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© Actex 2015 | Johnny Li and Andrew Ng | SoA Exam MLC

P-7Preface

Preface

Thank you for choosing ACTEX. A new version of Exam MLC is launched in Spring 2014. The new Exam MLC is significantly different from the old one, most notably in the following aspects:

(1) Written-answer questions are introduced and form a major part of the examination.

(2) The number of official textbooks is reduced from two to one. The new official textbook, Actuarial Mathematics for Life Contingent Risks 2nd edition (AMLCR), contains a lot more technical materials than other textbooks written on the same topic.

(3) The level of cognitive skills demanded from candidates is much higher. In particular, the new learning objectives require candidates to not only calculate numerical values but also, for example, interpret the results they obtain.

(4) Several new (and more advanced) topics, such as participating insurance, are added to the syllabus.

Because of these major changes, ACTEX have decided to bring you this new study manual, which is written to fit the new exam. We know very well that you may be worried about written-answer questions. To help you best prepare for the new exam, this manual contains some 150 written-answer questions for you to practice. Seven full-length mock exams, written in exactly the same format as that announced in SoA’s Exam MLC Introductory Note, are also provided. Many of the written-answer questions in our mock exams are highly challenging! We are sorry for giving you a hard time, but we do want you to succeed in the real exam. The learning outcomes of the new exam syllabus require candidates to be able to interpret a lot of actuarial concepts. This skill is drilled extensively in our written-answer practice problems, which often ask you to interpret a certain actuarial formula or to explain your calculation. Also, as seen in SoA’s Exam MLC Sample Written-Answer Questions (e.g., #9), you may be asked in the new exam to define or describe a certain insurance product or actuarial terminology. To help you prepare for this type of exam problems, we have prepared a special chapter (Chapter 0), which contains definitions and descriptions of various products and terminologies. The special chapter is written in a “fact sheet” style so that you can remember the key points more easily. Proofs and derivations are another key challenge. In the new exam, you are highly likely to be asked to prove or derive something. This is evidenced by, for example, problem #4 in SoA’s Exam MLC Sample Written-Answer Questions, which demands a mathematical derivation of the Kolmogorov forward differential equations for a certain transition probability. In this new study manual, we do teach (and drill) you how to prove or derive important formulas. This is in stark contrast to some other exam prep products in which proofs and derivations are downplayed, if not omitted. We have paid special attention to the topics that are newly introduced in the recent two syllabus updates. Seven full-length chapters (Chapters 0, 10, 12 – 16) and two sections (amount to more

Preface


P-8

than 300 pages) are especially devoted to these topics. Moreover, instead of treating the new topics as “orphans”, we demonstrate, as far as possible, how they can be related to the old topics in an exam setting. This is very important for you, because multiple learning outcomes can be examined in one single exam question. We have made our best effort to ensure that all topics in the syllabus are explained and practiced in sufficient depth. For your reference, a detailed mapping between this study manual and the official textbook is provided on pages P-10 to P-12. Besides the topics specified in the exam syllabus, you also need to know a range of numerical techniques in order to succeed. These techniques include, for example, Euler’s method, which is involved in SoA’s Exam MLC Sample Multiple-Choice Question #299. We know that quite a few of you have not even heard of Euler’s method before, so we have prepared a special chapter (Appendix 1, appended to the end of the study manual) to teach you all numerical techniques required for this exam. In addition, whenever a numerical technique is used, we clearly point out which technique it is, letting you follow our examples and exercises more easily. Other distinguishing features of this study manual include:

− We use graphics extensively. Graphical illustrations are probably the most effective way to explain formulas involved in Exam MLC. The extensive use of graphics can also help you remember various concepts and equations.

− A sleek layout is used. The font size and spacing are chosen to let you feel more comfortable in reading. Important equations are displayed in eye-catching boxes.

− Rather than splitting the manual into tiny units, each of which tells you a couple of formulas only, we have carefully grouped the exam topics into 17 chapters. Such a grouping allows you to more easily identify the linkages between different concepts, which, as we mentioned earlier, are essential for your success.

− Instead of giving you a long list of formulas, we point out which formulas are the most important. Having read this study manual, you will be able to identify the formulas you must remember and the formulas that are just variants of the key ones.

− We do not want to overwhelm you with verbose explanations. Whenever possible, concepts and techniques are demonstrated with examples and integrated into the practice problems.

− We write the practice problems and the mock exams in a similar format as the released exam and sample questions. This will enable you to comprehend questions more quickly in the real exam.

On page P-13, you will find a flow chart showing how different chapters of this manual are connected to one another. You should first study Chapters 0 to 10 in order. Chapter 0 will give you some background factual information; Chapters 1 to 4 will build you a solid foundation; and Chapters 5 to 11 will get you to the core of the exam. You should then study Chapters 12 to 16 in any order you wish. Immediately after reading a chapter, do all practice problems we provide for that chapter. Make sure that you understand every single practice problem. Finally, work on the mock exams.


P-9Preface

Before you begin your study, please download the exam syllabus from SoA’s website:

http://www.soa.org/Files/Edu/edu-2015-spring-exam-mlc-syllabus.pdf

On the last page of the exam syllabus, you will find a link to Exam MLC Tables, which are frequently used in the exam. You should keep a copy of the tables, as we are going to refer to them from time to time. You should also check the exam home page periodically for updates, corrections or notices. If you find a possible error in this manual, please let us know at the “Customer Feedback” link on the ACTEX homepage (www.actexmadriver.com). Any confirmed errata will be posted on the ACTEX website under the “Errata & Updates” link. Enjoy your study!

Preface


P-10

Syllabus Reference

Our Manual AMLCR Chapter 0: Some Factual Information

0.1 − 0.6 1

Chapter 1: Survival Distributions 1.1 2.1, 2.2 1.2 2.2 1.3 2.4 1.4 2.6 1.5 2.3

Chapter 2: Life Tables

2.1 3.1, 3.2 2.2 3.3 2.3 3.7, 3.8, 3.9 2.4 2.5, 2.6 2.5

Chapter 3: Life Insurances

3.1 4.4.1, 4.4.5, 4.4.7, 4.6 3.2 4.4.2, 4.4.6, 4.4.7, 4.6 3.3 4.4.3 3.4 4.4.8, 4.5 3.5 4.4.4 3.6 4.5 3.7

Chapter 4: Life Annuities

4.1 5.5 4.2 5.4.1, 5.4.2, 5.9, 5.10 4.3 5.4.3, 5.4.4 4.4 5.6 4.5 5.8 4.6 5.11.1 4.7 5.11.2, 5.11.3 4.8


P-11Preface

Our Manual AMLCR Chapter 5: Premium Calculation

5.1 6.1, 6.2 5.2 6.5 5.3 6.5 5.4 6.4 5.5 6.7

Chapter 6: Net Premium Reserves

6.1 7.1, 7.3.1, 7.8 6.2 7.3.3 6.3 7.4 6.4 7.7

Chapter 7: Insurance Models Including Expenses

7.1 6.6 7.2 7.3.2, 7.5 7.3 7.9 7.4 6.7, 7.3.4, 7.3.5

Chapter 8: Multiple Decrement Models: Theory

8.1 8.8 8.2 8.8, 8.9 8.3 8.8, 8.10 8.4 8.12

Chapter 9: Multiple Decrement Models: Applications

9.1 9.2 9.3 7.6 9.4

Chapter 10: Multiple State Models

10.1 8.13 10.2 8.2, 8.3, 8.11 10.3 8.4, 8.5 10.4 8.6 10.5 8.7

Preface


P-12

Our Manual AMLCR Chapter 11: Multiple Life Functions

11.1 9.2 − 9.4 11.2 9.2 – 9.4 11.3 9.5 − 9.7

Chapter 12: Interest Rate Risk

12.1 11.1 − 11.3 12.2 11.1 − 11.3 12.3 6.8, 11.4

Chapter 13: Profit Testing

13.1 12.2 − 12.4 13.2 12.5 13.3 12.6, 12.7

Chapter 14: Universal Life Insurance

14.1 13.1 – 13.2, 13.4.1, 13.4.2, 13.5 14.2 13.4.1, 13.4.2, 13.4.5, 13.4.6 14.3 13.4.7 14.4 13.4.3 14.5 13.4.4, 13.4.8

Chapter 15: Participating Insurance

15.1 13.3 15.2 13.3

Chapter 16: Pension Mathematics

16.1 10.3 16.2 10.1, 10.2 16.3 10.4 16.4 10.5, 10.6

Appendix 1: Numerical Techniques

A1.1 8.6 A1.2 7.5.2 A1.3 7.5.2


P-13Preface

Flow Chart

1. Survival Distributions

2. Life Tables

3. Life Insurances

4. Life Annuities

6. Net Premium Reserves

12. Interest Rate Risk

8. Multiple Decrement Models: Theory

14. Universal Life Insurance

10. Multiple State Models

11. Multiple Life Functions

A1. Numerical Techniques

5. Premium Calculation

7. Insurance Models Including Expenses

15. Participating Insurance

0. Some Factual Information

16. Pension Mathematics

9. Multiple Decrement Models: Applications

13. Profit Testing

Preface


P-14


C1-1Chapter 1: Survival Distributions

Chapter 1 Survival Distributions

1. To define future lifetime random variables 2. To specify survival functions for future lifetime random variables

3. To define actuarial symbols for death and survival probabilities and

develop relationships between them 4. To define the force of mortality

In Exam FM, you valued cash flows that are paid at some known future times. In Exam MLC,

by contrast, you are going to value cash flows that are paid at some unknown future times.

Specifically, the timings of the cash flows are dependent on the future lifetime of the underlying

individual. These cash flows are called life contingent cash flows, and the study of these cash

flows is called life contingencies.

It is obvious that an important part of life contingencies is the modeling of future lifetimes. In

this chapter, we are going to study how we can model future lifetimes as random variables. A

few simple probability concepts you learnt in Exam P will be used.

Let us begin with the age-at-death random variable, which is denoted by T0. The definition of T0

can be easily seen from the diagram below.

1. 1 Age-at-death Random Variable

OBJECTIVES

Chapter 1: Survival Distributions


C1-2

The age-at-death random variable can take any value within [0, ∞). Sometimes, we assume that

no individual can live beyond a certain very high age. We call that age the limiting age, and

denote it by ω. If a limiting age is assumed, then T0 can only take a value within [0, ω].

We regard T0 as a continuous random variable, because it can, in principle, take any value on

the interval [0, ∞) if there is no limiting age or [0, ω] if a limiting age is assumed. Of course, to

model T0, we need a probability distribution. The following notation is used throughout this

study guide (and in the examination).

− F0(t) = Pr(T0 ≤ t) is the (cumulative) distribution function of T0.

− f0(t) = 0d ( )d

F tt

is the probability density function of T0. For a small interval Δt, the product

f0(t)Δt is the (approximate) probability that the age at death is in between t and t + Δt.

In life contingencies, we often need to calculate the probability that an individual will survive to

a certain age. This motivates us to define the survival function:

S0(t) = Pr(T0 > t) = 1 – F0(t).

Note that the subscript “0” indicates that these functions are specified for the age-at-death

random variable (or equivalently, the future lifetime of a person age 0 now).

Not all functions can be regarded as survival functions. A survival function must satisfy the

following requirements:

1. S0(0) = 1. This means every individual can live at least 0 years.

2. S0(ω ) = 0 or )(lim 0 tSt ∞→

= 0. This means that every individual must die eventually.

3. S0(t) is monotonically decreasing. This means that, for example, the probability of surviving

to age 80 cannot be greater than that of surviving to age 70.

T0 0 Age

Death occurs



Summing up, f0(t), F0(t) and S0(t) are related to one another as follows.

