02
Contents
List of Figures
viii
List of Tables
viii
Preface
ix
About the Authors
xi
1 Introduction
1
2 Valuation of Financial Transactions
2.1 Simple Interest . . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Compound Interest . . . . . . . . . . . . . . . . . . . . . . . .
2.2.1 Valuation of a single payment . . . . . . . . . . . . . .
2.2.2 Valuation of a series of payments . . . . . . . . . . . .
2.2.3 The equation of value . . . . . . . . . . . . . . . . . . .
2.3 Common Compound Interest Transactions . . . . . . . . . . .
2.3.1 Fixed interest bonds . . . . . . . . . . . . . . . . . . .
2.3.2 Housing loans . . . . . . . . . . . . . . . . . . . . . . .
2.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
4
4
7
7
17
25
27
27
34
45
3 Demography
3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . .
3.2 Characteristics of a Population . . . . . . . . . . . . . . . . .
3.2.1 Sources of information . . . . . . . . . . . . . . . . . .
3.2.2 Classication of data . . . . . . . . . . . . . . . . . . .
3.2.3 Summary statistics . . . . . . . . . . . . . . . . . . . .
3.2.4 Rates of change: the population growth rate . . . . . .
3.2.5 Demographic transition . . . . . . . . . . . . . . . . . .
3.2.6 Development of world demographics . . . . . . . . . . .
3.3 Individual Characteristics: Mortality . . . . . . . . . . . . . .
3.3.1 The survival function . . . . . . . . . . . . . . . . . . .
50
50
51
51
53
53
59
60
64
66
66
v
vi
CONTENTS
3.3.2 The life table . . . . . . . . . . . . . . . . . . . . . . . 72
3.3.3 Characteristics, causes and trends of mortality experience 81
3.3.4 Mathematical models of mortality . . . . . . . . . . . . 84
3.4 Individual Characteristics: Fertility . . . . . . . . . . . . . . . 85
3.4.1 Denition of fertility . . . . . . . . . . . . . . . . . . . 85
3.4.2 Measures of fertility . . . . . . . . . . . . . . . . . . . . 86
3.4.3 Factors aecting fertility . . . . . . . . . . . . . . . . . 90
3.4.4 Modern trends in family formation . . . . . . . . . . . 91
3.4.5 Trends in fertility rates, and consequences . . . . . . . 92
3.5 Population Projections . . . . . . . . . . . . . . . . . . . . . . 93
3.5.1 Context . . . . . . . . . . . . . . . . . . . . . . . . . . 93
3.5.2 Projection models . . . . . . . . . . . . . . . . . . . . . 95
3.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
4 Actuarial Practice
109
4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109
4.2 Principles of Insurance . . . . . . . . . . . . . . . . . . . . . . 109
4.3 Life Insurance . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
4.3.1 Whole life insurance policy . . . . . . . . . . . . . . . . 110
4.3.2 Term insurance policy . . . . . . . . . . . . . . . . . . 111
4.3.3 Endowment insurance policy . . . . . . . . . . . . . . . 112
4.3.4 Unitised insurances . . . . . . . . . . . . . . . . . . . . 113
4.3.5 Trauma insurance . . . . . . . . . . . . . . . . . . . . . 116
4.3.6 Disability insurance . . . . . . . . . . . . . . . . . . . . 117
4.3.7 Reverse mortgages . . . . . . . . . . . . . . . . . . . . 119
4.3.8 The role of the actuary in life insurance . . . . . . . . . 120
4.4 Private Health Insurance . . . . . . . . . . . . . . . . . . . . . 125
4.5 Superannuation . . . . . . . . . . . . . . . . . . . . . . . . . . 127
4.5.1 Dened benet schemes . . . . . . . . . . . . . . . . . 128
4.5.2 Dened contribution schemes . . . . . . . . . . . . . . 132
4.5.3 The role of the actuary in superannuation . . . . . . . 133
4.6 General Insurance . . . . . . . . . . . . . . . . . . . . . . . . . 134
4.6.1 The role of the actuary in general insurance . . . . . . 137
4.7 National Insurance . . . . . . . . . . . . . . . . . . . . . . . . 138
4.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140
5 Valuation of Contingent Payments
142
5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142
5.2 Discrete Random Variables . . . . . . . . . . . . . . . . . . . . 143
5.3 Valuation of a Single Contingent Payment . . . . . . . . . . . 145
5.4 Valuation of a Series of Contingent Payments . . . . . . . . . 149
CONTENTS
vii
5.5 Premium Calculation . . . . . . . . . . . . . . . . . . . . . . . 154
5.5.1 Whole life insurance . . . . . . . . . . . . . . . . . . . 154
5.5.2 Endowment insurance . . . . . . . . . . . . . . . . . . 158
5.6 Relationships between Actuarial Functions . . . . . . . . . . . 164
5.7 Parameter Variability . . . . . . . . . . . . . . . . . . . . . . . 167
5.8 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 170
Further Reading and Acknowledgements
174
Appendix 1. Male Mortality Table
176
Appendix 2. Female Mortality Table
180
Appendix 3. Solutions to Exercises
184
References
229
Index
230
List of Figures
2.1 Time line diagram for a single payment . . . . . . . . . . . . 14
2.2 Time line diagram for multiple payments . . . . . . . . . . . . 14
3.1
3.2
3.3
3.4
3.5
3.6
3.7
3.8
Population pyramid for Ethiopia in 2009 . . . . . . . . . . . . 62
Population pyramid for Australia in 2009 . . . . . . . . . . . . 62
Population pyramid for France in 2010 . . . . . . . . . . . . . 63
Female mortality table — o{ . . . . . . . . . . . . . . . . . . . . 73
Female mortality table — g{ . . . . . . . . . . . . . . . . . . . 74
Male and female (dotted line) mortality rates . . . . . . . . . 82
Australia’s population from 1901 to 2009 . . . . . . . . . . . . 99
Population pyramid for Japan in 2009 . . . . . . . . . . . . . . 102
List of Tables
3.1 Population summary statistics . . . . . . . . . . . . . . . . . . 64
5.1 Spreadsheet output for Example 5.6 . . . . . . . . . . . . . . . 152
5.2 Spreadsheet output for Example 5.10 . . . . . . . . . . . . . . 164
viii
Preface
Preface to the 1st edition
This book was conceived while I was involved in lecturing a rst year university course providing an introduction to the concepts and practice of actuarial
work. Students were drawn to the course who already had the intention of
pursuing a professional career as an actuary, and also who merely intended
to taste the introduction for interest and pleasure. I believe that a signicant
number who entered the course as belonging to the latter group nished the
course belonging to the former, and continued with actuarial studies thereafter. This was most gratifying to see.
It became clear that there was no single introductory text available to
these students which gave an overview of the gamut of actuarial techniques
and practice. Such a view encompasses a broad landscape, indeed. My
intention has been to supply such a text, both to answer the specic need
of my students, and for its own sake. Having observed the interest and
enthusiasm of some ve hundred students or more I felt the need existed,
and the reward was real.
This text therefore is designed to take a slice of the actuarial landscape —
a slice o the top — providing a supercial, but not trivial, coverage of a broad
range of topics. As such, this text is accessible to many, requiring very little
in the way of technical pre-knowledge of the topics covered. It is my hope
that such a broad perspective will provide students with some understanding
of the complexity and variety of actuarial training and methodology, long
before they can hope to actually master the detail of the techniques.
The text assumes an elementary understanding of basic calculus, statistics and probability, such as a university entrant might already have encountered in a mathematical curriculum at high school, or in introductory courses
at tertiary level. The pre-requisites for understanding the text are a basic
mathematical knowledge and an ability for problem solving.
The basic principal characteristics of actuarial work are associated with
observing, analysing, modelling and learning to adapt. These are practices
ix
x
PREFACE
valuable in most walks of life, and I do believe that the actuary’s knack as a
problem solver is the gift which distinguishes him or her.
I am fortunate to have secured the help of David Dickson, a colleague
and friend, as co-author and co-editor in the production of this text. He
brings his particular gift for rigour to the organisation and presentation of
the material. I am thankful to have worked with him, and grateful for his
unfailing support.
I wish to thank the students of Actuarial Studies I at the University of
Melbourne, and at the Australian National University in 1998, for “roadtesting” the prototype of this text, and for their enthusiastic and helpful
response to it.
Margaret E. Atkinson
Melbourne, November 1999
Preface to the 2nd edition
Throughout the years that this textbook has been used, a recurring comment
from students has been that they would like more exercises. In this 2nd edition we have added a large number of exercises and have provided solutions.
However, this is not the only substantial change from the 1st edition. New
material has been added to the nal chapter, most notably in Sections 5.6
and 5.7. We have also taken the opportunity to update material, especially
in Chapter 3, and there are alterations and additions scattered throughout
the text. Our thanks go to sta and students at the University of Melbourne
who have given us feedback on the rst edition.
David C.M. Dickson
Melbourne, January 2011
About the Authors
Margaret Atkinson first graduated at the University of York in England, with
a B.A. in Mathematics and Philosophy and an M.Phil. in Applied Analysis,
and was awarded the K.M. Stott prize for Mathematics. Shortly after, she
began her actuarial studies with the Faculty of Actuaries in Edinburgh. She
emigrated to Australia in 1985 and in 1993 began work as a research assistant at the University of Melbourne, was appointed as Lecturer in Actuarial
Studies in 1995, and completed her Ph.D. there in 1998. She has since moved
on to alternative pursuits.
David Dickson gained his tertiary qualifications at Heriot-Watt University,
Edinburgh. After a couple of years at the Government Actuary’s Department, London, he returned to Heriot-Watt and lectured there for seven years.
During this period he qualified as a Fellow of the Faculty of Actuaries and was
heavily involved in educational activities for the British actuarial profession.
He took up an appointment at the University of Melbourne in 1993, where
he has been Professor of Actuarial Studies since 2000. His main research
interest is risk theory, and he has published extensively in this field.
xi
Chapter 1
Introduction
What is an actuary? This straightforward question does not have a straightforward answer. A dictionary might dene an actuary as a person who does
insurance or related calculations. However, there is much more to actuarial
work than either insurance or just doing calculations. In the current era,
we might dene an actuary as a nancial risk manager, but such a compact
denition hides the range of skills required for success in actuarial work.
