x + x2 + x3),

(x5 − 1) = (x − 1)(1 + x + x2 + x3 + x4).


24/??


Corollary 1

(i) If r 6= 1,∑n

j=0 r j = rn+1−1r−1 .

(ii) If r = 1,∑n

j=0 r j = n + 1.

Proof.(i) If r 6= 1, from rn+1 − 1 = (r − 1)

∑nj=0 r j , we get that

n∑j=0

r j =rn+1 − 1

r − 1.

(ii) If r = 1,n∑

j=0

r j =n∑

j=0

1 = n + 1.


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Corollary 1

(i) If r 6= 1,∑n

j=0 r j = rn+1−1r−1 .

(ii) If r = 1,∑n

j=0 r j = n + 1.

Proof.(i) If r 6= 1, from rn+1 − 1 = (r − 1)

∑nj=0 r j , we get that

n∑j=0

r j =rn+1 − 1

r − 1.

(ii) If r = 1,n∑

j=0

r j =n∑

j=0

1 = n + 1.


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Example 2

Find∑20

k=5 2k .

Solution:

20∑k=5

2k =20∑

k=0

2k−4∑

k=0

2k =221 − 1

2− 1− 25 − 1

2− 1= 221−25 = 2097120.


27/??


Example 2

Find∑20

k=5 2k .

Solution:

20∑k=5

2k =20∑

k=0

2k−4∑

k=0

2k =221 − 1

2− 1− 25 − 1

2− 1= 221−25 = 2097120.


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We also have that for 1 ≤ m ≤ n and r 6= 1,

n∑j=m

r j =n∑

j=0

r j −m−1∑j=0

r j =rn+1 − 1

r − 1− rm − 1

r − 1=

rn+1 − rm

r − 1.

Example 3

Find∑20

k=5 2k .

Solution:∑20

k=5 2k = 221−25

2−1 = 221 − 25 = 2097120.


29/??



n∑j=m

r j =n∑

j=0

r j −m−1∑j=0

r j =rn+1 − 1

r − 1− rm − 1

r − 1=

rn+1 − rm

r − 1.

Example 3

Find∑20

k=5 2k .

Solution:∑20

k=5 2k = 221−25

2−1 = 221 − 25 = 2097120.


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n∑j=m

r j =n∑

j=0

r j −m−1∑j=0

r j =rn+1 − 1

r − 1− rm − 1

r − 1=

rn+1 − rm

r − 1.

Example 3

Find∑20

k=5 2k .

Solution:∑20

k=5 2k = 221−25

2−1 = 221 − 25 = 2097120.


31/??


Theorem 5For |r | < 1,

∑∞j=0 r j = 1

1−r .

Proof.If |r | < 1, then ln(|r |) < 0 and |rn+1| ≤ |r |n+1 = e(n+1) ln(|r |) → 0,as n →∞. Hence,

∞∑j=0

r j = limn→∞

n∑j=0

r j = limn→∞

rn+1 − 1

r − 1=

1

1− r.


32/??


Theorem 5For |r | < 1,

∑∞j=0 r j = 1

1−r .

Proof.If |r | < 1, then ln(|r |) < 0 and |rn+1| ≤ |r |n+1 = e(n+1) ln(|r |) → 0,as n →∞. Hence,

∞∑j=0

r j = limn→∞

n∑j=0

r j = limn→∞

rn+1 − 1

r − 1=

1

1− r.


33/??


Corollary 2

For |r | < 1,∞∑j=n

r j =rn

1− r.

Proof.By the change of variables j = k + n,

∞∑j=n

r j =∞∑

k=0

rk+n = rn∞∑

k=0

rk =rn

1− r.


34/??


Corollary 2

For |r | < 1,∞∑j=n

r j =rn

1− r.

Proof.By the change of variables j = k + n,

∞∑j=n

r j =∞∑

k=0

rk+n = rn∞∑

k=0

rk =rn

1− r.


35/??


Example 4

Find∑∞

k=9 3−k .

Solution:

∞∑k=9

3−k =3−9

1− (1/3)= 0.0000762079.


36/??


Example 4

Find∑∞

k=9 3−k .

Solution:

∞∑k=9

3−k =3−9

1− (1/3)= 0.0000762079.


37/??


Theorem 6For r 6= 1,

n∑j=1

jr j =nrn+2 − (n + 1)rn+1 + r

(r − 1)2.

Proof.Taking derivatives with respect to r in the inequality∑n

j=0 r j = rn+1−1r−1 , we get that

∑nj=1 jr j−1 = (n+1)rn(r−1)−(rn+1−1)

(r−1)2= nrn+1−(n+1)rn+1

(r−1)2


38/??


Theorem 6For r 6= 1,

n∑j=1

jr j =nrn+2 − (n + 1)rn+1 + r

(r − 1)2.

Proof.Taking derivatives with respect to r in the inequality∑n

j=0 r j = rn+1−1r−1 , we get that

∑nj=1 jr j−1 = (n+1)rn(r−1)−(rn+1−1)

(r−1)2= nrn+1−(n+1)rn+1

(r−1)2


39/??


Corollary 3

For |r | < 1,∑∞

j=1 jr j = r(r−1)2

.

Proof.If |r | < 1, then

∞∑j=1

jr j = limn→∞

n∑j=1

jr j = limn→∞

nrn+2 − (n + 1)rn+1 + r

(r − 1)2=

r

(r − 1)2.


40/??


Corollary 3

For |r | < 1,∑∞

j=1 jr j = r(r−1)2

.

Proof.If |r | < 1, then

∞∑j=1

jr j = limn→∞

n∑j=1

jr j = limn→∞

nrn+2 − (n + 1)rn+1 + r

(r − 1)2=

r

(r − 1)2.


41/??


Example 5

Find∑∞

k=1 k4−k .

Solution:

∞∑k=1

k4−k =4−1

(1− 4−1)2=

4

9= 0.4444444444.


42/??


Example 5

Find∑∞

k=1 k4−k .

Solution:

∞∑k=1

k4−k =4−1

(1− 4−1)2=

4

9= 0.4444444444.


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Section 3.2. Level payment annuities.





2/82

Chapter 3. Annuities. Section 3.2. Level payment annuities.

An annuity is a sequence of payments made at equal intervals oftime. We have n periods of times[0, t], [t, 2t], [2t, 3t], . . . [(n − 1)t, nt] with the same length. By achange of units, we will assume that intervals have unit length. So,the intervals are [0, 1], [1, 2], [2, 3], . . . [n − 1, n]. We order theperiods as follows:

Time interval NameBeginning

of the periodEnd

of the period

[0, 1] 1st period time 0 time 1[1, 2] 2nd period time 1 time 2[2, 3] 3rd period time 2 time 3· · · · · · · · · · · ·

[n − 2, n − 1] (n − 1)–th period time n − 2 time n − 1[n − 1, n] n–th period time n − 1 time n


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An annuity is said to have level payments if all payments Cj areequal.

An annuity has non–level payments if some payments Cj aredifferent from other ones.The payments can be made either at the beginning or at the endof intervals of time.For an annuity–immediate payments are made at the end of theintervals of time.An annuity–immediate is a cashflow of the type:

Contributions 0 C1 C2 · · · Cn

Time 0 1 2 · · · n

For an annuity–due the payments are made at the beginning ofthe intervals of time.An annuity–due is a cashflow of the type:

Contributions C0 C1 · · · Cn−1 0

Time 0 1 · · · n − 1 n


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An annuity is said to have level payments if all payments Cj areequal.An annuity has non–level payments if some payments Cj aredifferent from other ones.

The payments can be made either at the beginning or at the endof intervals of time.For an annuity–immediate payments are made at the end of theintervals of time.An annuity–immediate is a cashflow of the type:


Time 0 1 2 · · · n



Time 0 1 · · · n − 1 n


5/82


An annuity is said to have level payments if all payments Cj areequal.An annuity has non–level payments if some payments Cj aredifferent from other ones.The payments can be made either at the beginning or at the endof intervals of time.

For an annuity–immediate payments are made at the end of theintervals of time.An annuity–immediate is a cashflow of the type:


Time 0 1 2 · · · n



Time 0 1 · · · n − 1 n


6/82


An annuity is said to have level payments if all payments Cj areequal.An annuity has non–level payments if some payments Cj aredifferent from other ones.The payments can be made either at the beginning or at the endof intervals of time.For an annuity–immediate payments are made at the end of theintervals of time.An annuity–immediate is a cashflow of the type:


Time 0 1 2 · · · n



Time 0 1 · · · n − 1 n


7/82


An annuity is said to have level payments if all payments Cj areequal.An annuity has non–level payments if some payments Cj aredifferent from other ones.The payments can be made either at the beginning or at the endof intervals of time.For an annuity–immediate payments are made at the end of theintervals of time.An annuity–immediate is a cashflow of the type:


Time 0 1 2 · · · n



Time 0 1 · · · n − 1 n


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The cashflow of an annuity–immediate with level payments of oneis

Contributions 0 1 1 · · · 1

Time 0 1 2 · · · n

If the time value of the money follows an accumulation functiona(t), then the present value of an annuity–immediate withlevel annual payments of one is

an| =1

a(1)+

1

a(2)+ · · ·+ 1

a(n)=

n∑j=1

1

a(j).

The accumulated value of an annuity–immediate with levelannual payments of one is

sn| =a(n)

a(1)+

a(n)

a(2)+ · · ·+ a(n)

a(n)=

n∑j=1

a(n)

a(j).


9/82


Example 1

You are given that δt = 18+t , t ≥ 0. Find an| and sn|.

Solution: We have that∫ t

0δs ds =

∫ t

0

1

8 + sds = ln(8 + s)

∣∣t0= ln

(8 + t

8

).

So, a(t) = eR t0 δs = 8+t

8 ,

an| =n∑

j=1

1

a(j)=

n∑j=1

8

8 + j

and

sn| =n∑

j=1

a(n)

a(j)=

n∑j=1

8 + n

8 + j.


10/82


Example 1

You are given that δt = 18+t , t ≥ 0. Find an| and sn|.

Solution: We have that∫ t

0δs ds =

∫ t

0

1

8 + sds = ln(8 + s)

∣∣t0= ln

(8 + t

8

).

So, a(t) = eR t0 δs = 8+t

8 ,

an| =n∑

j=1

1

a(j)=

n∑j=1

8

8 + j

and

sn| =n∑

j=1

a(n)

a(j)=

n∑j=1

8 + n

8 + j.


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The cashflow of an annuity–due with n level payments of one is

Contributions 1 1 1 · · · 1 0

Time 0 1 2 · · · n − 1 n

The present value of an annuity–due with n level annualpayments of one is

an| = 1 +1

a(1)+

1

a(2)+ · · ·+ 1

a(n − 1)=

n−1∑j=0

1

a(j).

The future value at time n of an annuity–due with levelannual payments of one is

sn| = a(n) +a(n)

a(1)+

a(n)

a(2)+ · · ·+ a(n)

a(n − 1)=

n−1∑j=0

a(n)

a(j).


12/82


Example 2

Suppose that the annual effective interest rate for year n isin = 2

n+4 . Find an| and sn|.

Solution: Since 1 + in = 1 + 2n+4 = n+6

n+4 ,

a(n) = (1 + i1)(1 + i1) · · · (1 + in) =1 + 6

1 + 4· 2 + 6

2 + 4· 3 + 6

3 + 4· · · n + 6

n + 4

=(n + 6)(n + 5)

30,

an| =n∑

j=1

1

a(j)=

n−1∑j=0

30

(j + 6)(j + 5)=

n−1∑j=0

(30)

(1

j + 5− 1

j + 6

)

=30

(1

0 + 5+ · · ·+ 1

n − 1 + 5

)− 30

(1

1 + 5+ · · ·+ 1

n + 5

)=(30)

(1

5− 1

n + 5

)=

6n

n + 5, sn| = a(n)an| =

n(n + 6)

5.


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Example 2

Suppose that the annual effective interest rate for year n isin = 2

n+4 . Find an| and sn|.

Solution: Since 1 + in = 1 + 2n+4 = n+6

n+4 ,

a(n) = (1 + i1)(1 + i1) · · · (1 + in) =1 + 6

1 + 4· 2 + 6

2 + 4· 3 + 6

3 + 4· · · n + 6

n + 4

=(n + 6)(n + 5)

30,

an| =n∑

j=1

1

a(j)=

n−1∑j=0

30

(j + 6)(j + 5)=

n−1∑j=0

(30)

(1

j + 5− 1

j + 6

)

=30

(1

0 + 5+ · · ·+ 1

n − 1 + 5

)− 30

(1

1 + 5+ · · ·+ 1

n + 5

)=(30)

(1

5− 1

n + 5

)=

6n

n + 5, sn| = a(n)an| =

n(n + 6)

5.


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Remember:The cashflow of an annuity–immediate with n level payments ofone is


Time 0 1 2 · · · n



Time 0 1 2 · · · n − 1 n


15/82


Remember:The cashflow of an annuity–immediate with n level payments ofone is


Time 0 1 2 · · · n



Time 0 1 2 · · · n − 1 n


16/82


Theorem 1If i 6= 0, the present value of an annuity–immediate with levelpayments of one is

an|i = an| = ν + ν2 + · · ·+ νn =ν(1− νn)

1− ν=

1− νn

1ν − 1

=1− νn

i

=1− (1 + i)−n

i,

where we have used that ν(1 + i) = 1.

If i 6= 0, the future value of an annuity–immediate with levelpayments of one at time n is

sn|i = sn| = (1+ i)n−1+(1+ i)n−2+ · · ·+(1+ i)+1 =(1 + i)n − 1

i.

If i = 0, an|i = sn|i = n.


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an|i = an| = ν + ν2 + · · ·+ νn =ν(1− νn)

1− ν=

1− νn

1ν − 1

=1− νn

i

=1− (1 + i)−n

i,

where we have used that ν(1 + i) = 1.If i 6= 0, the future value of an annuity–immediate with levelpayments of one at time n is

sn|i = sn| = (1+ i)n−1+(1+ i)n−2+ · · ·+(1+ i)+1 =(1 + i)n − 1

i.

If i = 0, an|i = sn|i = n.


18/82



an|i = an| = ν + ν2 + · · ·+ νn =ν(1− νn)

1− ν=

1− νn

1ν − 1

=1− νn

i

=1− (1 + i)−n

i,

where we have used that ν(1 + i) = 1.If i 6= 0, the future value of an annuity–immediate with levelpayments of one at time n is

sn|i = sn| = (1+ i)n−1+(1+ i)n−2+ · · ·+(1+ i)+1 =(1 + i)n − 1

i.

If i = 0, an|i = sn|i = n.


19/82


For example,

a1|i = ν,

a2|i = ν + ν2,

a3|i = ν + ν2 + ν3,

a4|i = ν + ν2 + ν3 + ν4.

s1|i = 1,

s2|i = 1 + 1 + i ,

s3|i = 1 + 1 + i + (1 + i)2,

s4|i = 1 + 1 + i + (1 + i)2 + (1 + i)3.


20/82


Example 3

Calculate the present value of $5000 paid at the end of each yearfor 15 years using an annual effective interest rate of 7.5%.

Solution: The present value is

(5000)a15|0.075 = (5000)1− (1.075)−15

0.075= 44135.59873.


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Example 3

Calculate the present value of $5000 paid at the end of each yearfor 15 years using an annual effective interest rate of 7.5%.

Solution: The present value is

(5000)a15|0.075 = (5000)1− (1.075)−15

0.075= 44135.59873.


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If every period is exactly one year, then i in the formulas above isthe annual effective rate of interest. If the length of a period is nota year, i in the formulas above is the effective rate of interest perperiod. If each period lasts t years, then the t–year interest factoris (1 + i)t and the t–year effective rate of interest is (1 + i)t − 1.So, if each period lasts t years, we use the previous formulas with ireplaced by (1 + i)t − 1, where the last i is the annual effectiverate of interest.


23/82


For example, suppose that the payments are made each 1/m years.Then, the interest factor for 1/m years is (1 + i)1/m and the 1/m

year interest rate is (1 + i)1/m − 1 = i (m)

m , where i (m) is the nominalannual rate of interest convertible m times a year. For example,the present value at time 0 of the annuity


Time (in years) 0 1/m 2/m · · · n/m

is an|(1+i)1/m−1 = an|i (m)/m. The future value at time n/m years ofthe previous cashflow is sn|(1+i)1/m−1 = sn|i (m)/m.


24/82


Example 4

John invest $500 into an account at the end of each month for 5years. The annual effective interest rate is 4.5%. Calculate thebalance of this account at the end of 5 years.

Solution: The number of payments made is (5)(12) = 60. Thecashflow is

Contributions 500 500 500 · · · 500

Time (in months) 1 2 2 · · · 60

The one–month effective interest rate is (1.045)1/12 − 1. Hence,the balance of this account at the end of 5 years is

(500)s60|(1.045)1/12−1 = (500)((1.045)1/12 − 1 + 1)60 − 1

(1.045)1/12 − 1

=(500)(1.045)5 − 1

(1.045)1/12 − 1= 33495.8784.


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Example 4

John invest $500 into an account at the end of each month for 5years. The annual effective interest rate is 4.5%. Calculate thebalance of this account at the end of 5 years.

Solution: The number of payments made is (5)(12) = 60. Thecashflow is

Contributions 500 500 500 · · · 500

Time (in months) 1 2 2 · · · 60

The one–month effective interest rate is (1.045)1/12 − 1. Hence,the balance of this account at the end of 5 years is

(500)s60|(1.045)1/12−1 = (500)((1.045)1/12 − 1 + 1)60 − 1

(1.045)1/12 − 1

=(500)(1.045)5 − 1

(1.045)1/12 − 1= 33495.8784.


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If payments are made every m years, the interest factor per periodis (1 + i)m − 1. For example, the future value at time nm years ofthe annuity


Time (in years) 0 m 2m · · · nm

is an|(1+i)m−1, where i is the annual effective rate of interest. Thefuture value at time nm years of the previous annuity is sn|(1+i)m−1.


27/82


Example 5

A cashflow pays $8000 at the end of every other year for 16 years.The first payment is made in two years. The annual effectiveinterest rate is 6.5%. Calculate the present value of this cashflow.


Contributions 8000 8000 8000 · · · 8000

Time (in years) 2 4 6 · · · 16

Notice that eight payments are made. The two–year effectiveinterest rate is (1.065)2 − 1. Hence, the present value of thecashflow is

(8000)a8|(1.065)2−1 = (8000)1− ((1.065)2 − 1 + 1)−8

(1.065)2 − 1

=(8000)1− (1.065)−16

(1.065)2 − 1= 37841.21717.


28/82


Example 5

A cashflow pays $8000 at the end of every other year for 16 years.The first payment is made in two years. The annual effectiveinterest rate is 6.5%. Calculate the present value of this cashflow.


Contributions 8000 8000 8000 · · · 8000

Time (in years) 2 4 6 · · · 16

Notice that eight payments are made. The two–year effectiveinterest rate is (1.065)2 − 1. Hence, the present value of thecashflow is

(8000)a8|(1.065)2−1 = (8000)1− ((1.065)2 − 1 + 1)−8

(1.065)2 − 1

=(8000)1− (1.065)−16

(1.065)2 − 1= 37841.21717.


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Theorem 2If i 6= 0, the present value of an annuity–due with levelpayments of one is

an|i = an| = 1 + ν + ν2 + · · ·+ νn−1 =1− νn

1− ν=

1− νn

d,

where we have used that ν = 1− d.If i 6= 0, the future value at time n of an annuity–due withlevel payments of one is

sn|i = sn|i = (1 + i)n + (1 + i)n−1 + · · ·+ (1 + i)

=(1 + i)n+1 − (1 + i)

i=

(1 + i)n − 1

d,

where we have used that 1− d = 11+i and d = i

1+i .If i = 0, an|i = sn|i = n.


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The annuities factors which have introduced give the present valueof the cashflow

Contributions 1 1 · · · 1

Time 1 2 · · · n

at different times.The present value of the cashflow at time 0 is an|i .The present value of the cashflow at time 1 is an|i .The present value of the cashflow at time n is sn|i .The present value of the cashflow at time n + 1 is sn|i .


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6 6 6 6

0 1 2 3 n − 1 n n + 1

1 1 1 1 1

an|i an|i sn|i sn|i

Figure 1: Present value of an annuity at different times


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Often, the contributions do not start at time 0. But, an|i is alwaysthe present value of a level unity annuity one period before the firstpayment. an|i is the present value of a level unity annuity at thetime of the first payment. sn|i is the future value of a level unityannuity at the time of the last payment. sn|i is the future value ofa level unity annuity one period after the last payment. Forexample, for the cashflow

Contributions 1 1 · · · 1

Time t + 1 t + 2 · · · t + n

The present value of the cashflow at time t is an|i .The present value of the cashflow at time t + 1 is an|i .The present value of the cashflow at time t + n is sn|i .The present value of the cashflow at time t + n + 1 is sn|i .


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6 6 6 6

t t + 1 t + 2 t + 3 t + n − 1 t + n t + n + 1

1 1 1 1 1

an|i an|i sn|i sn|i

Figure 2: Present value of an annuity at different times


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Example 6

Investment contributions of $4500 are made at the beginning ofthe year for 20 years into an account. This account pays an annualeffective interest rate is 7%. Calculate the accumulated value atthe end of 25 years.

Solution: The cashflow of payments is

Contributions 4500 4500 · · · 4500

Time 0 1 · · · 19

We can find the accumulated value at the end of 25 years doing


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Example 6



Contributions 4500 4500 · · · 4500

Time 0 1 · · · 19

We can find the accumulated value at the end of 25 years doingeither

a20|0.07(1 + 0.07)26 =1− (1 + 0.07)−20

0.07(1 + 0.07)26

=(1 + 0.07)26 − (1 + 0.07)6

0.07= 276854.31,


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Example 6



Contributions 4500 4500 · · · 4500

Time 0 1 · · · 19

We can find the accumulated value at the end of 25 years doingor

a20|0.07(1 + 0.07)25 =1− (1 + 0.07)−20

1− (1 + 0.07)−1(1 + 0.07)25

=(1 + 0.07)26 − (1 + 0.07)6

0.07= 276854.31,


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Example 6



Contributions 4500 4500 · · · 4500

Time 0 1 · · · 19


s20|0.07(1 + 0.07)6 =(1 + 0.07)20 − 1

0.07(1 + 0.07)6

=(1 + 0.07)26 − (1 + 0.07)6

0.07= 276854.31,


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Example 6



Contributions 4500 4500 · · · 4500

Time 0 1 · · · 19


s20|0.07(1 + 0.07)5 =(1 + 0.07)20 − 1

1− (1 + 0.07)−1(1 + 0.07)5

=(1 + 0.07)26 − (1 + 0.07)6

0.07= 276854.31.


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Theorem 3(Interest factor relations for annuities)

an|i = (1 + i)an|i , sn|i = (1 + i)nan|i , sn|i = (1 + i)n+1an|i

Proof.Consider the cashflow:


Time 0 1 2 · · · n

The present value at time 0 of the cashflow is an|i .The present value at time 1 of the cashflow is an|i .The present value at time n of the cashflow is sn|i .The present value at time n + 1 of the cashflow is sn|i .


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Theorem 3(Interest factor relations for annuities)

an|i = (1 + i)an|i , sn|i = (1 + i)nan|i , sn|i = (1 + i)n+1an|i

Proof.Consider the cashflow:


Time 0 1 2 · · · n

The present value at time 0 of the cashflow is an|i .The present value at time 1 of the cashflow is an|i .The present value at time n of the cashflow is sn|i .The present value at time n + 1 of the cashflow is sn|i .


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An application ofan|i = (1 + i)an|i .

Example 7

If an|i = 11.5174109 and an|i = 11.9205203, find n.

Solution: We have that an|i = (1 + i)an|i . So,

1 + i = 11.920520311.5174109 = 1.035 and i = 3.5%. Solving

11.5174109 = an|3.5% = 1−(1.035)−n

0.035 , we get(1.035)−n = 1− (11.5174109)(0.035) = 0.5968906185 andn = − log 0.5968906185

log 1.035 = 15.


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An application ofan|i = (1 + i)an|i .

Example 7

If an|i = 11.5174109 and an|i = 11.9205203, find n.

Solution: We have that an|i = (1 + i)an|i . So,

1 + i = 11.920520311.5174109 = 1.035 and i = 3.5%. Solving

11.5174109 = an|3.5% = 1−(1.035)−n

0.035 , we get(1.035)−n = 1− (11.5174109)(0.035) = 0.5968906185 andn = − log 0.5968906185

log 1.035 = 15.


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An application ofsn|i = (1 + i)sn|i .

Example 8

If sn|i = 21 and sn|i = 20, find i .


21 = sn|i = (1 + i)sn|i = (1 + i)(20).

So, i = 2120 − 1 = 5%.


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An application ofsn|i = (1 + i)sn|i .

Example 8

If sn|i = 21 and sn|i = 20, find i .


21 = sn|i = (1 + i)sn|i = (1 + i)(20).

So, i = 2120 − 1 = 5%.


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Theorem 4(Induction relations)

an|i = 1 + an−1|i and sn|i = 1 + sn−1|i .

Proof.Consider the cashflows

Cashflow 1 Contributions 1 0 0 · · · · · · 0 0



Time 0 1 2 · · · · · · n − 1 n

We have that the third cashflow is the sum of the first two. Thepresent value of the previous cashflows are 1 and an−1|i and an|i ,respectively. This implies the first relation. The second relationfollows similarly.


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Theorem 4(Induction relations)

an|i = 1 + an−1|i and sn|i = 1 + sn−1|i .

Proof.Consider the cashflows




Time 0 1 2 · · · · · · n − 1 n

We have that the third cashflow is the sum of the first two. Thepresent value of the previous cashflows are 1 and an−1|i and an|i ,respectively. This implies the first relation. The second relationfollows similarly.


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An application ofan|i = 1 + an−1|i .

Example 9

If a10|i = 8, find a9|i .

Solution: We have that a9|i = a10|i − 1 = 7.


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An application ofan|i = 1 + an−1|i .

Example 9

If a10|i = 8, find a9|i .

Solution: We have that a9|i = a10|i − 1 = 7.


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An application ofsn|i = 1 + sn−1|i

Example 10

If s10|i = 15, find s11|i .

Solution: We have that s11|i = 1 + s10|i = 16.


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An application ofsn|i = 1 + sn−1|i

Example 10

If s10|i = 15, find s11|i .

Solution: We have that s11|i = 1 + s10|i = 16.


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Theorem 5(Amortization relations)

1

an|i=

1

sn|i+ i and

1

an|i=

1

sn|i+ d .


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Proof of 1an|i

= 1sn|i

+ i : Suppose that a loan of $1 is paid in n

payments made at the end of the each period. Then, eachpayment should be 1

an|i. Suppose that we pay the loan as follows,

we pay i at the end of each period and put x in an extra accountpaying an effective rate of interest i . Since the initial loan is $1 theinterest accrued at the end of the first period is i . But, we pay i atthe end of the first period. Hence, immediately after this paymentwe owe $1. Proceeding in this way, we deduce that we owe $1immediately after each payment. The money in the extra accountaccumulates to xsn|i at time n. In order to pay the loan, we need

x = 1sn|i

. Our total payments at the end of the each period are1

sn|i+ i . Since this series of payments repays the loan of $1, we

must have that 1an|i

= 1sn|i

+ i . The proof of the second formula is

similar and it is omitted.


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An application of1

an|i=

1

sn|i+ i .

Example 11

If sn|i = 15.9171265 and an|i = 8.86325164, calculate n.

Solution: Using that 1an|i

= 1sn|i

+ i ,we get that

i = 18.86325164 −

115.9171265 = 5%. From an|5% = 8.86325164, we

get n = 12.


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An application of1

an|i=

1

sn|i+ i .

Example 11

If sn|i = 15.9171265 and an|i = 8.86325164, calculate n.

Solution: Using that 1an|i

= 1sn|i

+ i ,we get that

i = 18.86325164 −

115.9171265 = 5%. From an|5% = 8.86325164, we

get n = 12.


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In the calculator TI–BA–II–Plus we can use the time value ofmoney worksheet to solve problems with annuities. There are 5main financial variables in this worksheet:

I The number of periods N .

I The nominal interest for year I/Y .

I The present value PV .

I The payment per period PMT .

I The future value FV .

Recall that how to use the money worksheet was explained in

Section 1.3. It is recommended that you set–up P/Y =1 and

C/Y =1, by pressing:

2nd , P/Y , 1 , ENTER , ↓ , 1 , ENTER , 2nd , QUIT .

Unless it is said otherwise, we will assume that the entries forC/Y and P/Y are both 1. To check that this is so, do

2nd P/Y ↓ 2nd QUIT .


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When BGN is set–up at END (and C/Y and P/Y have value 1),

the value time of money formula in the calculator is

PV + PMT · 1− (1 + i)−N

i+ FV (1 + i)−N = 0.

Using the calculator, we can solve for any variable in the equation:

L + Pan|i + F (1 + i)−n = 0.

This equation is equivalent to

L(1 + i)n + Psn|i + F = 0.

In the calculator, we input the variables we know using: L as the

PV , P as the PMT , F as the FV , i as the I/Y , n as the N .

L, P and F can take negative values.


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To solve for any variable from the equation

L + Pan|i = 0

we input in the calculator: L as PV , P as PMT , 0 as FV , i as

the I/Y , n as N . If we are solving for either I/Y or N , PV

and PMT must have different signs.To solve for any variable from the equation

Psn|i + F = 0.

we input: F as the FV , P as the PMT , i as the I/Y , n as the

N and 0 as the PV . If we are solving for either I/Y or N ,

FV and PMT must have different signs.


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Example 12

What must you deposit at the end of each of the next 10 years inorder to accumulate 20,000 at the end of the 10 years assumingi = 5%.

Solution: We solve for P in 20000 = Ps10|0.05 and getP = 1590.091499. In the calculator TI–BA–II–Plus, press:

10 N 20000 FV 5 I/Y 0 PV CPT PMT

Note the display in the calculator is negative.


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Example 12

What must you deposit at the end of each of the next 10 years inorder to accumulate 20,000 at the end of the 10 years assumingi = 5%.

Solution: We solve for P in 20000 = Ps10|0.05 and getP = 1590.091499. In the calculator TI–BA–II–Plus, press:

10 N 20000 FV 5 I/Y 0 PV CPT PMT

Note the display in the calculator is negative.


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Example 13

If i (12) = 9% and $300 is deposited at the end of the each monthfor 1 year, what will the accumulated value be in 1 year?

Solution: We solve for FV in FV = 300s12|0.09/12 and getFV = 3752.275907. In the calculator, press:

12 N 300 PMT 0.75 I/Y 0 PV CPT FV

where we used that 9/12 = 0.75.Since the length of a period is a month:

I The monthly effective interest rate is i (12)

12 = 0.0912 = 0.75%.

I The number of periods is 12 months.


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Example 13

If i (12) = 9% and $300 is deposited at the end of the each monthfor 1 year, what will the accumulated value be in 1 year?

Solution: We solve for FV in FV = 300s12|0.09/12 and getFV = 3752.275907. In the calculator, press:

12 N 300 PMT 0.75 I/Y 0 PV CPT FV

where we used that 9/12 = 0.75.Since the length of a period is a month:

I The monthly effective interest rate is i (12)

12 = 0.0912 = 0.75%.

I The number of periods is 12 months.


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When BGN is set–up at BGN, the value time of money formula inthis calculator is

PV + PMT · (1 + i) · 1− (1 + i)−N

i+ FV (1 + i)−N = 0.

To solve for a variable from the equation:

L + Pan|i + F (1 + i)−n = 0,

we proceed as before, with payments set–up at beginning.Previous equation is equivalent to

L(1 + i)n + Psn|i + F = 0,

To change the setting of the payments (either at the beginning ofthe period, or at the end of the period), press:

2nd , BGN , 2nd , SET , 2nd , Quit .

If the calculator is set–up with payments at the end of the periods,there is no indicator in the screen. If the calculator is set–up withpayments at the beginning of the periods, the indicator ”BGN”appears in the screen.


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Example 14

A company purchases 100 acres of land for $200,000 and agrees toremit 10 equal annual installments of $27,598 each at thebeginning of the year. What is the annual interest rate on thisloan?

Solution: We solve for i in the equation 200000 = 27598a10|i toget i = 8%. In the calculator, set payments at the beginning of theperiod and press:

10 N −27598 PMT 200000 PV 0 FV CPT I/Y


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Example 14

A company purchases 100 acres of land for $200,000 and agrees toremit 10 equal annual installments of $27,598 each at thebeginning of the year. What is the annual interest rate on thisloan?

Solution: We solve for i in the equation 200000 = 27598a10|i toget i = 8%. In the calculator, set payments at the beginning of theperiod and press:

10 N −27598 PMT 200000 PV 0 FV CPT I/Y


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Example 15

An annuity pays $7000 at the end of the year for 7 years with thefirst payment made 5 years from now. The effective annual rate ofinterest is 6.5%. Find the present value of the this annuity.


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Example 15


Solution 1: The cashflow of payments is

Payments 7000 7000 7000 7000 7000 7000 7000

Time 5 6 7 8 9 10 11

Using an immediate annuity, (7000)a7|0.06 is the present value of theannuity, one period before the first payment, i.e. (7000)a7|0.06 is thepresent value of the annuity at time 4. So, the present value of theannuity is

(1.06)−4(7000)a7|0.06 = (0.7920937)(7000)(5.582381)

=30952.38.


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Example 15


Solution 2: The cashflow of payments is

Payments 7000 7000 7000 7000 7000 7000 7000

Time 5 6 7 8 9 10 11

Using a due annuity, (7000)a7|0.06 is the present value of the annuity,at the time of the first payment, i.e. (7000)a7|0.06 is the presentvalue of the annuity at time 5. So, the present value of the annuityis

(1.06)−5(7000)a7|0.06 = (0.7472582)(7000)(5.917324326)

=30952.38.


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Example 16

An annuity–immediate pays $7000 at the end of the year for 7years. The current annual effective rate of interest is 4.5% for thefirst three years and 5.5% thereafter. Find the present value of thisannuity.


Payments 7000 7000 7000 7000 7000 7000 7000

Time 1 2 3 4 5 6 7

Consider two cashflows: one with the first three payments andanother one with the last four payments. The present value attime 0 of the first three payments is (7000)a3|0.045. The presentvalue at time 3 of the last four payments is (7000)a4|0.055. Thepresent value at time 0 of the last four payments is(7000)(1.045)−3a4|0.055. The present value of the whole annuity is

(7000)a3|0.045 + (7000)(1.045)−3a4|0.055

=19242.75048 + 21500.85804 = 40743.60852.


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Example 16

An annuity–immediate pays $7000 at the end of the year for 7years. The current annual effective rate of interest is 4.5% for thefirst three years and 5.5% thereafter. Find the present value of thisannuity.


Payments 7000 7000 7000 7000 7000 7000 7000

Time 1 2 3 4 5 6 7

Consider two cashflows: one with the first three payments andanother one with the last four payments. The present value attime 0 of the first three payments is (7000)a3|0.045. The presentvalue at time 3 of the last four payments is (7000)a4|0.055. Thepresent value at time 0 of the last four payments is(7000)(1.045)−3a4|0.055. The present value of the whole annuity is

(7000)a3|0.045 + (7000)(1.045)−3a4|0.055

=19242.75048 + 21500.85804 = 40743.60852.


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Theorem 6Consider the cashflow

Contributions 0 1m

1m · · · 1

m1m · · · · · · 1

m

Time (in years) 0 1m

2m · · · m

mm+1m · · · · · · nm

m

The present value of this cashflow is

a(m)n|i =

1− νn

i (m).

The future value at time n of this cashflow is

s(m)n|i =

(1 + i)n − 1

i (m).

Proof: The present value of the cashflow is

1manm|i (m)/m = 1

m

1−„

1+ i(m)

m

«−mn

i(m)

m

= 1−νn

i (m) .


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Example 17

Suppose that Arthur takes a mortgage for L at an annual nominalrate of interest of 7.5% compounded monthly. The loan is paid atthe end of the month with level payments of $1200 for n years.Suppose at the last minute, Arthur changes the conditions of hisloan so that the payments will biweekly. The duration of the loanand the effective annual rate remain unchanged. Calculate theamount of the biweekly payment. Assume that there are 365/7weeks in a year.


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Solution: Let P be the biweekly payment. We have that

L = (12)(1200)a(12)n|i = (12)(1200)

1− νn

i (12)

and

L = (365/14)(P)a(365/14)n|i = (365/14)(P)

1− νn

i (365/14)

Hence,(12)(1200)

i (12)=

(365/14)(P)

i (365/14).

and

P =(12)(1200)i (365/14)

i (12)(365/14)=

(12)(1200)(0.07485856348)

(0.075)(365/14)= 551.2871743,

using that

i (365/14) = (365/14)

(1 +

0.075

12

)12(14/365)

= 7.485856348%.


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Contributions 1m

1m

1m · · · 1

m 0

Time (in years) 0 1m

2m · · · nm−1

mnmm


a(m)n|i =

1− νn

d (m).

The future value at time n of this cashflow is

s(m)n|i =

(1 + i)n − 1

d (m).

Proof: The present value of the cashflow is

1

manm|i (m)/m =

1

m

1−(1 + i (m)

m

)−mn

d (m)

m

=1− νn

d (m).


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Theorem 8For integers k, n ≥ 1, ank|i = an|i ak|(1+i)n−1.

Proof.We have that

ank|ian|i

=1−νnk

i1−νn

i

= 1 + νn + ν2n + · · ·+ νn(k−1) = ak|(1+i)n−1.


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Theorem 8For integers k, n ≥ 1, ank|i = an|i ak|(1+i)n−1.

Proof.We have that

ank|ian|i

=1−νnk

i1−νn

i

= 1 + νn + ν2n + · · ·+ νn(k−1) = ak|(1+i)n−1.


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An application of ank|i = an|i ak|(1+i)n−1.

Example 18

Carrie receives 200,000 from a life insurance policy. She uses thefund to purchase to two different annuities, each costing 100,000.The first annuity is a 24–year annuity–immediate paying k per yearto herself. The second annuity is a 8 year annuity–immediatepaying 2k per year to her boyfriend. Both annuities are based uponan annual effective interest rate of i , i > 0. Determine i .

Solution: We have that 100000 = ka24|i = 2ka8|i . So,

2 =a24|ia8|i

= a3|(1+i)8−1. Using the calculator, we get that

(1 + i)8 − 1 = 0.618034 and i = 6.1997%.


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An application of ank|i = an|i ak|(1+i)n−1.

Example 18

Carrie receives 200,000 from a life insurance policy. She uses thefund to purchase to two different annuities, each costing 100,000.The first annuity is a 24–year annuity–immediate paying k per yearto herself. The second annuity is a 8 year annuity–immediatepaying 2k per year to her boyfriend. Both annuities are based uponan annual effective interest rate of i , i > 0. Determine i .

Solution: We have that 100000 = ka24|i = 2ka8|i . So,

2 =a24|ia8|i

= a3|(1+i)8−1. Using the calculator, we get that

(1 + i)8 − 1 = 0.618034 and i = 6.1997%.


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Theorem 9For integers k, n ≥ 1,

ank|isn|i

= ak|(1+i)n−1.

Proof.By the previous theorem, ank|i = an|i ak|(1+i)n−1. The claim follows

noticing that an|i = (1 + i)−nsn|i andak|(1+i)n−1 = (1 + i)nak|(1+i)n−1.


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Theorem 9For integers k, n ≥ 1,

ank|isn|i

= ak|(1+i)n−1.

Proof.By the previous theorem, ank|i = an|i ak|(1+i)n−1. The claim follows

noticing that an|i = (1 + i)−nsn|i andak|(1+i)n−1 = (1 + i)nak|(1+i)n−1.


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An application of ank|i = sn|iak|(1+i)n−1.

Example 19

The present value of a 4n–year annuity–immediate of 1 at the endof every year is 16.663. The present value of a 4n–yearannuity–immediate of 1 at the end of every fourth year is 3.924.Find n and i.

Solution: We know that 16.663 = a4n|i and 3.924 = an|(1+i)4−1.Dividing the first equation over the second one, we get that

4.246432 =16.663

3.924=

a4n|i

an|(1+i)4−1= s4|i

and i = 4%. From the equation 16.663 = a4n|4%, we get n = 7.


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An application of ank|i = sn|iak|(1+i)n−1.

Example 19

The present value of a 4n–year annuity–immediate of 1 at the endof every year is 16.663. The present value of a 4n–yearannuity–immediate of 1 at the end of every fourth year is 3.924.Find n and i.

Solution: We know that 16.663 = a4n|i and 3.924 = an|(1+i)4−1.Dividing the first equation over the second one, we get that

4.246432 =16.663

3.924=

a4n|i

an|(1+i)4−1= s4|i

and i = 4%. From the equation 16.663 = a4n|4%, we get n = 7.


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Since an|i =∑n

j=1(1+ i)−j as a function of n, an|i increases with n.As a function of i , i ≥ 0, an|i decreases with i . We have thatan|0 = n and an|∞ = 0. So, n > an|i > 0 for i > 0; and an|i > n for0 > i > −1.It is proved in the manual that

an|i =∞∑j=1

(n + j − 1)!

(n − 1)!j!(−i)j−1.

The first order Taylor expansion of an|i on i is

an|i ≈ n − n(n + 1)

2i .

The second order Taylor expansion of an|i on i is

an|i ≈ n − n(n + 1)

2i +

n(n + 1)(n + 2)

6i2.


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Section 3.3. Level payment perpetuities.





2/16

Chapter 3. Annuities. Section 3.3. Level payment perpetuities.

A perpetuity is a series of payments made forever along equalintervals of time. By a change of units, we will assume thatintervals have unit length.The payments can be made either at the beginning or at the endof the intervals.A perpetuity has level payments if all payments Cj , j ≥ 0, areequal.A perpetuity has non–level payments if some payments Cj aredifferent from other ones.


3/16


For a perpetuity–immediate the payments are made at the endof the periods of time, i.e. at the times 1, 2, . . . . So, aperpetuity–immediate is a cashflow of the type:

Payments C1 C2 C3 · · ·Time 1 2 3 · · ·

For a perpetuity–due the payments are made at the beginning ofthe intervals of time, i.e. at the times 0, 1, 2, . . . . So, a perpetuityimmediate is a cashflow of the type:

Payments C0 C1 C2 · · ·Time 0 1 2 · · ·


4/16


Theorem 1The cashflow value of a perpetuity–immediate with level paymentsof one is

Payments 1 1 1 · · ·Time 1 2 3 · · ·

The present value of a perpetuity–immediate with level paymentsof one is

a∞|i = ν + ν2 + · · · = ν

1− ν=

1

i.


5/16


Example 1

The present value of a series of payments of 3 at the end of everyeight years, forever, is equal to 9.5. Calculate the effective annualrate of interest.

Solution: The 8–year interest factor is (1 + i)8. So, the 8–yeareffective interest rate is (1 + i)8 − 1. We have that9.5 = (3)a∞|(1+i)8−1 = 3

(1+i)8−1and

i =(1 + 3

9.5

)1/8 − 1 = 3.489979511%.


6/16


Example 1

The present value of a series of payments of 3 at the end of everyeight years, forever, is equal to 9.5. Calculate the effective annualrate of interest.

Solution: The 8–year interest factor is (1 + i)8. So, the 8–yeareffective interest rate is (1 + i)8 − 1. We have that9.5 = (3)a∞|(1+i)8−1 = 3

(1+i)8−1and

i =(1 + 3

9.5

)1/8 − 1 = 3.489979511%.


7/16


Theorem 2The cashflow value of a perpetuity–due with level payments of oneis

Payments 1 1 1 · · ·Time 0 1 2 · · ·

The present value of an perpetuity–due with level payments of oneis

a∞|i = 1 + ν + ν2 + · · · = 1

1− ν=

1

d=

1 + i

i.


8/16


Example 2

John uses his retirement fund to buy a perpetuity–due of 20,000per year based on an annual nominal yield of interest i = 8%compounded monthly. Find John’s retirement fund.

Solution: Since i (12) = 8%, i = 8.299950681%. John’s retirementfund is worth

(20000)a∞|i = (20000)1 + i

i= (20000)

1.08299950681

0.08299950681= 260963.8554.


9/16


Example 2

John uses his retirement fund to buy a perpetuity–due of 20,000per year based on an annual nominal yield of interest i = 8%compounded monthly. Find John’s retirement fund.Solution: Since i (12) = 8%, i = 8.299950681%. John’s retirementfund is worth

(20000)a∞|i = (20000)1 + i

i= (20000)

1.08299950681

0.08299950681= 260963.8554.


10/16


Example 3

A perpetuity pays $1 at the end of every year plus an additional $1at the end of every second year. The effective rate of interest isi = 5%. Find the present value of the perpetuity at time 0.


11/16


Solution: The cashflow of the perpetuity is

Payments 1 2 1 2 1 2 · · ·Time 1 2 3 4 5 6 · · ·

This cashflow can be decomposed into the cashflows:

Payments 1 1 1 · · ·Time 1 3 5 · · ·

and

Payments 2 2 2 · · ·Time 2 4 6 · · ·

The present value at time 0 of the first part of the cashflow is1+i

(1+i)2−1. The present value at time 0 of the second part of the

cashflow is 2(1+i)2−1

. Hence, the present value at time of the total

cashflow is

1 + i

(1 + i)2 − 1+

2

(1 + i)2 − 1=

3 + i

i(2 + i)=

3 + 0.05

0.05(2 + 0.05)= 29.7561.


12/16


Theorem 3The cashflow value of the perpetuity–immediate with levelpayments of one per year and m payments per year is

Payments 0 1m

1m · · ·

Time 0 1m

2m · · ·

The present value of the previous perpetuity–immediate is

a(m)∞|i =

1

m·(ν1/m + ν2/m + · · ·

)=

1

m· ν1/m · 1

1− ν1/m

=1

m(ν−1/m − 1)=

1

i (m).


13/16


Example 4

A perpetuity pays x at the end of each month. The nominalannual rate of interest compounded monthly is i (12). Calculate thepercentage of increase in the value of this perpetuity if the nominalannual rate of interest compounded monthly decreases by 10%.

Solution: At the rate i (12), the present value of the perpetuity isx

i (12) . At the rate i (12)(0.9), the present value of the perpetuity isx

i (12)(0.9). The percentage of increase in the value of this perpetuity

isx

i (12)(0.9)− x

i (12)

xi (12)

=1

0.9− 1 = 11.11111111%.


14/16


Example 4

A perpetuity pays x at the end of each month. The nominalannual rate of interest compounded monthly is i (12). Calculate thepercentage of increase in the value of this perpetuity if the nominalannual rate of interest compounded monthly decreases by 10%.

Solution: At the rate i (12), the present value of the perpetuity isx

i (12) . At the rate i (12)(0.9), the present value of the perpetuity isx

i (12)(0.9). The percentage of increase in the value of this perpetuity

isx

i (12)(0.9)− x

i (12)

xi (12)

=1

0.9− 1 = 11.11111111%.


15/16


Theorem 4The cashflow value of the perpetuity–due with level payments ofone per year and m payments per year is

Payments 1m

1m

1m · · ·

Time 0 1m

2m · · ·

The present value of an perpetuity–due with level payments of oneper year and m payments per year is

a(m)∞|i =

1

m·(1 + ν1/m + ν2/m + · · ·

)=

1

m· 1

1− ν1/m=

1

d (m).


16/16


Example 5

The present value of a series of payments of $500 at the beginningof every month years, forever, is equal to $10000. Calculate thenominal annual discount rate compounded monthly.


10000 = 5001

d (12).

and d (12) = 50010000 = 0.05.


1/37



Section 3.4. Non-level payment annuities and perpetuities.





2/37

Chapter 3. Annuities. Section 3.4. Non-level payment annuities and perpetuities.

Theorem 1(Geometric Annuity) Let i , r > −1. The present value of theannuity

Payments 1 1 + r (1 + r)2 · · · (1 + r)n−1

Time 1 2 3 · · · n

is

(Ga)n|i ,r =1

1 + ran| i−r

1+r=

1

1 + ian| i−r

1+r.


3/37


Proof: Let i ′ = i−r1+r . Then, 1+i

1+r = 1 + i ′. The present value of thecashflow at time 0 is

n∑j=1

(1 + r)j−1(1 + i)−j =1

1 + r

n∑j=1

(1 + i

1 + r

)−j

=1

1 + r

n∑j=1

(1 + i ′

)−j=

1

1 + ran|i ′ .

Using that an|i = (1 + i)an|i , we get that

1

1 + ran|i ′ =

1

1 + r

1

1 + i ′an|i ′ =

1

1 + ian|i ′ .


4/37


Example 1

An annuity provides for 10 annuals payments, the first payment ayear hence being $2600. The payments increase in such a way thateach payment is 3% greater than the previous one. The annualeffective rate of interest is 4%. Find the present value of thisannuity.


Payments 2600 2600(1.03) 2600(1.03)2 · · · 2600(1.03)9

Time 1 2 3 · · · 10

The present value at time 0 of the annuity is

(2600) (Ga)n|i ,r =2600

1.03a10−−| 0.04−0.03

1.03= 2524.271845a10|0.9708737864%

=23945.54454.


5/37


Example 1

An annuity provides for 10 annuals payments, the first payment ayear hence being $2600. The payments increase in such a way thateach payment is 3% greater than the previous one. The annualeffective rate of interest is 4%. Find the present value of thisannuity.


Payments 2600 2600(1.03) 2600(1.03)2 · · · 2600(1.03)9

Time 1 2 3 · · · 10


(2600) (Ga)n|i ,r =2600

1.03a10−−| 0.04−0.03

1.03= 2524.271845a10|0.9708737864%

=23945.54454.


6/37


Example 2

An annuity provides for 20 annuals payments, the first payment ayear hence being $4500. The payments increase in such a way thateach payment is 4.5% greater than the previous one. The annualeffective rate of interest is 4.5%. Find the present value of thisannuity.


Payments 4500 4500(1.045) 4500(1.045)2 · · · 4500(1.045)19

Time 1 2 3 · · · 20


(4500) (Ga)n|i ,r = (4500)1

1.045a20−−| 0.045−0.045

1.045= 4500

20

1.045= 86124.40191.


7/37


Example 2

An annuity provides for 20 annuals payments, the first payment ayear hence being $4500. The payments increase in such a way thateach payment is 4.5% greater than the previous one. The annualeffective rate of interest is 4.5%. Find the present value of thisannuity.


Payments 4500 4500(1.045) 4500(1.045)2 · · · 4500(1.045)19

Time 1 2 3 · · · 20


(4500) (Ga)n|i ,r = (4500)1

1.045a20−−| 0.045−0.045

1.045= 4500

20

1.045= 86124.40191.


8/37


Example 3

Chris makes annual deposits into a bank account at the beginningof each year for 10 years. Chris initial deposit is equal to 100, witheach subsequent deposit k% greater than the previous year deposit.The bank credits interest at an annual effective rate of 4.5%. Atthe end of 10 years, the accumulated amount in Chris account isequal to 1657.22. Calculate k.


Payments 100 100(1 + r) 100(1 + r)2 · · · 100(1 + r)9

Time 0 1 2 · · · 9

The accumulated amount at the end of 10 years is1657.22 = (100) 1

1+i an| i−r1+r

(1 + i)11. Hence,

a10| 0.045−r1+r

= (1657.22)(1.045)−10

100 = 10.67129833,0.045−r

1+r = −0.0141511755, r = 0.045+0.01415117551−0.0141511755 = 6% and k = 6.


9/37


Example 3

Chris makes annual deposits into a bank account at the beginningof each year for 10 years. Chris initial deposit is equal to 100, witheach subsequent deposit k% greater than the previous year deposit.The bank credits interest at an annual effective rate of 4.5%. Atthe end of 10 years, the accumulated amount in Chris account isequal to 1657.22. Calculate k.


Payments 100 100(1 + r) 100(1 + r)2 · · · 100(1 + r)9

Time 0 1 2 · · · 9

The accumulated amount at the end of 10 years is1657.22 = (100) 1

1+i an| i−r1+r

(1 + i)11. Hence,

a10| 0.045−r1+r

= (1657.22)(1.045)−10

100 = 10.67129833,0.045−r

1+r = −0.0141511755, r = 0.045+0.01415117551−0.0141511755 = 6% and k = 6.


10/37


Corollary 1

The present value of the perpetuity

Payments 1 1 + r (1 + r)2 · · · (1 + r)n−1 · · ·Time 1 2 3 · · · n · · ·

is

(Ga)∞|i ,r =

{1

i−r if i > r ,

∞ if i ≤ r .

Proof.If i > r , (Ga)∞|i ,r = 1

1+r a∞| i−r1+r

= 11+r

1i−r1+r

= 1i−r .

If i ≤ r , (Ga)∞|i ,r = 11+r a∞| i−r

1+r= ∞.


11/37


Corollary 1

The present value of the perpetuity

Payments 1 1 + r (1 + r)2 · · · (1 + r)n−1 · · ·Time 1 2 3 · · · n · · ·

is

(Ga)∞|i ,r =

{1

i−r if i > r ,

∞ if i ≤ r .

Proof.If i > r , (Ga)∞|i ,r = 1

1+r a∞| i−r1+r

= 11+r

1i−r1+r

= 1i−r .

If i ≤ r , (Ga)∞|i ,r = 11+r a∞| i−r

1+r= ∞.


12/37


Example 4

An perpetuity–immediate provides annual payments. The firstpayment of 13000 is one year from now. Each subsequent paymentis 3.5% more than the one preceding it. The annual effective rateof interest is i = 6%. Find the present value of this perpetuity.

Solution: The present value is(13000) (Ga)∞|i ,r = (13000) 1

i−r = (13000) 10.06−0.035 = 520000.


13/37


Example 4

An perpetuity–immediate provides annual payments. The firstpayment of 13000 is one year from now. Each subsequent paymentis 3.5% more than the one preceding it. The annual effective rateof interest is i = 6%. Find the present value of this perpetuity.

Solution: The present value is(13000) (Ga)∞|i ,r = (13000) 1

i−r = (13000) 10.06−0.035 = 520000.


14/37


Theorem 2(Increasing Annuity) The present value of the annuity

Payments 1 2 3 · · · n

Time 1 2 3 · · · n

is

(Ia)n|i =an|i (1 + i)− nνn

i=

an|i − nνn

i.

The accumulated value of this cashflow at time n is

(Is)n|i =sn|i − n

i.


15/37


Proof:The cashflow is the sum of the cashflows:

cashflow 1 1 1 1 · · · 1 1

cashflow 2 0 1 1 · · · 1 1

cashflow 3 0 0 1 · · · 1 1

· · · · · · · · · · · · · · · · · · · · ·· · · · · · · · · · · · · · · · · · · · ·

cashflow n 0 0 0 · · · 0 1

Time 1 2 3 · · · n − 1 n

The future value at time n of all these cashflows is

(Is)n|i =n∑

j=1

sj |i =n∑

j=1

(1 + i)j − 1

i=

sn|i − n

i.


16/37


In the calculator, it is possible to find an|i − nνn in onecomputation. With payments set–up at the beginning, we enter n

in N , i in I/Y , 1 in PMT and −n in FV . We ask the

calculator to compute PV .


17/37


Example 5

Find the present value at time 0 of an annuity–immediate such thatthe payments start at 1, each payment thereafter increases by 1until reaching 10, and they remain at that level until 25 paymentsin total are made. The effective annual rate of interest is 4%.


Payments 1 2 3 · · · 10 10 · · · 10

Time 1 2 3 · · · 10 11 · · · 25

The present value at time 0 of the perpetuity is

(Ia)10|0.04 + (1 + 0.04)−1010a15|0.04

=an|4% − 10(1 + 0.04)−10

0.04+ (1 + 0.04)−10(10)a15|0.04

=41.99224806 + 75.11184164 = 117.1040897.


18/37


Example 5

Find the present value at time 0 of an annuity–immediate such thatthe payments start at 1, each payment thereafter increases by 1until reaching 10, and they remain at that level until 25 paymentsin total are made. The effective annual rate of interest is 4%.


Payments 1 2 3 · · · 10 10 · · · 10

Time 1 2 3 · · · 10 11 · · · 25

The present value at time 0 of the perpetuity is

(Ia)10|0.04 + (1 + 0.04)−1010a15|0.04

=an|4% − 10(1 + 0.04)−10

0.04+ (1 + 0.04)−10(10)a15|0.04

=41.99224806 + 75.11184164 = 117.1040897.


19/37


Theorem 3(Decreasing Annuity) The present value of the annuity

Payments n n − 1 n − 2 · · · 1

Time 1 2 3 · · · n

is

(Da)n|i =n − an|i

i.

The accumulated value of this cashflow at time n is

(Ds)n|i =n(1 + i)n − sn|i

i.


20/37


Proof.The cashflow is the sum of the cashflows:

cashflow 1 1 1 1 · · · 1 1

cashflow 2 1 1 1 · · · 1 0

cashflow 3 1 1 1 · · · 0 0

· · · · · · · · · · · · · · · · · · · · ·· · · · · · · · · · · · · · · · · · · · ·

cashflow n 1 0 0 · · · 0 0

Time 1 2 3 · · · n − 1 n

The present value at time 0 of all these cashflows is

n∑j=1

aj |i =n∑

j=1

1− (1 + i)−j

i=

n − an|i

i.


21/37


In the calculator, it is possible to find n(1 + i)n − sn|i in one

computation. We enter n in N , i in I/Y , 1 in PMT and −n in

PV . We ask the calculator to compute FV .


22/37


Example 6

Find the present value of a 15–year decreasing annuity–immediatepaying 150000 the first year and decreasing by 10000 each yearthereafter. The effective annual interest rate of 4.5%.


Payments (15)(10000) (14)(10000) · · · (1)(10000)

Time 1 2 · · · 15


(10000) (Da)15|4.5% = (10000)15− a15|4.5%

0.045= 946767.616.