Note that because T0 is a continuous random variable, Pr(T0 = c) = 0 for any constant c. Now, let

us consider the following example.

You are given that S0(t) = 1 – t/100 for 0 ≤ t ≤ 100.

(a) Verify that S0(t) is a valid survival function.

(b) Find expressions for F0(t) and f0(t).

(c) Calculate the probability that T0 is greater than 30 and smaller than 60.

Solution

(a) First, we have S0(0) = 1 – 0/100 = 1.

Second, we have S0(100) = 1 – 100/100 = 0.

Third, the first derivative of S0(t) is −1/100, indicating that S0(t) is non-increasing.

Hence, S0(t) is a valid survival function.

(b) We have F0(t) = 1 – S0(t) = t/100, for 0 ≤ t ≤ 100.

Also, we have and f0(t) = ddt

F0(t) = 1/100, for 0 ≤ t ≤ 100.

(c) Pr(30 < T0 < 60) = S0(30) – S0(60) = (1 – 30/100) – (1 – 60/100) = 0.3.

F O R M U L A

Relations between f0(t), F0(t) and S0(t)

0 0 0d d( ) ( ) ( )d d

f t F t S tt t

= = − , (1.1)

)(1d)(1d)()( 0

0 0

00 tFuufuuftSt

t−=−== ∫∫

∞, (1.2)

Pr(a < T0 ≤ b) = )()()()(d)( 0000

0 bSaSaFbFuufb

a−=−=∫ . (1.3)

Example 1.1 [Structural Question]

[ END ]



C1-4

Consider an individual who is age x now. Throughout this text, we use (x) to represent such an

individual. Instead of the entire lifetime of (x), we are often more interested in the future

lifetime of (x). We use Tx to denote the future lifetime random variable for (x). The definition of

Tx can be easily seen from the diagram below.

[Note: For brevity, we may only display the portion starting from age x (i.e., time 0) in future

illustrations.]

If there is no limiting age, Tx can take any value within [0, ∞). If a limiting age is assumed, then

Tx can only take a value within [0, ω − x]. We have to subtract x because the individual has

attained age x at time 0 already.

We let Sx(t) be the survival function for the future lifetime random variable. The subscript “x”

here indicates that the survival function is defined for a life who is age x now. It is important to

understand that when modeling the future lifetime of (x), we always know that the individual is

alive at age x. Thus, we may evaluate Sx(t) as a conditional probability:

0 0

0 0 0 0

0 0 0

( ) Pr( ) Pr( | )Pr( ) Pr( ) ( ) .

Pr( ) Pr( ) ( )

x xS t T t T x t T xT x t T x T x t S x t

T x T x S x

= > = > + >

> + ∩ > > + += = =

> >

The third step above follows from the equation Pr( )Pr( | )Pr( )A BA B

B∩

= , which you learnt in

Exam P.

1. 2 Future Lifetime Random Variable

x + Tx 0 Age

Time from now Tx 0

x

Death occursNow



With Sx(t), we can obtain Fx(t) and fx(t) by using

Fx(t) = 1 – Sx(t) and fx(t) = d ( )d xF tt

,

respectively.

You are given that S0(t) = 1 – t/100 for 0 ≤ t ≤ 100.

(a) Find expressions for S10(t), F10(t) and f10(t).

(b) Calculate the probability that an individual age 10 now can survive to age 25.

(c) Calculate the probability that an individual age 10 now will die within 15 years.

Solution

(a) In this part, we are asked to calculate functions for an individual age 10 now (i.e., x = 10).

Here, ω = 100 and therefore these functions are defined for 0 ≤ t ≤ 90 only.

First, we have 010

0

(10 ) 1 (10 ) /100( ) 1(10) 1 10 /100 90

S t t tS tS

+ − += = = −

−, for 0 ≤ t ≤ 90.

Second, we have F10(t) = 1 – S10(t) = t/90, for 0 ≤ t ≤ 90.

Finally, we have 10 10d 1( ) ( )d 90

f t F tt

= = .

(b) The probability that an individual age 10 now can survive to age 25 is given by

Pr(T10 > 15) = S10(15) = 1 − 9015 =

65 .

(c) The probability that an individual age 10 now will die within 15 years is given by

Pr(T10 ≤ 15) = F10(15) = 1 − S10(15) = 61 .

F O R M U L A

Survival Function for the Future Lifetime Random Variable

0

0

( )( )( )x

S x tS tS x

+= (1.4)


[ END ]



C1-6

For convenience, we have designated actuarial notation for various types of death and survival

probabilities.

Notation 1: t px

We use t px to denote the probability that a life age x now survives to t years from now. By

definition, we have

t px = Pr(Tx > t) = Sx(t).

When t = 1, we can omit the subscript on the left-hand-side; that is, we write 1px as px.

Notation 2: t qx

We use t qx to denote the probability that a life age x now dies before attaining age x + t. By

definition, we have

t qx = Pr(Tx ≤ t) = Fx(t).

When t = 1, we can omit the subscript on the left-hand-side; that is, we write 1qx as qx.

Notation 3: t|u qx

We use t |u qx to denote the probability that a life age x now dies between ages x + t and x + t + u.

By definition, we have

t|u qx = Pr(t < Tx ≤ t + u) = Fx(t + u) − Fx(t) = Sx(t) − Sx(t + u).

When u = 1, we can omit the subscript u; that is, we write t |1 qx as t | qx.

Note that when we describe survival distributions, “p” always means a survival probability,

while “q” always means a death probability. The “|” between t and u means that the death

probability is deferred by t years. We read “t | u” as “t deferred u”. It is important to remember

the meanings of these three actuarial symbols. Let us study the following example.

1. 3 Actuarial Notation



Express the probabilities associated with the following events in actuarial notation.

(a) A new born infant dies no later than age 45.

(b) A person age 20 now survives to age 38.

(c) A person age 57 now survives to age 60 but dies before attaining age 65.

Assuming that S0(t) = e−0.0125t for t ≥ 0, evaluate the probabilities.

Solution

(a) The probability that a new born infant dies no later than age 45 can be expressed as 45q0.

[Here we have “q” for a death probability, x = 0 and t = 45.]

Further, 45q0 = F0(45) = 1 – S0(45) = 0.4302.

(b) The probability that a person age 20 now survives to age 38 can be expressed as 18p20. [Here

we have “p” for a survival probability, x = 20 and t = 38 – 20 = 18.]

Further, we have 18p20 = S20(18) = 0

0

(38)(20)

SS

= 0.7985.

(c) The probability that a person age 57 now survives to age 60 but dies before attaining age 65

can be expressed as 3|5q57. [Here, we have “q” for a (deferred) death probability, x = 57, t =

60 – 57 = 3, and u = 65 – 60 = 5.]

Further, we have 3|5q57 = S57(3) – S57(8) = 0

0

(60)(57)

SS

− 0

0

(65)(57)

SS

= 0.058357.

[ END ]

Other than their meanings, you also need to know how these symbols are related to one another.

Here are four equations that you will find very useful.

Equation 1: t px + t qx = 1

This equation arises from the fact that there are only two possible outcomes: dying within t

years or surviving to t years from now.

Example 1.3



C1-8

Equation 2: t+u px = t px × u px+t

The meaning of this equation can be seen from the following diagram.

Mathematically, we can prove this equation as follows:

0 0 0

0 0 0

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )t u x x x x t t x u x t

S x t u S x t S x t up S t u S t S u p pS x S x S x t+ + +

+ + + + += + = = = = ×

+.

Equation 3: t|u qx = t+u qx – t qx = t px – t+u px

This equation arises naturally from the definition of t|u qx.

We have t|u qx = Pr(t < Tx ≤ t + u) = Fx( t + u) − Fx( t ) = t|u qx = t+u qx – t qx.

Also, t|u qx = Pr(t < Tx ≤ t + u) = Sx(t) − Sx(t + u) = t px – t+u px.

Equation 4: t|u qx = t px × u qx+t

The reasoning behind this equation can be understood from the following diagram:

0 t t + u Time from now

Survive from time 0 to t: probability = t px

Survive from time t to t + u: probability = u px+ t

Survive from time 0 to t + u: probability = t+u px

(Age x) (Age x + t)

Death occurs: prob.= uqx+t Time from now

0 t t + u

Survive from time 0 to time t: probability = t px

t|u qx

(Age x) (Age x + t)



Mathematically, we can prove this equation as follows:

t|u qx = t px – t + u px (from Equation 3)

= t px – t px × u px+t (from Equation 2)

= t px (1 − u px+t )

= t px × u qx+t (from Equation 1)

Here is a summary of the equations that we just introduced.

Let us go through the following example to see how these equations are applied.

You are given:

(i) px = 0.99

(ii) px+1 = 0.985

(iii) 3px+1 = 0.95

(iv) qx+3 = 0.02

Calculate the following:

(a) px+3

(b) 2px

(c) 2px+1

(d) 3px

(e) 1|2qx

F O R M U L A

Relations between t px, t qx and t|u qx

t px + t qx = 1, (1.5)

t+u px = t px × u px+ t, (1.6)

t|u qx = t+u qx – t qx = t px – t+u px = t px × u qx+ t. (1.7)

Example 1.4



C1-10

Solution

(a) px+3 = 1 – qx+3

= 1 – 0.02 = 0.98

(b) 2px = px × px+1

= 0.99 × 0.985 = 0.97515

(c) Consider 3px+1 = 2px+1× px+3

⇒ 0.95 = 2px+1 × 0.98

⇒ 2px+1 = 0.9694

(d) 3px = px × 2px+1

= 0.99 × 0.9694 = 0.9597

(e) 1|2qx = px × 2qx+1

= px (1 – 2px+1)

= 0.99 (1 – 0.9694) = 0.0303

[ END ]

In practice, actuaries use Excel extensively, so a discrete version of the future lifetime random

variable would be easier to work with. We define

x xK T= ⎢ ⎥⎣ ⎦ ,

where y⎢ ⎥⎣ ⎦ means the integral part of y. For example, ⎣ ⎦ 11 = , 4.3⎢ ⎥⎣ ⎦ = 4 and 10.99⎢ ⎥⎣ ⎦ = 10. We

call Kx the curtate future lifetime random variable.

It is obvious that Kx is a discrete random variable, since it can only take non-negative integral

values (i.e., 0, 1, 2, …). The probability mass function for Kx can be derived as follows:

Pr(Kx = 0) = Pr(0 ≤ Tx < 1) = qx,

Pr(Kx = 1) = Pr(1 ≤ Tx < 2) = 1|1qx,

Pr(Kx = 2) = Pr(2 ≤ Tx < 3) = 2|1qx, …

1. 4 Curtate Future Lifetime Random Variable



Inductively, we have

The cumulative distribution function can be derived as follows:

Pr(Kx ≤ k) = Pr(Tx < k + 1) = k+1qx, for k = 0, 1, 2, … .

It is just that simple! Now, let us study the following example, which is taken from a previous

SoA Exam.

For (x):

(i) K is the curtate future lifetime random variable.

(ii) qx+k = 0.1(k + 1), k = 0, 1, 2, …, 9

Calculate Var(K ∧ 3).

(A) 1.1 (B) 1.2 (C) 1.3 (D) 1.4 (E) 1.5

Solution

The notation ∧ means “minimum”. So here K ∧ 3 means min(K, 3). For convenience, we let

W = min(K, 3). Our job is to calculate Var(W). Note that the only possible values that W can

take are 0, 1, 2, and 3.

To accomplish our goal, we need the probability function of W, which is related to that of K.

