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Actuarial Models

An Introductory Guide for Actuaries and other Business Professionals

Third Edition

Michael A. Gauger, Ph.D.

Associate of the Society of Actuaries

Julie Lewis, Ph.D. Affiliate of the Faculty of Actuaries

Mark Willder, Ph.D. Fellow of the Institute of Actuaries

Mike Lewry, MA Fellow of the Institute of Actuaries

BPP Professional Education Phoenix, AZ

Copyright 2011 by BPP Professional Education, Inc.

All rights reserved. No portion of this book may be reproduced in any form or by any means without the prior written permission of the copyright owner. Requests for permission should be addressed to: BPP Professional Education 4025 S. Riverpoint Parkway Phoenix AZ 85040 Mail Stop: CF-K909 Manufactured in the United States of America 10 9 8 7 6 5 4 3 2 1 First Printing ISBN: 978-0-9816081-1-2

i

Preface to the Third Edition

Welcome to this introductory guide to actuarial models.

Based on our experience as professional educators, our aim when writing and updating this text has been to produce a clear, practical and student-friendly guide in which theoretical derivations have been balanced with a helpful, structured approach to the material. We have supplemented the explanations with almost 400 worked examples and practice questions to provide ample opportunity to see how the theory is applied. The result—we hope—is a thorough but accessible introduction to actuarial models.

This text has been written by financial professionals with financial professionals in mind. It is of particular relevance to actuarial students who are preparing for the Life Contingencies segment of Exam M of the Society of Actuaries (Exam MLC), and Exam 3 of the Casualty Actuarial Society. Most examples are set in an insurance or risk management context.

For more information about an actuarial career, visit www.beanactuary.org or www.soa.org. Aspiring actuaries in the UK should visit www.actuaries.org.uk.

The numerical solutions to all of the end-of-chapter practice questions can be found at the end of this book. Detailed worked solutions to these practice questions can be downloaded free of charge from the BPP Professional Education website at www.bpptraining.com (look for Text Question Solutions in the left menu). Information about other useful study resources can also be found there, including an extensive supplemental Question and Answer Bank with hundreds of exam-style questions as well as online lectures and online multiple choice tests.

For students preparing for the SOA’s Exam MLC, it is critically important to work as many exam-style questions as possible. For such preparation, this text should be used in conjunction with our supplemental Q&A Bank which contains hundreds of multiple choice questions (including relevant questions from past exams). Practice Questions included in this book at the end of each chapter are designed to emphasize first principles and basic calculation whereas exam-style questions, such as those contained in the Q&A Bank, can be quite obtuse.

This text could not have been completed without the helpful contributions of several outstanding individuals. David Carr, Robert Chadburn, David Hopkins, Michael Hosking and David Wilmot deserve particular mention for their superb technical reviews, which have greatly improved the clarity and accuracy of each chapter. Any errors in this text are solely our own.

We hope that you find this text helpful in your studies, wherever these may lead you.

Julie Lewis October 2011

iii

Table of contents

Introduction Chapter 1 Introduction to Survival Models 1 1.1 The role of a survival model in a contingent payment model 2 1.2 The life table – a discrete survival model 3 1.3 The theory of continuous survival models 6

1.4 The continuous future lifetime after age x 15 1.5 The curtate future lifetime after age x 18 1.6 Important life table functions 21 1.7 Extending a life table to a continuous survival model 29 1.8 Select and ultimate life tables 33 1.9 Mortality laws 37 1.10 Applications of survival models 39

Chapter 1 practice questions 43

Chapter 2 Life Insurance Models 47 2.1 A discrete model of whole life insurance 49 2.2 A continuous model of whole life insurance 53 2.3 Other types of level-benefit life insurance 55 2.4 The variance of a random present value variable 65

2.5 Aggregate life insurance models 71 2.6 Formulas for APV’s for simple mortality laws 73 2.7 Insurance models with non-level benefits 75 2.8 Interest rates other than fixed level rates 79


Chapter 3 Life Annuity Models 87 3.1 A discrete model of a whole life annuity 88 3.2 A continuous model of a whole life annuity 94 3.3 Additional types of level life annuities 97 3.4 The variance of a random present value variable 102 3.5 Aggregate life annuity models 106

3.6 Special types of life annuities 107 3.7 Non-level life annuity models 113 3.8 Recursive calculations 116 3.9 Applications of life insurance and annuity principles 117 3.10 Calculating actuarial present values using tables 118 3.11 Second moments – annuities versus insurances 121 3.12 Random and actuarial present values 123


Table of Contents

iv

Chapter 4 Annual Benefit Premiums 127 4.1 An example illustrating premium calculation 128 4.2 Annual benefit premiums for whole life insurance models 131 4.3 Annual benefit premiums for other life insurance models 136 4.4 Limited payment plans and m-thly premiums 139 4.5 The insurer’s loss function 143 4.6 Percentile premiums 150 4.7 Aggregate loss functions 152 Chapter 4 practice questions 154

