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Angelos Dassios c2013

Investment Performance Bonds Makeham Redington Arbitrage Derivatives Term Structure Stochastic Rates

Actuarial Investigations: Financial

Part 2

Angelos Dassios

Department of StatisticsLondon School of Economics and Political Science

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Measures of Investment Performance

We already know two tools to measure the investmentperformance, the NPV and the Equation of value.Suppose we have a fund. We can invest more money any time (oreven take money out).

F0= the initial fund before new money comes in

New money (positive and negative) C1,C2, . . . , Cn are paid at timet1, t2, . . . , tn.

Ft1 = value of the fund before new money comes in at t1

Ft1 =Ft1+C1= value of the fund after new money coming in at t1.

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At time t2, a further injection ofC2 is made.

Ft2 = Ft2+C2...

...

Ftn = Ftn+Cn then it grows until T(end of project, it becomes FT

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TWRR (time weighted rate of return):This is equivalence to the Equation of Value if all investments

were to occur upfront and nothing else happens until time T.We have to find the growth of the fund.

Ft1

F0 Ft2

Ft1 Ft3

Ft2 Ftn

Ftn1 FT

Ftn= (1 +i)T

And then find i.Or look at it this way:

Ft1

F0 Ft2

Ft1

+C1 Ft3

Ft2

+C2 Ftn

Ftn1

+Cn

1 FT

Ftn+Cn= (1+i)T

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Example: Suppose the fund is initially $10, 000.At the end of year 1, it grows to $11, 000, $1, 500 is added;

At the end of year 2, it grows to$

10, 000,$

2, 000 is added;At the end of year 3, it grows to $14, 000. End of project.

MWRR (look at payments, then discount)

10000 + 1500(1 +i)1 + 2000(1 +i)2 = 14000(1 +i)3

So MWRR 1.4%TWRR

11, 000

10, 00010, 000

12, 50014, 000

12, 000= (1 +i)3

So i= [(1.1)( 11.25 )(1.41.2 )]

1/3 1So TWRR 0.88%

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I P f B d M k h R di A bi D i i T S S h i R

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Example: The prices of a unit in the fund on 1st January of eachyear are as follows:

Year Unit Price2006 1002007 1102008 952009 105

An investor invested 100 on the 1st of January of 2006, 2007 and2008 and withdraws everything on the 1st of January 2009.

The money weighted rate of return is more complicated thistime. The accumulated amount at the end is

100 105100

+ 100 105110

+ 100 10595

= 310.98

so the equation of value is

100 + 100 (1 +i)1 + 100 (1 +i)2 =310.98 (1+i)37/102

I t t P f B d M k h R di t A bit D i ti T St t St h ti R t
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Trying various values, we see that i= 0.018 satisfies theequation

The time weighted rate of return is a lot simpler.

1.05(1/3) 1 = 0.0164

Another example: 2009 Q2 will give you a good understanding of

why the TWRR and the MWRR are different.

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Bonds

Bonds are fixed interest securities. They have a nominal price of100. We pay a price to buy the bond and if we hold it toredemption, we will get capital back. In the meantime, there isregular interest.

Notation:n: redemption time

R: redemption price, which is usually R= 100 (redemption atpar). IfR>100, e.g. R= 110, this is above par. IfR

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D: coupon payable annually (usually at the end of the year, e.g.5% bond, D= 5 and zero-coupon bond, D= 0)

Dp: coupon payable pthly, e.g. if it is payable

12 yearly, then we

receive D2 every six months. A coupon payable say, monthlyrather than yearly is more profitable, since you are gettingmoney earlier on during the year and do not have to wait untilthe end of the year.

N: number of bonds we purchase/hold/sell etc

C: redemption money, C =NR. coupons we receive become NDor NDp

P: Price of a bond

A: Price of whole investment i.e. all the bonds you have bought ,So A= NP.

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Capital and interest are already separate (unlike loans we dealtwith earlier). R (orNR=C) is the capital to be refunded and D

(orND) is the interest.Suppose an investor wants a yield i. We have

P = Rvn +Dan (annual coupon)

NP = NRvn +NDan

A = Cvn +NDan

If coupon (D) is paid more frequently, say ptimes per year,

P=Rvn +Da(p)n

This equation can be used in 2 ways:

i- find P for given i

ii- find i for given P (by trial and error)

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We define the coupon yield g= DR

and we can rewrite theequation

P = Rvn +gRa(p)n

= R

vn +ga(p)n

= Rvn +g

1 vn

ip

Ifg=i(p) then P=R i.e. price of bond equals redemption priceof bond (rather intuitive).