The probability function of W is derived as follows:

Pr(W = 0) = Pr(K = 0) = qx = 0.1

Pr(W = 1) = Pr(K = 1) = 1|qx

= px × qx+1

F O R M U L A

Probability Mass Function for Kx

Pr(Kx = k) = k|1qx, k = 0, 1, 2, … (1.8)

Example 1.5 [Course 3 Fall 2003 #28]



C1-12

= (1 – qx)qx+1

= (1 – 0.1) × 0.2 = 0.18

Pr(W = 2) = Pr(K = 2) = 2|qx

= 2px × qx+ 2 = px × px+1 × qx+ 2

= (1 – qx)(1 – qx+1) qx+ 2

= 0.9 × 0.8 × 0.3 = 0.216

Pr(W = 3) = Pr(K ≥ 3) = 1 – Pr(K = 0) – Pr(K = 1) – Pr(K = 2) = 0.504.

From the probability function for W, we obtain E(W) and E(W 2

) as follows:

E(W) = 0 × 0.1 + 1 × 0.18 + 2 × 0.216 + 3 × 0.504 = 2.124

E(W 2 ) = 02 × 0.1 + 12 × 0.18 + 22 × 0.216 + 32 × 0.504 = 5.58

This gives Var(W) = E(W 2

) – [E(W)]2 = 5.58 – 2.1242 = 1.07. Hence, the answer is (A).

[ END ]

In Exam FM, you learnt a concept called the force of interest, which measures the amount of

interest credited in a very small time interval. By using this concept, you valued, for example,

annuities that make payouts continuously. In this exam, you will encounter continuous life

contingent cash flows. To value such cash flows, you need a function that measures the

probability of death over a very small time interval. This function is called the force of mortality.

Consider an individual age x now. The force of mortality for this individual t years from now is

denoted by μx+t or μx(t). At time t, the (approximate) probability that this individual dies within a

very small period of time Δt is μx+t Δt. The definition of μx+t can be seen from the following

diagram.

1. 5 Force of Mortality



From the diagram, we can also tell that fx(t) Δt = Sx(t)μx+t Δt. It follows that

fx(t) = Sx(t)μx+t = t px μx+ t .

This is an extremely important relation, which will be used throughout this study manual.

Recall that )()()( tStFtf xxx ′−=′= . This yields the following equation:

)()(

tStS

x

xtx

′−=+μ ,

which allows us to find the force of mortality when the survival function is known.

Recall that d ln 1d

xx x

= , and that by chain rule, )()(

d)(lnd

xgxg

xxg ′

= for a real-valued function g.

We can rewrite the previous equation as follows:

)].(lnd[dd

)](lnd[)()(

tStt

tStStS

xtx

xtx

x

xtx

=−

−=

′−=

+

+

+

μ

μ

μ

Replacing t by u,

.dexp)(

)0(ln)(lnd

)](ln[dd

)](lnd[d

0

0

0

0

⎟⎠⎞⎜

⎝⎛−=

−=−

=−

=−

∫

∫∫∫

+

+

+

+

t

uxx

xx

t

ux

x

tt

ux

xux

utS

StSu

uSu

uSu

μ

μ

μ

μ

This allows us to find the survival function when the force of mortality is known.

Time from now

0 t t + Δt

Survive from time 0 to time t: Prob. = Sx(t)

Death occurs during t to t + Δt:

Prob. ≈ μx+ t Δt

Death between time t and t + Δt: Prob. (measured at time 0) ≈ fx(t)Δt



C1-14

Not all functions can be used for the force of mortality. We require the force of mortality to

satisfy the following two criteria:

(i) μx+t ≥ 0 for all x ≥ 0 and t ≥ 0.

(ii) ∫∞

+ ∞=

0 duuxμ .

Criterion (i) follows from the fact that μx+t Δt is a measure of probability, while Criterion (ii)

follows from the fact that )(lim tS xt ∞→= 0.

Note that the subscript x + t indicates the age at which death occurs. So you may use μx to

denote the force of mortality at age x. For example, μ20 refers to the force of mortality at age 20.

The two criteria above can then be written alternatively as follows:

(i) μx ≥ 0 for all x ≥ 0.

(ii) ∫∞

∞=

0 dxxμ .

The following two specifications of the force of mortality are often used in practice.

Gompertz’ law

μx = Bcx

Makeham’s law

μx = A + Bcx

In the above, A, B and c are constants such that A ≥ −B, B > 0 and c > 1.

F O R M U L A

Relations between μx+t, fx(t) and Sx(t)

fx(t) = Sx(t)μx+ t = t px μx+ t, (1.9)

)()(

tStS

x

xtx

′−=+μ , (1.10)

.dexp)(

0 ⎟⎠⎞⎜

⎝⎛−= ∫ +

t

uxx utS μ (1.11)



Let us study a few examples now.

For a life age x now, you are given: 2(10 )( )

100xtS t −

= , 0 ≤ t < 10.

(a) Find μx+t .

(b) Find fx(t).

Solution

(a) tt

t

tStS

x

xtx −

=−

−−

−=′

−=+ 102

100)10(

100)10(2

)()(

2μ .

(b) You may work directly from Sx(t), but since we have found μx+ t already, it would be quicker

to find fx(t) as follows:

fx(t) = Sx(t)μx+t = 2(10 ) 2 10

100 10 50t t

t− −

× =−

.

[ END ]

For a life age x now, you are given

μx+t = 0.002t, t ≥ 0.

(a) Is μx+ t a valid function for the force of mortality of (x)?

(b) Find Sx(t).

(c) Find fx(t).

Solution

(a) First, it is obvious that μx+t ≥ 0 for all x and t.

Second, 2

00 0d 0.002 d 0.001x u u u u uμ

∞ ∞ ∞

+ = = = ∞∫ ∫ .





C1-16

Hence, it is a valid function for the force of mortality of (x).

(b) Sx(t) = )001.0exp(d002.0expdexp 2

0

0 tuuu

tt

ux −=⎟⎠⎞⎜

⎝⎛−=⎟

⎠⎞⎜

⎝⎛− ∫∫ +μ .

(c) fx(t) = Sx(t)μx+t = 0.002texp(−0.001t2).

[ END ]

You are given:

(i) ⎟⎠⎞⎜

⎝⎛−−= ∫ +

1

0 dexp1 tR txμ

(ii) ⎟⎠⎞⎜

⎝⎛ +−−= ∫ +

1

0 d)(exp1 tkS txμ

(iii) k is a constant such that S = 0.75R.

Determine an expression for k.

(A) ln((1 – qx) / (1 − 0.75qx))

(B) ln((1 – 0.75qx) / (1 − px))

(C) ln((1 – 0.75px) / (1 − px))

(D) ln((1 – px) / (1 − 0.75qx))

(E) ln((1 – 0.75qx) / (1 − qx))

Solution

First, R = 1 – Sx(1) = 1 − px = qx.

Second,

xk

xkt

txkt

tx peSeueukS −−+

−+ −=−=⎟

⎠⎞⎜

⎝⎛−−=⎟

⎠⎞⎜

⎝⎛ +−−= ∫∫ 1)1(1dexp1d)(exp1

0

0 μμ .

Since S = 0.75R, we have

Example 1.8 [Course 3 Fall 2002 #35]



1 0.751 0.75

.1 0.75

1ln ln1 0.75 1 0.75

kx x

k x

x

k x

x

x x

x x

e p qqe

ppe

q

p qkq q

−

−

− =−

=

=−

⎛ ⎞ ⎛ ⎞−= =⎜ ⎟ ⎜ ⎟− −⎝ ⎠ ⎝ ⎠

Hence, the answer is (A).

[ END ]

(a) Show that when μx = Bcx, we have )1( −=

tx ccxt gp ,

where g is a constant that you should identify.

(b) For a mortality table constructed using the above force of mortality, you are given that 10p50

= 0.861716 and 20p50 = 0.718743. Calculate the values of B and c.

Solution

(a) To prove the equation, we should make use of the relationship between the force of

mortality and tpx.

⎟⎠⎞

⎜⎝⎛ −

−=⎟

⎠⎞⎜

⎝⎛−=⎟

⎠⎞⎜

⎝⎛−= ∫∫ +

+ )1(ln

expdexpdexp00

txt sxt

sxxt cccBsBcsp μ .

This gives g = exp(−B/lnc).

(b) From (a), we have )1( 1050

861786.0 −= ccg and )1( 2050

718743.0 −=tccg . This gives

)861716.0ln()718743.0ln(

11

10

20

=−−

cc .

Solving this equation, we obtain c = 1.02000. Substituting back, we obtain g = 0.776856 and

B = 0.00500.

[ END ]




C1-18

Now, let us study a longer structural question that integrates different concepts in this chapter.

The function

1800011018000 2xx −−

has been proposed for the survival function for a mortality model.

(a) State the implied limiting age ω.

(b) Verify that the function satisfies the conditions for the survival function S0(x).

(c) Calculate 20p0.

(d) Calculate the survival function for a life age 20.

(e) Calculate the probability that a life aged 20 will die between ages 30 and 40.

(f) Calculate the force of mortality at age 50.

Solution

(a) Since

018000

11018000)(2

0 =−−

=ωωωS ,

We have ω2 + 110ω – 18000 = 0 ⇒ (ω – 90)(ω + 200) = 0 ⇒ ω = 90 or ω = −200 (rejected).

Hence, the implied limiting age is 90.

(b) We need to check the following three conditions:

(i) 118000

0011018000)0(2

0 =−×−

=S

(ii) 018000

11018000)(2

0 =−−

=ωωωS

(iii) 018000

1102)(dd

0 <+

−=xxS

x

Therefore, the function satisfies the conditions for the survival function S0(x).




(c) 85556.018000

202011018000)20(2

0020 =−×−

== Sp

(d)

.15400

1501540015400

)220)(70(18000

)20020)(2090(18000

)20020)(2090(

)20()20()(

2

0

020

xxxx

xx

SxSxS

−−=

+−=

+−

++−−

=+

=

(e) The required probability is

10|10q20 = 10p20 – 20p20

= 11688.077922.089610.015400

)22020)(2070(15400

)22010)(1070(=−=

+−−

+−

(f) First, we find an expression for μx.

)200)(90(1102

18000)200)(90(

180002110

)()(

0

0

+−+

=+−

−−

−=′

−=xx

xxx

x

xSxS

xμ .

Hence, μ50 = )20050)(5090(

110502+−

+×= 0.021.

[ END ]

You may be asked to prove some formulas in the structural questions of Exam MLC. Please

study the following example, which involves several proofs.

(d)



C1-20

Prove the following equations:

(a) xt ptd

d= −t pxμx+t

(b) ∫ +=t

sxxsxt spq

0 dμ

(c) 1d

0 =∫

−

+

x

txxt tpω

μ

Solution

(a) LHS = )(ddd)dexp()dexp(

dd

dd

0

0

0 txxt

t

sx

t

sx

t

sxxt pst

sst

pt ++++ −=⎟

⎠⎞

⎜⎝⎛−−=−= ∫∫∫ μμμμ = RHS

(b) LHS = t qx = Pr(Tx ≤ t) = spssft

sxxs

t

x dd)(

0

0 ∫∫ += μ = RHS

(c) LHS = ttftpx

xtx

x

xt d)(d

0

0 ∫∫−

+

−=

ωωμ = ω−xqx = 1 = RHS

[ END ]




1. [Structural Question] You are given:

01( )

1S t

t=

+, t ≥ 0.

(a) Find F0(t).

(b) Find f0(t).

(c) Find Sx(t).

(d) Calculate p20.

(e) Calculate 10|5q30. 2. You are given:

2

0(30 )( )

9000tf t −

= , for 0 ≤ t < 30

Find an expression for t p5. 3. You are given:

020( )200

tf t −= , 0 ≤ t < 20.

Find μ10. 4. [Structural Question] You are given:

1100x x

μ =−

, 0 ≤ x < 100.

(a) Find S20(t) for 0 ≤ t < 80.