Historically, actuarial work was concerned with life insurance. Nowadays, actuaries are employed in a range of other activities in the nancial
sector, working on problems in the areas of superannuation, general insurance, health insurance, stockbroking, banking and investments. Naturally,
the skills required vary across these areas. Nevertheless, there are some basic
ideas which can be applied to many problems tackled by actuaries.
The purpose of this text is to provide an introduction to some of the
techniques used in actuarial work, and to give an overview of some of the
areas in which actuaries are currently involved. It is perhaps most convenient
to use the most traditional of actuarial areas — life insurance — to illustrate
the multi-disciplinary nature of actuarial work and to provide the setting for
much of what follows in this text.
One of the most basic products oered by life insurance companies is a
whole life insurance policy. Under such a policy, the policyholder makes regular (say annual) payments, known as premiums, to the insurance company.
In return, the insurance company promises to pay a sum of money to the policyholder’s estate on the policyholder’s death. Such a simple contract gives
rise to a number of problems which the actuary must solve. For example, at
the time such a policy is issued, the following will be unknown:
(a) how many premiums the policyholder will pay,
(b) what sort of return the insurance company can earn when it invests the
premiums.
1
2
CHAPTER 1. INTRODUCTION
Actuaries use past experience to build mathematical models which allow
them to assess risks and price insurance products. For this insurance policy,
the actuary will need a model for investment returns. The duration of an
insurance policy can be anything from a few days up to more than seventy
years. Clearly, actuaries cannot take a very short term view of investment,
but must consider investing for considerable time periods. A simple model
that has served the actuarial community well over many years is compound
interest. In Chapter 2 we discuss the growth of money over time and describe
some common nancial transactions to which compound interest techniques
apply.
To price a whole life insurance policy, the actuary will also need a model
of mortality. Such a model would be used to estimate the probability that
a premium is paid when it is due (or that the policyholder will die in a
given year). One of the things that an actuary would do in formulating such
a model would be to consider data on insured lives, or, if that were not
available, data from a similar group of persons. The branch of statistics that
deals with such problems is known as survival modelling, and in Chapter 3
we not only discuss survival models, but also consider the broader study of
populations of individuals, known as demography.
In Chapter 4, we give an overview of some of the main areas of actuarial
practice, using examples current in Australia. Actuarial principles and techniques are not country dependent, but legislation (for example on taxation
or investments) may dictate how they are applied. We also describe some
products currently available in the Australian marketplace.
We conclude in Chapter 5 with a simple introduction to one of the cornerstones of actuarial work — the valuation of contingent payments. A contingent
payment is simply a payment that is made on condition that a specied event
occurs. In the case of a whole life insurance policy, the specied event is the
death of the policyholder, and the insurance company makes a payment only
when this occurs.
We have provided numerous worked examples, illustrating the principles
described in the text, and exercises with full solutions for you to establish
condence in using the techniques. In an introductory text, it is not possible to describe solutions to a great many problems. In addition to skills
in mathematics and statistics, actuaries must also understand the nancial
environment in which they work and be capable of making judgements on
the future well-being of an economy. Such topics are beyond the scope of this
text. Our aim is to provide a avour of actuarial mathematics and the areas
to which it can be applied. We hope it whets your appetite to learn more
about the actuarial world. Even if you do not intend to pursue a specically
actuarial career or course of study, these basic techniques are portable and
robust, and can be brought to an understanding and analysis of a broad range
3
of situations which are not specically ‘actuarial’. Any situation requiring an
assessment of uncertain future events benets from actuarial understanding
and methodology.
Chapter 2
Valuation of Financial
Transactions
2.1
Simple Interest
One of the fundamental concepts associated with valuing transactions which
involve monetary payments over a period of time is that the value of an
amount of money depends upon when that payment is made, and what value
the parties concerned place upon the timing of the payment.
For example, if you are oered $100 now or a payment at the end of one
year from now, under what conditions will waiting for an extra year be worth
it for you? Clearly, if you are prepared to accept $100 at the end of a year,
the delay is worth nothing to you.
If you would require $150 at the end of the year, then the delay is worth
50% of the original sum. Alternatively, perhaps you consider that the likelihood of the payment being made is not high, and so you require an even
higher payment to compensate for the risk of default. This represents another major factor involved in the valuation of a promised payment — the
risk associated with it. In practice a combination of these factors is likely.
For the moment we consider only the eect of the value placed upon time
in the evaluation of monetary transactions.
Consider a common and very simple monetary transaction. An investor
places an amount of money in a bank account earning interest, and after a
period of time, withdraws all the money from the account. Typically, we
may want to calculate how much the balance of the account will have grown
to at the end of the period. In order to calculate this, we need to know what
rate of interest is earned by the original investment.
Let us begin by dening some common terms.
4
2.1. SIMPLE INTEREST
5
Denition 2.1 Principal: the principal is the amount of money originally
invested, and which attracts interest.
Denition 2.2 Simple interest: let u be the rate of simple interest earned
by a unit investment over a unit period of time. Then, after a period of time
w, where w 0, the unit investment will have grown to 1 + uw.
In general, if an amount S , the principal, is invested for a time w> at a simple rate of interest of u per unit time, the investment will have accumulated
to
S (1 + uw)
at the end of that time.
Rates of interest are usually expressed per cent, and the commonly used
unit of time is one year.
Example 2.1 An amount of $1,000 is invested for 1 year at a rate of simple
interest of 8% per annum. How much has it accumulated to at the end of the
year?
Solution 2.1 1,000(1 + 0=08) = 1,080=
Example 2.2 An amount of $1,000 is invested for ve years at a rate of
simple interest of 8% per annum. How much has it accumulated to at the
end of the ve years?
Solution 2.2 1,000(1 + 0=08 × 5) = 1,400=
If we let D stand for the amount of the accumulation, we can write a
general statement relating items in a simple interest accumulation as
S (1 + uw) = D.
Most of the problems we consider involve
• calculating the accumulation of a given principal after a given time at
a given rate of interest, or
• calculating the amount of principal which will amount to a required
accumulation after a given time at a given rate of interest, or
• calculating the length of time required for a given principal to accumulate to a given amount at a given rate of interest, or
• calculating what rate of interest is required so that a given principal
will accumulate to a given amount after a given period of time.
6
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
That is, for a problem involving simple interest, we are generally trying to
evaluate one of S> D, u, or w> given that we know the other three.
Example 2.3 We must pay a debt of $10,000 at the end of two years, and
may invest money at a rate of simple interest of 7% per annum (p.a. hereafter). How much must we invest now in order to be able to repay the debt?
Solution 2.3 We need to calculate the principal which will accumulate to
$10,000 at simple rate 7% p.a. over two years. That is, we require S such
that
S (1 + 2 × 0=07) = 10,000
, S = 8,771=93=
Note that simple interest is added to the principal only at the end of
the term of the investment. In practice there are a few circumstances in
which this happens, for example, certain types of investment in the short
term money market are quoted as earning simple interest or simple discount.
Denition 2.3 Simple discount: a unit investment made for a unit period of
time at a rate of simple discount g per period of time accumulates to 1@(1g).
Equivalently, the cost now of purchasing a unit payment at the end of a
unit period of time, at a rate of simple discount g per unit time, is 1 g=
In general, the value now, denoted S> of an amount D payable at time
w, calculated at a rate of simple discount of g per unit time, is given by the
following expression:
S = D (1 gw).
Simple discount is commonly used, for example, in transactions involving
Bills of Exchange, but is not often encountered outside the nancial markets.
Indeed, simple rates of interest are used only in a limited way also. For
example, they are usually quoted in hire purchase, or consumer credit types
of arrangement.
Example 2.4 An amount of $1,000 is invested for 1 year at a rate of simple
discount of 8% per annum. How much has it accumulated to at the end of
the year?
Solution 2.4 1,000@(1 0=08) = 1,086=96.
Remark 2.1 Note that this is not the same accumulation that was calculated
for Example 2.1, and that a simple rate of discount of 8% p.a. has resulted
in a larger accumulation than a simple rate of interest of 8% p.a.
2.2. COMPOUND INTEREST
7
Example 2.5 An amount of $1,000 is invested for ve years at a rate of
simple discount of 8% per annum. How much has it accumulated to at the
end of the ve years?
Solution 2.5 1,000@(1 0=08 × 5) = 1,666=67.
Remark 2.2 Again, note how this solution diers from that in Example 2.2.
2.2
Compound Interest
2.2.1
Valuation of a single payment
Eective rate
In practice, simple interest applies only in a limited number of circumstances.
It is more usual for compound rates of interest to apply. Compound interest
diers from simple interest in that simple interest is added to the principal
only at the end of the period of investment, whereas compound interest is
added to the principal at regular intervals. Once the interest earned has been
added to the principal, it then attracts interest itself. Under the action of
simple interest, only the principal ever earns interest.
Let l denote a compound rate of interest per unit time. Then after one
unit of time, a unit investment will have accumulated to 1 + l> the original
unit investment plus the interest earned. However, the interest earned is now
added to the principal, and now earns interest also. Thus, at the end of the
second unit of time, the investment will have accumulated to
(1 + l)
the principal at the end of the rst period,
+ l (1 + l) plus the interest earned in the second period.
Thus, the accumulation at the end of the second period is
(1 + l) + l (1 + l) = (1 + l)2 .
This is now the amount of the principal, and this amount earns interest
during the next period of investment.
Denition 2.4 If l is the eective rate of interest per unit time, then a unit
investment accumulates to 1 + l at the end of one time period.
Example 2.6 Consider an investment of $100 made for three years at a
rate of 5% compound p.a. Construct a table showing the accumulation of
this investment over the three year period.
8
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
Solution 2.6 The investment will grow according to the following tabulation,
where amounts are shown to two decimal places.
Principal at
5% interest
Accumulation
Year start of year at end of year at end of year
1
100
5
105
2
105
5.25
110.25
3
110.25
5.51
115.76
Remark 2.3 Note how this diers from growth under simple interest. At
5% p.a. simple interest, for three years, the total interest earned would be
3 × 0=05 × 100 = 15= Here, the action of compound interest results in a
total interest earning of 15.76. The additional amount represents the interest
earned by the interest which is added during the term of the investment.
We can dene accumulation under compound interest as follows.
Denition 2.5 Suppose S is invested for q time periods at a rate of compound interest l per period. After q periods, the investment will have accumulated to S (1 + l)q .