23/37


Example 6

Find the present value of a 15–year decreasing annuity–immediatepaying 150000 the first year and decreasing by 10000 each yearthereafter. The effective annual interest rate of 4.5%.


Payments (15)(10000) (14)(10000) · · · (1)(10000)

Time 1 2 · · · 15


(10000) (Da)15|4.5% = (10000)15− a15|4.5%

0.045= 946767.616.


24/37


Theorem 4(Increasing Perpetuity) The present value of the perpetuity

Payments 1 2 3 · · ·Time 1 2 3 · · ·

is (Ia)∞|i = 1+ii2

.


25/37


Proof:The cashflow is the sum of the infinitely many cashflows:

cashflow 1 1 1 1 · · · 1 1 · · · · · ·cashflow 2 0 1 1 · · · 1 1 · · · · · ·cashflow 3 0 0 1 · · · 1 1 · · · · · ·· · · · · · · · · · · · · · · · · · · · · · · · · · ·· · · · · · · · · · · · · · · · · · · · · · · · · · ·

Time 1 2 3 · · · n − 1 n · · · · · ·

The present value at time 0 of all these cashflows is

∞∑j=1

1

i(1 + i)−(j−1) =

1

i

1

1− 11+i

=1 + i

i2.


26/37


Example 7

An investor is considering the purchase of 500 ordinary shares in acompany. This company pays dividends at the end of each year.The next payment is one year from now and it is $3 per share. Theinvestor believes that each subsequent payment per share willincrease by $1 each year forever. Calculate the present value of thisdividend stream at a rate of interest of 6.5% per annum effective.


27/37



Payments 3 4 5 · · ·Time 1 2 3 · · ·

The cashflow is the sum of the following cashflows

Payments 2 2 2 · · ·Payments 1 2 3 · · ·

Time 1 2 3 · · ·

Hence, the present value of this dividend stream is

(500)(2)a∞|6.5% + (500)(1) (Ia)∞|6.5%

=(500)(2)1

0.065+ (500)(1)

1.065

0.0652

=15384.6154 + 126035.503 = 141420.118.


28/37


Theorem 5(Rainbow Immediate) The present value of the annuity

Payments 1 2 · · · n − 1 n n − 1 · · · 2 1

Time 1 2 · · · n − 1 n n + 1 · · · 2n − 2 2n − 1

is an|ian|i .


29/37


Proof: The value of the annuity is

(Ia)n|i + νn (Da)n−1|i =an|i−nνn

i +νn(n−1−an−1|i )

i =an|i−νn(1+an−1|i )

i

=(1+i)an|i−νn(1+i)an|i

i =(1+i)an|i (1−νn)

i = (1 + i)an−−|ian|i = an|ian|i


30/37


Example 8

A 15 year annuity pays 1000 at the end of year 1 and increases by1000 each year until the payment is 8000 at the end of year 8.Payments then decrease by 1000 each year until a payment of 1000is paid at the end of year 15. The annual effective interest rate of6.5%. Compute the present value of this annuity.


Paym. 1000 2000 · · · 7000 8000 7000 · · · 2000 1000

Time 1 2 · · · 7 8 9 · · · 14 15

The present value is

1000a8|ia8|i = (1000)(6.08875096)(6.48451977) = 39482.626.


31/37


Example 8

A 15 year annuity pays 1000 at the end of year 1 and increases by1000 each year until the payment is 8000 at the end of year 8.Payments then decrease by 1000 each year until a payment of 1000is paid at the end of year 15. The annual effective interest rate of6.5%. Compute the present value of this annuity.


Paym. 1000 2000 · · · 7000 8000 7000 · · · 2000 1000

Time 1 2 · · · 7 8 9 · · · 14 15


1000a8|ia8|i = (1000)(6.08875096)(6.48451977) = 39482.626.


32/37


Theorem 6(Paused Rainbow Immediate) The present value of the annuity

Paym. 1 2 · · · n − 1 n n n − 1 · · · 2 1

Time 1 2 · · · n − 1 n n + 1 n + 2 · · · 2n − 1 2n

is an+1|ian+1|i .

Proof: The present value of the annuity is

(Ia)n|i + νn (Da)n−−|i =

an|i−nνn

i +νn(n−an|i )

i =an|i−νnan|i

i

=(1+i)an|i−(1−ian|i )an|i

i= an|i (1 + an|i ) = an|i an+1|i .


33/37


Theorem 6(Paused Rainbow Immediate) The present value of the annuity

Paym. 1 2 · · · n − 1 n n n − 1 · · · 2 1

Time 1 2 · · · n − 1 n n + 1 n + 2 · · · 2n − 1 2n

is an+1|ian+1|i .

Proof: The present value of the annuity is

(Ia)n|i + νn (Da)n−−|i =

an|i−nνn

i +νn(n−an|i )

i =an|i−νnan|i

i

=(1+i)an|i−(1−ian|i )an|i

i= an|i (1 + an|i ) = an|i an+1|i .


34/37


Example 9

A 20 year annuity pays 5000 at the end of year 1 and increases by5000 each year until the payment is 50000 at the end of year 10.The payment remains constant for one year. Payments thendecrease by 5000 each year until a payment of 5000 is paid at theend of year 20. The annual effective interest rate of 4%. Computethe present value of this annuity.


Paym. (1)(5000) · · · (10)5000 (10)5000 · · · (1)(5000)

Time 1 · · · 10 11 · · · 20


5000a11|ia10|i = (5000)(8.11089578)(9.11089578) = 369487.631


35/37


Example 9

A 20 year annuity pays 5000 at the end of year 1 and increases by5000 each year until the payment is 50000 at the end of year 10.The payment remains constant for one year. Payments thendecrease by 5000 each year until a payment of 5000 is paid at theend of year 20. The annual effective interest rate of 4%. Computethe present value of this annuity.


Paym. (1)(5000) · · · (10)5000 (10)5000 · · · (1)(5000)

Time 1 · · · 10 11 · · · 20


5000a11|ia10|i = (5000)(8.11089578)(9.11089578) = 369487.631


36/37


Theorem 7The present value of the annuity

Payments 0 1m

1m · · · 1

m2m · · · 2

m · · · · · · nm

Time 0 1m

2m · · · 1 1 + 1

m · · · 2 · · · · · · n

is

(Ia)(m)n|i =

an|i − nνn

i (m).

For the proof, see the manual.


37/37



Payments 0 1m2

2m2

3m2 · · · · · · n

m2

Time 0 1m

2m

3m · · · · · · n

is (I (m)a

)(m)

n|i=

a(m)n|i − nνn

i (m).

For the proof, see the manual.


1/15



Section 3.5. Continuous annuities.





2/15

Chapter 3. Annuities. Section 3.5. Continuous annuities.

Continuous annuities

Annuities with length of period very small are approximatelycontinuous annuities.For example, the cashflows

Inflow 0 1m

1m · · · 1

m1m · · · · · · 1

m

Time 0 1m

2m · · · m

mm+1m · · · · · · nm

m

and

Inflow 1m

1m

1m · · · 1

m1m · · · · · · 1

m 0

Time 0 1m

2m · · · m

mm+1m · · · · · · nm−1

mnmm

tend to a continuous cashflow with rate C (t) = 1, 0 ≤ t ≤ n, asm →∞.


3/15


Theorem 1The present value of a continuous annuity with rate C (t) = 1,0 ≤ t ≤ n, is

an−−|i =

1− νn

δ.

The future value at time n of a continuous annuity with rate ofone is

sn|i =(1 + i)n − 1

δ.

Proof: We have that∫ n

0νt dt =

et ln ν

ln ν

∣∣∣∣n0

=1− νn

− ln ν=

1− e−nδ

δ=

1− νn

δ.


4/15


Recall


Inflow 0 1m

1m · · · 1

m1m · · · · · · 1

m

Time 0 1m

2m · · · m

mm+1m · · · · · · nm

m

Then, the present value of this cashflow is

a(m)n|i =

1− νn

i (m),

where i (m) is the nominal annual rate of interest convertible mtimes at year. The future value at time n of this cashflow is

s(m)n|i =

(1 + i)n − 1

i (m).


5/15


Recall


Inflow 1m

1m

1m · · · 1

m1m · · · · · · 1

m 0

Time 0 1m

2m · · · m

mm+1m · · · · · · nm−1

mnmm


a(m)n|i =

1− νn

d (m),

where d (m) is the nominal annual rate of discount convertible mtimes at year. The future value at time n of this cashflow is

s(m)n|i =

(1 + i)n − 1

d (m).


6/15


Theorem 4

an|i = limm→∞

a(m)n|i = lim

m→∞a(m)n|i

andsn|i = lim

m→∞s(m)n|i = lim

m→∞s(m)n|i .

Proof:

limm→∞

a(m)n|i = lim

m→∞

1− νn

i (m)=

1− νn

δ= an|i .

limm→∞

a(m)n|i = lim

m→∞

1− νn

d (m)=

1− νn

δ= an|i .


7/15


Given a real number x , the integer part of x is the largest integersmaller than or equal to x , i.e. the integer k satisfyingk ≤ x < k + 1. The integer part of x is noted by [x ]. Nexttheorem considers the continuous annuity with rate equal to theinteger part.

Theorem 5The present value of a continuous annuity with C (t) = [t],0 ≤ t ≤ n, is

(I a)n|i =an|i − nνn

δ.


8/15


Proof.The present value of the continuous cashflow is

(I a)n|i =

∫ n

0C (s)νs ds =

n∑j=1

∫ j

j−1jνs ds =

n∑j=1

j(ν j − ν j−1)

ln ν

=n∑

j=1

j(e−jδ − e−(j−1)δ)

−δ=

n∑j=1

j(e−(j−1)δ − e−jδ)

δ

=1 + e−δ + · · ·+ e−(n−1)δ − ne−nδ

δ.

Now, e−δ = ν and

1 + e−δ + · · ·+ e−(n−1)δ = 1 + ν + · · ·+ νn−1 =1− νn

1− ν= an|i .

So, (I a)n|i =an|i−nνn

δ .


9/15


Recall


Payments 0 1m

1m · · · 1

m2m · · · 2

m · · · · · · nm

Time 0 1m

2m · · · 1 1 + 1

m · · · 2 · · · · · · n

is

(Ia)(m)n|i =

an|i − nνn

i (m).


10/15


Theorem 7

(I a)n|i = limm→∞

(Ia)(m)

n−−|i.

Proof.We have that

limm→∞

(Ia)(m)n|i = lim

m→∞

an|i − nνn

i (m)=

an|i − nνn

δ= (I a)n|i .


11/15


Theorem 7

(I a)n|i = limm→∞

(Ia)(m)

n−−|i.

Proof.We have that

limm→∞

(Ia)(m)n|i = lim

m→∞

an|i − nνn

i (m)=

an|i − nνn

δ= (I a)n|i .


12/15


Theorem 8The present value of a continuous annuity with C (t) = t,0 ≤ t ≤ n, is (

I a)n|i =

an|i − nνn

δ.

Proof.By the change of variables x = δs,

(I a

)n|i =

∫ n

0C (s)νs ds =

∫ n

0sνs ds =

∫ n

0se−sδ ds

=δ−2

∫ nδ

0xe−x dx = δ−2(−1− x)e−x

∣∣∣∣nδ

0

=δ−2 − δ−2e−nδ(1 + nδ) =1− e−nδ

δ2− ne−nδ

δ=

an|i − nνn

δ.


13/15


Recall


Payments 0 1m2

2m2

3m2 · · · · · · n

m2

Time 0 1m

2m

3m · · · · · · n

is (I (m)a

)(m)

n|i=

a(m)n|i − nνn

i (m).


14/15


Theorem 10

(I a

)n|i = lim

m→∞

(I (m)a

)(m)

n|i.

Proof.

limm→∞

(I (m)a

)(m)

n|i= lim

m→∞

a(m)n|i − nνn

i (m)= lim

m→∞

1− νn

i (m)d (m)− nνn

i (m)

=1− νn

δ2− nνn

δ=

an|i − nνn

δ=

(I a

)n|i .


15/15


Theorem 10

(I a

)n|i = lim

m→∞

(I (m)a

)(m)

n|i.

Proof.

limm→∞

(I (m)a

)(m)

n|i= lim

m→∞

a(m)n|i − nνn

i (m)= lim

m→∞

1− νn

i (m)d (m)− nνn

i (m)

=1− νn

δ2− nνn

δ=

an|i − nνn

δ=

(I a

)n|i .


1/52

Chapter 4. Amortization and sinking bonds.

Manual for SOA Exam FM/CAS Exam 2.Chapter 4. Amortization and sinking bonds.

Section 4.1. Amortization schedules.





2/52

Chapter 4. Amortization and sinking bonds. Section 4.1. Amortization schedules.

In this chapter, we study different problems related with thepayment of a loan. Suppose that a borrower (also called debtor)takes a loan from a lender. The borrower will make paymentswhich eventually will repay the loan. Payments made by theborrower can be applied to the outstanding balance or not.According with the amortization method, all the payments madeby the borrower reduce the outstanding balance of the loan.When a loan is paid usually, the total amount of loan paymentsexceed the loan amount. The finance charge is the total amountof interest paid (the total payments minus the loan payments).


3/52


The simplest way to pay a loan is by unique payment. Supposethat a borrower takes a loan with amount L at time zero and thelender charges an annual effective rate of interest of i . If theborrower pays the loan with a lump sum P at time n, thenP = L(1 + i)n. The finance charge in this situation is L(1 + i)n− L.


4/52


Example 1

Juan borrows $35,000 for four years at an annual nominal interestrate of 7.5% convertible monthly. Juan will pay the loan with aunique payment at the end of four years.(i) Find the amount of this payment.(ii) Find the finance charge which Juan is charged in this loan.

Solution: (i) The amount of the loan payment is

(35000)(1 + 0.075

12

)(12)(4)= 47200.97.

(ii) The finance charge which Juan is charged in this loan is47200.97− 35000 = 12200.97.


5/52


Example 1

Juan borrows $35,000 for four years at an annual nominal interestrate of 7.5% convertible monthly. Juan will pay the loan with aunique payment at the end of four years.(i) Find the amount of this payment.(ii) Find the finance charge which Juan is charged in this loan.

Solution: (i) The amount of the loan payment is

(35000)(1 + 0.075

12

)(12)(4)= 47200.97.

(ii) The finance charge which Juan is charged in this loan is47200.97− 35000 = 12200.97.


6/52


Suppose that a borrower takes a loan of L at time 0 and repays theloan in a series of payments C1, . . . ,Cn at times t1, . . . , tn, where0 < t1 < t2 < · · · < tn. The debtor cashflow is

Inflows L −C1 −C2 −Cn · · · −Cn

Time 0 t1 t2 t3 · · · tn

Assume that the loan increases with a certain accumulationfunction a(t), t ≥ 0. Since the loan will be repaid, the presentvalue a time zero (or any other time) of this cashflow is zero.Hence

L =n∑

j=1

Cj

a(tj).

The finance charge for this loan is

n∑j=1

Cj − L =n∑

j=1

Cj −n∑

j=1

Cj

a(tj)=

n∑j=1

Cj

(1− 1

a(tj)

).


7/52


According with the retrospective method, the outstandingbalance at certain point is the present value of the loan at thattime minus the present value of the payments made at that time.For the cashflow


Time 0 t1 t2 t3 · · · tn

the outstanding balance immediately after the k–th payment, is

Bk = La(tk)−k∑

j=1

a(tk)Cj

a(tj).

Of course, we have that B0 = L, Bn = 0.


8/52


According to the prospective method, the outstanding balanceafter the k–th payment is equal to the present value of theremaining payments.For the cashflow


Time 0 t1 t2 t3 · · · tn

the outstanding balance immediately after the k–th payment, is

Bk =n∑

j=k+1

a(tk)Cj

a(tj)

Of course, we have that

La(tk)−k∑

j=1

a(tk)Cj

a(tj)=

n∑j=k+1

a(tk)Cj

a(tj).


9/52


An inductive relation for the outstanding balance is

Bk = Bk−1a(tk)

a(tk−1)− Ck .

Previous relation says that the outstanding balance after the k–thpayment is the accumulation of the previous outstanding balanceminus the amount of the payment made.


10/52


During the period [tk−1, tk ], the amount of interest accrued is

Ik = Bk−1

(a(tk)

a(tk−1)− 1

).

Immediately before the k–th payment, the outstanding balance isBk−1 + Ik = Bk−1

a(tk )a(tk−1)

. Immediately after the k–th payment, the

outstanding balance is Bk = Bk−1 + Ik − Ck . The k–th paymentCk can be split as Ik plus Ck − Ik . Ik is called the interest portionof the k–th payment. Ck − Ik is called the principal portion of thek–the payment. If Ck − Ik < 0, then the outstanding balanceincreases during the k–th period. Notice that

Ck − Ik = Ck − Bk−1

(a(tk)

a(tk−1)− 1

)= Bk−1 − Bk

is the reduction on principal made during the the k–period. Thetotal amount of reduction on principal is equal to the loan amount:∑n

k=1(Bk − Bk−1) = Bn − B0 = L.


11/52


Under compound interest,

L =n∑

j=1

Cj(1 + i)−tj .

The outstanding balance immediately after the k–th payment is

Bk = L(1 + i)tk −k∑

j=1

Cj(1 + i)tk−tj =n∑

j=k+1

Cj(1 + i)tk−tj .

The inductive relation for outstanding balances is

Bk = Bk−1(1 + i)tk−tk−1 − Ck .

The amount of interest accrued during the period [tk−1, tk ] is

Ik = Bk−1

((1 + i)tk−tk−1 − 1

).

The principal portion of the k–the payment is

Ck − Ik = Ck − Bk−1

((1 + i)tk−tk−1 − 1

)= Bk−1 − Bk .

The finance charge is∑n

j=1 Cj − L =∑n

j=1 Cj (1− (1 + i)−tj ) .


12/52


Usually, we consider payments made at equally spaced intervals oftime and compound interest. Suppose that a borrower takes a loanL at time 0 and repays the loan in a series of level paymentsC1, . . . ,Cn at times t0, 2t0, . . . , nt0. By a change of units, we mayassume that t0 = 1. Hence, the debtor cashflow is


Time 0 1 2 3 · · · n

Let i be the effective rate of interest per period. Then, we havethat

L =n∑

j=1

Cj(1 + i)−j .


13/52


The outstanding balance immediately after the k–th payment, is

Bk = L(1 + i)k −k∑

j=1

Cj(1 + i)k−j =n∑

j=k+1

Cj(1 + i)k−j .

The amount of interest accrued during the k–th year is iBk−1. Theprincipal portion of the k–th payment is Ck − iBk−1 = Bk − Bk−1.Hence, the outstanding balance after the k–th payment is

Bk = Bk−1 − (Ck − iBk−1) = (1 + i)Bk−1 − Ck .

The finance charge is

n∑j=1

Cj − L =n∑

j=1

Cj(1− (1 + i)−j) =n∑

j=1

Cj(1− ν j).


14/52


Example 2

Roger buys a car for $25,000 by making level payments at the endof the month for three years. Roger is charged an annual nominalinterest rate of 8.5% compounded monthly in his loan.(i) Find the amount of each monthly payment.(ii) Find the total amount of payments made by Roger.(iii) Find the total interest paid by Roger during the duration ofthe loan.(iv) Calculate the outstanding loan balance immediately after the12–th payment has been made using the retrospective method.(v) Calculate the outstanding loan balance immediately after the12–th payment has been made using the prospective method.

Solution:


15/52


Example 2


Solution:(i) Let P be the monthly payment. We have that 25000 =Pa

36−−|0.085/12and P = 789.1884356.


16/52


Example 2


Solution:(ii) The total amount of payments made by Roger is(36)(789.1884356) = 28410.78368.


17/52


Example 2


Solution:(iii) The total interest paid by Roger during the duration of the loanis 28410.78368− 25000 = 3410.78368.


18/52


Example 2


Solution:(iv) The outstanding loan balance immediately after the 12–th pay-ment has been made using the retrospective method. is

(25000)(1+0.085/12)12−(789.1884356)s12−−|0.085/12

= 17361.71419.


19/52


Example 2


Solution:(v) The outstanding loan balance immediately after the 12–th pay-ment has been made using the prospective method is

(789.1884356)a24−−|0.085/12

= 17361.71419.


20/52


Example 3

A loan is being repaid with 10 payments of $3000 followed by 20payments of $5000 at the end of each year. The effective annualrate of interest is 4.5%.(i) Calculate the amount of the loan.(ii) Calculate the outstanding loan balance immediately after the15–th payment has been made by both the prospective and theretrospective method.(iii) Calculate the amounts of interest and principal paid in the16–th payment.

Solution:


21/52


Example 3


Solution:(i) The cashflow of payments is

Inflows 3000 3000 · · · 3000 5000 5000 · · · 5000

Time 1 2 · · · 10 11 12 · · · 30

The loan amount is 3000a10−−|4.5%

+ (1.045)−105000a20−−|4.5%

=23738.1545 + 41880.8518 = 65619.0063.


22/52


Example 3


Solution:(ii) The outstanding loan balance immediately after the 15–th pay-ment using the retrospective method is

65619.0063(1.045)15 − 3000(1.045)5s10−−|4.5%

− 5000s5−−|4.5%

= 126991.311− 45940.0337− 27353.5486 = 53697.7287.


23/52


Example 3


Solution:The outstanding loan balance immediately after the 15–th paymentusing the prospective method is 5000a

15−−|4.5%= 53697.7286.


24/52


Example 3


Solution:(iii) The amount of interest paid in the 16–th payment is(53697.7286)(0.045) = 2416.39779. The amount of interest paid inthe 16–th payment is 5000− 2416.39779 = 2583.60221.


25/52


Next, we consider the amortization method of repaying a loan withlevel payments made at the end of periods of the same length. LetL be the amount borrowed. Let P be the level payment. Let n bethe number of payments. Let i be the effective rate of interest perpayment period. The cashflow of payments is

Inflows P P P · · · P

Time 1 2 3 · · · n

We have that L = Pan−−|i .

The outstanding principal after the k–th payment is

Bk = L(1 + i)k − Psk−−|i = P(a

n−−|i (1 + i)k − sk−−|i ) = Pa

n−k−−|i .

The interest portion of the k–th payment is

iBk−1 = iPan+1−k−−|i = P(1− νn+1−k).

The principal reduction of the k–th payment is

Bk−1 − Bk = P − iBk−1 = Pνn+1−k .


26/52


Using that Bk−1 = Pan+1−k−−|i and Bk−1 − Bk = Pνn+1−k , we get

that Bk = P(an+1−k−−|i − νn+1−k). The outstanding principal after

the k–th payment can be found using all these formulas

Bk = L(1 + i)k − Psk−−|i = P(a

n−−|i (1 + i)k − sk−−|i )

=Pan−k−−|i = P(a

n+1−k−−|i − νn+1−k).


27/52


The following is the amortization schedule for a loan of L = Pan−−|i

with level payments of P.

Period Payment Interest paid Principal repaid Outstanding balance0 − − − L = Pan−−|i1 P P(1− νn) Pνn Pan−1−−|i2 P P(1− νn−1) Pνn−1 Pan−2−−|i3 P P(1− νn−2) Pνn−2 Pan−3−−|i· · · · · · · · · · · · · · ·· · · · · · · · · · · · · · ·k P P(1− νn+1−k) Pνn+1−k Pan−k−−|i· · · · · · · · · · · · · · ·· · · · · · · · · · · · · · ·

n − 1 P P(1− ν2) Pν2 Pa1−−|in P P(1− ν) Pν 0


28/52


Example 4The following is the amortization schedule of a loan of $20,000.00at an effective interest rate of 8% for 12 years.

TimePaymentamount

Interestpaid

Principalreduction

Balance

0 − − − 2000.001 2653.90 1600.00 1053.90 18946.102 2653.90 1515.69 1138.21 17807.893 2653.90 1424.63 1229.27 16578.624 2653.90 1326.29 1327.61 15251.015 2653.90 1220.08 1433.82 13817.196 2653.90 1105.38 1548.52 12268.677 2653.90 981.49 1672.41 10596.268 2653.90 847.70 1806.20 8790.069 2653.90 703.20 1950.70 6839.3610 2653.90 547.15 2106.75 4732.6111 2653.90 378.61 2275.29 2457.3212 2653.91 196.59 2457.32 0.00


29/52


Example 5

A loan of 100,000 is being repaid by 15 equal annual installmentsmade at the end of each year at 6% interest effective annually.(i) Find the amount of each annual installment.(ii) Find the finance charge of this loan.(iii) Find how much interest is accrued in the first year.(iv) Find how principal is repaid in the first payment.(v) Find the balance in the loan immediately after the firstpayment.

Solution:


30/52


Example 5


Solution:(i)We solve 100000 = Pa

15−−|6% and get P = 10296.2764.


31/52


Example 5


Solution:(ii) The finance charge is (15)(10296.2764)− 100000 = 54444.146.


32/52


Example 5


Solution:(iii) The amount of interest accrued in the first year is(100000)(0.06) = 6000.


33/52


Example 5


Solution:(iv) The amount of principal repaid in the first year is 10296.2764−6000 = 4296.2764.


34/52


Example 5


Solution:(v) The balance in the loan immediately after the first payment is100000− 4296.2764 = 95703.7236.


35/52


Example 6

A loan L is being paid with 20 equal annual payments at the endof each year. The principal portion of the 8–th payment is 827.65and the interest portion is 873.81. Find L.

Solution: We know that

827.65 = Pνn+1−k = Pν13, 873.81 = P(1−νn+1−k) = P(1−ν13).

Adding the two equations, we get thatP = 827.65 + 873.81 = 1701.46. From the equation827.65 = 1701.46(1 + i)−13, we get that i = 5.7%. Hence,L = 1701.46a

20−−|5.7%= 20000.


36/52


Example 6

A loan L is being paid with 20 equal annual payments at the endof each year. The principal portion of the 8–th payment is 827.65and the interest portion is 873.81. Find L.

Solution: We know that

827.65 = Pνn+1−k = Pν13, 873.81 = P(1−νn+1−k) = P(1−ν13).

Adding the two equations, we get thatP = 827.65 + 873.81 = 1701.46. From the equation827.65 = 1701.46(1 + i)−13, we get that i = 5.7%. Hence,L = 1701.46a

20−−|5.7%= 20000.


37/52


A way to pay a loan is to pay interest as it accrues and to pay theprincipal in level installments. Suppose that a loan of amount L ispaid at the end of each year for n years. At the end of each yeartwo payments are made: one paying the interest accrued andanother one making a principal payment of L

n . At the end of j years

the outstanding balance is L(n−j)n . Hence, the interest payment at

the end of j years is i L(n+1−j)n . The total payment made at the end

of the j–th year is Ln + i L(n+1−j)

n = Ln (1 + i(n + 1− j)).


38/52


Example 7

Samuel takes a loan of $175000. He will pay the loan in 15 yearsby paying the interest accrued at the end of each year, and payinglevel payments of the principal at the end of each year. The annualeffective rate of interest of the loan is 8.5%.


39/52


Example 7

Samuel takes a loan of $175000. He will pay the loan in 15 yearsby paying the interest accrued at the end of each year, and payinglevel payments of the principal at the end of each year. The annualeffective rate of interest of the loan is 8.5%.(i) Find the amount of each payment of principal.


40/52


Example 7

Samuel takes a loan of $175000. He will pay the loan in 15 yearsby paying the interest accrued at the end of each year, and payinglevel payments of the principal at the end of each year. The annualeffective rate of interest of the loan is 8.5%.(i) Find the amount of each payment of principal.Solution: (i) The annual payment of principal is 175000

15 = 11666.67.


41/52


Example 7

Samuel takes a loan of $175000. He will pay the loan in 15 yearsby paying the interest accrued at the end of each year, and payinglevel payments of the principal at the end of each year. The annualeffective rate of interest of the loan is 8.5%.(ii) Find the outstanding principal owed at the end of the ninth year.


42/52


Example 7

Samuel takes a loan of $175000. He will pay the loan in 15 yearsby paying the interest accrued at the end of each year, and payinglevel payments of the principal at the end of each year. The annualeffective rate of interest of the loan is 8.5%.(ii) Find the outstanding principal owed at the end of the ninth year.Solution: (ii) The outstanding principal owed at the end of the

ninth year is (175000)(15−9)15 = 70000.


43/52


Example 7

Samuel takes a loan of $175000. He will pay the loan in 15 yearsby paying the interest accrued at the end of each year, and payinglevel payments of the principal at the end of each year. The annualeffective rate of interest of the loan is 8.5%.(iii) Find the interest accrued during the tenth year.


44/52


Example 7

Samuel takes a loan of $175000. He will pay the loan in 15 yearsby paying the interest accrued at the end of each year, and payinglevel payments of the principal at the end of each year. The annualeffective rate of interest of the loan is 8.5%.(iii) Find the interest accrued during the tenth year.Solution: (iii) The amount of interest paid at the end of the tenthyear is (0.085)70000 = 5950.


45/52


Example 7

Samuel takes a loan of $175000. He will pay the loan in 15 yearsby paying the interest accrued at the end of each year, and payinglevel payments of the principal at the end of each year. The annualeffective rate of interest of the loan is 8.5%.(iv) Find the total amount of payments made at the end of the tenthyear.


46/52


Example 7

Samuel takes a loan of $175000. He will pay the loan in 15 yearsby paying the interest accrued at the end of each year, and payinglevel payments of the principal at the end of each year. The annualeffective rate of interest of the loan is 8.5%.(iv) Find the total amount of payments made at the end of the tenthyear.Solution: (iv) The total amount of payments made at the end ofthe tenth year is 11666.67 + 5950 = 17616.67.


47/52


Example 7

Samuel takes a loan of $175000. He will pay the loan in 15 yearsby paying the interest accrued at the end of each year, and payinglevel payments of the principal at the end of each year. The annualeffective rate of interest of the loan is 8.5%.(v) Find the total amount of payments which Samuel makes.


48/52


Example 7

Samuel takes a loan of $175000. He will pay the loan in 15 yearsby paying the interest accrued at the end of each year, and payinglevel payments of the principal at the end of each year. The annualeffective rate of interest of the loan is 8.5%.(v) Find the total amount of payments which Samuel makes.Solution: (v) The interest payment at the end of j years is

i L(n+1−j)n = (0.085) (175000)(16−j)

15 . Hence, the total interest pay-ments are

(0.085)15∑j=1

(175000)(16− j)

15

=(0.085)175000

15

((16)(15)− (16)(15)

2

)= 119000.

The total amount of payments which Samuel makes is 119000 +175000 = 294000.


49/52


Example 8

A loan of $150000 is going to be paid over 20 years with monthlypayments. The first payment is one month from now. During eachyear, the payments are constant. But, they increase by 3% eachyear. The annual effective rate of interest is 6%. Calculate thetotal amount of the payments made during the first year. Calculatethe outstanding loan balance on the loan ten years from now afterthe payment is made.


50/52


Solution: Let P be the monthly payment during the first year.During the k–th year, 12 payments of P(1.03)k−1 are made. Thevalue of these payments at the end of the k–th year isP(1.03)k−1s12|i (12)/12 = P(1.03)k−112.32652834, where we have

used that i (12) = 5.84106068%. So, the cashflow of payments isequivalent to

Payments P12.3265 P(1.03)12.3265 · · · P(1.03)1912.3265

Time 1 2 · · · 20

The present value of this cashflow is the loan amount:

150000 =12.3265P

1 + ran| i−r

1+r=

12.3265P

1.03a20| 0.06−0.03

1+0.03= 179.493145P

and P = 835.6865105.The outstanding loan balance on the loan ten years from now afterthe payment is made is

(835.6865105)(1.03)9(12.3265)a10| 0.06−0.031+0.03

= 201586.9934.


51/52


Example 9

Mary takes on a loan of $135,000. The loan is being repaid by a10–year increasing annuity–immediate. The initial payment is10000, and each subsequent payment is x larger than the precedingpayment. The annual effective interest rate is 6.5%. Determinethe principal outstanding immediately after the 5–th payment.


52/52



Contributions 10000 10000 + x 10000 + 2x · · · 10000 + 9x

Time 1 2 3 · · · 10

The present value of the payments is

135000 = (10000− x)a10|6.5% + x(Ia)10|6.5%

=(10000− x)(7.188830223) + 35.82836665x .

So, x = 135000−(7.188830223)(10000)35.82836665−7.188830223 = 2203.656401.

The payments to be made after the 5-th payment are

Contributions 10000 + (5)(2203.656401) · · · 10000 + (9)(2203.656401)

Time 6 · · · 10

Its present value at time 5 is

(10000 + (4)(2203.656401))a5|6.5% + (2203.656401)(Ia)5|6.5%

=78187.55276 + 26321.63894 = 104509.1917.


1/17



Section 4.2. Sinking funds.





2/17

Chapter 4. Amortization and sinking bonds. Section 4.2. Sinking funds.

An alternate way to repay a loan is to make two payments, onedirectly to the lender and another to an auxiliary fund. Theauxiliary fund is called a sinking fund. The payments madedirectly to the lender apply to the principal. The deposits madeinto the sinking fund do not. Usually, the sinking fund accumulateswith a different interest rate than the rate charged by the lender.At the end of the duration of the loan, the borrower withdraws thetotal accumulated in the sinking fund and uses this money to paythe loan to the lender.


3/17


Usually, we consider the case of payments made at the end of eachof n periods. The simplest case is the one when all the paymentsare equal. Let i be the periodic effective rate charged by the lenderon the loan. At the end of each period, the borrower pays Pdirectly to lender. The borrower deposits Q into a sinking fundearning a rate of interest j . Usually, j < i . The cashflow ofpayments to the principal is

Contributions 0 P P · · · P P + R

Time 0 1 2 · · · n − 1 n

where R is the lump–sum payment obtained by withdrawing thetotal accumulated in the sinking fund at the end of n periods. Thecashflow of deposits in the sinking fund is

Contributions 0 Q Q Q · · · Q

Time 0 1 2 3 · · · n

Hence, the accumulated value in the sinking fund at time n isR = Qs

n−−|j .


4/17


The borrower’s total cashflow is

Contributions L −P − Q −P − Q −P − Q · · · −P − Q

Time 0 1 2 3 · · · n

The lender cashflow is

Contributions −L P P P · · · P + R

Time 0 1 2 3 · · · n

In order to the loan to be repaid:

L = Pan−−|i + R(1 + i)−n = Pa

n−−|i + Qsn−−|j(1 + i)−n.


5/17


Example 1

Dave borrows 150,000 from a trust company at an annual effectiverate of interest of 9.5%. He agrees to pay the interest annually atthe end of the each year, and build up a sinking fund which willrepay the loan at the end of 15 years. The sinking fundaccumulates at annual effective rate of interest of 4.5%. Calculate:


6/17


Example 1

Dave borrows 150,000 from a trust company at an annual effectiverate of interest of 9.5%. He agrees to pay the interest annually atthe end of the each year, and build up a sinking fund which willrepay the loan at the end of 15 years. The sinking fundaccumulates at annual effective rate of interest of 4.5%. Calculate:(i) The annual interest payments.


7/17


Example 1

Dave borrows 150,000 from a trust company at an annual effectiverate of interest of 9.5%. He agrees to pay the interest annually atthe end of the each year, and build up a sinking fund which willrepay the loan at the end of 15 years. The sinking fundaccumulates at annual effective rate of interest of 4.5%. Calculate:(i) The annual interest payments.Solution: (i) The annual interest payment is 150000(0.095) =14250.


8/17


Example 1

Dave borrows 150,000 from a trust company at an annual effectiverate of interest of 9.5%. He agrees to pay the interest annually atthe end of the each year, and build up a sinking fund which willrepay the loan at the end of 15 years. The sinking fundaccumulates at annual effective rate of interest of 4.5%. Calculate:(ii) The annual sinking fund payment.


9/17


Example 1

Dave borrows 150,000 from a trust company at an annual effectiverate of interest of 9.5%. He agrees to pay the interest annually atthe end of the each year, and build up a sinking fund which willrepay the loan at the end of 15 years. The sinking fundaccumulates at annual effective rate of interest of 4.5%. Calculate:(ii) The annual sinking fund payment.Solution: (ii) 150000 = Qs

15−−|4.5%and Q = 7217.071217.


10/17


Example 1

Dave borrows 150,000 from a trust company at an annual effectiverate of interest of 9.5%. He agrees to pay the interest annually atthe end of the each year, and build up a sinking fund which willrepay the loan at the end of 15 years. The sinking fundaccumulates at annual effective rate of interest of 4.5%. Calculate:(iii) Dave’s total annual outlay.


11/17


Example 1

Dave borrows 150,000 from a trust company at an annual effectiverate of interest of 9.5%. He agrees to pay the interest annually atthe end of the each year, and build up a sinking fund which willrepay the loan at the end of 15 years. The sinking fundaccumulates at annual effective rate of interest of 4.5%. Calculate:(iii) Dave’s total annual outlay.Solution: (iii) Dave’s total annual outlay is 14250+7217.071217 =21467.07122.


12/17


Example 1

Dave borrows 150,000 from a trust company at an annual effectiverate of interest of 9.5%. He agrees to pay the interest annually atthe end of the each year, and build up a sinking fund which willrepay the loan at the end of 15 years. The sinking fundaccumulates at annual effective rate of interest of 4.5%. Calculate:(iv) The annual effective rate of interest i ′ for which the paymentsmade at the end of the year will be equal to Dave’s total annualoutlay.


13/17


Example 1

Dave borrows 150,000 from a trust company at an annual effectiverate of interest of 9.5%. He agrees to pay the interest annually atthe end of the each year, and build up a sinking fund which willrepay the loan at the end of 15 years. The sinking fundaccumulates at annual effective rate of interest of 4.5%. Calculate:(iv) The annual effective rate of interest i ′ for which the paymentsmade at the end of the year will be equal to Dave’s total annualoutlay.Solution: (iv) 150000 = (21467.07122)a

15−−|i ′ and i ′ =

11.52440895%. Notice that although Dave borrows at a rate 9.5%,by getting only 4.5% in his sinking fund, the actual rate of interestDave which is paying is higher.


14/17


Example 1

Dave borrows 150,000 from a trust company at an annual effectiverate of interest of 9.5%. He agrees to pay the interest annually atthe end of the each year, and build up a sinking fund which willrepay the loan at the end of 15 years. The sinking fundaccumulates at annual effective rate of interest of 4.5%. Calculate:(v) Find i + n−1

n+1(i − j), where i = 9.5%, j = 4.5% and n = 15, andcompare with i ′.


15/17


Example 1

Dave borrows 150,000 from a trust company at an annual effectiverate of interest of 9.5%. He agrees to pay the interest annually atthe end of the each year, and build up a sinking fund which willrepay the loan at the end of 15 years. The sinking fundaccumulates at annual effective rate of interest of 4.5%. Calculate:(v) Find i + n−1

n+1(i − j), where i = 9.5%, j = 4.5% and n = 15, andcompare with i ′.Solution: (v) i + n−1

n+1(i− j) = 0.095+ (15−1)(0.95−0.045)15+1 = 0.13875.

which is sort of close to 0.1152440895.


16/17


Example 2

Steve repays a loan of $18,000 by making interest payment at theend of the year for 15 years and equal deposits at the end of eachyear into a sinking fund for 15 years. At the end of the 15 years,Steve withdraws the balance from the sinking fund and pays theloan. The sinking fund earns 6% effective annually. Immediatelyafter the fourth payment, the yield on the sinking fund increases to7% effective annually. At that time, Steve adjusts his sinking fundpayment to x so that the sinking fund will accumulate to $18,000,15 years after the original loan date. Find x.

Solution: Let Q be the initial payment Joe makes to the sinkingfund. Then, 18000 = Qs

15−−|6%. Hence, Q = 773.3297512. The

balance in the sinking fund immediately after the fourth payment is773.3297512s

4−−|6% = 3383.020703. The final accumulation in the

sinking fund is 18000. So, 18000 = 3383.020703(1.07)11 + xs11−−|7%

and x = 689.2751, which can be found doing:

11 N 7 I/Y 3383.020703 PV −18000 FV CPT PMT


17/17


Example 2

Steve repays a loan of $18,000 by making interest payment at theend of the year for 15 years and equal deposits at the end of eachyear into a sinking fund for 15 years. At the end of the 15 years,Steve withdraws the balance from the sinking fund and pays theloan. The sinking fund earns 6% effective annually. Immediatelyafter the fourth payment, the yield on the sinking fund increases to7% effective annually. At that time, Steve adjusts his sinking fundpayment to x so that the sinking fund will accumulate to $18,000,15 years after the original loan date. Find x.

Solution: Let Q be the initial payment Joe makes to the sinkingfund. Then, 18000 = Qs

15−−|6%. Hence, Q = 773.3297512. The

balance in the sinking fund immediately after the fourth payment is773.3297512s

4−−|6%= 3383.020703. The final accumulation in the

sinking fund is 18000. So, 18000 = 3383.020703(1.07)11 + xs11−−|7%

and x = 689.2751, which can be found doing:

11 N 7 I/Y 3383.020703 PV −18000 FV CPT PMT


1/6



Section 4.3. Reinvestment.





2/6

Chapter 4. Amortization and sinking bonds. Section 4.3. Reinvestment.

Suppose that a bank account pays interest in the original deposit,but not in the obtained interest. Then, it will wise to withdraw theinterest and invest it in another account. In other situations, itmakes sense to reinvest the earned interest in a differentinvestment. For example, a stock pays dividends and/or capitalgains, which can be invested somewhere else. We obtain a flow ofinterest payments which are invested at different rate from the onein the initial investment. Suppose that a mortgage company makesa loan to a customer. To know the mortgage company’s return inits investment we need to take in account the interest rate chargedto the customer in the loan and the interest rate which themortgage company gets in the monthly payments it receives.


3/6


Example 1

Payments of $2500 are invested at the end of each quarter for 5years. The payments earn interest at an annual nominal interestrate converted quarterly of 14% and the interest payments arereinvested at an annual nominal interest rate converted quarterlyof 10%. Find the total accumulation for both accounts at the endof 5 years.


4/6


Solution: The balance in the first account is

Balance in the 1st account 2500 2(2500) · · · · · · (20)(2500)

Time (in quarters) 1 2 · · · · · · 20

Since the rate of interest per quarter is 14%/4 = 3.5%, theinterest payments are

Cashflow of interest (1)(0.035)(2500) · · · 19(0.035)(2500)

Time (in quarters) 2 · · · 20

The total accumulation for both accounts at the end of the 5–thyear is

(20)(2500) + (0.035)(2500) (Is)19−−|0.025

= 50000 + (87.5)

((1.025)s

19−−|0.025

−19

0.025

)= 50000 + (87.5)

(24.54465761−19

0.025

)= 69406.30164.


5/6


Example 2

John invests 1000 at the beginning of each year for 5 years at anannual effective interest rate of 10% and reinvests the interest atan annual effective interest rate of 8%. Calculate the total value ofhis investment at the end of 5 years.


6/6


Solution: The balance in John’s account is

Balance in John’s account 1000 2000 3000 4000 5000 5000

Time 0 1 2 3 4 5

The interest payments which John gets are

John’s interest payments 0 100 200 300 400 500

Time 0 1 2 3 4 5

The total accumulation of John’s investments at time 5 is

5000 + 100 (Is)5−−|0.08

= 5000 + 100

((1.08)s

5−−|0.08− 5

0.08

)

=5000 + 100

(6.335929037− 5

0.08

)= 6669.911296.


1/??

Chapter 5. Bonds.

Manual for SOA Exam FM/CAS Exam 2.Chapter 5. Bonds.

Section 5.1. Securities.





2/??

Chapter 5. Bonds. Section 5.1. Securities.

Securities

When a corporation or a public institution needs to raise money, itarranges contracts with investors. These contracts are calledfinancial assets or securities. From the investor’s view point,securities are investment instruments, endorsed by a corporation,government, or other organization. The borrower offers either debtor ownership (equity). In the case of debt, the borrower agrees tomake a series of payments to the investor. In the case of equitysecurities, the borrower gives a part of the ownership of thecorporation to the investor.


3/??


Stock

For a corporation, claims to its ownership are determined by sharesof stock. The proportion of a company which an investor owns isthe fraction of its shares owned over the total number of sharesoutstanding.There are several types of stocks. Some stocks provide votingrights to the holder. Some stocks entitle the holder to receivepayments, such as dividends and/or capital appreciation. Differenttypes of stocks have different order of preference to the company’sassets in the case of liquidation (the company is forced to sells itsassets, pay outstanding debts, and distribute the remainder toshareholders).Common stock entitles the holder to payments of dividends andcapital appreciation. Common stock gives voting rights.A preferred stock is a stock which provides a fixed dividend thatdoes not fluctuate. Preferred stock shareholders do not enjoyvoting rights.


4/??


Bonds

A way for a government or corporation to raise money is to issuebonds. A bond is a certificate issued from a borrower to a lenderagreeing to make payment(s) in a loan. The price of a bond P isthe amount that the lender pays (loans) to the government orcorporation for the bond. Types of available bonds are: U.S.government securities, municipal bonds, corporate bonds,mortgage and asset–backed securities, federal agency securities andforeign government bonds.


1/17

Chapter 5. Bonds.


Section 5.2. Price of a bond.





2/17

Chapter 5. Bonds. Section 5.2. Price of a bond.

Bonds

There are two kind of bonds: accumulation bonds and bondswith coupons. The time at which the loan is repaid is called thematurity date (or redemption date).

I In the case of accumulation bonds, the borrower agrees to paythe loan plus interest at a unique date, called the redemptiontime. An accumulation bond is also called a zero couponbond.

I The most common bonds are bonds with coupons. For bondswith coupons, the borrower agrees to make period payments(coupons) plus a balloon payment (the redemption value)C at the maturity date.


3/17


Cashflow of a bond with coupons.

Every bond has a face value (or par value) F . The couponpayment is Fr . Here, r is the coupon rate per interest period.Often, the payments are semiannually and 2r is the annual nominalrate of interest convertibly semiannually. A bond is calledredeemable at par if C = F . Unless said otherwise we assumethat a bond is redeemable at par. Let n be the number of interestperiods until the redemption date. Let i be the yield rate perinterest period. The cashflow for the borrower is

Contributions P −Fr −Fr −Fr · · · −Fr − C

Time 0 1 2 3 · · · n


4/17


Variables for a bond.

I P = the price of a bond.I F = the par or face value of a bond.I C = the redemption value of a bond.I r = the coupon rate of a bond.I Fr = the amount of a coupon.I i = the yield rate of the bond per coupon period.I ν = 1

1+i = the discount factor per coupon period.I n = the number of coupon payment periods.I g = Fr

C = the modified coupon rate of a bondI G = Fr

i = the base amount of a bond.I K = Cνn = the present value, compounded at the yield rate,

of the redemption value of a bondI P − C = the premium (if P > C ).I C − P = the discount (if C > P).I k = P−C

C premium as a fraction of redemption value.


5/17


Price of a bond

The basic formula for the price of a bond is

P = Fran−−|i + C (1 + i)−n = Fra

n−−|i + K .

The premium/discount formula for the price of a bond is

P = Fran−−|i + Cνn = Fra

n−−|i + C (1− ian−−|i )

=C + (Fr − Ci)an−−|i = C + C (g − i)a

n−−|i .

The base amount formula for the price of a bond is

P = Fran−−|i+Cνn = Gia

n−−|i+Cνn = G (1−νn)+Cνn = G+(C−G )νn.

The Makehan formula for the price of a bond is

P = Fran−−|i + Cνn = Cg

1− νn

i+ Cνn

=g

i(C − Cνn) + Cνn =

g

i(C − K ) + K .


6/17


Example 1

Find the price of a 10–year bond, redeemable at par, with facevalue of $10,000 and coupon rate of 10%, convertible quarterly,that will yield 8%, convertible quarterly.

Solution: We know that F = C = 10000, n = (10)(4) = 40,r = 0.10

4 = 0.025, Fr = (10000)(0.025) = 250. andi = 0.08

4 = 0.02. So, the price of the bond is

P = Fran−−|i + C (1 + i)−n = 250a

40−−|0.02+ 10000(1.02)−40

=11367.77396.

In the calculator, do:40 N 2 I/Y 250 PMT 10000 PMT CPT PV .


7/17


Example 1

Find the price of a 10–year bond, redeemable at par, with facevalue of $10,000 and coupon rate of 10%, convertible quarterly,that will yield 8%, convertible quarterly.

Solution: We know that F = C = 10000, n = (10)(4) = 40,r = 0.10

4 = 0.025, Fr = (10000)(0.025) = 250. andi = 0.08

4 = 0.02. So, the price of the bond is

P = Fran−−|i + C (1 + i)−n = 250a

40−−|0.02+ 10000(1.02)−40

=11367.77396.

In the calculator, do:40 N 2 I/Y 250 PMT 10000 PMT CPT PV .


8/17


Example 2

A 30 year bond matures at its face value of 10,000. It payssemiannual coupons of 600. Calculate the price of the bond if theannual nominal interest rate convertible semiannually is 7.5%.

Solution: We know that F = C = 10000, n = (30)(2) = 60,Fr = 600, and i = 7.5%

2 = 3.75%. The price of the bond is

(600)a60−−|0.0375

+ (10000)(1 + 0.0375)−60 = 15341.03109.


9/17


Example 2

A 30 year bond matures at its face value of 10,000. It payssemiannual coupons of 600. Calculate the price of the bond if theannual nominal interest rate convertible semiannually is 7.5%.

Solution: We know that F = C = 10000, n = (30)(2) = 60,Fr = 600, and i = 7.5%

2 = 3.75%. The price of the bond is

(600)a60−−|0.0375

+ (10000)(1 + 0.0375)−60 = 15341.03109.


10/17


Remember that unless said otherwise a bond is redeemable at par.

Example 3

What is the price of a 5–year 100 par–value bond having quarterlycoupons at a quarter rate of 1.5% that is bought to yield anominal annual rate of 12% convertible monthly?

Solution: Solution: We know that F = C = 100,n = (5)(4) = 20, r = 0.015 and Fr = (100)(0.015) = 1.5. Let j bethe effective yield rate per quarter. We have that i (12) = 12%,i = 12.68250301% and i (4) = 12.1204%, j = i (4)/4 = 3.0301%.Hence,

P = 1.5a20−−|3.0301%

+ 100(1.030301)−20 = 77.29919664.


11/17


Remember that unless said otherwise a bond is redeemable at par.

Example 3

What is the price of a 5–year 100 par–value bond having quarterlycoupons at a quarter rate of 1.5% that is bought to yield anominal annual rate of 12% convertible monthly?

Solution: Solution: We know that F = C = 100,n = (5)(4) = 20, r = 0.015 and Fr = (100)(0.015) = 1.5. Let j bethe effective yield rate per quarter. We have that i (12) = 12%,i = 12.68250301% and i (4) = 12.1204%, j = i (4)/4 = 3.0301%.Hence,

P = 1.5a20−−|3.0301%

+ 100(1.030301)−20 = 77.29919664.


12/17


Example 4

The price of a zero coupon 1000 face value bond is 599.4584. Theyield rate convertible semi–annually is 6.5%. Calculate thematurity date.

Solution: Let n be the maturity date in years. We have that

599.4584 = (1000)(1 + 0.065

2

)2nand n = 8 years.


13/17


Example 4

The price of a zero coupon 1000 face value bond is 599.4584. Theyield rate convertible semi–annually is 6.5%. Calculate thematurity date.

Solution: Let n be the maturity date in years. We have that

599.4584 = (1000)(1 + 0.065

2

)2nand n = 8 years.


14/17


Example 5

What is the yield as an annual effective rate of interest on a 100par–value 10–year bond with coupon rate 6%, convertible monthly,that is selling for 90?

Solution: We know that P = 90, C = F = 100,n = (10)(12) = 120, r = 0.06

12 = 0.005 and

Fr = (100)(0.005) = 0.5. Let j = i (12)/12. Then,

90 = P = Fran−−|j + C (1 + j)−12n = (0.5)a

120−−|j + 100(1 + j)−120.

Hence, j = 0.618181404%, i (12) = 12j = 7.419376846% andi = 7.676949087%.


15/17


Example 5

What is the yield as an annual effective rate of interest on a 100par–value 10–year bond with coupon rate 6%, convertible monthly,that is selling for 90?

Solution: We know that P = 90, C = F = 100,n = (10)(12) = 120, r = 0.06

12 = 0.005 and

Fr = (100)(0.005) = 0.5. Let j = i (12)/12. Then,

90 = P = Fran−−|j + C (1 + j)−12n = (0.5)a

120−−|j + 100(1 + j)−120.

Hence, j = 0.618181404%, i (12) = 12j = 7.419376846% andi = 7.676949087%.


16/17


Example 6

A 1000 par value 10–year bond with semiannual coupons andredeemable at 1200 is purchased to yield 8% convertiblesemiannually. The first coupon is 50. Each subsequent coupon is3% greater than the preceding coupon. Find the price of the bond.

Solution: The cashflow of coupons is

Coupons 50 (50)(1.03) · · · (50)(1.03)19

Time (in half–years) 1 2 · · · 20


(50)1

1.03a20−−| 0.04−0.03

1.03+ (1200)(1.04)−20

=878.5721 + 547.6643 = 1426.2364.


17/17


Example 6

A 1000 par value 10–year bond with semiannual coupons andredeemable at 1200 is purchased to yield 8% convertiblesemiannually. The first coupon is 50. Each subsequent coupon is3% greater than the preceding coupon. Find the price of the bond.

Solution: The cashflow of coupons is

Coupons 50 (50)(1.03) · · · (50)(1.03)19

Time (in half–years) 1 2 · · · 20


(50)1

1.03a20−−| 0.04−0.03

1.03+ (1200)(1.04)−20

=878.5721 + 547.6643 = 1426.2364.


1/??

Chapter 5. Bonds.


Section 5.3. Book value and amortization schedules.





2/??

Chapter 5. Bonds. Section 5.3. Book value and amortization schedules.

Book value

The book value Bk of a bond at time k of a bond is the presentvalue of the payments to be made, i.e. the present value at time kof the remaining n − k coupons and the redemption value C . Thisis also the outstanding balance of the loan at that time. So, Bk

can be found using any of the following expressions:

Bk = Fran−k−−|i +Cνn−k = C +(Fr−Ci)a

n−k−−|i = C +C (g−i)an−k−−|i .

Notice that

Fran−k−−|i + Cνn−k = Fra

n−k−−|i + C (1− ian−k−−|i )

=C + (Fr − Ci)an−k−−|i = C + C (g − i)a

n−k−−|i .


3/??


Of course, we have that B0 = P and Bn = C . The previousformula for Bk is that of the outstanding balance of a loan usingthe prospective method. Using the retrospective method, the bookvalue of a bond is

Bk = P(1 + i)k − Frsk−−|i .

The previous formula is equivalent to

P = Bk(1 + i)−k + Frak−−|i .

A way to interpret the previous formula is as follow. Bk is thebalance in the loan after the first k payments are made. Thepresent value of Bk and the payments made until that momentequals the initial balance.


4/??


Example 1

Zack buys a 20 year bond with a par value of 4000 and 10%semiannual coupons. He attains an annual yield of 5% convertiblesemiannually. The redemption value of the bond is 1200. Find thebook value of the bond at the end of the 12–th year.

Solution: We have that F = 4000, r = 5%,Fr = (4000)(0.05) = 200, n = 40, k = 24 and C = 1200. Hence,

B24 = Fran−k−−|i + Cνn−k = (200)a

16−−|5%+ (1200)(1.05)−16

=3419.350452.


5/??


Example 1

Zack buys a 20 year bond with a par value of 4000 and 10%semiannual coupons. He attains an annual yield of 5% convertiblesemiannually. The redemption value of the bond is 1200. Find thebook value of the bond at the end of the 12–th year.

Solution: We have that F = 4000, r = 5%,Fr = (4000)(0.05) = 200, n = 40, k = 24 and C = 1200. Hence,

B24 = Fran−k−−|i + Cνn−k = (200)a

16−−|5% + (1200)(1.05)−16

=3419.350452.


6/??


Example 2

An n–year 5000 par value bond pays 6% annual coupons. Atannual yield of 3%, the book value of the bond at the end of year7 is 5520. Calculate the price of the bond.

Solution: We have that F = C = 5000, r = 6%,Fr = (5000)(0.06) = 300, B7 = 5520. Hence,

P = Bk(1+ i)−k +Frak−−|i = 5520(1.03)−7 +(300)a

7−−|3%= 6357.35.


7/??


Example 2

An n–year 5000 par value bond pays 6% annual coupons. Atannual yield of 3%, the book value of the bond at the end of year7 is 5520. Calculate the price of the bond.

Solution: We have that F = C = 5000, r = 6%,Fr = (5000)(0.06) = 300, B7 = 5520. Hence,

P = Bk(1+ i)−k +Frak−−|i = 5520(1.03)−7 +(300)a

7−−|3% = 6357.35.


8/??


Inductive relation for book value

Theorem 1

Bk+1 = Bk(1 + i)− Fr .

Proof:

Bk(1 + i)− Fr = (Fran−k−−|i + Cνn−k)(1 + i)− Fr

=Fr((1 + i)an−k−−|i − 1) + Cνn−k−1

=Fran−k−1−−|i + Cνn−k−1 = Bk+1.

Bk is the outstanding balance at time k. One period later, theprincipal has increased to Bk(1 + i), i.e. Bk(1 + i) is theoutstanding balance immediately before the (k + 1)–th payment ismade. Immediately after the (k + 1)–th payment to principal of Fris made, the outstanding balance is Bk+1 = Bk(1 + i)− Fr .


9/??



Theorem 1

Bk+1 = Bk(1 + i)− Fr .

Proof:


=Fr((1 + i)an−k−−|i − 1) + Cνn−k−1

=Fran−k−1−−|i + Cνn−k−1 = Bk+1.



10/??



Theorem 1

Bk+1 = Bk(1 + i)− Fr .

Proof:


=Fr((1 + i)an−k−−|i − 1) + Cνn−k−1

=Fran−k−1−−|i + Cνn−k−1 = Bk+1.



11/??