(b) Compute 40p20.

(c) Find f20(t) for 0 ≤ t < 80. 5. You are given:

2100x x

μ =−

, for 0 ≤ x < 100.

Find the probability that the age at death is in between 20 and 50. 6. You are given:

(i) S0(t) =α

ω⎟⎠⎞

⎜⎝⎛ −

t1 0 ≤ t < ω, α > 0.

(ii) μ40 = 2μ20.

Find ω.

Exercise 1



C1-22

7. Express the probabilities associated with the following events in actuarial notation.

(a) A new born infant dies no later than age 35.

(b) A person age 10 now survives to age 25.

(c) A person age 40 now survives to age 50 but dies before attaining age 55.

Assuming that S0(t) = e−0.005t for t ≥ 0, evaluate the probabilities. 8. You are given:

2

0 ( ) 1100

tS t ⎛ ⎞= −⎜ ⎟⎝ ⎠

, 0 ≤ t < 100.

Find the probability that a person aged 20 will die between the ages of 50 and 60. 9. You are given:

(i) 2px = 0.98

(ii) px+2 = 0.985

(iii) 5qx = 0.0775


(a) 3px

(b) 2px+3

(c) 2|3qx 10. You are given:

qx+k = 0.1(k + 1), k = 0, 1, 2, …, 9.


(a) Pr(Kx = 1)

(b) Pr(Kx ≤ 2) 11. [Structural Question] You are given μx = μ for all x ≥ 0.

(a) Find an expression for Pr(Kx = k), for k = 0, 1, 2, …, in terms of μ and k.

(b) Find an expression for Pr(Kx ≤ k), for k = 0, 1, 2, …, in terms of μ and k.

Suppose that μ = 0.01.

(c) Find Pr(Kx = 10).

(d) Find Pr(Kx ≤ 10).



12. Which of the following is equivalent to ∫ +

t

uxxu up

0 dμ ?

(A) t px

(B) t qx

(C) fx(t)

(D) – fx(t)

(E) fx(t)μx+t

13. Which of the following is equivalent to dd t xpt

?

(A) –t px μx+t

(B) μx+t

(C) fx(t)

(D) –μx+t

(E) fx(t)μx+t 14. (2000 Nov #36) Given:

(i) μx = F + e2x, x ≥ 0

(ii) 0.4p0 = 0.50

Calculate F.

(A) –0.20

(B) –0.09

(C) 0.00

(D) 0.09

(E) 0.20 15. (CAS 2004 Fall #7) Which of the following formulas could serve as a force of mortality?

(I) μx = Bcx, B > 0, C > 1

(II) μx = a(b + x)−1, a > 0, b > 0

(III) μx = (1 + x)−3, x ≥ 0 (A) (I) only

(B) (II) only

(C) (III) only

(D) (I) and (II) only

(E) (I) and (III) only



C1-24

16 (2002 Nov #1) You are given the survival function S0(t), where

(i) S0(t) = 1, 0 ≤ t ≤ 1

(ii) S0(t) 1001

te−= , 1 ≤ t ≤ 4.5

(iii) S0(t) = 0, 4.5 ≤ t

Calculate μ 4.

(A) 0.45

(B) 0.55

(C) 0.80

(D) 1.00

(E) 1.20

17. (CAS 2004 Fall #8) Given 1/ 2

0 ( ) 1100

tS t ⎛ ⎞= −⎜ ⎟⎝ ⎠

, for 0 ≤ t ≤ 100, calculate the probability that

a life age 36 will die between ages 51 and 64. (A) Less than 0.15

(B) At least 0.15, but less than 0.20

(C) At least 0.20, but less than 0.25

(D) At least 0.25, but less than 0.30

(E) At least 0.30 18. (2007 May #1) You are given:

(i) 3p70 = 0.95

(ii) 2p71 = 0.96

(iii) 107.0d 75

71=∫ xxμ

Calculate 5p70. (A) 0.85

(B) 0.86

(C) 0.87

(D) 0.88

(E) 0.89



19. (2005 May #33) You are given:

0.05 50 60

0.04 60 70x

xx

μ≤ <⎧

= ⎨ ≤ <⎩

Calculate 4|14q50 .

(A) 0.38

(B) 0.39

(C) 0.41

(D) 0.43

(E) 0.44 20. (2004 Nov #4) For a population which contains equal numbers of males and females at birth:

(i) For males, mxμ = 0.10, x ≥ 0

(ii) For females, fxμ = 0.08, x ≥ 0

Calculate q60 for this population. (A) 0.076

(B) 0.081

(C) 0.086

(D) 0.091

(E) 0.096 21. (2001 May #28) For a population of individuals, you are given:

(i) Each individual has a constant force of mortality.

(ii) The forces of mortality are uniformly distributed over the interval (0, 2).

Calculate the probability that an individual drawn at random from this population dies within one year. (A) 0.37

(B) 0.43

(C) 0.50

(D) 0.57

(E) 0.63



C1-26

22. [Structural Question] The mortality of a certain population follows the De Moivre’s Law; that is

xx −=

ωμ 1 , x < ω.

(a) Show that the survival function for the age-at-death random variable T0 is

ωxxS −=1)(0 , 0 ≤ x < ω.

(b) Verify that the function in (a) is a valid survival function.

(c) Show that

xpxt −

−=ω

11 , 0 ≤ t < ω – x, x < ω.

23. [Structural Question] The probability density function for the future lifetime of a life age 0

is given by

10 )()( ++

= α

α

λαλ

xxf , α, λ > 0

(a) Show that the survival function for a life age 0, S0(x), is α

λλ

⎟⎠⎞

⎜⎝⎛

+=

xxS )(0 .

(b) Derive an expression for μx.

(c) Derive an expression for Sx(t).

(d) Using (b) and (c), or otherwise, find an expression for fx(t). 24. [Structural Question] For each of the following equations, determine if it is correct or not.

If it is correct, prove it.

(a) t|uqx = tpx + uqx+t

(b) t+uqx = tqx × uqx+t

(c) )(dd

txxxtxt ppx +−= μμ



Solutions to Exercise 1

1. (a) 01( ) 1

1 1tF t

t t= − =

+ +.

(b) 0 0 2 2

d 1 1( ) ( )d (1 ) (1 )

t tf t F tt t t

+ −= = =

+ +.

(c) 0

0

1( ) 11( ) 1( ) 1

1

xS x t xx tS t

S x x tx

+ ++ += = =+ +

+

.

(d) p20 = S20(1) = 21/22.

(e) 10|5q30 = 10p30 – 15p30 = S30(10) – S30(15) = 1 30 1 30 31 311 30 10 1 30 15 41 46

+ +− = −

+ + + + = 0.0822.

2. S0(t) [ ]

27000)30(

27000)30(

9000

d)30(d)(

330330

230

0tuuu

uuf ttt

−=

−−=

−== ∫

∫ .

If follows that 33

05 5 3

0

(5 ) (30 5 )( ) 1(5) (30 5) 25t

S t t tp S tS

+ − − ⎛ ⎞= = = = −⎜ ⎟− ⎝ ⎠.

3. S0(t) [ ]

400)20(

400)20(

200

d)20(d)(

220220

20

0tuuu

uuf ttt

−=

−−=

−== ∫

∫ .

tt

t

tStf

t −=

−

−

==20

2

400)20(

20020

)()(

20

0μ .

Hence, μ10 = 2/(20 – 10) = 0.2.

4. (a) First, note that 201 1

100 20 80t t tμ + = =

− − −. We have

.80

1)80

80exp(ln))]80exp([ln(

d80

1expdexp)(

0

0

0 20200

ttu

uu

utS

t

tt

u

−=−

=−=

⎟⎠⎞

⎜⎝⎛

−−=⎟

⎠⎞⎜

⎝⎛−= ∫∫ +μ

(b) 40p20 = S20(40) = 1 – 40/80 = 1/2.

(c) f20(t) = S20(t)μ20+t = 1 1180 80 80t

t⎛ ⎞⎛ ⎞− =⎜ ⎟⎜ ⎟−⎝ ⎠⎝ ⎠

.



C1-28

5. Our goal is to find Pr(20 < T0 < 50) = S0(20) – S0(50).

Given the force of mortality, we can find the survival function as follows:

2

0

0

0 0

1001)

100100ln2exp())]100[ln(2exp(

d100

2expdexp)(

⎟⎠⎞

⎜⎝⎛ −=

−=−=

⎟⎠⎞

⎜⎝⎛

−−=⎟

⎠⎞⎜

⎝⎛−= ∫∫

ttu

uu

utS

t

tt

uμ

So, the required probability is (1 – 20/100)2 – (1 – 50/100)2 = 0.82 – 0.52 = 0.39.

6. xx

x

xSxS

x −=

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ −⎟

⎠⎞

⎜⎝⎛−−

−=′

−=

−

ωα

ω

ωωα

μ α

α

1

11

)()(

1

0

0 .

We are given that μ40 = 2μ20. This implies 240 20

α αω ω

=− −

, which gives ω = 60.

7. (a) The probability that a new born infant dies no later than age 35 can be expressed as 35q0.

[Here we have “q” for a death probability, x = 0 and t = 35.]

Further, 35q0 = F0(35) = 1 – S0(35) = 0.1605.

(b) The probability that a person age 10 now survives to age 25 can be expressed as 15p10. [Here we have “p” for a survival probability, x = 10 and t = 25 – 10 = 15.]

Further, we have 15p10 = S10(15) = =)15()25(

0

0

SS

0.9277.

(c) The probability that a person age 40 now survives to age 50 but dies before attaining age 55 can be expressed as 10|5q40. [Here, we have “q” for a (deferred) death probability, x = 40, t = 50 – 40 = 10, and u = 55 – 50 = 5.]

Further, we have 10|5q40 = S40(10) – S40(15) = 0

0

(50)(40)

SS

− 0

0

(55)(40)

SS

= 0.0235.

8. The probability that a person aged 20 will die between the ages of 50 and 60 is given by

30|10q20 = 30p20 – 40p20 = S20(30) – S20(40).

2

2

2

0

020 80

1

100201

100201

)20()20(

)( ⎟⎠⎞

⎜⎝⎛ −=

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ +

−=

+=

tt

StS

tS .

So, S20(30) =6425

80301

2

=⎟⎠⎞

⎜⎝⎛ − , S20(40) =

6416

80401

2

=⎟⎠⎞

⎜⎝⎛ − . As a result, 30|10q20 = 9/64.



9. (a) 3px = 2px × px+2 = 0.98 × 0.985 = 0.9653.

3 2 3 5 5

52 3

3

11 1 0.0775 0.95566

0.9653

x x x x

xx

x

p p p qqp

p

+

+

× = = −

− −⇒ = = =

(c) 2|3qx = 2px – 5px = 0.98 – (1 – 0.0775) = 0.0575. 10. (a) Pr(Kx = 1) = 1|qx = px × qx+1 = (1 – qx)qx+1 = (1 – 0.1) × 0.2 = 0.18

(b) Pr(Kx = 0) = qx = 0.1

Pr(Kx = 2) = 2|qx = 2px × qx+ 2 = px × px+1 × qx+2 = (1 – qx)(1 – qx+1)qx+ 2

= 0.9 × 0.8 × 0.3 = 0.216.

Hence, Pr(Kx ≤ 2) = 0.1 + 0.18 + 0.216 = 0.496. 11. (a) Given that μx = μ for all x ≥ 0, we have t px = e−μt, px = e−μ and qx = 1 – e−μ.

Pr(Kx = k) = k|qx = kpx qx+k = e− kμ (1 – e−μ).

(b) Pr(Kx ≤ k) = k+1qx = 1 − k+1px = 1 − e−(k + 1)μ.