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g g

Income Tax

Now we consider the situation that we have to pay income tax atthe rate t1 on the interest we earn, so

P = Rvn + (1 t1)gRa(p)n= Rvn + (1

t1)gR

1 vn

i(p)

= R

vn + (1 t1)g1 v

n

i(p)

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If (1 t1)g=i(p) then P=R. (1 t1)D is the net coupon.(1 t1)g is the net yield.

NP = NRvn + (1 t1)gNRa(p)nA = Cvn + (1 t1)gCa(p)n

where A is a function ofn and i, A(n, i). As i increases, A(n, i)

decreases.

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Moreover, since

a

(p)

n =

1

vn

i(p)

vn = 1 i(p)a(p)n

we have

A(n, i) = C[vn + (1 t1)ga(p)n ]= C[1 i(p)a(p)

n + (1 t1)ga(p)n ]

= C+C[(1 t1)g i(p)]a(p)nIf i(p) C.If i(p) >(1 t1)g, A(n, i) is an decreasing function ofn. AndA(n, i)

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If the price is given and A>C (i.e. i(p) (1 t1)g, A(n, i)

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Recall, we had two problems.

1. For a given yield, find the price.

2. For a given price, find the yield.We used the same equation to solve both problems, but now it isslightly different as we need to consider CGT.

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Example: Suppose an investor paid $95 for a bond. It has a 5%coupon payable annually, redeemable at par after 10 years. Incometax rate=t

1= 20%. CGT rate=t

2= 30%. What is the yield?

Net coupon = (1 t1)D= (1 0.2)0.05 100= 4

P=Rvn +Dan

Without CGT95 = 100v10 + 4an

With CGT (we need to subtract the CGT)

95 = 100v10 + 4an (100 95)0.3v10

95 = 98.5v10 + 41 v10

i

i 0.045 18/102
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In general:

A = price

C = redemption price

gC(1 t1) = net coupont2 = CGT rate

A= (a)C vn +g(b)C(1 t1)an t2((a)C A)vn

Note:

(a) may not always be 100N, could sell it for more/less;

(b) is always 100N, since your coupon (a) is say 5% of the 100N.Solve for A or i, depending on which is given.

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Example (Indexation of Capital Gains): Suppose the situationis the same as before. 10-year security, redemption at par, 5%coupon,t1= 25%, t2= 30%, want a yield of 6%. Now given alsothat the cost is inflated by 1% per year, what is the price?

P= 100v10 + (0.75)5a10

0.3[100 P(1.01)10](1.06)10

After you are solved for P,check P(1.01)10 100 then there is no capital gain, soyou would have to solve the equation again, which simply becomes

P= 100v10 + (0.75)5a10

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Example (more than 1 redemption dates): Suppose we have asecurity of$300 (redemption price redeemable at par). There is a5% coupon, t1= 40%, t2= 30%, want a 6% yield. What is theprice if we redeem:$100 after 5 years;$100 after 10 years;$100 after 15 years.

Capital costs are also indexed by 2% annually. CGT is applicable

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throughout.

A = 100(1.06)5 + (0.6)15 a5

at 6%

0.3[100

A

3(1.02)5](1.06)

5

+ 100(1.06)10 + (0.6)10 5|a5

at 6%0.3[100 A

3(1.02)10](1.0610)

+ +100(1.06)15 + (0.6)5 10|a5 at 6%

0.3[100 A3

(1.02)15](1.06)15

Solve for A and check:

If A3 (1.02)15

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0.3[100 A3 (1.02)15](1.06)15 since there is no CGT for thatchunk, get A.

Is

A

3(1.02)10

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Makehams formula

Assume there is no CGT and IR is constant. We have a simple

security.

A = Cvn + (1 t1)gCan (ora(p)n )

A = Cvn + (1 t1)gC1 vn

i (recall (1 t1)g= is the net coupon)

Define K =Cvn =P.V. of redemption price, I=net P.V. interest ofwe have

A = K+I

A = Cvn +(1 t1)gi

C(1 vn)

= Cvn +(1 t1)g

i (C K) (Makehams Formula)

(orA= K+ (1

t1)g

i(p) (C K) if we were given i(p)

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Example: Recall the previous example of the $300 security where$100 is redeemed at 5 years;$100 is redeemed at 10 years;$100 is redeemed at 15 years.There is a 5% coupon, t1= 40% and there is no CGT. Find A:

A= 100v5+100v10+100v15+(0.6)15a5+(0.6)105|a5+(0.6)510|a5

or we can use the Makehams formula (even though it doesnt saveus a lot of time here):C= 300, K= 100v5 + 100v10 + 100v15

A = K+(1 t1)gi

(C K)

A = K+(0.05)(0.6)

i (C K) (1 t1= 0.6 g= 0.05 i.e. 5% coupon)

Check both formulas are equal. 27/102

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Since Makehams formula works for arbitrary maturity dates, itworks for arbitrary payments, so for example it works for loans aswell.Example: (using the Makehams formula will be quite beneficialhere) 2001 exam Q11. A loan is to be repaid by 10 annualinstalments of capital and interest (need to sort out which bits arecapital and which bits are interest) at 7%. The size of all the

instalments are$

1000 except the 2nd and 7th years, when theinstalments are $2000.