We can use this denition to link payments, or the value of an investment, made at dierent times under the action of compound interest. We
have described above the growth of an amount S at time w = 0 attracting
compound interest at rate l per period for q periods, that is, at time w = q=
Denition 2.6 The present value at time w = 0 of the accumulation in
Denition 2.5 is S=
The term present value is commonly used in actuarial calculations, and
denotes the value of a transaction at a particular time. Commonly the time
it refers to is w = 0> but it need not be. The present value in the above
denitions at time w = q is S (1 + l)q = Hereafter, we adopt the convention
that present value implies w = 0 unless otherwise stated.
Suppose that we wish to calculate the present value S of a payment of D
to be made in q years’ time. We know, from the above, that after q periods
the accumulation of S is S (1 + l)q = Thus,
S (1 + l)q = D
, S = D (1 + l)3q .
Denition 2.7 Terminology: y = (1 + l)31 = That is, y is the present value
of a unit payment at time w = 1, evaluated at an eective rate of interest l
per period. This is standard actuarial notation.
2.2. COMPOUND INTEREST
9
Example 2.7 How much must be invested now in order to accumulate an
amount of $200 at the end of three years, given a compound rate of interest
of 5% p.a.?
Solution 2.7 Let [ denote the amount to be invested now. Then
1=053 [ = 200
, [ = 200@1=053 = 172=77=
That is, the present value at rate 5% p.a. of $200, paid at the end of three
years, is $172.77.
For simplicity, we adopt the convention that interest rates quoted are
compound rates, unless otherwise stated.
Example 2.8 An investor is about to invest an amount of $1,000 for ten
years. There are two alternative investments available. The rst oers accumulation at 5% p.a. eective, and the second oers accumulation at 6%
p.a. simple interest. Which investment should the investor choose?
Solution 2.8 At 5% p.a. eective, $1,000 will accumulate to
1,000 × 1=0510 = 1,628=89
after ten years. At 6% p.a. simple, $1,000 will accumulate to
1,000 × (1 + 0=06 × 10) = 1,600=
Thus, over a period of ten years, the compounding eect means that 5% p.a.
compound is more valuable than 6% p.a. simple interest.
Fractional periods
We have dened an eective rate of compound interest l per period, such
that principal S invested for q periods will accumulate to S (1 + l)q = This
relationship holds for non-integral values of q. Thus, suppose that q is an
integer, q 0, and 0 i ? 1. Then, if S is invested at compound rate l
per period for q + i periods, the accumulated value of S at the end of this
time is
S (1 + l)q+i .
Similarly, the present value of an amount D payable at the end of q + i
periods, calculated at compound rate l per period is
D (1 + l)3(q+i ) .
10
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
Nominal rates of interest
There are several ways of describing the rate of compound growth of an
investment. We have dened the eective rate of interest l per period as
being such that if an amount S is invested for one period, it will accumulate
to amount D at the end of the period where
D = (1 + l) S=
Thus, given D and S we can calculate the eective rate of interest which
applies over the period.
Another commonly used way of describing compound growth is the nominal rate of interest.
Denition 2.8 If l(p) is the nominal rate of interest payable (or convertible)
p-thly per unit time, then a unit investment accumulates to D at the end of
one period, where
μ
¶p
l(p)
D= 1+
.
p
Splitting the time period into p new periods of equal length, we see that
l @p is the eective rate per new period, i.e. per 1@p-th of the original time
period. For example, a nominal interest rate of 12% p.a. payable monthly is
equivalent to an eective rate of 1% per month.
(p)
Example 2.9 The nominal rate of interest is 6% p.a. payable half yearly.
(a) What is the accumulated value of $1,000 after two years?
(b) What is the eective annual rate of interest earned by this investment?
Solution 2.9
(a) 1,000(1 + 0=06
)4 = 1,125=51=
2
(b) The eective annual rate of interest l p.a. is such that
1,000(1 + l)2 = 1,125=51=
We can solve this equation for l> to obtain the eective rate earned:
s
l = 1=12551 1 = 0=0609=
2.2. COMPOUND INTEREST
11
Note that in the above example, the eective annual rate of 6.09% is
greater than 6%. This is because 3% interest is added to the principal at the
end of half a year, and it then accrues interest itself. Thus, by the end of
the rst year, the total interest accumulated includes interest on the interest
paid in the rst half year. The more frequently the interest is compounded,
(that is the greater p is, for l(p) ), the higher the eective rate of interest
achieved will be. The following examples illustrate this.
Example 2.10 The nominal rate of interest is 10% p.a., payable p-thly.
Calculate the eective rate of interest p.a. which is equivalent to this, for
values of p of 1, 2, 4, 6 and 12.
Solution 2.10
p
1
2
4
6
12
³
´p
l(p)
1+ p
1
0=10
0=1025
0=10381
0=10426
0=10471
Example 2.11 The eective annual rate of interest is 10%. What is the
equivalent nominal rate l(p) p.a., when p is equal to 1, 2, 4, 6 and 12?
Solution 2.11 The accumulation at the end of one year of a unit payment
is 1.1 at an eective rate of interest of 10% p.a. Therefore, setting
μ
¶p
l(p)
1=1 = 1 +
,
p
the equivalent nominal rates for dierent values of p are:
p p(1=11@p 1)
1
0=10
2
0=09762
4
0=09645
6
0=09607
12
0=09569
The force of interest
We now know two ways of describing growth under compound interest, the
eective rate of interest, and the nominal rate of interest, per period. We can,
12
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
given one description of compound growth, express it in an equivalent but
dierent way. For example, if l represents the compound annual eective rate
of interest, then l(p) is the equivalent nominal rate of interest p.a., payable
p times per year, provided that
¶p
μ
l(p)
.
1+l= 1+
p
That is, these ways of expressing growth are equivalent if, and only if, an
investment grows to the same accumulation in the same time under each of
them.
The force of interest is yet another way of describing growth under compound interest. It is commonly denoted by and represents the instantaneous rate of growth, like a nominal rate of interest which is compounding
continuously. Under an eective annual rate of interest l, interest compounds
annually; under the nominal rate of interest l(p) , interest compounds every
1
wk of a year; and under the force of interest , interest compounds continp
uously.
The force of interest is dened as
= lim l(p) ,
p<"
where l(p) is the nominal rate of interest corresponding to an eective rate
of interest of l per period, i.e.
¶p
μ
l(p)
.
1+l= 1+
p
Starting from
{2 {3
+
+ ===
2!
3!
we can show that = log(1 + l), as follows. We start with the denition of
, then use the fact that
¡
¢
l(p) = p (1 + l)1@p 1 =
h{ = 1 + { +
It is convenient to dene G = log(1 + l) so that hG = 1 + l. Thus,
lim l(p)
¡
¢
= lim p (1 + l)1@p 1
p<"
¡
¢
= lim p hG@p 1
=
p<"
p<"
2.2. COMPOUND INTEREST
=
=
=
=
13
μ
¶
G2
G
G3
+
+ === 1
lim p 1 + +
p<"
p 2p2 6p3
μ
¶
G3
G2
+
lim G +
+ ===
p<"
2p 6p2
G
log(1 + l)=
Thus,
1 + l = h
(2.1)
and y = h3 .
It follows from equation (2.1) that as the value of l> the eective rate of
interest, increases, the equivalent force of interest also increases.
It may not be immediately obvious why we are interested in dening the
force of interest. However, there are cases which may be most conveniently
dealt with by viewing a series of events as happening continuously, for example, the receipt of tolls on a toll bridge. In this case, the rate at which
tolls may be taken may vary from hour to hour, and we might model this as
a continuous function which varies over time.
Example 2.12 Calculate the force of interest p.a. which is equivalent to
each of the following:
(a) a nominal rate of interest of 10% p.a., payable half yearly;
(b) an eective rate of interest of 10% p.a.;
(c) an eective rate of interest of 10% per half year.
Solution 2.12 A simple method is useful for calculating equivalent rates of
interest. In each case, consider how much a unit investment at the start of
a year would accumulate to at the end of that year, and use expression (2.1)
to calculate the equivalent force of interest.
¡
¢2
(a) 1 + 0=1
= h , = log 1=1025 = 0=09758.
2
(b) 1=1 = h , = log 1=1 = 0=09531.
(c) 1=12 = h , = log 1=21 = 0=19062.
Remark 2.4 Note, from part (b) of the above example, that the value of is
less than the equivalent eective rate of interest p.a. This is a general result,
and the equivalent force of interest is less than the rate of interest p.a., or any
nominal rate of interest. This is intuitively reasonable, since the continuous
compounding under a force of interest means that the equivalent eective
rate will be higher than = The same reasoning applies to compounding under
a nominal rate of interest.
14
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
Time line diagrams
So far, the examples we have considered are very simple to imagine. However,
in most cases, problems will be more complicated than this, for example,
involving several payments made at dierent times. An invaluable tool in
considering all but the very simplest problem, is the time line diagram. A
time line diagram is a visual representation of the events and conditions
which are involved in the problem being considered, which serves to clarify
the problem by placing the events in order over time. Suppose an amount
S0 is invested at time 0 for fteen years at compound rate l p.a. Figure 2.1
shows a time line diagram representing this situation.
Time
0
15
Investment S0
S0 (1 + l)15
Accumulation
Figure 2.1: Time line diagram for a single payment
Let us introduce other elements to the transaction. Suppose that there is
also a payment of S2 made at time w = 2 and S5 made at time w = 5= Figure
2.2 shows how Figure 2.1 would be adjusted to account for this.
Time
0
2
5
Investment S0
S2
S5
Accumulation
15
?
Figure 2.2: Time line diagram for multiple payments
Suppose we want to evaluate the accumulated amount at time w = 15=
This can be calculated as the sum of the accumulation of each of the separate
payments, i.e.
S0 (1 + l)15 + S2 (1 + l)13 + S5 (1 + l)10 .
(2.2)
2.2. COMPOUND INTEREST
15
Alternatively, but equivalently, we can accumulate S0 to time w = 2> then
add S2> accumulate the sum to time w = 5> add the third payment, then
accumulate that sum to the end of the period of the investment, giving
¡¡
¢
¢
S0 (1 + l)2 + S2 (1 + l)3 + S5 (1 + l)10 .
Simplifying this results in the same expression as (2.2).
There are generally many ways of solving problems involving compound
interest, and it is often a matter for the student to devise the most streamlined
approach to the problem, and the one which suits the student best.