(c) When μ = 0.01, Pr(Kx = 10) = e−10 × 0.01(1 – e−0.01) = 0.0090.

(d) When μ = 0.01, Pr(Kx ≤ 10) = 1 – e−(10 + 1) × 0.01 = 0.1042. 12. First of all, note that upx μx+u in the integral is simply fx(u).

)Pr(d)(d

0

0 tTuufup x

t

x

t

uxxu ≤== ∫∫ +μ = Fx(t) = t qx.

Hence, the answer is (B).

13. Method I: We use t px = 1 − t qx. Differentiating both sides with respect to t,

)()(dd

dd

dd tftF

tq

tp

t xxxtxt −=−=−= .

Noting that fx(t) = t px μx+t, the answer is (A).

Method II: We differentiate t px with respect to t as follows:

.ddddexp

dexpdd)(

dd

dd

0

0

0

⎟⎠⎞⎜

⎝⎛−⎟

⎠⎞⎜

⎝⎛−=

⎟⎠⎞⎜

⎝⎛−==

∫∫

∫

++

+

t

ux

t

ux

t

uxxxt

ut

u

ut

tSt

pt

μμ

μ

Recall the fundamental theorem of calculus, which says that )(d)(dd

tguug

tt

c=∫ . Thus

txxttx

t

uxxt pupt +++ −=−⎟

⎠⎞⎜

⎝⎛−= ∫ μμμ )(dexp

dd

0 .

Hence, the answer is (A).

(b)



C1-30

14. First, note that

0.4 0.4 20 0

d ( )d

0.4 0 0.5u

u u F e up e e

μ− − +∫ ∫= = = .

The exponent in the above is 0.4

0.4 2 2

00

1( )d2

0.4 1.11277 0.50.4 0.61277

u uF e u Fu e

FF

⎛ ⎞− + = − +⎜ ⎟⎝ ⎠

= − − += − −

∫

As a result, 0.5 = e−0.4F−0.61277, which gives F = 0.2. Hence, the answer is (E). 15. Recall that we require the force of mortality to satisfy the following two criteria:

(i) μx ≥ 0 for all x ≥ 0, (ii) 0

dx xμ∞

= ∞∫ .

All three specifications of μx satisfy Criterion (i). We need to check Criterion (ii).

We have

00

dln

xx BcBc x

c

∞∞

= = ∞∫ ,

00d ln( )a x a b x

b x∞ ∞= + = ∞

+∫ ,

and

3 200

1 1 1d(1 ) 2(1 ) 2

xx x

∞∞ −

= =+ +∫ .

Only the first two specifications can satisfy Criterion (ii). Hence, the answer is (D).

[Note: μx = Bcx is actually the Gompertz’ law. If you knew that you could have identified that μx = Bcx can serve as a force of mortality without doing the integration.]

16. Recall that )()(

tStS

x

xtx

′−=+μ .

Since we need μ4, we use the definition of S0(t) for 1 ≤ t ≤ 4.5:

0 ( ) 1100

teS t = − , 100

)(0

tetS =′− .

As a result,

4

4

4 4 4100 1.203

1001100

ee

e eμ = = =

−−. Hence, the answer is (E).



17. The probability that a life age 36 will die between ages 51 and 64 is given by

S36(15) – S36(28).

We have 8

6464

64

100361

100361

)36()36(

)(2/1

2/1

2/1

0

036

ttt

StS

tS −=⎟

⎠⎞

⎜⎝⎛ −

=

⎟⎠⎞

⎜⎝⎛ −

⎟⎠⎞

⎜⎝⎛ +

−=

+= .

This gives S36(15)87

= and S36(28) 86

= . As a result, the required probability is

S36(15) – S36(28) = 1/8 = 0.125. Hence, the answer is (A). 18. The computation of 5p70 involves three steps.

First, 3 7070

2 71

0.95 0.98960.96

ppp

= = = .

Second, 75

71 d 0.107

4 71 0.8985x xp e e

μ− −∫= = = .

Finally, 5p70 = 0.9896 × 0.8985 = 0.889. Hence, the answer is (E). 19. 4p50 = e−0.05 × 4 = 0.8187

10p50 = e−0.05 × 10 = 0.6065

8p60 = e−0.04 × 8 = 0.7261

18p50 = 10p50 × 8p60 = 0.6065 × 0.7261 = 0.4404

Finally, 4|14q50 = 4p50 – 18p50 = 0.8187 – 0.4404 = 0.3783. Hence, the answer is (A). 20. For males, we have

0 0d 0.10d 0.10

0 ( )t tm

u u um tS t e e eμ− − −∫ ∫= = = .

For females, we have

0 0d 0.08d 0.08

0 ( )t tf

u u uf tS t e e eμ− − −∫ ∫= = = .

For the overall population,

0.1 60 0.08 60

0 (60) 0.0053542

e eS− × − ×+

= = ,

and

0.1 61 0.08 61

0 (61) 0.004922

e eS− × − ×+

= = .



C1-32

Finally, 060 60

0

(61)1 1 0.081(60)

Sq pS

= − = − = . Hence, the answer is (B).

21. Let M be the force of mortality of an individual drawn at random, and T be the future

lifetime of the individual. We are given that M is uniformly distributed over (0, 2). So the density function for M is fM(μ) = 1/2 for 0 < μ < 2 and 0 otherwise.

This gives

0

2

0

2

2

Pr( 1)E[Pr( 1| )]

Pr( 1| ) ( )d

1(1 ) d2

1 (2 1)21 (1 )20.56767.

M

TT M

T M f

e

e

e

μ

μ μ μ

μ

∞

−

−

−

≤= ≤

= ≤ =

= −

= + −

= +

=

∫

∫

Hence, the answer is (D). 22. (a) We have, for 0 ≤ x < ω,

ωω

ωμ ω xess

ssxS

xxxx

s −==−=−

−=−=−

∫∫ 1)]exp([ln()d1exp()dexp()()1ln(

0

0

0 0 .

(b) We need to check the following three conditions:

(i) S0(0) = 1 – 0/ω = 1

(ii) S0(ω) = 1 – ω/ω = 0

(iii) S ′0(ω) = −1/ω < 0 for all 0 ≤ x < ω, which implies S0(x) is non-increasing.

Hence, the function in (a) is a valid survival function.

(c) x

tx

txx

tx

xStxSpxt −

−=−

−−=

−

+−

=+

=ωω

ω

ω

ω 11

1

)()(

0

0 , for 0 ≤ t < ω – x, x < ω.

23. (a) ∫ ∫ +=

+−=−=−= +

x x

xs

sssfxFxS

0 0 1000 )(d

)(1d)(1)(1)( α

α

α

α

λλ

λαλ .

(b) xxS

xfx +

==λ

αμ)()(

0

0 .



(c) .)(

)()(0

0α

α

α

λλ

λλ

λλ

⎟⎠⎞

⎜⎝⎛

+++

⎟⎠⎞

⎜⎝⎛

+

⎟⎠⎞

⎜⎝⎛

++=+

=tx

x

x

txxS

txStSx

(d) txtx

xtStf txxx ++⎟⎠⎞

⎜⎝⎛

+++

== + λα

λλμ

α

)()( .

24. (a) No, the equation is not correct. The correct equation should be t|uqx = tpx × uqx+t. (b) No, the equation is not correct. The correct equation should be t+upx = tpx × upx+t. (c) Yes, the equation is correct. The proof is as follows:

)(

)()(

)()(

)()(

)()(

)()(

)()(

)()(

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

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

)()(

dd

dd

0

0

0

0

0

0

0

0

0

0

0

0

0

0

20

0000

20

0000

0

0

txxxt

xxtxttx

xt

ppp

xSxf

xStxS

xStxS

txStxf

xSxf

xStxS

xStxf

xSxftxStxfxS

xSxStxStxSxS

xStxS

xp

x

+

+

−=

+−=

++

+++−

=

++

+−=

−+−+−=

+−+=

+=

μμμμ



C1-34


T7-1Mock Test 7

Mock Test 7

Section A: Multiple Choice Questions. Each question is worth 2 points. 1. Let Y be the present value random variable for a special three-year temporary life annuity.

You are given:

(i) The life annuity pays 2 + k at time k, for k = 0, 1, and 2.

(ii) Payment ceases after 3 years.

(iii) v = 0.9

(iv) px = 0.8, px+1 = 0.75, px+2 = 0.5

Calculate the standard deviation of Y. (A) 1.2

(B) 1.8

(C) 2.4

(D) 3.0

(E) 3.6 2. For a special fully continuous insurance on (x) and (y), you are given:

(i) Tx and Ty are independent.

(ii) μx+t = 0.08, t > 0

(iii) μy+t = 0.04, t > 0

(iv) δ = 0.06

(v) π is the annual rate of net premium payable until the first of (x) and (y) dies.

(vi) The insurance pays 1 at the moment when the first of (x) and (y) dies, and 3π when the second of (x) and (y) dies.

Calculate π. (A) 0.06

(B) 0.08

(C) 0.10

(D) 0.12

(E) 0.14


Mock Test 7

T7-2

3. Consider a permanent disability model with three states:

State 0: Healthy, State 1: Permanently disabled, State 2: Dead.

The transition intensities are =01xμ 0.02, =02

xμ 0.03, =12xμ 0.05.

For a person who is healthy at age 50, calculate the probability that he cannot survive to age 70, and he has entered the state of permanent disability before reaching 60.

(A) 0.06

(B) 0.08

(C) 0.10

(D) 0.12

(E) 0.14 4. For a universal life insurance policy, you are given:

(i) The death benefit is $50,000 plus the account value at the end of the year of death. The benefit is payable at the end of the year of death.

(ii) The account value at the beginning of year 1 (before premium payment and deductions) is $12,000.

(iii) The account value at the beginning of year 2 (before premium payment and deductions) is $13,900.

(iv) Level premiums of $2,000 each are made at the beginning of each year.

(v) The annual expense charge is $50 plus 5% of the annual premium.

(vi) The credited interest rate for years 1, 2 and 3 is 5% per annum effective.

(vii) There is no corridor factor requirement for this policy.

(viii) In calculating the cost of insurance, it is assumed that the force of mortality is constant for all ages. The interest rate used is 6% per annum effective.

Assume that the policy is still in force at the end of year 2. Project the account value at the end of year 2.

(A) 14,695

(B) 14,995

(C) 15,295

(D) 15,595

(E) 15,895


T7-3Mock Test 7

5. For a special 20-year term insurance on (38), you are given:

(i) The death benefit is 10,000.

(ii) The death benefit is payable at the moment of death.

(iii) During the 5th year and the 6th year the gross premium is 145 per annum, paid continuously at a constant rate.

(iv) The force of mortality follows Makeham’s law with A = 0.00022, B = 0.00004 and c = 1.1.

(v) The force of interest is 4%.

(vi) Expenses are: − 5% of premium payable continuously − 100 payable at the moment of death

(vii) At the end of the 5th year the expected value of the present value of future losses random variable is 950.

Euler’s method with steps of h = 1/12 years is used to calculate a numerical solution to Thiele’s differential equation.

Calculate the expected value of the present value of future losses random variable at the

end of 615 years.

(A) 975

(B) 962

(C) 949

(D) 936

(E) 923 6. You are given:

(i) Mortality follows the Illustrative Life Table.

(ii) Deaths between integral ages are uniformly distributed.

Calculate the 75-percentile of the age at death for a life aged 40 years. (A) 27

(B) 44

(C) 53

(D) 68

(E) 85


Mock Test 7

T7-4

7. For a double decrement table for disabled, there are two modes of decrement: (r) for recovery, and (d) for death. It is given:

(i) ⎩⎨⎧

≥<≤−

=+ 3030)3(1.0)(

22 tttr

tμ

(ii) tdt 1.0)(

22 =+μ .