Example 3

Consider a 30–year $50,000 par–value bond with semiannualcoupons, with r = 0.03, and yield rate 10%, convertiblesemiannually.(i) Find the book value of the bond immediately after the 25–thcoupon payment.(ii) Find book price immediately before the 26–th coupon payment.(iii) Find the book value of the bond immediately after the 26–thcoupon payment.

Solution: (i) We have that F = C = 50000, i = 0.05,Fr = 50000(0.03) = 1500 and n = 60. So,

B25 = Fran−k−−|i+Cνn−k = 1500a

35−−|0.05+50000(1.05)−35 = 33625.80571.

(ii) Just before the next coupon payment the book value is33625.80571(1.05) = 35307.096.(iii) The book value of the bond immediately after the 26–thcoupon payment is 35307.096− 1500 = 33807.096.


12/??


Example 3




35−−|0.05+50000(1.05)−35 = 33625.80571.



13/??


Example 3




35−−|0.05+50000(1.05)−35 = 33625.80571.

(ii) Just before the next coupon payment the book value is33625.80571(1.05) = 35307.096.

(iii) The book value of the bond immediately after the 26–thcoupon payment is 35307.096− 1500 = 33807.096.


14/??


Example 3




35−−|0.05+50000(1.05)−35 = 33625.80571.



15/??


General inductive relation for book value

By induction from the previous formula, we get that

Bk+m = Bk(1 + i)m − Frsm−−|i .

Notice that:

I Bk+m is the outstanding balance immediately after the k + mpayment.

I Bk(1 + i)m is the accrued balance at time k + m of theoutstanding balance immediately after the k payment.

I Frsm−−|i is the future value at time k + m of the coupon

payments k + 1, k + 2, . . . , k + m, i.e. the coupons paymentsfrom when the outstanding was Bk until when theoutstanding is Bk+m.


16/??


Example 4

An n–year 4000 par value bond with 9% semiannual coupons hasan annual nominal yield of i , i > 0, convertible semiannually. Thebook value of the bond at the end of year 4 is 3812.13 and thebook value at the end of year 7 is 3884.27. Calculate i .

Solution: We have that F = C = 4000, r = 4.5% andFr = 4000(0.045) = 180. The end of year 4 is the end of the 8–thperiod. The end of year 7 is the end of the 14–th period. Hence,B8 = 3812.13 and B14 = 3884.27 and

3884.27 = 3812.13(1 + i)6 − (180)s6−−|i .

using the calculator with

6 N −3812.13 PV 180 PMT 3884.27 FV CPT I/Y

we get that i = 5% (this is the effective interest rate per period).The annual nominal rate of interest convertible semiannually isi = 10%.


17/??


Example 4

An n–year 4000 par value bond with 9% semiannual coupons hasan annual nominal yield of i , i > 0, convertible semiannually. Thebook value of the bond at the end of year 4 is 3812.13 and thebook value at the end of year 7 is 3884.27. Calculate i .

Solution: We have that F = C = 4000, r = 4.5% andFr = 4000(0.045) = 180. The end of year 4 is the end of the 8–thperiod. The end of year 7 is the end of the 14–th period. Hence,B8 = 3812.13 and B14 = 3884.27 and

3884.27 = 3812.13(1 + i)6 − (180)s6−−|i .

using the calculator with

6 N −3812.13 PV 180 PMT 3884.27 FV CPT I/Y

we get that i = 5% (this is the effective interest rate per period).The annual nominal rate of interest convertible semiannually isi = 10%.


18/??


Amount of interest contained in the k–th coupon.

The amount of interest contained in the k–th coupon isIk = iBk−1, which can obtained using any of the followingformulas:

Ik = iBk−1 = Fr − (Fr − Ci)νn−k+1 = Cg − C (g − i)νn−k+1.

Notice that

iBk−1 = iFran+1−k−−|i + iCνn−k+1 = Fr(1− νn−k+1) + iCνn−k+1

=Fr − (Fr − Ci)νn−k+1 = Cg − (Cg − Ci)νn−k+1

=Cg − C (g − i)νn−k+1.


19/??


Principal portion in the the k–th coupon.

The principal portion in the k–th coupon isPk = Bk−1 − Bk = Fr − Bk−1i = Fr − Ik and it can be obtainedusing any of the following formulas:

Pk = (Fr − Ci)νn−k+1 = C (g − i)νn−k+1.

Pk is the change in the book value of the bond (principaladjustment) between times k − 1 and k. Pk could be eithernegative, or zero or positive. Pk is the amortization in the k–thpayment.


20/??


Example 5

Kendal buys a 5000 par–value 10 year bond with 8% semiannualcoupons to yield 4% converted semiannually. Find the amount ofinterest and principal in the 5–th coupon.

Solution: We have that F = C = 5000, r = g = 0.04, Fr = 200,n = 20 and i = 2%. Using that Ik = Fr − (Fr − Ci)νn+1−k andPk = (Fr − Ci)νn+1−k , we get that

I5 = 200− (200− (5000)(0.02))(1.02)−(20+1−5) = 127.1554

and

P5 = (200− (5000)(0.02))(1.02)−(20+1−5) = 72.8446.


21/??


Example 5

Kendal buys a 5000 par–value 10 year bond with 8% semiannualcoupons to yield 4% converted semiannually. Find the amount ofinterest and principal in the 5–th coupon.

Solution: We have that F = C = 5000, r = g = 0.04, Fr = 200,n = 20 and i = 2%. Using that Ik = Fr − (Fr − Ci)νn+1−k andPk = (Fr − Ci)νn+1−k , we get that

I5 = 200− (200− (5000)(0.02))(1.02)−(20+1−5) = 127.1554

and

P5 = (200− (5000)(0.02))(1.02)−(20+1−5) = 72.8446.


22/??


Example 6

For a bond pays annual coupons. The principal amortized in thefirst coupon is half that amortized in the 10–th coupon. Calculatethe annual yield rate.

Solution: We know that (Fr − Ci)νn = 12(Fr − Ci)νn−9. Hence,

(1 + i)9 = 2 and i = 8.005973889%.


23/??


Example 6

For a bond pays annual coupons. The principal amortized in thefirst coupon is half that amortized in the 10–th coupon. Calculatethe annual yield rate.

Solution: We know that (Fr − Ci)νn = 12(Fr − Ci)νn−9. Hence,

(1 + i)9 = 2 and i = 8.005973889%.


24/??


The amortization schedule of a bond is

Time Payment Interest paid (iBk−1) Principal repaid Book value (Bk)

0 − − − C + (Fr − Ci)an−−|i

1 Fr Fr − (Fr − Ci)νn (Fr − Ci)νn C + (Fr − Ci)an−1−−|i

2 Fr Fr − (Fr − Ci)νn−1 (Fr − Ci)νn−1 C + (Fr − Ci)an−2−−|i

· · · · · · · · · · · · · · ·k Fr Fr − (Fr − Ci)νn−k+1 (Fr − Ci)νn+1−k C + (Fr − Ci)a

n−k−−|i

· · · · · · · · · · · · · · ·n − 1 Fr Fr − (Fr − Ci)ν2 (Fr − Ci)ν2 C + (Fr − Ci)a

1−−|i

n Fr + C Fr − (Fr − Ci)ν (Fr − Ci)ν + C 0

The total payments in a bond are nFr + C .The total coupon interest (sum of the column of interestpayments) is nFr + C − P.The total coupon principal (sum of the column of payments toprincipal) is P.


25/??


Notice that for each time k the interest paid plus the principal paidequal to the total payment. We also have that

Ik = iBk−1 = i(C + (Fr − Ci)a

n−k+1−−|i

)= iC + i(Fr − Ci)a

n−k+1−−|i

=Fr + (Fr − Ci)(1− νn−k+1) = Fr − (Fr − Ci)νn−k+1.

and

Bk = Bk−1 − (Fr − Ci)νn+1−k

=C + (Fr − Ci)an−k+1−−|i − (Fr − Ci)νn+1−k

=C + (Fr − Ci)(an−k+1−−|i − νn+1−k) = C + (Fr − Ci)a

n−k−−|i .


26/??


Example 7

A 1000 par–value 3–year bond pays 6%, convertible semiannually,and has a yield rate of 8%, convertible semiannually.(i) What is the interest paid in the 3rd coupon?(ii) What is the change in book value contained in the 3rd coupon?(iii) Construct a bond amortization schedule.

Solution:


27/??


Example 7


Solution:(i) We know that F = C = 1000, r = 0.03, Fr = 30, n = 6,i = 0.04 and Ci = 40. The interest paid in the 3rd coupon is

I3 = Fr−(Fr−Ci)νn−k+1 = 30−(30−40)(1.04)−4 = 38.54804191.

We also can do

B2 = 30a4−−|0.04

+ 1000(1.04)−4 = 963.7010478

and I3 = B2(0.04) = 38.54804191.


28/??


Example 7


Solution:(ii) Since I3 > Fr , the book value increases in the 3rd coupon.The increase in the book value in the 3rd coupon is I3 − Fr =38.54804191− 30 = 8.54804191. We also can do

B3 = 30a3−−|0.04

+ 1000(1.04)−3 = 972.2490897

and B3 − B2 = 972.2490897− 963.7010478 = 8.5480419.


29/??


(iii) Here is the bond amortization schedule:

Principal BookTime Coupon Interest Adjustment Value

0 947.58

1 30 37.90 7.90 955.48

2 30 38.22 8.22 963.70

3 30 38.59 8.59 972.25

4 30 38.89 8.89 981.14

5 30 39.25 9.25 990.37

6 30 39.62 9.62 1000.00


30/??


The price of a typical bond changes in the opposite direction froma change in interest rates.As interest rates rise, the price of a bond falls. A bond assures adetermined number of payments in the future, if interest rates rise,the present value of these payments decreases. We make anunrealized capital loss, if the market value is less than the bookvalue.Reciprocally, if interest rates decline, the price of a bond rises. Themarket value of a bond is the price at which a bond isbought/sold. When rates of interest change, the market value of abond changes. We make an unrealized capital gain, if the marketvalue is bigger than the book value.


31/??


Example 8

Oliver buys a ten–year 5000 face value bond with semiannualcoupons at annual rate of 6%. He buys his bond to yield 8%compounded semiannually and immediately sell them to aninvestor to yield 4% compounded semiannually. What is Oliver’sprofit in this investment?

Solution: We have that F = C = 5000, r = 0.03, Fr = 150 andn = 20. Oliver buys his bond for

(150)a20−−|0.04

+ 5000(1.04)−20 = 4320.484

Oliver sells his bond for

(150)a20−−|0.02

+ 5000(1.02)−20 = 5817.572

Oliver’s profit is 5817.572− 4320.484 = 1497.088.


32/??


Example 8

Oliver buys a ten–year 5000 face value bond with semiannualcoupons at annual rate of 6%. He buys his bond to yield 8%compounded semiannually and immediately sell them to aninvestor to yield 4% compounded semiannually. What is Oliver’sprofit in this investment?

Solution: We have that F = C = 5000, r = 0.03, Fr = 150 andn = 20. Oliver buys his bond for

(150)a20−−|0.04

+ 5000(1.04)−20 = 4320.484

Oliver sells his bond for

(150)a20−−|0.02

+ 5000(1.02)−20 = 5817.572

Oliver’s profit is 5817.572− 4320.484 = 1497.088.


33/??


Example 9

On January 1, 2000, Maxwell bought a 10–year $5000 non–callablebond with coupons at 7% convertible semiannually. Maxwellbought the bond to yield 7%, compounded semiannually. On July1, 2005, the market value of bonds is based on a 5% interest rate,compounded semiannually. Calculate the unrealized capital gain onJuly 1, 2005.

Solution: We have that F = 5000, r = 0.035 and Fr = 175. OnJuly 1, 2005, there are nine remaining coupons. On July 1, 2005,the book value of the bond is

175a9−−|3.5%

+ 5000(1.035)−9 = 5000.

On July 1, 2005, the market value of the bond is

175a9−−|2.5%

+ 5000(1.025)−9 = 5398.543276.

The unrealized capital gain is 5398.543276− 5000 = 398.543276.


34/??


Example 9

On January 1, 2000, Maxwell bought a 10–year $5000 non–callablebond with coupons at 7% convertible semiannually. Maxwellbought the bond to yield 7%, compounded semiannually. On July1, 2005, the market value of bonds is based on a 5% interest rate,compounded semiannually. Calculate the unrealized capital gain onJuly 1, 2005.

Solution: We have that F = 5000, r = 0.035 and Fr = 175. OnJuly 1, 2005, there are nine remaining coupons. On July 1, 2005,the book value of the bond is

175a9−−|3.5%

+ 5000(1.035)−9 = 5000.

On July 1, 2005, the market value of the bond is

175a9−−|2.5%

+ 5000(1.025)−9 = 5398.543276.

The unrealized capital gain is 5398.543276− 5000 = 398.543276.c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2.

35/??


Premium.

If the investor is paying more than the redemption value, i.e. ifP > C , we say that the bond has being bought at premium. Thepremium is P − C . We have that

P − C = (Fr − Ci)an−−|i = C (g − i)a

n−−|i .

A bond has been bought at premium if and only if g > i . For abond bought at premium Fr > Ci andP = B0 > B1 > B2 > · · · > Bn = C . Notice thatBk − C = (Fr − Ci)a

n−k−−|i . Pk = Bk−1 − Bk = (Fr − Ci)νn+1−k is

the write–up in premium in the k–th coupon. The premium isP − C =

∑nj=1(Bj−1 − Bj).


36/??


Discount.

If the investor is paying less than the redemption value, i.e. ifP < C , we say that the bond has being bought at discount. Thediscount is C − P. We have that

C − P = (Ci − Fr)an−−|i = C (i − g)a

n−−|i .

A bond has been bought at discount if and only if g < i . For abond bought at discount Fr < Ci andP = B0 < B1 < B2 < · · · < Bn = C .|Pk | = Bk − Bk−1 = (Ci − Fr)νn+1−k is the write–up in discountin the k–th coupon. The discount is C − P =

∑nj=1(Bj − Bj−1).


37/??


Example 10

A 10 year 50000 face value bond pays semi–annual coupons of3000. The bond is bought to yield a nominal annual interest rateof 6% convertible semi–annually. Calculate the premium paid forthe bond.

Solution: We know that F = C = 50000, Fr = 3000, n = 20, andi = 3%. The price of the bond is

P = Fran−−|i + C (1 + i)−n = (3000)a

20−−|3%+ (50000)(1.03)−20

=72316.21229.

The premium of the bond isP − C = 72316.21229− 50000 = 22316.21229.


38/??


Example 10

A 10 year 50000 face value bond pays semi–annual coupons of3000. The bond is bought to yield a nominal annual interest rateof 6% convertible semi–annually. Calculate the premium paid forthe bond.

Solution: We know that F = C = 50000, Fr = 3000, n = 20, andi = 3%. The price of the bond is

P = Fran−−|i + C (1 + i)−n = (3000)a

20−−|3%+ (50000)(1.03)−20

=72316.21229.

The premium of the bond isP − C = 72316.21229− 50000 = 22316.21229.


39/??


Example 11

Oprah buys a 10000 par–value 15 year bond with 9% semiannualcoupons to yield 5% converted semiannually.(i) Find the premium in the bond.(ii) Find the write up in premium in the 8–th coupon.

Solution: (i) We have that F = C = 10000, r = 0.045, Fr = 450and n = 30. Oprah buys her bond for

(450)a30|2.5% + 10000(1.025)−30 = 14186.06

The premium of the bond is P −C = 14186.06− 10000 = 4186.06.(ii) The write–up in premium in the 8–th coupon is

(Fr − Ci)νn+1−k = (10000)(0.045− 0.025)(1.025)−23 = 113.34.

We also can do the problem in the following way:

B7 = (450)a23−−|2.5%

+ 10000(1.025)−23 = 13466.42

B8 = (450)a22−−|2.5%

+ 10000(1.025)−22 = 13353.08

and B7 − B8 = 13466.42− 13353.08 = 113.34.


40/??


Example 11



(450)a30|2.5% + 10000(1.025)−30 = 14186.06

The premium of the bond is P −C = 14186.06− 10000 = 4186.06.

(ii) The write–up in premium in the 8–th coupon is

(Fr − Ci)νn+1−k = (10000)(0.045− 0.025)(1.025)−23 = 113.34.


B7 = (450)a23−−|2.5%

+ 10000(1.025)−23 = 13466.42

B8 = (450)a22−−|2.5%

+ 10000(1.025)−22 = 13353.08

and B7 − B8 = 13466.42− 13353.08 = 113.34.


41/??


Example 11



(450)a30|2.5% + 10000(1.025)−30 = 14186.06

The premium of the bond is P −C = 14186.06− 10000 = 4186.06.(ii) The write–up in premium in the 8–th coupon is

(Fr − Ci)νn+1−k = (10000)(0.045− 0.025)(1.025)−23 = 113.34.


B7 = (450)a23−−|2.5%

+ 10000(1.025)−23 = 13466.42

B8 = (450)a22−−|2.5%

+ 10000(1.025)−22 = 13353.08

and B7 − B8 = 13466.42− 13353.08 = 113.34.c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2.

1/10

Chapter 5. Bonds.


Section 5.4. Book value between coupon payments dates.





2/10

Chapter 5. Bonds. Section 5.4. Book value between coupon payments dates.

Book value between coupon payments dates

We know that the book value of a bond immediately after acoupon payment is given by

Bk = Fran−k|i +Cνn−k = C +(Fr−Ci)an−k|i = C +C (g− i)an−k|i .

We want to determine the book value of a bond betweensuccessive coupon dates. We use as a unit of time coupon periods.We want to determine the value of a bond at time k + t periods,where k is an integer and 0 ≤ t < 1. The present value at timek + t of the remaining payments is Bk(1 + i)t . Immediately beforethe payment of the (k + 1)–th coupon, the price of the bond isBk(1 + i). Immediately after the payment of the (k + 1)–thcoupon, the price of the bond is Bk(1 + i)− Fr . Hence, the priceof a bond is a discontinuous function (see Figure 1).


3/10


Figure 1: Theoretical flat and market values


4/10


During the time period (k, k + 1) a bond accrues a couponpayment. The accrued value at time t ∈ [0, 1) of the next couponis denoted by Frt . We have that Fr0 = 0, limt→1− Frt = Fr . Theflat price of a bond is the money that actually changes hands atthe date of sale. The market price Bm

k+t is the price of a bondexcluding the accrued value of the next coupon. Hence, we havethat B f

k+t = Bmk+t + Frt .

The flat price (also known as the dirty price) is the book value of abond. It is the price that an investor pays for a bond. The marketprice (also known as the clean price) is the price of bond quoted ina newspaper.


5/10


There are three methods to determine the flat value, accruedcoupon and market value of a bond:

I Theoretical method:

B fk+t = Bk(1 + i)t , Frt = Fr

((1 + i)t − 1

i

)Bm

k+t = Bk(1 + i)t − Fr

((1 + i)t − 1

i

).

I Practical method:

B fk+t = Bk(1 + ti), Frt = tFr , Bm

k+t = Bk(1 + ti)− tFr

I Semi–theoretical method:

B fk+t = Bk(1 + i)t , Frt = tFr , Bm

k+t = Bk(1 + i)t − tFr


6/10


The practical method assumes that the flat price accrues undersimple interest. It assumes that the accrued coupon is proportionalto the time since the last coupon payment.As to the theoretical method, Bk(1 + i)t is the present value of thepayments to be made. This is actual outstanding balance in theloan at time k + t. We have that

Bmk+t = Bk(1 + i)t − Fr

((1 + i)t − 1

i

)=

(Fr

(1− νn−k

i

)+ Cνn−k

)ν−t − Fr

(ν−t − 1

i

)=Fr

(1− νn−k−t

i

)+ Cνn−k−t = Fran−k−t|i + Cνn−k−t ,

where as|i = 1−νs

i , s > 0 and s is not necessarily a positive integer.The market price according with the theoretical method has acontinuous function.


7/10


Example 1

Find the flat price, the accrued interest and the market price of a1000 10–year bond with 4% annual coupons, bought to yield 3%,four months after the second coupon has been issued. Use allthree methods.

Solution: We have that F = 1000, n = 10, r = 0.04, Fr = 40,i = 3%, k = 2 and t = 1

3 . So,

B2 = Fran−k−−|i + Cνn−k = 40a

8−−|3% + (1000)(1.03)−8 = 1070.20.


8/10


Example 1



3 . So,


8−−|3%+ (1000)(1.03)−8 = 1070.20.

Using the theoretical method,

B f2+(1/3) = B2(1 + i)1/3 = (1070.20)(1.03)1/3 = 1080.80,

Fr(1/3) = Fr

[(1 + i)1/3 − 1

i

]= (40)

[(1 + 0.03)1/3 − 1

0.03

]= 13.20

Bm2+(1/3) = B f

2+(1/3) − Fr(1/3) = 1080.80− 13.20 = 1067.60.


9/10


Example 1



3 . So,


8−−|3%+ (1000)(1.03)−8 = 1070.20.

Using the practical method,

B f2+(1/3) = B2(1 + (1/3)i) = (1070.20)(1 + (0.03)/3)) = 1080.90,

Fr(1/3) = (1/3)Fr = (1/3)(40) = 13.33

Bm2+(1/3) = B f

2+(1/3) − Fr(1/3) = 1080.90− 13.33 = 1067.57


10/10


Example 1



3 . So,


8−−|3%+ (1000)(1.03)−8 = 1070.20.

Using the semi–theoretical method:

B f2+(1/3) = B2(1 + i)1/3 = (1070.20)(1.03)1/3 = 1080.80,

Fr(1/3) = (1/3)Fr = (1/3)(40) = 13.33,

Bm2+(1/3) = B f

2+(1/3) − Fr(1/3) = 1080.79− 13.33 = 1067.46.


1/10

Chapter 5. Bonds.


Section 5.5. Callable bonds.





2/10

Chapter 5. Bonds. Section 5.5. Callable bonds.

Callable bonds

A callable bond is a bond which gives the issuer (not the investor)the right to redeem prior to its maturity date, under certainconditions. When issued, the call provisions explain when thebond can be redeemed and what the price will be. In most cases,there is some period of time during which the bond cannot becalled. This period of time is named the call protection period.The earliest time to call the bond is named the call date. The callprice is the amount of money the insurer must pay to buy thebond back.Usually, bonds can be only called immediately after the payment ofa coupon. We will study the computation of the yield rate ofreturn for the investor in this situation. Since the investor does notknow the cashflow obtained from his investment, he will assumethat the issuer calls the bond under the worst possible situation (inthe sense of lowest possible interest rates).


3/10


If the bond is called immediately after the payment of k–thcoupon, the present value of the obtained payments is

Pk = Frak−−|i + Cνk = C + (Fr − Ci)a

k−−|i = C + C (g − i)ak−−|i .

Pk is the price which the investor would pay for the bond assumingthat the bond is called immediately after the k coupon. As smalleras Pk is, as worst for the lender is. Between all possible choices torecall a bond, the borrower will choose the option with the smallestprice. Assuming that the redemption value is a constant and that abond can be called after any coupon payment:(i) if Fr > Ci (bond sells at a premium), Pk increases with k, andwe assume the redemption date is the earliest possible.(ii) if Fr < Ci (bond sells at a discount), Pk decreases with k, andwe assume that the redemption date is the latest possible.If one investor wants to get an effective rate of interest of i perperiod, then the maximum price which the investor should pay isthe lowest possible Pk under that particular rate i .


4/10


Example 1

Consider a 100 par–value 8% bond with semiannual couponscallable at 120 on any coupon date starting 5 years after issue forthe next 5 years, at 110 starting 10 years after issue for the next 5years and maturing at 105 at the end of 15 years. What is thehighest price which an investor can pay and still be certain of ayield of 9% converted semiannually.

Solution: We have that F = 100, r = 4%, Fr = 4, n = 30 andi = 4.5%. Let k be the number of the half year when the bond iscalled.


5/10


Example 1


Solution: We have that F = 100, r = 4%, Fr = 4, n = 30 andi = 4.5%. Let k be the number of the half year when the bond iscalled.If 10 ≤ k ≤ 19, then C = 120 and Ci = (120)(0.045) = 5.4. So,the bond sells at a discount. We have that lowest price which theinvestor can get is

P19 = 4a19−−|4.5%

+ 120(1.045)−19 = 102.37.


6/10


Example 1


Solution: We have that F = 100, r = 4%, Fr = 4, n = 30 andi = 4.5%. Let k be the number of the half year when the bond iscalled.If 20 ≤ k ≤ 29, then C = 110 and Ci = (110)(0.045) = 4.95. So,the bond sells at a discount. We have that lowest price which theinvestor can get is

P29 = 4a29−−|4.5%

+ (110)(1.045)−29 = 94.78.


7/10


Example 1


Solution: We have that F = 100, r = 4%, Fr = 4, n = 30 andi = 4.5%. Let k be the number of the half year when the bond iscalled.If k = 30, then the price is

P30 = 4a30−−|4.5%

+ 105(1.045)−30 = 93.19.


8/10


Example 1


Solution: We have that F = 100, r = 4%, Fr = 4, n = 30 andi = 4.5%. Let k be the number of the half year when the bond iscalled.We conclude that the highest price which an investor can pay andstill be certain of a yield of 9% converted semiannually is 93.19.


9/10


Example 2

Joshua paid 800 for a 15-year 1000 par value bond withsemiannual coupons at a nominal annual rate of 4% convertiblesemiannually. The bond can be called at 1300 on any coupon datestarting at the end of year 7. What is the minimum annualnominal rate convertible semiannually yield that Joshua couldreceive? Answer: 7.3521%.

Solution: We have that F = 1000, P = 800, C = 1300, r = 0.02and n = 30. Since P < C , the bond was bought at discount. Weassume that the redemption value is as late as possible. From theequation

800 = 20a30−−|i + 1300(1 + i)−30

we get that i = 3.6760 and i (2) = 7.3521%.


10/10


Example 2

Joshua paid 800 for a 15-year 1000 par value bond withsemiannual coupons at a nominal annual rate of 4% convertiblesemiannually. The bond can be called at 1300 on any coupon datestarting at the end of year 7. What is the minimum annualnominal rate convertible semiannually yield that Joshua couldreceive? Answer: 7.3521%.

Solution: We have that F = 1000, P = 800, C = 1300, r = 0.02and n = 30. Since P < C , the bond was bought at discount. Weassume that the redemption value is as late as possible. From theequation

800 = 20a30−−|i + 1300(1 + i)−30

we get that i = 3.6760 and i (2) = 7.3521%.


1/11

More securities


Section 5.6. More securities.





2/11

More securities

Serial bonds

A serial bond is a collection of bonds issued at the same time butwith different redemption dates. The price of a serial bond is thesum of the prices of the individual bonds.Let Pk ,Ck ,Kk be the price value, the redemption value and thepresent value of redemption value of the k–th bond. Let P ′,C ′,K ′

be the price value, redemption value and present value ofredemption value of the serial bond. For each k we have that

Pk = Kk +g

i(Ck − Kk).

Let P ′ =∑k

j=1 Pj , C ′ =∑k

j=1 Cj and K ′ =∑k

j=1 Kj . Hence,

P ′ = K ′ +g

i(C ′ − K ′).


3/11

More securities

Example 1

A 12% serial bond with semiannual coupons and par value of 1000will be redeemed by the following schedule:(i) 100 at the end of years 10 through 14; and(ii) 500 at the end of year 15.Calculate the price of the bond on the issue date to yield 10% perannum convertible semiannually.

Solution: We have that g = FrC = 6% and i = 5%. Since the

annual nominal rate compounded semiannually is 10%, the annualeffective rate of interest is 10.25%. The redemption values andtimes of redemption are given by the following table:

Redemption value 100 100 100 100 100 500

Time in years 10 11 12 13 14 15


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More securities

Example 1

A 12% serial bond with semiannual coupons and par value of 1000will be redeemed by the following schedule:(i) 100 at the end of years 10 through 14; and(ii) 500 at the end of year 15.Calculate the price of the bond on the issue date to yield 10% perannum convertible semiannually.




Time in years 10 11 12 13 14 15


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More securities




Time in years 10 11 12 13 14 15

Hence,

K ′ =14∑

k=10

(100)(1 + 0.1025)−k + (500)(1 + 0.1025)−15

=(100)(1.1025)−9a5−−|10.25%

+ (500)(1.1025)−15

=156.51 + 115.69 = 272.20.

Hence,

P ′ = K ′ +g

i(C ′−K ′) = 272.20+

0.06

0.05(1000− 272.20) = 1145.56.


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More securities

Preferred stock

Preferred stock is like a perpetual bond. This stock pays dividendsforever. The price of the stock is the present value of futuredividends. If a preferred stock pays an annual dividend D, then theprice of this stock is

P = Da∞−−|i =D

i,

where i the annual effective rate of interest.


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More securities

Example 2

You have acquired some Microsot preferred stock that pays $3000per year forever. What is the present value of this investment?Assume that i = 6%.

Solution: 30000.06 = 50000.


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More securities

Example 2

You have acquired some Microsot preferred stock that pays $3000per year forever. What is the present value of this investment?Assume that i = 6%.

Solution: 30000.06 = 50000.


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More securities

Common stock

For common stock the dividends are not known in advance. Soone has to project what these dividends will be in the future. Forexample, let D be the dividend at the end of the current period andassume that the next dividends change geometrically with commonratio 1 + k with −1 < k < i , then the cashflow of dividends is:

Contributions D D(1 + k) D(1 + k)2 · · ·Time, in years 1 2 3 · · ·

The price of the common stock is

P =D

1 + ka∞−−| i−k

1+k= D

1

i − k.

Recall that

P = D1

1 + i+D

1 + k

(1 + i)2+D

(1 + k)2

(1 + i)3+. . . =

D

1 + i· 1

1− 1+k1+i

= D1

i − k.


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More securities

Example 3

Each quarter the corporation plans to pay 45% of its earnings as astock dividend. The earnings of a corporation increase at 1% perquarter indefinitely. At the start of a quarter, an investor purchasesthe stock to yield a nominal rate of 5% compounded quarterly.The first stock dividend is 2.4 payable at the end of the quarter.Calculate the theoretical price of the stock.

Solution: Since the earnings of a corporation increase at 1% perquarter, the dividends also increase at 1% per quarter. Thecashflow of dividends is

Dividends 2.4 (2.4)(1.01) (2.4)(1.01)2 · · ·Time 1 2 3 · · ·

The present value of the cashflow of dividends isP

(i (4)/4)−k= 2.4

(0.05/4)−0.01 = 960.


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More securities

Example 3

Each quarter the corporation plans to pay 45% of its earnings as astock dividend. The earnings of a corporation increase at 1% perquarter indefinitely. At the start of a quarter, an investor purchasesthe stock to yield a nominal rate of 5% compounded quarterly.The first stock dividend is 2.4 payable at the end of the quarter.Calculate the theoretical price of the stock.

Solution: Since the earnings of a corporation increase at 1% perquarter, the dividends also increase at 1% per quarter. Thecashflow of dividends is

Dividends 2.4 (2.4)(1.01) (2.4)(1.01)2 · · ·Time 1 2 3 · · ·

The present value of the cashflow of dividends isP

(i (4)/4)−k= 2.4

(0.05/4)−0.01 = 960.


1/6

Chapter 6. Variable interest rates and portfolio insurance.

Manual for SOA Exam FM/CAS Exam 2.Chapter 6. Variable interest rates and portfolio insurance.

Section 6.1. Inflation.





2/6

Chapter 6. Variable interest rates and portfolio insurance. Section 6.1. Inflation.

Inflation

Inflation is the fall in the purchasing power of money over time. Itis usually measured with reference to an index representing thecost of certain goods and services. One the most frequently usedindex is the consumer index price. The consumer index price isreleased by the Bureau of Labor Statistics monthly. The Bureau ofLabor Statistics finds the consumer index price by averaging thechanges of prices of a market basket of goods and services.Sometimes, it is convenient for all the calculations related with aninvestment to find the units of real purchasing power rather thanunits of ordinary currency.


3/6


Example 1

Peter uses his 450000 in his retirement fund to buy aperpetuity–immediate. The perpetuity is expected to pay dividendsat the end of each year forever. The next payment (payable oneyear from now) is x, and is expected to increase at a rate of 3%per year. This increase is made to take in account for the inflation.The current annual effective rate of interest is 4.5%. Calculate x.


Payments x x(1.03)1 x(1.03)2 · · ·Time 1 2 3 · · ·

Using the formula for the present value of a geometric perpetuity,450000 = x

0.045−0.03 and x = (450000)(0.045− 0.03) = 6750.


4/6


Example 1

Peter uses his 450000 in his retirement fund to buy aperpetuity–immediate. The perpetuity is expected to pay dividendsat the end of each year forever. The next payment (payable oneyear from now) is x, and is expected to increase at a rate of 3%per year. This increase is made to take in account for the inflation.The current annual effective rate of interest is 4.5%. Calculate x.


Payments x x(1.03)1 x(1.03)2 · · ·Time 1 2 3 · · ·

Using the formula for the present value of a geometric perpetuity,450000 = x

0.045−0.03 and x = (450000)(0.045− 0.03) = 6750.


5/6


Example 2

Richard invests $10,000 at the end of each year for 10 years intoan account earning an effective rate of interest is 6.5%. Theannual rate of inflation is 3.5% over the 10 year period. Calculatethe value at the end of 10 years of Richard’s investment in today’sdollars.

Solution: The balance at the end of ten years is(10000)s10|6.5% = 134944.2254. The value at the end of 10 yearsof Richard’s investment in today’s dollars is134944.2254(1.035)−10 = 95664.50019.


6/6


Example 2

Richard invests $10,000 at the end of each year for 10 years intoan account earning an effective rate of interest is 6.5%. Theannual rate of inflation is 3.5% over the 10 year period. Calculatethe value at the end of 10 years of Richard’s investment in today’sdollars.

Solution: The balance at the end of ten years is(10000)s10|6.5% = 134944.2254. The value at the end of 10 yearsof Richard’s investment in today’s dollars is134944.2254(1.035)−10 = 95664.50019.


1/2



Section 6.2. Arbitrage.





2/2

Chapter 6. Variable interest rates and portfolio insurance. Section 6.2. Arbitrage.

Arbitrage

In economics, arbitrage is the practice of taking advantage of astate of imbalance between two (or possibly more) markets by acombination of matching deals to make a profit. A simple case ofarbitrage consists in buying something in one place and selling it inanother place at the same time. Suppose that the exchange rates(after taking out the fees for making the exchange) in London are£5 = U10 and the exchange rates in Tokyo are £6 = U10.Converting U10 to £6 in Tokyo and converting that £6 into U12in London, for a profit of U2, would be arbitrage.


1/25



Section 6.3. Term structure of interest rates.





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Chapter 6. Variable interest rates and portfolio insurance. Section 6.3. Term structure of interest rates.

Term structure of interest rates

The relationship between yield and time to mature is called theterm structure of interest rates.As larger as money is tied up in an investment as more likely adefault is. Usually, interest rates increase with maturity date.For US Treasury zero–coupons bonds, different interest rates aregiven according with the maturity date.

Definition 1A yield curve is a graph that shows interest rates (vertical axis)versus (maturity date) duration of a investment/loan (horizontalaxis).

Yield curves are studied to predict of changes in economic activity(economic growth, inflation, etc.).


3/25


Example 1A bank offers CD’s with the following interest rates

length of the investment (in years) 1 year 2 years 3 years 4 years 5 years

Interest rate 7% 8% 8.5% 9% 9.25%

Graph the yield curve for these interest rates.

Solution: Assuming that the interest rate is a linear functionbetween the given points, the yield curve is given by the graph inFigure 1.


4/25


Example 1A bank offers CD’s with the following interest rates

length of the investment (in years) 1 year 2 years 3 years 4 years 5 years

Interest rate 7% 8% 8.5% 9% 9.25%

Graph the yield curve for these interest rates.

Solution: Assuming that the interest rate is a linear functionbetween the given points, the yield curve is given by the graph inFigure 1.


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Figure 1: Yield curve for Example 1


6/25


The interest rates appearing in the yield curve are called the spotrates. Thus, for Example 1, the spot rates are 7%, 8%, 8.5%, 9%and 9.25%.

Definition 2The j year spot rate sj is the rate of interest charged in a loanpaid with a unique payment at the end of j years.

Note that is the j year spot rate sj is as an effective annual rateof interest, the current j year interest factor is (1 + sj)

j . Moneyinvested now multiply by (1 + sj)

j in j years. The price of a zerocoupon j–year bond with face value F is P = F (1 + sj)

−j .


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Spot rates refer to a fixed maturity date. Usually, bonds havecoupon payments over time. But, often strip bonds are traded.Strip or zero coupon bonds are bonds that have being”separated” into their component parts (each coupon payment andthe face value). Often strip bonds are obtained from US Treasurybonds. A financial trader (strips) ”separates” the coupons from aUS Treasury bond, by accumulating a large number of US Treasurybonds and selling the rights of obtaining a particular payment toan investor. In this way, the investor can buy a strip bond as anindividual security. The strip bond market consists of coupons andresiduals, with coupons representing the interest portion of theoriginal bond and the residual representing the principal portion.An investor will get a unique payment from a strip bond. In thissituation, interest rates of a strip bond depend on the maturitydate. The yield rate of a zero–coupon bond is called its spot rate.


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The following table consists of the Daily Treasury Yield CurveRates, which can be found athttp://www.treas.gov/offices/domestic-finance/debt-management/interest-rate/yield.html

Date 1 mo 3 mo 6 mo 1 yr 2 yr 3 yr 5 yr 7 yr 10 yr 20 yr07/01/04 1.01 1.22 1.64 2.07 2.64 3.08 3.74 4.18 4.57 5.3107/02/04 1.07 1.30 1.61 2.02 2.54 2.96 3.62 4.08 4.48 5.2207/06/04 1.11 1.34 1.68 2.15 2.56 2.99 3.65 4.10 4.49 5.2407/07/04 1.16 1.30 1.64 2.00 2.56 2.99 3.67 4.10 4.50 5.2407/08/04 1.14 1.27 1.63 1.99 2.55 2.97 3.65 4.09 4.49 5.2407/09/04 1.14 1.28 1.63 2.00 2.55 2.96 3.64 4.08 4.49 5.23


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Suppose that a zero–coupon bond with a face value of F andmaturity j years has a price of Pj . Then, Pj(1 + sj)

j = F , where sjis the j year spot rate. Note that a payment of Pj now isexchanged by a payment of F in j years. (1 + sj)

j is the interestfactor from year zero to year j . If the current interest rates followthe accumulation function is a(t), t ≥ 0, then a(j) = (1 + sj)

j , i.e.sj = (a(j))1/j − 1.


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Example 2

The following table lists prices of zero-coupon $1000 bonds withtheir respective maturities:

Number of years to maturity Price

1 $980.39

2 $957.41

5 $888.18

10 $781.20

Calculate the 1–year, 2–year, 5–year, and 10–year spot rates ofinterest.

Solution: Since (1000)(1 + s1)−1 = 980.39,

(1000)(1 + s2)−2 = 957.41, (1000)(1 + s5)

−5 = 888.17, and(1000)(1 + s10)

−10 = 781.1984, we get s1 = 2.00%, s2 = 2.20% ,s5 = 2.40%, s10 = 2.50%.


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Example 2



1 $980.39

2 $957.41

5 $888.18

10 $781.20

Calculate the 1–year, 2–year, 5–year, and 10–year spot rates ofinterest.

Solution: Since (1000)(1 + s1)−1 = 980.39,

(1000)(1 + s2)−2 = 957.41, (1000)(1 + s5)

−5 = 888.17, and(1000)(1 + s10)

−10 = 781.1984, we get s1 = 2.00%, s2 = 2.20% ,s5 = 2.40%, s10 = 2.50%.


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The present value at time zero of a cashflow


Time 0 1 2 · · · n

following the spot rates

spot rate s1 s2 · · · snmaturity time 1 2 · · · n

is given by the formula

PV =n∑

j=1

(1 + sj)−jCj .


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Example 3(i) Find the price of a 2–year 1000 par value 6% bond withsemi–annual coupons using the spot rates:

nominal annual interestrate convertible semiannually

4% 5% 6% 7%

maturity time (in half years) 1 2 3 4


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4% 5% 6% 7%


Solution: (i) The cashflow of payments is

Payments 30 30 30 1030Time (in half years) 1 2 3 4

The present value of these payments is

PV = (30)

(1 +

0.04

2

)−1

+ (30)

(1 +

0.05

2

)−2

+ (30)

(1 +

0.06

2

)−3

+ (1030)

(1 +

0.07

2

)−4

=29.41176 + 28.55443 + 27.45425 + 897.5855 = 983.0059c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2.

15/25




4% 5% 6% 7%


(ii) Find the annual effective yield rate of the previous bond, ifbought at the price in (i).


16/25




4% 5% 6% 7%


(ii) Find the annual effective yield rate of the previous bond, ifbought at the price in (i).Solution: (ii) To find the yield rate, we solve for i (2) in 983.0059 =30a

4−−|i (2)/2+1000(1+ i (2)/2)−4, to get i (2) = 6.92450%. The annual

effective yield rate is i = 7.0443%.Note that i (2) = 6.92450% is a sort of average of the spot ratesused to find the price of the bond. Since the biggest payment is attime t = 4 half years, i (2) = 6.92450% is close to 7%.


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The one year forward rate for the j–th year fj is defined as

fj =(1 + sj)

j

(1 + sj−1)j−1− 1.

fj is also called the 1 year forward rate from time j − 1 to time j .fj is also called the 1 year forward rate from the j–th year.fj is also called the (j − 1)–year forward rate.fj is also called the (j − 1)–year deferred 1–year forward rate.fj is also called the (j − 1)–year forward rate, 1–year interest rate.1 + fj is the interest factor from year j − 1 to year j .


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(1 + sj−1)j−1 is the interest factor from year zero to year j − 1.

(1 + sj)j is the interest factor from year zero to year j .

Hence, we have that (1 + sj−1)j−1(1 + fj) = (1 + sj)

j .Notice that by the definition of the one year forward rate

(1 + s2)2 = (1 + s1)(1 + f2),

(1 + s3)3 = (1 + s2)

2(1 + f3) = (1 + s1)(1 + f2)(1 + f3),

(1 + s4)4 = (1 + s3)

3(1 + f4) = (1 + s1)(1 + f2)(1 + f3)(1 + f4).

In general,

(1 + sn)n = (1 + s1)(1 + f2)(1 + f3) · · · (1 + fn).


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Example 4

Suppose that the following spot rates are given:

maturity time(in years)

1 2 3 4 5

Interest rate 12.00% 11.75% 11.25% 10.00% 9.25%

Calculate the one–year forward rates for years 2 through 5.

Solution:f2 = (1.1175)2

1.12 − 1 = 0.115006

f3 = (1.1125)3

(1.1175)2− 1 = 0.102567

f4 = (1.1)4

(1.1125)3− 1 = 0.063336

f5 = (1.0925)5

(1.1)4− 1 = 0.063008

The one–year forward rate for year 2 is 11.5006%.The one–year forward rate for year 3 is 10.2567%.The one–year forward rate for year 4 is 6.3336%.The one–year forward rate for year 5 is 6.3008%.


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Example 4

Suppose that the following spot rates are given:

maturity time(in years)

1 2 3 4 5

Interest rate 12.00% 11.75% 11.25% 10.00% 9.25%

Calculate the one–year forward rates for years 2 through 5.

Solution:f2 = (1.1175)2

1.12 − 1 = 0.115006

f3 = (1.1125)3

(1.1175)2− 1 = 0.102567

f4 = (1.1)4

(1.1125)3− 1 = 0.063336

f5 = (1.0925)5

(1.1)4− 1 = 0.063008

The one–year forward rate for year 2 is 11.5006%.The one–year forward rate for year 3 is 10.2567%.The one–year forward rate for year 4 is 6.3336%.The one–year forward rate for year 5 is 6.3008%.


21/25


Example 5


Number of yearsto maturity

Price

1 $96.15

2 $92.10

3 $87.63

4 $82.27


22/25


Example 5



Price

1 $96.15

2 $92.10

3 $87.63

4 $82.27

(i) Calculate the 1–year, 2–year, 3–year, and 4–year spot rates ofinterest.


23/25


Example 5



Price

1 $96.15

2 $92.10

3 $87.63

4 $82.27

(i) Calculate the 1–year, 2–year, 3–year, and 4–year spot rates ofinterest.Solution: (i) Note that the price of j–th bond is Pj = (100)(1 +sj)

−j . To get the j year spot rate sj , we solve (100)(1 + s1)−1 =

96.15, (100)(1 + s2)−2 = 92.10, (100)(1 + s3)

−3 = 87.63, and(100)(1 + s4)

−4 = 82.27, to get s1 = 4.00%, s2 = 4.20% , s3 =4.50%, s4 = 5.00%.


24/25


Example 5



Price

1 $96.15

2 $92.10

3 $87.63

4 $82.27

(ii) Calculate the 1–year, 2–year, and 3–year forward rates of interest.


25/25


Example 5



Price

1 $96.15

2 $92.10

3 $87.63

4 $82.27

(ii) Calculate the 1–year, 2–year, and 3–year forward rates of interest.Solution: (ii) To get the j − 1 year forward rate fj , we do fj =

(1+sj )j

(1+sj−1)j−1 − 1 =Pj−1

Pj− 1, where Pj is the price of the j–th bond.

We get that:f2 = 96.15

92.10 − 1 = 4.397394%,f3 = 92.10

87.63 − 1 = 5.100993%,f4 = 87.63

82.27 − 1 = 6.515133%.c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2.

1/90



Section 6.4. Duration, convexity.





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Chapter 6. Variable interest rates and portfolio insurance. Section 6.4. Duration, convexity.

Duration

Next we will assume that the rate of interest is constant overmaturity.

Definition 1The duration (or Macaulay’s duration) of a cashflow

Contributions C1 C2 · · · Cn

Time in years 1 2 · · · n

with Cj ≥ 0 for each 1 ≤ j ≤ m, is defined as

d =

∑nj=1 jCjν

j∑nj=1 Cjν j

=

∑nj=1 jCj(1 + i)−j∑nj=1 Cj(1 + i)−j

=n∑

j=1

jCjν

j∑nk=1 Ckνk

.


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Main Properties of volatility

I The duration is an average of the times when the payments ofthe cashflow are made:

d =n∑

j=1

jCjν

j∑nk=1 Ckνk

=n∑

j=1

jwj ,

where wj =Cjν

jPnk=1 Ckνk satisfy wj ≥ 0 and

∑nj=1 wj = 1.

I wj is the fraction of the present value of contribution at timet over the present value of the whole cashflow.

I If Cj0 > 0 and Cj = 0, for each j 6= j0, then d = j0, for eachrate of interest i .

I The units of the duration are years.

I The Macaulay duration is a measure of the price sensitivity ofa cashflow to interest rate changes.


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Example 1

An investment pays 1000 at the end of year two and 1000 at theend of year 12. The annual effective rate of interest is 8%.Calculate the Macaulay duration for this investment.

Solution:

d =

∑nj=1 jCjν

j∑nj=1 Cjν j

=(2)(1000)(1.08)−2 + (12)(1000)(1.08)−12

(1000)(1.08)−2 + (1000)(1.08)−12

=5.165633881 years.


5/90


Example 1

An investment pays 1000 at the end of year two and 1000 at theend of year 12. The annual effective rate of interest is 8%.Calculate the Macaulay duration for this investment.

Solution:

d =

∑nj=1 jCjν

j∑nj=1 Cjν j

=(2)(1000)(1.08)−2 + (12)(1000)(1.08)−12

(1000)(1.08)−2 + (1000)(1.08)−12

=5.165633881 years.


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Theorem 1Let r > 0. If the Macaulay duration of the cashflow



is d , then the Macaulay duration of the cashflow

Contributions rC1 rC2 · · · rCn


is d .

Proof: The duration of the modified cashflow is∑nj=1 jrCjν

j∑nj=1 rCjν j

=

∑nj=1 jCjν

j∑nj=1 Cjν j

= d .


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Theorem 1Let r > 0. If the Macaulay duration of the cashflow




Contributions rC1 rC2 · · · rCn


is d .

Proof: The duration of the modified cashflow is∑nj=1 jrCjν

j∑nj=1 rCjν j

=

∑nj=1 jCjν

j∑nj=1 Cjν j

= d .


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Example 2

The Macaulay duration of a 10–year annuity–immediate withannual payments of $1000 is 5.6 years. Calculate the Macaulayduration of a 10–year annuity–immediate with annual payments of$50000.

Solution: Both cashflows have duration 5.6 years.


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Example 2

The Macaulay duration of a 10–year annuity–immediate withannual payments of $1000 is 5.6 years. Calculate the Macaulayduration of a 10–year annuity–immediate with annual payments of$50000.

Solution: Both cashflows have duration 5.6 years.


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Theorem 2If the Macaulay duration of the cashflow





Time in years t + 1 t + 2 · · · t + n

is d + t.

Proof:∑nj=1(t + j)Cjν

j∑nj=1 Cjν j

=t∑n

j=1 Cjνj +∑n

j=1 jCjνj∑n

j=1 Cjν j= t +

∑nj=1 jCjν

j∑nj=1 Cjν j

.


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Theorem 2If the Macaulay duration of the cashflow





Time in years t + 1 t + 2 · · · t + n

is d + t.

Proof:∑nj=1(t + j)Cjν

j∑nj=1 Cjν j

=t∑n

j=1 Cjνj +∑n

j=1 jCjνj∑n

j=1 Cjν j= t +

∑nj=1 jCjν

j∑nj=1 Cjν j

.


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Example 3

The Macaulay duration of a 10–year annuity–immediate withannual payments of $1000 is 5.6 years. Calculate the Macaulayduration of a 10–year annuity–due with annual payments of $5000.

Solution: The Macaulay duration of the two annuities does notdependent on the amount of the payment. So, we may assumethat the two annual payments agree. Since the cashflow of anannuity–due is obtained from the cashflow of an annuity–immediateby translating payments 1 year, the answer is 5.6− 1 = 4.6 years.


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Example 3

The Macaulay duration of a 10–year annuity–immediate withannual payments of $1000 is 5.6 years. Calculate the Macaulayduration of a 10–year annuity–due with annual payments of $5000.

Solution: The Macaulay duration of the two annuities does notdependent on the amount of the payment. So, we may assumethat the two annual payments agree. Since the cashflow of anannuity–due is obtained from the cashflow of an annuity–immediateby translating payments 1 year, the answer is 5.6− 1 = 4.6 years.


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Theorem 3Suppose that two cashflows have durations d1 and d2, respectively,present values P1 and P2, respectively. Then, the duration of thecombined cashflow is

d =P1d1 + P2d2

P1 + P2.

By induction the previous formula holds for a combination offinitely many cashflows. Suppose that we have n cashflows. Thej–the cashflow has present value Pj and duration dj . Then, theduration of the combined cashflow is

d =

∑nj=1 Pj(i)dj∑nj=1 Pj(i)

.


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Proof: Suppose that the considered cashflows are


Time 0 1 2 · · · n

and

Contributions 0 D1 D2 · · · Dn

Time 0 1 2 · · · n

Then, the combined cashflow is

Contributions 0 C1 + D1 C2 + D2 · · · Cn + Dn

Time 0 1 2 · · · n


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We have that P1 =∑n

j=1 Cjνj and P2 =

∑nj=1 Djν

j . By definitionof duration,

d1 =

∑nj=1 jCjν

j∑nj=1 Cjν j

=

∑nj=1 jCjν

j

P1

and

d2 =

∑nj=1 jDjν

j∑nj=1 Djν j

=

∑nj=1 jDjν

j

P2.

Hence,

d =

∑nj=1 j(Cj + Dj)ν

j∑nj=1(Cj + Dj)ν j

=

∑nj=1 jCjν

j +∑n

j=1 jDjνj∑n

j=1 Cjν j +∑n

j=1 Djν j=

d1P1 + d2P2

P1 + P2.


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Example 4

An insurance has the following portfolio of investments:(i) Bonds with a value of $1,520,000 and duration 4.5 years.(ii) Stock dividends payments with a value of $1,600,000 andduration 14.5 years.(iii) Certificate of deposits payments with a value of $2,350,000and duration 2 years.Calculate the duration of the portfolio of investments.

Solution: The duration of the portfolio is

d =


=(4.5)(1520000) + (14.5)(1600000) + (2)(2350000)

1520000 + 1600000 + 2350000

=6.351005484 years.


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Example 4

An insurance has the following portfolio of investments:(i) Bonds with a value of $1,520,000 and duration 4.5 years.(ii) Stock dividends payments with a value of $1,600,000 andduration 14.5 years.(iii) Certificate of deposits payments with a value of $2,350,000and duration 2 years.Calculate the duration of the portfolio of investments.

Solution: The duration of the portfolio is

d =


=(4.5)(1520000) + (14.5)(1600000) + (2)(2350000)

1520000 + 1600000 + 2350000

=6.351005484 years.


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Theorem 4The Macaulay duration of a level payments annuity–immediate is

d =(Ia)

n−−|i

an−−|i

.

Proof.We have that

d =

∑nj=1 jPν j∑nj=1 Pν j

=(Ia)

n−−|i

an−−|i

.


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Theorem 4The Macaulay duration of a level payments annuity–immediate is

d =(Ia)

n−−|i

an−−|i

.

Proof.We have that

d =

∑nj=1 jPν j∑nj=1 Pν j

=(Ia)

n−−|i

an−−|i

.


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Example 5

Calculate Macaulay the duration of a 15–year annuity immediatewith level payments if the current effective interest rate per annumis 5%.

Solution: The Macaulay the duration is

d =(Ia)

n−−|i

an−−|i

=(Ia)

15−−|5%

a15−−|5%

=73.66768937

10.37965804= 7.097313716.


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Example 5

Calculate Macaulay the duration of a 15–year annuity immediatewith level payments if the current effective interest rate per annumis 5%.

Solution: The Macaulay the duration is

d =(Ia)

n−−|i

an−−|i

=(Ia)

15−−|5%

a15−−|5%

=73.66768937

10.37965804= 7.097313716.


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Theorem 5The duration of a level payments perpetuity–immediate is

d =1 + i

i.

Proof.We have that

d =

∑∞j=1 jPν j∑∞j=1 Pν j

=(Ia)∞−−|ia∞−−|i

=1+ii2

1i

=1 + i

i.


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Example 6

Suppose that the Macaulay duration of a perpetuity immediatewith level payments of 1000 at the end of each year is 21. Find thecurrent effective rate of interest.

Solution: We have that 21 = d = 1+ii . So, i = 1

20 = 5%.


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Example 6

Suppose that the Macaulay duration of a perpetuity immediatewith level payments of 1000 at the end of each year is 21. Find thecurrent effective rate of interest.

Solution: We have that 21 = d = 1+ii . So, i = 1

20 = 5%.


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Theorem 6The duration of n year bond with r% annual coupons, face value Fand redemption value C is

d =Fr(Ia)

n−−|i + Cnνn

Fran−−|i + Cνn

.

Proof.Since the cashflow

Contributions Fr Fr · · · Fr Fr + C

Time 1 2 · · · n − 1 n

the duration is

d =Fr∑n

j=1 jν j + Cnνn

Fr∑n

j=1 ν j + Cνn=

Fr(Ia)n−−|i + Cnνn

Fran−−|i + Cνn

.


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Example 7

Megan buys a 10–year 1000–face–value bond with a redemptionvalue of 1200 which pay annual coupons at rate 7.5%. Calculatethe Macaulay duration if the effective rate of interest per annum is8%.


d =Fr(Ia)

n−−|i + Cnνn

Fran−−|i + Cνn

=(1000)(0.075)(Ia)

10−−|8%+ (1200)(10)(1.08)−10

(1000)(0.075)a10−−|8% + (1200)(1.08)−10

=(75)(32.68691288) + 5558.321857

(75)(6.710081399) + 555.8321857= 7.562958059.


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Example 7

Megan buys a 10–year 1000–face–value bond with a redemptionvalue of 1200 which pay annual coupons at rate 7.5%. Calculatethe Macaulay duration if the effective rate of interest per annum is8%.


d =Fr(Ia)

n−−|i + Cnνn

Fran−−|i + Cνn

=(1000)(0.075)(Ia)

10−−|8% + (1200)(10)(1.08)−10

(1000)(0.075)a10−−|8% + (1200)(1.08)−10

=(75)(32.68691288) + 5558.321857

(75)(6.710081399) + 555.8321857= 7.562958059.


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We have the following table:

Cashflow Duration d

zero–coupon bond d = n

level payments annuity–immediate d =(Ia)

n−−|i

an−−|i

level payments perpetuity–immediate d = 1+ii

regular bond d =Fr(Ia)

n−−|i

+Cnνn

Fran−−|i

+Cνn


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Volatility

Consider a cashflow



Definition 2The quantity ν = −d ln P(i)

di = −P′(i)P(i) is called the volatility or

modified duration.

Notice that P(i) =∑n

j=1 Cjνj =

∑nj=1 Cj(1 + i)−j and

P ′(i) =∑n

j=1 Cj(−j)(1 + i)−j−1. So,

ν =

∑nj=1 jCjν

j+1∑nj=1 Cjν j

.


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Since

d =

∑nj=1 jCjν

j∑nj=1 Cjν j

and

ν =

∑nj=1 jCjν

j+1∑nj=1 Cjν j

,

ν = νd .


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I ν = −P′(i)P(i) .

I The volatility measures the loss of present value of thecashflow as i increases relative to the PV of the cashflow.

I

ν =

∑nj=1 jCjν

j+1∑nj=1 Cjν j

.

I If Cj ≥ 0, for each 1 ≤ j ≤ n, ν > 0.

I ν = νd .

I Volatility is measured in years.


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The present value P(i) of the above cashflow as a function on i ,i.e.

P(i) =n∑

j=1

Cjνj =

n∑j=1

Cj(1 + i)−j .

It is easy to see that

P ′(i) =n∑

j=1

Cj(−j)(1+i)−j−1, and P ′′(i) =n∑

j=1

Cj j(j+1)(1+i)−j−2.