The above example is a simple one, with a constant rate of interest and
a small number of payments. However, even here, a time line diagram can
help by illustrating the problem. When more complicated circumstances
exist, these diagrams can be invaluable.
The student is recommended to use such diagrams as a matter of course.
Changing rates in consecutive periods
The rules and denitions we have considered hold good when we consider
more complicated situations involving compound interest. Essentially, any
compound interest problem can be evaluated from rst principles using the
fundamental relationships already considered. One simple variation on the
very straightforward case we have already seen, is when the interest rate
changes at one or more time points during the transaction. There are more
or less complicated ways in which this can happen. We begin by considering
the case where the interest rate remains constant for a period of time, then
changes to another constant rate for a dierent consecutive period of time.
For example, the eective rate may be 5% p.a. for ve years, and then
6% p.a. for the next ve years. Suppose an amount S is invested at time
w = 0> what is the amount of the accumulation after six years?
In general, the most useful approach is to consider separately each of the
periods during which the interest rate is constant. Thus, in this example,
we calculate the value of the accumulation of S to the end of the ve year
period during which the rate is 5% p.a., and then accumulate for the next
year at the rate of 6% p.a. This gives the accumulation at the end of the six
year period as
S × 1=055 × 1=06.
Example 2.13 An amount of $100 is invested now and accumulated for a
period of ten years. Calculate the amount of the accumulation at the end of
ten years under each of the following interest rate scenarios.
(a) 5% p.a. eective for the rst three years, then 6% p.a. eective
for the next three years, and 7% p.a. eective for the remainder
of the period.
16
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
(b) 6% p.a. eective for the rst three years, then 5% p.a. eective
for the next three years, then 7% p.a. eective for the remainder
of the period.
Solution 2.13
(a) At the end of the ten years, the amount of the accumulation is
100(1=05)3 (1=06)3 (1=07)4 = 180=73.
(b) At the end of the ten years, the amount of the accumulation is
100(1=06)3 (1=05)3 (1=07)4 = 180=73.
Remark 2.5 Comparing the above two answers, we see that when valuing
the accumulation of a single payment the order of the periods at dierent rates
does not aect the total accumulation at the end of the period of investment.
This is true in general only for a single payment.
Example 2.14 Investments of $100 now, and a further $100 three years
from now, are to be accumulated until the end of the tenth year from the rst
investment. Calculate the amount of the accumulation at the end of ten years
under each of the following interest rate scenarios.
(a) 5% p.a. eective for the rst three years, then 6% p.a. eective
for the next three years, and 7% p.a. eective for the remainder
of the period.
(b) 6% p.a. eective for the rst three years, then 5% p.a. eective
for the next three years, and 7% p.a. eective for the remainder
of the period.
Solution 2.14
(a) Under the rst sequence of interest rates, the total accumulation
ten years after the rst investment is made is
(100(1=05)3 + 100)(1=06)3 (1=07)4 = 336=84.
(b) Under the second sequence of interest rates, the total accumulation
ten years after the rst investment is made is
(100(1=06)3 + 100)(1=05)3 (1=07)4 = 332=47.
2.2. COMPOUND INTEREST
17
Remark 2.6 Note that here, where the payments which are being accumulated occur on more than one date, the nal accumulation is dierent for the
two scenarios. If we expand the expressions above it is easy to see why the
dierence occurs. Compare:
(100(1=05)3 + 100)(1=06)3 (1=07)4
= 100(1=05)3 (1=06)3 (1=07)4 + 100(1=06)3 (1=07)4
with:
(100(1=06)3 + 100)(1=05)3 (1=07)4
= 100(1=06)3 (1=05)3 (1=07)4 + 100(1=05)3 (1=07)4 =
The rst term is identical in each expression, but the second one diers. The
second investment of $100 accumulates for three years at either 6% or 5%
p.a. eective, and then for 4 years at 7% p.a. eective. The accumulation of
this second payment will be greater, therefore, in the case where it is invested
during the three year period at 6% p.a. eective, than in the case where it
attracts 5% p.a. eective for three years.
2.2.2
Valuation of a series of payments
Present value of an annuity
In principle, once you know how to value a single payment at any time under
compound interest, you can evaluate payments under any complex project
or nancial transaction. In practice, there are commonly occurring types of
transaction for which we would prefer to nd convenient and fast ways to
evaluate payments, without resorting to rst principles.
One of the commonly occurring patterns of transaction is the annuity. An
annuity is simply a series of payments made at regular intervals. Common
examples of such series are the premiums payable on a life insurance policy,
or the monthly mortgage payments to repay a housing loan.
Denition 2.9 The symbol dq represents the present value of a series of
unit payments made at the end of each of the next q periods, evaluated at
an eective rate of interest l per period. Such an annuity is referred to as
an annuity in arrear. Sometimes we write dlq to emphasise that the rate of
interest is l.
Thus, the present value of a payment of 1 at the end of each of the next
q periods, evaluated at l per period, is dlq .
So, how do we evaluate dlq ? We can derive an expression to evaluate dlq
from rst principles, by calculating the present value of each of the annuity
payments, and summing for all q payments. Thus
dlq = y + y2 + y3 + = = = + yq
18
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
where the present value of the payment at the end of the rst period is y> the
present value of the payment at the end of the second period is y 2 and so on.
Since the terms y> y 2 > = = = > yq are in geometric progression, they can easily be
summed.
Since
(1 + l)dlq
= (1 + l)(y + y 2 + y3 + = = = + yq )
= 1 + y + y 2 + y3 + = = = + yq31 ,
we have
(1 + l)dlq dlq = 1 y q
giving
1 yq
.
l
This is the expression commonly used to evaluate dlq > where y = (1 + l)31
and l is the valuation rate.
We dene dlq only for positive integer values of q=
dlq =
Example 2.15 You are required to pay $100 at the end of each year, for the
next ten years. What is the present value of this series of payments, if the
interest rate is 8% p.a. eective?
Solution 2.15 The present value of the series of payments is
= 100
100 d0=08
10
1 1=08310
= 671=01.
0=08
Characteristics of the present value of an annuity
It is useful to develop the habit of performing a rough check of your calculations, and to aid this it is useful to notice some characteristics of the basic
functions. For example, dlq > the present value of an annuity of 1 per period
for q periods, is always worth rather less than q= To see this, note that
dlq
=
=
y + y2 + y3 + = = = + yq
y + y + y + = = = + y for all l
qy
q
since y 1. Note that this is a strict inequality except in the trivial case
l = 0= Also, the higher the interest rate l, the smaller is y> and the faster
each term in y w becomes small as w increases. This means that the higher the
interest rate is, the poorer is qy as an approximation to dlq .
2.2. COMPOUND INTEREST
19
This very simple approximation gives a quick and easy means of testing
whether your answer is reasonable. Note that this cannot conrm that your
answer is correct, but it may indicate if it is incorrect.
For any non-zero value of l> as the term, q, of the annuity increases, dlq
tends towards a limiting value. Consider the following:
dlq =
1 yq
.
l
Since 0 ? y ? 1 for l A 0> it follows that
y q $ 0 as q $ 4.
Hence
1
as q $ 4.
l
Thus, the maximum value that dlq can have, for a given value of l, is 1@l, no
matter for how long the annuity is payable.
So, for example, if the interest rate is 10% p.a. eective, the present value
of $100 p.a. in arrear cannot exceed 100@0=1 regardless of the term of the
annuity.
dlq $
Denition 2.10 A perpetuity is an annuity payable for an unlimited term.
Denition 2.11 The present value at an eective rate of l per period of unit
payments at the end of each period is denoted by dl" , or simply d" .
From the above, it is clear that the present value of a perpetuity of 1 p.a.
in arrear at eective rate l p.a. is 1@l.
Example 2.16 What is the maximum amount you would be prepared to pay
for a series of payments of $1,000 annually in arrear, if the valuation rate of
interest is 8% p.a. eective?
Solution 2.16 The maximum value of the series of payments would be the
value of a perpetuity of $1,000 annually in arrear. The value of this at the
rate 8% p.a. is given by
1,000 d" =
1,000
= 12,500.
0=08
So the maximum value of this series of payments, no matter how many payments are to be made, is $12,500.
20
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
Other annuities
The symbol dlq denotes the present value of a unit payment made at the
end of each of the next q periods, evaluated at eective rate l per period.
There are other standard actuarial terms used to denote the present value of
annuities payable in dierent ways.
Denition 2.12 An annuity payable at the start of each period is known as
an annuity-due or an annuity payable in advance.
Denition 2.13 The symbol d̈q (called ‘d due q’) represents the present
value of a series of unit payments made at the beginning of each of the next
q periods, evaluated at an eective rate of interest l per period. Sometimes
we write d̈lq to emphasise that the rate of interest is l.
Thus, an annuity-due is a stream of payments similar to an annuity in
arrear, but where each payment is made one period earlier.
We can derive an expression to evaluate d̈lq quite simply by analogy to
that for dlq above, and relate the present values of the two types of annuity
as follows:
d̈lq
= 1 + y + y2 + y3 + = = = + yq31
= (1 + l)(y + y 2 + y3 + = = = + yq )
= (1 + l) dlq =
Also, we have
d̈lq = 1 + y + y2 + y3 + = = = + y q31
= 1 + dlq31 .
Denition 2.14 A deferred annuity is a series of payments which do not
begin immediately, or in the rst period, but only after a period of deferment
during which no payments are made.
For example, p |dlq denotes the present value at an eective rate of l per
period of an annuity certain payable for q periods in arrear, deferred for p
periods. That is, the rst unit payment is made at time p + 1, and not at
time 1. Again, it is a simple matter to derive an expression to evaluate this,
and relate it to dlq :
l
p |dq
= yp+1 + y p+2 + y p+3 + = = = + yp+q
= yp (y + y 2 + y3 + = = = + yq )
= yp dlq
1 yq
= yp
l
2.2. COMPOUND INTEREST
21
Also,
l
p |dq =
=
y p y p+q
l
1 y p+q (1 yp )
l
= dlp+q dlp .
Denition 2.15 An annuity certain payable p-thly in arrear is a series of
payments of p1 made at the end of each p1 th of a period.
(p)
Denition 2.16 The symbol dq represents the present value of an annuity certain payable p-thly in arrear for the next q periods, evaluated at an
eective rate of interest l per period.