It is also given that the prices of three zero-coupon bonds, all of which will pay 1,000 at maturity, are as follows:

Maturity (in years) Current price 1 980 2 955 3 900

Mark is disabled and is scheduled to begin receiving disability payments today, his 22nd birthday. On every birthday up to and including his 25th, he will receive $1,000 as long as he has not yet recovered or died. There will be no payments beyond his 25th birthday.

Calculate the expected present value of Mark’s disability payments. (A) 2,250

(B) 2,620

(C) 2,690

(D) 2,780

(E) 2,840 8. For a fully discrete 2-year term insurance of 100 on (x), you are given:

(i) i = 0.05

(ii) The net premium is 19.

(iii) The time-1 net premium reserve is 4.5.

Calculate the standard deviation of the loss-at-issue random variable. (A) 46

(B) 47

(C) 48

(D) 49

(E) 50


T7-5Mock Test 7

9. You are given:

(i) δ = 0.05

(ii) xA = 0.44

(iii) xA2 = 0.22

Consider a portfolio of 100 fully continuous whole life insurances. The ages of the all insureds are x, and their lifetimes are assumed to be independent. The face amount of the policies, the premium rate and the number of policies are as follows:

Face amount Premium rate Number of Policies

100 4.3 75 400 17.5 25

By using a normal approximation, calculate the probability that the present value of the aggregate loss-at-issue for the insurer’s portfolio will exceed 700.

(A) 1 − Φ(2.28)

(B) 1 − Φ(0.17)

(C) Φ(0)

(D) Φ(0.17)

(E) Φ(2.28) 10. For a fully discrete 15-year endowment insurance of 1,000 on (50) that has been in force

for 5 years, you are given:

(i) Mortality follows the Illustrative Life Table.

(ii) i = 0.06

(iii) At issue, the net premium was calculated using the equivalence principle.

(iv) When the insured decides to stop paying premiums after 5 years, the death benefit remains at 1,000 but the pure endowment value is reduced such that the expected prospective loss at age 55 is unchanged.

Calculate the reduced pure endowment value. (A) 260

(B) 280

(C) 300

(D) 320

(E) 340


Mock Test 7

T7-6

11. The expense-loaded premium, G, for a fully-discrete 3-year endowment insurance of 1,000 on (x) is calculated using the equivalence principle. Expenses are paid at the beginning of each year. You are given:

(i) The net premium for the insurance is 323.12.

(ii) G = 390.20

(iii) Renewal expenses of 10% of each gross premium and per policy expense of 3 are payable at the beginning of the second and the third policy year.

Calculate the expense reserve at the end of the second year. (A) −25

(B) −35

(C) −45

(D) −55

(E) −60 12. You are given the following about a homogeneous discrete-time Markov chain with 3

states:

State 0: Employed full-time State 1: Employed part-time State 2: Unemployed

(i) Workers transit through the labour force according to the following matrix:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

6.015.025.015.08.005.003.007.09.0

(ii) Workers transit through the labour force independently at the end of the year.

Consider two workers X and Y, one employed full-time at time 0, another employed part-time at time 0.

Calculate the probability that either one of them, but not both, will be unemployed at the end of year 2.

(A) 0.18

(B) 0.20

(C) 0.22

(D) 0.24

(E) 0.26


T7-7Mock Test 7

13. You are given:

(i) Deaths are uniformly distributed over each year of age.

(ii) q70 = 0.02

(iii) i = 0.07

(iv) .071 =A 445

Calculate )4(70A .

(A) 0.402

(B) 0.415

(C) 0.423

(D) 0.431

(E) 0.443 14. An employee aged exactly 35 on January 1, 2010 has an annual salary of 26,000 during

that year. Salaries are revised annually on December 31 each year. Future salaries are estimated using the salary scale given in the table below, where Sy / Sx, y > x denotes the ratio of salary earned in the year of age from y to y + 1 to the salary earned in the year of age x to x + 1, for a life in employment over the entire period (x, x + 1).

x Sx

35 1.130 36 1.210 M M

62 3.589 63 3.643 64 3.698 65 3.751

If the final average salary is defined as the average salary in the three years before

retirement, compute the member’s expected final average salary. Assume the retirement age is 65.

(A) 74,812

(B) 79,447

(C) 83,829

(D) 85,071

(E) 96,131


Mock Test 7

T7-8

15. Which of the following statement(s) is/are true for a universal life insurance sold with a no-lapse guarantee?

(I) The policy must be discontinued if the account value drops to zero.

(II) The no-lapse guarantee is more likely to come into effect if the interest rate used to calculate the cost of insurance is smaller.

(III) The no-lapse guarantee is more likely to come into effect if the death probability used to calculate the cost of insurance is smaller.

(A) (I) only

(B) (I) and (II) only

(C) (I) and (III) only

(D) (II) and (III) only

(E) (I), (II) and (III)

16. Which of the following is/are strictly increasing function(s) of Tx for all Tx ≥ 0?

(I) The present value random variable for a continuous whole life annuity of $1 on (x)

(II) The present value random variable for a continuous n-year temporary life annuity of $1 on (x)

(III) The net future loss at issue random variable for a fully continuous whole life insurance of $1 on (x).

(A) (I) only

(B) (III) only

(C) (I), (II) only

(D) (I), (III) only

(E) (I), (II) and (III)


T7-9Mock Test 7

17. An insurer issues a 10-year deferred whole life annuity to a life aged 40. The annuity pays $10,000 at the beginning of each year, starting in the 11th policy year, as long as the policyholder survives. Level premiums are paid at the beginning of each year during the first 10 years.

If the policyholder dies within the first 10 years, premiums paid will be returned to the policyholder without interest.

The insurer incurs initial expenses of $1,000 plus 50% of the first premium, and renewal expenses of 1% of each subsequent premium.

You are given:

(i) 1|40:10( ) 1.6IA =

(ii) |40:10 8.5a =&&

(iii) 40 19.5a =&&

Calculate the gross annual premium using the equivalence principle. (A) 17,550

(B) 17,650

(C) 17,750

(D) 17,850

(E) 17,950 18. Let Y be the present value random variable for a fully continuous 10-year deferred 10-year

payment life annuity with a payment rate of 10 per annum on (22), with the following cumulative distribution function:

F22(t) = 1 − (1 + 0.04t)e−0.04t.

The force of interest is 0.06.

Find Pr(Y ≤ 25). (A) 0.13

(B) 0.22

(C) 0.31

(D) 0.40

(E) 0.49


Mock Test 7

T7-10

19. For a participating fully discrete whole life insurance of 10,000 on (50), you are given:

(i) The gross premium is 160.

(ii) Reversionary bonuses are based on 90% of profits before dividends, calculated using this information:

Item Value

)(69

dq′ 0.019 )(

69wq′ 0.075

i 0.06 Expenses, payable at the start of the year 10 Cumulative face amount of reversionary

bonuses credited in years 1 − 19 3165.5

Bonus declared at end of year 19 204 Reserve end of year 19 4765.0 Reserve end of year 20 5210.5

(iii) Lapses occur immediately before a premium is due.

(iv) Reserve at the end of year k + 1 in the table above includes reserves for reversionary bonuses credited in years 1 to k.

(v) Cash value is 80% of the reserve.

(vi) Reversionary bonuses face amounts are determined using 1000A69 = 499.7 and 1000A70 = 515.0.

(vii) Policyholders who die during year 20 receive their share of year 20 profits as a dividend.

Calculate the compound reversionary bonus rate for year 20. (A) 0.15%

(B) 0.25%

(C) 0.35%

(D) 0.45%

(E) 0.55%


T7-11Mock Test 7

20. A multiple decrement table has 2 decrements, death (d) and withdrawal (w). Withdrawals occur once a year two-thirds of the way through the year of age. Deaths in the associated single-decrement table are uniformly distributed over each year of age.

You are given:

(i) )(τxl = 1000

(ii) )(wxq′ = 0.2

(iii) )(1

τ+xl = 750

Calculate )(dxd .

(A) 54

(B) 58

(C) 62

(D) 68

(E) 71 Section B: Written Answer 1. Consider a Markov model with three states, Healthy (0), Disabled (1), and Dead (2). The

annual transition matrix in the first three years is given by

,10025.075.0005.025.07.0

210

2 1 0

0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=P

.10025.075.001.02.07.0

210

2 1 0

21

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡== PP

Transitions (if any) happen at the end of each year.

(a) Calculate the probability that an insured who is healthy at time 0 and is already deceased by time 3 has never been to the state of disability before death.

[5 points]

(b) You are given that i = 0.08. Calculate the net level premium for a 3-year insurance that pays 100 at the end of the year of death when the insured dies after becoming disabled.

Assume that and that premium is payable at the beginning of every policy year when the insured is healthy.

[3 points]

(c) What does it mean when a stochastic process is Markov? [3 points]


Mock Test 7

T7-12

2. For a population, the death probabilities for healthy individuals follow the life table below:

k 0 1 2 qx+k 0.13 0.15 0.17

(i) 1000 lives aged x are observed at the beginning of a study. Future lifetimes during the first year are originally independent before they are exposed to a virus.

(ii) Immediately after the 1000 lives are observed at the beginning, all lives are exposed to a virus. Each life has a probability of 0.3 to become infected.

(iii) The death probabilities for each year are 150% of the death probabilities of healthy individuals of the same age.

Let kLH be the number of healthy survivors at time k, originated from the 1000 lives, and kLI be the number of infected survivors at time k for k = 0, 1, 2.

(a) What is the distribution of 0LH? [1 point]

(b) Find E(2LH + 2LI ), the expected number of survivors at time 2. [3 points]

It is anticipated that a vaccine would be available to all infected survivors at low cost at time 2, with a probability of 0.3. If the vaccine is available, the infected individual would immediately recover and their death probabilities are again given by the life table above.

(c) Calculate the expected number of survivors at the end of year 3. [6 points]

3. A term life insurance policy issued to a life aged 60 has the following expected profit at

the end of each year given that the policy is in force at the beginning of the year:

Time in years 0 1 2 3 Profit −400 210 −190 450

The expected profits calculated above are without allowances for reserves.

Under the profit test basis, the mortality rates are given by the Illustrative Life Table, but the age is set 2 years lower than the true age of the policyholder. The interest rate is 5.5% per year. The insurance company sets reserves by zeroization.

(a) Calculate the zeroized reserves and the expected profits at issue of the policy.

[6 points]

(b) Find the annual rate of return for the policy with allowance for reserves computed in (a).

[1 point]


T7-13Mock Test 7

4. (a) Let x be an integer. Show that

xx vqi

iAI ⎟⎠⎞

⎜⎝⎛ −=

11)( 1|1: δδ

if deaths are uniformly distributed over each year of age.

[4 points]

(b) Suppose that mortality follows the Illustrative Life Table and that deaths are uniformly distributed over each year of age.

Find 1000 1|2:45

)( AI under i = 5%.

[6 points] 5. You are given:

(i) δ = 0.05

(ii) μx = 0.0008 + 0.00005 × 1.1x

(iii) μ[x]+s = 0.92−s μx+s for 0 ≤ s ≤ 2 and μ[x]+s = μx+s for s > 2.

(a) Find 5p60 and 2p[55]. [5 points]

(b) By using the trapezium rule with 5 subintervals, compute |5:]55[

a .

[5 points]

(c) A life selected at age 55 purchases a fully continuous deferred life annuity that starts payments at his retirement age 60, with 5 years of certain payments. The payment rate is 1000 per annum. The life makes continuous premium of rate P per annum for 5 years or until he dies, whichever occurs first.

You are also given that 65a = 9.38707. Using the equivalence principle, compute P.

[4 points] (d) It is further given that

|3:]55[a = 2.7546.