If Cj > 0, for each 1 ≤ j ≤ n, then P ′(i) < 0 and P ′′(i) > 0, foreach i ≥ 0. This implies that P(i) is a decreasing convex functionon i .


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Since P(i), i ≥ 0, is a decreasing function of i , so is lnP(i). Hence,

0 < −d lnP(i)

di= −P ′(i)

P(i)= −

∑nj=1 Cj(−j)(1 + i)−j−1∑n

j=1 Cj(1 + i)−j

=

∑nj=1 jCjν

j+1∑nj=1 Cjν j

.

Hence, ν > 0.


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Since ν = νd , we have that the volatility satisfies some of theproperties of the duration. Suppose that we have n cashflows. Thej–the cashflow has present value Pj and duration νj . Then, theduration of the combined cashflow is

ν =

∑nj=1 Pj(i)νj∑nj=1 Pj(i)

.


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Example 8

A perpetuity pays 100 immediately. Each subsequent payment inincreased by inflation. The current annual effective rate of interestis 6.5%. Calculate the modified duration of the perpetuityassuming that inflation will be 5% annually.

Solution: The present value of the perpetuity is P(i) = 100i−0.05 , if

i > 0.05. Hence, P ′(i) = −100(i−0.05)2

,

ν = −P′(0.065)P(0.065) = 1

0.065−0.05 = 66.66666667.


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Example 8

A perpetuity pays 100 immediately. Each subsequent payment inincreased by inflation. The current annual effective rate of interestis 6.5%. Calculate the modified duration of the perpetuityassuming that inflation will be 5% annually.

Solution: The present value of the perpetuity is P(i) = 100i−0.05 , if

i > 0.05. Hence, P ′(i) = −100(i−0.05)2

,

ν = −P′(0.065)P(0.065) = 1

0.065−0.05 = 66.66666667.


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Example 9

A portfolio consists of four bonds. The prices and modifieddurations of the four bonds are given by the table:

Bond Present value Modified duration in years

Bond A $15050 4.3Bond B $10350 10.4Bond C $67080 7.6Bond D $16750 6.5

Find the volatility of the whole portfolio.


ν =


=(15050)(4.3) + (10350)(10.4) + (67080)(7.6) + (16750)(6.5)

15050 + 10350 + 67080 + 16750

=7.241948183 years.


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Example 9

A portfolio consists of four bonds. The prices and modifieddurations of the four bonds are given by the table:

Bond Present value Modified duration in years

Bond A $15050 4.3Bond B $10350 10.4Bond C $67080 7.6Bond D $16750 6.5

Find the volatility of the whole portfolio.


ν =


=(15050)(4.3) + (10350)(10.4) + (67080)(7.6) + (16750)(6.5)

15050 + 10350 + 67080 + 16750

=7.241948183 years.


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Let P(i) be the present value of a portfolio, when i is the effectiverate of interest. By a Taylor expansion, for h close to zero,

P(i + h) ≈ P(i) + P ′(i)h = P(i)(1− νdh

)= P(i) (1− νh) .

Example 10

A portfolio of bonds is worth 535000 at the current rate of interestof 4.75%. Its Macaulay duration is 6.375. Estimate the value ofthe portfolio if interest rates decrease by 0.10%.

Solution: We have that P(i + h) ≈ P(i)(1− νdh

). In our case,

P(0.0475− 0.0010) ≈ 535000(1− (1.0475)−1(6.375)(−0.001))

=538255.9666.


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P(i + h) ≈ P(i) + P ′(i)h = P(i)(1− νdh

)= P(i) (1− νh) .

Example 10



). In our case,

P(0.0475− 0.0010) ≈ 535000(1− (1.0475)−1(6.375)(−0.001))

=538255.9666.


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P(i + h) ≈ P(i) + P ′(i)h = P(i)(1− νdh

)= P(i) (1− νh) .

Example 10



). In our case,

P(0.0475− 0.0010) ≈ 535000(1− (1.0475)−1(6.375)(−0.001))

=538255.9666.


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If interest rates change from i into i + h, the percentage of changein the present value of the portfolio is

P(i + h)− P(i)

P(i)≈ P(i) + P ′(i)h − P(i)

P(i)= −νdh = −νh.

Example 11

A bond has a volatility of 4.5 years, at the current annual interestrate of 5%. Calculate the percentage of loss of value of the bond ifthe annual effective interest rate increase 250 basis points.

Solution: The percentage of change is−νh = −(4.5)(0.025) = −0.1125 = −11.25%. The bond loses11.25% of its value.


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P(i + h)− P(i)

P(i)≈ P(i) + P ′(i)h − P(i)


Example 11




45/90



P(i + h)− P(i)

P(i)≈ P(i) + P ′(i)h − P(i)


Example 11




46/90


Duration is a measurement of how long in years it takes for thepayments to be made. Mainly, we will consider applications to thebond market. Duration is an important measure for investors toconsider, as bonds with higher durations are riskier and have ahigher price volatility than bonds with lower durations. We havethe following rules of thumb:

I Higher coupon rates lead to lower duration.

I Longer terms to maturity usually lead to longer duration.

I Higher yields lead to lower duration.


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The price of a bond decreases as the rate of interest increases.Suppose that you believe that interest rates will drop soon. Youwant to make a benefit by buying a bond today and selling it laterfor a higher price. The profit you make is P(i + h)− P(i), where iis the interest you buy the bond and i + h is the interest rate whenyou sell the bond. Notice that you make a benefit if h < 0. Therate of return in your investment is

P(i + h)− P(i)

P(i)≈ −νh.

So, between all possible bonds, you will make a biggest profitinvesting in the bond with the highest possible volatility.


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Example 12

Suppose that you are comparing two five–year bonds with a facevalue of 1000, and are expecting a drop in yields of 1% almostimmediately. The current yield is 8%. Bond 1 has 6% annualcoupons and bond 2 has annual 12% coupons. You would like toinvest 100,000 in the bond giving you the biggest return.(i) Which would provide you with the highest potential gain if youroutlook for rates actually occurs?(ii) Find the duration of each bond.


49/90


Solution: (i) The price of the bond 1 is

(60)a5−−|8%

+ 1000(1.08)−5 = 920.15.

Its price after the change of interest rates is

(60)a5−−|7%

+ 1000(1.07)−5 = 959.00.

The gain is 38.85. The percentage of change is959.00−920.15

920.15 = 4.22%

The price of the bond 2 is

(120)a5−−|8% + 1000(1.08)−5 = 1159.71.


(120)a5−−|7% + 1000(1.07)−5 = 1205.01.


1159.71 = 3.91%.Bond 1 is a better investment than bond 2 (if we believe that theinterest rates are going to fall to 1%).


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Solution: (i) The price of the bond 1 is

(60)a5−−|8%

+ 1000(1.08)−5 = 920.15.


(60)a5−−|7%

+ 1000(1.07)−5 = 959.00.


920.15 = 4.22%The price of the bond 2 is

(120)a5−−|8%

+ 1000(1.08)−5 = 1159.71.


(120)a5−−|7% + 1000(1.07)−5 = 1205.01.


1159.71 = 3.91%.Bond 1 is a better investment than bond 2 (if we believe that theinterest rates are going to fall to 1%).


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(ii) For the first bond, F = C = 1000, Fr = 60, i = 8% and n = 5.Its price is

Fran−−|i + Fνn = 60a

5−−|8%+ 1000(1.08)−5 = 920.15.

and its Macaulay duration is

d =Fr(Ia)

n−−|i + Fnνn

Fran−−|i + Fνn

=60(Ia)

5−−|0.08+ (1000)(5)(1.08)−5

920.15

=60(11.3651) + 3402.92

920.15= 4.4393.

For the second bond, F = 1000, Fr = 120, i = 8% and n = 5. Itsprice is


5−−|8%+ 1000(1.08)−5 = 1159.71.


d =120(Ia)

5−−|0.08+ (1000)(5)(1.08)−5

1159.71=

120(11.3651) + 3402.92

1159.71=4.1103.


52/90


(ii) For the first bond, F = C = 1000, Fr = 60, i = 8% and n = 5.Its price is


5−−|8%+ 1000(1.08)−5 = 920.15.


d =Fr(Ia)

n−−|i + Fnνn

Fran−−|i + Fνn

=60(Ia)

5−−|0.08+ (1000)(5)(1.08)−5

920.15

=60(11.3651) + 3402.92

920.15= 4.4393.

For the second bond, F = 1000, Fr = 120, i = 8% and n = 5. Itsprice is


5−−|8%+ 1000(1.08)−5 = 1159.71.


d =120(Ia)

5−−|0.08+ (1000)(5)(1.08)−5

1159.71=

120(11.3651) + 3402.92

1159.71=4.1103.


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Convexity

Definition 3The convexity of the cashflow



is defined as

c =P ′′(i)

P(i)=

∑nj=1 Cj j(j + 1)(1 + i)−j−2∑n

j=1 Cj(1 + i)−j=

∑nj=1 Cj j(j + 1)ν−j−2∑n

j=1 Cjν−j.

Convexity is measured in years2.


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I Convexity measures the rate of change of the volatility:

d

diν =

d

di

P ′(i)

P(i)=

P ′′(i)P(i)− P ′(i)P ′(i)

(P(i))2= c − (ν)2.

I The second order Taylor expansion of the present value withrespect to the yield is:

P(i + h) ≈ P(i) + P ′(i)h +h2

2P ′′(i) = P(i)

(1− νh +

h2

2c

).

I Convexity is a measure of the curvature of the price–yieldcurve for a bond. Convexity is related with the second term inthe Taylor expansion of the PV.

I Using duration and convexity, we measure of how sensitive thepresent value of a cashflow is to interest rate changes.


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I Using duration and convexity, we have the following Taylorexpansion:

P(i + h) ≈ P(i)

(1− νh +

h2

2c

).

I The percentage change in the PV of a cashflow is

P(i + h)− P(i)

P(i)≈ −νh +

h2

2c .

I Convexity can be used to compare bonds. If two bonds offerthe same duration and yield but one exhibits greaterconvexity, the bond with greater convexity is more affected byinterest rates.


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Example 13

A portfolio of bonds is worth 350000 at the current rate of interestof 5.2%. Its modified duration is 7.22. Its convexity is 370.Estimate the value of the portfolio if interest rates increase by0.2%.

Solution: We have that P(i + h) ≈ P(i)(1− νh + h2

2 c). In our

case,

P(0.052 + 0.002) = 350000

(1− (7.22)(0.002) + (370)

(0.002)2

2

)=345205.


57/90


Example 13

A portfolio of bonds is worth 350000 at the current rate of interestof 5.2%. Its modified duration is 7.22. Its convexity is 370.Estimate the value of the portfolio if interest rates increase by0.2%.

Solution: We have that P(i + h) ≈ P(i)(1− νh + h2

2 c). In our

case,

P(0.052 + 0.002) = 350000

(1− (7.22)(0.002) + (370)

(0.002)2

2

)=345205.


58/90


Example 14

Calculate the duration, the modified duration and the convexity ofa $5000 face value 15–year zero–coupon bond if the currenteffective annual rate of interest is 7.5%.

Solution: Since P(i) = (5000)(1 + i)−15,P ′(i) = (5000)(−15)(1 + i)−16,P ′′(i) = (5000)(−15)(−16)(1 + i)−17, we have that

ν = −P′(0.75)P(0.75) = (15)(1 + 0.075)−1 = 13.95348837 years,

d = (1 + i)ν = (1.075)(13.95348837) = 15 years andc = (−15)(−16)(1 + 0.75)−2 = 78.36734694 years2.


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Example 14

Calculate the duration, the modified duration and the convexity ofa $5000 face value 15–year zero–coupon bond if the currenteffective annual rate of interest is 7.5%.

Solution: Since P(i) = (5000)(1 + i)−15,P ′(i) = (5000)(−15)(1 + i)−16,P ′′(i) = (5000)(−15)(−16)(1 + i)−17, we have that

ν = −P′(0.75)P(0.75) = (15)(1 + 0.075)−1 = 13.95348837 years,

d = (1 + i)ν = (1.075)(13.95348837) = 15 years andc = (−15)(−16)(1 + 0.75)−2 = 78.36734694 years2.


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Example 15

Calculate the duration, the modified duration and the convexity ofa level payments perpetuity–immediate with payments at the endof the year if the current effective annual rate of interest is 5%.

Solution: Since P(i) = Ci , P ′(i) = −C

i2, P ′′(i) = 2C

i3, we have that

ν = −P′(0.05)P(0.05) = 1

0.05 = 20 years, d = (1 + i)ν = (1.05)(20) = 21

ears and c = 2i2

= 2(0.05)2

= 800 years2.


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Example 15

Calculate the duration, the modified duration and the convexity ofa level payments perpetuity–immediate with payments at the endof the year if the current effective annual rate of interest is 5%.

Solution: Since P(i) = Ci , P ′(i) = −C

i2, P ′′(i) = 2C

i3, we have that

ν = −P′(0.05)P(0.05) = 1

0.05 = 20 years, d = (1 + i)ν = (1.05)(20) = 21

ears and c = 2i2

= 2(0.05)2

= 800 years2.


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Example 16

A 100 par value 3 year bond pays annual coupons at a rate 7%coupon rate (with annual coupon payments). The current annualeffective interest rate is 7%.


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Example 16

A 100 par value 3 year bond pays annual coupons at a rate 7%coupon rate (with annual coupon payments). The current annualeffective interest rate is 7%.(i) Calculate the duration, the modified duration and the convexityof the bond.


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Example 16

A 100 par value 3 year bond pays annual coupons at a rate 7%coupon rate (with annual coupon payments). The current annualeffective interest rate is 7%.(i) Calculate the duration, the modified duration and the convexityof the bond.

Solution: (i) The cashflow isContributions 7 7 107

Time 1 2 3The duration is

d =(7)(1.07)−1 + 2(7)(1.07)−2 + 3(107)(1.07)−3

100= 2.808018.

The modified duration is ν = 2.8080181.07 = 2.6243. The convexity is

c =(7)(1)(2)(1.07)−3 + (7)(2)(3)(1.07)−4 + (107)(3)(4)(1.07)−5

100=9.58944.


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Example 16

A 100 par value 3 year bond pays annual coupons at a rate 7%coupon rate (with annual coupon payments). The current annualeffective interest rate is 7%.(ii) If the interest rate change from 7% to 8%, what is the percentagechange in the price of the bond?


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Example 16

A 100 par value 3 year bond pays annual coupons at a rate 7%coupon rate (with annual coupon payments). The current annualeffective interest rate is 7%.(ii) If the interest rate change from 7% to 8%, what is the percentagechange in the price of the bond?Solution: (ii) If i = 7%, the price of the bond is

7a3−−|7% + 100(1.07)3 = 100.

If i = 8%, the price of the bond is

7a3−−|8% + 100(1.08)3 = 97.4229.

The change in percentage is 97.4229100 − 1 = −2.5771%.


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Example 16

A 100 par value 3 year bond pays annual coupons at a rate 7%coupon rate (with annual coupon payments). The current annualeffective interest rate is 7%.(iii) Using the duration rule, including convexity, what is the per-centage change in the bond price?


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Example 16

A 100 par value 3 year bond pays annual coupons at a rate 7%coupon rate (with annual coupon payments). The current annualeffective interest rate is 7%.(iii) Using the duration rule, including convexity, what is the per-centage change in the bond price?Solution: (iii) The estimation in the change in percentage is

− νh +h2

2c = −(2.6243)(0.01) +

(0.01)2

2(9.58944)

=− 0.025760528 = −2.576353%.


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Theorem 7Suppose that we have n different investments. The j–thinvestment has present value Pj and convexity cj . Then, theconvexity of the combined investments is

c =

∑nj=1 Pj(i)cj∑nj=1 Pj(i)

.

Proof.Notice that

c =

∑nj=1 P ′′

j (i)

Pj(i)=

∑nj=1 Pj(i)cj

Pj(i).


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Theorem 7Suppose that we have n different investments. The j–thinvestment has present value Pj and convexity cj . Then, theconvexity of the combined investments is

c =

∑nj=1 Pj(i)cj∑nj=1 Pj(i)

.

Proof.Notice that

c =

∑nj=1 P ′′

j (i)

Pj(i)=

∑nj=1 Pj(i)cj

Pj(i).


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Example 17

A company has issued debt using the following bonds:

Bond Present value Macaulay’s duration convexity

Bond A 100000 5.3 1.2Bond B 50000 3.4 3.2Bond C 120000 12.2 6.2Bond D 80000 2.3 3.6

Find the Macaulay’s duration and the convexity for the entireportfolio.


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Solution: Let Pj , dj and cj be the present value, Macaulay’sduration and convexity, respectively, of the j–th bond, 1 ≤ j ≤ 4.Then, the Macaulay’s duration of the whole portfolio is

d =

∑nj=1 Pj dj∑nj=1 Pj

=100000(5.3) + 50000(3.4) + 120000(12.2) + 80000(2.3)

100000 + 50000 + 120000 + 80000

=6.708571429.

The convexity of the whole portfolio is

c =

∑nj=1 Pj cj∑nj=1 Pj

=100000(1.2) + 50000(3.2) + 120000(6.2) + 80000(3.6)

100000 + 50000 + 120000 + 80000

=3.748571429.


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In the case of payments made every 1m years, it is usual to use the

nominal rate of interest i (m) as the variable. The present value ofthe cashflow


Time (in years) 1m

2m · · · n

m

is

P(i (m)) =n∑

j=1

Cj

(1 +

i (m)

m

)−j

.


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The duration (or Macaulay’s duration) of the cashflow is

d =

∑nj=1

jmCj

(1 + i (m)

m

)−j

∑nj=1 Cj

(1 + i (m)

m

)−j=

1

m

∑nj=1 jCj

(1 + i (m)

m

)−j

∑nj=1 Cj

(1 + i (m)

m

)−jyears.

The volatility is

ν = −P ′(i (m))

P(i (m))= −

∑nj=1 Cj(−j)(1 + i (m)

m )−j−1 1m∑n

j=1 Cj(1 + i (m)

m )−j=

(1 +

i (m)

m

)−1

d .

The convexity is

c =P ′′(i (m))

P(i (m))=

∑nj=1 Cj(−j)(−j − 1)(1 + i (m)

m )−j−2 1m2∑n

j=1 Cj(1 + i (m)

m )−j

=1

m2

∑nj=1 Cj j(j + 1)(1 + i (m)

m )−j−2∑nj=1 Cj(1 + i (m)

m )−j.


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Example 18

You are given the following information about a bond:

I The term–to–maturity is 2 years.

I The bond has a 9% annual coupon rate, paid semiannually.

I The annual bond–equivalent yield–to–maturity is 8%.

I The par value is $100.


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Example 18






(i) Calculate the current price of the bond.


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Example 18






(i) Calculate the current price of the bond.Solution: (i) Since F = 100, Fr = 4.5, i = 4% and n = 4, theprice is is

Fran−−|i + Pνn = (4.5)a

4−−|4% + 100(1.04)−4 = 101.8149.


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Example 18






(ii) Calculate the Macaulay duration of the bond.


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Example 18






(ii) Calculate the Macaulay duration of the bond.Solution: (ii) The Macaulay duration is (in years)

d = 12

Fr(Ia)n−−|i

+nFνn

Fran−−|i

+Fνn = 12

(4.5)(8.896856)+(4)(100)(0.854804)101.8149 = 1.875744.


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Example 18






(iii) Calculate the convexity of the bond.


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Example 18






(iii) Calculate the convexity of the bond.Solution: (iii) The convexity is (in years2)

c =1

4×

(4.5)(1)(2)(1.04)−3+(4.5)(2)(3)(1.04)−4+(4.5)(3)(4)(1.04)−5+(104.5)(4)(5)(1.04)−6

101.8149

=1

4

8.0009 + 23.0797 + 44.3841 + 1651.7574

101.8149= 4.241083

where we have used 14 because the time is in half years.


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Example 18






(iv) For a 200 basis point increase in yield, determine the amount oferror in using duration and convexity to estimate the price change.


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(iv) For a 200 basis point increase in yield, determine the amount oferror in using duration and convexity to estimate the price change.Solution: (iv) We need to find

P(i+h)−P(i)−P ′(i)h−P ′′(i)h2

2= P(i+h)−P(i)

(1− νh + c

h2

2

).

The price of the bond after the change in interest rates is

P(i+h) = Fran−−|i+h+P(1+i+h)−n = (4.5)a4−−|6%+100(1.06)−4 = 94.8023.

So, the error is

P(i + h)− P(i)

(1− νh + c

h2

2

)=94.8023− 101.8149

(1− (1.875744)(1.04)−1(0.02) + 4.241083

(0.02)2

2

)=94.8023− 101.8149(0.9647762) = −3.426292.


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Example 18

Find the price and Macaulay duration of the followingfixed–income securities, given the annual effective rate of interest4.75% and par value of each bond is $1,000.


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Example 18


(i) 3–year bond with 5.00% annual coupons


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Example 18


(i) 3–year bond with 5.00% annual couponsSolution: (i) We have F = 1000, r = 0.05, i = 4.75%, Fr = 50and n = 3. The price of the bond is


3−−|4.75%+ 1000(1.0475)−3 = 1006.84.

The Macaulay duration is (in years)

d =Fr(Ia)

n−−|i + Fnνn

Fran−−|i + Fνn

=50(Ia)

3−−|0.0475+ (1000)(3)(1.0475)−3

1006.84

=50(5.3875) + 2610.11

1006.84= 2.8599.


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Example 18


(ii) 3–year bond with 5.00% semiannual coupons


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Example 18


(ii) 3–year bond with 5.00% semiannual coupons

Solution: (ii) We have i = 0.0475, i (2) = 0.046949, i (2)

2 =0.0234745, F = 1000, r = 0.025, Fr = 25 and n = 6. The price ofthe bond is

25a6−−|0.0234745

+ 1000(1.0234745)−6 = 1008.45.


d =1

2

Fr(Ia)n−−|i + Fnνn

Fran−−|i + Fνn=

1

2

25(Ia)6−−|0.0234745 + (1000)(6)(1.0234745)−6

1008.45

=(12.5)(19.0026) + 2610.11

1008.45= 2.823782.


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Example 18


(iii) 3–year bond with 5.00% quarter coupons.


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Example 18


(iii) 3–year bond with 5.00% quarter coupons.

Solution: (iii) We have i = 0.0475, i (4) = 0.046677, i (4)

4 =0.0116692, F = 1000, Fr = 12.5 and n = 6. The price of thebond is

12.5a12−−|0.0116692

+ 1000(1.0116692)−12 = 1009.25.


d =1

4

Fr(Ia)n−−|i + Fnνn

Fran−−|i + Fνn=

1

4

12.5(Ia)12−−|0.0116692 + (1000)(12)(1.0116692)−12

1009.25

=(3.125)(70.8531) + 2610.11

1009.25= 2.8056.


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Section 6.5. Asset–liability management.





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Chapter 6. Variable interest rates and portfolio insurance. Section 6.5. Asset–liability management.

Hedging

The money collected by insurance companies is invested andsubject to risk (possible losses). For example, bonds can drop invalue if the rate of interest changes. Several methods have beingdeveloped to minimize the risk of investing. A method, oftensophisticated, employed to minimize investment risk is calledhedging. In this section, we study several hedging methods.


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Match assets and liabilities

If possible we would like to match assets and liabilities. i.e. thetotal amount of contributions in assets equals the total amount ofcontributions in liabilities.

I If an insurance company has more liabilities than assets, itmay fail to meet its commitments to its policyholders.

I If an insurance company has more assets than liabilities, it willnot the make the appropriate profit for the capital available.


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Example 1

A company has liabilities of 2000, 5000 and 10000 payable at theend of years 1, 2 and 5 respectively. The investments available tothe company are the following zero–coupon 1000 par value bonds:

Bond Maturity (years) Effective Annual Yield

Bond A 1 year 4.5%Bond B 2 years 5.0Bond C 3 years 5.5Bond D 4 years 6.0Bond E 5 years 6.5

Determine the cost for matching these liabilities exactly.

Solution: We need to buy 2 Bonds A, 5 Bonds B and 10 Bonds E.The cost is

2000(1.045)−1 + 5000(1.05)−2 + 10000(1.065)−5 = 13747.83136.


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Example 1

A company has liabilities of 2000, 5000 and 10000 payable at theend of years 1, 2 and 5 respectively. The investments available tothe company are the following zero–coupon 1000 par value bonds:

Bond Maturity (years) Effective Annual Yield

Bond A 1 year 4.5%Bond B 2 years 5.0Bond C 3 years 5.5Bond D 4 years 6.0Bond E 5 years 6.5

Determine the cost for matching these liabilities exactly.

Solution: We need to buy 2 Bonds A, 5 Bonds B and 10 Bonds E.The cost is

2000(1.045)−1 + 5000(1.05)−2 + 10000(1.065)−5 = 13747.83136.


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Example 2

A bond portfolio manager in a pension fund is designing a bondportfolio. His company has an obligation to pay 50000 at the endof each year for 3 years. He can purchase a combination of thefollowing three bonds in order to exactly match its obligation:(i) 1–year 5% annual coupon bond with a yield rate of 6%.(ii) 2–year 7% annual coupon bond with a yield rate of 7%.(ii) 3–year 9% annual coupon bond with a yield rate of 8%.(i) How much of each bond should you purchase in order toexactly match the liabilities?(ii) Find the cost of such a combination of bonds.


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Solution: (i) Suppose that the face values of the bonds we buy arex , y , z , in each of the bonds, respectively. Then, the cashflows are

Liabilities 50000 50000 50000

Bond 1 (1.05)x 0 0

Bond 2 (0.07)y (1.07)y 0

Bond 3 (0.09)z (0.09)z (1.09)z

Time 1 2 3

In order to match assets and liabilities,

50000 = (1.05)x + (0.07)y + (0.09)z ,50000 = (1.07)y + (0.09)z ,50000 = (1.09)z .

Hence,

z = 500001.09 = 45871.56,

y = 50000−(0.09)(45871.56)1.07 = 42870.61645,

x = 50000−(0.07)(42870.61645)−(0.09)(45871.56)1.05 = 40829.15852.


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(ii) Find the cost of such a combination of bonds.Solution: (ii) The total price of the bonds is

(40829.15852)(1.05)(1.06)−1

+ (42870.61645)((0.07)a2−−|7%

+ (1.07)−2)

+ (45871.56)((0.09)a3−−|8%

+ (1.08)−3)

=40035.97426 + 42870.61645 + 47053.71459 = 129960.3053.


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Theory of immunization

Fluctuation in interest rates can cause losses to a financialinstitution. Suppose that a financial institution has a cashflow ofassets and a cashflow of liabilities. The present values of thesecashflows is sensitive to changes of interest rates. If interest ratesfall, the present value of the cashflow of liabilities increases. Ifinterest rates increase, the present value of the cashflow of assetsdecreases. To mitigate the risk associated with a change in theinterest rates, Redington (1954) introduced the theory ofimmunization. Immunization is a hedging method against the riskassociated with changes in interest rates.


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According with the traditional immunization theory, a portfolio isimmunized against fluctuations in interest rates if the 3 criteria aresatisfied:

1. The present value of the assets must equal the present valueof the liabilities.

2. The duration of the assets must equal the duration of theliabilities.

3. The convexity of the assets must be greater than that of theliabilities.

The first of the previous 3 conditions is an efficiency condition.The last conditions are imposed so that interest rate risk for theassets offsets the interest rate risk for the liabilities.


11/20


If the the immunization conditions are satisfied, then

PA(i) = PL(i), νA = νL, cA > cL.

The approximations to the present value of assets and liabilities are:

PA(i + h) ≈ PA(i)

(1− νAh +

h2

2cA

)and

PL(i + h) ≈ PL(i)

(1− νLh +

h2

2cL

).

Hence

PA(i + h)− PL(i + h) = PA(i)h2

2(c − cL) > 0.

For small variation in interest rates the previous approximations areaccurate. Since high variations in interest rates in short periods oftime are unlikely, it is possible to hedge against interest ratevariations by immunizing periodically.


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Example 3

An actuarial department needs to set–up an investment program topay for a loan of $20000 due in 2 years. The only availableinvestments are:(i) a money market fund paying the current rate of interest.(ii) 5–year zero–coupon bonds earning 4%.Assume that the current rate of interest is 4%. Develop aninvestment program satisfying the theory of immunization. Graphthe present value of asset minus liabilities versus interest rates.


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Solution: The investment program invest x in the money makerfund, and y in the zero coupon bond. The PV of the cashflow ofassets and liabilities isP(i) = x + y(1.04)5(1 + i)−5 − 20000(1 + i)−2. We solve for xand y such that P(0.04) = 0 and P ′(0.04) = 0. SinceP ′(i) = −(5)y(1.04)5(1 + i)−6 + (2)(20000)(1 + i)−3, we need tosolve

x+y = 20000(1.04)−2, and −(5)y(1.04)−1+(2)(20000)(1.04)−3 = 0.

We get y = (2)(20000)(1.04)−2

5 = 7396.45 andx = 20000(1.04)−2 − y = 11094.67. Since

P ′′(i) = (5)(6)y(1.04)5(1 + i)−7 − (2)(3)(20000)(1 + i)−4,

P ′′(0.04) = (5)(6)7396.45(1.04)−2−(2)(3)(20000)(1.04)−4 = 102576.5.

The convexity of the cashflow is positive. The investment strategyconsisting in allocate 11094.67 in the money market account and7396.45 in bonds satisfies the immunization requirements.


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The graph of the present value of asset minus liabilities versusinterest rates is Figure 1.

Figure 1: Present value of assets minus liabilities


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Example 4

An actuarial department has determined that the company has aliability of $10,000 that will be payable in seven years. Thecompany has two choices of assets to invest in: a 5–yearzero-coupon bond and a 10–year zero coupon bond. The interestrate is 5%. How can the actuarial department immunize itsportfolio?


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Solution: The investment program invest x in the 5–year bond,and y in the 10–year bond. The PV of the cashflow of assets andliabilities is

P(i) = x(1.05)5(1 + i)−5 + y(1.05)10(1 + i)−10 − 10000(1 + i)−7.

So,

P ′(i) = −(5)x(1.05)5(1 + i)−6 − (10)y(1.05)10(1 + i)−11

+ (7)10000(1 + i)−8

and

P ′′(i) = (5)(6)x(1.05)5(1 + i)−7 + (10)(11)y(1.05)10(1 + i)−12

− (7)(8)10000(1 + i)−9.

We solve for x and y in the equations P(0.05) = 0 andP ′(0.05) = 0, i.e.

x + y = (10000)(1.05)−7, and 5x + 10y = (70000)(1.05)−7.

We get x = (6000)(1.05)−7 = 4264.08 andy = (4000)(1.05)−7 = 2842.72.


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We have that

P ′′(0.05) = (5)(6)(6000)(1.05)−7(1.05)5(1.05)−7

+ (10)(11)(4000)(1.05)−7(1.05)10(1.05)−12 − (7)(8)(10000)(1.05)−9

=(60000)(1.05)−9 > 0.

So, the investment program satisfies the immunizationrequirements.


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Example 5

An actuarial department has determined that the company has aliability of $10,000 that will be payable in two years. The companyhas two choices of assets to invest in: a one–year zero–couponbond and a three–year zero–coupon bond. The interest rate is 6%.(i) Find an investment portfolio which immunizes this portfolio.(ii) Find the interval of interest rates at which the present value ofassets is bigger than the present value of liabilities.


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Solution: (i) The investment program invest x in the one–yearzero–coupon bond, and y in the three–year zero–coupon bond.The PV of the cashflow of assets and liabilities is

P(i) = x(1.06)(1 + i)−1 + y(1.06)3(1 + i)−3 − 10000(1 + i)−2.

So,

P ′(i) = x(1.06)(−1)(1 + i)−2 + y(1.06)3(−3)(1 + i)−4

− 10000(−2)(1 + i)−3

and

P ′′(i) = x(1.06)(−1)(−2)(1 + i)−3 + y(1.06)3(−3)(−4)(1 + i)−5

− 10000(−2)(−3)(1 + i)−4.

We solve for x and y in the equations P(0.06) = 0 andP ′(0.06) = 0, i.e.

x + y = (10000)(1.06)−2, and x + 3y = (20000)(1.06)−2.

We get x = y = (5000)(1.06)−2 = 4449.98222.c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2.

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We have that

P ′′(0.06) = (10000)(1.06)−2 > 0.

So, the investment program satisfies the immunizationrequirements.(ii) The present value of assets is bigger than the present value ofliabilities if

0 ≤(5000)(1.06)−1(1 + i)−1 + (5000)(1.06)(1 + i)−3 − 10000(1 + i)−2

which is equivalent to

(5000)(1.06)−1(1 + i)2 + (5000)(1.06)− 10000(1 + i)

=(5000)(1.06)−1((1 + i)2 − (2)(1.06) + (1.06)2

)=(5000)(1.06)−1(i − 0.06)2 ≥ 0.

The present value of assets is bigger than the present value ofliabilities for any interest rate.


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Chapter 7. Derivatives markets.

Manual for SOA Exam FM/CAS Exam 2.Chapter 7. Derivatives markets.

Section 7.1. Derivatives.





2/52

Chapter 7. Derivatives markets. Section 7.1. Derivatives.

Risk sharing

I A risk is a contingent financial loss. Changes in commodityprices, currency exchange rates and interest rates are potentialrisks for a business. A farmer faces the possible fall of theprice of his/her crop. Surging oil prices can wipe out airlines’profits. Manufacturing companies face high rising prices ofcommodities. These changes in prices could hurt the viabilityof a business.

I Many of the risks faced by business are diversifiable. A risk isdiversifiable if it is unrelated to another risk. Markets permitdiversifiable risks to be widely shared.

I Risk is nondiversifiable when it does vanish when spreadacross many investors.

I A way to do risk sharing for companies is to do contracts toavoid risks.


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Risk sharing






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Risk sharing






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Risk sharing






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Derivatives

Definition 1A derivative is a contract which specifies the right or obligation toreceive or deliver certain asset for a certain price. The value of aderivative contract depends on the value of another asset.

They are several possible reasons to enter into a derivative market:

I Risk management. Parties enter derivatives to avoid risks.I Speculation. Parties enter derivatives to make money. A

market–maker enters into derivatives to make money.I Reduce transaction costs. Derivatives can be used to reduce

commodity costs, borrowing costs, etc.I Arbitrage. When derivatives are miss priced, investors can

make a profit.I Regulatory arbitrage. Sometimes business enter into

derivatives to get around regulatory limitations, accountingregulations and taxes.


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Derivatives









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Derivatives



I Risk management. Parties enter derivatives to avoid risks.

I Speculation. Parties enter derivatives to make money. Amarket–maker enters into derivatives to make money.

I Reduce transaction costs. Derivatives can be used to reducecommodity costs, borrowing costs, etc.

I Arbitrage. When derivatives are miss priced, investors canmake a profit.

I Regulatory arbitrage. Sometimes business enter intoderivatives to get around regulatory limitations, accountingregulations and taxes.


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Derivatives




market–maker enters into derivatives to make money.

I Reduce transaction costs. Derivatives can be used to reducecommodity costs, borrowing costs, etc.




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Derivatives





commodity costs, borrowing costs, etc.




11/52


Derivatives






make a profit.



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Derivatives









13/52


Example 1

Suppose that a farmer grows wheat and a baker makes bread usingwheat and other ingredients. If the price of the wheat decreases,the farmer loses money. If the price of the wheat increases, thebaker loses money. In order to avoid possible financial losses whichmay jeopardy their businesss profitability, the farmer and the bakercan agree to sell/buy wheat one year from now at a certain price.The contract they enter is a derivative. It is a win–win contract forboth of them. The two risks are diversifiable.

Usually, the contract is not made directly between them. Amarket–maker or scalper makes a contract with the farmer andanother with the baker. The farmer and the baker enter into thiscontract to do hedging.


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Example 1

Suppose that a farmer grows wheat and a baker makes bread usingwheat and other ingredients. If the price of the wheat decreases,the farmer loses money. If the price of the wheat increases, thebaker loses money. In order to avoid possible financial losses whichmay jeopardy their businesss profitability, the farmer and the bakercan agree to sell/buy wheat one year from now at a certain price.The contract they enter is a derivative. It is a win–win contract forboth of them. The two risks are diversifiable.Usually, the contract is not made directly between them. Amarket–maker or scalper makes a contract with the farmer andanother with the baker. The farmer and the baker enter into thiscontract to do hedging.


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Derivatives in Practice.

I Derivatives are often traded on commodities, stock, stockindexes, currency exchange rates and interest rates. Commoncommodities in derivatives are agricultural (corn, soybeans,wheat, live cattle, cattle feeder, hogs lean, sugar, coffee,orange juice), metals (gold, silver, copper, lead, aluminum,platinum) and energy (crude oil, ethanol, natural gas,gasoline).

I Derivative contracts for agricultural commodities have beentraded in the U.S. for more than 100 years. The biggestmarkets in derivatives are the Chicago Board for Trade, theChicago Mercantile Exchange, the New York MercantileExchange, and the Eurex (Frankfurt, Germany).

I The market in derivatives is regulated by the (SEC) Securitiesand Exchange Commission and the (CFTC) CommodityFutures Trading Commission.


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17/52







18/52


Buying an asset

I (Scalpers) Market–makers buy/sell stock and derivativesmaking transactions possible. Usually, scalpers are financialinstitutions.

I In order to make a living, scalpers buy low and sell high.

I The price at which the scalper buys is called the bid price.The bid price of an asset is the price at which a scalper takesbids for an asset (a bid is an offer).

I The price at which the scalper sells is called the offer price orask price.

I The bid price is lower than the offer price. The differencebetween the bid price and the offer price is called the bid–askspread.

I bid–ask percentage spread is the bid–ask spread dividedover the ask price. Scalpers also ask for commissions.


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Buying an asset








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Buying an asset








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Buying an asset








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Buying an asset








23/52


Buying an asset








24/52


bid price at which the scalper buys

ask or offer price price at which the scalper sells

The bid price is lower than the offer price.


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Example 2

An online currency exchange service’s bid rate for Japanese yens is$0.00838 and its ask rate is $0.00847. Find the bid–ask percentagespread.

Solution: The ask percentage spread is0.00847−0.00838

0.00847 = 1.062574%.


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Example 2

An online currency exchange service’s bid rate for Japanese yens is$0.00838 and its ask rate is $0.00847. Find the bid–ask percentagespread.

Solution: The ask percentage spread is0.00847−0.00838

0.00847 = 1.062574%.


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Example 3

$1 million face value six month T bill is traded by a governmentsecurity dealer who give the following annual nominal discountyield convertible semiannually:

Bid Ask

4.96% 4.94%

(i) Find the price the dealer is asking for the T bill(ii) Find the price the dealer is buying the T bill.(iii) Find the bid–ask percentage spread.

Solution: (i) (1000000)(1− (0.0494/2)) = 975300(ii) (1000000)(1− (0.0496/2)) = 975200.Notice that the dealer sells the T bill for money than he buys it.(iii) The bid–ask percentage spread is975300−975200

975300 = 0.01025325541%.


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Example 3


Bid Ask

4.96% 4.94%


Solution: (i) (1000000)(1− (0.0494/2)) = 975300

(ii) (1000000)(1− (0.0496/2)) = 975200.Notice that the dealer sells the T bill for money than he buys it.(iii) The bid–ask percentage spread is975300−975200

975300 = 0.01025325541%.


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Example 3


Bid Ask

4.96% 4.94%


Solution: (i) (1000000)(1− (0.0494/2)) = 975300(ii) (1000000)(1− (0.0496/2)) = 975200.

Notice that the dealer sells the T bill for money than he buys it.(iii) The bid–ask percentage spread is975300−975200

975300 = 0.01025325541%.


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Example 3


Bid Ask

4.96% 4.94%


Solution: (i) (1000000)(1− (0.0494/2)) = 975300(ii) (1000000)(1− (0.0496/2)) = 975200.Notice that the dealer sells the T bill for money than he buys it.

(iii) The bid–ask percentage spread is975300−975200

975300 = 0.01025325541%.


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Example 3


Bid Ask

4.96% 4.94%


Solution: (i) (1000000)(1− (0.0494/2)) = 975300(ii) (1000000)(1− (0.0496/2)) = 975200.Notice that the dealer sells the T bill for money than he buys it.(iii) The bid–ask percentage spread is975300−975200

975300 = 0.01025325541%.


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Long and short positions

I When some one owns an asset, we say to this person has along position in this asset. Later, he may sell the asset andreceive cash. Buying an asset means to make an investment.Buying an asset is like lending money.

I When some one needs to buy an asset in the future, it is saidthat this person has a short position in the asset.


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Long and short positions

I When some one owns an asset, we say to this person has along position in this asset. Later, he may sell the asset andreceive cash. Buying an asset means to make an investment.Buying an asset is like lending money.

I When some one needs to buy an asset in the future, it is saidthat this person has a short position in the asset.


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Short sale

A short sale of an asset entails borrowing an asset and thenimmediately selling the asset receiving cash. Later, the short sellermust buy back the asset paying cash for it and return it to thelender. The act of buying the replacement asset and return it tothe lender is called to close or cover the short position.

believe derivative desired outcome

price will increase purchase buy low now and sell high later

price will decrease short sale sell high now and buy low later

Table: An investor makes money buying low and selling high.


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Short sale

A short sale of an asset entails borrowing an asset and thenimmediately selling the asset receiving cash. Later, the short sellermust buy back the asset paying cash for it and return it to thelender. The act of buying the replacement asset and return it tothe lender is called to close or cover the short position.

believe derivative desired outcome

price will increase purchase buy low now and sell high later

price will decrease short sale sell high now and buy low later

Table: An investor makes money buying low and selling high.


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There are three main reasons to short sell:

I Speculation. An investor enters a short sale because hebelieves that the price of the stock will drop and desires toprofit from this price movement.

I Financing. A short sale is a way to borrow money.

I Hedging. A short sale can be undertaken to offset the risk ofowning an asset.


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In order to a short sale to take place several conditions must betaken into account:

I Availability of a lender of stock. A lender must interested inmaking some money and losing temporarily his voting rightson the company issuing the stock.

I Credit risk of the short seller. The short seller make berequired to set–up a bank account with a deposit as collateral.

I Scarcity of shares.

If the stock pays dividends, the short seller must return the paiddividend payments to the stock lender.


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Example 4

Mary short sells 200 shares of XYZ stock which has a bid price of$18.12 and ask price of $18.56. Her broker charges her a 0.5%commission to take on the short sale. Mary covers her position 6months later when the bid price is $15.74 and the ask price is$15.93. The commission to close the short sale is $15. XYZ stockdid not pay any dividends in those six months. How much doesMary earn in this short sale?

Solution: Mary short sells the stock for

(200)(18.12)(1− 0.005) = $3605.88.

Mary covers her position for

(200)(15.93) + 15 = $3201.

Mary earns 3605.88− 3201 = $404.88.


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Example 4

Mary short sells 200 shares of XYZ stock which has a bid price of$18.12 and ask price of $18.56. Her broker charges her a 0.5%commission to take on the short sale. Mary covers her position 6months later when the bid price is $15.74 and the ask price is$15.93. The commission to close the short sale is $15. XYZ stockdid not pay any dividends in those six months. How much doesMary earn in this short sale?

Solution: Mary short sells the stock for

(200)(18.12)(1− 0.005) = $3605.88.

Mary covers her position for

(200)(15.93) + 15 = $3201.

Mary earns 3605.88− 3201 = $404.88.


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I Often, short sellers are required to make a deposit ascollateral into an account with the lender. This account iscalled the margin account. This deposit is called the marginrequirement. Usually, the margin requirement is a percentageof the current price of the stock. The margin requirementcould be bigger than the current asset price. If this is so, theexcess of the margin over the current asset price is calledhaircut.

I This margin account generates an interest for the investor.Demand on short sales is a factor to determine the marginaccount interest rate. The margin interest rate is calledthe repo rate for bonds and the short rebate for stock.

I When borrowing, the lender can require payment of certainbenefits lost by lending the asset. This payment requirementis called the lease rate of the asset. Usually, the lease rate ofa stock is the payment of the dividends obtained while thestock was shorted. Usually, the lease rate of a bond is thepayment of the coupons obtained while the bond was shorted.


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Sometimes short sales are subject to a margin requirement (ordeposit). The investor has to set–up an account with a percentageof the current price of the stock. This margin account generatesan interest for the investor. If the stock which the investor borrowspays a dividend, the investor must pay the dividend to thebrokerage firm making the loan.We have the following variables in a short sale of a stock:

I P = profit on sale=price sold−price bought.

I M = margin requirement= deposit on the short sale.

I D = dividend paid by the short seller to the security’s owner.

I j = rate of interest earned in the margin account.

I I = Mj interest earned by the short seller on the margindeposit.

I i = yield rate on the short sale.


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We have that the net profit is

Net profit = gain on short sale+interest on margin−dividend on stock = P+I−D.

Hence, the yield rate earner in a short sale is

i =net profit

margin=

P + I − D

M.


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Example 5

Jason sold short 1,000 shares of FinanTech at $75 a share onJanuary 2, 2006. Jason is required to hold a margin account withhis broker equal to 50% of the short security’s initial value. Jason’smargin account earns an annual effective interest rate of i . Thereis a $0.25 per share dividend paid on December 31, 2006. OnJanuary 2, 2007, Jason buys back stock to cover his position at aprice of $70 per share. Jason’s annual effective yield in thisinvestment is 17.6667%. Find i .

Solution: We have that P = (75)(1000)− (70)(10000) = 5000,M = (75)(1000)(0.50) = 37500, I = 37500i andD = (1000)(0.25) = 250. Jason’s annual effective yield is

0.176667 =P + I − D

M=

5000 + 37500i − 250

37500.

Hence, i = (0.176667)(37500)−5000+25037500 = 5%.


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Example 5

Jason sold short 1,000 shares of FinanTech at $75 a share onJanuary 2, 2006. Jason is required to hold a margin account withhis broker equal to 50% of the short security’s initial value. Jason’smargin account earns an annual effective interest rate of i . Thereis a $0.25 per share dividend paid on December 31, 2006. OnJanuary 2, 2007, Jason buys back stock to cover his position at aprice of $70 per share. Jason’s annual effective yield in thisinvestment is 17.6667%. Find i .

Solution: We have that P = (75)(1000)− (70)(10000) = 5000,M = (75)(1000)(0.50) = 37500, I = 37500i andD = (1000)(0.25) = 250. Jason’s annual effective yield is

0.176667 =P + I − D

M=

5000 + 37500i − 250

37500.

Hence, i = (0.176667)(37500)−5000+25037500 = 5%.


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Section 7.2. Forwards.





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Chapter 7. Derivatives markets. Section 7.2. Forwards.

Forwards

Definition 1A forward is a contract between a buyer and seller in which theyagree upon the sale of an asset of a specified quality for a specifiedprice at a specified future date.

Forward contracts are privately negotiated and are notstandardized.Common forwards are in commodities, currency exchange, stockshares and stock indices.


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A forward contract states the following:

I forces the seller to sell and the buyer to buy.

I spells out the quantity, quality and exact type of asset to besold.

I states the delivery price and the time, date and place for thetransfer of ownership of the asset.

I specify the time, date, place for payment.

Usually a forward contract has more terms.Sometimes instead of the asset to be delivered, there exists a cashsettlement between the parties engaging in a forward contract.Either physical settlement or cash settlement can be used tosettle a forward contract.When entering in a forward contract, parties must checkcounterparts for credit risk (sometimes using a collateral, bankletters, real state guarantee, etc).


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Usually a forward contract has more terms.

Sometimes instead of the asset to be delivered, there exists a cashsettlement between the parties engaging in a forward contract.Either physical settlement or cash settlement can be used tosettle a forward contract.When entering in a forward contract, parties must checkcounterparts for credit risk (sometimes using a collateral, bankletters, real state guarantee, etc).


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Usually a forward contract has more terms.Sometimes instead of the asset to be delivered, there exists a cashsettlement between the parties engaging in a forward contract.Either physical settlement or cash settlement can be used tosettle a forward contract.

When entering in a forward contract, parties must checkcounterparts for credit risk (sometimes using a collateral, bankletters, real state guarantee, etc).


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I The asset in which the forward contract is based is called theunderlier or underlying asset.

I The nominal amount (also called notional amount) of aforward contract is the quantity of the asset traded in theforward contract.

I The price of the asset in the forward contract is called theforward price.

I The time at which the contract settles is called the expirationdate.

For example, if a forward contract involves 10,000 barrels of oil tobe delivered in one year, oil is the underlying asset, 10000 barrels isthe notional amount and one year is the expiration date.


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The two main reasons why an investor might be interested inforward contracts are: speculation and (hedging) reduceinvestment risk.


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Apart from commission, a forward contract requires no initialpayment. The current price of an asset is called its spot price.Besides the spot price, to price a forward contract, several factors,such as delivery cost and time of delivery must be taken intoaccount.The difference between the spot and the forward price is called theforward premium or forward discount.If the forward price is higher than the spot price, the asset isforwarded at a premium. The premium is the forward priceminus the current spot price.If the forward price is lower than the spot price, the asset isforwarded at a discount. The discount is the current spot priceminus the forward price.


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The buyer of the asset in a forward contract is called the longforward. The long forward benefits when prices rise.The seller of the asset in a forward contract is called the shortforward. The short forward benefits when prices decline.


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We will denote by ST the spot price of an asset at time T . S0 isthe current price of the asset. S0 is a fixed quantity. For T > 0,ST is a random variable. We will denote by F0,T to the price attime zero of a forward with expiration time T paid at time T .The payoff of a derivative is the value of this position at expiration.The payoff of a long forward contract is ST − F0,T . Notice thatthe bearer of a long forward contract buys an asset at time T withvalue ST for F0,T . Assuming that there are no expenses setting theforward contract, the profit for a long forward is ST − F0,T .The payoff of a short forward is F0,T − ST . The holder of a shortforward contract sells at time T an asset with value ST for F0,T .Assuming that there are no expenses setting the forward contract,the profit for a short forward contract is F0,T − ST .


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The profits of the long and the short in a forward contract are theopposite of each other. The sum of their profits is zero. A forwardcontract is a zero–sum game.

I The minimum long forward’s profit F0,T , which is attainedwhen ST = 0.

I The maximum long forward’s profit is infinity, which isattained when ST =∞.

I The minimum short forward’s profit is −∞, which is attainedwhen ST =∞.

I The maximum short forward’s profit is F0,T , which is attainedwhen ST = 0.

long forward’s profit = ST − F0,T

short forward’s profit = F0,T − ST

minimum profit maximum profit

long forward −F0,T ∞short forward −∞ −F0,T


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Example 1

A gold miner enters a forward contract with a jeweler to sell him200 ounces of gold in six months for $600 per ounce.(i) Find the jeweler’s payoff in the forward contract if the spotprice at expiration of a gold ounce is $590, $595, $600, $605,$610. Graph the jeweler’s payoff.(ii) Find the gold miner’s payoff in the forward contract if the spotprice at expiration of a gold ounce is $590, $595, $600, $605,$610. Graph the gold miner’s payoff.


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Solution: (i) The jeweler’s payoff is (200)(ST − 600). A table ofthe jeweler’s payoff is

ST 590 595 600 605 610

Payoff −2000 −1000 0 1000 2000

The graph of the jeweler’s payoff is given in Figure 1.(ii) The gold miner’s payoff is (200)(600− ST ). A table of thegold miner’s payoff is

ST 590 595 600 605 610

Payoff 2000 1000 0 −1000 −2000


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Figure 1: Example 1. (Long forward) Jeweler’s payoff


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Figure 2: Example 1. (Short forward) Gold miner’s payoff

Notice how figures 1 and 2 are the opposite of each other. Thegold miner’s payoff equals minus the jeweler’s payoff.


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Usually, a commissions has to be paid to enter a forward contract.Suppose that the long forward has to paid CL to the market–makerat negotiation time to enter into the forward contract. Then, theprofit for a long forward is

Profit of a long forward = ST − F0,T − CL(1 + i)T ,

where i is the annual effective rate of interest.If the short forward has to paid CS at negotiation time to themarket–maker to enter the forward contract, then profit for a shortforward is

Profit of a short forward = F0,T − ST − CS(1 + i)T .


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Sometimes, instead of using the annual effective rate of interest,we will use the annual interest rate compounded continuously.This is another name for the force of interest. This rate is alsocalled the annual continuous interest rate. If r is the annualcontinuously compounded interest rate, then the future value attime T of a payment of P made at time zero is PerT .


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Alternative ways to buy an asset.

Suppose that you want to buy an asset. Suppose that the buyer’spayment can be made at either time zero or time T . Suppose thatthe transfer of ownership of an asset can be made either at timezero or at time T . There are four possible ways to buy an asset(see Table 1):

Pay at timeReceivethe assetat time

0 T

0 Outright purchase Fully leveraged purchaseT Prepaid forward contract Forward contract

Table 1: Alternative ways to buy an asset.


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1. Outright purchase. Both the payment and the transfer ofownership are made at time zero. The price paid per share isthe current spot price S0.

2. Fully leveraged purchase. The transfer of ownership is madeat time zero. The payment is made at time T . The paymentis S0e

rT , where S0 is the current spot price and r is therisk–free continuously compounded annual interest rate.

3. Prepaid forward contract. A payment of FP0,T is made at

time zero. The transfer of ownership is made at time T . Thepayment FP

0,T is not necessarily the current spot price S0.

4. Forward contract. Both the payment and the transfer ofownership happen at time T . The price of a forward contractis denoted by F0,T . We have that F0,T = erTFP

0,T , where r isthe risk–free annual interest rate continuously compounded.


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Pricing a forward contract.

Whenever an asset is delivered and paid at time zero, the fair priceof the asset is its (spot) market price. The market of an outrightpurchase is S0. Commissions and bid–ask spreads must be takeninto account.The case of a fully leverage purchase, the price of a forwardcontract is just the price of a loan of S0. The price of a fullyleveraged purchase is S0e

rT , which is the price of a loan of S0

taken at time zero and paid at time T .


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We are interested in determining F0,T , the price of a forwardcontract. Many different factors such as the cost of storing,delivering, the convenience yield and the scarcity of the asset.Some commodities like oil have high storage costs. Theconvenience yield measures the cost of not having the asset, but aforward contract on it. For example, if instead of having a forwardon gasoline, we have the physical asset, we may use it in case ofscarcity. In the case of stock paying dividends, an stock ownerreceives dividend payments, and a long forward holder does not.


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Pricing a prepaid forward contract.

To find FP0,T , we make three cases.

1. Price of prepaid forward contract if there are nodividends. We consider an asset with no cost/benefit in holdingthe asset. This applies to the the price of a stock which does payany dividends. It is irrelevant whether the transfer of ownershiphappens now or laterThe no arbitrage price of a prepaid forward contract is FP

0,T = S0,where S0 is the price of an asset today.


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Example 2

XYZ stock costs $55 per share. XYZ stock does not pay anydividends. The risk–free interest rate continuously compounded8%. Calculate the price of a prepaid forward contract that expires30 months from today.

Solution: The prepaid forward price is FP0,T = S0 = 55.


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Example 2

XYZ stock costs $55 per share. XYZ stock does not pay anydividends. The risk–free interest rate continuously compounded8%. Calculate the price of a prepaid forward contract that expires30 months from today.

Solution: The prepaid forward price is FP0,T = S0 = 55.


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2. Price of prepaid forward contract when there are discretedividends. Suppose that the stock is expected to make a dividendpayment of DTi

at the time ti , i = 1 . . . , n. A prepaid forwardcontract will entitle you receive the stock at time T withoutreceiving the interim dividends. The prepaid forward price is

FP0,T = S0 −

n∑i=1

Dti e−rti .


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Example 3

XYZ stock cost $55 per share. It pays $2 in dividends every 3months. The first dividend is paid in 3 months. The risk–freeinterest rate continuously compounded 8%. Calculate the price ofa prepaid forward contract that expires 18 months from today,immediately after the dividend is paid.

Solution: The prepaid forward price is

FP0,T = S0 −

n∑i=1

Dti e−rti = 55−

6∑j=1

2e−(0.08)j(1/4)

=55−6∑

j=1

2e−(0.02)j = 55− 2a6−−|e0.02−1

= 43.80474631.

Recall that an−−|i =

∑nj=1(1 + i)−j .


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Example 3

XYZ stock cost $55 per share. It pays $2 in dividends every 3months. The first dividend is paid in 3 months. The risk–freeinterest rate continuously compounded 8%. Calculate the price ofa prepaid forward contract that expires 18 months from today,immediately after the dividend is paid.

Solution: The prepaid forward price is

FP0,T = S0 −

n∑i=1

Dti e−rti = 55−

6∑j=1

2e−(0.08)j(1/4)

=55−6∑

j=1

2e−(0.02)j = 55− 2a6−−|e0.02−1

= 43.80474631.

Recall that an−−|i =

∑nj=1(1 + i)−j .


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3. Price of prepaid forward contract when there arecontinuous dividends. In the case of an index stock, dividendsare given almost daily. We may model the dividend payments as acontinuous flow. Let δ be the rate of dividends given per unit oftime. Suppose that dividends payments are reinvested into stock.Let tj = jT

n , 1 ≤ j ≤ n. If Aj is the amount of shares at time tj ,

then Aj+1 = Aj(1 + δTn ). Hence, the total amount of shares

multiplies by 1 + δTn in each period. Hence, one share at time zero

grows to(1 + δT

n

)nat time T . Letting n→∞, we get that one

share at time zero grows to eδT shares at time T . With $K attime 0, we can buy K

S0shares in the market at time 0. These

shares grow to KS0

eδT at time T . With $K at time 0, we can buyK

FP0,T

shares to be delivered at time T using a prepaid forward.

Hence, is there exists no arbitrage, KS0

eTδ = KFP

0,T

and

FP0,T = S0e

−δT .


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Example 4

An investor is interested in buying XYZ stock. The current price ofstock is $45 per share. This stock pays dividends at an annualcontinuous rate of 5%. Calculate the price of a prepaid forwardcontract which expires in 18 months.