Chapter 5 Benefit Reserves 157 5.1 Different uses of reserves 158 5.2 Reserving methods 159 5.3 An example illustrating the calculation of benefit reserves 161 5.4 Benefit reserves for whole life insurance models 163 5.5 Benefit reserves for term and endowment insurance models 170 5.6 Alternative reserve formulas 172 5.7 Reserves with m-thly premiums and limited payments 174 5.8 The prospective loss function at duration t 177 5.9 Allocation of risk to policy years 181 5.10 Reserves at fractional durations 189 Chapter 5 practice questions 192

Chapter 6 Multiple Life Theory 195 6.1 The notion of a multiple life status 196 6.2 The joint life status / independent lifetimes 197 6.3 The last survivor status / independent lifetimes 202 6.4 Insurances for multiple life statuses 206 6.5 Annuities for multiple life statuses 211 6.6 Multiple life statuses with dependent lifetimes 216

6.7 The common shock model for dependent lifetimes 222 6.8 Special joint life probabilities and annuities 224 6.9 Contingent functions 226


Chapter 7 Multiple Decrement Models 233 7.1 A comparison of single and multiple decrement models 234 7.2 The theory of continuous time multiple decrement models 238 7.3 Extending a multiple decrement table to a continuous model 250 7.4 Discrete decrement patterns 261 7.5 Benefits depending on the time and mode of decrement 264 7.6 Stochastic versus deterministic 268 Chapter 7 practice questions 270

Chapter 8 Insurance Models with Expenses and Profit Loadings 275 8.1 Types of expenses 276 8.2 An example including expenses 278 8.3 Fully discrete whole life insurance with expenses 286 8.4 A multiple decrement model with expenses 293 Chapter 8 practice questions 297

Table of Contents

v

Chapter 9 Discrete-time Markov Chains 301 9.1 Multiple-state models 302 9.2 A simple two-state example: the up-down model 304 9.3 Markov chains 306 9.4 The APV of a cash flow while in a state 315 9.5 The APV of a cash flow upon transition 317 9.6 Premiums and reserves with multi-state models 320


Chapter 10 Continuous-time Markov Chains 325 10.1 Introduction 326 10.2 The Poisson process 326 10.3 Inter-arrival and waiting times in a Poisson process 328 10.4 Adding and thinning Poisson processes 331 10.5 Time-homogeneous processes 333 10.6 Time-inhomogeneous processes 343 10.7 Calculating probabilities using discrete approximations 348 10.8 Valuing benefits using a multiple-state model 350 10.9 Calculating reserves using a multiple-state model 353


Chapter 11 Universal Life Insurances 363 11.1 Universal life insurances 364 11.2 A multiple decrement model for universal life insurance 369 11.3 A Markov chain model for universal life insurance 374 11.4 Reserves for universal life insurance 376


Answers to practice questions 383

Bibliography 389

Index 391

vi

Introduction

Before we start the main subject matter in this text, we should take care of a little housekeeping.

Assumed knowledge

We assume that the reader has knowledge of calculus, probability theory, and interest theory. Ideally, your familiarity with these topics should be at a level sufficient for success with Exams P (probability theory) and FM (financial mathematics) of the SOA. The necessary background information can be found in BPP textbooks written for these two examinations. You can download a free sample chapter from these texts or order a copy by visiting the BPP Professional Education website at www.bpptraining.com.

Notation and rounding

We have tried hard to ensure that all new notation is explained clearly. We sometimes use

exp( )x in place of xe , especially when this avoids complicated superscripts that might otherwise be difficult to read.

Rounding poses a particular dilemma. Our standard policy in this text has been to keep full accuracy within intermediate calculations even though an intermediate result may be shown as a rounded value. So, you may occasionally disagree with the last significant figure or two in a calculation if you calculate the result using the rounded values shown.

Solutions to practice questions

Short numerical answers to all of the end-of-chapter practice questions can be found at the end of this book. Detailed worked solutions to these practice questions can be downloaded free of charge from the BPP Professional Education website at www.bpptraining.com (look for Text Question Solutions in the left menu). Other useful study resources can also be found here.

Errors in this text

If you find an error in this text, we’ll be pleased to hear from you so that we can publish an errata list for students on our website and correct these errors in the next edition. Please email details of any errors to [email protected]. A current errata list is maintained in the Student Mailbag on the Exam MLC page of the BPP website. Thank you.

1

Introduction to survival models

Overview A survival model is a probabilistic model of a random variable that represents the time until the occurrence of an unpredictable event. For example, we may wish to study the life expectancy of a newborn baby, or the future working lifetime of a machine until it fails. In both cases, we study how long the subject may be expected to survive.

The theory that we will develop throughout this course can be applied in a wide range of situations, in which the concept of “survival” may not be immediately obvious, for example:

the time until a claim is made on an automobile insurance policy

the time until a patient in a coma recovers from the coma, given that he recovers

the time until a worker leaves employment.