There are several points to note here. The total payment per period adds
up to 1, no matter what the value of p is, and each payment is made at the
end of p1 th of a period, so that there are a total of p × q payments made,
each of amount p1 =
(p)
We can evaluate dq in dierent ways. For example, consider the annuity
in terms of a new unit period of p1 , evaluated at an eective rate per p1 th of
an old period as follows. Let m = (1 + l)1@p 1 be the eective rate per new
period (i.e. per p1 th of an old period), and let ym = 1@(1 + m). Then
(p)
dq
1
(ym + ym2 + ym3 + = = = + ympq ) @ m
p
1 m
d .
=
p pq
=
(We use the symbol @ to mean ‘at’.)
Alternatively, we can express the present value of this series of payments
in terms of dq and an eective annual rate. Consider one period at a time —
in each period the value at the start of that period of the payments in that
(p)
period is d1 . Hence, the value of the whole series of payments is given by
(p)
(p)
dq
= d1
(p)
= (1 + l)
(p)
d̈q = (1 + l) d1
dq .
Now
(1 + l) d1
= (1 + l)
¢
1 ¡ 1@p
y
+ y 2@p + = = = + y
p
1 1@p 1 y
y
.
p
1 y 1@p
22
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
Since (1 + l)(1 y) = l and
1
1
1 y1@p
1
= (p) ,
=
1@p
1@p
p1y
p (1 + l)
1 l
we have
(p)
(1 + l)d1
and so
(p)
=
l
l(p)
l
dq .
l(p)
This expression is commonly used to relate the present value of an annuity
payable p times p.a. in arrear to that of an annuity payable annually in
arrear. Note that the ratio l@l(p) is greater than 1 for equivalent values of l
and l(p) = Intuitively, it is not di!cult to understand why the present value
of the p-thly annuity should exceed the present value of the annuity paid in
arrear.
Consider the p-thly annuity by comparison with the annuity in arrear,
where, for convenience, we will take the unit of time to be one year. There
is the same total amount paid in each year, that is 1 per year, but the pthly annuity makes the rst payment of p1 only p1 th of a year from now, the
valuation date, whereas the annual annuity makes its rst payment of 1 in
one year’s time. Half of the payments are made in the rst half of the year,
and so on. Thus, on average, the amounts paid under the p-thly annuity
occur half a year earlier than those made under the annual annuity.
dq
=
The accumulated value of an annuity
We have considered how to evaluate the present value of a series of regular
payments, and it is also useful to consider how to calculate the accumulated
value of a series of regular payments. In practice we often encounter situations which involve the accumulation of a series of payments over a period
of time to give a target amount at the end of the period. Examples include
regular payments made into a bank account attracting a rate of compound
interest which are made in order to pay o a loan, or saving up in order to
pay a bill, or make a purchase. We might have a target sum which we wish to
accumulate, and need to know what regular payment is required in order to
achieve it at a given rate of interest. Alternatively, we may wish to estimate
how much a series of known payments will accumulate to after a given period
of time.
Denition 2.17 The symbol vq represents the accumulation or accumulated
value at time q of a series of unit payments made at the end of each of the next
2.2. COMPOUND INTEREST
23
q periods, evaluated at an eective rate of interest of l per period. Sometimes
we write vlq to emphasise that the rate of interest is l.
Consider the series of unit payments at the end of each of the q periods.
Then, at evaluation rate of interest l, dq is the value of this series of payments
at time w = 0, and vq is the value of this same series of payments at time
w = q= Both expressions give a value for the same series of unit payments,
but each gives the value at a dierent time.
We can derive expressions to evaluate vq in much the same way as we
did for dq =
Consider the sum of the accumulated value of each single unit payment
at time w = q, that is, at the time of the last payment. We can sum this as
the terms in the sum are in a geometric progression, and we can also relate
it to the expressions we have derived for the present value of the annuity:
vq = 1 + (1 + l) + (1 + l)2 + (1 + l)3 + = = = + (1 + l)q31
=
(1 + l)q 1
.
l
Alternatively,
vq
= (1 + l)q (y + y 2 + y 3 + = = = + y q )
= (1 + l)q dq =
This last relationship makes intuitive sense, since vq gives the accumulated
value of the annuity whose present value is dq = That is, at time 0, the value
of the series of payments is the same as dq > and so the accumulated value at
time q is the same as the accumulated value of a single payment of dq made
at time 0. At this rate of interest, these are equivalent in value.
There are expressions and symbols which give the accumulated value of
a series of payments such as we have already obtained for the present value
of the same series of payments.
Denition 2.18 The symbol v̈q represents the accumulation, or accumulated value, at time q of a series of unit payments made at the beginning of
each of the next q periods, evaluated at an eective rate of interest of l per
period. Sometimes we write v̈lq to emphasise that the rate of interest is l.
That is, v̈lq is the accumulated value of a series of payments, evaluated
at an eective rate of interest of l per period, which has a present value of
24
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
d̈lq = Thus:
v̈lq
= (1 + l) + (1 + l)2 + (1 + l)3 + = = = + (1 + l)q
= (1 + l)
(1 + l)q 1
l
or
v̈lq
= (1 + l)q (1 + y + y 2 + y3 + = = = + yq31 )
= (1 + l)q d̈lq =
Evaluation of annuities with changes in interest rate
The denitions of dlq and of vlq are based on the assumption that the rate
of interest l is constant throughout the q periods. We can use these expressions to evaluate series of payments over a period during which interest rates
change. In general, the best way to deal with such situations is to treat
each period of constant interest rate separately, and also to use an eective
interest rate which matches the frequency of the annuity payments.
Example 2.17 A sum of $100 is to be paid at the end of each year for
twenty years. The eective interest rate is 10% p.a. for the rst ve years,
5% per half year for the next ten years, and 3% per quarter year for the
remaining ve years. What is the accumulated value of the payments at the
end of twenty years?
Solution 2.17 The payments are made annually in arrear throughout the
twenty years. We can express each interest rate in terms of an eective
annual rate:
10% p.a. eective for ve years, followed by
1=052 1 10=25% p.a. eective for ten years, followed by
1=034 1 12=551% p.a. eective for ve years.
Now we can calculate the accumulated value of the annuity paid in each of
these constant interest rate periods, and then accumulate each to the end of
the twenty years. Consider a series of unit payments over the twenty years.
The accumulated value at the end of ve years of the rst ve payments is
v0=1
=
5
1=15 1
= 6=1051.
0=1
2.2. COMPOUND INTEREST
25
Hence, the value at time 20 is
1=102510 × 1=125515 × v0=1
= 29=2566.
5
Next, the value at the end of fteen years of the payments in years six to
fteen is
1=102510 1
v0=1025
= 16=1297.
=
10
0=1025
Hence, their value at time 20 is
1=125515 × v0=1025
= 29=1321=
10
Lastly,
1=125515 1
0=12551
v5
= 6=4227,
=
0=12551
is the value at time 20 of the payments in the last ve years. Thus, the
accumulated value of the whole twenty years’ payments is $100 multiplied by
the sum of these three values, namely
100 (29=2566 + 29=1321 + 6=4227) = 6,481=14.
Note that this is not a unique method of solution, and that there are other
techniques which could be used to calculate the accumulated value. However,
this method is to be recommended because it is simple, robust, and can be
applied in most circumstances.
2.2.3
The equation of value
The equation of value is the equation which we construct to describe a nancial transaction. For example, it may equate, at a particular date, the value
of the cost of the purchase of an investment with the value of the proceeds
of the investment. It may, as in the previous sections, describe the value at
a particular time of a number of payments made at dierent times. All the
equations in previous sections are equations of value, since they describe a
nancial transaction, and state an equality that is true at a given rate of
interest, at a particular time.
Equations of value have a number of essential constituents, and the equality makes sense and is true only if these constituents are provided. In order
to be well dened, the equation must be accompanied by a statement of
the conditions under which the equality holds, and all of the items in the
equation must be consistent with these conditions, and with one another.
The equation of value holds at a given rate of interest, and at a particular
time. Each term in the equation must therefore be an expression which
26
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
represents the value of a nancial ‘event’ at the particular date specied.
Symbols and standard actuarial expressions which are used must be rendered
in terms of this valuation rate of interest, or else they must be described as
holding for a dierent, but specied rate of interest. In order to be sensible
the rate of interest must be described as applying over a particular unit of
time.
Thus, to be complete and in order to make sense, you must ensure that
the equation of value is described as holding at a given date, and at a given
rate of interest per dened period, and each term in the equation must be
consistent with this.
Consider the previous example. A sum of $100 is paid at the end of each
year for twenty years. The eective interest rate is 10% p.a. for the rst ve
years, 5% per half year for the next ten years, and 3% per quarter year for
the remaining ve years. What is the accumulated value of the payments at
the end of twenty years?
We constructed an equation of value which assumed a unit of time of one
year, and which was expressed in terms of eective interest rates per year:
³
´
5
0=1025
0=12551
100 1=102510 × 1=125515 × v0=1
+
1=12551
×
v
+
v
5
10
5
= accumulation at time twenty years.
Interest rates were expressed in terms of dierent units of time. The rates
were 10% p.a. eective for ve years, then 5% eective per half year for
ten years, and then 3% eective per quarter year for ve years. We converted each of these rates to an eective annual rate in order to arrive at the
above equation of value. Since the annuity payments are made annually, it
is simplest to express the vq functions in terms of a unit of time of one year,
matching the frequency of the payments, and also, therefore, in terms of an
eective rate of interest per year. However, we could have expressed the other
terms in the equation dierently without losing consistency or simplicity.
The value of the payments at the end of twenty years is
³
´
5
0=1025
0=12551
100 1=102510 × 1=125515 × v0=1
>
+
1=12551
×
v
+
v
5
10
5
but we could have written
³
´
20
0=1025
0=12551
100 1=0520 × 1=0320 × v0=1
.
+
1=03
×
v
+
v
5
10
5
These two expressions are equivalent, and both expressions give the required
information in order to be correctly interpreted. Provided the rate of interest
associated with a term in the expression is consistent with the unit of time
which the term assumes, it is quite in order for the dierent units of time
2.3. COMMON COMPOUND INTEREST TRANSACTIONS
27
to be used in the same expression. However, each item in the expression
must represent a value at the same specied date as all the other terms in
the equation, or expression. Thus, while it is acceptable to manipulate the
unit of time to suit the expression of the rate of interest, it is necessary
always to evaluate each term at a single consistent date. Provided the terms
are consistent with one another, it does not matter, however, which single
evaluation date is chosen.