(a) Find the amount of the loan

(b) An investor, not liable to tax, requires a 6% yield p.a. Whatprice should she pay?

(c) Suppose she is liable to income tax at 40%, what price shouldshe pay? (use clever trick).

(d) Suppose also that there is CGT as well, at 40%, what priceshould she pay?

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(a) Amount of the loan

L0 = 1000 a10at 7%

+1000(1.07)2 + 1000(1.07)7

= 1000(1 v10

i ) + 1000(1.07)2 + 1000(1.07)7

= 8519.77

(b) Price

Price = 1000 a10

at 6%+1000(1.06)2 + 1000(1.06)7

= 1000

1 v10i

+ 1000(1.06)2 + 1000(1.06)7

= 8915.14

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(c) The long way - need to break everything up, so we cancalculate income tax on the interest earned

t Xt Lt

1 It=iLt

1 Ct=Xt

It Lt=Lt

1

Ct

1 1000 8519.77 596.13 403.62 8116.152 2000 8116.15 568.13 1421.87 6684.28

3 1000 6684.28 467.90 532.10 6152.18

4 1000 6152.18 430.65 569.35 5582.83

5 1000 5582.83 390.80 609.20 4973.636 1000 4973.63 348.15 651.87 4321.78

7 2000 4321.63 302.52 1697.48 2624.31

8 1000 2624.31 183.70 816.30 1808.01

9 1000 1808.01 126.56 873.44 934.57

10 1000 934.57 65.42 934.58

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equation becomes:

A = K+g

i(C

K) then substituting values in:

8915.15 = K+0.07

0.06(8519.77 K)

K = 6147.49

I = A K= 2767.66

So now the price with income tax is:

A= K+ 0.6I= 7808.09

(d) Finding the price when there is a CGT rate of 40%.First, always ask whether there has been a capital gain. Therehas been, since 6%> 0.6 7%.We are now using the formula

A= Cvn

+gC(1 t1)an t2(C A)vn

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where

Cvn = K=P.V. of capital repayments,Ct

g = coupon rate

t1 = income tax

t2 = CGT

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We need to be careful with t2(C A)vn. We need to scalethe price, A, since we are making payments Xtat timest1, . . . , t10 (previously we were redeeming at different times).

We scale by CtL0 . So we have

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A =10

t=1

Ct(1.06)

t + 0.6

10

t=1

It(1.06)

t

0.410t=1

Ct A

Ct

L0

(1.06)t

= K+ 0.6I 0.410t=1

L0Ct ACtL0

(1.06)t

= K+ 0.6I 0.4 L0 AL0

10

t=1Ct(1.06)

t

= K+ 0.6I 0.4 L0 AL0

K

A = 6147.49 + 1660.60 3407.91 0.4A8519.77

(6147.49)

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Further reading: 2009 Q9 is a very nice example where the onlypossible solution is to use Makehams formula. The calculationsare very simple, so the use of the formula is very clear.

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Redingtons theory of immunisation

Suppose we have a series of payments (negative and positive ones)

C1, C2, . . . , Cn at times t1, t2, . . . , tn.The equation of value wasn

j=1

Cjvtj = 0

or nj=1

Cjetj = 0

We will call all the positive payments assets, A1,A2, . . . , Am and

all the negative payments liabilities, Lm+1, Lm+2, Ln (taking theabsolute values ofLi. It may not be the case that we get all thenegative payments after the positive ones. We may haveA1,A2, L3, L4,A5, etc). So the equation becomes:

Ajetj = Ljetj 37/102
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We can define V() =VA() VL(), thenV() =VA() VL() = 0 at . In this case, we have a localturning point. In addition, ifV()> 0, we have a localminimum. If there is a small change in the force of interest, we willnot lose money. If the following three conditions are satisfied, weareimmunisedagainst small changes in the IR:

VA(

) = VL()VA() = VL()VA(

) > VL()

If the local minimum is a global minimum, then we have a fullyimmunisedposition (rare occasion).