Solution: The price of the prepaid forward contract is

FP0,T = S0e

−δT = 45e−(0.05)(18/12) = 41.74845688.


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Example 4

An investor is interested in buying XYZ stock. The current price ofstock is $45 per share. This stock pays dividends at an annualcontinuous rate of 5%. Calculate the price of a prepaid forwardcontract which expires in 18 months.

Solution: The price of the prepaid forward contract is

FP0,T = S0e

−δT = 45e−(0.05)(18/12) = 41.74845688.


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Example 5

XYZ stock costs $55 per share. The annual continuous interestrate is 0.055. This stock pays dividends at an annual continuousrate of 3.5%. A one year prepaid forward has a price of $52.60. Isthere any arbitrage opportunity? If so, describe the position anarbitrageur would take and his profit per share.

Solution: The no arbitrage prepaid forward price is

FP0,T = S0e

−δT = 55e−0.035 = 53.10829789.

An arbitrage portfolio consists of entering a prepaid long forwardcontract for one share of stock and shorting e−0.035 shares ofstock. The return of this transaction is55e−0.035 − 52.60 = 0.5082978942. At redemption time, thearbitrageur covers his short position after executing the prepaidforward contract.


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Example 5



FP0,T = S0e

−δT = 55e−0.035 = 53.10829789.

An arbitrage portfolio consists of entering a prepaid long forwardcontract for one share of stock and shorting e−0.035 shares ofstock. The return of this transaction is55e−0.035 − 52.60 = 0.5082978942. At redemption time, thearbitrageur covers his short position after executing the prepaidforward contract.


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Example 6



FP0,T = S0e

−δT = 55e−0.055 = 52.05668314.

An arbitrage portfolio consists of entering a prepaid short forwardcontract for one share of stock and buying e−0.055 shares of stock.The return of this transaction is52.60− 55e−0.055 = 0.5433168626. At redemption time, we usethe bought stock to meet the short forward.

Notice that in the previous questions, we can make arbitragewithout making any investment of capital. The total price ofsetting the portfolios at time zero is zero.


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Example 6



FP0,T = S0e

−δT = 55e−0.055 = 52.05668314.

An arbitrage portfolio consists of entering a prepaid short forwardcontract for one share of stock and buying e−0.055 shares of stock.The return of this transaction is52.60− 55e−0.055 = 0.5433168626. At redemption time, we usethe bought stock to meet the short forward.

Notice that in the previous questions, we can make arbitragewithout making any investment of capital. The total price ofsetting the portfolios at time zero is zero.


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Pricing a forward contract.

Both the payment and the transfer of ownership happen at timeT . The price of a forward contract is the future value of theprepaid forward contract, i.e. F0,T = erTFP

0,T . So,

I The price of a forward contract for a stock with no dividendsis F0,T = erTS0.

I The price of a forward contract for a stock with discretedividends is F0,T = erTS0 −

∑ni=1 Dti e

r(T−ti ).

I The price of a forward contract for a stock with continuousdividends is F0,T = e(r−δ)TS0.


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Example 7

The current price of one share of XYZ stock is 55.34. The price ofa nine–month forward contract on one share of XYZ stock is 57.6.XYZ stock is not going to pay any dividends on the next 2 years.(i) Calculate the annual compounded continuously interest rateimplied by this forward contract.(ii) Calculate the price of a two–year forward contract on one shareof XYZ stock.

Solution: (i) Since F0,T = erTS0, 57.6 = e(3/4)r55.34 andr = (4/3) ln(57.6/55.34) = 0.05336879112.(ii) We have thatF0,2 = er2S0 = e(0.05336879112)(2)55.34 = 61.57362151.


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Example 7


Solution: (i) Since F0,T = erTS0, 57.6 = e(3/4)r55.34 andr = (4/3) ln(57.6/55.34) = 0.05336879112.

(ii) We have thatF0,2 = er2S0 = e(0.05336879112)(2)55.34 = 61.57362151.


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Example 7


Solution: (i) Since F0,T = erTS0, 57.6 = e(3/4)r55.34 andr = (4/3) ln(57.6/55.34) = 0.05336879112.(ii) We have thatF0,2 = er2S0 = e(0.05336879112)(2)55.34 = 61.57362151.


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Example 8

A stock is expected to pay a dividend of $1 per share in 2 monthsand again in 5 months. The current stock price is $59 per share.The risk free effective annual rate of interest is 6%.(i) What is the fair price of a 6–month forward contract?(ii) Assume that 3 months from now the stock price is $57 pershare, what is the fair price of the same forward contract at thattime?

Solution: (i) The forward price is the future value of the paymentsassociated with owning the stock in six months:F0,0.5 = (59)(1.06)0.5 − (1)(1.06)4/12 − (1)(1.06)1/12 = 58.71974.(ii) (57)(1.06)3/12 − (1)(1.06)1/12 = 56.83154.


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Example 8


Solution: (i) The forward price is the future value of the paymentsassociated with owning the stock in six months:F0,0.5 = (59)(1.06)0.5 − (1)(1.06)4/12 − (1)(1.06)1/12 = 58.71974.

(ii) (57)(1.06)3/12 − (1)(1.06)1/12 = 56.83154.


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Example 8


Solution: (i) The forward price is the future value of the paymentsassociated with owning the stock in six months:F0,0.5 = (59)(1.06)0.5 − (1)(1.06)4/12 − (1)(1.06)1/12 = 58.71974.(ii) (57)(1.06)3/12 − (1)(1.06)1/12 = 56.83154.


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Example 9

An investor is interested in buying XYZ stock. The current price ofstock is $30 per share. The risk–free annual interest ratecontinuously compounded is 0.03. The price of a fourteen–monthforward contract is 30.352. Calculate the continuous dividend yieldδ.


30.352 = F0,T = S0e(r−δ)T = 30e(0.03−δ)(14/12).

and

δ = 0.03− (12/14) ln(30.352/30) = 0.02000140155.


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Example 9

An investor is interested in buying XYZ stock. The current price ofstock is $30 per share. The risk–free annual interest ratecontinuously compounded is 0.03. The price of a fourteen–monthforward contract is 30.352. Calculate the continuous dividend yieldδ.


30.352 = F0,T = S0e(r−δ)T = 30e(0.03−δ)(14/12).

and

δ = 0.03− (12/14) ln(30.352/30) = 0.02000140155.


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Example 10

An investor is interested in buying XYZ stock. The current price ofstock is $30 per share. This stock pays dividends at an annualcontinuous rate of 0.02. The risk–free annual effective rate ofinterest is 0.045.(i) What is the price of prepaid forward contract which expires in18 months?(ii) What is the price of forward contract which expires in 18months?

Solution: (i) The prepaid forward price is

FP0,T = S0e

−δT = 30e−(0.02)(18/12) = 29.11336601.

(ii) The 18–month forward price is29.11336601(1.045)18/12 = 31.1004631.


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Example 10



FP0,T = S0e

−δT = 30e−(0.02)(18/12) = 29.11336601.



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Example 10



FP0,T = S0e

−δT = 30e−(0.02)(18/12) = 29.11336601.



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off–market forward contract

A forward contract where either you pay a premium or you collecta premium for entering into the deal is called an off–marketforward contract.


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Example 11

Suppose that the current value of a certain amount of acommodity is $45000. The annual effective rate of interest is 4.5%.(i) You are offered a 2–year long forward contract at a forwardprice of $50000. How much would you need to be paid to enterinto this contract?(ii) You are offered a 2–year long forward contract at a forwardprice of $48000. How much would you need be willing to pay toenter into this contract?


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Example 11

Suppose that the current value of a certain amount of acommodity is $45000. The annual effective rate of interest is 4.5%.(i) You are offered a 2–year long forward contract at a forwardprice of $50000. How much would you need to be paid to enterinto this contract?(ii) You are offered a 2–year long forward contract at a forwardprice of $48000. How much would you need be willing to pay toenter into this contract?Solution: (i) Let x be how much you need to be paid to enter intothis contract. The current value of the commodity should be equalto the present value of the expenses needed to get the commodityusing the long forward contract. Hence, 50000(1.045)−2 − x =45000. So, x = 50000(1.045)−2 − 45000 = 786.4976.


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Example 11

Suppose that the current value of a certain amount of acommodity is $45000. The annual effective rate of interest is 4.5%.(i) You are offered a 2–year long forward contract at a forwardprice of $50000. How much would you need to be paid to enterinto this contract?(ii) You are offered a 2–year long forward contract at a forwardprice of $48000. How much would you need be willing to pay toenter into this contract?Solution: (ii) Let y be how much would you need be willing topay to enter into this contract. The current value of the commod-ity should be equal to the present value of the expenses neededto get the commodity using the long forward contract. Hence,48000(1.045)−2 + y = 45000. So, y = 45000− 48000(1.045)−2 =1044.962341.


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Arbitrage

If the price of a forward contract does not follow the previousformulas, an arbitrageur can do arbitrage.


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Example 12

XYZ stock pays no dividends and has a current price of $42.5 pershare. A long position in a forward contract is available to buy1000 shares of stock six months from now for $43 per share. Abank pays interest at the rate of 5% per annum (continuouslycompounded) on a 6–month certificate of deposit. Describe astrategy for creating an arbitrage profit and determine the amountof the profit.

Solution: The no arbitrage price of a forward contract isS0e

rT = (42.5)e0.05(0.5) = 43.57589262. Hence, it is possible to doarbitrage by entering into the long forward position. An arbitrageurcan: sell 1000 shares of stock for (1000)(42.5) = 42500, deposit42500 in the bank for six months, and sign up a forward contractfor a long position for 1000 shares of stock. In six months, the CDreturns (1000)(42.5)e0.05(0.5) = 43575.89262. The cost of theforward is (1000)(43) = 430000. Hence, the profit is43575.89262− 430000 = 575.89262.


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Example 12

XYZ stock pays no dividends and has a current price of $42.5 pershare. A long position in a forward contract is available to buy1000 shares of stock six months from now for $43 per share. Abank pays interest at the rate of 5% per annum (continuouslycompounded) on a 6–month certificate of deposit. Describe astrategy for creating an arbitrage profit and determine the amountof the profit.

Solution: The no arbitrage price of a forward contract isS0e

rT = (42.5)e0.05(0.5) = 43.57589262. Hence, it is possible to doarbitrage by entering into the long forward position. An arbitrageurcan: sell 1000 shares of stock for (1000)(42.5) = 42500, deposit42500 in the bank for six months, and sign up a forward contractfor a long position for 1000 shares of stock. In six months, the CDreturns (1000)(42.5)e0.05(0.5) = 43575.89262. The cost of theforward is (1000)(43) = 430000. Hence, the profit is43575.89262− 430000 = 575.89262.


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Example 13

Suppose that the risk–free effective rate of interest is 5% perannum. XYZ stock is currently trading for $45.34 per share. XYZstock is expected to pay a dividend of $1.20 per share six monthsfrom now. The price of a nine–month forward contract on oneshare of XYZ stock is $47.56. Is there an arbitrage opportunity onthe forward contract? If so, describe the strategy to realize profitand find the arbitrage profit.

Solution: The no arbitrage forward price is

F0,T = erTS0 −n∑

i=1

Dti er(T−ti ) = 45.34(1.05)9/12 − 1.2(1.05)3/12

=45.81511211.

We can make arbitrage by buying stock and entering a shortforward contract. The profit per share at expiration is47.56− 45.81511211 = 1.74488789.


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Example 13

Suppose that the risk–free effective rate of interest is 5% perannum. XYZ stock is currently trading for $45.34 per share. XYZstock is expected to pay a dividend of $1.20 per share six monthsfrom now. The price of a nine–month forward contract on oneshare of XYZ stock is $47.56. Is there an arbitrage opportunity onthe forward contract? If so, describe the strategy to realize profitand find the arbitrage profit.

Solution: The no arbitrage forward price is

F0,T = erTS0 −n∑

i=1

Dti er(T−ti ) = 45.34(1.05)9/12 − 1.2(1.05)3/12

=45.81511211.

We can make arbitrage by buying stock and entering a shortforward contract. The profit per share at expiration is47.56− 45.81511211 = 1.74488789.


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Suppose that a stock pays dividends at the continuous rate δ. Inthe absence of arbitrage, entering a forward for one share for F0,T

is equivalent to buying e−δT shares of stock for S0e−δT and

(borrowing S0e−δT ) selling a zero–coupon bond for S0e

−δT withexpiration in T years. In both cases, at time T we have one shareof stock after we make a payment of F0,T . There exists noarbitrage if S0e

−δT = F0,T e−rT . If S0e−δT 6= F0,T e−rT , we can

make arbitrage.


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If F0,T < e(r−δ)TS0, we can enter into a long forward for one shareof stock, and short e−δT shares of stock. At redemption time, wecover the short position by paying F0,T for the stock. It is like wehave borrowed S0e

−δT and pay the loan for F0,T . In some sensewe have created a zero–coupon bond. The position is called asynthetic zero–coupon bond. Let r ′ be the continuous annualrate of interest of the synthetic bond. This rate is called theimplied repo rate. We have that

S0e−δT er ′T = F0,T .

Hence, if F0,T < e(r−δ)TS0,

r ′ =1

Tlog

(F0,T

S0e−δT

)<

1

Tlog

(S0e

(r−δ)T

S0e−δT

)< r .

By doing an arbitrage, we are able to reduce the interest rate atwhich we borrow. Technically, this is not call arbitrage. It is calledquasi–arbitrage. We benefit from this portfolio, only if we arealready borrowing.


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Notice that the synthetic bond is created observing that

Long forward = Buy stock + Issue a bond

implies that

Issue a bond = Short stock + Long forward


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Reciprocally, if F0,T > e(r−δ)TS0, we can create a portfolio earninga rate of interest bigger than the risk–free interest rate. We canenter a short forward contract for one share of stock and buy e−δT

shares of stock. At redemption time, we get an inflow of F0,T .Since we invested S0e

−δT , the continuous annual interest rate r ′,which we earned in the investment satisfies

S0e−δT er ′T = F0,T .

Hence, if F0,T > e(r−δ)TS0,

r ′ =1

Tln

(F0,T

S0e−δT

)>

1

Tln

(S0e

(r−δ)T

S0e−δT

)= r .

Again this rate is called the implied repo rate.


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Example 14

XYZ stock costs $123.118 per share. This stock pays dividends atan annual continuous rate of 2.5%. A 18 month forward has aprice of $130.242. You own 10000 shares of XYZ stock. Calculatethe annual continuous rate of interest at which you can borrow byshorting your stock.


r ′ = δ+1

Tln

(F0,T

S0

)= 0.025+

1

1.5ln

(130.242

123.118

)= 6.250067554%.


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Example 14

XYZ stock costs $123.118 per share. This stock pays dividends atan annual continuous rate of 2.5%. A 18 month forward has aprice of $130.242. You own 10000 shares of XYZ stock. Calculatethe annual continuous rate of interest at which you can borrow byshorting your stock.


r ′ = δ+1

Tln

(F0,T

S0

)= 0.025+

1

1.5ln

(130.242

123.118

)= 6.250067554%.


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Example 15

XYZ stock costs $124 per share. This stock pays dividends at anannual continuous rate of 1.5%. A 2–year forward has a price of$135.7 per share. Calculate the annual continuous rate of interestwhich you earn by buying stock and entering into a short forwardcontract, both positions for the same nominal amount.


r ′ = δ +1

Tln

(F0,T

S0

)= 0.015 +

1

2ln

(135.7

124

)= 6%.


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Example 15

XYZ stock costs $124 per share. This stock pays dividends at anannual continuous rate of 1.5%. A 2–year forward has a price of$135.7 per share. Calculate the annual continuous rate of interestwhich you earn by buying stock and entering into a short forwardcontract, both positions for the same nominal amount.


r ′ = δ +1

Tln

(F0,T

S0

)= 0.015 +

1

2ln

(135.7

124

)= 6%.


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The forward premium isF0,T

S0. Notice that this is not a price.

Prices of options are called premiums. But, here the nomenclatureis different. If a stock index pays dividends according with a

continuous rate δ, thenF0,T

S0= eT (r−δ).


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Example 16

XYZ stock cost $55 per share. A four–month forward on XYZstock costs $57.5. XYZ stock pays dividends according acontinuous rate.(i) Calculate the four–month forward premium.(ii) Calculate the eight–month forward premium.(iii) Calculate the eight–month forward price.

Solution: (i) The four–month forward premium isF0,4/12

S0= 57.5

55 = 1.045454545.(ii) The eight–month forward premium is

F0,8/12

S0= e(8/12)(r−δ) =

(e(4/12)(r−δ)

)2=

(57.5

55

)2

= 1.092975206.

(iii) The eight–month forward price isF0,8/12 = (55)(1.092975206) = 60.11363633.


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Example 16



S0= 57.5

55 = 1.045454545.

(ii) The eight–month forward premium is

F0,8/12

S0= e(8/12)(r−δ) =

(e(4/12)(r−δ)

)2=

(57.5

55

)2

= 1.092975206.



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Example 16



S0= 57.5


F0,8/12

S0= e(8/12)(r−δ) =

(e(4/12)(r−δ)

)2=

(57.5

55

)2

= 1.092975206.



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Example 16



S0= 57.5


F0,8/12

S0= e(8/12)(r−δ) =

(e(4/12)(r−δ)

)2=

(57.5

55

)2

= 1.092975206.



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The annualized forward premium is 1T ln

(F0,T

S0

). Note that in

the case of continuous dividends, the annualized forward premiumis r − δ.


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Example 17

XYZ stock cost $55 per share. A four–month forward on XYZstock costs $57.5.(i) Calculate the annualized forward premium(ii) Calculate the twelve–month forward price.

Solution: (i) The annualized forward premium is

r − δ =1

Tln

(F0,T

S0

)=

1

4/12ln

(57.5

55

)= 0.1333552877.

(ii) The twelve–month forward price is

F0,T = S0er−δ = 55e0.1333552877 = 62.84607438.


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Example 17



r − δ =1

Tln

(F0,T

S0

)=

1

4/12ln

(57.5

55

)= 0.1333552877.


F0,T = S0er−δ = 55e0.1333552877 = 62.84607438.


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Example 17



r − δ =1

Tln

(F0,T

S0

)=

1

4/12ln

(57.5

55

)= 0.1333552877.


F0,T = S0er−δ = 55e0.1333552877 = 62.84607438.


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Hedging a forward contract.

A (scalper) market maker must be able to offset the risk of tradingforward contracts. Assume continuous dividends. Suppose that ascalper enters into a short forward contract. The profit atexpiration for a long forward position is ST − F0,T . In order toobtain this same payoff a scalper can borrow S0e

−δT and use thismoney to get e−δT shares of stock. At time T , he sells the stockwhich he owns for S0e

(r−δ)T = F0,T . Notice that by investing thedividends, e−δT shares of stock have grown to one share at timeT . Borrowing S0e

−δT and buying e−δT shares of stock is called asynthetic long forward. So, if a scalper enters into a shortforward contract with a client, the scalper either matches thisposition with another client’s long forward contract or creates asynthetic long forward


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The profit of a short forward position is F0,T − ST . A scalper canget this profit, by (lending) buying a zero–coupon bond for S0e

−δT

and shorting e−δT shares receiving S0e−δT . Buying a zero–coupon

bond for S0e−δT and shorting a tailed position for e−δT shares is

called a synthetic short forward. Again, a scalper may need tocreate this position to match a client’s long forward position.


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Using these strategies, a market–maker can hedge his clientspositions.A transaction in which you buy the asset and short the forwardcontract is called cash–and–carry (or cash–and–carry hedge). Itis called cash–and–carry, because the cash is used to buy the assetand the asset is kept. A cash–and–carry has no risk. You haveobligation to deliver the asset, but you also own the asset. Anarbitrage that involves buying the asset and selling it forward iscalled cash–and–carry arbitrage. A (reverse cash–and–carryhedge) reverse cash–and–carry involves short–selling and assetand entering into a long forward position.


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An arbitrageur can make money if F0,T 6= S0e(r−δ)T . But, in the

real world, transaction costs have to be taken into account.Suppose that:(i) The stock bid and ask prices are Sb

0 and Sa0 , where Sb

0 < Sa0 .

(ii) The forward bid and ask prices are F b0,T < F a

0,T .(iii) The cost of a transaction in the stock is KS .(iv) The cost of a transaction in the forward is KF .(v) The interest rates for borrowing and lending are rb > rl ,respectively.


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Suppose that the arbitrageur believes that the observed forwardprice F0,T is too high. Then, he could:(a) contract a short forward for F b

0,T

(b) buy a tailed position in stock for Sa0e−δT .

(c) borrow Sae−δT

0 + KF + KS .The payoff of this combined transaction is:

F b0,T − ST + ST − (Sa

0e−δT + Kf + KS)erbT

=F b0,T − (Sa

0e−δT + Kf + KS)erbT .

The scalper makes money if F b0,T > (Sa

0e−δT + Kf + KS)erbT . Theprevious strategy is cash–and–carry arbitrage.


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Suppose that the scalper believes that the observed forward priceF0,T is too low. Then, he could:(a) enter a long forward for F a

0,T

(b) short a tailed position in stock for Sb0 e−δT .

(c) lend Sb0 e−δT − KF − KS .

The payoff of this combined transaction is:

ST − F a0,T − ST + (Sa

0e−δT − KF − KS)erlT

=− F a0,T + (Sa

0e−δT − KF − KS)erlT .

The arbitrageur makes money if (Sa0e−δT − KF − KS)erlT > F a

0,T .The previous strategy is reverse cash–and–carry arbitrage.


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Example 18

Suppose that an arbitrageur would like to enter a cash–and–carryfor 10000 barrels of oil for delivery in six months. Suppose that hecan borrow at an annual effective rate of interest of 4.5%. Thecurrent price of a barrel of oil is $55.(i) What is the minimum forward price at which he would make aprofit?(ii) What is his profit if the forward price is $57?

Solution: (i) He would make a profit ifF0,T > 55(1.045)1/2 = 56.22388283.(ii) The profit is (10000)(57− (55)(1.045)1/2) = 7761.171743.


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Example 18


Solution: (i) He would make a profit ifF0,T > 55(1.045)1/2 = 56.22388283.

(ii) The profit is (10000)(57− (55)(1.045)1/2) = 7761.171743.


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Example 18


Solution: (i) He would make a profit ifF0,T > 55(1.045)1/2 = 56.22388283.(ii) The profit is (10000)(57− (55)(1.045)1/2) = 7761.171743.


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Manual for SOA Exam FM/CAS Exam 2.Chapter 7. Derivative markets.

Section 7.3. Futures.





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Chapter 7. Derivatives markets. Section 7.3. Futures.

Futures

A future is a standardized contract in which two counterpartsagree to buy/sell an asset for a specified price (the future price) ata specified date (the delivery date).The buyer in the future contract is called the long future. Theseller in the future contract is called the short future.The main reasons to enter into a future contract are hedging andspeculation.At difference of futures, forward contracts are privately negotiatedand are not standardized. Forward contracts are entirely flexible.Forward contracts are tailor–made contracts.


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Futures are bought and sold in organized futures exchanges. Thebiggest future exchanges are the Chicago Mercantile Exchange, theChicago Board of Trade, the International Petroleum Exchange ofLondon, the New York Mercantile Exchange, the London MetalExchange and the Tokyo Commodity Exchange.Futures transactions in the USA are regulated by the (CFTC)Commodity Futures Trading Commission, an agency of the USAgovernment. The CFTC also regulates option markets.A future contract is negotiated through a brokerage firm that holdsa seat on the exchange. A future contract is settled by aclearinghouse owned by or associated with the exchange. Theclearinghouse matches the purchases and the sales which takeplace during the day. By matching trades, the clearinghouse nevertakes market risk because it always has offsetting positions withdifferent counterparts. By having the clearinghouse as counterpart,an individual entering a future contract does not face the possiblecredit risk of its counterpart.


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Let us consider some common futures.Crude oil futures trade in units of 1,000 U.S. barrels (42,000gallons). The underlying is a US barrel. The notional amount is1000 barrels. The current price is $70/barrel. Hence, the currentvalue of a future contract on crude oil is $70000.S & P 500 future contracts trade on 250 units of the index. Theyare cash settled. At expiration time, instead of a sale, one of thefuture counterpart receive a payment according with S & P 500spot price at expiration. The current price of S & P 500 is 1500.The current value of a future contract on S & P 500 is(250)(1500) = $375000.


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Suppose that two parties agree in a future contact for crude oil fordelivery in 18 months. The contract is worth $70000. Each(investor) party makes a trade with the clearinghouse. Thiscontract has two risks: market risk and credit risk. The marketrisk is related with the volatility of the price of the asset. Thecredit risk is related with the solvency of each party. To avoidcredit risk, an individual or corporation entering a future contractmust make a deposit into an account called the margin account.This deposit is called the initial margin. The margin accountearns interest. The amount of the initial margin is determined bythe exchange. It is usually a fraction of the market value of thefutures’ underlying asset. Usually future positions are settled intothe margin account either every day or every week. By every daywe mean every day which the market is open. Let us suppose thata clearinghouse settles accounts daily. Suppose that the annualcontinuously compounded interest rate is r .


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Every day, the profit or loss is calculated on the investor’s futuresposition. If there exists a loss, the investor’s broker transfers thatamount from the investor’s margin account to the clearinghouse. Ifa profit, the clearinghouse transfers that amount to investor’sbroker who then deposits it into the investor’s margin account.The profit for a long position in a future contract is

Mt−(1/365)(er/365 − 1) + N(St − St−(1/365)),

where Mt−(1/365) is the yesterday’s balance in the margin account,N is the nominal amount, St is the current price, St−(1/365) is theyesterday price. Hence, after the settlement, the balance in theinvestor’s margin account is

Mt = Mt−(1/365)er/365 + N(St − St−(1/365)).

The profit for a short position in a future contract is

Mt−(1/365)(1− er/365) + N(St−(1/365) − St).

Marking–to–market is to calculate the value of a future contractaccording with the current value of the asset.


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Example 1

On July 5, 2007, John enters a long future contract for 1,000 U.S.barrels of oil at $71.6 a barrel. The margin account is 50% of themarket value of the futures’ underlier. The annual continuouslycompounded rate of return is 0.06.(i) On July 6, 2007, the price of oil is $70.3. What is the balancein John’s margin account after settlement?(ii) On July 7, 2007, the price of oil is $72.1. What is the balancein John’s margin account after settlement?


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Solution: (i) The initial balance in John’s margin account is

(0.50)(1000)(71.6) = 35800.

The balance in John’s margin account on July 6, 2007, aftersettlement, is

Mt−(1/365)er/365 + N(St − St−(1/365))

=(35800)e0.06/365 + (1000)(70.3− 71.6) = 35105.89.

Since the price of the oil decreases, the value of having 1000barrels in 18 months decreases.


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Solution: (ii) The balance in John’s margin account on July 6,2007, after settlement, is

Mt−(1/365)er/365 + n(St − St−(1/365))

=(35105.89)e0.06/365 + (1000)(72.1− 70.3) = 35711.56.

Notice that this balance is different from

(35800)e(0.06)(2/365) + (1000)(72.1− 71.6) = 36311.77.

In the first day, John’s account balance was smaller. So, John lostinterest because the drop on price on July 6, 2007.


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If the balance in the margin account falls the clearinghouse has lessprotection against default. Investors are required to keep themargin account to a minimum level. This level is a fraction of theinitial margin. The maintenance margin is the fraction of theinitial margin which participants are asked to hold in theiraccounts. If the balance in the margin account falls below thislevel, an investor’s broker will require the investor to deposit fundssufficient to restore the balance to the initial margin level. Such ademand is called a margin call. If an investor fail to the deposit,the investor’s broker will immediately liquidate some or all of theinvestor’s positions.


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Example 2

A company enters into a short futures contract to sell 100,000pounds of frozen orange juice for $1.4 cents per pound. The initialmargin is 30% and the maintenance margin is 20%. The annualeffective rate of interest is 4.5%. The account is settled everyweek. What is the minimum next week price which would lead to amargin call?


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Solution: The initial balance in the margin account is(0.30)(100000)(1.4) = 42000. The minimum balance in themargin account is (0.20)(100000)(1.4) = 28000. After settlementnext week balance is

42000(1.045)1/52 + 100000(1.4− S1/52).

A margin call happens if

28000 > 42000(1.045)1/52 + 100000(1.4− S1/52),

or

S1/52 > 1.4− 28000− 42000(1.045)1/52

100000= 1.540355672.


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Besides holding the contract until expiration, there are two ways toclose a future contract: offset the contract and exchange forphysicals. To offset the contract means to enter a reverse positionwith the same broker. Since future contracts are standardized, it ispossible to find a reserve position on a contract. Exchange forphysicals consists selling/buying the commodity.


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The two main advantages of futures versus forwards are liquidityand counter–party risk. It is much easier to cancel beforeexpiration a future contract than a forward contract. Since thetrade is made against a clearinghouse, a participant does facecredit risk. At the same time, the margin and the marking tomarket reduces the default risk.


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Having a margin account makes the profit/losses of the investmenthigher for a future than for a forward. The oscillations of the priceof the asset make the earnings in the margin account morevariable. Usually, if there is a profit from the change of the price ofthe asset, there is also a profit in the balance account.Reciprocally, if there is a loss from the price change, there exists aloss in the margin account.


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Section 7.4. Call options.





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Chapter 7. Derivatives markets. Section 7.4. Call options.

Minimums and maximums

Definition 1Given two real numbers a and b,(i) min(a, b) denotes the (minimum) smallest of the two numbers.(ii) max(a, b) denotes the (maximum) biggest of the two numbers.

Example 1

min(10, 5) = 5, max(10, 5) = 10, min(−1, 5) = −1,max(−1, 5) = 5, min(−2,−100) = −100, max(−2,−100) = −2.


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Definition 2Given real numbers a1, . . . , an,(i) min(a1, . . . , an) denotes the (minimum) smallest of thesenumbers.(ii) max(a1, . . . , an) denotes the (maximum) biggest of thesenumbers.

Example 2

min(−1, 5, 3,−6) = −6, max(−1, 5, 3,−6) = 5,min(−2,−100,−50) = −100 and max(−2,−100,−50) = −2.


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Theorem 1For each a, b, c ∈ R and each λ ≥ 0,

I min(a, b) = min(b, a).

I max(a, b) = max(b, a).

I min(min(a, b), c) = min(a,min(b, c)) = min(a, b, c).

I max(max(a, b), c) = max(a,max(b, c)) = max(a, b, c).

I min(a + c , b + c) = min(a, b) + c.

I max(a + c , b + c) = max(a, b) + c.

I min(λa, λb) = λ min(a, b).

I max(λa, λb) = λ max(a, b).

I min(−a,−b) = −max(a, b).

I max(−a,−b) = −min(a, b).


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Definition 3Given a real number a, |a| = a, if a ≥ 0; and |a| = −a, if a ≤ 0

Example 3

|23| = 23, | − 4| = 4.

Theorem 2For each a, b ∈ R, min(a, b) + max(a, b) = a + b.

Proof.min(a, b) and max(a, b) are a and b in some order. Hence,min(a, b) + max(a, b) = a + b.

Theorem 3For each a ∈ R, |a| = max(a, 0)−min(a, 0).

Proof.If a ≥ 0, then max(a, 0) = a, min(a, 0) = 0, andmax(a, 0)−min(a, 0) = a = |a|. If a ≤ 0, then max(a, 0) = 0,min(a, 0) = a, and max(a, 0)−min(a, 0) = −a = |a|.


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Call options

Definition 4A call option is a financial contract which gives the owner theright, but not the obligation, to buy a specified amount of a givenasset at a specified price during a specified period of time.

The call option owner exercises the option by buying the asset atthe specified call price from the call writer. A call option isexecuted only if the call owner decides to do so. A call optionowner executes a call option only when it benefits him, i.e. whenthe specified call price is smaller than the current (market value)spot price. Since the owner of a call option can make money if theoption is exercised, call options are sold. The owner of the calloption must pay to its counterpart for holding a call option. Theprice of a call option is called its premium.


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Call options

Definition 4A call option is a financial contract which gives the owner theright, but not the obligation, to buy a specified amount of a givenasset at a specified price during a specified period of time.

The call option owner exercises the option by buying the asset atthe specified call price from the call writer. A call option isexecuted only if the call owner decides to do so. A call optionowner executes a call option only when it benefits him, i.e. whenthe specified call price is smaller than the current (market value)spot price. Since the owner of a call option can make money if theoption is exercised, call options are sold. The owner of the calloption must pay to its counterpart for holding a call option. Theprice of a call option is called its premium.


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I The (owner) buyer of a call option is called the option callholder. The holder of a call option is said to have a long callposition.

I The seller of a call option is called the option call writer.The writer of a call is said to have a short call position.

I Assets used in call options are in commodities, currencyexchange, stock shares and stock indices.

I A call option needs to specify the type and quality of theunderlying.

I The asset used in the call option is called the underlier orunderlying asset.

I The amount of the underlying asset to which the call optionapplies is called the notional amount.

I The specified price of an asset in a call option is called thestrike price, or exercise price.

I A forward contract forces the buyer and seller to execute thesale. A call option is executed only if the call holder decidesto do so.


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I For an European option, the exercise of the option mustoccur at a certain time (the expiration date).

I For an American option, the exercise of the option mustoccur any time by the expiration date.

I For a Bermudan option, the buyer can exercise the calloption during specified periods.

Unless say otherwise, we will assume that an option is an Europeanoption. European options are simpler and easier to study.


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Example 4

Suppose that an investor buys a call option of 100 shares of XYZstock with a strike price of $76 per share. The exercise date is oneyear from now.(i) If the spot price at expiration is $70 per share, the call optionholder does not exercise the option. The option is worthless. Thecall option holder can buy stock in the market for a price smallerthan the call option price.(ii) If the (the market price) spot price at expiration is $80 pershare, the call option holder exercises the call option, i.e. he buys100 shares of XYZ stock for $76 from the option seller. Since thecall option holder can sell these shares for $80 per share, the calloption holder gets a payoff of 100(80− 76) = $400.


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Let K be the strike price of a call option. Let ST be the price ofthe asset at expiration.

I The call option holder’s payoff is{0 if ST < K ,

ST − K if ST ≥ K .

We also can write this as max(0,ST − K ).

I The payoff for the call option writer is the opposite of theholder’s payoff. The payoff for the call option writer is−max(0,ST − K ).

I A call–option is a zero–sum game. The sum of the twopayoffs is zero.

Figure 1 shows a graph of the call option payoff as a function ofST .


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6

-��

��

��

max(ST − K , 0)

STK

Payoff for the call option holder

6

-@

@@

@@

−max(ST − K , 0)

ST

Payoff for the call option writer

K

Figure 1: Payoffs of a call option


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Recall:

I The call option holder’s payoff is

max(0,ST − K ).

I The call option writer’s payoff is

−max(0,ST − K ).

We get from Figure 1 that:

I The minimum payoff for the call option holder is 0. Themaximum payoff for the call option holder is ∞.

I The minimum payoff for the call option writer is −∞. Themaximum payoff for the call option writer is 0.

minimum payoff maximum payoff

call option holder 0 ∞call option writer −∞ 0


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Example 5

Andrew buys a 45–strike call option for XYZ stock with a nominalamount of 2000 shares. The expiration date is 6 months from now.(i) Calculate Andrew’s payoff for the following spot prices per shareat expiration: 35, 40, 45, 55, 60.(ii) Calculate Andrew’s minimum and maximum payoffs.

Solution: (i) Andrew’s payoff is (2000)max(ST − 45, 0). Thecorresponding payoffs are:

if ST = 35, payoff = (2000) max(35− 45, 0) = 0,

if ST = 40, payoff = (2000) max(40− 45, 0) = 0,

if ST = 45, payoff = (2000) max(45− 45, 0) = 0,

if ST = 50, payoff = (2000) max(50− 45, 0) = 10000,

if ST = 55, payoff = (2000) max(55− 45, 0) = 20000.

(ii) Andrew’s minimum payoff is 0. Andrew’s maximum payoff is∞.


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Example 5

Andrew buys a 45–strike call option for XYZ stock with a nominalamount of 2000 shares. The expiration date is 6 months from now.(i) Calculate Andrew’s payoff for the following spot prices per shareat expiration: 35, 40, 45, 55, 60.(ii) Calculate Andrew’s minimum and maximum payoffs.

Solution: (i) Andrew’s payoff is (2000)max(ST − 45, 0). Thecorresponding payoffs are:

if ST = 35, payoff = (2000) max(35− 45, 0) = 0,

if ST = 40, payoff = (2000) max(40− 45, 0) = 0,

if ST = 45, payoff = (2000) max(45− 45, 0) = 0,

if ST = 50, payoff = (2000) max(50− 45, 0) = 10000,

if ST = 55, payoff = (2000) max(55− 45, 0) = 20000.

(ii) Andrew’s minimum payoff is 0. Andrew’s maximum payoff is∞.


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Example 6

Madison sells a 45–strike call option for XYZ stock with a nominalamount of 2000 shares. The expiration date is 6 months from now.(i) Calculate Madison’s payoff for the following spot prices atexpiration: 35, 40, 45, 55, 60.(ii) Calculate Madison’s minimum and maximum payoffs.

Solution: (i) Madison’s payoff is −(2000)max(ST − 45, 0). Thecorresponding payoffs are:

if ST = 35, payoff = −(2000)max(35− 45, 0) = 0,

if ST = 40, payoff = −(2000)max(40− 45, 0) = 0,

if ST = 45, payoff = −(2000)max(45− 45, 0) = 0,

if ST = 50, payoff = −(2000)max(50− 45, 0) = −10000,

if ST = 55, payoff = −(2000)max(55− 45, 0) = −20000.

(ii) Madison’s payoff is (2000) max(ST − 45, 0). Madison’sminimum payoff is −∞. Madison’s maximum payoff is 0.


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Example 6

Madison sells a 45–strike call option for XYZ stock with a nominalamount of 2000 shares. The expiration date is 6 months from now.(i) Calculate Madison’s payoff for the following spot prices atexpiration: 35, 40, 45, 55, 60.(ii) Calculate Madison’s minimum and maximum payoffs.

Solution: (i) Madison’s payoff is −(2000)max(ST − 45, 0). Thecorresponding payoffs are:

if ST = 35, payoff = −(2000)max(35− 45, 0) = 0,

if ST = 40, payoff = −(2000)max(40− 45, 0) = 0,

if ST = 45, payoff = −(2000)max(45− 45, 0) = 0,

if ST = 50, payoff = −(2000)max(50− 45, 0) = −10000,

if ST = 55, payoff = −(2000)max(55− 45, 0) = −20000.

(ii) Madison’s payoff is (2000) max(ST − 45, 0). Madison’sminimum payoff is −∞. Madison’s maximum payoff is 0.


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Let Call(K ,T ) be the premium per unit paid by the buyer of a calloption with strike price K and expiration time T years. Notice thatCall(K ,T ) > 0. The premium of a call option for N units isNCall(K ,T ). Let i be the risk–free annual effective rate ofinterest.

I The call option holder’s profit per unit is

max(ST − K , 0)− Call(K ,T )(1 + i)T

=

{−Call(K ,T )(1 + i)T if ST < K ,

ST − K − Call(K ,T )(1 + i)T if ST ≥ K .

I The call option seller’s profit per unit is

Call(K ,T )(1 + i)T −max(0,ST − K )

=

{Call(K ,T )(1 + i)T if ST < K ,

Call(K ,T )(1 + i)T − (ST − K ) if ST ≥ K ..


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The call option holder profit max(ST −K , 0)−Call(K ,T )(1 + i)T

as a function of ST is nondecreasing. The call option holderbenefits from the increase of the spot price.

I The minimum call option holder profit is

−Call(K ,T )(1 + i)T .

I The maximum call option holder profit is ∞.

I The profit for the call option holder is positive if

ST > K + Call(K ,T )(1 + i)T .

I If ST < K + Call(K ,T )(1 + i)T , the call option holder profitis negative.


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The call option writer’s profit Call(K ,T )(1+ i)T −max(0,ST −K )as a function of ST is nonincreasing. The call option writerbenefits from the decrease of the spot price.

I The minimum call option writer profit is −∞. The call optionwriter position is riskier than his counterpart. A call optionwriter can assumed unbounded loses.

I The maximum call option writer profit is Call(K ,T )(1 + i)T .

I The profit for the call option writer is positive if

ST < K + Call(K ,T )(1 + i)T .

I The profit for the call option writer is negative if

ST > K + Call(K ,T )(1 + i)T .


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profit

call option holder max(ST − K , 0)− Call(K ,T )(1 + i)T

call option writer −max(ST − K , 0) + Call(K ,T )(1 + i)T


call option holder −Call(K ,T )(1 + i)T ∞call option writer −∞ Call(K ,T )(1 + i)T

Figure 2 shows the graph of the profit of a call option as a functionof ST .


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6

-��

��

�

max(ST − K , 0)− C (1 + i)T

ST

K−C (1 + i)T

Profit for the call option holder

6

-@

@@

@@

C (1 + i)T −max(ST − K , 0)

ST

C (1 + i)T

K

Profit for the call option writer

Figure 2: Profit of a call option


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If r is the annual interest rate compounded continuously, then theprofit for the call option holder is

max(0,ST − K )− Call(K ,T )erT

and the profit of the call option writer is

Call(K ,T )erT −max(0,ST − K ).


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Example 7

Ethan buys a 35–strike call option for XYZ stock for 4.337 pershare. The nominal amount of this call option is 2000 shares. Theexpiration date of this option is 18 months. The annual effectiveinterest rate is 5.5%.


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Example 7

Ethan buys a 35–strike call option for XYZ stock for 4.337 pershare. The nominal amount of this call option is 2000 shares. Theexpiration date of this option is 18 months. The annual effectiveinterest rate is 5.5%.(i) Calculate Ethan’s profit function.


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Example 7

Ethan buys a 35–strike call option for XYZ stock for 4.337 pershare. The nominal amount of this call option is 2000 shares. Theexpiration date of this option is 18 months. The annual effectiveinterest rate is 5.5%.(i) Calculate Ethan’s profit function.Solution: (i) Ethan’s profit function is

(2000)(max(ST − 35, 0)− 4.337(1.055)1.5)

=(2000)max(ST − 35, 0)− 9400.


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Example 7

Ethan buys a 35–strike call option for XYZ stock for 4.337 pershare. The nominal amount of this call option is 2000 shares. Theexpiration date of this option is 18 months. The annual effectiveinterest rate is 5.5%.(ii) Calculate Ethan’s profit for the following spot prices at expira-tion: 25, 30, 35, 40, 45.


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Example 7

Ethan buys a 35–strike call option for XYZ stock for 4.337 pershare. The nominal amount of this call option is 2000 shares. Theexpiration date of this option is 18 months. The annual effectiveinterest rate is 5.5%.(ii) Calculate Ethan’s profit for the following spot prices at expira-tion: 25, 30, 35, 40, 45.Solution: (ii) Since Ethan’s profit is (2000) max(ST−35, 0)−9400,Ethan’s profit for the considered spot prices is:

if ST = 25, profit = (2000) max(25− 35, 0)− 9400 = −9400,

if ST = 30, profit = (2000) max(30− 35, 0)− 9400 = −9400,

if ST = 35, profit = (2000) max(35− 35, 0)− 9400 = −9400,

if ST = 40, profit = (2000) max(40− 35, 0)− 9400 = 600,

if ST = 45, profit = (2000) max(45− 35, 0)− 9400 = 10600.


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Example 7

Ethan buys a 35–strike call option for XYZ stock for 4.337 pershare. The nominal amount of this call option is 2000 shares. Theexpiration date of this option is 18 months. The annual effectiveinterest rate is 5.5%.(iii) Calculate Ethan’s minimum and maximum profits.


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Example 7

Ethan buys a 35–strike call option for XYZ stock for 4.337 pershare. The nominal amount of this call option is 2000 shares. Theexpiration date of this option is 18 months. The annual effectiveinterest rate is 5.5%.(iii) Calculate Ethan’s minimum and maximum profits.Solution: (iii) Since Ethan’s profit is (2000)max(ST−35, 0)−9400,Ethan’s minimum profit is −9400 and Ethan’s maximum profit is∞.


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Example 7

Ethan buys a 35–strike call option for XYZ stock for 4.337 pershare. The nominal amount of this call option is 2000 shares. Theexpiration date of this option is 18 months. The annual effectiveinterest rate is 5.5%.(iv) Find the spot prices at which Ethan’s profit is positive.


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Example 7

Ethan buys a 35–strike call option for XYZ stock for 4.337 pershare. The nominal amount of this call option is 2000 shares. Theexpiration date of this option is 18 months. The annual effectiveinterest rate is 5.5%.(iv) Find the spot prices at which Ethan’s profit is positive.Solution: (iv) Since Ethan’s profit is (2000) max(ST−35, 0)−9400,Ethan’s profit is positive if (2000) max(ST − 35, 0)− 9400 > 0, i.e.if ST > 35 + 9400

2000 = 39.7.


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Example 7

Ethan buys a 35–strike call option for XYZ stock for 4.337 pershare. The nominal amount of this call option is 2000 shares. Theexpiration date of this option is 18 months. The annual effectiveinterest rate is 5.5%.(v) Calculate the spot price at expiration at which Ethan does notmake or lose money on this contract.


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Example 7

Ethan buys a 35–strike call option for XYZ stock for 4.337 pershare. The nominal amount of this call option is 2000 shares. Theexpiration date of this option is 18 months. The annual effectiveinterest rate is 5.5%.(v) Calculate the spot price at expiration at which Ethan does notmake or lose money on this contract.Solution: (v) Since Ethan’s profit is (2000) max(ST−35, 0)−9400,Ethan breaks even if (2000)(ST − 35) − 9400 = 0, i.e. if ST =35 + 9400

2000 = 39.7.


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Example 7

Ethan buys a 35–strike call option for XYZ stock for 4.337 pershare. The nominal amount of this call option is 2000 shares. Theexpiration date of this option is 18 months. The annual effectiveinterest rate is 5.5%.(vi) Find the spot price at expiration at which Ethan makes an annualeffective yield of 4.75%.


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Example 7

Ethan buys a 35–strike call option for XYZ stock for 4.337 pershare. The nominal amount of this call option is 2000 shares. Theexpiration date of this option is 18 months. The annual effectiveinterest rate is 5.5%.(vi) Find the spot price at expiration at which Ethan makes an annualeffective yield of 4.75%.Solution: (vi) Ethan invests (2000)(4.337) = 8674. If his yield is4.75%, his payoff is

(8674)(1.0475)18/12 = 9300 = (2000)max(ST − 35, 0)

and

ST = 35 +9300

2000= 39.65.


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Example 7

Ethan buys a 35–strike call option for XYZ stock for 4.337 pershare. The nominal amount of this call option is 2000 shares. Theexpiration date of this option is 18 months. The annual effectiveinterest rate is 5.5%.(vii) Find the annual effective rate of return earned by Ethan if thespot price at expiration is 38.


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Example 7

Ethan buys a 35–strike call option for XYZ stock for 4.337 pershare. The nominal amount of this call option is 2000 shares. Theexpiration date of this option is 18 months. The annual effectiveinterest rate is 5.5%.(vii) Find the annual effective rate of return earned by Ethan if thespot price at expiration is 38.Solution: (vii) Let i be Ethan’s annual effective rate of re-turn. Ethan invests (2000)(4.337) = 8674. His payoff is(2000) max(38 − 35, 0) = 6000. Hence, 8674(1 + i)1.5 = 6000and i = −21.78538923%.


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Example 8

Hannah sells a 35–strike call option for XYZ stock for 4.337 pershare. The nominal amount of this call option is 2000 shares. Theexpiration date of this option is 18 months. The annual effectiveinterest rate is 5.5%. Hannah invests the proceeds of the sale in azero–coupon bond.


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Example 8


(i) Calculate Hannah’s profit function.


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Example 8


(i) Calculate Hannah’s profit function.Solution: (i) Hannah’s profit is

− (2000)(max(ST − 35, 0)− 4.337(1.055)1.5)

=9400− (2000)max(ST − 35, 0).


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Example 8


(ii) Calculate Hannah’s profit for the following spot prices at expi-ration: 25, 30, 35, 40, 45.


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Example 8


(ii) Calculate Hannah’s profit for the following spot prices at expi-ration: 25, 30, 35, 40, 45.Solution: (ii) Since Hannah’s profit is 9400 − (2000)max(ST −35, 0), Hannah’s profit for the considered spot prices is:

if ST = 25, profit = 9400− (2000)max(25− 35, 0) = 9400,

if ST = 30, profit = 9400− (2000)max(30− 35, 0) = 9400,

if ST = 35, profit = 9400− (2000)max(35− 35, 0) = 9400,

if ST = 40, profit = 9400− (2000)max(40− 35, 0) = −600,

if ST = 45, profit = 9400− (2000)max(45− 35, 0) = −10600.


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Example 8


(iii) Calculate Hannah’s minimum and maximum profits.


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Example 8


(iii) Calculate Hannah’s minimum and maximum profits.Solution: (iii) Since Hannah’s profit is 9400 − (2000)max(ST −35, 0), Hannah’s minimum profit is −∞ and Hannah’s maximumprofit is 9400.


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Next we consider the pricing of a call option. The profit of a calloption depends on ST , which is random. In the case of uncertainscenarios, an arbitrage portfolio consists of a zero investmentportfolio, which shows non–negative payoffs in all scenarios. Thisimplies that if there exists no arbitrage, the profit function of aportfolio is either constantly zero, or its minimum is negative andits maximum positive.


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Theorem 4If there exist no arbitrage, then

max(S0 − (1 + i)−TK , 0) < Call(K ,T ) < S0.


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Proof: Consider the portfolio consisting of selling a call option andbuying the asset. The profit per unit at expiration is

ST −max(ST − K , 0)− (S0 − Call(K ,T ))(1 + i)T

=ST + K −max(ST ,K )− (S0 − Call(K ,T ))(1 + i)T

=min(ST ,K )− (S0 − Call(K ,T ))(1 + i)T .

The profit is nondecreasing on ST . The minimum of this portfoliois −(S0 − Call(K ,T ))(1 + i)T . The maximum of this portfolio isK − (S0 −Call(K ,T ))(1 + i)T . If there exists no arbitrage and theprofit function is not constant, the minimum profit is negative andthe maximum profit is positive. Hence,

−(S0 −Call(K ,T ))(1 + i)T < 0 < K − (S0 −Call(K ,T ))(1 + i)T


S0 − (1 + i)−TK < Call(K ,T ) < S0.


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If the bounds in Theorem 4 do not hold, we can make arbitrage.For example, if the price of the call is bigger than the spot price, wecan make money by buying the asset, selling the call and investingthe proceeds in a zero–coupon bond. At redemption time, we havethe asset which can use to satisfy the requirements of the call.


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Example 9

Consider an European call option on a stock worth S0 =32, withexpiration date exactly one year from now, and with strike price$30. The risk–free annual rate of interest compoundedcontinuously is r = 5%.


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Example 9


(i) If the call is worth $3, find an arbitrage portfolio.


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Example 9


(i) If the call is worth $3, find an arbitrage portfolio.Solution: (i) We have that

Call(K ,T )−S0+(1+i)−TK = 3−32+e−0.0530 = −0.463117265 < 0.

We can do arbitrage by buying the call and shorting stock. If thespot price at expiration is more than 30, we buy the stock using thecall option. If the spot price at expiration is less than 30, we buythe stock at market price. Any case, we buy stock for min(ST , 30).Hence, the profit is

(32− 3)e0.05 −min(ST , 30) ≥ (32− 3)e0.05 − 30 = 0.4868617949.


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Example 9


(i) If the call is worth $35, find an arbitrage portfolio.


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Example 9


(i) If the call is worth $35, find an arbitrage portfolio.Solution: (ii) In this case Call(K ,T ) > S0. We can do arbitrageby selling the call and buying stock. If the spot price at expiration ismore than 30, we sell the stock to the call option holder. If the spotprice at expiration is less than 30, we sell the stock at the marketprice. In any case, we sell stock for min(ST , 30). The profit is

(35− 32)e0.05 + min(ST , 30) ≥ (35− 32)e0.05 = 3.153813289.


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A call option is a way to buy stock in the future. A long forward isanother way to buy stock in the future. Buying a call option, youare guaranteed that the price you pay is not bigger than the strikeprice. If you buy a call option, you can buy the asset at expirationfor min(ST ,K ). The baker in the example in Section 7.1, insteadof buying a long forward for F0,T , he can buy a call option to hedgeagainst high wheat prices. Doing this we will be able to buy wheatat time T for min(ST ,K ). The cost of this investment strategy is

Call(K ,T )erT + min(ST ,K ).

Recall that F0,T is the price of a forward contract with delivery inT years. The profit of a long forward is ST − F0,T . The minimumprofit of a long forward is −F0,T . The maximum profit of a longforward is ∞.


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Example 10

Joseph buys a one–year long forward for 100 shares of a stock at$74 per share. Samantha buys a call option of 100 shares of XYZstock for $76 per share. The exercise date is one year from now.The risk free effective annual interest rate is 6%. The premium ofthis call is $6.4133 per share.


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Example 10


(i) Make a table with Joseph’s profit and Samantha’s profit whenthe spot price at expiration is $50, $70, $90 and $110. Comparethese profits.


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Example 10


(i) Make a table with Joseph’s profit and Samantha’s profit whenthe spot price at expiration is $50, $70, $90 and $110. Comparethese profits.Solution: (i) Joseph’s profit is given by the formula

(100)(ST − 74).

Samantha’s profit is

100 max(0,ST − K )− 100Call(K ,T )(1 + i)T

= 100 max(0,ST − 76)− (100)(6.4133)(1.06)= 100 max(0,ST − 76)− 679.81.


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Example 10


(i) Make a table with Joseph’s profit and Samantha’s profit whenthe spot price at expiration is $50, $70, $90 and $110. Comparethese profits.Solution: (i) (continuation)

Joseph’s profit −2400 −400 1600 3600

Samantha’s profit −679.81 −679.81 720.19 2720.19

Spot Price 50 70 90 110

For high spot prices at expiration, Samantha’s profits are smallerthan John’s profits. For low prices, Samantha’s losses are smallerthan Joseph’s losses.


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Example 10


(ii) Calculate Joseph’s profit and Samantha’s minimum and maxi-mum payoffs.


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Example 10


(ii) Calculate Joseph’s profit and Samantha’s minimum and maxi-mum payoffs.Solution: (ii) Joseph’s minimum profit is −7400. Joseph’s maxi-mum profit is ∞. Samantha’s minimum profit is −679.81. Saman-tha’s maximum profit is ∞.


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Example 10


(iii) Which is the minimum spot price at expiration at which Josephmakes a profit? Which is the minimum spot price at expiration atwhich Samantha makes a profit?


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Example 10


(iii) Which is the minimum spot price at expiration at which Josephmakes a profit? Which is the minimum spot price at expiration atwhich Samantha makes a profit?Solution: (iii) Joseph is even if ST = 74. Samantha is even if100(ST−76)−679.81 = 0, i.e. ST = 76+(679.81/100) = 82.7981.


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Example 10


(iv) Draw the graph of the profit versus the spot price at expirationfor Joseph and Samantha.


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Example 10


(iv) Draw the graph of the profit versus the spot price at expirationfor Joseph and Samantha.Solution: (iv) The graphs of (long forward) Joseph’s profit and(purchased call) Samantha’s profit are in Figure 3.


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Example 10


(v) Find the spot price at redemption at which both profits are equal.


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Example 10


(v) Find the spot price at redemption at which both profits are equal.Solution: (v) We solve (100)(ST − 74) = 100 max(0,ST − 76) −679.81 for ST . There is not solution with ST ≥ 76. If ST < 76we have the equation (100)(ST − 74) = −679.81, or ST = 74 −6.7981 = 67.2019.


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Figure 3: Example 10. Profit for long forward and purchased call.


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A purchased call option reduces losses over a long forward. Noticethat in Figure 3 the losses for a long forward holder can be big ifthe spot price at redemption is small. A call option is an insuredlong position in an asset. In return for not having large losses, thepossible profits in a call option are smaller. The spot price neededto make money is bigger for a purchased call than for a longforward. The profit for the call option holder is positive if

ST > K + Call(K ,T )(1 + i)T .

The profit for the long forward is positive if ST > F0,T . ByTheorem 5,

K + Call(K ,T )(1 + i)T > F0,T .

To make a positive profit, a call option holder needs a biggerincrease on the spot price than a long forward holder.


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Theorem 5If there exists no arbitrage, then

(1 + i)−T max(F0,T − K , 0) < Call(K ,T ) < (1 + i)−TF0,T .


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Proof: Suppose that you enter into a short forward contract andyou buy a call option. Both contracts have the same expirationtime and nominal amount. At expiration, the profit of this strategyis

F0,T − ST + max(ST − K , 0)− (1 + i)TCall(K ,T )

=F0,T + max(−K ,−ST )− (1 + i)TCall(K ,T )

=F0,T −min(K ,ST )− (1 + i)TCall(K ,T ).

This profit function is increasing on ST and it not constant. Theminimum profit of this portfolio is

F0,T − K − (1 + i)TCall(K ,T ).

The maximum profit of this portfolio is

F0,T − (1 + i)TCall(K ,T ).

If there is no arbitrage,

F0,T − K − (1 + i)TCall(K ,T ) < 0 < F0,T − (1 + i)TCall(K ,T ).


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Example 11

The current price of a forward contract for 1000 units of an assetwith expiration date two years from now is $120000. The risk–freeannual rate of interest compounded continuously is 5%. The priceof a two–year 100–strike European call option for 1000 units of theasset is $15000. Find an arbitrage portfolio and its minimum profit.

Solution: Since

e−rT (F0,T − K ) = e−(2)(0.05)(120000− (100)(1000))

=18096.74836 > 15000,

the call option is under priced. Consider the portfolio consisting ofbuying the call and entering into a short forward. The profit is

1000 max(ST − 100, 0)− 15000e(2)(0.05) + 120000− 1000ST

=1000max(−100,−ST ) + 103422.4362

=103422.4362− 1000 min(100,ST ).