The focus of our study—the time until the specified event—is known as a waiting time or a random time-to-event variable. Probabilities associated with these models play a central role in actuarial calculations such as pricing insurance contracts.

Introduction to survival models Chapter 1

2

1.1 The role of a survival model in a contingent payment model

Let’s start by considering the most basic contingent payment model, in which a specified amount is paid if and only if a particular event occurs.

Suppose that an amount P is to be paid in n years if a random event E occurs. Otherwise, if the complementary event occurs, then nothing is to be paid.

At an effective annual rate of interest i , the random present value of the payment is:

if occurs

0 if occurs

nP v EZ

E

where 1(1 )v i is the one-year present value discount factor.

The random present value of the payment, Z , is a discrete random variable. Its expected value is known as the actuarial present value of the payment, which incorporates the amount of the payment, the discount factor associated with the timing of the payment, and the probability of the payment being made:

amount discount probability

Pr 0 Pr

Pr

n

n

E Z P v E E

P v E

Throughout Chapters 2 through 8 of this course, we will be concerned with the distribution of random present value of payment variables. We will frequently calculate the mean, the variance, a percentile, or the probability of some event regarding Z such as Pr( )Z E Z .

In Chapters 2 and 3 we will introduce contingent payment models that arise in the context of life insurance and life annuities. For these models, we will need to compute probabilities of events that are expressed in terms of the random future lifetime after age x , when an insurance contract has just been issued. For example, in order to calculate the appropriate life insurance premium for a 30 year old policyholder, we need to calculate the probability that the policyholder will die before age 31, or 32, or 33, and so on.

In this first chapter, our aim is to familiarize you with the theory of survival models, the notation employed in these models, and standard terminology.

There are three principal variables, all of which are measured in years:

the random lifetime (ie time until death) of a newborn life is denoted X

the complete future lifetime at age x , given that a newborn has survived to age x , is denoted T x , ie ( )T x X x X x

the curtate future lifetime at age x , given that a newborn has survived to age x , is the complete number of years of future lifetime at age x and is denoted ( )K x , ie

K x T x (greatest integer)

The variables X and T are assumed to be continuous random variables, whereas K is obviously discrete. For example, suppose a newborn life eventually dies at age 74.72. Then 74.72X ,

30 74.72 30 44.72T , and 30 [ 30 ] [44.72] 44K T .

Notice that T is a function of X , and K is a function of T . So, the distributions of these three variables are closely related. These relationships will be developed over the rest of this chapter.

Chapter 1 Introduction to survival models

3

1.2 The life table – a discrete survival model

We begin by studying the life table, a discrete survival model commonly used in insurance applications. This model gives us the opportunity to gain an intuitive understanding of some of the most fundamental concepts before we study continuous time models.

We start by defining xl as the number of lives expected to survive to age x from a group of 0l newborn lives. A life table displays in a table format the values of xl at ages x equal to 0,1, 2, , , where is the first whole number age at which there are no remaining lives in the group. If we are modeling human mortality, we may choose a value of of around 120 years.

We’ll start by taking a deterministic view of future mortality. By this, we mean that the table tells us exactly how many of the 0l lives will be surviving at ages 1, 2, and so on.

Here is a portion of a hypothetical life table:

x 0 1 2 3 4 5 6 7 8 9

xl 1,000 991 985 982 979 976 972 968 964 959

xd 9 6 3 3 3 4 4 4 5 6

The row labeled xd represents the number of lives among 0l newborn lives that die in the age range [ , 1)x x . It is computed as:

1x x xd l l

For example, since 2 985l lives survive to age 2, and 3 982l lives survive to age 3, exactly 2 2 3 3d l l lives must die between age 2 and age 3.

A number of probabilities can be computed from the entries in such a table. Let’s begin by introducing the standard notation for the most significant types of probabilities that you will see in the actuarial models throughout Chapters 2 – 8.

The probability that a life currently age x will survive n years is denoted n xp . With our deterministic interpretation of the life table, along with the point of view that the probability of an event is the relative frequency with which it occurs, we have:

x n

n xx

lp

l

It is a standard convention to omit the n subscript when 1n , so the probability that a life currently age x will survive 1 year is:

1xx

x

lp

l

For example, the probability that a life age 5 survives for 2 years to age 7 is:

72 5

5

968976

lp

l


4

The probability that a life currently age x will die within n years is denoted n xq , and we have:

1 1x n x x n

n x n xx x

l l lq p

l l

Intuitively, this is the probability that a life is one of the ( )x x nl l lives to die between age x and age x n , out of the xl lives age x .