Equations of value are not unique, and there are always a number of ways
to describe any transaction or situation. Which equation is to be preferred is
partly a matter of personal taste, and partly a matter of good design. Some
equations are simpler, clearer and easier to use than others. Provided that
the conditions governing it are stated, and the appropriate information given
regarding the evaluation rate(s), the unit of time, and the evaluation date of
the whole equation, then you may construct any equation to suit your own
taste, and to ease the solution of the problem.
2.3
Common Compound Interest Transactions
There are many situations in everyday life where individuals are involved in
transactions involving compound interest. The most common perhaps is the
use of a bank account which attracts income in the form of interest, and the
use of loan or credit facilities, which also attract interest costs. In the following sections we look at examples of two common xed interest transactions,
the xed interest security and the xed rate home loan or mortgage contract.
2.3.1
Fixed interest bonds
Terminology
A xed interest bond (also referred to as a debenture, stock or security)
is a monetary vehicle used to borrow money from the purchasers of the
bond. Bonds are commonly sold by commercial bodies and by local, state
or national government in order to raise capital. Bonds are often largely
bought by nancial institutions, but also by private and other investors. The
purchasers of the bond provide a capital sum to the borrower in exchange for
a series of payments of coupon, also known as dividend or interest payments,
plus a payment of capital at the end of a xed period of time. The coupon
and capital repayments are made by the borrowers, i.e. those who are seeking
to raise a capital sum by issuing the bond. The capital payment at the end
of the term of the bond represents a return of the original capital raised.
There are many variations available amongst dierent xed interest bonds,
28
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
but, basically, they contain just a few key elements. The total amount initially raised, or issued, is described in terms of a nominal amount. Often
the nominal amount is described in terms of $100 ‘bundles’. The amount
of coupon and capital to be paid to the buyers, or holders, of the bond is
described in terms of these bundles of nominal. Thus, the nominal amount
is merely a description of the rights to income and capital repayment which
the holder of the bond owns, and does not represent the actual cash value of
the bond.
Bonds are described in terms of the amount and timing of coupon and
capital payments per nominal amount of a bond. For example, a xed interest
security may be described as 10% 12th July 2025, redeemable at 100. This
implies that for every $100 nominal that you hold, you will receive $10 on
12th July each year (including 12th July 2025) plus $100 on 12th July 2025.
Here we see the essential elements in describing the bond:
• The coupon or interest rate: the rate per $100 nominal and frequency
at which the coupon is paid.
• The dividend date: the date of the coupon payments each year.
• The redemption or maturity date: the date on which the bond is redeemed, or matures.
• The redemption price: the amount per $100 nominal which is paid on
redemption, or maturity.
These items adequately describe the stream of payments which a buyer
of stock will receive for a given nominal amount. While the price or value
of an investor’s holding of bonds may change with changes in stock market
conditions, the nominal amount which they hold remains constant, unless
the bond is sold or redeemed. The amount of coupon received on the coupon
dates and the amount paid on the maturity date depend only upon the
nominal amount of stock which is held, and not upon the price of the stock
in the market place. The price may vary from minute to minute in fact, but
the nominal size of your investment, and the payments to which that nominal
amount entitles you, remain the same. When you sell your stock, you also
sell your right to receive further coupon payments.
There are special cases of this general formula describing a xed interest
security and many variations in the detail of bonds. In particular, there is a
zero coupon bond, which, as the name suggests, consists only of a redemption
of capital and has no interest payments during the term of the bond. Also,
there exist bonds in the market which have no redemption date. These are
perpetuities, since the coupon payments are made indenitely, and the holder
has no explicit right to a return of capital, or redemption. These are also
known as irredeemable bonds.
2.3. COMMON COMPOUND INTEREST TRANSACTIONS
29
Calculating the price
The coupon, interest or dividend payment describes the amount of income
which the holder receives per $100 nominal of bond which they hold, and this
should not be confused with the valuation rate of interest. The valuation
rate is the rate at which we evaluate the stream of income, and it varies
with economic conditions and assumptions, while the dividend rate actually
describes the amount which is received, but does not value it.
Example 2.18 An investor buys $10,000 nominal of an 8% p.a. xed interest bond which is redeemable at $100 in ten years’ time. What price does
she pay per $100 nominal in order to realise a yield of 10% p.a. eective?
Solution 2.18 We can calculate the price per $100 nominal by constructing
an equation of value, equating the cost of the bond with the present value of
the income and capital payments which the buyer receives, per $100 nominal.
Let S be the price per $100 nominal. Then
S = 8 d10 + 100 y 10 .
If the buyer realises a yield of 10% p.a., then solving this equation for S at
rate 10% p.a. will give us the price she paid per $100 nominal:
S = 8 d10 + 100 y10
= 8
@ 10%
¡
¢
1 1=1310
+ 100 1=1310
0=1
= 87=71=
Remark 2.7 $87.71 per $100 nominal is the value calculated, or price paid,
ten years from the maturity date. At a dierent date, the price or value would
be dierent.
Remark 2.8 10% p.a. represents the yield to the investor, or the purchaser,
and also represents the cost of borrowing to the issuer, or seller, of the stock.
Example 2.19 An investor buys $10,000 nominal of an 8% p.a. xed interest bond which is redeemable at $100 in ten years’ time. What price does
she pay per $100 nominal in order to realise a yield of 8% p.a. eective?
30
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
Solution 2.19 The equation of value is the same as in the previous example,
except that the valuation rate of interest is 8% p.a. eective. If S is the price
per $100 nominal, then
S = 8 d10 + 100 y10
= 8
@ 8%
¡
¢
1 1=08310
+ 100 1=08310
0=08
= 100=
That is, for a stock which is redeemable at $100 per $100 nominal, if the
coupon rate is the same as the valuation rate, the price is $100 per $100
nominal.
Remark 2.9 Note that if the yield is greater than the coupon rate, then the
purchase price will be less than the redemption price.
Remark 2.10 Similarly, if the yield is less than the coupon rate, then the
purchase price will be greater than the redemption price.
It is straightforward to give an intuitive explanation for this. Purchasers
of the stock receive only two kinds of return for the purchase price which
they pay. They receive a return in the form of interest or coupon payments,
and they receive a return in the form of a capital payment. Their prot, or
yield on the investment, can come only from either interest or capital prot.
Thus, if they purchase the bond for a price equal to the redemption price,
their yield will be exactly equal to the coupon rate. If the purchase price is
less than the redemption price, then a capital prot is made on redemption,
and so the yield on the whole transaction will be greater than the coupon
rate. If the purchase price is greater than the redemption price, then a loss is
made on the capital redemption, and so the total rate of return on the whole
transaction will be less than the coupon rate.
This relationship between rate of return, coupon rate, purchase price
and redemption price, is a very useful one to understand, and gives a quick
method of estimation for the price or yield of a transaction.
Common variations
There are two common variations in the structure of a xed interest security. The interest, or coupon, payments can be made other than yearly, and
the redemption of the stock may be other than at par, where par means
redemption at $100 per $100 nominal.
2.3. COMMON COMPOUND INTEREST TRANSACTIONS
31
Example 2.20 What is the price on 1st June 2010 of a 10% 1st June 2012
Treasury stock, to yield 8% p.a. eective? Treasury stocks pay dividends
twice yearly, and are redeemable at par.
Solution 2.20 If the coupon is payable other than yearly, the simplest method
to use in calculating the purchase price is to use the period between coupon
payments as the unit of time when constructing the equation of value. That
is, let the frequency of the coupon payments determine the unit of time. The
coupon is payable half yearly, so we may construct an equation of value using
a half year as the unit of time, and evaluated at an eective rate per half
year. If S is the price per $100 nominal on 1 June 2010 then
S = 5 d4 + 100 y4
@l
where l is the eective rate of interest per half year which is equivalent to the
required valuation rate of 8% p.a. eective. That is, l = 1=081@2 1 = 0=03923
per half year eective. Thus
S = 5 d4 + 100 y 4
= 103=92=
@ 3=923%
Note that the yield is lower than the coupon rate, and so the price paid is
higher than the redemption price.
Example 2.21 A stock pays dividends at 10% twice yearly, and is redeemable
at $110 per $100 nominal on 1st June 2012. What is the price per $100 nominal on 1st June 2010 to yield 8% p.a. eective?
Solution 2.21 This stock is the same as the Treasury stock in the previous example except that the redemption price is $110 instead of $100. We
calculate S , the price per $100 nominal, using the following equation.
S = 5 d4 + 110 y 4
= 112=49.
@ 3=923%
Here, the yield is less than the coupon rate and so the price, as in the previous
example, must be higher than the redemption price.
Tax on income
If the purchaser of the bond is liable to pay tax on income then they will not
receive the full coupon payments. In order to calculate the value of the bond
to an income tax payer, an adjustment must be made to allow for the fact
that coupons are received net of tax, rather than gross as above.
32
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
Assume that the rate of tax on income is w per unit of income, and that
it is due to be paid at the same time as the income is received. In general, if
S is the net purchase price, L the coupon or interest payment per unit time,
and U is the redemption price payable at the end of q units of time, we can
write the following equation, evaluated at the yield l per unit time:
S = L(1 w) dq + U y q .
We say that l is the net yield per unit time on the investment, after tax at
rate w on income. In practice, tax may be paid some time after the receipt
of the coupon payment. Suppose that tax is paid p units of time after each
payment of coupon is received. Then the net price is
S = L(1 wyp ) dq + U y q .
(2.3)
Example 2.22 The government is about to issue a bond with term 20 years
which pays dividends at 12% quarterly, and is redeemable at par. What price
should an investor, subject to tax on dividends at 30%, pay per $100 nominal
of the bond at the issue date to obtain an eective yield of 2.5% per quarter
if
(a) the government pays dividends net of tax, and
(b) the investor pays tax exactly 1 month after each dividend payment.
Solution 2.22 In each case, it is convenient to take a quarter of the year
as the time unit, so that the term of the bond is 80 time units.
(a) The net (of tax) dividend payment per $100 nominal is 0=7 × 3 =
2=1. Hence the price is
2=1 d80 + 100 y80 = 86=22 @ 2=5%.
(b) From (2.3), with tax being deferred by 1@3rd of a unit of time, the
price per $100 nominal is
¡
¢
3 1 0=3y1@3 d80 + 100 y 80 = 86=47 @ 2=5%.