The minimum profit is 103422.4362− 1000(100) = 3422.4362.


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Example 11

The current price of a forward contract for 1000 units of an assetwith expiration date two years from now is $120000. The risk–freeannual rate of interest compounded continuously is 5%. The priceof a two–year 100–strike European call option for 1000 units of theasset is $15000. Find an arbitrage portfolio and its minimum profit.

Solution: Since

e−rT (F0,T − K ) = e−(2)(0.05)(120000− (100)(1000))

=18096.74836 > 15000,

the call option is under priced. Consider the portfolio consisting ofbuying the call and entering into a short forward. The profit is

1000 max(ST − 100, 0)− 15000e(2)(0.05) + 120000− 1000ST

=1000max(−100,−ST ) + 103422.4362

=103422.4362− 1000 min(100,ST ).

The minimum profit is 103422.4362− 1000(100) = 3422.4362.c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2.

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Another motive to buy call options is to speculate. Call optionsallow betting in the increase of the price of a particular asset for asmall cash outlay. Buying a call option, a speculator achievesleverage. Call options provide price exposure without having topay, hold and warehouse the underlying asset. If a speculatorbelieves that an asset price is going to increase and it is right, hecan get a much higher yield of return buying a call option thanbuying the asset.


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Example 12

Rachel is a speculator. She anticipates XYZ stock to appreciatefrom its current level of $130 per share in four months. Rachelbuys a four–month 1000–share call option with a strike price of$150 per share and a premium of $1.8074 per share. Luke is also aspeculator. He also expects XYZ stock to appreciate and buysXYZ stock at the current market price.


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Example 12


(i) Find Rachel’s annual effective rate of return in her investmentfor the following spot prices at expiration 130, 150, 160 and 170.


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Example 12


(i) Find Rachel’s annual effective rate of return in her investmentfor the following spot prices at expiration 130, 150, 160 and 170.Solution: (i) Rachel invests (1000)(1.8074) = 1807.4. Four monthslater, she receives (1000)max(ST − 150, 0).If ST ≤ 150, Rachel loses all her money and her yield of returnis −100%. If ST = 160, Rachel receives (1000)(160 − 150) =

10000 at expiration. Rachel’s annual rate of return is(

100001807.4

)3 −1 = 168.3702647 = 16837.02647%. If ST = 170, Rachel receives(1000)(170 − 150) = 20000 at expiration. Rachel’s annual rate of

return is(

200001807.4

)3 − 1 = 1353.962117 = 135396.2117%.


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Example 12


(ii) Luke sells his stock at the end of four months. Find Luke’sannual effective rate of return in his investment for the spot pricesin (i).


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Example 12


(ii) Luke sells his stock at the end of four months. Find Luke’sannual effective rate of return in his investment for the spot pricesin (i).Solution: (ii) Luke invests 130 per share. His annual rate of return

j satisfies ST = 130(1 + j)1/3. So, j =(

ST130

)3− 1.

If ST = 130, j = 0%.If ST = 150, j = 53.61857078%.If ST = 160, j = 86.43604916%.If ST = 170, j = 123.6231224%.


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Example 12


(iii) Compare the rates in (i) and (ii).


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Example 12


(iii) Compare the rates in (i) and (ii).Solution: (iii) In the case that XYZ stock does not appreciate,Rachel loses all her money. But in the cases where XYZ stockappreciates, Rachel makes a much higher yield than Luke.


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Let jC be the rate of return which an investor makes buying a calloption. Since the payoff per share is max(ST −K , 0), we have that

max(ST − K , 0)

Call(K,T)= (1 + jC )T .

Let jB be the rate of return which an investor makes buying anasset and holding it for T years. Since the payoff per share is ST ,we have that

ST = (1 + jB)TS0.

We have that jC > jB if

max(ST − K , 0)

Call(K,T)>

ST

S0,

which is equivalent to S0 > Call(K,T) and

ST >

KCall(K,T)

1Call(K,T) −

1S0

=KS0

S0 − Call(K,T).

We conclude that if ST is large enough, investing in an option callgives a larger yield than buying an asset.


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Next we consider call options with different strike prices. If0 < K1 < K2, then

max(ST − K2, 0) ≤ max(ST − K1, 0),

i.e. the payoff of a K1–strike call option is higher than the payoffof a K2–strike call option (see Figure 4). Hence, the price of thecall is bigger for the call with smaller strike price (see Theorem 6).


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Example 13

The current price of XYZ stock is $75 per share. The annualeffective rate of interest is 5%. The redemption time is one yearfrom now. Draw the payoff and profit diagrams for the buyer of:


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Example 13


(i) a $70 strike call option with a premium of $10.755.


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Example 13


(i) a $70 strike call option with a premium of $10.755.Solution: (i) The payoff is max(ST − 70, 0). The diagram of thispayoff is in Figure 4. The profit is

max(ST − 70, 0)− (10.755)(1.05) = max(ST − 70, 0)− 11.29275.

The diagram of this profit is in Figure 5.


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Example 13


(ii) a $80 strike call option with a premium of $5.445.


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Example 13


(ii) a $80 strike call option with a premium of $5.445.Solution: (ii) The payoff is max(ST − 80, 0). The diagram of thispayoff is in Figure 4. The profit is

max(ST − 80, 0)− (5.445)(1.05) = max(ST − 80, 0)− 5.71725.

The diagram of this profit is in Figure 5.


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Example 13


(iii) Find the spot price at redemption at which both profits areequal.


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Example 13


(iii) Find the spot price at redemption at which both profits areequal.Solution: (iii) The profit amounts are equal for some ST ∈ (70, 80).So,

ST − 70− 11.29275 = max(ST − 70, 0)− 11.29275

=max(ST − 80, 0)− 5.71725 = −5.71725

and ST = 70 + 11.29275− 5.71725 = 75.5755.


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Figure 4: Example 13. Payoff for two calls with different strikes.


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Figure 5: Example 13. Profit for two calls with different strikes.


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Theorem 6If 0 < K1 < K2, then

Call(K2,T ) ≤ Call(K1,T ) ≤ Call(K2,T ) + (K2 − K1)e−rT .


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Proof.We have that

max(ST − K2, 0)

≤max(ST − K1, 0) = K2 − K1 + max(ST − K2,K1 − K2)

≤K2 − K1 + max(ST − K2, 0).

In other words,(i) The payoff for a K2–strike call is smaller than the payoff for aK1–strike call.(ii) The payoff for a K1–strike call is smaller than K2 − K1 plus thepayoff for a K2–strike call.Hence, if there exist no arbitrage, then

Call(K2,T ) ≤ Call(K1,T ) ≤ Call(K2,T ) + (K2 − K1)e−rT .


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Example 14

Consider two European call options on a stock worth S0 =32, bothwith expiration date exactly two years from now and the samenominal amount. The risk–free annual rate of interestcompounded continuously is 5%. One call option has strike price$30 and the other one $35. The price of the 30–strike call is 7.


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Example 14


(i) Suppose that the price of the 35–strike call option is 8, find anarbitrage portfolio.


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Example 14


(i) Suppose that the price of the 35–strike call option is 8, find anarbitrage portfolio.Solution: (i) Here, Call(35,T ) ≤ Call(30,T ) does not hold. Wecan do arbitrage by a buying a 30–strike call option and selling a35–strike call option, both for the same nominal amount. The profitper share is

max(ST − 30, 0)−max(ST − 35, 0) + (8− 7)e0.05

≥(8− 7)e0.05 = 1.051271096.


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Example 14


(ii) Suppose that the price of the 35–strike call option is 1, find anarbitrage portfolio.


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Example 14


(ii) Suppose that the price of the 35–strike call option is 1, find anarbitrage portfolio.Solution: (ii) We have that

Call(K2,T )− Call(K1,T ) + (K2 − K1)e−rT

=1− 7 + (35− 30)e−0.05 = −1.243852877 < 0.

We can do arbitrage by buying a 35–strike call option and selling a30–strike call option.


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Example 14


(ii) Suppose that the price of the 35–strike call option is 1, find anarbitrage portfolio.Solution: (ii) (continuation) We can do arbitrage by buying a 35–strike call option and selling a 30–strike call option. The profit pershare is

max(ST − 35, 0)−max(ST − 30, 0) + (7− 1)e0.05

=− 5 + max(ST − 30, 5)−max(ST − 30, 0) + (7− 1)e0.05

≥− 5 + (7− 1)e0.05 = −5 + 6.307626578 = 1.307626578.


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The premium of a call option of an asset depends on severalfactors, like asset price, interest rate, expiration time, strike price,and asset price variability. We have the following rules of thumpfor the price of a call:

I Higher asset prices lead to higher call option prices.

I Higher strike prices lead to lower call option prices.

I Higher interest rates lead to higher call option prices.

I Higher expiration time leads to higher call option prices.

I Higher variation of an asset price leads to higher call optionprices.


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Since the call option buyer’s payoff decreases as the strikeincreases, the (price) premium of a call option decrease as thestrikes increases. Hence, between two call options with differentstrike prices:(i) The call option with smaller strike price has a bigger premium.(ii) If the spot price is low enough, both call options substain a loss.The loss is bigger for the call option with the smaller strike price.(iii) If the spot price is high enough, both call options have apositive profit. The profit is bigger for the call option with thesmaller strike price.We can check the previous assertions analytically using Theorem 6.


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The strike price is paid at the expiration time, as higher theinterest rate is as higher the call option premium is. As higher theexpiration time as higher the call option premium is. The greaterthe past variability of the price of an asset is as more likely is thatthe option will be exercised. So, higher variation of an asset priceleads to higher call option prices.


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The common method to find the price of a call option of a stock isto use the Black–Scholes formula1

Call(K ,T ) = S0e−δTΦ(d1)− Ke−rTΦ(d2)

where

d1 =log(S0/K ) + (r − δ + σ2/2)T

σ√

T;

d2 = d1 − σ√

T ;

S0 is the current price of the stock; K is the strike price; r is therisk free continuously compounded annual interest rate; δ is thecontinuous rate of dividend payments; T is the expiration time inyears of the option; σ is the implied volatility for the underlyingasset and Φ the cumulative distribution function of a standardnormal distribution.

1In 1973, Fischer Black and Myron Scholes published a paper presenting thepricing formula for call and put options.


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Table 1 shows the premium of a call option for different strikeprices. We have used S0 = 75, T = 1, σ = 0.20, δ = 0,r = ln(1.05).

Table 1:

K 65 70 75 80 85

Call(K ,T ) 14.31722 10.75552 7.78971 5.444947 3.680736


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Example 15

The current price of XYZ stock is $75 per share. The annualeffective rate of interest is 5%. The redemption time is one yearfrom now. The price of stock one year from now is $73.5.Calculate the profit per share at expiration for the holder of eachone of the call options in Table 1.


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Solution: The profit is

max(ST − K , 0)− (1.05)Call(K ,T )

=max(73.5− K , 0)− (1.05)Call(K ,T ).

The corresponding profits are:

if K = 65,max(73.5− 65, 0)− (1.05)(14.31722) = −6.533081,

if K = 70,max(73.5− 70, 0)− (1.05)(10.75552) = −7.793296,

if K = 75,max(73.5− 75, 0)− (1.05)(7.78971) = −8.1791955,

if K = 80,max(73.5− 80, 0)− (1.05)(5.444947) = −5.71719435,

if K = 85,max(73.5− 85, 0)− (1.05)(3.680736) = −3.8647728.


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If K is very small, the call option will almost certainly be executed.Hence, if K is very small, Call(K ,T ) = F0,T , i.e.lim

K→0+Call(K ,T ) = F0,T . If K is very large, the call option will

almost certainly not be executed. Hence, limK→∞

Call(K ,T ) = 0. As

a function on K , Call(K ,T ) is a decreasing function withlim

K→0+Call(K ,T ) = F0,T and limK→∞Call(K ,T ) = 0. Figure 6

shows the graph of Call(K ,T ) as a function of T .


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Figure 6: Example 16. Graph of Call(K ,T ) as a function of K .


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Example 16Using the Black–Scholes formula with T = 1, S0 = 100, T = 1,σ = 0.25, r = ln(1.06) and δ = 0.0, the following table of calloption premiums was obtained:

Call(K , T ) 76.4150 52.8366 30.0399 12.7562 4.1341 0.8417 0.2672 0.0605K 25 50 75 100 125 150 175 200

Figure 6 shows the graph of this function.

When we consider Call(K ,T ) as function of T . If T is smallenough, then the option will be exercised if S0 > K with a profit ofS0 − K . Hence, if S0 > K , lim

T→0+Call(K ,T ) = S0 − K . Notice

that by buying the call option for Call(K ,T ), we buy an assetworth S0 for K . If T is small enough and S0 < K , the option isnot exercised and his value is zero, i.e. lim

T→0+Call(K ,T ) = 0.


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An option is in–the–money option if it would have a positivepayoff if exercised immediately. An option is out–the–moneyoption if it would have a negative payoff if exercised immediately.An option is at–the–money option if it would have a zero payoff ifexercised immediately. The previous definition hold for both calland put options. Put options will considered shortly. For apurchased call option, we have


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I The purchased call option is in–the–money, if S0 > K .I The purchased call option is out–the–money, if S0 < K .I The purchased call option is at–the–money, if S0 = K .


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Section 7.5. Put options.





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Chapter 7. Derivatives markets. Section 7.5. Put options.

Put options

Definition 1A put option is a financial contract which gives the (holder)owner the right, but not the obligation, to sell a specified amountof a given security at a specified price at a specified time.

The put option owner exercises the option by selling the asset atthe specified call price to the put writer. A put option is executedonly if the put owner decides to do so. A put option ownerexecutes a put option only when it benefits him, i.e. when thespecified call price is bigger than the current (market value) spotprice. Since the owner of a put option can make money if theoption is exercised, put options are sold. The owner of the putoption must pay to its counterpart for holding a put option. Theprice of a put option is called its premium.


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Put options

Definition 1A put option is a financial contract which gives the (holder)owner the right, but not the obligation, to sell a specified amountof a given security at a specified price at a specified time.

The put option owner exercises the option by selling the asset atthe specified call price to the put writer. A put option is executedonly if the put owner decides to do so. A put option ownerexecutes a put option only when it benefits him, i.e. when thespecified call price is bigger than the current (market value) spotprice. Since the owner of a put option can make money if theoption is exercised, put options are sold. The owner of the putoption must pay to its counterpart for holding a put option. Theprice of a put option is called its premium.


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I The (owner) buyer of a put option is called the option putholder. The holder of a put option is said to have a long putposition.

I The seller of a put option is called the option put writer.The writer of a put is said to have a short put position.

I Assets used in put options are in commodities, currencyexchange, stock shares and stock indices.

I A put option needs to specify the type and quality of theunderlying.

I The asset used in the put option is called the underlier orunderlying asset.

I The amount of the underlying asset to which the put optionapplies is called the notional amount.

I The specified price of an asset in a put option is called thestrike price, or exercise price.

I A forward contract forces the buyer and seller to execute thesale. A put option is executed only if the put holder decides todo so.


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I For an European option, the exercise of the option mustoccur at a certain time (the expiration date).

I For an American option, the exercise of the option mustoccur any time by the expiration date.

I For a Bermudan option, the buyer can exercise the calloption during specified periods.

Unless say otherwise, we will assume that an option is an Europeanoption. European options are simpler and easier to study.


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Example 1

John buys a six–month put option for 150 shares with a strike priceof $45 per share.(i) If the price per share six months from now is $40, John sells150 shares to the put option writer for (150)(45) = 6750. Sincethe market value of these 150 shares is (150)(40) = 6000. Johnmakes (before expenses) 6750− 6000 = 750 on this contract.(ii) If the price per share six months from now is $50, John doesnot exercise the put option.


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I The put option buyer’s payoff per share is

max(K − ST , 0) =

{K − ST if ST < K ,

0 if ST ≥ K ,

where K is the strike price and ST is the spot price atredemption.

I The put option writer’s payoff per share is

−max(K − ST , 0) =

{−(K − ST ) if ST < K ,

0 if ST ≥ K ,

Figure 1 shows the graph of the payoff of a put option. A putoption is a zero–sum game.


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6

-

@@

@@

max(0,K − ST )

ST

K

Payoff for the put option holder

6

-

��

��

−max(0,K − ST )

ST

−K

Payoff for the put option writer

Figure 1: Payoffs of a put option


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Recall:

I The put option holder’s payoff is

max(0,K − ST ).

I The put option writer’s payoff is

−max(0,K − ST ).

We get from Figure 1 that:

I The minimum payoff for the put option holder is 0. Themaximum payoff for the put option holder is K .

I The minimum payoff for the put option writer is −K . Themaximum payoff for the put option writer is 0.

minimum payoff maximum payoff

put option holder 0 K

put option writer −K 0


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Example 2

Daniel buys a 55–strike put option on XYZ stock with a nominalamount of 5000 shares. The expiration date is 6 months from now.The nominal amount of the put option is 5000 shares of XYZstock.(i) Calculate Daniel’s payoff for the following spot prices per shareat expiration: 40, 45, 55, 60, 60.(ii) Calculate Daniel’s minimum and maximum payoffs.

Solution: (i) Daniel’s payoff is 5000 max(55− ST , 0). Thecorresponding payoffs are:

if ST = 40, payoff = (5000) max(55− 40, 0) = 75000,

if ST = 45, payoff = (5000) max(55− 45, 0) = 50000,

if ST = 50, payoff = (5000) max(55− 50, 0) = 25000,

if ST = 55, payoff = (5000) max(55− 55, 0) = 0,

if ST = 60, payoff = (5000) max(55− 60, 0) = 0.

(ii) Daniel’s minimum payoff is zero. Daniel’s maximum payoff is(5000)(55) = 275000.


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Example 2

Daniel buys a 55–strike put option on XYZ stock with a nominalamount of 5000 shares. The expiration date is 6 months from now.The nominal amount of the put option is 5000 shares of XYZstock.(i) Calculate Daniel’s payoff for the following spot prices per shareat expiration: 40, 45, 55, 60, 60.(ii) Calculate Daniel’s minimum and maximum payoffs.

Solution: (i) Daniel’s payoff is 5000 max(55− ST , 0). Thecorresponding payoffs are:

if ST = 40, payoff = (5000) max(55− 40, 0) = 75000,

if ST = 45, payoff = (5000) max(55− 45, 0) = 50000,

if ST = 50, payoff = (5000) max(55− 50, 0) = 25000,

if ST = 55, payoff = (5000) max(55− 55, 0) = 0,

if ST = 60, payoff = (5000) max(55− 60, 0) = 0.

(ii) Daniel’s minimum payoff is zero. Daniel’s maximum payoff is(5000)(55) = 275000.c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2.

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Example 3

Isabella sells a 55–strike put option on XYZ stock. The expirationdate is 18 months from now. The nominal amount of the putoption is 10000 shares of XYZ stock.(i) Calculate Isabella’s payoff for the following spot prices per shareat expiration: 40, 45, 55, 60, 60.(ii) Calculate Isabella’s minimum and maximum payoffs.

Solution: (i) Isabella’s payoff is −10000 max(55− ST , 0). Thecorresponding payoffs are:

if ST = 40, payoff = −(10000)max(55− 40, 0) = −150000,

if ST = 45, payoff = −(10000)max(55− 45, 0) = −100000,

if ST = 50, payoff = −(10000)max(55− 50, 0) = −50000,

if ST = 55, payoff = −(10000)max(55− 55, 0) = 0,

if ST = 60, payoff = −(10000)max(55− 60, 0) = 0.

(ii) Isabella’s minimum payoff is −(10000)(55) = −550000.Isabella’s maximum payoff is zero.


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Example 3

Isabella sells a 55–strike put option on XYZ stock. The expirationdate is 18 months from now. The nominal amount of the putoption is 10000 shares of XYZ stock.(i) Calculate Isabella’s payoff for the following spot prices per shareat expiration: 40, 45, 55, 60, 60.(ii) Calculate Isabella’s minimum and maximum payoffs.

Solution: (i) Isabella’s payoff is −10000 max(55− ST , 0). Thecorresponding payoffs are:

if ST = 40, payoff = −(10000)max(55− 40, 0) = −150000,

if ST = 45, payoff = −(10000)max(55− 45, 0) = −100000,

if ST = 50, payoff = −(10000)max(55− 50, 0) = −50000,

if ST = 55, payoff = −(10000)max(55− 55, 0) = 0,

if ST = 60, payoff = −(10000)max(55− 60, 0) = 0.

(ii) Isabella’s minimum payoff is −(10000)(55) = −550000.Isabella’s maximum payoff is zero.


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Let Put(K ,T ) be the premium per unit paid of a put option withstrike price K and expiration time T years. Notice thatPut(K ,T ) > 0. Let i be the risk free annual effective rate ofinterest. The put option holder’s profit is

max(K − ST , 0)− Put(K ,T )(1 + i)T

=

{K − ST − Put(K ,T )(1 + i)T if ST < K ,

−Put(K ,T )(1 + i)T if ST ≥ K .

Put(K ,T )(1 + i)T is the future value at time T of the purchaseprice. The put option writer’s profit is

−max(K − ST , 0) + Put(K ,T )(1 + i)T

=

{−K + ST + Put(K ,T )(1 + i)T if ST < K ,

Put(K ,T )(1 + i)T if ST ≥ K .

Figure 2 shows a graph of the put profit.


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profit

put holder max(K − ST , 0)− Put(K ,T )erT

put writer −max(K − ST , 0) + Put(K ,T )erT


put holder −Put(K ,T )erT K − Put(K ,T )erT

put writer −K + Put(K ,T )erT Put(K ,T )erT

Figure 2 shows a graph of the put profit as a function of ST .


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6

-

@@

@@

K

max(K − ST , 0)− P(1 + i)T

ST

K − P(1 + i)T

−P(1 + i)T

Profit for the put option holder

6

-

��

��

K

P(1 + i)T −max(K − ST , 0)

ST

P(1 + i)T − K

P(1 + i)T

Profit for the put option writer

Figure 2: Profit of a put option


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Notice that the put option holder’s profit as a function of ST isnonincreasing. The put option holder benefits from a decrease onthe spot price. The minimum of the put option holder’s profit is

−Put(K ,T )(1 + i)T .

The maximum of the put option’s holder profit isK − Put(K ,T )(1 + i)T . If there exists no arbitrage

−Put(K ,T )(1 + i)T < 0 < K − Put(K ,T )(1 + i)T ,


0 < Put(K ,T ) < K (1 + i)−T .


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max((1 + i)−TK − S0, 0) < Put(K ,T ) < K (1 + i)−T .

Proof.Consider the portfolio consisting of buying an asset and a putoption on this asset, both for the same notional amount. Theprofit at expiration is

ST + max(K − ST , 0)− S0(1 + i)T − Put(K ,T )(1 + i)T

=max(K ,ST )− (Put(K ,T ) + S0)(1 + i)T .

The maximum profit is ∞. The minimum profit isK − (Put(K ,T ) + S0)(1 + i)T . If there exists no arbitrageK − (Put(K ,T ) + S0)(1 + i)T < 0, which is equivalent to(1 + i)−TK − S0 < Put(K ,T ). From this bound and the boundsbefore the theorem, the claim follows.


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Example 4

The current price of XYZ stock is 160 per share. The annualeffective interest rate is 7%. The price of a one–year European200–strike put option for XYZ stock is $20 per share. Find anarbitrage strategy and the minimum profit per share.


Put(K ,T ) + S0 − (1 + i)−TK = 20 + 160− (200)(1.07)−1

=− 6.91588785 < 0.

The put premium is too low. Consider the portfolio consisting ofbuying the put and the stock, both for the same nominal amount.The profit per share is

max(200−ST , 0)−20(1.07)+ST−160(1.07) = max(200,ST )−192.6.

The minimum profit per share is 200− 192.6 = 7.4.


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Example 4



Put(K ,T ) + S0 − (1 + i)−TK = 20 + 160− (200)(1.07)−1

=− 6.91588785 < 0.

The put premium is too low. Consider the portfolio consisting ofbuying the put and the stock, both for the same nominal amount.The profit per share is

max(200−ST , 0)−20(1.07)+ST−160(1.07) = max(200,ST )−192.6.

The minimum profit per share is 200− 192.6 = 7.4.


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Example 5



(1 + i)−TK − Put(K ,T ) = (200)(1.07)−1 − 190

=− 3.08411215 < 0.

The put is overpriced. Consider the portfolio consisting of sellingthe put. The profit per share is 190(1.07)−max(200− ST , 0).The minimum profit per share is 190(1.07)− 200 = 3.3.


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Example 5



(1 + i)−TK − Put(K ,T ) = (200)(1.07)−1 − 190

=− 3.08411215 < 0.

The put is overpriced. Consider the portfolio consisting of sellingthe put. The profit per share is 190(1.07)−max(200− ST , 0).The minimum profit per share is 190(1.07)− 200 = 3.3.


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The profit of a put option holder is positive if

max(K − ST , 0)− Put(K ,T )(1 + i)T > 0,


K − Put(K ,T )(1 + i)T > ST .

If K − Put(K ,T )(1 + i)T < ST , the put option holder’s profit isnegative.


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Example 6

Ashley buys a 85–strike put option for 4.3185816 per share. Thenominal amount of the put option is 2500 shares of XYZ stock.The expiration date of this option is one year. The annual interestrate compounded continuously is 5%.


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Example 6


(i) Calculate Ashley’s profit function.


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Example 6


(i) Calculate Ashley’s profit function.Solution: (i) Ashley’s profit is

(2500)(max(85− ST , 0)− 4.3185816e0.05)

=(2500)max(85− ST , 0)− 11350.


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Example 6


(ii) Calculate Ashley’s profit for the following spot prices at expira-tion: 75, 80, 85, 90, 95.


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Example 6


(ii) Calculate Ashley’s profit for the following spot prices at expira-tion: 75, 80, 85, 90, 95.Solution: (ii) The profits corresponding to the considered spotprices are:

if ST = 75, profit = (2500)max(85− 75, 0)− 11350 = 13650,

if ST = 80, profit = (2500)max(85− 80, 0)− 11350 = 1150,

if ST = 85, profit = (2500)max(85− 85, 0)− 11350 = −11350,

if ST = 90, profit = (2500)max(85− 90, 0)− 11350 = −11350,

if ST = 95, profit = (2500)max(85− 95, 0)− 11350 = −11350.


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Example 6


(iii) Calculate Ashley’s minimum and maximum profits.


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Example 6


(iii) Calculate Ashley’s minimum and maximum profits.Solution: (iii) Ashley’s minimum profit is −11350. Ashley’s maxi-mum profit is (2500)(85)− 11350 = 201150.


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Example 6


(iv) Calculate the spot prices at which Ashley’s profit is positive.


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Example 6


(iv) Calculate the spot prices at which Ashley’s profit is positive.Solution: (iv) Ashley’s profit is positive if (2500) max(85−ST , 0)−11350 > 0, which is equivalent to ST < 85− 11350

2500 = 80.46.


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Example 6


(v) Calculate the spot price at expiration at which Ashley breakseven.


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Example 6


(v) Calculate the spot price at expiration at which Ashley breakseven.Solution: (v) Ashley breaks even if (2500) max(85 − ST , 0) −11350 = 0, i.e. if ST = 85− 11350

2500 = 80.46.


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Example 6


(vi) Calculate Ashley’s annual yield in her investment for the spotprices at expiration in (ii).


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Example 6


(vi) Calculate Ashley’s annual yield in her investment for the spotprices at expiration in (ii).Solution: (vi) Ashley invests (2500)(4.3185816) = 10796.454.Ashley’s return is (2500) max(85 − ST , 0). Let j be Ashley’s an-nual yield. Then, 10796.454(1 + j)0.5 = (2500)max(85 − ST , 0).

Hence, j =(

(2500) max(85−ST ,0)10796.454

)2− 1. Therefore,

if ST = 75, j =(

(2500) max(85−75,0)10796.454

)2− 1 = 436.1888022%,

if ST = 80, j =(

(2500)max(85−80,0)10796.454

)2− 1 = 34.04720055%,

if ST = 85, or 90, or 95, j = −100%.


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Example 7

William sells a 85–strike put option for 4.3185816 per share. Thenominal amount of the put option is 2500 shares of XYZ stock.The expiration date of the option is one year. The annual interestrate compounded continuously is 5%.


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Example 7


(i) Calculate William’s profit function.


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Example 7


(i) Calculate William’s profit function.Solution: (i) William’s profit is

(2500)(4.3185816e0.05 −max(85− ST , 0))

=11350− (2500)max(85− ST , 0).


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Example 7


(ii) Calculate William’s profit for the following spot prices at expi-ration: 75, 80, 85, 90, 95.


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Example 7


(ii) Calculate William’s profit for the following spot prices at expi-ration: 75, 80, 85, 90, 95.Solution: (ii) William’s profits for the considered spot prices are

if ST = 75, profit = 11350− (2500) max(85− 75, 0) = −13650,

if ST = 80, profit = 11350− (2500) max(85− 80, 0) = −1150,

if ST = 85, profit = 11350− (2500) max(85− 85, 0) = 11350,

if ST = 90, profit = 11350− (2500) max(85− 90, 0) = 11350,

if ST = 95, profit = 11350− (2500) max(85− 95, 0) = 11350.


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Example 7


(iii) Calculate William’s minimum and maximum profits.


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Example 7


(iii) Calculate William’s minimum and maximum profits.Solution: (iii) William’s minimum profit is 11350 − (2500)(85) =−201150. William’s maximum profit is 11350.


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Example 7


(iv) Calculate the spot prices at expiration at which William makesa positive profit.


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Example 7


(iv) Calculate the spot prices at expiration at which William makesa positive profit.Solution: (iv) William makes a positive profit if 11350 −(2500) max(85− ST , 0) > 0, i.e. if ST > 85− 11350

2500 = 80.46.


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A put option is a way to sell an asset in the future. A shortforward is another way to sell an asset in the future.


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Example 8

Rachel enters into a short forward contract for 500 shares of XYZstock for $26.35035 per share. The exercise date is three monthsfrom now. The risk free effective annual interest rate is 5.5%. Thecurrent price of XYZ stock is $26 per share. Dan buys a putoption of 500 shares of XYZ stock with a strike price of $26 pershare. The exercise date is three months from now. The premiumof this put option is $1.377368 per share.


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Example 8


(i) Find Rachel’s and Dan’s profits as a function of the spot priceat expiration.


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Example 8


(i) Find Rachel’s and Dan’s profits as a function of the spot priceat expiration.Solution: (i) The no arbitrage price of XYZ stock is(26)(1.055)0.25 = 26.35035454. Rachel’s profit is

(500)(26.35035− ST ) = 13175.17727− 500ST .


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Example 8


(i) Find Rachel’s and Dan’s profits as a function of the spot priceat expiration.Solution: (i) (continuation) Dan’s profit is

500 max(0,K − ST )− 500Put(K ,T )(1 + i)T

= 500 max(0, 26− ST )− (500)(1.377368)(1.055)0.25

= 500 max(0, 26− ST )− 697.9641.


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Example 8


(ii) Make a table with Rachel’s and Dan’s profits when the spot priceat expiration is $18, $20, $22, $24, $26, $28, $30.


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Example 8


(ii) Make a table with Rachel’s and Dan’s profits when the spot priceat expiration is $18, $20, $22, $24, $26, $28, $30.Solution: (ii)

Rachel’s profit 4175.18 3175.18 2175.18 1175.18 175.18 −824.82 −1824.82Dan’s profit 3302.04 2302.04 1302.04 302.04 −697.96 −697.96 −697.96Spot Price 18 20 22 24 26 28 30


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Example 8


(iii) Calculate Rachel’s and Dan’s minimum and maximum payoffs.


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Example 8


(iii) Calculate Rachel’s and Dan’s minimum and maximum payoffs.Solution: (iii) Rachel’s minimum profit is −∞. Rachel’s maximumprofit is 13175.18. Dan’s minimum profit is −697.96. Dan’s maxi-mum profit is (500)(26)− 697.9641 = 12302.04.


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Example 8


(iv) Calculate the spot price at expiration at which Dan and Rachelmake the same profit.


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Example 8


(iv) Calculate the spot price at expiration at which Dan and Rachelmake the same profit.Solution: (iv) Since the profits are equal for some ST > 26, wesolve −697.9641 = (500)(26.35035−ST ) and get ST = 26.35035+697.9641/500 = 27.74628.


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Example 8


(v) Draw the graphs of the profit versus the spot price at expirationfor Dan and Rachel.


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Example 8


(v) Draw the graphs of the profit versus the spot price at expirationfor Dan and Rachel.Solution: (v) The graphs of (short forward) Rachel’s profit and(purchased put option) Dan’s profit are in Figure 3.


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Figure 3: Profit for short forward and purchased put option.


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A purchased put option reduces losses over a short forward. Theprofit per unit of a short forward contract is F0,T − ST . Theminimum profit for a short forward contract is −∞. The maximumprofit for a short forward contract is F0,T . The profit for the putoption holder is max(K − ST , 0)− Put(K ,T )(1 + i)T . Theminimum of the put option holder’s profit is −Put(K ,T )(1 + i)T .The maximum of the put option holder’s profit isK − Put(K ,T )(1 + i)T . A put option is an insured position in anasset. In return for not having large losses, the possible returns fora put option are smaller than those for a short forward.


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max((1 + i)−T (K − F0,t), 0) ≤ Put(K ,T ) ≤ K (1 + i)−T .

Proof.Consider the portfolio consisting of entering a long forwardcontract and buying a put option. The profit at expiration is

ST − F0,T + max(K − ST , 0)− Put(K ,T )(1 + i)T

=max(K ,ST )− F0,T − (Put(K ,T ) + S0)(1 + i)T .

The maximum profit is ∞. The minimum profit isK − F0,T − (Put(K ,T ) + S0)(1 + i)T . If there exists no arbitrageK − F0,T − Put(K ,T )(1 + i)T < 0. The claim follows from thisbound and the bounds in Theorem 1.


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max((1 + i)−T (K − F0,t), 0) ≤ Put(K ,T ) ≤ K (1 + i)−T .

Proof.Consider the portfolio consisting of entering a long forwardcontract and buying a put option. The profit at expiration is

ST − F0,T + max(K − ST , 0)− Put(K ,T )(1 + i)T

=max(K ,ST )− F0,T − (Put(K ,T ) + S0)(1 + i)T .

The maximum profit is ∞. The minimum profit isK − F0,T − (Put(K ,T ) + S0)(1 + i)T . If there exists no arbitrageK − F0,T − Put(K ,T )(1 + i)T < 0. The claim follows from thisbound and the bounds in Theorem 1.


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Example 9

The current price of a forward of corn is $3.3 per bushel. Theannual effective interest rate is 7.5%. The price of a one–yearEuropean 3.5–strike put option for corn is $0.18 per bushel. Findan arbitrage strategy and its minimum profit per bushel.


Put(K ,T )− ((1 + i)−T (K − F0,t)

=0.18− (1.075)(3.5− 3.3) = −0.035 < 0.

The put premium is too low. Consider the portfolio consisting ofentering into a long forward contract and buying a put option,both for the same nominal amount. The profit is

ST − 3.3 + max(3.5− ST , 0)− (0.18)(1.075)

=max(3.5,ST )− 3.3− (0.18)(1.075)

The minimum profit per share is3.5− 3.3− (0.18)(1.075) = 0.0065.


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Example 9

The current price of a forward of corn is $3.3 per bushel. Theannual effective interest rate is 7.5%. The price of a one–yearEuropean 3.5–strike put option for corn is $0.18 per bushel. Findan arbitrage strategy and its minimum profit per bushel.


Put(K ,T )− ((1 + i)−T (K − F0,t)

=0.18− (1.075)(3.5− 3.3) = −0.035 < 0.

The put premium is too low. Consider the portfolio consisting ofentering into a long forward contract and buying a put option,both for the same nominal amount. The profit is

ST − 3.3 + max(3.5− ST , 0)− (0.18)(1.075)

=max(3.5,ST )− 3.3− (0.18)(1.075)

The minimum profit per share is3.5− 3.3− (0.18)(1.075) = 0.0065.


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Example 10

The payoff of a purchase put option is similar to the one on apolicy insurance of some asset. Suppose that you car is worth$20000. In the case of an accident, the insurance company paysyou max(20000− ST , 0), where ST is the price of the car after theaccident. If you own stock valued at K and buy a put option withstrike price K, the payoff of the put option at expiration time ismax(K − ST , 0). The payoff is precisely the loss in value of thestock.There are minor differences between these two examples. In thecase of the insurance of car, usually a deductible is applied. If thedeductible in your car insurance is $500, the payment by theinsurance company is max(20000− 500− ST , 0). Since cost of anaccident is always positive, ST < 20000.


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Theorem 3If 0 < K1 < K2, then

Put(K1,T ) ≤ Put(K2,T ) ≤ Put(K1,T ) + (K2 − K1)e−rT .


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Proof.We have that

max(K1 − ST , 0)

≤max(K2 − ST , 0) = K2 − K1 + max(K1 − ST ,K1 − K2)

≤K2 − K1 + max(K1 − ST , 0).

In other words,(i) The payoff for a K1–strike put is smaller than or equal to thepayoff for a K2–strike put.(ii) The payoff for a K2–strike put is smaller than or equal to(K2 − K1) plus the payoff for K1–strike put.Hence, if there exist no arbitrage, then

Put(K1,T ) ≤ Put(K2,T ) ≤ Put(K1,T ) + (K2 − K1)e−rT .


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Example 11

The price of a one–year European 3.5–strike put option for corn is$0.18 per bushel. The price of a one–year European 3.75–strike putoption for corn is $0.15 per bushel. The annual effective interestrate is 7.5%. Find an arbitrage strategy and it minimum profit.

Solution: In this case, Put(K1,T ) ≤ Put(K2,T ) does not hold.We can do arbitrage by a buying a 3.75–strike put option andselling 3.5–strike put option, both for the same nominal amount.The profit per share is

max(3.75− ST , 0)−max(3.5− ST , 0)− (0.15− 0.18)(1.075)

≥max(3.75,ST )−max(3.5,ST )− (0.15− 0.18)(1.075)

≥(0.18− 0.15)(1.075) = 0.03225.


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Example 11

The price of a one–year European 3.5–strike put option for corn is$0.18 per bushel. The price of a one–year European 3.75–strike putoption for corn is $0.15 per bushel. The annual effective interestrate is 7.5%. Find an arbitrage strategy and it minimum profit.

Solution: In this case, Put(K1,T ) ≤ Put(K2,T ) does not hold.We can do arbitrage by a buying a 3.75–strike put option andselling 3.5–strike put option, both for the same nominal amount.The profit per share is

max(3.75− ST , 0)−max(3.5− ST , 0)− (0.15− 0.18)(1.075)

≥max(3.75,ST )−max(3.5,ST )− (0.15− 0.18)(1.075)

≥(0.18− 0.15)(1.075) = 0.03225.


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Example 12

Consider two European put options on a stock, both withexpiration date exactly two years from now. One put option hasstrike price $85 and the other one $95. The price of the 85–strikeput is 8. The price of the 95–strike put option is 20. The risk–freeannual rate of interest compounded continuously is 5%. Find anarbitrage portfolio and its minimum profit.

Solution: In this case Put(K2,T ) ≤ Put(K1,T ) + (K2 − K1)e−rT

does not hold. Notice that

Put(K1,T )+(K2−K1)e−rT = 8+(95−85)e−(2)(0.05) = 17.04837418.

We can do arbitrage by buying a 85–strike put option and selling a95–strike put option. The profit per share is

max(85− ST , 0)−max(95− ST , 0) + (20− 8)e(2)(0.05)

=max(85− ST , 0)− 10−max(85− ST ,−10) + 13.26205102

≥− 10 + 13.26205102 = 3.26205102.


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Example 12

Consider two European put options on a stock, both withexpiration date exactly two years from now. One put option hasstrike price $85 and the other one $95. The price of the 85–strikeput is 8. The price of the 95–strike put option is 20. The risk–freeannual rate of interest compounded continuously is 5%. Find anarbitrage portfolio and its minimum profit.

Solution: In this case Put(K2,T ) ≤ Put(K1,T ) + (K2 − K1)e−rT

does not hold. Notice that

Put(K1,T )+(K2−K1)e−rT = 8+(95−85)e−(2)(0.05) = 17.04837418.

We can do arbitrage by buying a 85–strike put option and selling a95–strike put option. The profit per share is

max(85− ST , 0)−max(95− ST , 0) + (20− 8)e(2)(0.05)

=max(85− ST , 0)− 10−max(85− ST ,−10) + 13.26205102

≥− 10 + 13.26205102 = 3.26205102.


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We have the following rules of thump for price of a put option ofan asset:

I Higher asset prices lead to higher put option prices.

I Higher strike prices lead to higher put option prices.

I Higher interest rates lead to higher put option prices.

I Higher expiration time lead to higher put option prices.

I Higher variation of an asset price lead to higher put optionprices.


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Since the put option buyer’s payoff increases as the strike increases,the premium of a put option increases as the strike price increases.Hence, between two put options with different strike prices:(i) The put option with smaller strike price has a smaller price.(ii) If the spot price is low enough, both put options have a positiveprofit. The loss is bigger for put option with the higher strike price.(iii) If the spot price is high enough, both put options have a loss.The profit is bigger for put option with the bigger strike price.


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Since the strike price is paid at the expiration time, as higher theinterest rate is as higher the put option price is. As higher theexpiration time is as higher the put option price is.Usually, the price of a put option is found using the Black—Scholesformula. The Black–Scholes formula for the price of a put option is

Put(K ,T ) = Ke−rTΦ(−d2)− S0e−δTΦ(−d1).


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Table 1 shows the premium of a put option for different strikeprices. We have used the Black–Scholes formula with S0 = 26,t = 0.25, σ = 0.3, δ = 0 and r = ln(1.055).

Table 1:

Spot Price 22 24 26 28 30Premium ofa put option

0.1983495 0.6038701 1.377368 2.546227 4.045966


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Example 13

Use the put premiums from Table 1, i = 5.5% and T = 0.25. Aninvestor buys a put option for 500 shares. Find the profit functionfor the buyer of a put option with the following strike prices: $24,$26, $28.

Solution: Profit = 500max(0,K − ST )− 500Put(K ,T )(1 + i)T .If K = 24, the profit is

500 max(0, 24− ST )− (500)(0.6038701)(1.055)0.25

=500 max(0, 24− ST )− 306.0036775.

If K = 26, the profit is

500 max(0, 26− ST )− (500)(1.377368)(1.055)0.25

=500max(0, 26− ST )− 697.9641.


500 max(0, 28− ST )− (500)(2.546227)(1.055)0.25

=500max(0, 28− ST )− 1290.268926.


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Example 13

Use the put premiums from Table 1, i = 5.5% and T = 0.25. Aninvestor buys a put option for 500 shares. Find the profit functionfor the buyer of a put option with the following strike prices: $24,$26, $28.

Solution: Profit = 500max(0,K − ST )− 500Put(K ,T )(1 + i)T .If K = 24, the profit is

500 max(0, 24− ST )− (500)(0.6038701)(1.055)0.25

=500 max(0, 24− ST )− 306.0036775.


500 max(0, 26− ST )− (500)(1.377368)(1.055)0.25

=500max(0, 26− ST )− 697.9641.


500 max(0, 28− ST )− (500)(2.546227)(1.055)0.25

=500max(0, 28− ST )− 1290.268926.


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When ST is low, the higher the strike price is, the higher the profitis. When ST is high, the higher the strike price is, the higher theloss is.

Figure 4: Profit for three puts.


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I A purchased put option is in–the–money, if ST < K .I A purchased put option is out–the–money, if ST > K .I A purchased put option is at–the–money, if ST = K .


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If K is very small, the put option will almost certainly not beexecuted. Hence, if K is very small, Put(K ,T ) = 0, i.e.lim

K→0+Put(K ,T ) = 0. If K is very large, the put option will almost

certainly be executed. If a put is executed its profit is K − ST .Hence, lim

K→∞Put(K ,T ) =∞. As a function on K , Put(K ,T ) is

an increasing function with limK→0+

Put(K ,T ) = 0 and

limK→∞

Put(K ,T ) =∞.

Figure 5 shows the graph of Put(K ,T ) as a function of K .Put(K ,T ) was found using the Black–Scholes formula withT = 1, S0 = 100, T = 1, σ = 0.25, r = ln(1.06) and δ = 0.0.


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Figure 5: Graph of Put(K ,T ) as a function of K .


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We have the following strategies to speculate on the change of anasset

volatilitywill decrease

no volatility infovolatility

will increaseprice

will decreasesell a call sell asset buy a put

pricewill increase

sell a put buy asset buy a call


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If the price of asset will decrease, we can make a profit by eitherselling a call, or selling asset, or buying a put. If the volatility willdecrease, the chances than option is executed decrease. Hence, ifthe price of asset and volatility will decrease, then the preferredstrategy is to sell a call.By a similar argument, we have that:If the price of asset will decrease and the volatility will increase,then the preferred strategy is to buy a put.If the price of asset will increase and the volatility will decrease,then the preferred strategy is to sell a put.If the price of asset and volatility will increase, then the preferredstrategy is to buy a call.


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Section 7.6. Put–call parity.





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Chapter 7. Derivatives markets. Section 7.6. Put–call parity.

Put–call parity

Recall that the actions and payoffs corresponding to a call/put are:

If ST < K If K < ST

long call no action buy the stock

short call no action sell the stock

long put sell the stock no action

short put buy the stock no action

If ST < K If K < ST

long call 0 ST − K

short call 0 −(ST − K )

long put K − ST 0

short put −(K − ST ) 0


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I If we have a K–strike long call and a K–strike short put, weare able to buy the asset at time T for K . Hence, havingboth a K–strike long call and a K–strike short put isequivalent to have a K–strike long forward contract withprice K .

I Entering into both a K–strike long call and a K–strike shortput is called a synthetic long forward.

I Reciprocally, if we have a K–strike short call and a K–strikelong put, we are able to sell the asset at time T for K .Having both a K–strike short call and a K–strike longput is equivalent to have a short forward contract withprice K .

I Entering into both a K–strike short call and a K–strike longput is called a synthetic short forward.


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The no arbitrage cost at time T of buying an asset using a longforward contract is F0,T . The cost at time T for buying an assetusing a K–strike long call and a K–strike short put is

(Call(K ,T )− Put(K ,T ))erT + K .

If there exists no arbitrage, then:

Theorem 1(Put–call parity formula)

(Call(K ,T )− Put(K ,T ))erT + K = F0,T .

If we use effective interest, the put–call parity formula becomes:

(Call(K ,T )− Put(K ,T ))(1 + i)T + K = F0,T .


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Often, F0,T = S0(1 + i)T . This forward price applies to assetswhich have neither cost nor benefit associated with owning them.In the absence of arbitrage, we have the following relation betweencall and put prices:

Theorem 2(Put–call parity formula) For a stock which does not pay anydividends,

(Call(K ,T )− Put(K ,T ))erT + K = S0erT .


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Proof.Consider the portfolio consisting of buying one share of stock anda K–strike put for one share; selling a K–strike call for one share;and borrowing S0 − Call(K ,T ) + Put(K ,T ). At time T , we havethe following possibilities:1. If ST < K , then the put is exercised and the call is not. Wefinish without stock and with a payoff for the put of K .2. If ST > K , then the call is exercised and the put is not. Wefinish without stock and with a payoff for the call of K .In any case, the payoff of this portfolio is K . Hence, K should beequal to the return in an investment ofS0 + Put(K ,T )− Call(K ,T ) in a zero–coupon bond, i.e.K = (S0 + Put(K ,T )− Call(K ,T ))erT .


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Example 1

The current value of XYZ stock is 75.38 per share. XYZ stockdoes not pay any dividends. The premium of a nine–month80–strike call is 5.737192 per share. The premium of a nine–month80–strike put is 7.482695 per share. Find the annual effective rateof interest.

Solution: The put–call parity formula states that

(Call(K ,T )− Put(K ,T ))(1 + i)T + K = S0(1 + i)T .

So,

(5.737192− 7.482695)(1 + i)3/4 + 80 = 75.38(1 + i)T .

80 = (75.38− (5.737192− 7.482695))(1 + i)3/4 =(77.125503)(1 + i)3/4, and i = 5%.


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Example 1

The current value of XYZ stock is 75.38 per share. XYZ stockdoes not pay any dividends. The premium of a nine–month80–strike call is 5.737192 per share. The premium of a nine–month80–strike put is 7.482695 per share. Find the annual effective rateof interest.



So,

(5.737192− 7.482695)(1 + i)3/4 + 80 = 75.38(1 + i)T .

80 = (75.38− (5.737192− 7.482695))(1 + i)3/4 =(77.125503)(1 + i)3/4, and i = 5%.


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Example 2

The current value of XYZ stock is 85 per share. XYZ stock doesnot pay any dividends. The premium of a six–month K–strike callis 3.329264 per share and the premium of a one year K–strike putis 10.384565 per share. The annual effective rate of interest is6.5%. Find K.



So, (3.329264− 10.384565)(1.065)0.5 + K = 85(1.065)0.5 and

K = (85− 3.329264 + 10.384565)(1.065)0.5 = 95.


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Example 2

The current value of XYZ stock is 85 per share. XYZ stock doesnot pay any dividends. The premium of a six–month K–strike callis 3.329264 per share and the premium of a one year K–strike putis 10.384565 per share. The annual effective rate of interest is6.5%. Find K.



So, (3.329264− 10.384565)(1.065)0.5 + K = 85(1.065)0.5 and

K = (85− 3.329264 + 10.384565)(1.065)0.5 = 95.


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Example 3

XYZ stock does not pay any dividends. The price of a one yearforward for one share of XYZ stock is 47.475. The premium of aone year 55–strike put option of XYZ stock is 9.204838 per share.The annual effective rate of interest is 5.5%. Calculate the price ofa one year 55–strike call option for one share of XYZ stock.



So, (Call(55, 1)− 9.204838)(1.055) + 55 = 47.475 and

Call(55, 1) = 9.204838 + (47.475− 55)(1.055)−1 = 2.072136578.


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Example 3

XYZ stock does not pay any dividends. The price of a one yearforward for one share of XYZ stock is 47.475. The premium of aone year 55–strike put option of XYZ stock is 9.204838 per share.The annual effective rate of interest is 5.5%. Calculate the price ofa one year 55–strike call option for one share of XYZ stock.



So, (Call(55, 1)− 9.204838)(1.055) + 55 = 47.475 and

Call(55, 1) = 9.204838 + (47.475− 55)(1.055)−1 = 2.072136578.


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If prices of put options and call options do not satisfy the put–callparity, it is possible to do arbitrage.

I If(S0 − Call(K ,T ) + Put(K ,T ))erT > K ,

we can make a profit by buying a call option, selling a putoption and shorting stock. The profit of this strategy is

=− K + (S0 − Call(K ,T ) + Put(K ,T ))erT .

I If(S0 − Call(K ,T ) + Put(K ,T ))erT < K ,

we can do arbitrage by selling a call option, buying a putoption and buying stock. At expiration time, we get rid of thestock by satisfying the options and make

K − (S0 − Call(K ,T ) + Put(K ,T ))erT .


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Example 4

XYZ stock trades at $54 per share. XYZ stock does not pay anydividends. The cost of an European call option with strike price$50 and expiration date in three months is $8 per share. The riskfree annual interest rate continuously compounded is 4%.


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Example 4


(i) Find the no–arbitrage price of a European put option with thesame strike price and expiration time.


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Example 4


(i) Find the no–arbitrage price of a European put option with thesame strike price and expiration time.Solution: (i) With continuous interest, the put–call parity formulais

(Call(K ,T )− Put(K ,T ))erT + K = S0erT .

Hence,

(8− Put(50, 0.25))e0.04(0.25) + 50 = 54e0.04(0.25)

and Put(50, 0.25) = 8− 54 + 50e−0.04(0.25) = 3.502491687.


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Example 4


(ii) Suppose that the price of an European put option with the samestrike price and expiration time is $3, find an arbitrage strategy andits profit per share.


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Example 4


(ii) Suppose that the price of an European put option with the samestrike price and expiration time is $3, find an arbitrage strategy andits profit per share.Solution: (ii) A put option for $3 per share is undervalued. An ar-bitrage portfolio consists in selling a call option, buying a put optionand stock and borrowing $−8 + 3 + 54 =$49, with all derivativesfor one share of stock. At redemption time, we sell the stock anduse it to execute the option which will be executed. We also repaidthe loan. The profit is 50− (49)e(0.04)(0.25) = 50− 49.49245819 =0.50754181.


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Example 5

Suppose that the current price of XYZ stock is 31. XYZ stock doesnot give any dividends. The risk free annual effective interest rateis 10%. The price of a three–month 30–strike European call optionis $3. The price of a three–month 30–strike European put option is$2.25. Find an arbitrage opportunity and its profit per share.


(S0 + Put(K ,T )− Call(K ,T ))(1 + i)T

=(31 + 2.25− 3)(1.1)0.25 = 30.97943909 > 30.

We conclude that the put is overpriced relatively to the call. Wecan sell a put, buy a call and short stock. The profit per share is

(2.25− 3 + 31)(1.1)0.25 − 30 = 0.9794390948.

Notice that at expiration time one of the options is executed andwe get back the stock which we sold.


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Example 5

Suppose that the current price of XYZ stock is 31. XYZ stock doesnot give any dividends. The risk free annual effective interest rateis 10%. The price of a three–month 30–strike European call optionis $3. The price of a three–month 30–strike European put option is$2.25. Find an arbitrage opportunity and its profit per share.


(S0 + Put(K ,T )− Call(K ,T ))(1 + i)T

=(31 + 2.25− 3)(1.1)0.25 = 30.97943909 > 30.

We conclude that the put is overpriced relatively to the call. Wecan sell a put, buy a call and short stock. The profit per share is

(2.25− 3 + 31)(1.1)0.25 − 30 = 0.9794390948.

Notice that at expiration time one of the options is executed andwe get back the stock which we sold.


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Synthetic forward.

Definition 1A synthetic long forward is the combination of buying a call andselling a put, both with the same strike price, amount of the assetand expiration date.

The payments to get a synthetic long forward are

I (Call(K ,T )− Put(K ,T )) paid at time zero.

I K paid at time T .

The future value of these payments at time T is

K + (Call(K ,T )− Put(K ,T ))(1 + i)T .

The payment of long forward is F0,T paid at time T .In the absence of arbitrage (put–call parity)

F0,T = K + (Call(K ,T )− Put(K ,T ))(1 + i)T .


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The premium of a synthetic long forward, i.e. the cost of enteringthis position, is

(Call(K ,T )− Put(K ,T )).

I If F0,T = K , the premium of a synthetic long forward is zero.You will buying the asset at the estimated future value of theasset.

I If F0,T > K , the premium of a synthetic long forward ispositive. You will buying the asset lower than the estimatedfuture value of the asset.

I If F0,T < K , the premium of a synthetic long forward isnegative. You will buying the asset higher than the estimatedfuture value of the asset.


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Constructive sale.

An investor owns stock. He would like to sell his stock. But, hedoes not want to report capital gains to the IRS this year. So,instead of selling this stock, he holds the stock, buys a K–strikeput, sells a K–strike call, and borrows K (1 + i)−T . The payoffwhich he gets at time T is

ST + max(K − ST , 0)−max(ST − K , 0)− K

=max(ST ,K )−max(ST ,K ) = 0.

At time zero, the investor gets

Call(K ,T )− Put(K ,T ) + K (1 + i)−T = F0,T (1 + i)−T .

At expiration time, the investor can use the stock to meet theoption which will be executed. Practically, the investor sold hisstock at time zero for F0,T (1 + i)−T . According with current USAtax laws, this is considered a constructive sale. He will have todeclare capital gains when the options are bought.


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Floor.

Suppose that you own some asset. If the asset losses value in thefuture, you lose money. A way to insure this long position is to buya put position. The purchase of a put option is called a floor. Afloor guarantees a minimum sale price of the value of an asset.

I The profit of buying an asset is ST − S0(1 + i)T , which is thesame as the profit of a long forward. The minimum profit ofbuying an asset is −S0(1 + i)T .

I The profit for buying an asset and a put option is

ST − S0(1 + i)T + max(K − ST , 0)− Put(K ,T )(1 + i)T

=max(ST ,K )− (S0 + Put(K ,T ))(1 + i)T .

The minimum profit for buying an asset and a put option is

K − (S0 + Put(K ,T ))(1 + i)T .

We know that −S0(1 + i)T < K − (S0 + Put(K ,T ))(1 + i)T .c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2.

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Here is the graph of the profit for buying an asset and a putoption:

Figure 1: Profit for a long position and a long put.

Notice that this is the graph of the profit of a purchased call.c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2.

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Here is a joint graph for the profits of the strategies: (i) buying anasset, (ii) buying an asset and a put.

Figure 2: Profit for long forward and buying an asset and a put.


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Table 1 was found using the Black–Scholes formula with S0 = 75,t = 1, σ = 0.20, δ = 0, r = log(1.05).

Table 1: Prices of some calls and some puts.

K 65 70 75 80 85

Call(K ,T ) 14.31722 10.75552 7.78971 5.444947 3.680736Put(K ,T ) 1.221977 2.422184 4.218281 6.635423 9.633117


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Example 6

The current price of one share of XYZ stock is $75. Use the Table1 for prices of puts and calls. Steve buys 500 shares of XYZ stock.Nicole buys 500 shares of XYZ stock and a put option on 500shares of XYZ stock with a strike price 65 and an expiration dateone year from now. The premium of a 65–strike put option is$1.221977. The risk free annual effective rate of interest is 5%.


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Example 6

The current price of one share of XYZ stock is $75. Use the Table1 for prices of puts and calls. Steve buys 500 shares of XYZ stock.Nicole buys 500 shares of XYZ stock and a put option on 500shares of XYZ stock with a strike price 65 and an expiration dateone year from now. The premium of a 65–strike put option is$1.221977. The risk free annual effective rate of interest is 5%.(i) Calculate the profits for Steve and Nicole.