For example, the probability that a life age 1 dies within 3 years is:

1 43 1

1

991 979 12991 991

l lq

l

Again, we omit the n subscript when 1n , so the probability that a life currently age x will die within 1 year is:

1 or x x x

x xx x

l l dq q

l l

Finally, the probability that a life currently age x will survive for m years and then die within the following n years is denoted xm nq , and we have:

x m x m n

xm nx

l lq

l

Intuitively, xm nq is the probability that a life age x survives for m years, multiplied by the

probability that a life age x m dies within n years:

x m x m x m n x m x m n

x m x n x mm nx x m x

l l l l lq p q

l l l

Again, we omit the n subscript when 1n , so the probability that a life currently age x will survive for m years and then die within 1 year is:

1

or x m x m x m

x xm mx x

l l dq q

l l

For example, the probability that a life age 4 survives for 3 years and then dies within the following 2 years is:

7 943 2

4

968 959 9979 979

l lq

l

Example 1.1

Compute the following probabilities from the life table above:

(a) 5 0p

(b) 05 q

(c) 14 2 q

(d) 1p

(e) 2q


5

Solution

(a) 55 0

0

9761,000

lp

l

(b) 505

0

41,000

dq

l

(c)

5 714 2

1

976 968 8991 991

l lq

l

(d) 21

1

985991

lp

l

(e) 22

2

3985

dq

l

We can also take a stochastic (ie random) view of future mortality. Under a stochastic approach, the number of survivors at age x is a random variable L x , and the life table function xl

represents the expected number of survivors, ie:

xl E L x

The random variable L x follows a binomial distribution. Each life is viewed as an

independent Bernoulli trial. The number of trials is 0n l . We define “success” as survival to age x , with probability 0xp p .

For example, let’s consider the distribution of 5L , the random number of survivors at age 5 in

this life table from the 1,000 newborn lives.

The random variable 5L follows a binomial distribution with 0 1,000n l trials and a

probability of success of 5 0 976 /1,000 0.976p p .

The mean and variance of 5L are:

0 5 0 5

5 0 5 0

5 1,000 0.976 976 (same as )

var 5 1,000 0.976 0.024 23.424

E L np l p l

L npq n p q

Finally, we can deduce little from the life table about the continuous random lifetime X , but we can identify the distribution of the curtate lifetime of a newborn, (0)K .

The probability Pr( (0) )K k is the probability that a newborn life dies in the age range [ , 1)k k , which we have already defined as 0k q . Hence:

00

Pr (0) kk

dK k q

l

For example, the probability that a newborn life has a curtate lifetime of 3 years is:

303

0

3Pr (0) 3 0.003

1,000d

K ql


6

1.3 The theory of continuous survival models

In this section we will study five different mathematical functions that can all be employed to specify the distribution of X , the random lifetime (ie age at death) of a newborn life:

the cumulative distribution function of X

the probability density function of X

the survival function

the life table function

the force of mortality.

We will focus on the relations between these functions as well as their meaning.

The PDF and CDF of the random lifetime

The random lifetime (ie age at death) of a newborn life, X , is assumed to be a continuous random variable. We will review the basic properties of continuous random variables and explain their interpretation in the context of the random lifetime.

Let’s begin with the cumulative distribution function (CDF):

( ) Pr( )XF x X x

The CDF XF x represents the probability that a newborn life will die at or before age x .

XF x is continuous and non-decreasing with 0 0XF and 1XF where is the first age

at which death is certain to have occurred for a newborn life.

The probability density function (PDF) is:

( ) ( )X Xf x F x wherever the derivative exists

The PDF Xf x is non-negative and continuous on the interval [0 , ) .

Recall that a value of Xf x is not a probability in itself. The probability that a newborn life dies

between ages a and b is:

Pr ( ) ( ) ( )b

Xaa X b f x dx F b F a

You should note that since X is assumed to be a continuous random variable, all of the intervals ,a b , [ , )a b , ( , ]a b , and ,a b have the same probability, F b F a .

Hence:

0

1Xf x dx

and:

0

Prx

X XF x X x f u du

Finally, the probability that a newborn life dies in the interval [ , ]x x x can be estimated as:

Pr ( )Xx X x x f x x


7

Example 1.2

Suppose that the lifetime X of a newborn life is uniformly distributed on the interval 0 ,100 .

(a) Identify the probability density function.

(b) Identify the cumulative distribution function.

(c) Calculate the probability of death occurring between ages 60 and 80.

Solution

(a) The PDF for a uniform distribution is constant and equal to the reciprocal length of the interval:

1

0.01 for 0 100100Xf x x

(b) The CDF is:

0 00.01 0.01 for 0 100

x xX XF x f u du du x x

(c) The probability of death between ages 60 and 80 is:

Pr 60 80 80 60 0.01(80 60) 0.20X XX F F

Note that the uniform distribution is not particularly well suited as a model of human mortality, but it is useful as a simple context to illustrate the theory. This mortality model is commonly known as de Moivre’s law. It was actually the first mortality model to be used in insurance practice.

Example 1.3

Suppose that the lifetime X of a newborn life is exponentially distributed with mean 75 years.