Remark 2.11 Note that the price is higher when the payment of tax is delayed, in part (b) above. This illustrates the eect of timing on the value of
money. The sooner that a gain, or income, is realised, the more valuable it is
to us, and the higher the price we are prepared to pay for it. Similarly, if we
can delay losses this increases the value of the transaction. The longer we can
delay the payment of tax, the higher the value of the bond to us. Check this
eect further by repeating the calculation, but assuming that the tax payment
is made one year after the dividend payment is received.
2.3. COMMON COMPOUND INTEREST TRANSACTIONS
33
Calculating the yield on a xed interest security
Commonly in actuarial work of any kind, one is likely to be either calculating
the value of a nancial transaction of some kind at a given date and interest
rate, or else calculating the yield on a transaction, given the value at some
date. So far we have looked at examples which consider the rst of these
possibilities. In either case, the initial step is the same. An equation of value
at some specied date must be constructed, e.g.
S = L(1 wy p ) dq + U yq @ l=
If we know all the other elements in the above equation, we can solve for l> and
this will be the net eective yield per unit time obtained on the investment
if the bond is bought at price S=
In general, we can solve for l by linear interpolation. In order to choose
useful values of l between which to interpolate, it is wise to take the preliminary step of calculating an approximate value for l. The next example
illustrates this. However, we rst give a reminder of the workings of linear interpolation. Suppose that we want to solve an equation of the form
j(l) = 0. The idea is to nd two values l1 and l2 which are close together
and are such that j(l1 ) and j(l2 ) have dierent signs. We then assume that
j(l) is of the form d + e l for l1 l l2 . Under this assumption, there is a
value m such that j(m) = 0 (since j(l1 ) and j(l2 ) have dierent signs), and
setting d + e m = 0 gives m = d@e= The values of d and e are found from the
equations
j(l1 ) = d + e l1
and
j(l2 ) = d + e l2 =
Example 2.23 An 8 year, 10% p.a., xed interest bond is bought at the date
of issue, for a price of $95 per $100 nominal. The bond is redeemable at par.
What is the gross annual eective yield to maturity on the investment?
Solution 2.23 The equation of value at the date of purchase, per $100 nominal, describing the transaction is as follows:
95 = 10 d8 + 100 y8
@ l=
(2.4)
We can solve this by interpolation, but rst need to nd an approximate
value for l= We can do this by assigning a notional income in each year, and
dividing this by the price of the bond. The income comes from two sources,
there is an annual coupon of $10 per $100 nominal, and also a capital gain
of $5 per $100 nominal (that is, 100 95) realised at the end of the term of
8 years. As a rough estimate, we can say this is an average of 58 per year
per $100 nominal. This overstates the value of the capital gain somewhat,
because all of it is in fact realised at the end of the 8 years, whereas our rough
34
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
estimate will allow for one eighth of the gain to be realised in each year. We
nd an approximation to the yield by rst adding this averaged gain to the
coupon, then dividing by the purchase price. Hence, our approximation is
l
10 + 5@8
= 0=11184.
95
This overstates the yield slightly since we have assumed that the capital gain
occurs earlier than it in fact does, and so 11% would be a good starting point
for interpolation. Let’s re-arrange equation (2.4). We want l such that
j(l) = 95 (10 d8 + 100 y8 ) = 0=
At rate l = 11% we get
j(0=11) = 0=14612=
For the yield l> j(l) = 0, and so our rst estimate of 11% is a little too high.
Evaluate j(l) at l = 10=5%:
j(0=105) = 2=3804=
Thus, we can see that the value for l which gives j(l) = 0 must lie between
10.5% and 11%. Setting
0=14612 = d + 0=11e
and
2=3804 = d + 0=105e
we obtain e = 505=306 and d = 55=438, giving our solution for l as d@e =
10=971% s=d=
2.3.2
Housing loans
Terminology
A typical housing loan contract enables an individual to borrow a capital
sum now, in exchange for a series of level repayments over a specied period
of time. There are many variations on the basic housing loan contract now
available in the market, in response to consumer demand, and changing economic and commercial conditions. Variations include the ability to combine
variable rates of interest and xed rates of interest, and to make irregular,
temporary, repayments of capital. Discussion here is conned to consideration of a xed interest loan, repaid by level regular instalments of capital
and interest over a xed term.
Housing loans are generally made by a nancial institution to an individual. The amount of capital borrowed by the individual, or lent by the
2.3. COMMON COMPOUND INTEREST TRANSACTIONS
35
institution, is called the principal. The repayments are made at regular intervals, often fortnightly (that is, every two weeks) or monthly, but in principle, the time period may be anything, and payments are made in arrear.
Each repayment is used rst to pay the interest which is owing on the capital
outstanding, and then any remainder is used to repay the capital.
When a loan contract is rst agreed, a loan schedule may be drawn up.
This schedule traditionally shows the amount of the capital outstanding at
the beginning of each of the payment periods, the amount of interest repaid,
the amount of capital repaid, and the amount of the loan outstanding at
the end of each payment period. It shows the history of the loan and the
repayments.
Calculating the level repayment
Consider a loan of O to be repaid by q level annual instalments [ of capital
and interest, at rate l p.a. eective. The amount of the loan O is equal to the
present value of the series of repayments. This gives the following equation
of value:
O = [dq @ l=
This equation gives the relationship between the basic constituents of the
loan contract.
Example 2.24 What is the level annual repayment required to repay an
amount of $100,000 over twenty years, at an eective annual rate of interest
of 7%?
Solution 2.24 The present value of the repayments equals the amount of
the loan. Let [ be the annual repayment. Then
100,000 = [d20
@ 7%
@ 7%
, [ = 100,000@d20
, [ = 100,000@10=594
, [ = 9,439=29=
That is, the annual payment in arrear is $9,439.29 in each of the twenty years
of the contract. At the end of the term of the loan, the capital outstanding
has been reduced to zero.
Example 2.25 What is the level annual repayment required to repay an
amount of $100,000 over twenty years, at an eective annual rate of interest
of 3%?
36
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
Solution 2.25 If [ is the level annual payment, then
100,000 = [d20
@ 3%
@ 3%
, [ = 100,000@d20
, [ = 100,000@14=8775
, [ = 6,721=57.
Example 2.26 What is the level annual repayment required to repay an
amount of $100,000 over ten years, at an eective annual rate of interest
of 7%?
Solution 2.26 If [ is the level annual payment, then
100,000 = [d10
@ 7%
@ 7%
, [ = 100,000@d10
, [ = 100,000@7=0236
, [ = 14,237=75.
Example 2.27 What is the level annual repayment required to repay an
amount of $100,000 over ten years, at an eective annual rate of interest
of 3%?
Solution 2.27 If [ is the level annual payment, then
100,000 = [d10
@ 3%
@ 3%
, [ = 100,000@d10
, [ = 100,000@8=5302
, [ = 11,723=05.
Example 2.28 What is the level monthly repayment required to repay an
amount of $100,000 over twenty years, at an eective monthly rate of interest
of 1%?
Solution 2.28 If [ is the level monthly payment, then
100,000 = [d240
@ 1%
@ 1%
, [ = 100,000@d240
, [ = 100,000@90=8194
, [ = 1,101=09.
Example 2.29 What is the level monthly repayment required to repay an
amount of $100,000 over twenty years, at an eective annual rate of interest
of 7%?
2.3. COMMON COMPOUND INTEREST TRANSACTIONS
37
Solution 2.29 The repayments are made monthly, and so it is convenient
to express the interest rate as an eective monthly rate, l> say, such that l is
dened by the following relationship:
1=07 = (1 + l)12
, l = 0=5654% per month.
If [ is the level monthly payment, then
@ 0=5654%
100,000 = [d240
@ 0=5654%
, [ = 100> 000@d240
, [ = 100> 000@131=1570
, [ = 762=44.
Remark 2.12 The preceding exercises illustrate some basic characteristics
of the relationship between the amount of the regular repayment, and other
elements of the loan contract.
1. The amount of the level repayment rises if the interest rate rises.
2. The amount of the level repayment rises if the term is reduced.
3. The amount of the level repayment per period decreases if the frequency
of repayments per period increases.
The loan repayment schedule
Consider a loan of O to be repaid over q periods by a level repayment of [ per
period, at eective rate l per period. We can construct a schedule showing
the progress of the repayment, and the amount of interest and capital paid
in each period. The columns in the schedule correspond to:
1. the time period, w;
2. the loan outstanding at the beginning of this time period, Ow31 ;
3. the amount of the repayment made at the end of this period, [w ;
4. the amount of interest due on the loan outstanding at the start of this
period, Lw ;
5. the amount of capital repaid at the end of this time period, Fw ;
38
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
6. the loan outstanding at the end of this time period, Ow .
This gives six column headings for the schedule, as follows:
(1) (2)
w Ow31
(3) (4) (5) (6)
[w Lw Fw Ow
Thus, column (3) is equal to the sum of columns (4) and (5). Also,
column (6) gives the value which is entered in column (2) in the next row.
That is, the loan outstanding at the end of a period, is the same as the loan
outstanding at the beginning of the next period.
Note that w is measured in periods which correspond to the frequency
of the repayments. If repayments are made annually, then w is measured in
years; if the repayments are made monthly, then w is measured in months.
Consider the rst row of the schedule. The loan outstanding at the beginning of the rst period is O> and so the interest due on this is lO since l is
the eective rate of interest per period. The total payment made at the end
of the period is [. Thus, the amount available for repayment of capital is
[ lO= This gives us the information for the rst row of the schedule. The
amount of the loan outstanding at the end of the rst period then provides
the amount of the loan outstanding at the beginning of the next period.
(1)
(2)
w
Ow31
1
O
2 O(1 + l) [
(3) (4)
(5)
(6)
[w Lw
Fw
Ow
[ lO [ lO O(1 + l) [
[ ===
===
===
The above table shows how the schedule is constructed, with the amount in
column (6) providing the value in the next row of column (2), and so on.
The complete schedule continues for the term of the loan, q periods, in this
case, and in the nal row the loan outstanding in column (6) is reduced to
nil.
Clearly it is not always convenient to construct an entire schedule in order
to examine the condition of the loan at some point in time. We can derive
analytical expressions to describe various constituents of the schedule.
The essential elements describing the loan are the principal, O> the term
q> the rate of interest per period l> and the level regular repayment [= These
are related to one another by the equation of value
O = [dq
@ l=
We can use this relationship to express the items in the loan schedule in
dierent ways.