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Redingtons Theory of Immunisation:

Ajetj = Ljetj tjAjetj = tjLjetjt2jAje

tj >

t2jLjetj

=

tjAje

tj

Aje

tj

=

tjLje

t

Lje

tj

t2jAje

tj

Aje

tj>

t2jLje

t

Lje

tj

Total convexity = convexity of assets convexity of liabilitiesIf we had negative convexity, no matter what, we are losing moneyCaution - In the core reading, they differentiated w.r.t i ratherthan . Similar results, but slightly different formula

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Example: Suppose we have a liability at time 5 of$1000 and youwant to match it with assets at time 3 and 10. Zero coupon bondsare great for that. If the asset at time 3 is A1, and at time 10 is

A2, and i= 5%, then

A1(1.05)3 +A2(1.05)10 = 1000(1.05)5

3A1(1.05)3 + 10A2(1.05)10 = 1000 5(1.05)5

Solve for A1 and A2 and check if9A1(1.05)

3 + 100A2(1.05)10 >1000 25 (1.05)5 (it isactually a global minimum).

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Note: the assets may have been quoted as prices of zero-couponbonds

X1 = A1(1.05)3X2 = A2(1.05)

10

Then we have

X1+X2 = A1(1.05)3 +A2(1.05)10 = 1000(1.05)53X1+ 10X2 = 3A1(1.05)

3 + 10A2(1.05)10 = 1000 5(1.05)5

Check 9X1+ 100X2 >1000 25 (1.05)5.Another interesting example 2006 Q2

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Arbitrage

We have a market that consists of equities which are instruments

that are not possible to price with our techniques. However, themarket does this for us. The market has cash which attracts aconstant interest rate, says force or iannually. We can borrow orsave at these rate.We can buy or sell as much of each equity as we like (we can even

hold negative amounts, i.e. sell shares we dont have). Also we cansave as much in the bank as we like or borrow as much as we like.We assume there is no arbitrage (a risk free profit) which couldoccur in the following ways:

(a) We can make a deal with immediate profit and no risk offuture loss;

(b) We can start with no money and make a deal with zeroprobability of future loss and non-zero probability of futureprofit;

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E l 1 Th IR i 5% ff i A i l h i
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Example 1: The IR is 5% p.a. effective. A particular share isworth $100. In a years time, it will be worth either $107 or $106.We have arbitrage.

We can borrow $100 today and buy a share. In a years time, wesell the share and pay back $105. We make a profit of either$2(= 107 105) or $1(= 106 105). This is an example of case(b).Or, if price of share is worth $107 or $105 in a years time, we havean example of case (b). we either make a profit $2(= 107 105)or we make $0(= 105 105), i.e. if the price of the bond in oneyears time is 105, we make no loss or profit. This is still arbitrage.

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E l 2 IR i 5% ff ti A ti l h th
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Example 2: IR is 5% p.a. effective. A particular share now worth$100. In a years time it will be worth $105 or $103. Do we havearbitrage?

We could sell one share now and put the money in to the bank. Inone years time we get $105. You have to buy back the share(since you sold one you never had in the first place). If the shareprice is 105, you make $0(= 105 105), but if the price is $103,you make a profit of$2(= 105

103).

So there is arbitrage since, there is a probability of profit but a zeroprobability of loss.

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E l 3 D h bit i l 2 if th h ill b
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Example 3: Do we have arbitrage in example 2 if the share will beworth either $107 or $103?No, since there is a chance of making a loss of$2 = (107

105).

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Example 4: We have share A and B An announcement is going

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Example 4: We have share A and B. An announcement is goingto be made tonight affecting the current share price of$6 and $11respectively. If the news is true, then the share prices become $7

and $14. If the news is negative, then the share prices become $5and $10. IR is 0%, since we are only talking about one night(interest is negligible).We can sell 2 units of A and buy 1 unit of B. So now we make$1(= 12

11). Then tomorrow we buy 2 of A and sell 1 of B.

If the news is good, we have $0(= 14 2 7);If the news is bad, we have $0(= 10 2 5).So we have arbitrage (an example of (a)). We made $1 at thebeginning with no risk of loss.

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Example 5: Suppose we have 2 shares both worth $100 now IR
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Example 5: Suppose we have 2 shares, both worth $100 now. IRis 5% p.a. effective.If the year is good, share A is worth 108 and if the year is bad,

share A becomes 103.If the year is good, share B is worth 120, if the year is bad, share Bis worth 100(1 +b) =X.What is X, so that we have no arbitrage?To solve for X, we first look at the arbitrage portfolio, i.e. allrisked is eliminated.