(a) Identify the probability density function.

(b) Identify the cumulative distribution function.

(c) Calculate the probability of death between ages 60 and 80.

Solution

The PDF for an exponential distribution with mean is

/1for 0x

Xf x e x . Hence:

(a) The PDF is:

/751for 0

75x

Xf x e x

(b) The CDF is:

/75 /75 /750 0

11

75

xx u u xXF x e du e e

(c) The probability of death between ages 60 and 80 is:

60/75 80 /75Pr 60 80 80 60 0.10518X XX F F e e


8

The survival function

In actuarial mathematics it is common to describe a survival model by giving the survival function rather that the density function or distribution function. The survival function is denoted ( )Xs x and is defined as:

PrXs x X x

The survival function gives the probability that a newborn life dies after age x . This is the same as saying that the newborn survives to age x , or is alive at age x .

From the preceding discussion of the lifetime variable X , we can deduce the following properties of the survival function.

Key properties of the survival function

1. Xs x is continuous and non-increasing with 0 1Xs and 0Xs

2. 1X Xs x F x

3. Prb

X X Xaa X b f x dx s a s b

4. X Xf x s x

Example 1.4

Suppose that the lifetime X of a newborn is exponentially distributed with mean 75 years.

(a) Identify the survival function ( )Xs x .

(b) Calculate the probability that a newborn is still alive at age 100.

(c) Calculate the probability that a newborn dies between ages 60 and 75.

Solution

(a) In Example 1.3 we saw that the CDF is /751 xXF x e . So, we have:

/751 for 0xX Xs x F x e x

(b) The probability a newborn is still alive at age 100 is:

100/75100 0.26360Xs e

(c) The probability a newborn dies between ages 60 and 75 is:

60/75 75/7560 (75) 0.08145X Xs s e e


9

The life table function

In contrast with the discussion of the discrete case in Section 1.2, here we will define the life table function xl for all ages between 0 and w .

As before, let L x denote the random number of survivors at any age x from a group of 0l

newborn lives. The random variable L x follows a binomial distribution with 0n l trials. We

define “success” as survival to age x , with probability Pr Xp X x s x .

The life table function xl is defined as the expected number of survivors at age x . Hence:

0x Xl E L x np l s x

Example 1.5

Suppose that the lifetime X of a newborn is uniformly distributed on 0 ,100 .

(a) Identify the survival function ( )Xs x .

(b) Identify the life table function xl if 0 100l .

Solution

(a) The survival function is given by:

100 100

Pr 0.01

0.01 100 1 0.01 for 0 100

X Xx xs x X x f u du du

x x x

(b) The life table function is:

0 ( ) 100(1 0.01 ) 100 for 0 100x Xl l s x x x x

Let’s summarize some of the key properties of the life table function, xl .

Key properties of the life table function

1. xl is the expected number of survivors at age x from a group of 0l newborn lives

2. 0x Xl l s x is continuous and non-increasing with 0l

3. 0

xX

ls x

l

Note that the value of 0l (sometimes called the radix of the life table) is not important to the survival model, since (by property 3) the survival function is independent of this quantity. So, we can choose the value of 0l for convenience. The survival function will be identical whether we choose 0 100l or 0 1,000,000l .


10

Example 1.6

Suppose that the life table function is 210,000 100 for 0 100xl x x . Identify the

cumulative distribution function and the probability density function for the associated lifetime variable X .

Solution

It is elementary to compute the survival function from the life table by property 3:

2 2

2 20

10,000 100 100 for 0 100

10,000 100 100x

Xx xl

s x xl

We can then calculate ( )XF x using the relationship 1X XF x s x :

2

2100

1 for 0 100100

Xx

F x x

Finally, we can calculate ( )Xf x using the relationship X Xf x F x :

2

2 2100 2 100 100

1 for 0 1005,000100 100

X Xx x x

f x F x x

Example 1.7

Suppose that there are 1,000 newborn lives whose lifetime follows the survival model given in Example 1.6. Determine the interval that lies within two standard deviations either side of the mean for 10L , the random number of survivors at age 10.

Solution

10L follows a binomial distribution with:

0 1,000n l

2

2

100 1010 0.81

100Xp s

So we have:

10

10 1,000 0.81 810

var 10 1,000 0.81 (1 0.81) 153.90

153.90 12.406L

E L np

L npq

Hence the required interval is:

810 2 12.406, 810 2 12.406 785.19, 834.81


11

The force of mortality

We can also specify a survival model in terms of the force of mortality.

The force of the mortality is denoted x . It is an instantaneous measure of mortality at age x ,

and it can be defined in several equivalent ways:

lnX X xX

X X x

f x s x lx s x

s x s x l

These equalities can be verified using simple calculus.

For example, using the information in Example 1.6, we have:

2

2 2

2 100 /100 2 for 0 100

100100 /100X

X

xs xx x

s x xx

Or, using the information in Example 1.4, we have:

1ln for 0

75 75Xx

x s x x

Let’s now see how to calculate the survival function from the force of mortality.