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Example 6

The current price of one share of XYZ stock is $75. Use the Table1 for prices of puts and calls. Steve buys 500 shares of XYZ stock.Nicole buys 500 shares of XYZ stock and a put option on 500shares of XYZ stock with a strike price 65 and an expiration dateone year from now. The premium of a 65–strike put option is$1.221977. The risk free annual effective rate of interest is 5%.(i) Calculate the profits for Steve and Nicole.Solution: (i) Steve’s profit is (500)(ST − 75). Nicole’s profit is

(500)(ST − 75 + max(65− ST , 0)− 1.221977(1.05))

=(500)(max(65,ST )− 76.28307585).


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Example 6

The current price of one share of XYZ stock is $75. Use the Table1 for prices of puts and calls. Steve buys 500 shares of XYZ stock.Nicole buys 500 shares of XYZ stock and a put option on 500shares of XYZ stock with a strike price 65 and an expiration dateone year from now. The premium of a 65–strike put option is$1.221977. The risk free annual effective rate of interest is 5%.(ii) Find a table with Steve’s and Nicole’s profits for the followingspot prices at expiration: 60, 65, 70, 75, 80 and 85.


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Example 6

The current price of one share of XYZ stock is $75. Use the Table1 for prices of puts and calls. Steve buys 500 shares of XYZ stock.Nicole buys 500 shares of XYZ stock and a put option on 500shares of XYZ stock with a strike price 65 and an expiration dateone year from now. The premium of a 65–strike put option is$1.221977. The risk free annual effective rate of interest is 5%.(ii) Find a table with Steve’s and Nicole’s profits for the followingspot prices at expiration: 60, 65, 70, 75, 80 and 85.Solution: (ii)

Spot Price 60 65 70 75 80 85

Steve’s prof −7500 −5000 −2500 0 2500 5000

Nicole’s prof −5641.54 −5641.54 −3141.54 −641.54 1858.46 4358.46


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Example 6

The current price of one share of XYZ stock is $75. Use the Table1 for prices of puts and calls. Steve buys 500 shares of XYZ stock.Nicole buys 500 shares of XYZ stock and a put option on 500shares of XYZ stock with a strike price 65 and an expiration dateone year from now. The premium of a 65–strike put option is$1.221977. The risk free annual effective rate of interest is 5%.(iii) Calculate the minimum and maximum profits for Steve andNicole.


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Example 6

The current price of one share of XYZ stock is $75. Use the Table1 for prices of puts and calls. Steve buys 500 shares of XYZ stock.Nicole buys 500 shares of XYZ stock and a put option on 500shares of XYZ stock with a strike price 65 and an expiration dateone year from now. The premium of a 65–strike put option is$1.221977. The risk free annual effective rate of interest is 5%.(iii) Calculate the minimum and maximum profits for Steve andNicole.Solution: (iii) The minimum and maximum of Steve’s profit are(500)(−75) = −37500 and ∞, respectively. The minimum Nicole’sprofit is (500)(65− 76.28307585) = −5641.537925. The maximumNicole’s profit is ∞.


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Example 6

The current price of one share of XYZ stock is $75. Use the Table1 for prices of puts and calls. Steve buys 500 shares of XYZ stock.Nicole buys 500 shares of XYZ stock and a put option on 500shares of XYZ stock with a strike price 65 and an expiration dateone year from now. The premium of a 65–strike put option is$1.221977. The risk free annual effective rate of interest is 5%.(iv) Find spot prices at expiration at which each of Steve makes aprofit. Answer the previous question for Nicole.


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Example 6

The current price of one share of XYZ stock is $75. Use the Table1 for prices of puts and calls. Steve buys 500 shares of XYZ stock.Nicole buys 500 shares of XYZ stock and a put option on 500shares of XYZ stock with a strike price 65 and an expiration dateone year from now. The premium of a 65–strike put option is$1.221977. The risk free annual effective rate of interest is 5%.(iv) Find spot prices at expiration at which each of Steve makes aprofit. Answer the previous question for Nicole.Solution: (iv) Steve’s profit is positive if (500)(ST − 75) > 0,i.e. if ST > 75. Nicole’s profit is positive if (500)(max(65,ST ) −76.28307585) > 0, ie. if ST > 76.28307585.


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Next, we proof that the profit of buying an asset and a put is thesame as the profit of buying a call option. The payoff for buying acall option and a zero–coupon bond which pays the strike price atexpiration date is

max(0,ST − K ) + K = max(ST ,K )

The payoff for buying stock and a put option is

ST + max(K − ST , 0) = max(ST ,K ).

Since the two strategies have the same payoff in the absence ofarbitrage, they have the same profit, i.e.

max(ST ,K )− K − (Call(K ,T ))(1 + i)T

=max(ST ,K )− (S0 + Put(K ,T ))(1 + i)T .

This equation is equivalent to the put–call parity:

(S0 + Put(K ,T ))(1 + i)T = K + Call(K ,T )(1 + i)T .


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Example 7

Michael buys 500 shares of XYZ stock and a 45–strike four–yearput for 500 shares of XYZ stock. Rita buys a 45–strike four–yearcall for 500 shares of XYZ stock and invests P into a zero–couponbond. The annual rate of interest continuously compounded is4.5%. Find P so that Michael and Rita have the same payoff atexpiration.

Solution: Michael’s payoff at expiration is

500(S4 + max(45− S4, 0)) = 500 max(45,S4).

Rita’s payoff at expiration is

500 max(S4 − 45, 0) + Pe(0.045)(4)

=500 max(45,S4)− (500)(45) + Pe(0.045)(4).

Hence, P = (500)(45)e−(0.045)(4) = 18793.57976.


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Example 7

Michael buys 500 shares of XYZ stock and a 45–strike four–yearput for 500 shares of XYZ stock. Rita buys a 45–strike four–yearcall for 500 shares of XYZ stock and invests P into a zero–couponbond. The annual rate of interest continuously compounded is4.5%. Find P so that Michael and Rita have the same payoff atexpiration.

Solution: Michael’s payoff at expiration is

500(S4 + max(45− S4, 0)) = 500 max(45,S4).

Rita’s payoff at expiration is

500 max(S4 − 45, 0) + Pe(0.045)(4)

=500 max(45,S4)− (500)(45) + Pe(0.045)(4).

Hence, P = (500)(45)e−(0.045)(4) = 18793.57976.


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Cap.

If you have an obligation to buy stock in the future, you have ashort position on the stock. You will experience a loss, when theprice of the stock price rises. You can insure a short position bypurchasing a call option. Buying a call option when you are in ashort position is called a cap. The payoff of having a short positionand buying a call option is

−ST + max(ST − K , 0) = max(−ST ,−K ) = −min(ST ,K ).

The payoff of having a purchased put combined with borrowing thestrike price at closing is

max(K − ST , 0)− K = max(−ST ,−K ) = −min(ST ,K ),

which is the same as before.


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Suppose that you have a short position on a stock. Consider thefollowing two strategies:1. Buy a call option. The payoff at expiration is

−ST + max(ST − K , 0) = max(−ST ,−K ).

The profit at expiration is

max(−ST ,−K )− Call(K ,T )(1 + i)T .

2. Buy stock (to cover the short) and a put option. The payoff atexpiration is

max(K − ST , 0) = K + max(−ST ,−K ).

The profit at expiration is

K + max(−ST ,−K )− (S0 + Put(K ,T ))(1 + i)T .

By the put–call parity formula, both strategies have the sameprofit.


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Example 8

Heather has a short position in 1200 shares of stock XYZ. She issupposed to return this stock one year from now. To insure herposition, she buys a $65–strike call. The premium of a $65–strikecall is $14.31722. The current price of one share of XYZ stock is$75. The risk free annual effective rate of interest is 5%.


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Example 8

Heather has a short position in 1200 shares of stock XYZ. She issupposed to return this stock one year from now. To insure herposition, she buys a $65–strike call. The premium of a $65–strikecall is $14.31722. The current price of one share of XYZ stock is$75. The risk free annual effective rate of interest is 5%.(i) Make a table with Heather’s profit when the spot price at expi-ration is $40, $50, $60, $70, $80, $90.


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Example 8

Heather has a short position in 1200 shares of stock XYZ. She issupposed to return this stock one year from now. To insure herposition, she buys a $65–strike call. The premium of a $65–strikecall is $14.31722. The current price of one share of XYZ stock is$75. The risk free annual effective rate of interest is 5%.(i) Make a table with Heather’s profit when the spot price at expi-ration is $40, $50, $60, $70, $80, $90.Solution: (i) Heather’s profit is

(1200)(max(−ST ,−K )− Call(K ,T )(1 + i)T

)=1200(max(−ST ,−65)− 14.31722(1.05)1)

=1200(max(−ST ,−65)− 15.033081200).

Spot Price 40 50 60 70 80 90H’s profit −66039.70 −78039.70 −90039.70 −96039.70 −96039.70 −96039.70


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Example 8

Heather has a short position in 1200 shares of stock XYZ. She issupposed to return this stock one year from now. To insure herposition, she buys a $65–strike call. The premium of a $65–strikecall is $14.31722. The current price of one share of XYZ stock is$75. The risk free annual effective rate of interest is 5%.(ii) Assuming that she does not buy the call, make a table with herprofit when the spot price at expiration is $40, $50, $60, $70, $80,$90.


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Example 8

Heather has a short position in 1200 shares of stock XYZ. She issupposed to return this stock one year from now. To insure herposition, she buys a $65–strike call. The premium of a $65–strikecall is $14.31722. The current price of one share of XYZ stock is$75. The risk free annual effective rate of interest is 5%.(ii) Assuming that she does not buy the call, make a table with herprofit when the spot price at expiration is $40, $50, $60, $70, $80,$90.Solution: (ii) Heather’s profit is −(1200)ST .

Spot Price 40 50 60 70 80 90

H’s profit −48000 −60000 −72000 −84000 −96000 −108000


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Example 8

Heather has a short position in 1200 shares of stock XYZ. She issupposed to return this stock one year from now. To insure herposition, she buys a $65–strike call. The premium of a $65–strikecall is $14.31722. The current price of one share of XYZ stock is$75. The risk free annual effective rate of interest is 5%.(iii) Draw the graphs of the profit versus the spot price at expirationfor the strategies in (i) and in (ii).


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Example 8

Heather has a short position in 1200 shares of stock XYZ. She issupposed to return this stock one year from now. To insure herposition, she buys a $65–strike call. The premium of a $65–strikecall is $14.31722. The current price of one share of XYZ stock is$75. The risk free annual effective rate of interest is 5%.(iii) Draw the graphs of the profit versus the spot price at expirationfor the strategies in (i) and in (ii).Solution: (iii) The graph of profits is on Figure 3.


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Notice that the possible loss for the short position can be arbitrarilylarge. By buying the call Heather has limited her possible losses.

Figure 3: Profit for a long position and a long put.


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Selling calls and puts.

Selling a option when there is a corresponding long position in theunderlying asset is called covered writing or option overwriting.Naked writing occurs when the writer of an option does not havea position in the asset.

I A covered call is achieved by writing a call against a longposition on the stock. An investor holding a long position inan asset may write a call to generate some income from theasset.

I A covered put is achieved by writing a put against a shortposition on the stock.


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An arbitrageur buys and sells call and puts. A way to limit risks inthe sales of options is to cover these positions. He also can matchopposite options. For example,

I A purchase of K–strike call option, a sale of a K–strike putoption and a short forward cancel each other.

I A sale of K–strike call option, a purchase of a K–strike putoption and a long forward cancel each other.


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7.7. Equity linked certificates of deposit.





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Chapter 7. Derivatives markets. Section 7.7. Equity linked certificates of deposit.

Equity linked certificates of deposit

To estimate the evolution of the stock market, stock indexes areused. A stock market index is a listing of stocks and a way toobtain the composite value of its components. The three mostused stock indexes are the (Dow Jones) Dow Jones IndustrialAverage, (S & P 500) Standard & Poor’s 500 index, and theNASDAQ Composite Index. Usually the composite value of anindex is a sort of average of the stocks in the index. However,there are different ways to find this average. Every stock marketindex has its rules to find its composite value.


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The S & P 500 index consisting of 500 stocks of large corporationsselected by Standard & Poor. Standard & Poor is a financialcompany which specializes in providing independent credit ratingand index evaluation. The stocks on the S & P 500 trade in stockmarkets, like the (NYSE) NYSE New York Stock Exchange and(National Association of Securities Dealers Automated Quotationssystem) NASDAQ. The New York Stock Exchange is the largestequities marketplace in the world. NASDAQ is an electronic stockexchange.


4/21


The Dow Jones Industrial Average is obtained ”averaging” thevalue of 30 stocks selected by the Dow Jones & Company. These30 stocks are selected from largest and the most widely held publiccompanies in the USA across a range of industries except fortransport and utilities. Dow Jones & Company publishes the TheWall Street Journal. The Wall Street Journal editors have a lotinput on the selection of the stocks in the Dow Jones IndustrialAverage.The NASDAQ Composite Index consists of all securities listed onNASDAQ. It contains mainly stocks of technology and growthcompanies.Roughly, the difference between the three indexes is on the type ofstocks which they represent. The S & P 500 focuses on alllarge–cap stocks in the market. The Dow Jones Industrial Averagefocuses on a very selected group of large companies. NASDAQfocuses on technology and fast growing companies.


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An (ELCD) equity linked CD (or equity linked note, or equityindexed CD, or market index linked CD) is an FDIC–insuredcertificate of deposit that ties the rate of return to theperformance of a stock index such as the S & P 500 and guaranteea certain payment. Chase Manhattan Bank first introducedELCD’s in 1987. But, now many financial institutions offerELCD’s. Usually, the guarantee payment is the original principal.The investor at expiration also gets a payment depending on theperformance of the stock index. Usually, there exists aparticipation rate r , 0 < r ≤ 1, such that the investor gets the

guarantee payment plus rP max(

STS0− 1, 0

), where P is the

principal invested, ST is the index price at expiration, S0 is thespot price. Hence, usually, the payoff of an ELCD is

P

(1 + r max

(ST

S0− 1, 0

)).


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An investor is attracted to ELCD’s because it has the potential formarket appreciation and diversification without risking capital.Diversification is attained by using market indices, which combineseveral stocks. Usually the return of the capital is FDIC insured.One disadvantage is the possible loss of interest on the investedprincipal. Notice that the smallest payoff which the investor mayget is his invested principal, i.e. he does not get any interest. Theinstrument is appropriate for conservative equity investors or fixedincome investors who desire equity exposure with controlled risk.


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Example 1

Aaron deposits $15000 in an ELCD, which provides 100% principalprotection and pays 80% of the appreciation of the S & P 500three years from now. The index closes at 1500 on the day theELCD is issued.


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Example 1

Aaron deposits $15000 in an ELCD, which provides 100% principalprotection and pays 80% of the appreciation of the S & P 500three years from now. The index closes at 1500 on the day theELCD is issued.(i) Find Aaron’s payoff as a function of S3.


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Example 1

Aaron deposits $15000 in an ELCD, which provides 100% principalprotection and pays 80% of the appreciation of the S & P 500three years from now. The index closes at 1500 on the day theELCD is issued.(i) Find Aaron’s payoff as a function of S3.Solution: (i) Aaron’s payoff is

P

(1 + r max

(ST

S0− 1, 0

))=(15000)

(1 + (0.80) max

(ST

1500− 1, 0

)).


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Example 1

Aaron deposits $15000 in an ELCD, which provides 100% principalprotection and pays 80% of the appreciation of the S & P 500three years from now. The index closes at 1500 on the day theELCD is issued.(ii) Find the Aaron’s payoff in the forward contract if S3 is $1300,$1400, $1500, $1600, $1700, $1800.


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Example 1

Aaron deposits $15000 in an ELCD, which provides 100% principalprotection and pays 80% of the appreciation of the S & P 500three years from now. The index closes at 1500 on the day theELCD is issued.(ii) Find the Aaron’s payoff in the forward contract if S3 is $1300,$1400, $1500, $1600, $1700, $1800.Solution: (ii) Using that Aaron’s payoff is

(15000)

(1 + (0.80) max

(ST

1500− 1, 0

)),

Payoff 15000 15000 15000 15800 16600 17400

S3 1300 1400 1500 1600 1700 1800


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Example 1

Aaron deposits $15000 in an ELCD, which provides 100% principalprotection and pays 80% of the appreciation of the S & P 500three years from now. The index closes at 1500 on the day theELCD is issued.(iii) Graph Aaron’s payoff as a function of S3.


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Example 1

Aaron deposits $15000 in an ELCD, which provides 100% principalprotection and pays 80% of the appreciation of the S & P 500three years from now. The index closes at 1500 on the day theELCD is issued.(iii) Graph Aaron’s payoff as a function of S3.Solution: (iii) Aaron’s payoff is Figure 1.


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Figure 1: Example 1. ELCD payoff.


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The payoff of an ELCD is a combination of the payoff of a long calloption and a long bond position. It is possible to create a syntheticELCD by buying a long call option and a zero–coupon bond.


16/21


The return on a ELCD is

P

(1 + r max

(ST

S0− 1, 0

))= P +

Pr

S0max(ST − S0, 0),

Suppose that an investor buys a zero–coupon bond with face valueP and a S0–strike call option for Pr

S0shares. His payoff at

expiration date is

P +Pr

S0max(ST − S0, 0).

The cost of making this investment is

P(1 + i)−T +Pr

S0Call(S0,T ).

If there exist no arbitrage,

P = P(1 + i)−T +Pr

S0Call(S0,T ).

Hence, r = (1−(1+i)−T )S0

Call(S0,T ) .


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Example 2

The risk–free effective rate of interest is 7%. The current price ofthe S & P 500 is 1500. The price of an European call with strike1500 and expiration date in 3 years is 400. Find the participationrate of a three–year ELCD which provides 100% principalprotection.


r =(1− (1 + i)−T )S0

Call(S0,T )

=(1− (1.07)−3)1500

400= 0.6888829617 = 68.88829617%.


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Example 2

The risk–free effective rate of interest is 7%. The current price ofthe S & P 500 is 1500. The price of an European call with strike1500 and expiration date in 3 years is 400. Find the participationrate of a three–year ELCD which provides 100% principalprotection.


r =(1− (1 + i)−T )S0

Call(S0,T )

=(1− (1.07)−3)1500

400= 0.6888829617 = 68.88829617%.


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Suppose that an ELCD return all the principal invested atexpiration, offers a guaranteed rate of return g , where g < i , and aparticipation rate r on the S & P 500 stock index. The payoff ofthis ELCD is

P(1+g)T+Pr max

(ST − S0

S0, 0

)= P(1+g)T+

Pr

S0max(ST−S0, 0).

We can get this payoff by buying a zero–coupon bond with facevalue P(1 + g)T and a S0–strike call option for Pr

S0shares. The

cost of this portfolio is

P(1 + g)T (1 + i)−T +Pr

S0Call(S0,T ).

If there exist no arbitrage,

P = P(1 + g)T (1 + i)−T +Pr

S0Call(S0,T )

and

r =(1− (1 + g)T (1 + i)−T )S0

Call(S0,T ).


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Example 3

The risk–free effective rate of interest is 5%. The current price ofthe S & P 500 is 1500. The price of an European call with strike1500 and expiration date in six month is 125. Find theparticipation rate of a ELCD which provides 100% principalprotection and a guaranteed annual interest rate of 1.5%.

Solution:

r =(1− (1 + g)T (1 + i)−T )S0

Call(S0,T )

=(1− (1.015)0.5(1.05)−0.5)1500

125= 0.201695 = 20.1695%.


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Example 3

The risk–free effective rate of interest is 5%. The current price ofthe S & P 500 is 1500. The price of an European call with strike1500 and expiration date in six month is 125. Find theparticipation rate of a ELCD which provides 100% principalprotection and a guaranteed annual interest rate of 1.5%.

Solution:

r =(1− (1 + g)T (1 + i)−T )S0

Call(S0,T )

=(1− (1.015)0.5(1.05)−0.5)1500

125= 0.201695 = 20.1695%.


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Section 7.8. Spreads.





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Chapter 7. Derivatives markets. Section 7.8. Spreads.

Spreads

An option spread (or a vertical spread) is a combination of onlycalls or only puts, in which some options are bought and someothers are sold. By buying/selling several call/puts we can createportfolios useful for many different objectives. A ratio spread is acombination of buying m calls at one strike price and selling n callsat a different strike price.


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Speculating on the increase of an asset price. Bull spread.

Definition 1A bull spread consists on buying a K1–strike call and selling aK2–strike call, both with the same expiration date T and nominalamount, where 0 < K1 < K2.

A way to speculate on the increase of an asset price is buying theasset. This position needs a lot of investment. Another way tospeculate on the increase of an asset price is to buy a call option.A bull spread allows to speculate on increase of an asset price bymaking a limited investment.


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The payoff for buying a K1–strike call is max(ST − K1, 0).The payoff for selling a K2–strike call is −max(ST − K2, 0).

6

-��

��

��

��max(ST − K1, 0)

STK1 K2

6

-@

@@

@@

−max(ST − K2, 0)

STK1 K2

Figure 1: Payoff of the calls in a bull spread


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The bull spread payoff is

max(ST − K1, 0)−max(ST − K2, 0)

=max(ST − K1, 0) + K2 − K1 −max(ST − K1,K2 − K1)

=max(ST − K1, 0) + K2 − K1 −max(ST − K1, 0,K2 − K1)

=max(ST − K1, 0) + K2 − K1 −max(max(ST − K1, 0),K2 − K1)

=min(max(ST − K1, 0),K2 − K1)

=

0 if ST < K1,

ST − K1 if K1 ≤ ST < K2,

K2 − K1 if K2 ≤ ST .


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6

-��

��

��

max(ST − K1, 0)−max(ST − K2, 0)

STK1 K2

Figure 2: Payoff of a bull spread


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The profit of a bull spread is

min(max(ST − K1, 0),K2 − K1)

− (Call(K1,T )− Call(K2,T ))(1 + i)T

=

−(Call(K1,T )− Call(K2,T ))(1 + i)T if ST < K1,

ST − K1 − (Call(K1,T )− Call(K2,T ))(1 + i)T if K1 ≤ ST < K2,

K2 − K1 − (Call(K1,T )− Call(K2,T ))(1 + i)T if K2 ≤ ST .

Figure 3 shows a graph of the profit of a bull spread. Notice thatthe profit is positive for values of ST large enough.


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Figure 3: Profit for a bull spread.


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In this section, we will use the values of calls/puts in Table 1.

Table 1: Prices of some calls and some puts.

K 65 70 75 80 85

Call(K ,T ) 14.31722 10.75552 7.78971 5.444947 3.680736Put(K ,T ) 1.221977 2.422184 4.218281 6.635423 9.633117


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Example 1

(Use Table 1) Ronald buys a $70–strike call and sells a $85–strikecall for 100 shares of XYZ stock. Both have expiration date oneyear from now. The current price of one share of XYZ stock is$75. The risk free annual effective rate of interest is 5%. Thepremium per share of $70–strike and $85–strike calls are 10.75552and 3.680736 respectively.


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Example 1


(i) Find the profit at expiration as a function of the strike price.


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Example 1


(i) Find the profit at expiration as a function of the strike price.Solution: (i) Ronald’s profit is

(100) (min(max(ST − 70, 0), 85− 70)− (10.75552− 3.680736)(1.05))

=(100) (min(max(ST − 70, 0), 85− 70)− 7.428523)

=

−(100)(7.428523) if ST < 70,

(100)(ST − 70− 7.428523) if 70 ≤ ST < 85,

(100)(85− 70− 7.428523) if 85 ≤ ST .


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Example 1


(ii) Make a table with Ronald’s profit when the spot price at expi-ration is $65, $70, $75, $80, $85, $90.


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Example 1


(ii) Make a table with Ronald’s profit when the spot price at expi-ration is $65, $70, $75, $80, $85, $90.Solution: (ii)

Spot Price 65 70 75

Ronald’s profit −742.8523 −742.8523 −242.8523

Spot Price 80 85 90

Ronald’s profit 257.1477 757.1477 757.1477


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Example 1


(iii) Draw the graph of Ronald’s profit.


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Example 1


(iii) Draw the graph of Ronald’s profit.Solution: (ii) The graph of the profit is Figure 3.


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Let 0 < K1 < K2. We have that

[long K1 − strike call ] + [short K1 − strike put ]

≡ [buying asset for K1 at time T ],

[long K2 − strike call ] + [short K2 − strike put ]

≡ [buying asset for K2 at time T ].

Since the two investment strategies differ by a constant, they havethe same profit. Hence, so do the following strategies,

[long K1 − strike call ] + [short K2 − strike call ],

[long K1 − strike put ] + [short K2 − strike put ].

In other words, buying a K1–strike call and selling a K2–strike callhas the same profit as buying a K1–strike put and selling aK2–strike put.We can form a bull spread either buying a K1–strike call and sellinga K2–strike call, or buying a K1–strike put and selling a K2–strikeput.


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Example 2

The current price of XYZ stock is $75 per share. The effectiveannual interest rate is 5%. Elizabeth, Daniel and Catherine believethat the price of XYZ stock is going to appreciate significantly inthe next year. Each person has $10000 to invest. The premium ofa one–year 85–strike call option is 3.680736 per share. Thepremium of a one–year 75–strike call option is 7.78971 per share.Elizabeth buys a one–year zero–coupon bond for $10000. She alsoenters into a one–year forward contract on XYZ stock worth equalto the her bond payoff at redemption. Daniel buys a one–year85–strike call option which costs $10000. Catherine buys aone–year 75–strike call option and sells a one–year 85–strike calloption. The nominal amounts on both calls are the same. Thedifference between the cost of the 85–strike call option and the75–strike call option is 10000. Suppose that the stock price atredemption is 90 per share. Calculate the profits and the yieldrates for Elizabeth, Daniel and Catherine. Which one makes abigger profit?


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Solution: The price of a long forward is 75(1.05) = 78.75. So,

Elizabeth long forward is for (10000)(1.05)78.75 = 133.333333 shares.

Elizabeth’s profit is 133.333333(90− 78.75) = 1500. Her yield rateis 1500

10000 = 15%.The nominal amount in Daniel’s call is 10000

3.680736 = 2716.847935shares. Daniel’s profit is 2716.847935(90− 85) = 13584.24.Daniel’s yield of return is 13584.24

10000 = 135.8424%.The nominal amount in Catherine’s calls is

100007.78971−3.680736 = 2433.697561 shares. Catherine’s profit is2433.697561(85− 75) = 24336.97561. Catherine’s yield of returnis 24336.97561

10000 = 243.3697561%.Catherine’s profit is biggest.


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Speculating on the decrease of an asset price. Bear spread.

A bear spread is precisely the opposite of a bull spread. Supposethat you want to speculate on the price of an asset decreasing. Let0 < K1 < K2. Consider selling a K1–strike call and buying aK2–strike call, both with the same expiration date T . The profit is

−min(max(ST−K1, 0),K2−K1)+(Call(K1,T )−Call(K2,T ))(1+i)T

or(Call(K1,T )− Call(K2,T ))(1 + i)T if ST < K1,

K1 − ST + (Call(K1,T )− Call(K2,T ))(1 + i)T if K1 ≤ ST < K2,

K1 − K2 + (Call(K1,T )− Call(K2,T ))(1 + i)T if K2 ≤ ST .

A graph of this profit is in Figure 4. Notice that the profit ispositive for values of ST small enough.


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Figure 4: Profit for a bear spread.


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Example 3

(Use Table 1) Rebecca sells a $65–strike call and buys a $80–strikecall for 1000 shares of XYZ stock. Both have expiration date oneyear from now. The current price of one share of XYZ stock is$75. The risk free annual effective rate of interest is 5%. Thepremium per share of $65–strike and $80–strike calls are 14.31722and 5.444947 respectively.


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Example 3


(i) Find Rebecca’s profit as a function of the spot price at expiration.


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Example 3


(i) Find Rebecca’s profit as a function of the spot price at expiration.Solution: (i) Rebecca’s profit is

− (100)max(ST − 65, 0) + (100) max(ST − 80, 0)

+ (100)(14.31722− 5.444947)(1.05)

=− (100)min(max(ST − 65, 0), 80− 65) + (100)(9.31588665)

=

(100)(9.31588665) if ST < 65

(100)(65− ST + 9.31588665) if 65 ≤ ST < 80

(100)(65− 80 + 9.31588665) if 80 ≤ ST .


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Example 3


(ii) Make a table with Rebecca’s profit when the spot price at expi-ration is $60, $65, $70, $75, $80, $85, $90.


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Example 3


(ii) Make a table with Rebecca’s profit when the spot price at expi-ration is $60, $65, $70, $75, $80, $85, $90.Solution: (ii)

Spot Price 60 65 70 75

Rebecca’s profit 931.59 931.59 431.59 −68.41


Rebecca’s profit −68.41 −568.41 −568.41 −568.41


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Example 3


(iii) Draw the graph of Rebecca’s profit.


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Example 3


(iii) Draw the graph of Rebecca’s profit.Solution: (iii) Figure 4 shows the graph of the profit.


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Collar

A collar is the purchase of a put option at a strike price and thesale of a call option at a higher strike price. Let K1 be the strikeprice of the put option. Let K2 be the strike price of the calloption. Assume that K1 < K2. The profit of this strategy is

max(K1 − ST , 0)−max(ST − K2, 0)

− (Put(K1,T )− Call(K2,T ))(1 + i)T

=

K1 − ST − (Put(K1,T )− Call(K2,T ))(1 + i)T if ST < K1,

−(Put(K1,T )− Call(K2,T ))(1 + i)T if K1 ≤ ST < K2,

K2 − ST − (Put(K1,T )− Call(K2,T ))(1 + i)T if K2 ≤ ST .

A collar can be use to speculate on the decrease of price of anasset.The collar width is the difference between the call strike and theput strike.


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Example 4

(Use Table 1) Toto buys a $65–strike put option and sells a$80–strike call for 100 shares of XYZ stock. Both have expirationdate one year from now. The current price of one share of XYZstock is $75. The risk free annual effective rate of interest is 5%.The premium of a $65–strike put option is 1.221977 per share.The premium of a $80–strike call option is 5.444947 per share.


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Example 4


(i) Find Toto’s profit as a function of the spot price at expiration.


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Example 4


(i) Find Toto’s profit as a function of the spot price at expiration.Solution: (i) Toto’s profit is

(100) (max(65− ST , 0)−max(ST − 80, 0)

−(1.221977− 5.444947)(1.05))

=(100) (max(65− ST , 0)−max(ST − 80, 0) + 4.4341185)

=

(100)(65− ST + 4.4341185) if ST < 65,

100(4.4341185) if 65 ≤ ST < 80,

100(80− ST + 4.4341185) if 80 ≤ ST .


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Example 4


(ii) Make a table with Toto’s profit when the spot price at expirationis $55, $60, $65, $70, $75, $80, $85.


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Example 4


(ii) Make a table with Toto’s profit when the spot price at expirationis $55, $60, $65, $70, $75, $80, $85.Solution: (ii) A table with Toto’s profit is


Toto’s profit 1443.41 943.41 443.41 443.41


Toto’s profit 443.41 443.41 −56.59 −556.59


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Example 4


(iii) Draw the graph of Toto’s profit.


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Example 4


(iii) Draw the graph of Toto’s profit.Solution: (iii) The graph of the profit is on Figure 5.


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Figure 5: Profit for a collar.


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Written collar

A written collar is a reverse collar.


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Collars are used to insure a long position on a stock. This positionis called a collared stock. A collared stock involves buying theindex, buy a K1–strike put option and selling a K2–strike calloption, where K1 < K2. The payoff per share of this strategy is

ST + max(K1 − ST , 0)−max(ST − K2, 0)

=max(K1,ST )−max(ST ,K2) + K2

=max(K1,ST )−max(ST ,K1,K2) + K2

=min(max(K1,ST ),K2)

The profit per share of this portfolio is

min(max(K1,ST ),K2)− (S0 + Put(K1,T )− Call(K2,T ))(1 + i)T

=

K1 − (S0 + Put(K1,T )− Call(K2,T ))(1 + i)T if ST < K1,

ST − (S0 + Put(K1,T )− Call(K2,T ))(1 + i)T if K1 ≤ ST < K2,

K2 − (S0 + Put(K1,T )− Call(K2,T ))(1 + i)T if K2 ≤ ST .


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Example 5

(Use Table 1) Maggie buys 100 shares of XYZ stock, a $65–strikeput option and sells a $80–strike call. Both options are for 100shares of XYZ stock and have expiration date one year from now.The current price of one share of XYZ stock is $75. The risk freeannual effective rate of interest is 5%. The premium per share of a$65–strike put option is 1.221977. The premium per share of a$80–strike call is 5.444947.


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Example 5

(Use Table 1) Maggie buys 100 shares of XYZ stock, a $65–strikeput option and sells a $80–strike call. Both options are for 100shares of XYZ stock and have expiration date one year from now.The current price of one share of XYZ stock is $75. The risk freeannual effective rate of interest is 5%. The premium per share of a$65–strike put option is 1.221977. The premium per share of a$80–strike call is 5.444947.(i) Find Maggie’s profit as a function of the spot price at expiration.


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Example 5

(Use Table 1) Maggie buys 100 shares of XYZ stock, a $65–strikeput option and sells a $80–strike call. Both options are for 100shares of XYZ stock and have expiration date one year from now.The current price of one share of XYZ stock is $75. The risk freeannual effective rate of interest is 5%. The premium per share of a$65–strike put option is 1.221977. The premium per share of a$80–strike call is 5.444947.(i) Find Maggie’s profit as a function of the spot price at expiration.Solution: (i) Maggie’s profit is

(100)(min(max(65,ST ), 80)− (75 + 1.221977− 5.444947)(1.05)1

)=(100) (min(max(65,ST ), 80)− 74.315882)

=

100(65− 74.315882) if ST < 65,

100(ST − 74.315882) if 65 ≤ ST < 80,

100(80− 74.315882) if 80 ≤ ST .


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Example 5

(Use Table 1) Maggie buys 100 shares of XYZ stock, a $65–strikeput option and sells a $80–strike call. Both options are for 100shares of XYZ stock and have expiration date one year from now.The current price of one share of XYZ stock is $75. The risk freeannual effective rate of interest is 5%. The premium per share of a$65–strike put option is 1.221977. The premium per share of a$80–strike call is 5.444947.(ii) Make a table with Maggie’s profit when the spot price at expi-ration is $60, $65, $70, $75, $80, $85.


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Example 5

(Use Table 1) Maggie buys 100 shares of XYZ stock, a $65–strikeput option and sells a $80–strike call. Both options are for 100shares of XYZ stock and have expiration date one year from now.The current price of one share of XYZ stock is $75. The risk freeannual effective rate of interest is 5%. The premium per share of a$65–strike put option is 1.221977. The premium per share of a$80–strike call is 5.444947.(ii) Make a table with Maggie’s profit when the spot price at expi-ration is $60, $65, $70, $75, $80, $85.Solution: (ii) A table with Maggie’s profit for the considered spotprices is

Spot Price 60 65 70

Maggie’s profit −931.59 −931.59 −431.59

Spot Price 75 80 85

Maggie’s profit 68.41 568.41 568.41


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Example 5

(Use Table 1) Maggie buys 100 shares of XYZ stock, a $65–strikeput option and sells a $80–strike call. Both options are for 100shares of XYZ stock and have expiration date one year from now.The current price of one share of XYZ stock is $75. The risk freeannual effective rate of interest is 5%. The premium per share of a$65–strike put option is 1.221977. The premium per share of a$80–strike call is 5.444947.(iii) Draw the graph of Maggie’s profit.


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Example 5

(Use Table 1) Maggie buys 100 shares of XYZ stock, a $65–strikeput option and sells a $80–strike call. Both options are for 100shares of XYZ stock and have expiration date one year from now.The current price of one share of XYZ stock is $75. The risk freeannual effective rate of interest is 5%. The premium per share of a$65–strike put option is 1.221977. The premium per share of a$80–strike call is 5.444947.(iii) Draw the graph of Maggie’s profit.Solution: (iii) The graph of the profit is on Figure 6.


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Figure 6: Profit for a collar plus owning the stock.


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Suppose that you worked five years for Microsoft and have$100000 in stock of this company. You would like to insure thisposition buying a collar with expiration T years from now. You canchoose K1 and K2 so that the combined premium is zero. In thiscase, no matter what the price of the stock T years form now, youwill have $100000 or more. The cost of buying this insurance iszero. Notice that you have not got anything from free, you haveloss interest in the stock. You also can make a collar with premiumzero, by taking K1 = K2 = F0,T . In this case you will receive F0,T

at time T , i.e. you are entering into a synthetic forward.Suppose that a zero–cost collar consists of buying a K1–strike putoption and selling a K2–strike call option, where K1 < K2. Theprofit of a zero–cost collar is

max(K1−ST , 0)−max(ST−K2, 0) =

K1 − ST if ST < K1,

0 if K1 ≤ ST < K2,

K2 − ST if K2 ≤ ST .


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Speculating on volatility. Straddle

A straddle consists of buying a K–strike call and a K–strike putwith the same time to expiration. The payoff of this strategy is

max(K − ST , 0) + max(ST − K , 0)

=max(K − ST , 0)−min(K − ST , 0)

=|ST − K |.

Its profit is

|ST − K | − (Put(K ,T ) + Call(K ,T ))(1 + i)T .

A straddle is used to bet that the volatility of the market is higherthan the market’s assessment of volatility. Notice that the prices ofthe put and call use the market’s assessment of volatility.


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Example 6

(Use Table 1) Pam buys a $80–strike call option and a $80–strikeput for 100 shares of XYZ stock. Both have expiration date oneyear from now. The current price of one share of XYZ stock is$75. The risk free annual effective rate of interest is 5%. Thepremium per share of a $80–strike call option is 5.444947. Thepremium per share of a $80–strike put option is 6.635423.


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Example 6


(i) Calculate Pam’s profit as a function of the spot price at expira-tion.


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Example 6


(i) Calculate Pam’s profit as a function of the spot price at expira-tion.Solution: (i) The future value per share of the cost of entering theoption contracts is

(Call(80,T ) + Put(80,T ))(1 + i)T

=(5.444947 + 6.635423)(1.05) = 12.6843885.

Pam’s profit is

(100)(|ST − 80| − 12.6843885).c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2.

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Example 6


(ii) Make a table with Pam’s profit when the spot price at expirationis $65, $70, $75, $80, $85, $90, $95.


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Example 6


(ii) Make a table with Pam’s profit when the spot price at expirationis $65, $70, $75, $80, $85, $90, $95.Solution: (ii) Pam’s profit table is


Pam’s profit 231.56 −268.44 −768.44 −1268.44


Pam’s profit −1268.44 −768.44 −268.44 231.56


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Example 6


(iii) Draw the graph of Pam’s profit.


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Example 6


(iii) Draw the graph of Pam’s profit.Solution: (iii) The graph of the profit is on Figure 7.


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Example 6


(iv) Find the values of the spot price at expiration at which Pammakes a profit.


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Example 6


(iv) Find the values of the spot price at expiration at which Pammakes a profit.Solution: (iv) Pam makes a profit if (100)(|ST − 80| −12.6843885) > 0, i.e. if |ST − 80| > 12.6843885. This can hap-pen if either ST − 80 < −12.6843885, or ST − 80 > 12.6843885.We have that ST − 80 < −12.6843885 is equivalent to ST <80 − 12.6843885 = 67.3156115. ST − 80 > 12.6843885 is equiv-alent to ST > 92.6843885. Hence, Pam makes a profit if eitherST < 67.3156115 or ST > 92.6843885.


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Figure 7: Profit for a straddle.


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Strangle

A strangle consists of buying a K1–strike put and a K2–strike callwith the same expiration date, where K1 < K2. The profit of thisportfolio is

max(K1 − ST , 0) + max(ST − K2, 0)

− ((Put(K1,T ) + Call(K2,T ))(1 + i)T

=

K1 − ST − ((Put(K1,T ) + Call(K2,T ))(1 + i)T if ST < K1,

−((Put(K1,T ) + Call(K2,T ))(1 + i)T if K1 ≤ ST < K2,

ST − K2 − ((Put(K1,T ) + Call(K2,T ))(1 + i)T . if K2 ≤ ST .


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Example 7

(Use Table 1) Beth buys a $75–strike put option and a $85–strikecall for 100 shares of XYZ stock. Both have expiration date oneyear from now. The current price of one share of XYZ stock is$75. The risk free annual effective rate of interest is 5%. Thepremium per share of a $75–strike put option is 4.218281. Thepremium per share of a $85–strike call option is 3.680736.


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Example 7


(i) Calculate Beth’s profit as a function of ST .


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Example 7


(i) Calculate Beth’s profit as a function of ST .Solution: (i) Beth’s profit is

100(max(75− ST , 0) + max(ST − 85, 0)

− (4.218281 + 3.680736)(1.05))

=100(max(75− ST , 0) + max(ST − 85, 0)− 8.29396785)

=

100(75− ST − 8.29396785) if ST < 75,

100(−8.29396785) if 75 ≤ ST < 85,

100(ST − 85− 8.29396785) if 85 ≤ ST .


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Example 7


(ii) Make a table with Beth’s profit when the spot price at expirationis $50, $60, $70, $80, $90, $100, $110.


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Example 7


(ii) Make a table with Beth’s profit when the spot price at expirationis $50, $60, $70, $80, $90, $100, $110.Solution: (ii)


Beth’s profit 1670.60 670.60 −329.40 −829.40

Spot Price 90 100 110

Beth’s profit −329.40 670.60 1670.60


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Example 7


(iii) Draw the graph of Beth’s profit.


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Example 7


(iii) Draw the graph of Beth’s profit.Solution: (iii) The graph of the profit is Figure 8.


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Figure 8: Profit for a strangle.


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The straddle and the strangle bet in volatility of the market in asimilar way. Suppose that a straddle and a strangle are centeredaround the same strike price. The maximum loss of the strangle issmaller than the maximum loss of the straddle. However, thestrangle needs more volatility to attain a profit. The possible profitof the strangle is smaller than that of the straddle. See Figure 9.


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Figure 9: Profit for a straddle and a strangle.


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Written strangle

A written strangle consists of selling a K1–strike call and aK2–strike put with the same time to expiration, where0 < K1 < K2. A written strangle is a bet on low volatility.


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Speculating on low volatility: Butterfly spread

Given 0 < K1 < K2 < K3, a butterfly spread consists of:(i) selling a K2–strike call and a K2–strike put, all options for thenotional amount.(ii) buying a K1–strike call and a K3–strike put, all options for thenotional amount.The profit per share of this strategy is

max(ST − K1, 0) + max(K3 − ST , 0)

−max(ST − K2, 0)−max(K2 − ST , 0)− FVP,

where

FVP = (Call(K1,T )− Call(K2,T )

−Put(K2,T ) + Put(K3,T )) (1 + i)T .


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The profit per share of a butterfly spread is

max(ST − K1, 0) + max(K3 − ST , 0)

−max(ST − K2, 0)−max(K2 − ST , 0)− FVP

=

K3 − K2 − FVP if ST < K1,

ST − K1 − K2 + K3 − FVP if K1 ≤ ST < K2,

−ST − K1 + K2 + K3 − FVP if K2 ≤ ST < K3,

K2 − K1 − FVP if K3 ≤ ST ,

The graph of this profit is Figure 10


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Figure 10: Profit for a butterfly.


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Example 8

(Use Table 1) Steve buys a 65–strike call and a 85–strike put andsells a 75–strike call and a 75–strike put. All are for 100 shares andhave expiration date one year from now. The current price of oneshare of XYZ stock is $75. The risk free annual effective rate ofinterest is 5%.


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Example 8

(Use Table 1) Steve buys a 65–strike call and a 85–strike put andsells a 75–strike call and a 75–strike put. All are for 100 shares andhave expiration date one year from now. The current price of oneshare of XYZ stock is $75. The risk free annual effective rate ofinterest is 5%.(i) Find Steve’s profit as a function of the spot price at expiration.


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Example 8

(Use Table 1) Steve buys a 65–strike call and a 85–strike put andsells a 75–strike call and a 75–strike put. All are for 100 shares andhave expiration date one year from now. The current price of oneshare of XYZ stock is $75. The risk free annual effective rate ofinterest is 5%.(i) Find Steve’s profit as a function of the spot price at expiration.Solution: (i) We have that

(Call(K1,T )− Call(K2,T )− Put(K2,T ) + Put(K3,T )) (1 + i)T

=(14.31722− 9.633117− 7.78971 + 4.218281)(1.05) = 12.539459.

Steve’s profit is

(100)max(ST − 65, 0) + (100) max(85− ST , 0)

− (100)max(ST − 75, 0)− (100)max(75− ST , 0)− (100)(12.539459)


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Example 8

(Use Table 1) Steve buys a 65–strike call and a 85–strike put andsells a 75–strike call and a 75–strike put. All are for 100 shares andhave expiration date one year from now. The current price of oneshare of XYZ stock is $75. The risk free annual effective rate ofinterest is 5%.(i) Find Steve’s profit as a function of the spot price at expiration.Solution: (i) (continuation)

(100)max(ST − 65, 0) + (100) max(85− ST , 0)

− (100)max(ST − 75, 0)− (100)max(75− ST , 0)− (100)(12.539459)

=

100(10− 12.539459) if ST < 65,

100(ST − 55− 12.539459) if 65 ≤ ST < 75,

100(−ST + 95− 12.539459) if 75 ≤ ST < 85,

100(10− 12.539459) if 85 ≤ ST .


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Example 8

(Use Table 1) Steve buys a 65–strike call and a 85–strike put andsells a 75–strike call and a 75–strike put. All are for 100 shares andhave expiration date one year from now. The current price of oneshare of XYZ stock is $75. The risk free annual effective rate ofinterest is 5%.(ii) Make a table with Steve’s profit when the spot price at expirationis $60, $65, $70, $75, $80, $85, $90.


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Example 8

(Use Table 1) Steve buys a 65–strike call and a 85–strike put andsells a 75–strike call and a 75–strike put. All are for 100 shares andhave expiration date one year from now. The current price of oneshare of XYZ stock is $75. The risk free annual effective rate ofinterest is 5%.(ii) Make a table with Steve’s profit when the spot price at expirationis $60, $65, $70, $75, $80, $85, $90.Solution: (ii) table with Steve’s profit is


Steve’s profit −253.95 −253.95 246.05 746.05


Steve’s profit 746.05 246.05 −253.95 −253.95


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Example 8

(Use Table 1) Steve buys a 65–strike call and a 85–strike put andsells a 75–strike call and a 75–strike put. All are for 100 shares andhave expiration date one year from now. The current price of oneshare of XYZ stock is $75. The risk free annual effective rate ofinterest is 5%.(iii) Draw the graph of Steve’s profit.


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Example 8

(Use Table 1) Steve buys a 65–strike call and a 85–strike put andsells a 75–strike call and a 75–strike put. All are for 100 shares andhave expiration date one year from now. The current price of oneshare of XYZ stock is $75. The risk free annual effective rate ofinterest is 5%.(iii) Draw the graph of Steve’s profit.Solution: (iii) The graph of the profit is Figure 10.


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Asymmetric Butterfly spread.

Given 0 < K1 < K2 < K3 and 0 < λ < 1, a butterfly spreadconsists of: buying λ K1–strike calls, buying (1− λ) K3–strikecalls; and selling one K2–strike call. The profit of this strategy is

(λ) max(ST − K1, 0) + (1− λ) max(ST − K3, 0)−max(ST − K2, 0)

=

0− FVPrem if ST < K1,

λ(ST − K1)− FVPrem if K1 ≤ ST < K2,

−λK1 + K2 − (1− λ)ST − FVPrem if K2 ≤ ST < K3,

−λK1 + K2 − (1− λ)K3 − FVPrem if K3 ≤ ST ,

where

FVPrem = (λCall(K1,T )− Call(K2,T ) + (1− λ)Call(K3,T )) (1+i)T .


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The profit of an symmetric butterfly spread is

(λ) max(ST − K1, 0) + (1− λ) max(ST − K3, 0)−max(ST − K2, 0)

=

0− FVPrem if ST < K1,

λ(ST − K1)− FVPrem if K1 ≤ ST < K2,

−λK1 + K2 − (1− λ)ST − FVPrem if K2 ≤ ST < K3,

−λK1 + K2 − (1− λ)K3 − FVPrem if K3 ≤ ST ,

For an asymmetric butterfly spread, the profit functions increasesin one interval and decreases in another interval. The rate ofincrease is not necessarily equal to the rate of decrease.


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Example 9

(Use Table 1) Karen buys seven 65–strike calls, buys three85–strike calls and sells ten 75–strike calls. Each option involves100 shares of XYZ stock. Both have expiration date one year fromnow. The risk free annual effective rate of interest is 5%.


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Example 9

(Use Table 1) Karen buys seven 65–strike calls, buys three85–strike calls and sells ten 75–strike calls. Each option involves100 shares of XYZ stock. Both have expiration date one year fromnow. The risk free annual effective rate of interest is 5%.(i) Find Karen’s profit as a function of the spot price at expiration.


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Example 9

(Use Table 1) Karen buys seven 65–strike calls, buys three85–strike calls and sells ten 75–strike calls. Each option involves100 shares of XYZ stock. Both have expiration date one year fromnow. The risk free annual effective rate of interest is 5%.(i) Find Karen’s profit as a function of the spot price at expiration.Solution: (i) The future value per share of the cost of entering theoption contracts is

(7)(14.31722)(1.05) + (3)(7.78971)(1.05)− (3.680736)(1.05)

=35.0339184.

Karen’s profit is

(100) ((7) max(ST − 65, 0) + (3) max(ST − 85, 0)

−(10)max(ST − 75, 0)− (35.0339184))


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Example 9

(Use Table 1) Karen buys seven 65–strike calls, buys three85–strike calls and sells ten 75–strike calls. Each option involves100 shares of XYZ stock. Both have expiration date one year fromnow. The risk free annual effective rate of interest is 5%.(i) Find Karen’s profit as a function of the spot price at expiration.Solution: (i) (continuation)

(100) ((7) max(ST − 65, 0) + (3) max(ST − 85, 0)

−(10)max(ST − 75, 0)− (35.0339184))

=

100(−35.0339184) if ST < 65,

100((7)(ST − 65)− 35.0339184) if 65 ≤ ST < 75,

100((7)(ST − 65)− (10)(ST − 75)

−35.0339184) if 75 ≤ ST < 85,

100((7)(ST − 65)− (10)(ST − 75)

+(3)(ST − 85)− 35.0339184) if 85 ≤ ST ,


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Example 9

(Use Table 1) Karen buys seven 65–strike calls, buys three85–strike calls and sells ten 75–strike calls. Each option involves100 shares of XYZ stock. Both have expiration date one year fromnow. The risk free annual effective rate of interest is 5%.(ii) Make a table with Karen’s profit when the spot price at expirationis $60, $65, $70, $75, $80, $85, $90.


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Example 9

(Use Table 1) Karen buys seven 65–strike calls, buys three85–strike calls and sells ten 75–strike calls. Each option involves100 shares of XYZ stock. Both have expiration date one year fromnow. The risk free annual effective rate of interest is 5%.(ii) Make a table with Karen’s profit when the spot price at expirationis $60, $65, $70, $75, $80, $85, $90.Solution: (ii) A table with Karen’s profit is


Karen’s profit −3503.39 −3503.39 −3.39 3496.61


Karen’s profit 3496.61 1996.61 496.61 496.61


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Example 9

(Use Table 1) Karen buys seven 65–strike calls, buys three85–strike calls and sells ten 75–strike calls. Each option involves100 shares of XYZ stock. Both have expiration date one year fromnow. The risk free annual effective rate of interest is 5%.(iii) Draw the graph of Karen’s profit.


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Example 9

(Use Table 1) Karen buys seven 65–strike calls, buys three85–strike calls and sells ten 75–strike calls. Each option involves100 shares of XYZ stock. Both have expiration date one year fromnow. The risk free annual effective rate of interest is 5%.(iii) Draw the graph of Karen’s profit.Solution: (iii) The graph of the profit is on Figure 11.


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Figure 11: Profit for an asymmetric butterfly.


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Box spreads.

A box spread is a combination of options which create a syntheticlong forward at one price and a synthetic short forward at adifferent price. Let K1 be the price of the synthetic long forward.Let K2 be the price of the synthetic short forward. With a boxspread, you are able to buy an asset for K1 at time T and sell it forK2 at time T not matter the spot price at expiration. At time T , apayment of K2 − K1 per share is obtained. A box spread can beobtained from:(i) buy a K1–strike call and sell a K1–strike put.(ii) sell a K2–strike call and buy a K2–strike put.If put–call parity holds, the premium per share to enter theseoption contract is

Call(K1,T )− Put(K1,T )− (Call(K2,T )− Put(K2,T ))

=(K2 − K1)(1 + i)−T .


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I If K1 < K2, a box spread is a way to lend money. Aninvestment of (K2 − K1)(1 + i)−T per share is made at timezero and a return of K2 − K1 per share is obtained at time T .

I If K1 > K2, a box spread is a way to borrow money. A returnof (K1 −K2)(1 + i)−T per share is received at time zero and aloan payment of K1 − K2 per share is made at time T .


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Example 10

Mario needs $50,000 to open a pizzeria. He can borrow at theannual effective rate of interest of 8.5%. Mario also can buy/sellthree–year options on XYZ stock with the following premiums pershare:

Call(K ,T ) 14.42 7.78

Put(K ,T ) 7.37 17.29

K 70 90

Mario buys a 90–strike call and a 70–strike put, and sells a90–strike put and a 70–strike call. All the options are for the samenominal amount. Mario receives a total of $50000 from thesesales. Find Mario’s cost at expiration time to settle these options.Find the annual rate of return that Mario gets on this ”loan”.


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Solution: The price per share of Mario’s portfolio is7.78 + 7.37− 17.29− 14.42 = −16.56. So, the nominal amount ofeach option is 50000

16.56 = 3019.3237. Initially, Mario gets $50000 forentering these option contracts. In three years, Mario buys at $90per share and sells at $70 per share. Hence, Mario pays(3019.3237)(90− 70) = 60386.474 for settling the optioncontracts. Let i be the annual effective rate of interest on this loan.We have that 50000(1 + i)3 = 60386.474 and i = 6.493529844%.


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Manual for SOA Exam FM/CAS Exam 2.Chapter 7. Derivatives markets.Section 7.9. Risk Management.





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Chapter 7. Derivatives markets. Section 7.9. Risk Management.

Risk Management

The main reasons why firms enter derivatives are: to hedge, tospeculate, and to reduce transactions costs. Managers usederivatives taking in account their view of the market. So, aspeculative component is added to each decision.Firms convert inputs, such a labor and raw materials, into goodsand services. A firm makes money if its income exceeds its costs.A change in the price of raw materials could make the firmunprofitable. A firm can use derivatives to alter its risk and protectits profitability. To do this is to do risk management.


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Firms do risk management to attain financial stability. There aremany reasons to avoid large losses. After a large loss, interest rateson loans will be obtained at a higher rate. Large losses can causebankruptcy and distress costs. After a large loss, a company canface low cashflows and difficulty making fixed obligations such aswages and payments to banks and suppliers. This makes morecostly to find employees, debtors and suppliers.Suppose that the profit can be modeled by a random variable X .Let f (X ) be the the profit after taking in account the effects ofthis profit. Because of the reasons before, f can be modeled usinga concave function.


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Example 1

Suppose that the financial total impact of the profit is

f (x) =

1.5x if x ≤ −2,

1.1x if − 2 < x ≤ 0

x if x > 0.

Since

f ′(x) =

1.5 if x ≤ −2,

1.1 if − 2 < x ≤ 0

1 if x > 0.

is a nonincreasing function, f is a concave function.


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Example 2

Suppose that the a company profit before taxes can be modeled bya random variable X . The company pays 35% of its profit in taxes,if the profit is positive. It does not pay any taxes if its profit isnegative. The company’s profit after taxes is f (X ), where

f (x) =

{(0.65)x if x > 0,

x if x ≤ 0.

f (x) is a concave function.


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If the company is able to hedge and attain a constant profit ofE [X ], then after taxes its profit is f (E [X ]). If the company doesnot hedge its profit, its expected profit after taxes is E [f (X )].Since f is a concave function, by the Jensen’s inequality,

E [f (X )] ≤ f (E [X ]).

Theorem 1(Jensen’s inequality) Let X be a random variable. Let f : R→ Rbe a function. Then,(i) If f is convex, then f (E [X ]) ≤ E [f (X )].(ii) If f is concave, then E [f (X )] ≤ f (E [X ]).

By using hedging, the random profit X is changed into a randomprofit with less variation (taking big losses with less probability).The expectation of this new random profit is larger than theexpectation of the original profit.


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Example 3

Hank is a wheat farmer. He will produce 50000 bushels of wheatat the end of one year. The cost of producing this wheat is $5.7per bushel. The price of wheat per bushel in one year will be:

S1 5.5 6.5

Probability 0.5 0.5

A speculator offers short forward contracts for a price equal to 1%less than the expected value of the price of the wheat. It alsooffers a 6–strike put option with a price equal to 1% more than thepresent value of the expected payoff of this put. The annualeffective rate of interest is 5%. Hank pays 30% of its profits ontaxes and has no tax benefits for losses.


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(i) Calculate the prices of the forward contract and the put option.

Solution: (i) The expected price of wheat is

E [S1] = (0.5)(5.5) + (0.5)(6.5) = 6.00/bush.

The price of a forward contract is 6(0.99) = 5.94/bush.The expected payoff of the put is

E [max(6− S1, 0)] = (0.5) max(6− 5.5, 0) + (0.5) max(6− 6.5, 0)

=(0.5)(6− 5.5) + (0.5)(0) = 0.25.

The price of a put contract is(0.25)(1.01)(1.05)−1 = 0.2404761905/bush.


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(i) Calculate the prices of the forward contract and the put option.Solution: (i) The expected price of wheat is

E [S1] = (0.5)(5.5) + (0.5)(6.5) = 6.00/bush.