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Suppose I have $1000 I invest 1000a in share A and 1000(1 a)

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Suppose I have $1000. I invest 1000a in share A and 1000(1 a)in share B. ais any real number, in such a way that there is no risk.At the end of a good year we have

1080a+ 1200(1 a) = 1200 120a;At the end of a bad year we have1030a+ 1000(1 +b)(1 a) = 1030a+ 10X(1 a).we will choose a such that both outcomes are equal

1200 120a = 1030a+ 10X(1 a)a =

1200 10X1150 10X

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Note that a > 1 so 1 a < 0 so I will short sell share B The
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Note that a>1, so 1 a

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Why shall we assume there is no arbitrage? If there wasarbitrage, someone would have eliminated it. Everyone would buyshares to get a risk free profit. As a result, demand increases, so

prices adjust until there is no arbitrage.

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Derivatives
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Derivatives

The market consists of a share whose price at time t is St, current

price is S0 and cash invested or borrowed at an effective IR of iannual or being the force. This is a no risk investment. $1 willbe $et at time t.A Derivative contractis an financial instrument whose valuedepends on the values of other more basic underlying assets (e.g.

share price Stand perhaps on time tas well).

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A forward contract is an agreement to sell or buy an asset at a
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A forward contractis an agreement to sell or buy an asset at acertain time, T, in the future for a certain price (delivery priceK).

Assume that a forward contract on a non-dividend paying stockmatured in 3 months, i.e. the contract involves delivery of thestock in 3 months time. The stock price is $100 now, thethree-month risk-free interest is 12% p.a. Suppose the forwardcontract price is $105, i.e. the stock will delivered for payment of$105 in 3 months time.

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Questions to consider:
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Q

Is this the correct price for the forward contract today?

What do we mean by correct price?

Can I make money without any risk?

Consider the following trading strategy:

Time 0:

Borrow $100 at 12%

Buy a share at $100Sell a forward at $105

3 months:

Get $105 for the sharePay back $103 (including $3 interest payment at the risk freerate of 12%)Get profit $2

So this has yielded a riskless profit of$2.

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What about the case where the forward price is $102? Consider an
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pinvestor who has a portfolio with one share of the stock. Can hemake money without any risk?

Time 0:Sell the share at $100Invest the $100 at 12%Buy a forward at $102

3 months:

Get $103Pay $102 for the shareGet profit $1

So this has yielded a riskless profit of$1.

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Clearly neither of the two forward prices are sustainable in a
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y pmarket without arbitrage opportunities. They cannot be thecorrect price for the forward contract in such a market.

In fact, whenever the forward price is higher or lower than $103there is an arbitrage opportunity.

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An alternative way is to price the FC by replication. If we can

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replicate the derivative by a portfolio (invest in cash + share)which initial value we know, then we can find the price of FC.

Suppose we long a FC at time 0. At time T, we will pay share Kand have a share. We can replicate the Long FC by borrow KeT

cash at time 0 and buy a share at price S0 so that at time T, wewill owe KeTeT =Kcash and have a share. We assume thereis no arbitrage, i.e. the present value of the FC is 0, so the present

value of the portfolio should be 0:

S0 KeT = 0So we get

K=S0eT

Value of a FC at time 0 is zero. But as time evolves, the price ofthe long FC is not going to be zero any more, eg at time T, it isworth f =ST K=ST S0eT(= 0). The value of the short FCis going to be worth f =K

ST =S0e

T

ST.

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Suppose I am at time rwhere 0

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Sr. At time 0, I had entered into a Long FC. What is is worthnow? At time T, I will get one share and pay K. How can I

replicate this at time r?I can buy the share now and borrow Ke(Tr). At time T, I willstill have the share and owe Ke(Tr)e(Tr) =K. So

Value of the Long FC at r = Sr

Ke(Tr)

(K =S0eT) = Sr S0eTe(Tr)= Sr S0er

Value of the Short FC at r =

(Sr

S0er)

= S0er Sr

All we said so far doesnt depend on the dynamics ofStas long asthere is no arbitrage.

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At a fixed time T, I will be paid h(ST) where h is a prearranged
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function. How much should I pay for this now?Ifh(ST) =ST, S0 is what I should pay now.

The forward contract was a special case where h(ST) =ST K (IfK is chosen appropriately we pay nothing now).A call optionis an option (but not an obligation) to buy a certainasset by a certain date (expiration date) for a certain price(strike price). So in exchange for a fee, C, to be paid now we

have the right, but not the obligation to go into the contract(recall in the Long FC, I had the obligation to go into thecontract). A European optioncan be exercised only on theexpiration date itself.