Suppose that at time 3, the life becomes unemployed and he decides to surrender the policy. The insurance company applies a surrender charge of 10% to the net premium reserve built up from the policy and reduces the payment of the 5-year certain whole life annuity accordingly.

Calculate the new payment rate. [4 points]

*** End of Examination ***


Mock Test 7

T7-14

Solutions to Mock Test 7 Section A

1. C 11. A 2. E 12. D 3. B 13. C 4. E 14. C 5. A 15. D 6. E 16. A 7. B 17. A 8. D 18. A 9. A 19. D 10. C 20. B

1. [Chapter 4] Answer: (C)

⎪⎩

⎪⎨

⎧

≥=++==+=

=294.743217.43202

2x

x

x

KvvKvK

Y

The corresponding probabilities are

Pr(Kx = 0) = 0.2, Pr(Kx = 1) = 0.8 × 0.25 = 0.2, Pr(Kx ≥ 2) = 0.8 × 0.75 = 0.6.

As a result,

E(Y ) = 2 × 0.2 + 4.7 × 0.2 + 7.94 × 0.6 = 6.104,

E(Y 2) = 22 × 0.2 + 4.72 × 0.2 + 7.942 × 0.6 = 43.04416.

Var(Y) = 43.04416 − 6.1042 = 5.785344, giving a standard deviation of 2.4. 2. [Chapter 11] Answer: (E)

Since lifetimes are independent, Txy has a force of failure of

12.0=+= +++ tytxtxy μμμ ,

a constant, so that 32

06.012.012.0

=+

=xyA .

Also, 74

06.008.008.0

=+

=xA and 52

06.004.004.0

=+

=yA and hence by symmetric relation,

10532

=−+= xyyxxy AAAA .

By the relation between insurance and annuity, 9

50100/6

3/21δ

1=

−=

−= xy

xy

Aa .

By the equivalence principle,


T7-15Mock Test 7

1436.0105/3239/50

3/23

3

=×−

=−

=

+=

xyxy

xy

xyxyxy

AaA

AAa

π

ππ

3. [Chapter 10] Answer: (B)

Let the time for the person to enter state 1 be S, and the time for the person to enter state 2 be T. The joint density of the random vector (S, T ) is given by

tststsstss eeeppts 05.0)(05.005.012

501150

0150

0050 001.005.002.0 ),(f −−−−

++−+ =×××== μμ ,

for 0 < s < t. The event under consideration is given by the figure below:

So, the answer can be obtained from

08381.0

1005.0

102.0

d)(02.0

dd001.0

15.0

10

0

105.0

10

0

20

05.0

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

−=

−=

−−

−−

−

∫∫ ∫

ee

see

ste

s

s

t

4. [Chapter 14] Answer: (E)

According to Statement (viii), it is assumed in calculating the cost of insurance that the force of mortality is constant for all ages. Hence, *tq = 1 – e−μ. This is a specified amount plus the account value (Type B) policy. Because there is no corridor factor requirement, we can write the cost of insurance for year t as

CoIt = *tq vq X = (1 – e−μ )(1.06−1)(50000),

which is a constant that is independent of t. We let CoI = CoI1 = CoI2. The account value at the end of year 1 is given by

AV1 = (AV0 + P1 − EC1 − CoI)(1 + 1ci ).

Hence, we have

13900 = (12000 + 0.95 × 2000 – 50 − CoI)(1.05),

which gives CoI = 611.904762.

The account value at the end of year 2 is given by

AV2 = (AV1 + P2 − EC2 − CoI)(1 + 2ci )

= (13900 + 0.95 × 2000 – 50 – 611.904762)(1.05) = 15895.

s

t

10

20


Mock Test 7

T7-16

5. [Chapter 7] Answer: (A)

Thiele’s differential equation says

txttg

ttxtttt

gt EbVecG

tV

++ +−++−−= μμ )()δ()1(d

d.

For x = 38, t = 5, we have μx+t = μ43 = 0.00022 + 0.00004 × 1.143 = 0.0026296,

68916.1510026296.0)10010000(9500426296.0)05.01(145d

d

5

=×+−×+−==t

gt

tV

.

So, 5+1/12Vg ≈ 950 + 151.68916 × 1/12 = 962.64076.

Similarly, for x = 38, t = 5 + 1/12, and we have

μx+t = μ43+1/12 = 0.00022 + 0.00004 × 1.143+1/12 = 0.0026488,

05259.1520026488.0)10010000(0426488.0)05.01(145d

d12/15

12/15

=×+−+−= +

+=

g

t

gt V

tV

.

So, 5+1/6Vg ≈ 962.64076 + 152.05259 × 1/12 = 975.3.

6. [Chapter 2] Answer: (E)

We have l40 = 9,313,166.

Let t be the 75-percentile of T40. Then we want Pr(T40 ≤ t) = 0.75, and t would be the 75-percentile for the future lifetime T40. This means the 75-percentile for the age at death is 40 + t.

1 − 40

40

ll t+ = 0.75 ⇒ l40+t = 0.25l40 = 2,328,291.5.

Since l85 = 2,358,246 and l86 = 2,066,090, we have 85 ≤ 40 + t ≤ 86. This means the answer must be (E).

7. [Chapters 9 and 12] Answer: (B)

The total force of decrement is 3.01.0)3(1.0)(22

)(22

)(22 =+−=+= +++ ttd

tr

tt μμμ τ for 0 ≤ t < 3, which is constant. As a result, t

t ep 3.0)(22

−=τ for 0 ≤ t ≤ 3. The expected present value of the disability payments is

03.2616)9.0955.098.01(1000

))3()2()1(1(000,19.06.03.0

)(223

)(222

)(22

=+++=

+++−−− eee

pvpvpv τττ


T7-17Mock Test 7

8. [Chapter 6]: Answer: (D)

By the recursive relation for reserves,

(hV + πh)(1 + i) = h+1V + (bh+1 − h+1V)qx+h.

Putting h = 0,

19 × 1.05 = 4.5 + (100 − 4.5)qx ⇒ qx = 0.161780.

Putting h = 1,

(4.5 + 19) × 1.05 = 0 + (100 − 0)qx+1 ⇒ qx+1 = 0.24675.

So, ⎪⎩

⎪⎨

⎧

≥−=+−==+−==−

=209524.37)1(19160771.53)1(19100023810.7619100

20

x

x

x

KvKvvKv

L , with probabilities

Pr(Kx = 0) = 0.161780, Pr(Kx = 1) = 0.83822 × 0.24675 = 0.206831 Pr(Kx = 2) = 0.83822 × 0.75325 = 0.631389

Since E(0L) = 0,

Var(0L) = E(0L2) = 76.23812 × 0.16178 + 53.607712 × 0.206831 + 37.095242 × 0.631389 = 2403.521,

giving SD(0L) = 49. 9. [Chapters 6 and 12] Answer: (A)

The mean of the loss of a policy with face 100 is

100(0.44) − 4.305.0

44.01−× = –4.16.

The variance of the loss of a policy with face 100 is

3344.913)44.022.0(05.03.4100 2

2

=−⎟⎠⎞

⎜⎝⎛ + .

The mean of the loss of a policy with face 400 is

400(0.44) − 17.505.0

44.01−× = –20.

The variance of the loss of a policy with face 400 is

14850)44.022.0(05.05.17400 2

2

=−⎟⎠⎞

⎜⎝⎛ + .

As a result, the expected present value of aggregate loss is

75(−4.16) + 25(–20) = −812,

and the variance is 75(913.3344) + 25(14850) = 439750.08. The required probability is


Mock Test 7

T7-18

)28.2(1)28.2Pr()08.439750

812700Pr()700Pr( Φ−=>=+

>≈> ZZL .


The reserve at the end of 5 years, which is the expected prospective loss at age 55, is

3336.238790894.9457395.711000

06.1/8969.9)8950901/7533964(2668.138969.948686.02758.1211000

110001000

15

|15:50

|10:55|15:505

=⎟⎠⎞

⎜⎝⎛ −=

⎟⎟⎠

⎞⎜⎜⎝

⎛×−

×−−=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−=

a

aV

&&

&&

Let S be the reduced pure endowment. When the policy is paid up after 5 years, the reserve is used to support

SSESA

48686.001897.9148686.0)8.43948686.014.305( 1000 5510

1|10:55

+=

+×−=+

Setting the above to 238.3336 and solving, we get S = 302.6. 11. [Chapter 7] Answer: (A)

The level expense premium is Pe = 390.20 − 323.12 = 67.08.

The expense reserve at time 3 is 0. By using the recursion relation for expense reserves,

(2Vg + Pe − 0.1G − 3)(1 + i) = qx+2 × 0 + px+2 × 3Ve

2Vg = −Pe + 0.1G + 3 = −67.08 + 39.02 + 3 = −25.06. 12. [Chapter 10] Answer: (D)

P2 =

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡2115.0666.01225.00555.01235.0821.0

6.015.025.015.08.005.003.007.09.0

6.015.025.015.08.005.003.007.09.0

P(X: 0 to 2) = 0.0555, P(X: 0 not to 2) = 0.9445

P(Y: 0 to 2) = 0.2115, P(Y: 0 not to 2) = 0.7885

The required probability is 0.9445 × 0.2115 + 0.0555 × 0.7885 = 0.2435. 13. [Chapter 3] Answer: (C)

By UDD, 7171 AiAδ

= , and hence 445.007.007.1ln

71 ×=A = 0.4301157.

By recursive relation for life insurances,


T7-19Mock Test 7

A70 = 07.1

4301157.098.002.0717070

×+=+ Avpvq = 0.4126293.

Then by i(m) = m[(1 + i)1/m − 1], i(4) = 0.0682341, and hence 70)4()4(

70 AiiA = = 0.42331.


9.8382813.13

93.10260003

2600035

646362 =×

×=++

×S

SSS.

15. [Chapter 14] Answer: (D)

For a universal life policy with a no lapse guarantee, the death benefit cover continues even if the account value declines to zero, provided that the policyholder pays a pre-specified minimum premium at each premium date. Hence, Statement (I) is FALSE.

This guarantee comes into effect when the account value (AVt−1) and the premium (Pt) are not sufficient to cover the cost of insurance (CoIt). Hence, the guarantee is more likely to come into effect if the cost of insurance is higher.

If the interest rate used to calculate the cost of insurance is smaller, the resulting cost of insurance is higher, and hence the no-lapse guarantee is more likely to come into effect. Statement (II) is TRUE.

If the death probability used to calculate the cost of insurance is smaller, the resulting cost of insurance is smaller, and hence the no-lapse guarantee is less likely to come into effect. Statement (III) is TRUE.

16. [Chapters 4, 5] Answer: (A)

The present value random variable for a continuous whole life annuity of $1 on (x) is given by

|

1 x

x

T

T

vY aδ

−= = , Tx ≥ 0,

which is clearly a strictly increasing function of Tx for all Tx ≥ 0.

The present value random variable for a continuous n-year temporary life annuity of $1 on (x) is

⎪⎪⎩

⎪⎪⎨

⎧

>−

=

≤−

==

.1

1

|

|

nTva

nTvaY

x

n

n

x

T

T

x

x

δ

δ

This is not a strictly increasing function of Tx, because when Tx > n, the present value random variable is a constant.

The net future loss at issue random variable for a fully continuous whole life insurance of $1 on (x) is given by


Mock Test 7

T7-20

δδδδδ PePPvPaPvL xx

x

x TTT

T −⎟⎠⎞

⎜⎝⎛ +=−⎟

⎠⎞

⎜⎝⎛ +=−= −11

|,

which is clearly a strictly decreasing function of Tx for all Tx ≥ 0.

Hence, the answer is (I) only. 17. [Chapters 7] Answer: (A)

Let P be the gross annual premium.