0 0

ln

ln ln ln 0

X

xxX X X

x s x

y dy s y s x s

But since 0 1 and ln 1 0Xs , we have:

0

0

ln

exp

xX

xX

y dy s x

s x y dy

Example 1.8

Suppose that the force of mortality for a survival model is given by the formula:

0.9for 0 90

90x x

x

Calculate the survival function.

Solution

The survival function is calculated as:

0 0

0

0.9

0.9exp exp

90

90exp 0.9 ln 90 exp 0.9 ln

90

90 for 0 90

90

x xX

x

s x y dy dyy

xy

xx


12

Since we have already studied simple relationships between the survival function, the life table function, and the PDF and CDF of the lifetime function, we can easily calculate any of these functions from the force of mortality.

For example, using the information in Example 1.8, we can calculate the life table function (with

0 1,000l ) as:

0.9

090

( ) 1,000 for 0 9090x X

xl l s x x

It is clear from the definition that the force of mortality is not a probability, so how should it be interpreted? In order to understand the meaning of x , it is useful to rewrite the defining

formula in the form:

X Xf x s x x

Now, recall that:

PrXf x x x X x x

Rewriting the probability term using a conditional probability, we have:

Pr

Pr Pr

Pr ( )

X

X

f x x x X x x

X x x X x X x

X x x X x s x

Substituting X Xf x s x x , we have:

Pr ( )

Pr

X Xs x x x X x x X x s x

x x X x x X x

So, x x is approximately equal to the conditional probability that a newborn that has

survived to age x subsequently dies during the next x years. For example, 20 multiplied

by 1/365x (a day), is approximately equal to the conditional probability that a newborn that has survived to exact age 20 will then die during the next day.

Example 1.9


0.9for 0 90

90x x

x

Calculate the approximate probability that a life age 40 dies within the next week.

Solution

Setting 7 /365x , the required probability is:

0.9 740 0.00035

90 40 365x


13

Let’s summarize the main properties of the force of mortality.

Key properties of the force of mortality

1.

0

ln limX X h xxX

hX X x

f x s x qlx s x

s x s x l h

2. 0

expx

Xs x y dy

3. Pr |x x X x x X x

4. x is non-negative and piece-wise continuous where defined

5. 0

y dy in order that 0Xs

Standard probabilities in a continuous survival model

Let’s now reconsider the ideas we met in Section 1.2 (in the context of a discrete life table) in the form of a continuous algebraic function defined for all x in 0 , .

Standard probabilities

1. Probability that a life currently age x survives t years:

PrPr |

PrXx t

t xx X

s x t X x tlp X x t X x

l s x X x

2. Probability that a life currently age x dies in the next t years:

Pr PrPr |

Prx x t

t xx

X x X x tl lq X x t X x

l X x

3. Probability that a life currently age x survives s years but dies in the following t years:

Pr |x s x s t

xs tx

l lq x s X x s t X x

l

These three functions are defined for all ages x in 0 , and for 0 t x .

It should be emphasized now that all of these probabilities are conditional, ie we are given that a newborn has survived to age x.

The symbol ( )x is commonly used to denote a newborn life that has survived to age x . So, 5 xp is the probability that ( )x will still be alive in 5 years’ time, at age 5x , and 5 xq is the probability that ( )x will die within the next 5 years.

As in Section 1.2, the general convention is to drop the subscript t from the symbol when 1t . So, for example, 3| xq is the probability that ( )x will die between ages 3x and 4x .


14

Example 1.10


0.9for 0 90

90x x

x

Compute the following probabilities:

(a) 2.5 20p

(b) 202.5q

(c) 202.5|q

Solution

Note that this is the force of mortality in Example 1.8. The life table function is:

0.9

0 090

( ) for 0 9090x X

xl l s x l x

Recall that we can choose any convenient value of 0l without changing the distribution of X . So

let’s simplify our computations by choosing 0.90 90l , which gives:

0.90 ( ) (90 ) for 0 90x Xl l s x x x

We can now calculate the required probabilities as follows.

(a)

0.922.5

2.5 20 0.920

90 22.50.96780

90 20

lp

l

(b)

20 22.52.5 20 2.5 20

201 0.03220

l lq p

l

(c)

0.9 0.9

22.5 23.52.5 20 0.9

20

67.5 66.5| 0.01291

70

l lq

l

The challenge in dealing with these standard probabilities is that there are so many relationships that involve them. The key relations are listed below without proof. Most of the proofs rely on simple probability theory – you may like to attempt them to improve your understanding.