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Let bbe the prearranged strike price (the price at which I have the)
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right to buy the share at time T), when time T comes, ifST>b,I will exercise the contract and realise a profit ofSt

b. If on the

other hand ST b, I walk away and get and pay nothing. (Notethere is no arbitrage, since you actually paid a fee at the beginning)There is no unique way to find that without knowing a bit of thedynamics and

Payoff =hC(ST) = max(ST b, 0) = (ST b)+

A put optionis an option (but not an obligation) to sell a certainasset by a certain date (expiration date) for a certain price(strike price). So in exchanges for a fee, P, which is payable now,

we have the right, but not the obligation to sell a share at theprearranged price b. IfST

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Suppose I buy one call option (strike price b) and sell one putoption (strike price b). So my payoff at time T is

Call Put = (ST b)+ (b ST)+

IfST>b, payoff is (ST b), since you would only exercise thecall option.IfST b, payoff is(b ST) =ST b, since the holder of theput option would exercise the call option.So whatever happens, your payoff would be ST bwhich has aP.V. ofS0

beT.

Call Put Parity = Share Money =C P=S0 beT

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Pricing by replication portfolio
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Example: Pricing a simple call option. The price of the share now

is $100. In a years time, it will be either $110 or $95. The IR is5% p.a. effective. Suppose we buy a one year European Calloption with a strike price of$106. What is the current price/value,X, of the contract?Payoff function if things go well: 110

106 = 4;

Payoff function if things go badly: 0 (since you dont exercise thecontract).

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Pricing by eliminating risk (hedging)

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Let us now work with the previous example in a different manner.

Suppose an investor invests a proportion of her capital in theoption and 1 in the stock. Let us assume the capital is 100. Ifthings go well, she will end up with

1004

X

+ 110 (1

)

and if things go badly

95(1 )

Eliminating all risk means that the two outcomes should be equal,but then in order to avoid arbitrage they should be equal to 105.Therefore,

1004

X

+ 110 (1

) = 95 (1

)= 105

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From the second equality = 0.10526 and substituting into thefirst X = 2 54
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first X = 2.54Further examples 2009 Q6, 2006 Q4

American Options: There is also an American Call and PutOptions which are more complicated than the European ones, sinceyou dont have to buy or sell shares at the time T. Note thatEuropean and American Call and Put Option are traded in both

Europe and America.

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Term Structure of Interest Rates
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So far we have assumed that a 2-year zero-coupon bond has now a

price of 100(1 +i)2

and a 10-year zero coupon bond has a priceof 100(1 +i)10. So the same yield is used for both. This is notnecessarily true in practice. For many reasons investors mightrequired a different yield for the 10-year and the 2-year bonds.Suppose we require a yield y2 from the 2-year bond and y10 from

the 10-year bond. Price of the 2-year bond is 100(1 +y2)2; priceof the 10-year bond is 100(1 +y10)

10.In general, a t-year zero coupon bond will have a price of100(1 +yt)

t where yt (tis any real number) is the yield of thet-year zero coupon bond. ytis also called the t-year spot rate,which is a curve we observe in the market. All the curves you aregoing to see in the next few pages have happened in the past.

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If liquidity was my only concern I would I get an increasing curve.
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In practice, we can also see a situation like this decreasing function(contradicts liquidity)

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(contradicts liquidity).

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The following can happen as well
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Consistency: Consider the accumulation of$1 from time t1 to t2,where t1

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t1 < t2, (t1, t2) gA(t1, t3) =A(t1, t2)A(t2, t3) then the whole structure is consistent.

Given 0 t1 t2, we need A(0, t2) =A(0, t1)A(t1, t2) forconsistency, i.e.

(1 +yt2 )t2 = (1 +yt1 )

t1 A(t1, t2)

Define ft1,t2t1 such that A(t1, t2) = (1 +ft1,t2t1 )t2t1

. So

(1 +yt2 )t2 = (1 +yt1 )

t1 (1 +ft1,t2t1 )t2t1

where ft1,t2t1 is the t2 t1 year forward rate at time t1.The r-year forward rateat time s is defined as fs,r(=

A(0,s+r)

A(0,s) ).So we have

(1 +ys+r)s+r = (1 +ys)

s(1 +fs,r)r

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Par yields: It is rarely used, but worth to know. The end year paryield, represents the annual coupon of a bond redeemable at par

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y p p psuch that its price now is par. Let Cbe the coupon.

100 = C(1 +y1)1 +C(1 +y2)2 + . . . +C(1 +yn)n + 100(1 +yn)

C = 100(1 (1 +yn))n

(1 +y1)1 + (1 +y2)2 + . . . + (1 +yn)n

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Why is there a term structure of IR? Why are IRs not constant andthe yield curves flat? There are many explanatory theories, such as
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Expectation Theory, Liquidity, Market Segmentation, all of which

are questionable.