The price of a forward contract is 6(0.99) = 5.94/bush.The expected payoff of the put is

E [max(6− S1, 0)] = (0.5) max(6− 5.5, 0) + (0.5) max(6− 6.5, 0)

=(0.5)(6− 5.5) + (0.5)(0) = 0.25.

The price of a put contract is(0.25)(1.01)(1.05)−1 = 0.2404761905/bush.


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(ii) Calculate Hank’s expected profit before taxes if (a) he does notbuy any derivative, (b) he enters a short forward contract, (c) hebuys a put option. Which strategy has the biggest expected profitbefore taxes?

Solution: (ii) For an uninsured position, Hank’s profit before taxesis 50000(S1 − 5.7). Hank’s expected profit before taxes is

50000(E [S1]− 5.7) = 50000(6− 5.7) = 15000.

Entering the short forward contract, Hank’s profit before taxes is

50000(S1 − 5.7 + 5.94− S1) = (50000)(5.94− 5.7) = 12000.

Entering the short forward contract, Hank’s expected profit beforetaxes is 12000.


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(ii) Calculate Hank’s expected profit before taxes if (a) he does notbuy any derivative, (b) he enters a short forward contract, (c) hebuys a put option. Which strategy has the biggest expected profitbefore taxes?Solution: (ii) For an uninsured position, Hank’s profit before taxesis 50000(S1 − 5.7). Hank’s expected profit before taxes is

50000(E [S1]− 5.7) = 50000(6− 5.7) = 15000.

Entering the short forward contract, Hank’s profit before taxes is

50000(S1 − 5.7 + 5.94− S1) = (50000)(5.94− 5.7) = 12000.

Entering the short forward contract, Hank’s expected profit beforetaxes is 12000.


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Buying the put option, Hank’s profit before taxes is

50000 (S1 − 5.7 + max(6− S1, 0)− (0.2404761905)(1.05))

=50000 (max(6,S1)− 5.7− (0.2404761905)(1.05))

=50000(max(6,S1)− 5.9525).

Buying the put option, Hank’s expected profit before taxes is

E [50000 (max(6,S1)− 5.9525)]

=50000 (E [max(6,S1)]− 5.9525)

=(50000)((0.5)(6) + (0.5)(6.5)− 5.9525)) = 14875.

The biggest expected profit before taxes is attained for anuninsured position.


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(iii) Calculate Hank’s expected profit after taxes for each of threestrategies in (ii). Which strategy has the biggest expected profitafter taxes?

Solution: (iii) For an uninsured position, Hank’s profit beforetaxes is

50000(S1 − 5.7) =

{50000(5.5− 5.7) if S1 = 5.5,

50000(6.5− 5.7) if S1 = 6.5,

Hank’s profit after taxes is

50000(S1 − 5.7) =

{50000(5.5− 5.7) if S1 = 5.5,

50000(6.5− 5.7)(0.7) if S1 = 6.5,

Hank’s expected profit after taxes is

(0.5)(50000)(5.5− 5.7) + (0.5)(50000)(6.5− 5.7)(0.7) = 9000.

Entering the forward contract, Hank’s expected profit after taxes is12000(0.7) = 8400.


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(iii) Calculate Hank’s expected profit after taxes for each of threestrategies in (ii). Which strategy has the biggest expected profitafter taxes?Solution: (iii) For an uninsured position, Hank’s profit beforetaxes is

50000(S1 − 5.7) =

{50000(5.5− 5.7) if S1 = 5.5,

50000(6.5− 5.7) if S1 = 6.5,


50000(S1 − 5.7) =

{50000(5.5− 5.7) if S1 = 5.5,

50000(6.5− 5.7)(0.7) if S1 = 6.5,


(0.5)(50000)(5.5− 5.7) + (0.5)(50000)(6.5− 5.7)(0.7) = 9000.

Entering the forward contract, Hank’s expected profit after taxes is12000(0.7) = 8400.


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Buying the put option, Hank’s profit before taxes is

=50000 (max(6,S1)− 5.9525)

= =

{50000(6− 5.9525) if S1 = 5.5,

50000(6.5− 5.9525) if S1 = 6.5,.


50000 (max(6,S1)− 5.9525)

=

{50000(6− 5.9525)(0.7) if S1 = 5.5,

50000(6.5− 5.9525)(0.7) if S1 = 6.5,


(0.5)(50000)(6−5.9525)(0.7)+(0.5)(50000)(6.5−5.9525)(0.7) = 10412.5.

The biggest expected profit after taxes is attained when buying theput.


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There are several reasons not to hedge:

I paying transaction costs.

I need expertise to asses costs and benefits of a given strategy.

I need expertise to do accounting and taxes in derivativetransactions.

In the real world, small companies are discouraged to doderivatives because the reasons above. However, large companieshave financial, accounting and legal departments which allow themto take advantage of the opportunities on market derivatives. Thefinancial department of a large company can asses derivatives aswell or better than the market does. Their legal and accountingdepartments allow them to take advantage of the current tax laws.


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A producer perspective on risk management.

(URMC) Utah Red Mountain Company is a copper–miningcompany. It plans to mine and sell 1,000,000 pounds of copperover the next year. Suppose that it sells all the next year’sproduction, precisely one year from today. Suppose that the firmincurs in two types of costs: fixed costs and variable costs. Fixedcosts are assumed whether its mine is open or closed. Variablecosts are assumed only its mine is open. Suppose that the totalfixed costs add to $0.75/lb. and the total variable costs add to$2.25/lb. Let S1 be the price of copper per pound in one year.

I The net income per pound if the mine is open isS1 − 0.75− 2.25 = S1 − 3.00.

I The net income per pound if the mine is closed is −0.75/lb.


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Utah Red Mountain Company would be better to close its mine ifS1 − 3 ≤ −0.75, which is equivalent to S1 ≤ 2.25.URMC would be better keep its mine open if S1 ≥ 2.25. But, if2.25 ≤ S1 ≤ 3, URMC assumes the loss 3− S1. If S1 ≥ 3 andURMC keeps its mine open, its (positive) net income is S1−3. Thefollowing table shows the net income of URMC (see also Figure 1).


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Figure 1: Profit according whether the mine is closed, insured open andopen insured with a forward.


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Suppose that URMC can enter a short forward contract agreeingto sell its copper one year from now. Suppose that F0,1 = 3.5/lb.If Utah Red Mountain Company enters this contract, its profit is$0.5/lb. In this way Utah Red Mountain Company reduces risk.Figure 1 shows the graph of the profit under the three consideredalternatives. For an uninsured position, the possible losses can bevery high. However, by entering the forward, the company has afixed benefit.


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However, URMC would like to benefit if the price of the coppergoes higher than $3.5/lb. It could use options. Suppose that thecontinuous rate of interest is 0.05 and the variability is σ = 0.25.Using the Black–Scholes formula, we have the following table ofpremiums of options:

Table 1:

K 3.3 3.4 3.5 3.6 3.7 3.8 3.9Call(K , T ) 0.42567 0.3762 0.33119 0.29048 0.25386 0.22111 0.19196Put(K , T ) 0.23542 0.28107 0.33119 0.3856 0.44411 0.50648 0.57245


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Utah Red Mountain Company could buy a 3.7–strike put. URMC’sprofit per pound is

S1 − 3 + max(3.7− S1, 0)− (0.44411)e0.05

=max(3.7,S1)− 3.466880007

=

{0.2331199934 if S1 < 3.7,

S1 − 3.466880007 if 3.7 ≤ S1.


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Under this strategy, URMC does not benefit much if the price ofcopper is high. Another strategy is to buy a 3.4–strike put. Theprofit per pound is

S1 − 3 + max(3.4− S1, 0)− (0.28107)e0.05

=max(3.4,S1)− 3.295480767

=

{0.104519233 if S1 < 3.4,

S1 − 3.295480767 if 3.4 ≤ S1.

Under this strategy, the company makes $0.17139924/lb morethan before if the price of copper is over $3.7/oz. However, itsguaranteed profit is 0.104519233, which is smaller than theguaranteed profit under a 3.7–strike put.


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See Figure 2 for the profit under a short forward, a 3.7–strike putand 3.4–strike put.

Figure 2: Profit for forward, 3.4–strike put and 3.7–strike put.


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In other words, buying a put is like buying insurance against smallspot prices. As bigger the strike price as bigger the price of theinsurance. As bigger the strike price as bigger the obtainedpayment when the insurance is needed. A producer company needsto buy a put with a strike large enough strike to cover low spotprices. If the strike put is too large, the company will be wastingmoney in insurance which it does not need.


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Example 4

Utah Red Mountain Company has the following profits:(i) If it does not insure, S1 − 3.00.(ii) If it enters a short forward, 0.5.(iii) If it buys a 3.4–strike put, max(3.4,S1)− 3.295480767.Suppose that an actuary consulting for Utah Red MountainCompany estimates that the price of cooper in one year will beeither 2.6, or 3.6, or 4.6, with respective probabilities 0.36, 0.33and 0.31.(i) Compute the URMC’s expected profit before taxes for each ofthe above strategies. Find the strategy with the biggest expectedprofit before taxes.(ii) The URMC pays a 40% tax rate, and has no tax benefits forlosses. Compute the URMC’s expected profit after taxes for eachof the above strategies. Find the strategy with the biggestexpected profit after taxes.


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Solution: (i)

uninsured profit −0.4 0.6 1.6profit under a short forward 0.5 0.5 0.5profit under a 3.4–strike put 0.104519233 0.304519233 1.304519233ST 2.6 3.6 4.6Probability 0.36 0.33 0.31

The expected profit under an uninsured position is

(−0.40)(0.36) + (0.60)(0.33) + (1.60)(0.31) = 0.55.

The expected profit under a short forward is 0.5. The expectedprofit under a 3.7–strike put is

(0.104519233)(0.36) + (0.304519233)(0.33)(1.304519233)(0.31)

=0.542519233.

The strategy with the biggest expected profit before taxes is theuninsured position.


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Solution: (ii) After taxes, the profits are given by the followingtable:

uninsured profit −0.40 0.36 0.96profit with a forward 0.3 0.3 0.3

profit with a 3.4–strike put 0.0627115398 0.1827115398 0.7827115398Probability 0.36 0.33 0.31

The expected profit after taxes for an uninsured position is

(−0.40)(0.36) + (0.36)(0.33) + (0.96)(0.31) = 0.2724.

The expected profit after taxes under a short forward is(0.6)(0.5) = 0.3. The expected profit after taxes under a a3.4–strike put is (0.6)(0.542519233) = 0.3255115398.


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Another strategy for URMC is to buy a 3.6–strike put and sell a3.9–strike call option. Buying these two options, URMS will sellcooper at min(max(3.6,S1), 3.9). This strategy is called a collar.URMC’s profit per pound is

S1 − 3 + max(3.6− S1, 0)−max(S1 − 3.9, 0)

− (0.3856− 0.19196)e0.05

=− 3 + 3.9 + max(3.6,S1)−max(S1, 3.9)− (0.3856− 0.19196)e0.05

=− 3 + 3.9 + max(3.6,S1)−max(S1, 3.9, 3.6)− (0.3856− 0.19196)e0.05

=min(max(3.6,S1), 3.9)− 3

− (0.3856− 0.19196)e0.05

=min(max(3.6,S1), 3.9)− 3.203568135

=

0.396431865 if S1 < 3.6,

S1 − 3.203568135 if 3.6 ≤ S1 < 3.9,

0.696431865 if 3.9 < ST .


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Under this strategy the profit of the company is very close to thatof a forward contract. But, instead of winning a constant of$0.5/lb. in the forward contract, the company’s profit varies withthe future price of copper, although not much.

Figure 3: Profit for a 3.6–3.9 collar.c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2.

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Another type of strategies are the ones called paylater strategies.By a buying a put, URMC hedges against low prices. But, if theprices are high, its profit is not as high as when it does not hedge.A paylater strategy allows to have the usual profit for high enoughspot prices. It is like the price of the insurance does not need to bepaid. Consider the strategy of buying two 3.6–strike puts andbuying a K–strike call, such that the cost of the portfolio is zero.The cost of two 3.6–strike puts is (2)(0.3856) = 0.7712. Bynumerical methods, we have that the cost of a 4.1763–strike put is0.7712. Hence, K = 4.1763. The profit of this strategy is

S1 − 3 + 2 max(3.6− S1, 0)−max(4.1763− S1, 0)

=− 3 + 2 max(3.6,S1)−max(4.1763,S1)

=

0.0237 if S1 < 3.6,

2S1 − 7.1763 if 3.6 ≤ S1 < 4.1763,

S1 − 3 if 4.1763 < S1,


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Notice that under the previous strategy Company URMC alwayshas a positive profit and if the spot price is bigger than 4.1763, itsprofit is the same as it would not hedge.

Figure 4: Profit for a paylater.


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The buyer’s perspective on risk management.

Toughminum makes fridges. Suppose that:(i) The fixed cost per fridge is $100.(ii) Toughminum sells fridges for $350.(iii) To manufacture a fridge Toughminum need 5 pounds ofaluminum.Let ST be the price of a pound of aluminum at time T . The profitone year from now is

350− (5)(S1)− 100 = 250− 5S1.


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Figure 5 shows the graph of this profit (”uninsured” line).

Figure 5: Profit uninsured, long forward and call.


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Toughminum could faces severe loses if the price of the aluminumgoes very high. To hedge risk, it could enter a long forward. Ifaluminum is selling at $40 a pound in the forward market, theprofit of entering a long forward is

350− (5)(40)− 100 = 50.

Figure 5 shows the graph of this profit (”forward” line).


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Another alternative is to buy a call. Suppose that Toughminumbuys a 35–strike call with an expiration date one year from nowand nominal amount 5 lbs. Assume that σ = 0.35 and r = 0.06.Then, Call(35, 1) = 7.609104. The profit is

250− 5S1 + 5 max(S1 − 35, 0)− 5Call(35, 1)er

=250 + 5 max(−35,−S1)− 5(7.609104)e0.06

=209.6018764− 5 min(35,S1)

=

{209.6018764− 5S1 if S1 < 35,

34.6018764 if 35 ≤ S1.

Figure 5 shows the graph of this profit (”call” line).


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Example 5

Toughminum has the following profits:(i) If it does not insure, 250− 5S1.(ii) If it enters a short forward 50.(iii) If it buys a 35–strike call, 209.6018− 5 min(35,S1).Suppose that an actuary consulting for Toughminum estimatesthat the price of aluminum in one year will be either 30, 35, or 40,or 55, with respective probabilities 0.25, 0.25, 0.25, and 0.25.(i) Compute the Toughminum’s expected profit before taxes foreach of the above strategies. Find the strategy with the biggestexpected profit before taxes.(ii) The Toughminum pays a 35% tax rate, and has no tax benefitsfor losses. Compute the Toughminum’s expected profit after taxesfor each of the above strategies. Find the strategy with the biggestexpected profit after taxes.


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Solution: (i)

uninsured profit 100 75 50 −25

profit under a long forward 50 50 50 50

profit under a 35–strike call 59.6019 34.6019 34.6019 34.6019

ST 30 35 40 55

Probability 0.25 0.25 0.25 0.25

The expected profit under an uninsured position is

(100)(0.25) + (75)(0.25) + (50)(0.25) + (−25)(0.25) = 50.

The expected profit under a long forward is 50. The expectedprofit under a 35–strike call is

(59.6019)(0.25) + (34.6019)(0.25) + (34.6019)(0.25) + (34.6019)(0.25) = 40.8519.

The strategies with the biggest expected payoff are the uninsuredposition and the long forward position.


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Solution: (ii) After taxes, the profits are given by the followingtable:

uninsured profit 65.00 48.75 32.50 −25.00

profit with a long forward 32.5 32.5 32.5 32.5

profit under a 35–strike call 38.7412 22.4912 22.4912 22.4912

ST 30 35 40 55

Probability 0.25 0.25 0.25 0.25

The expected profit for an uninsured position is

(65)(0.25)+(48.75)(0.25)+(32.5)(0.25)+(−25)(0.25) = 30.3125.

The expected profit under a long forward is 32.5. The expectedprofit under a 35–strike call is

(38.7412)(0.25) + (22.4912)(0.25) + (22.4912)(0.25) + (22.4912)(0.25) = 26.5537.

The strategy with the biggest expected payoff is the long forwardposition.


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Toughminum could sell a 30–strike put option and buy a 45–strikecall option. Both with nominal amount 5 lbs. This position iscalled a 30–45 reverse collar. Under this position, Toughminumwill buy aluminum at min(max(30,S1), 45) per lb. The cost of a30–strike put is $1.308289/lb. The cost of a 45–strike call is$3.514166/lb. The profit per ounce is

350− 100− 5S1 − 5 max(30− S1, 0) + 5 max(S1 − 45, 0)

+ 5Put(30, 1)er − 5Call(45, 1)er

=250− 5 max(30,S1)− 5(45) + 5 max(S1, 45)

+ 5(1.308289)e0.06 − 5(3.514166)e0.06

=238.2886− 5 max(30,S1)− 5(45) + 5 max(S1, 45, 30)

=238.2886− 5 min(max(30,S1), 45)

or

profit =

88.2886 if S1 < 30,

238.2886− 5S1 if 30 ≤ S1 < 45,

13.2886 if 45 ≤ S1.c©2009. Miguel A. Arcones. All rights reserved. Manual for SOA Exam FM/CAS Exam 2.

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The graph of this profit in Figure 6.

Figure 6: Profit for a 30–45 reverse collar.


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Importer/exporter’s perspective.

Suppose that Company ABCD imports electronics to Canada. Itgets paid in Canadian dollars. Whenever this company gets apayment, it needs to exchange Canadian dollars into US dollars.Suppose that in two months, ABCD expects to get 100,000Canadian dollars. If the price of a Canadian dollar at time T is ST ,the amount of US dollars the company ABCD will get is 100000

S2/12.

To hedge against possible changes in exchange rates, companyABCD can sell a forward on 100000 Canadian dollars at thecurrent price. It also can buy a put on Canadian dollars.


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Example 6

Suppose that Company ABCD buys a put on (Canadian $’s)CAD100000 with a strike price of (U.S.A. dollar) USD0.85 perCAD for USD0.01 per CAD. Two months later, ABCD receivesCAD100000. At this moment the exchange rate is USD0.845 perCAD.(i) How many US dollars does company ABCD gets in thistransaction?(ii) How many US dollars would company ABCD have gotten inthe exchange if it would not have signed the put?

Solution: (i) Since the strike price is bigger than the current spotprice, the company exercises the put. It gets(100000)(0.85− 0.01) = 84000 US dollars.(i) Company ABCD would have got 100000(0.845) = 84500 USdollars


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Example 6

Suppose that Company ABCD buys a put on (Canadian $’s)CAD100000 with a strike price of (U.S.A. dollar) USD0.85 perCAD for USD0.01 per CAD. Two months later, ABCD receivesCAD100000. At this moment the exchange rate is USD0.845 perCAD.(i) How many US dollars does company ABCD gets in thistransaction?(ii) How many US dollars would company ABCD have gotten inthe exchange if it would not have signed the put?

Solution: (i) Since the strike price is bigger than the current spotprice, the company exercises the put. It gets(100000)(0.85− 0.01) = 84000 US dollars.(i) Company ABCD would have got 100000(0.845) = 84500 USdollars


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Section 7.10. Swaps.





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Chapter 7. Derivatives markets. Section 7.10. Swaps.

Swaps

Definition 1A swap is a contract between two counterparts to exchange twosimilar financial quantities which behave differently.

I The two things exchanged are called the legs of the swap.

I A common type of swap involves a commodity. Anothercommon type of swap is an interest rate swap of a fixedinterest rate in return for receiving an adjustable rate.

I Usually, one leg involves quantities that are known in advance,known as the fixed leg. The other involves quantities that are(uncertain) not known in advance, known as the floating leg.

I Usually, a swap entails the exchange of payments over time.


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LIBOR.

The (London Interbank office rate) LIBOR is the most widelyused reference rate for short term interest rates world–wide. TheLIBOR is published daily the (British Bankers Association) BBA. Itis based on rates that large international banks in London offereach other for inter–bank deposits. Rates are quoted for 1–month,3–month, 6–month and 12–month deposits.


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The following table shows the LIBOR interest rates for a loan indollars during a week on June, 2007:

Date 1–month 3–month 6–month 9–month 12–month

6–18–2007 5.3200% 5.3600% 5.4000% 5.4400% 5.4741%6–19–2007 5.3200% 5.3600% 5.3981% 5.4369% 5.4650%6–20–2007 5.3200% 5.3600% 5.3931% 5.4194% 5.4387%6–21–2007 5.3200% 5.3600% 5.3934% 5.4273% 5.4531%6–22–2007 5.3200% 5.3600% 5.3900% 5.4200% 5.4494%

The previous rates are from http://www.bba.org.uk/bba.


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A LIBOR loan is an adjustable loan on which the interest rate istied to a specified Libor. The interest rate is the most recent valueof the LIBOR plus a margin, subject to any adjustment cap.LIBOR is used in determining the price of interest rate futures,swaps and Eurodollars. The most important financial derivativesrelated to LIBOR are Eurodollar futures. Traded at the ChicagoMercantile Exchange (CME), Eurodollars are US dollars depositedat banks outside the United States, primarily in Europe. Theinterest rate paid on Eurodollars is largely determined by LIBOR.Eurodollar futures provide a way of betting on or hedging againstfuture interest rate changes.


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Structure of interest rates.

Let P(0, t) be the price of a $1–face value zero coupon bondmaturing on date t. Notice that 1

P(0,t) is the interest factor from

time zero to time t, i.e. $1 invested at time 0 accumulates to1

P(0,t) at time t. P(0, t) is the discount factor from time zero totime t. The implied interest factor from time tj−1 to time tj isP(0,tj−1)P(0,tj )

. The implied forward rate from time tj−1 to time tj is

r0(tj−1, tj) =P(0,tj−1)P(0,tj )

− 1. Let sn be the n–year spot rate. Then,

(1 + sn)n = 1

P(0,n) .


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A quantity which appear later is the coupon rate R for a n–yearbond with annual coupons and face value, redemption value andprice all equal to one. The price of this bond is

1 =n∑

j=1

RP(0, j) + P(0, n).

Hence,

R =1− P(0, n)∑n

j=1 P(0, j).


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Example 1

The following table lists prices of zero–coupon $1–face value bondswith their respective maturities:


1 $0.956938

2 $0.907029

3 $0.863838

4 $0.807217

(i) Calculate the 1–year, 2–year, 3–year, and 4–year spot rates ofinterest.(ii) Calculate the 1–year, 2–year, and 3–year forward rates ofinterest.(iii) Calculate the coupon rate R for a j–year bond with annualcoupons whose face value, redemption value and price are all one.


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Solution: (i) Note that the price of j–th bond isP(0, j) = (1 + sj)

−j . Hence, sj = 1P(0,j)1/j − 1. In particular,

s1 =1

P(0, 1)− 1 =

1

0.956938− 1 = 4.499967135%

s2 =1

P(0, 2)1/2− 1 =

1

0.9070291/2− 1 = 5.000027694%

s3 =1

P(0, 3)1/3− 1 =

1

0.8638381/3− 1 = 4.999983734%

s4 =1

P(0, 4)1/4− 1 =

1

0.8072171/4− 1 = 5.499991613%


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Solution: (ii) To get the j − 1 year forward rate fj , we do

fj =(1+sj )

j

(1+sj−1)j−1 − 1 = P(0,j−1)P(0,j) − 1. We get that:

f2 = 0.9569380.907029 − 1 = 5.502470153%,

f3 = 0.9070290.863838 − 1 = 4.999895814%,

f4 = 0.8638380.807217 − 1 = 7.014346824%.


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Solution: (iii) We have that Rj = 1−P(0,j)Pjk=1 P(0,k)

. Hence, we get that:

R1 = 1−0.9569380.956938 − 1 = 4.499978055%,

R2 = 1−0.9070290.956938+0.907029 − 1 = 4.987802896%,

R3 = 1−0.8638380.956938+0.907029+0.863838 − 1 = 4.991632466%,

R4 = 1−0.8072170.956938+0.907029+0.863838+0.807217 − 1 = 5.453516272%.


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Example 2

Suppose the current LIBOR discount factors P(0, tj) are given bythe table below.

LIBORdiscout rates

P(0, tj)0.986923 0.973921 0.961067 0.948242

time in months 3 6 9 12

Calculate the annual nominal interest rate compounded quarterlyfor a loan for the following maturity dates: 3, 6, 9 and 12 months.


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Solution: Let s(4)j be the annual nominal interest rate

compounded quarterly for a loan maturing in 3j months,

j = 1, 2, 3, 4. Then, (1 + s(2)j /4)j = 1

P(0,j/4) . So,

s(4)1 = 4

(1

0.986923− 1

)= 5.300109532%,

s(4)2 = 4

((1

0.973921

)1/2

− 1

)= 5.320086032%,

s(4)3 = 4

((1

0.961067

)1/3

− 1

)= 5.330019497%,

s(4)4 = 4

((1

0.948242

)1/4

− 1

)= 5.350015985%.


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We denote by rt0(t1, t2) to the nonannualized interest rate for theperiod from t1 to t2 using the interest rates at t0, i.e. 1 + rt0(t1, t2)is the interest factor for the period from t1 to t2 using the interestrates at t0. We denote by Pt0(t0, t1) to the price at time t0 of azero–coupon bond with face value $1 and redemption time t1. So,

1 + rt0(t1, t2) =Pt0(t0, t1)

Pt0(t0, t2).

Notice that we abbreviate P0(0, t) = P(0, t).


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Forward rate agreement.

Suppose that a borrower plans to take a loan of $L at time t1,where t1 > 0. He will repaid the loan at time t2, where t2 > t1.The amount of the loan payment depends on the term structure ofinterest rates at time t1. Let rt1(t1, t2) be interest rate from t1 tot2 with respect to the structure of interest rates at time t1, i.e. azero–coupon bond with face value F and redemption time t2 costs

F1+rt1 (t1,t2)

at time t1. To pay the loan, the borrower needs to pay

$L(1 + rt1(t1, t2)) at time t2.


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Since rt1(t1, t2) is unknown at time zero, the borrower does notknow how much it will have to pay for the loan. In order to hedgeagainst increasing interest rates, the borrower can enter into a(FRA) forward rate agreement. A FRA is a financial contract toexchange interest payments for a notional principal on settlementdate for a specified period from start date to maturity date.Usually one of the interest payments is relative to a benchmarksuch as the LIBOR. This is a floating interest rate, which wasdescribed in Subsection 2. The other interest payment is withrespect to a fixed rate of interest. An FRA contract is settled incash. The settlement can be made either at the beginning or atthe end of the considered period, i.e. either at the borrowing timeor at the time of repayment of the loan.


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The two payments involved in an FRA are called legs. Bothpayments are made at time t2. Usually FRA’s arefloating–against–fixed. One leg consists of an interest paymentwith respect to a floating rate. The interest payment of thefloating rate leg is Lrt1(t1, t2). The side making the floating–rateleg payment is called either the floating–rate leg party, or thefloating–rate side, or the floating–rate payer. The interestpayment of the fixed rate leg is LrFRA, where rFRA is an interestrate specified in the contract. The side making the fixed–rate legpayment is called either the fixed–rate leg party, or thefixed–rate side, or the fixed–rate payer.


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If the FRA is settled at the time of the repayment of the loan, wesay that the FRA is settled in arrears.Suppose that the FRA is settled at time t2 (in arrears). The FRAis settled in two different ways:1. If L(rFRA − rt1(t1, t2)) > 0, the fixed–rate side makes a paymentof L(rFRA − rt1(t1, t2)) to the floating–rate side.2. If L(rFRA − rt1(t1, t2)) < 0, the floating–rate side makes apayment of L(rt1(t1, t2)− rFRA) to the fixed–rate side.


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Usually an FRA is mentioned as an exchange of (interestpayments) legs. By interchanging their legs, it is meant that:1. The floating–rate leg party makes a payment of Lrt1(t1, t2) toits counterpart.2. The fixed–rate leg party makes a payment of LrFRA to itscounterpart.The combination of these two payments is: the fixed–rate leg partymakes a payment of L(rFRA − rt1(t1, t2)) to the floating–rate legparty. This means that if L(rFRA − rt1(t1, t2)) is a negativenumber, the floating–rate side makes a payment ofL(rt1(t1, t2)− rFRA) to the fixed–rate side.


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Example 3

Company A pays $75,000 in interest payments at the end of oneyear. Company B pays the then–current LIBOR plus 50 basispoints on a $1,000,000 loan at the end of the year. Suppose thatthe two companies enter into an interest payment swap. Supposethat in one year the current LIBOR rate is 6.45%. Find whichcompany is making a payment at the end of year and its amount.

Solution: Company B’s interest payment is(1000000)(0.0645 + 0.0050) = 69500. To settle the forwardinterest agreement, company A must make a payment of75000− 69500 = 5500 to Company B.


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Example 3

Company A pays $75,000 in interest payments at the end of oneyear. Company B pays the then–current LIBOR plus 50 basispoints on a $1,000,000 loan at the end of the year. Suppose thatthe two companies enter into an interest payment swap. Supposethat in one year the current LIBOR rate is 6.45%. Find whichcompany is making a payment at the end of year and its amount.

Solution: Company B’s interest payment is(1000000)(0.0645 + 0.0050) = 69500. To settle the forwardinterest agreement, company A must make a payment of75000− 69500 = 5500 to Company B.


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The fixed–rate side payment is L(rFRA − rt1(t1, t2)). Assumingthat the fixed–rate side borrows L at time t1 his total interestpayment at time t2 is

Lrt1(t1, t2) + L(rFRA − rt1(t1, t2)) = LrFRA.

A borrower can enter into an FRA as a fixed–rate side to hedgeagainst increasing interest rates.If the FRA is settled at time t1 (at borrowing time), to settle the

FRA the floating–rate side makes a payment ofL(rt1 (t1,t2)−rFRA)

1+rt1 (t1,t2)to

the fixed–rate side. This number could be negative. IfrFRA > rt1(t1, t2), the fixed–rate side makes a (positive) payment

ofL(rFRA−rt1 (t1,t2))

1+rt1 (t1,t2)to the floating–rate side.


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Since the interest factor from time t1 to time t2 is 1 + rt1(t1, t2),the previous payoffs are equivalent to the ones for an FRA paid inarrears. In this case, the fixed–rate side can apply the FRApayment to the principal he borrows. He takes a loan of

L +L(rFRA − rt1(t1, t2))

1 + rt1(t1, t2)=

L(1 + rFRA)

1 + rt1(t1, t2)

at time t1. The principal of the loan at time t2 is

(1 + rt1(t1, t2))L(1 + rFRA)

1 + rt1(t1, t2)= L(1 + rFRA).

Again, it is like the fixed–rate side is able to borrow at the raterFRA.


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Usually the floating–rate side is a market maker. The FRAagreement transfers interest rate risk from the fixed–rate side tothe floating–rate side. In order to hedge this interest rate risk, themarket maker could create a synthetic reverse FRA.


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Suppose that the FRA is settled in arrears. The scalper buys azero–coupon bond maturing at t1 with face value L and short sellsa zero–coupon bond with face value LP(0,t1)

P(0,t2)maturing at t2. The

scalper cashflow at time zero is

LP(0, t1)−LP(0, t1)

P(0, t2)P(0, t2) = 0.

At time t1, the scalper gets L from the first bond, which invests atthe current interest rate. He gets L(1 + rt1(t1, t2)) at time t2 fromthis investment. His total cashflow at time t2 is

L(1 + rt1(t1, t2)) + L(rFRA − rt1(t1, t2))−LP(0, t1)

P(0, t2)

=L(1 + rt1(t1, t2)) + L(rFRA − rt1(t1, t2))− L(1 + r0(t1, t2))

=L(rFRA − r0(t1, t2)).

Hence, the no arbitrage rate of an FRA is r0(t1, t2), which is thecurrent nonannualized interest rate from t1 to t2.


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Example 4

Suppose that the current spot rates are given in the following table(as annual nominal rates convertible semiannually)

spot rate 6% 7.5%

maturity (in months) 6 12

Timothy and David enter into separate forward rate agreements asfixed–rate sides for the period of time between 6 months and 12months. Both FRA’s are for a notional amount $10000. Timothy’sFRA is settled in 12 months. David’s FRA is settled in 6 months.In six months, the annual nominal interest rate compoundedsemiannually for a six month loan is 7%.(i) Find the no arbitrage six month rate for an FRA for the periodof time between 6 months and 12 months.(ii) Calculate Timothy’s payoff from his FRA.(iii) Calculate David’s payoff from his FRA.


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Solution: (i) The no arbitrage six month rate for a FRA for theperiod of time between 6 months and 12 months is

rFRA =

(1 + 0.075

2

)21 + 0.06

2

− 1 = 4.5054612%.

(ii) Timothy’s payoff is

(10000)(0.035− 4.5054612) = 447.04612

(iii) David’s payoff is

(10000)(0.035− 4.5054612)

1 + 0.072 )

= 431.9286184.


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Interest rate swaps.

An interest rate swap is a contract in which one party exchangesa stream of interest payments for another party’s stream. Interestrate swaps are normally ”fixed–against–floating”. Interest rateswaps are valued using a notional amount. This nominal amountcan change with time. We only consider constant nominalamounts. The fixed stream of payments are computed with respectto a rate determined by the contract. The floating stream ofpayments are determined using a benchmark, such as the LIBOR.


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Suppose that a firm is interested in borrowing a large amount ofmoney for a long time. One way to borrow is to issue bonds.Unless its credit rating is good enough, the firm may have troublefinding buyers. Lenders are unwilling to absorb long term loansfrom a firm with a so and so credit rating. So, the firm may haveto borrow short term. Even if a company does not need to borrowshort term, usually short term interest rates are lower than longterm interest rates. As longer the maturity as more likely thedefault. Anyhow, suppose that a firm is interested in borrowingshort term, but needs the cash long term. The firm takes a shortterm loan. At maturity, the firm pays this loan and takes anothershort term loan. This process will be repeated as many times asneeded. Current short term rates are known. But, the short terminterest rates which the firm may need to take in the future areuncertain. The firm has an interest rate risk. If short term ratesincrease, the company may get busted. To hedge this risk, the firmmay enter into an interest rate swap, which we describe next.


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Suppose that a borrower takes a loan of L paying a floatinginterest rate according a benchmark such as the LIBOR. Supposethat the interest is paid at times t1 < t2 < · · · < tn. The principalowed after each payment is L. This means that at time tj , theborrower pays Lrtj−1(tj−1, tj) in interest, where 1 + rtj−1(tj−1, tj) isthe interest factor from time tj−1 to time tj , calculated using theLIBOR at time tj−1. This rate is the rate which the borrowerwould pay, if he borrows at time tj−1, pays this loan at time tj andtakes a new loan for L at time tj . The borrower is paying a streamof floating interest rate payments. The borrower can hedge bytaking several FRA’s. If the borrower would like to have a singlecontract, he enters into a interest rate swap.


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The borrower would like to enter into an interest rate swap so thatthe current interest payments plus the payments to the swap willadd to a fixed payment. This situation is similar to that of havingseveral FRA’s. The borrower can enter an interest rate swap withnotional amount L. The borrower would like to have a fixed–rateleg in its contract:

Payment LR LR · · · LR


where R is the swap interest rate in the contract. The borrowerwould like that its counterpart has a floating–rate leg:

Payment Lr0(0, t1) Lrt1(t1, t2) · · · Lrtn−1(tn−1, tn)



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An interest rate swap consists of an interchange of interestpayments. The total outcome of this interchange is that at everytime tj the floating–leg side makes a payment ofL(rtj−1(tj−1, tj)− R) to the fixed–leg side. AgainL(rtj−1(tj−1, tj)− R) could be negative. IfL(rtj−1(tj−1, tj)− R) < 0, the fixed–leg side makes a payment ateach time tj of L(R − rtj−1(tj−1, tj)) to the floating–leg side.


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By exchanging legs, the borrower makes a payment at each time tjwhen L(R − rtj−1(tj−1, tj)) to his counterpart. The borrower is alsomaking interest payments of Lrtj−1(tj−1, tj). The total borrower’sinterest payments add to LR. By entering a swap, a borrower ishedging against increasing interest rates.The total borrower’s cashflow is that of a company issuing bonds.Often the borrower has poor credit rating and it is unable to issuebonds. In some sense, some borrowers enter into an interest rateswap so that its counterpart issues a ”bond” to them. Sometimesthe borrower uses a interest rate swap to avoid to issue a fixed ratelong term loan. Takers of this loan could require a higher interestto borrow.


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Usually, the borrower’s counterpart is a market–maker, which musthedge its interest rate risk. The (market–maker) fixed–rate payergets a payment of L(rtj−1(tj−1, tj)− R) at each time tj . Thefixed–rate payer profit by entering the swap is

n∑j=1

P(0, tj)L(rtj−1(tj−1, tj)− R).


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By using bonds, a market–maker can create a synthetic cashflow ofpayments equal to the swap payments. The cost of these bonds isits present value according with the current term structure ofinterest rates. Hence, if there is no arbitrage, a market–maker canarrange so that the cost of payments he receives is

n∑j=1

LP(0, tj)(r0(tj−1, tj)− R),

i.e. instead of using the uncertain rates rtj−1(tj−1, tj), the scalpercan use the current forward rates. Therefore, the no arbitrage swaprate is

R =

∑nj=1 P(0, tj)r0(tj−1, tj)∑n

j=1 P(0, tj).

The swap rate R when there is no arbitrage is called the par swaprate. Notice that the par swap rate R is a weighted average ofimplied forward rates r0(tj−1, tj). The weights depend on thepresent value of a payment made at time tj .


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Using thatP(0,tj−1)P(0,tj )

= 1 + r0(tj−1, tj), we get that

R =

∑nj=1 P(0, tj)

(P(0,tj−1)P(0,tj )

− 1)

∑nj=1 P(0, tj)

=

∑nj=1 P(0, tj−1)−

∑ni=1 P(0, tj)∑n

j=1 P(0, tj)=

1− P(0, tn)∑nj=1 P(0, tj)

.

Notice that R is the coupon rate for a bond with price, face valueand redemption all equal, using the current term structure ofinterest rates. It is like that the floating–rate party enters the swapto use the market maker credit rating to issue a bond.


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Example 5

Suppose the LIBOR discount factors P(0, tj) are given in the tablebelow. Consider a 3–year swap with semiannual payments whosefloating payments are found using the LIBOR rate compiled asemester before the payment is made. The notional amount of theswap is 10000.

LIBORdiscount

ratesP(0, tj)

0.9748 0.9492 0.9227 0.8960 0.8687 0.8413

time (months) 6 12 18 24 30 36

(i) Calculate the par swap rate.(ii) Calculate net payment made by the fixed–rate side in 18months if the six–month LIBOR interest rate compiled in 12months is 2.3%.


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Solution: (i) The par swap rate is

R = 1−P(0,tn)Pnj=1 P(0,tj )

= 1−0.84130.9748+0.9492+0.9227+0.8960+0.8687+0.8413

= 0.02910484714 = 2.910484714%.

(ii) The payment made by the fixed–rate side is

L(R−rtj−1(tj−1, tj)) = 10000(0.02910484714−0.023) = 61.0484714.


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Notice that the floating interest payment use the interest ratescompiled one period before the payment. These interest rates arecalled realized interest rates.Since interest rates change daily, we may be interested in themarket value of a swap contract. One of the parties in the swapcontract may sell/buy his position in the contract. The marketvalue of a swap contract for the fixed–rate payer is the presentvalue of the no arbitrage estimation of the payments which he willreceive. The market value of a swap contract for the fixed–ratepayer immediately after the k–the payment is

n∑j=k+1

P(tk , tj)L(rtk (tj−1, tj)− R).

If this value is positive, the fixed–rate payer has exposure tointerest rates. Current interest rates are higher than when the swapwas issued. The market value of a swap is the no arbitrage price toenter this contract. One of the counterparts in the contract maybe interested in selling/buying his position on the contract.


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Example 6

Suppose current LIBOR discount factors P(0, tj) are given by thetable below. An interest rate swap has 6 payments left. The swaprate is 3.5% per period. The notional principal is two milliondollars. The floating payments of this swap are the realized LIBORinterest rates.

LIBORdiscount

ratesP(0, tj)

0.9748 0.9492 0.9227 0.8960 0.8687 0.8413

time (months) 6 12 18 24 30 36

Calculate market value of this swap for the fixed–rate payer.


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Solution: The market value of the swap per dollar is

n∑j=k+1

P(0, tj)(rtk (tj−1, tj)− R)

=0.9748

(1

0.9748− 1− 0.035

)+ 0.9492

(0.9748

0.9492− 1− 0.035

)+ 0.9227

(0.9492

0.9227− 1− 0.035

)+ 0.8960

(0.9227

0.8960− 1− 0.035

)+ 0.8687

(0.8960

0.8687− 1− 0.035

)+ 0.8413

(0.8687

0.8413− 1− 0.035

)=0.9748 (0.02585145671− 0.035) + 0.9492 (0.02697008007− 0.035)

+ 0.9227 (0.02872006069− 0.035) + 0.8960 (0.02979910714− 0.035)

+ 0.8687 (0.03142626914− 0.035) + 0.8413 (0.03256864377− 0.035)

=− 0.0321445.

The market value of the swap is(2000000)(−0.0321445) = −64289.


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A deferred swap is a swap which begins in k periods. The swappar rate is computed as

R =

∑nj=k P(0, tj)r0(tj−1, tj)∑n

j=k P(0, tj)=

∑nj=k P(0, tj)

(P(0,tj−1)P(0,tj )

− 1)

∑nj=k P(0, tj)

=

∑nj=k P(0, tj−1)−

∑nj=k P(0, tj)∑n

j=k P(0, tj)=

P(0, tk−1)− P(0, tn)∑nj=k P(0, tj)

.


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Example 7

Suppose the current annual nominal interest rates compoundedquarterly from the LIBOR are given by the table below. Twocounterparts enter into a fixed against floating swap using theLIBOR rate compiled a quarter before the payment is made. Thenotional principal is $50000. The times of the swap are in 12, 15and 18 months.

i (4) 4.5% 4.55% 4.55% 4.6% 4.6% 4.65%

maturation timein months

3 6 9 12 15 18

(i) Calculate the par swap rate.(ii) Calculate the payment made by the fixed–rate party in 18months if in 15 months the spot annual nominal interest ratecompounded quarterly is 4.65%.


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Solution: (i) The par swap rate is

R =P(0, tk−1)− P(0, tn)∑n

j=k P(0, tj)

=(1 + 0.0455/4)−3 − (1 + 0.0465/4)−6

(1 + 0.046/4)−4 + (1 + 0.046/4)−5 + (1 + 0.0465/4)−6

=0.01187359454 = 1.187359454%.

(ii) The payment made by the fixed–rate party is

L(R−rtj−1(tj−1, tj)) = (50000)(0.01187359454−0.0465/4) = 12.429727.


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The par swap rate R = 1−P(0,tn)Pnj=1 P(0,tj )

is a weighted average of implied

forward rates. If the current interest rate does depend on thematuring time, then P(0, t) = (1 + i)−t , for some constant i > 0.In this case,

R =1− P(0, tn)∑n

j=1 P(0, tj)=

1− (1 + i)−tn∑nj=1(1 + i)−tj

.

If the periods in the swap have the same length, then tj = jh,1 ≤ j ≤ n, for some h > 0, and

R =1− (1 + i)−nh∑n

j=1(1 + i)−nj=

1− (1 + i)−nh

(1+i)−h−(1+i)−(n+1)h

1−(1+i)−h

=1

(1+i)−h

1−(1+i)−h

=1− (1 + i)−h

(1 + i)−h= (1 + i)h − 1.

(1 + i)h − 1 is the effective rate for a period of length h. Noticethat the assumption P(0, t) = (1 + i)−t , for some constat i > 0,almost never happens.


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Example 8

Suppose the current annual nominal interest rates compoundedquarterly from the LIBOR is 5.4% independently of the maturity ofthe loan. Two counterparts enter into a fixed against floating swapusing realized LIBOR rates. The times of the swap are in 3, 6 and12 months. Calculate the par swap rate.

Solution: The par swap rate is

R =1− P(0, tn)∑n

j=1 P(0, tj)

=1− (1 + 0.054/4)−4

(1 + 0.054/4)−1 + (1 + 0.054/4)−2 + (1 + 0.054/4)−4

=0.017959328 = 1.7959328%.


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Example 8

Suppose the current annual nominal interest rates compoundedquarterly from the LIBOR is 5.4% independently of the maturity ofthe loan. Two counterparts enter into a fixed against floating swapusing realized LIBOR rates. The times of the swap are in 3, 6 and12 months. Calculate the par swap rate.

Solution: The par swap rate is

R =1− P(0, tn)∑n

j=1 P(0, tj)

=1− (1 + 0.054/4)−4

(1 + 0.054/4)−1 + (1 + 0.054/4)−2 + (1 + 0.054/4)−4

=0.017959328 = 1.7959328%.


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Notice that in the previous problem the swap periods do not havethe same length. Even if annual interest rate is constant over time,the interest payments vary over time. If the annual interest rateremains constant over time, the floating–rate payments are

L ((1 + i)t1 − 1) L ((1 + i)t2−t1 − 1) · · · L ((1 + i)tn−tn−1 − 1)

t1 t2 · · · tn

Suppose that we take a loan of L at time zero. We make paymentsof LR at tj , for 1 ≤ j ≤ n. The par swap rate R is the constantperiodic rate such that the final outstanding in this loan is L.Notice that the present value of the payments is LR

∑nj=1 P(0, tj).

If the final principal is L =L−LR

Pnj=1 P(0,tj )

P(0,tn), then, R = 1−P(0,tn)Pn

j=1 P(0,tj ).


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Commodity swaps.

A commodity swap is a swap where one of the legs is acommodity and the other one is cash. Hence, there are twocounterparts in a swap: a party with a commodity leg and anotherparty with a cash leg. Since the spot price of a commodity changesover time, the commodity leg is floating. Certain amount of acommodity is delivered at certain times. In some sense is like tocombine several forward contracts. But, swaps are valuatedconsidering the total deliveries. Hence, changes in interest rateschange the value of a swap. Usually, the swap payment for thiscommodity is constant.The commodity leg party is called the short swap side. The cashleg party is called the long swap side.


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Suppose that a party would like to sell a commodity at times0 < t1 < t2 < · · · < tn. Suppose that the nominal amounts of thecommodity are Q1,Q2, · · · ,Qn, respectively. The commodity leg is

Amount of delivered commodity Q1 Q2 · · · Qn


Usually the cash leg is either

Payments C0 0 0 · · · 0

Time 0 t1 t2 · · · tn

or

Payments C1 C2 · · · Cn



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If two parties enter into a commodity swap, one will be designedthe party with the commodity leg and the other the party with thecash leg. The party with the commodity leg will deliver commodityto its counterpart according with the table above. The party withthe cash leg will pay cash payments to its counterpart. From apractical point of view, the party with the commodity leg is aseller. The party with the cash leg is a buyer. A commodity swapcontract needs to specify the type and quality of the commodity,how to settle the contract, etc. A swap can be settled either byphysical settlement or by cash settlement. If a swap is settledphysically, the commodity leg side delivers the stipulated notionalamount to the cash leg side, and the cash leg side pays to thecommodity leg side. If a swap is cash settled, one of the partieswill make a payment to the other party.


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Suppose that the current forward price of this commodity withdelivery in T years is F0,T . Let P(0,T ) be the price of azero–coupon with face value $1 and expiration time T . Then, thepresent value of the commodity delivered is

n∑j=1

P(0, tj)QjF0,tj .

In a prepaid swap the buyer makes a unique payment at timezero. If there exists no arbitrage, the price of a prepaid swap is∑n

j=1 P(0, tj)QjF0,tj .


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Usually the cash leg consists of series of payments made at thetimes when the commodity is delivered. Usually each swappayment per unit of commodity is a fixed amount. Hence, thecashflow of payments is

Payment Q1R Q2R · · · QnR

Time t1 t2 · · · t2

where R is the swap price per unit of commodity. The presentvalue of the cashflow of payments is

∑nj=1 P(0, tj)QjR. The no

arbitrage price of a swap per unit of commodity is

R =

∑nj=1 P(0, tj)QjF0,tj∑n

j=1 P(0, tj)Qj.

R is a weighted average of forward prices.


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Suppose that at time of delivery, the buyer pays a level payment ofR at each of the delivery times. Then, the present value of thecashflow of payments is

∑nj=1 P(0, tj)R. The no arbitrage level

payment is

R =


j=1 P(0, tj).

A commodity swap allows to lock the price of a sale. It can be usedby a producer of a commodity to hedge by fixing the price that hewill get in the future for this commodity. It also can be used by amanufacturer to hedge by fixing the price that he will pay in thefuture for a commodity. A commodity swap can be used instead ofseveral futures/forwards. Since the price of a swap involves all thedeliveries, a commodity swap involves loaning/lending somehow.


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Example 9

Suppose that an airline company must buy 10,000 barrels of oilevery six months, for 2 years, starting 6 months from now. Thecompany enters into a long oil swap contract to buy this oil. Thecash leg swap consists of four level payments made at the deliverytimes. The following table shows the annual nominal rateconvertible semiannually of zero–coupon bonds maturing in6, 12, 18, 24 months and the forward price of oil with delivery atthose times.

F0,T $50 $55 $55 $60

annual nominal rate 4.5% 5% 5% 5.5%

expiration in months 6 12 18 24


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(i) Find the no arbitrage price of a prepaid swap with the commodityleg which the airline needs.


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(i) Find the no arbitrage price of a prepaid swap with the commodityleg which the airline needs.Solution: (i) The present value of the cost of oil is

n∑j=1

QjF0,tj P(0, tj)

=(10000)50

1 + 0.0452

+ (10000)55(

1 + 0.052

)2 + (10000)55(

1 + 0.052

)3+ (10000)

60(1 + 0.055

2

)4=488997.555 + 523497.9179 + 510729.676 + 538299.4402 = 2061524.589.


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(ii) Suppose that the airline company pays the swap by a level pay-ment of R at each of the delivery times. Calculate R.


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(ii) Suppose that the airline company pays the swap by a level pay-ment of R at each of the delivery times. Calculate R.Solution: (ii) We have that∑n

j=1 P(0, tj)

= 11+ 0.045

2

+ 1

(1+ 0.052 )

2 + 1

(1+ 0.052 )

3 + 1

(1+ 0.0552 )

4

= 0.97799511 + 0.9518143962 + 0.9285994109 + 0.8971657337= 3.755574651

and

R =

∑nj=1 QjF0,tj P(0, tj)∑n

j=1 P(0, tj)=

2061524.589

37555.74651= 548923.8747.


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(iii) Suppose that the airline company pays the swap by unique priceper barrel. Calculate the price per barrel.


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(iii) Suppose that the airline company pays the swap by unique priceper barrel. Calculate the price per barrel.Solution: (iii) The price of the swap per barrel is 548923.8747

10000 =54.89238747.


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Example 10

Suppose that an airline company must buy 10,000 barrels of oil intwo months, 12,000 barrels of oil in four months and 15,000barrels of oil in six months. The company enters into a long oilswap. The payment of the swap will be made at the delivery times.The following table shows the annual nominal rate convertiblemonthly of zero–coupon bonds maturing in 6, 12, 18, 24 monthsand the forward price of oil with delivery at those times.

Barrels of oil $10000 $12000 $15000

F0,T $55 $56 $58

annual nominal rate 4.5% 4.55% 4.65%

expiration in months 2 4 6


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(i) Suppose that the cash leg swap consists of a level payment of Rat each of the delivery times. Calculate R.


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Solution: (i) We have that

n∑j=1

QjF0,tj P(0, tj)

=(10000)55(

1 + 0.04512

)2 + (12000)56(

1 + 0.045512

)4 + (15000)58(

1 + 0.046512

)6=545898.0877 + 661903.8839 + 850044.0252 = 2057845.997,∑n

j=1 P(0, tj)

= 1

(1+ 0.04512 )

2 + 1

(1+ 0.045512 )

4 + 1

(1+ 0.046512 )

6

= 0.9925419775 + 0.9849760176 + 0.9770620979 = 2.954580093

and

R =

∑nj=1 QjF0,tj P(0, tj)∑n

j=1 P(0, tj)=

2057845.997

2.954580093= 696493.5564.


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(ii) Suppose that the cash leg swap consists of a unique payment perbarrel made at each of the delivery times. Calculate the no arbitrageswap price per barrel.


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(ii) Suppose that the cash leg swap consists of a unique payment perbarrel made at each of the delivery times. Calculate the no arbitrageswap price per barrel.Solution: (ii) We have that∑n

j=1 P(0, tj)Qj

= 10000

(1+ 0.04512 )

2 + 12000

(1+ 0.045512 )

4 + 15000

(1+ 0.046512 )

6

= 9925.419775 + 11819.712212 + 14655.931469 = 36401.06346

and the swap price per barrel is

R =


j=1 P(0, tj)Qj=

2057845.997

36401.06346= 56.53257903.


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If a swap is cashed settled, then the commodity is valued at thecurrent spot price. If the current value of the commodity is biggerthan the value of the cash payment, then the (cash leg side) longswap pays this difference to the (commodity leg side) short swap.Reciprocally, if the current value of the commodity is smaller thanthe value of the cash payment, the (commodity leg side) shortswap side pays this difference to the (cash leg side) long swap.


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Suppose that the swap involves the sale of a commodity at timest1 < t2 < · · · < tn. with notional amounts of Q1,Q2, · · · ,Qn,respectively. Suppose the swap payment is a fixed amount per unitof commodity. We saw that this amount is

R =


j=1 P(0, tj)Qj.

When a swap is cash settled, the value of the commodity is foundusing the current spot price Stj . At time tj the long swap paysQjR − QjStj to its counterpart. Notice that QjR − QjStj could bea negative real number. If QjR − QjStj < 0, the short swap paysQjStj − QjR to its counterpart.


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Changes in the forward contracts and interest rates alter the valueof the swap. The market value of a swap is found using the presentvalues of its legs using the current structure of interest rates. Themarket value of a long swap immediately after the settlement attime tk is

n∑j=k+1

P(tk , tj)Qj(Ftk ,tj − R).

This is the price which an investor would pay to enter the swap asa long swap side. Immediately after the swap is undertaken themarket value of the contract is

n∑j=1

Qj(P(0, tj)F0,tj − P(0, tj)F0,tj ),

where P(0, tj) and F0,tj are the new market values.


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Example 11

Suppose that an airline company must buy 1,000 barrels of oilevery six months, for 3 years, starting 6 months from now. Insteadof buying six separate long forward contracts, the company entersinto a long swap contract. According with this swap the companywill pay a level payment per barrel at delivery. The following tableshows the annual nominal rate convertible semiannually ofzero–coupon bonds maturing in 6, 12, 18, 24 months and theforward price of oil at those times.

F0,T $55 $57 $57 $60 $62 $64

annual nominal rate 5.5% 5.6% 5.65% 5.7% 5.7% 5.75%

expiration in months 6 12 18 24 30 36


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(i) Find the price per barrel of oil using the swap.


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(i) Find the price per barrel of oil using the swap.Solution: (i) We have that

n∑j=1

P(0, tj)QjF0,tj

=(1000)55

1 + 0.0552

+ (1000)57(

1 + 0.0562

)2 + (1000)57(

1 + 0.05652

)3+ (1000)

60(1 + 0.057

2

)4 + (1000)62(

1 + 0.0572

)5 + (1000)64(

1 + 0.05752

)6=54254.00740 + 55436.90114 + 54651.28637

+ 56698.44979 + 57765.24340 + 58747.41357 = 337553.3017,


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(i) Find the price per barrel of oil using the swap.Solution: (i)

n∑j=1

P(0, tj)Qj

=1000

1 + 0.0552

+1(

1 + 0.0562

)2 +1000(

1 + 0.05652

)3+

1000(1 + 0.057

2

)4 +1000(

1 + 0.0572

)5 +1000(

1 + 0.05752

)6=986.4364982 + 972.5772129 + 958.7944977

+ 944.9741632 + 931.6974742 + 917.9283370 = 5712.408183.

We have that

R =


j=1 P(0, tj)Qj=

337553.3017

5712.408183= 59.09124329.


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(ii) Suppose the swap is settled in cash. Assume that the spot ratefor oil in 18 months is $57. Calculate the payment which the airlinereceives.


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(ii) Suppose the swap is settled in cash. Assume that the spot ratefor oil in 18 months is $57. Calculate the payment which the airlinereceives.Solution: (ii) The airline gets a payment of

QjStj −QjR = (1000)(57)− (1000)(59.09124329) = −2091.24329.


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(iii) Suppose that immediately after the swap is signed up, the futureprices of oil are given by the following table

F0,T $55 $58 $59 $61 $62 $63

expiration in months 6 12 18 24 30 36

Calculate the value of the swap for the cash leg party.


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Solution: (iii) The market value of the swap for the long party is

n∑j=1

P(0, tj)(Qj Ft0,tj − QjR)

=1000

1 + 0.0552

(55− 59.09124329) +1(

1 + 0.0562

)2 (58− 59.09124329)

+1000(

1 + 0.05652

)3 (59− 59.09124329) +1000(

1 + 0.0572

)4 (61− 59.09124329)

+1000(

1 + 0.0572

)5 (62− 59.09124329) +1000(

1 + 0.05752

)6 (63− 59.09124329)

=− 4035.7517041− 1061.3183576− 87.4835644

+ 1803.7257747 + 2710.0812797 + 3587.9585464 = 2917.211975.


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Suppose that immediately after the forward is signed, every futureprice increases by K . Then, the market value of the swap is

n∑j=1

P(0, tj)(Qj(F0,tj + K )− QjR) =n∑

j=1

P(0, tj)QjK .

The swap counterpart is a scalper which hedges the commodityrisk resulting from the swap. The scalper has also interest rateexposure. The scalper needs to hedge changes in the price of thecommodity and in interest rates.


SOA exam FM CAS

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