Key relations concerning standard probabilities

1. 1t x t xp q

2. s t x s x t x sp p p

3. |s t x s x t x s s x s t x s t x s xq p q p p q q

4. 1 1 when is an integern x x x x np p p p n

5. 0 1 1| | | when is an integern x x x n xq q q q n


15

For example, if the probability that 20 survives for 10 years is 0.97, and if the probability that

30 survives for 10 years is 0.95, then the probability that 20 is still alive at age 40 is:

20 20 10 20 10 30 0.97 0.95 0.92150p p p

Or, the probability that 20 dies between ages 30 and 40 is:

10 10 20 10 20 10 30| 0.97 1 0.95 0.04850q p q

On the other hand, if 0 1 20.99 , 0.98 , and 0.97p p p , then the probability that a newborn dies within three years is:

3 0 3 0 0 1 21 1 1 0.99 0.98 0.97 0.05891q p p p p

Or the probability that a newborn dies during the second year of life is:

1 0 0 1| 0.99 1 0.98 0.01980q p q

Relations 4 and 5 are useful in constructing a discrete life table for human lives. A statistical study conducted over a time span of several years could be used to produce estimates of the mortality rates 0 1 2, ,q q q , … and so on. Values of xl at whole number ages can then be produced as follows:

0 0 0 0 0 1 1 0 0 1 10

1 1 1nn n n n n

lp l l p l p p p l q q q

l

1.4 The continuous future lifetime after age x

Let the continuous random variable X again denote the random lifetime of a newborn. Now suppose that we are given that a newborn has survived to age x , that is, X x . The future time lived after age x is X x .

The conditional distribution of the time lived after age x , given survival to age x, is:

| T x X x X x complete future lifetime :

The continuous random variable ( )T x is a survival model defined on the interval [0 , ]x . As such, it can also be specified in the same ways that we specified the survival model for a newborn life. It should be clear that the distribution of ( )T x is closely related to the distribution of X .

The quickest way to see the relation between the distributions of ( )T x and X is to calculate the

survival function for ( )T x , PrTT xs t s t T x t . In fact, we have already computed

this survival function in terms of the distribution of X , since the event T x t is equivalent to

saying that ( )x is alive at age x t . The probability of this event is simply t xp . So, we have:

0( ) Pr since ( )Xx t

t x x XT xx X

s x tls t T x t p l l s x

l s x

Note: When there is no ambiguity, we will write T for T x . However, subscripts are often

important, for example to distinguish 20Xs , the probability that a newborn survives to age 20,

from 10 20Ts , the probability that 10 survives to age 30.


16

Example 1.11

Suppose that the life table function is 210,000 100 for 0 100xl x x .

(a) Compute the survival function for newborn lives.

(b) Compute the survival function for lives currently aged 20.

Solution

(a) The survival function for newborn lives, ( )Xs x , is:

2 2

0 20

10,000 100 100for 0 100

10010,000( 100 0)x

X xx xl

s x p xl

(b) The survival function for lives currently aged 20, (20)( )Ts t , is:

2220

2020 220

10,000(100 (20 )) 80for 0 80

8010,000(100 20)t

tTtl t

s t p tl

When we deal with the future lifetime after age x , we’ll frequently see expressions of the type

x tl and ( )x t . In these expressions the value of x is fixed, and the value of t is allowed to

vary so that we can view the expressions as being functions of t .

For example, let’s see how to relate the PDF and CDF for the distributions of X and T .

Pr Pr |

Pr

Pr 1

T

X X

X

F t T x t X x t X x

x X x t F x t F xX x F x

1 1X X X X

T TX X X

F x t F x f x t f x tdf t F t

dt F x F x s x

In the figure below, we have areas X XA F x t F x and XA B s x .

x x+t

Area Area = A = B

fX(x)

Notice that we have:

X X

TX

F x t F x AF t

s x A B

If we view the age x as being fixed, then as the value of t increases, the value of A increases, the value of B decreases, while the sum A B remains constant.


17

Furthermore, if we examine the relation:

for 0XT

X

f x tf t t x

s x

in light of the figure above, we can see how the graph of the PDF of T is related to the graph of the PDF of X .

If we take the portion of the graph of Xf x to the right of x , divide by the total area under the

remainder of the graph, XA B s x , and then relabel the horizontal axis as t and the vertical

axis as Tf t , we have a graph of the PDF of T :

0 t -x

Area = Area = A / sX(x) B / sX(x)

fT(t)

One final point worth noting is the similarity between the PDF’s for the distributions of X and T .

For X we have:

0X

X X xX

f xx f x s x x p x

s x

and for T we have:

0 0

0 0

X x t x t xT t x

X x x

f x t p x t p p x tf t p x t

s x p p

Example 1.12

Suppose that the life table function is 210,000 100 for 0 100xl x x .

(a) Compute the distribution function for the future lifetime of a life aged 20.

(b) Compute the density function for the future lifetime of a life aged 20.

Solution

(a) In Example 1.11 we computed:

2

20 2080

for 0 8080t T

tp s t t

Hence, the distribution function for 20T is:

2

20 2080

1 1 for 0 8080T T

tF t s t t


18

(b) We have two options for calculating the density function.