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A No Arbitrage explanation:We will assume no arbitrage is allowed and yield curves a flat and
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will always be so. We start with $0. If I can engineer a situation

with some probability of profit and no possibility of loss, then thisis arbitrage, which is conflicting with our no arbitrage assumption(and which means our assumption about yield curves is wrong).Invest an amount Xin a 1-year zero-coupon bond and anotheramount X in a 3-year zero-coupon bond. How are you going to

finance it? We sell 2Xworth of a 2-year zero-coupon bond.Let 0 (constant) be the current force of interest. Nominalamounts areXe0 for the 1-year bond;Xe30 for the 3-year bond;2Xe20 for the 2-year bond.The P.V. of my position is X+X 2X = 0.

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Let us recall Redington and his immunisation theory. At somestage, pretty soon, 0 might move to 1. The curve is still assumed
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to be flat, but it will shift up or down.

So in a very short time, the P.V. of my investment becomes

Xe0 e1 +Xe30 e31 2Xe20 e21= Xe01 (1 2e01 +e2(01))= Xe01 (1

e01 )2 >0

According to Redington, we are fully immunised against any movein the IR. In fact the more 0 moves, the higher our P.V. so

(i) If IR moves, I make a profit

(ii) If IR stay at 0, I make no loss.And obviously, this is arbitrage.

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Conclusion: Yield curves can not always be flat. They may be flattoday, but cannot stay flat, since, as known above, there will be
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arbitrage. So

(i) Term structure exists(ii) Redington was wrong

(iii) Theory of immunisation is useless, since it makes theassumption of flat yield curves.

Further reading 2000 Q8 (solution does not exist but given above).Think about the last part.Another example 2011 Q8

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Stochastic Interest Rates

$1 i d ill l $(1 I ) i i
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$1 invested now will accumulate to $(1 +I1) in a years time,

where I1 is a random variable (r.v.). What can the distribution ofI1 be? Anything greater than1 will do, i.e. anything such that(1 +I1) 0.Expected accumulation will be

E(1 +I1) = 1 +E(I1)

Variance will beE(1 +I1)

2 [E(1 +I1)]2

We might even be interested in third and fourth order moments.

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Let I1, I2, I3, . . . be a sequence of r.v. indicating the yield each year.$1 invented now will accumulate to (1 +I1)(1 +I2) . . . (1 +In) ati

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time n.

The expected value is

E[(1 +I1)(1 +I2) . . . (1 +In)]

The variance is

E[(1 +I1)2(1 +I2)2 . . . (1 +In)2] E[(1 +I1)(1 +I2) . . . (1 +In)]

83/102 Investment Performance Bonds Makeham Redington Arbitrage Derivatives Term Structure Stochastic Rates

Example (Naive model): I1, I2 . . . are independent

E (accumulation) = E (1 + I )E (1 + I ) E (1 + I )
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E(accumulation) =E(1 +I1)E(1 +I2) E(1 +In)

second moment =E[(1 +I1)2]E[(1 +I2)2] E[(1 +In)2]variance = second moment [E(accumulation)]2

Make things more naive by assuming they are also identically

distributed: E(It) =j

Var(It) =s2

E(I2t) =s2 +j2

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1 +Itcan pretty much have any positive distribution. Onepossibility, Itcan take values i1, i2, . . . , in with probabilities

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p1, p2, . . . , pn:

E(It) =i1p1+i2p2+ +ikpkE(1 +It)

2 = (1 +i1)2p1+ (1 +i2)

2p2+ + (1 +ik)2pkIfIt is a continuous r.v. (1 + It) has a density function f(x), x>0:

E(1 +It) =

0

(1 +x)f(x)dx

E(1 +It)2 =

0

(1 +x)2f(x)dx

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Important Special case: (1 +It) follows the log-normaldistribution (1 +It) =e

Yt, where Ytis normal with mean andvariance 2 We have

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variance . We have

E(1 +It) =E(eYt) =e+ 122 = 1 +m

E(It) =e+ 1

22 1 =m

E(1 +It)2 =E(e2Yt) =e2+2

2

Var(It) =Var(1 +It) =e2+22 (e+ 122 )2 =e2+22 e2+2

=e2+2(e

2 1) =s2

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Ytis normal with mean and variance 2, i.e. Yt N(, 2) and

Y1+Y2+ +Y2 N(n, n2). We have

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P(A x) =P(eY

1+Y

2+

+Yn x) =P(Y1+Y2+ +Yn ln x) = ln

where

ln xnn

is the distribution function of the standard

normal table.

In all questions, try to establish whether you are given , 2 orm,s2.