APV of benefits = APV of the return of premium benefit + APV of the deferred annuity APV of benefit = 40|10

1|10:40 10000)( aIAP &&+ .

APV of expenses = Initial expenses + APV of renewal expenses = 1000 + 0.5P + 0.01P( |10:40a&& − 1) .

APV of premiums = P |10:40a&& .

Using the equivalence principle, we find P such that APV of benefits + APV of expenses = APV of premiums. This gives

41.175496.146.05.899.0

1000)5.85.19(10000)(49.099.0

1000100001

|10:40|10:40

40|10 =−−×+−

=−−

+=

IAaa

P&&

&&

18. [Chapter 4] Answer: (A)

The definition of Y is

⎪⎪⎩

⎪⎪⎨

⎧

≥=

<≤=<

=−

−−

−×−

2027.411020101010

100

22|106.0

22|106.0

|101006.0

22

2222

TaeTaeae

TY TT

.

1261.0612844.11

)3211.15Pr()7266822.0Pr(

)555297.406.0

1Pr()251.0Pr()25Pr(

61284.022

)10(06.0

)10(06.06.0

|10

22

22

22

=−=

≤=≥=

≤−

=≤=≤

−

−−

−−

−

eTe

eeaYT

T

T

19. [Chapter 15] Answer: (D)

The sum insured during policy year 20 is S + RB19 = 10000 + 3165.5 = 13165.5.

We are given 19−V = 4765.0 and 20−V = 5210.5.

Also, Div19 = B19A69 = 204 × 0.4997 = 101.9388.

Hence, 19V = 4765.0 + 101.9388 = 4866.9388.

The profit before the distribution of bonus is


T7-21Mock Test 7

Pr20− = 1.06(4866.9388 + 160 − 10) − 0.019 × 13165.5 − 0.981 × 0.075 × 0.8 × 5210.5 − 0.981 × 0.925 × 5210.5

= 32.9826

The dividend is Div20 = 32.9826 × 0.9 = 29.6844.

The extra death benefit that can be purchased using the bonus is

B20 = 29.6844 / A70 = 29.6844 / 0.515 = 57.64.

The compound reversionary bonus rate for year 20 is

b20 = B20 / S20 = 57.64 / 13165.5 = 0.44%. 20. [Chapter 8] Answer: (B)

From (i) and (iii), we have )(τxp = 0.75. Then by

)()()( dx

wxx ppp ′′=τ ,

we get )(dxp′ = 0.9375. Since death is UDD under its associated single decrement table,

1917.02.0)0625.0321(2.0)

321(2.0 )()(

3/2)( =××−=×′−=×′= d

xd

xw

x qpq .

Hence, =×= 1917.01000)(wxd 191.7, and )(d

xd = 1000 – 750 – 191.7 = 58.3. Section B 1. [Chapter 10]

(a) We first compute the probability that the insured is dead by time 3.

2P0 = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

1000

1825.03275.049.0

10025.075.001.02.07.0

10025.075.0005.025.07.0

,

3P0 =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−−

1000

313375.0343625.0343.0

10025.075.001.02.07.0

1000

1825.03275.049.0.

As a result, Pr(Y3 = 2 | Y0 = 0) = 0.313375.

Then we calculate the probability that the insured is dead by time 3, and he has never become disabled. The possible paths and the respective probabilities are shown below:

Path Probability 0 → 2 → 2 → 2 0.05 0 → 0 → 2 → 2 0.7 × 0.1 = 0.07 0 → 0 → 0 → 2 0.7 × 0.7 × 0.1 = 0.049

As a result, Pr(dead by time 3 and never become disabled before) = 0.169.


Mock Test 7

T7-22

The answer required is 0.169 / 0.313375 = 0.54.

(b) We first calculate the APV of the benefit:

Path Probability 0 → 1 → 2 0.25 × 0.25 = 0.0625

0 → 0 → 1 → 2 0.7 × 0.2 × 0.25 = 0.035 0 → 1 → 1 → 2 0.25 × 0.75 × 0.25 = 0.046875

APV of benefit = 85787.1108.1

081875.008.10625.0100 32 =⎟

⎠⎞

⎜⎝⎛ + .

The APV of the premium is PP 068244.208.149.0

08.17.01 2 =⎟

⎠⎞

⎜⎝⎛ ++ .

By the equivalence principle, the level annual premium is

11.85787 / 2.068244 = 5.733.

(c) A stochastic process has the Markov property if the conditional probability distribution of future states of the process, given both past and present values of states, depends only upon the present state. Loosely speaking, given the present, you can forget about the past.

2. [Chapter 1]

For a person who is infected, the death probabilities are

k 0 1 2 qx+k 0.195 0.225 0.255

(a) 0LH is binomial distributed with parameters n = 1000 and p = 0.7.

(b) 2px for healthy individuals = (1 − 0.13)(1 − 0.15) = 0.87 × 0.85 = 0.7395 2px for infected individuals = (1 − 0.195)(1 − 0.225) = 0.805 × 0.775 = 0.623875

E(2LH | 0LH) = 0.73950LH.

Expected number of survivors out of the 0LI = 1000 − 0LH infected people is

E(2LI | 0LH) = 0.623875(1000 − 0LH).

So, given 0LH, the expected number of survivors is

E(2L | 0LH) = 0.73950LH + 0.623875(1000 − 0LH) = 623.875 + 0.1156250LH.

So, the expected number of survivors is

E(2L) = 623.875 + 0.115625 × 1000 × 0.7 = 704.8125.

(c) Let 3LI be the number of survivors at time 3 originated from the infected survivors at time 2. Given that the vaccine is available and the value of 2LI, 3LI ~ B(2LI, 0.83). So,

E(3LI | 2LI, vaccine is available) = 0.832LI


T7-23Mock Test 7

Given that the vaccine is not available and the value of 2LI, 3LI ~ B(3LI, 0.745). So,

E(3LI | 2LI, vaccine is not available) = 0.7452LI

E(3LI | 2LI) = (0.3 × 0.83 + 0.7 × 0.745)2LI = 0.77052LI .

Obviously, E(3LH | 0LH) = 3px 0LH = 0.83 × 0.7395 0LH = 0.6137850LH

E(3LH) = 0.83 × 0.7395 × 700 = 0.613785 × 700 = 429.6495

E(3LI | 0LI) = E[E(3LI | 2LI) | 0LI] = 0.7705E(2LI | 0LI) = 0.7705 × 0.6238750LI

E(3LI) = 0.7705 × 0.623875 × 300 = 144.208706

So, E(3L) = 144.208706 + 429.6495 = 573.858. 3. [Chapter 13]

(a) The survival probabilities under the profit test basis are as follows

k p60+k = p58+k under ILT 0 0.98842 1 0.98738 2 0.98624 3 0.98499

Since Pr3 > 0, 3VZ = 2VZ = 0. We look at Pr2:

−190 + 1.0551VZ = 0 1VZ = 180.09479.

Then 0)0,3.30max()0,055.1

21009479.18098842.0max(0 =−=−×

=ZV .

This revised profit vector is

Pr0 = –400, Pr1 = 210 − 0.98842 × 180.09479 = 31.99071, Pr2 = 0, Pr3 = 450

We also have 2p60 = 0.9759461.

The profit signature is (−400, 31.99071, 0, 439.1757)′.

Time in years 0 1 2 3 Profit −400 31.99071 0 439.1757

(b) Using a financial calculator, the IRR for the policy is 5.8995%. 4. [Chapter 3]

(a) ∫∫∫ −−− ===1

0

1

0

1

0

1|1:

ddd)()( tteqtqtettfteAI txx

tx

tx

δδδ .

The integral is just the PV of a 1-year continuously increasing annuity certain:

δδ

vaaItte t

−==∫ − |1

|1

1

0 )(d .


Mock Test 7

T7-24

So,

⎟⎠⎞

⎜⎝⎛ −=

−=

−=

−=

iivqivq

svq

vaqAI xxxxx

111/1)( |1|11

|1: δδδδ

δδ.

[Alternatively, if you cannot recognize the integral as an annuity certain, you may simply compute it using integration by parts, as follow:

22

1

0

1

0

1

0

11d1d1dδδδδδδδ

δδδ

δδδ vveeteeettte ttt −

+−=−

+−=+−=−=−−

−−

−− ∫∫∫

which is the same as δ

va −|1 .]

(b) It is not difficult to see that we have the recursion-like formula

])[()()( 1|1:46

1|1:46451

1|1:45

1|2:45

AAIEAIAI ++= .

Applying UDD on 1|1:46

A , we have 1|1:46

1|1:46

AiAδ

= .

⎟⎠⎞

⎜⎝⎛ −

05.01

05.1ln1

05.1ln05.0 = 0.508232

So, ⎟⎠⎞

⎜⎝⎛ +×+×=

05.105.1ln05.00.508232

05.105.10.508232

05.1)( 464645451

|2:45

qqpqAI .

With q45 = 0.004, p45 = 0.996, q46 = 0.00431, we get 1000 1|2:45

)( AI = 7.9052.

5. [Chapters 2 + 9]

(a) Under Makeham’s law of mortality, )]1(ln

exp[ −−−= tx

xt cc

BcAtp .

Putting x = 60 and t = 5, we get

5p60 = exp(−0.004 − ))11.1(1.1ln

1.100005.0 560

−×

= 0.903465.

For 2p[55], we note

0250437.0

]1)9/11[()9/11ln(1.100005.0]1)9/10[(

)9/10ln(0008.081.0

d)9/11(1.100005.0d)9/10(0008.081.0

d)1.100005.00008.0(9.0

255

2

2

0

552

0

2

0

552

=

⎟⎟⎠

⎞⎜⎜⎝

⎛−

×+−=

⎥⎦⎤

⎢⎣⎡ ×+=

×+

∫∫∫ +−

ss

s

ss

ss

which gives 2p[55] = exp(−0.0250437) = 0.97991897

(b) ∫ −=5

0 ]55[05.0

|5:]55[dspea s

s . Let f (s) = e−0.05s s p[55] = sE[55].

Applying the trapezium rule,


T7-25Mock Test 7

2)]4()3()2()1([2)5()0(

|5:]55[

ffffffa +++++≈ .

f (0) = 1, f (2) = e−0.1 × 0.97991897 = 0.88666735

f (1) = e–0.05 × exp[ ⎟⎟⎠

⎞⎜⎜⎝

⎛−

×+−− )19/11(

)9/11ln(1.100005.0)19/10(

)9/10ln(0008.081.0

55

]

= 0.9425535

p57 = 0.987281, 2p57 = 0.973554, 3p57 = 0.958752

f (3) = e−0.15 × 0.97991897 p57 = 0.832696

f (4) = e−0.2 × 0.97991897 2p57 = 0.781073

f (5) = e−0.25 × 0.97991897 3p57 = 0.731683

Then ≈|5:]55[

a 4.30883.

(c) The expected present value of the benefits is

⎟⎟⎠

⎞⎜⎜⎝

⎛××+

−=+ −

−

38707.9903465.005.0

1)5(1000)(1000 25.025.0

65655|5]55[5 eefaEaE

= 1000 × 0.731683 × 11.02891 = 8069.666.

So, P = 8069.666 / 4.30883 = 1872.82.

(d) The net premium reserve at time 3, calculated using the retrospective formula, is

=×

==832696.0

7546.282.1872

]55[3

|3:]55[

|3:]55[ E

aP

sP

6195.382.

After applying the surrender charge, the policy would be supported by 5557.844.

By the equivalence principle,

5557.844 = e−0.1 2p58 × R × 11.02891,

where 2p58 = 0.971104. On solving, R = 575.3627.


Mock Test 7

T7-26

MLCSM ACTEX

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