The first option is to differentiate the distribution function:

2

20 20 28080

1 280 80

80 for 0 80

3, 200

T Ttt

f t F t

tt

The other option is to use the relation:

T t xf t p x t

It is consistent with the formula for xl that:

2100

xx

Hence, we have:

2

(20) 2080 2 80

2080 80 3, 200T t

t tf t p t

t

Key results concerning the relation of the distributions of X and T(x)

1.

Pr Xx tT t x

x X

s x tls t T x t p

l s x

2.

1 X XT t x T

X

F x t F xF t q s t

s x

3. X

T t xX

f x tf t p x t

s x

1.5 The curtate future lifetime after age x

In addition to computing the distribution of the continuous future lifetime T x , we may also

wish to derive the distribution of the curtate future lifetime after age x .

The curtate lifetime is a discrete random variable that is defined by:

the integer part (or greatest integer) of ( )K x T x ie T x

Since it is a function of T x , it is simple to calculate the probability function of K x from what

we know about T x . The possible values of K x are the numbers 0,1,2, , 1x .

For example, if 70x and 90 , then the possible values of 70K are the twenty whole

numbers 0 through 19. If the life 70 eventually dies at age 85.8, then the continuous future

lifetime is (70) 15.8T and the curtate lifetime is (70) 15.8 15K .


19

The key observation is that if K x k , then we must have:

( ) 1k T x k

This leads to the following formula for the probability function:

Pr ( ) Pr 1

| for 0 ,1 ,2, , 1x kk x

x

K x k k T x k

dq k x

l

Example 1.13

Suppose that the life table function is given by the formula:

100 for 0 100xl x x

Compute the probability function for 75K .

Solution

The probability function for 75K is:

75 75 175

75 75Pr 75

100 75 100 75 1 1100 75 25

k kk l ldK k

l l

k k

So 75K has 25 possible values ( 0,1, 2, , 24 ) that are equally likely to occur.

Example 1.14


0.0151,000 for 0xxl e x

Compute the probability function for 75K .

Solution

The probability function for 75K is:

75 75 175

75 75

0.015 75 10.015 750.015 0.015

0.015 75

Pr 75

1

k kk

kkk

l ldK k

l l

e ee e

e

Note that this is a geometric distribution.

It is also useful to develop formulas for the cumulative distribution function and survival function of the curtate future lifetime.


20

Recall that for any random variable ( ) Pr( )XF x X x , hence:

11 2

1

( ) Pr Pr 0 Pr 1 Pr

for 0 ,1 , , 1

K x

x x kx x x x k

x x x x x

k x

F k K x k K x K x K x k

l ld d d dl l l l l

q k x

The survival function of the curtate future lifetime is then easily derived as:

1 1Pr 1 1 for 0 ,1 ,... , 1k x k xK x K xs k K x k F k q p k x

Example 1.15


100 for 0 100xl x x

Compute the survival function for 75K .

Solution

The survival function for 75K is:

75 1

1 757575

100 (75 1) 24for 0 ,1 , , 24

100 75 25k

kKkl k

s k p kl

Example 1.16


0.0151,000 for 0xxl e x

Compute the survival function for 75K .

Solution

The survival function for 75K is:

0.015(75 1)

0.015 175 11 7575 0.015(75)

75for 0 ,1 ,2 ,

kkk

kKl e

s k p e kl e

Let’s conclude this section with a summary of the key relations concerning the curtate future lifetime, ( )K x .

Key relations concerning the curtate future lifetime

1. Pr | for 0 ,1 ,... , 1x kk xK x

x

df k K k q k x

l

2. 11 for 0 ,1 ,... , 1

x kk xK x

x

ls k p k x

l

3. 1 11 for 0 ,1 ,... , 1k x k xK xF k q p k x


43

Chapter 1 Practice Questions

Question 1.1

Compute the following probabilities from the life table in Section 1.2:

2 0 2| 0 4| 2 3 4 5, , , ,p q q p q

Question 1.2

Suppose that 2 1| 0.015q . Discuss the distribution of the random number of deaths between ages 3 and 4 for a group of 20 lives currently age 1.

Question 1.3

The life table below is a survival model for a group of newborns suffering from a certain heart impairment.

x 0 1 2 3 4

xl 100 46 19 6 0

Give the probability function for the curtate future lifetime of a member of this group of impaired lives.

Question 1.4

Suppose that 3/ 2x x for 0x . Determine the PDF, CDF, and survival function for the

lifetime of a newborn. Check your work by verifying properties such as 0

1f x dx

..

Question 1.5

Using the survival model in Question 1.4, determine the values of 1p and 2 1|q .

Question 1.6

Suppose that 0.5/ 100x x for 0 100x . Determine the PDF, CDF, and survival

function for the lifetime of a newborn.

Question 1.7

Using the survival model in Question 1.6, determine the values of 20 40p and 20 20 40| q .

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