Dependence models

Here Its are not independent. It all depends on the structure and
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one usually works from first principles. One possibility is

1 +It=eYt. Yts can be normally distributed but they also have acorrelation (see Ex 8, Q1 (b)).Another possibility is the so called Markov Chain Model.Suppose Itcan take values of 0.03, 0.05 and 0.07. Suppose at

0.03, the probability of it being 0.03 is 0.5; at 0.03, the probabilityof it being 0.07 is 0.2, etc. The rest is described by the table below:0.03 0.05 0.07

0.03 0.5 0.3 0.20.05 0.2 0.7 0.1

0.07 0 0.3 0.7It is called transition matrix, e.g P(It= 0.05|It1= 0.07) = 0.3.Suppose in 2000 IR was 0.05. Suppose $1 is invested on01/01/01. How much does it accumulate to on 01/01/03 (or31/12/02 - same 2 years).


All Possibilities Probabilities(1.03)(1.03) 0.2*0.5(1.03)(1.05) 0.2*0.3
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( )( )(1.03)(1.07) 0.2*0.2(1.05)(1.03) 0.7*0.2(1.03)(1.05) 0.7*0.7(1.05)(1.07) 0.7*0.1(1.07)(1.03) 0.1*0

(1.07)(1.05) 0.1*0.3(1.07)(1.07) 0.1*0.7


E[(1 +I1)(1 +I2)] = (1.03)(1.03)(0.2)(0.5) +

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[( )( )] ( )( )( )( )

+ (1.07)(1.07)(0.1)(0.7)E[(1 +I1)

2(1 +I2)2] = (1.03)2(1.03)2(0.2)(0.5) +

+ (1.07)2(1.07)2(0.1)(0.7)

What does P(A> 1.12) equal to?

P(A> 1.12) = (0.7)(0.1) + (0.1)(0.3) + (0.1)(0.7)

(1.05)(1.07) (1.07)(1.05) (1.07)(1.07)

The answer is 0.17.


Investing at Various Times

Example: Suppose $1 is invested at time 0, 1, 2, . . . , n 1.
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p ppWhat is the accumulated value at time n?

An = (1+I1)(1+I2) (1+In)+(1+I2)(1+I3) (1+In)+ +(1+In)

Suppose Its are iid, E(It) =j and Var(It) =s2.

E(An) = E[(1 +I1)(1 +I2) (1 +In)] + +E[(1 +In)]= (1 +j)n + (1 +J)n1 + + (1 +j)= (1 +j)

(1 +j)n

j 1

= sn (at j)

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We can work out E(An) from first principles. E(A2n) is a bit

difficult but can be done for a decently small n.We assume Its are iid, where E(It) =j and Var(It) =s

2 so that

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( ) j ( )

E(An) =x0(1 +j)n +x1(1 +j)n1 + +xn1(1 +j)

Var(An) =E(A2n) [E(An)]2


Lets find the second moment ifn is small, say n= 3:

A3=x0(1 +I1)(1 +I2)(1 +I3) +x1(1 +I2)(1 +I3) +x2(1 +I3)
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E(A23) =

E[x0(1 +I1)(1 +I2)(1 +I3) +x1(1 +I2)(1 +I3) +x2(1 +I3)]2

= E[(1 +I3)2(x0(1 +I1)(1 +I2) +x1(1 +I2) +x2)

2]

= E[(1 +I3)2]E[(x0(1 +I1)(1 +I2) +x1(1 +I2) +x2)2]

= [(1 +j)2 +s2][x20 ((1 +j)2 +s2)2 +x21 ((1 +j)

2 +s2)

+ x22 + 2x0x1(1 +j)((1 +j)2 +s2) + 2x0x2(1 +j)

2 + 2x1x2(1 +j)]

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A2n = [An1(1 +In) +xn1(1 +In)]2

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= A2

n1(1 +In)

2

+ 2An1

xn1(1 +

In)

2

+x2n1(1 +

In)

2

E(A2n) =

E(A2n1)E((1+In)2)+2xn1E(An1)E((1+In)2)+x2n1E((1+In)

2)

=kn1[(1+j)2 + s2] + 2xn1n1[(1+j)2 + s2] + x2n1[(1+j)2 + s2]

We can apply the recursion repeatedly. In the special case, we justsubstitute x0=x1= =xn1= 1. However, we can not get thedistribution. We can devise a similar procedure to get higher ordermoments, but not the distribution of the function.

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Dependence Structure: What if there is a dependence structurebetween the Its? Then one has to act from first principles(seeEx8, Q1). However, we may have simple but unrealistic Markov

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Chain type models. One in principle, can write out possibilities andgo quite a long way - even calculate probabilities (see Ex8, Q3)Further reading: 2010 Q8 2007 Q8 2006 Q10

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THE END

THANK YOU SO MUCH!

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