2.3. COMMON COMPOUND INTEREST TRANSACTIONS
39
Substituting the above expression for O we get
F1 =
=
=
=
[ lO
[ l[dq
[ [(1 y q )
[y q
so that
O1 = O [yq
= [dq [yq
= [dq31 .
Similarly,
F2 = [ lO1
= [ l[dq31
= [ [(1 y q31 )
= [y q31
so that
O2 = O1 F2
= [dq31 [yq31
= [dq32 .
Continuing in this fashion we nd that
Ou = [dq3u
for u = 1> 2> ===> q 1. This result assumes that none of the conditions of the
original loan contract has been changed.
This result can be interpreted as showing that at any time during the
course of the loan, the value of the contract to the borrower (that is, the
present value of the annuity of future repayments) is equal to the value
of the contract to the lender (that is, the present value of the money still
to be repaid). This makes sense intuitively, since if the present value of
the future repayments were greater than the loan outstanding, the borrower
could borrow a larger sum, and similarly, if the present value of the future
payments were less than the amount still owed, the lender would be able to
secure a higher rate of repayment elsewhere.
40
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
The amount of interest paid in a period can be expressed in terms of
[ and y only, using the original equation of value and substituting in the
expression above. At time w = 1 the interest due is:
L1 = lO
= l[dq
= [(1 y q ).
Generalising to time w = u> we obtain the following relationship.
Lu = lOu31
= l[dq3u31
1 y q3u+1
)
l
= [(1 yq3u+1 ).
= l[(
The interest due at time w = u> is the interest on the loan outstanding at time
w = u 1> which is the present value of the remaining q (u 1) payments.
We can use this expression to derive an expression for the amount of
capital repaid at time w = u> since this is simply the remainder of the repayment, [= So, the capital repaid at the end of the u-th period is given by the
following expression.
Fu = [ [(1 y q3u+1 )
= [y q3u+1 .
Example 2.30 A housing loan of $120,000 is repayable over twenty years
by level annual instalments, calculated at an eective annual rate of interest
of 10%.
(a) What is the annual repayment?
(b) What is the capital outstanding after ve years?
(c) What is the amount of interest paid in the 6th year?
(d) How much capital is repaid in the 6th year?
Solution 2.30
(a) Let [ be the annual repayment. Then
120,000 = [d20
@ 10%
, 120,000 = 8=5136[
, [ = 120,000@8=5136 = 14> 095=15.
2.3. COMMON COMPOUND INTEREST TRANSACTIONS
41
(b) The capital outstanding after ve years is
[d15
= 14,095=09 × 7=6061 @ 10%
= 107,208=83=
(c) This is the interest paid on the amount outstanding at the beginning of the 6th year, which is the same as the amount outstanding
at the end of the 5th year, above. The amount outstanding at that
time is [d15 and hence the amount of interest is
0=1 [ d15 = 10,720=88.
(d) This can easily be calculated using the previous result, since it is
the balance of the total annual payment. Thus, the capital payment
is
[ 10,720=88 = 3,374=27.
Alternatively, we could compare the capital outstanding at the end
of the 5th year, and the capital outstanding at the end of the 6th
year. The dierence is the amount of capital which is repaid in
the 6th year:
[d15 [d14 = [y15 = 3,374=27.
Prospective and retrospective valuations
We can now construct a loan schedule in terms of these simplied expressions
in [, as follows:
(1)
(2)
w
Ow31
1
O
===
u [dq3u+1
===
q
[d1
(3)
[w
[
[
[
(4)
Lw
lO
(5)
Fw
[ lO
[(1 y q3u+1 ) [y q3u+1
[(1 y)
[y
(6)
Ow
O(1 + l) [
===
0
We can calculate the amount of the loan outstanding at some time during
the course of the loan, either by using the above expression, which takes a
prospective view, or by taking a retrospective view.
The prospective view evaluates payments to be made in the future. The
retrospective view evaluates the past payments in the course of the loan.
The retrospective method of calculating the loan outstanding at time w = u>
42
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
reduces the accumulation of O at rate l for u periods, by the accumulation of
the u repayments of [ which have been received. This gives the following:
Ou = O(1 + l)u [vu @ l
= [dq (1 + l)u [(1 + l)u du
¶
μ
1 yq 1 yu
u
= [(1 + l)
l
l
¶
μ u
q
y y
= [(1 + l)u
l
μ
¶
q3u
1y
= [
l
= [dq3u >
the same as the prospective valuation. Provided that none of the conditions
of the loan has been changed, the prospective valuation and the retrospective
valuation are equivalent.
Changes in the conditions of the loan
In practice, the conditions associated with long term contracts like housing
loans often do change during the course of the loan. Currently it is common
for lenders to advertise themselves competitively, and to attempt to persuade
borrowers to withdraw from their current contracts and shift their business,
attracted by lower rates or better terms or the opportunity to consolidate
several loans into one contract.
The circumstances of the borrower may change, leading them to seek
changes in the conditions of the loan. For example, the borrower may wish to
make additional repayments of capital at some time, or may wish to increase
their borrowing to nance improvements to the property. These changes may
be accommodated in a number of ways, for example by changing the amount
of the repayments, the frequency of the repayments, or the remaining term
of the loan.
It may also happen that circumstances may cause the lender to change
the conditions of the loan. In uctuating economic conditions, interest rates
change, and these changes are periodically passed on to borrowers. Generally,
only in exceptional circumstances do interest rates change, and commercial
rivalries inhibit lenders from changing rates very frequently. However, when
rates do change, the elements of the loan must change in some way to accommodate this.
Example 2.31 A couple take out a housing loan for $100,000 to be repaid
over fteen years by level monthly instalments of capital and interest calculated at a rate of 12% p.a. payable monthly.
2.3. COMMON COMPOUND INTEREST TRANSACTIONS
43
(a) What is the total amount of instalments payable each year?
(b) After two complete years, the couple wish to make a special repayment of $15,000, continue to pay the same monthly instalment, but
reduce the outstanding term of the loan. What is the remaining
term? What is the nal reduced monthly payment?
(c) After a further two years, the couple wish to reduce their payments
such that the loan will be repaid at the original date, that is, in a
further eleven years time. What is the new monthly repayment?
Solution 2.31
(a) Let [ be the amount of the monthly repayment. The payments are
made for a total of 180 months, and the eective monthly interest
rate is 1%. Thus
100,000 = [d180
@ 1%
gives
[ = 100,000@83=32166 = 1,200=17=
Since [ is the regular monthly payment, 12[ = 14,402=02 is the
total annual payment.
(b) First, we must calculate the amount of the loan outstanding after
two years, call this O24 . This equals the value of the remaining
payments, so
O24 = [d156
@ 1%
= 1,200=17 × 78=82294
= 94,600=93=
This amount is reduced by the $15,000 special repayment. The
remaining loan outstanding of $79,600.93 is to be repaid over a
number of months, say p> by level monthly instalments of [= The
following equation of value, as at the time of the special repayment,
describes this transaction:
,
,
,
,
,
79,600=93 = 1,200=17 dp
@ 1%
p
1y
= 66=32471 @ 1%
0=01
y p = 0=3367529
p log 1=01 = log 0=3367529
p = 1=0884058@0=00995033
p = 109=38=
44
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
The term of the annuity must be an integral number of periods,
as an annuity is not dened for partial periods. Thus, in order
to repay the remaining capital, a series of 109 complete monthly
payments of [ are needed, plus a single reduced nal payment,
say \> after 110 months. This last payment of \ is simply the
balance required to reduce the loan to nil after an integral number
of months. We can use the following equation of value to solve for
\:
79,600=93 = 1,200=17 d109 + \ y 110
@ 1%
110
, \ = 1=01 (79,600=93 79,446=25)
, \ = 462=15=
Note that the special repayment of capital during the course of the
loan has made it possible to reduce the term of the loan since the
amount of the monthly repayment remains the same. The earlier
such payments are made in the course of the loan, the greater is
the reduction in the total term.
(c) Since the terms of the loan contract have changed it is necessary
to use a retrospective method to calculate the amount of the loan
outstanding at the time this change is made. We know the amount
of the loan outstanding 24 months before this, at the time of the
extra payment. It is su!cient to accumulate this and deduct the
accumulated payments made since then, to derive the loan outstanding 48 months from the outset. The following equation gives
us the loan outstanding after 48 months:
O48 = 79,600=93 × 1=0124 1,200=17 v24
= 101,072=06 32,372=74
= 68,699=32.
@ 1%
This is the amount of the loan which must be paid o by the new
monthly payment of, say, ]> in 11 years, i.e. 132 months. The
equation of value that gives ] is
@ 1%
68,699=32 = ] d132
, 68,699=32 = 73=11075 ]
, ] = 939=66=
Note that because the number of future instalments has been increased, from 86 to 132, the amount of the monthly repayment can
be reduced.
2.4. EXERCISES
2.4
45
Exercises
1. Calculate the values which are missing from the following table. S is the
principal invested, u is the rate of simple interest p.a., w is the number
of years for which the investment is made, and D is the accumulated
amount.
Principal, S
1,000
1,000
1,500
3,000
1,200
1,000
u
Years, w Accumulation, D
8% p.a.
10
4.5% p.a
5
7% p.a
2,100
4% p.a.
3,200
2
1,500
6% p.a.
10
2,500
3
1,200
8% p.a.
5
1,500
3% p.a.
4
2,300
2. An amount of $1,000 is invested for 1 year at a rate of simple discount
of 8% p.a. Show that the rate of simple discount of 8% is equivalent to
a rate of simple interest of 8.7% p.a.
3. You are about to invest $1,000 for a period of ve years. You can
choose from a range of investments which give returns as follows:
simple discount of 6% p.a., or
simple interest of 7% p.a., or
simple interest of 3% every half year.
Calculate the accumulation which each of the alternatives will give, and
say which one you will choose, and why.
4. An investor is about to invest $1,000. There are two alternative investments available. The rst gives a return of 5% p.a. eective, and
the second gives a return of 6% p.a. simple interest. Calculate the
minimum number of complete years for which the investment must
accumulate in order for 5% p.a. compound to be the more valuable
rate.
5. Calculate
(a) when l(6) = 0=1>
(b) l when l(4) = 0=08>
(c) d̈10 when l(2) = 0=07>
46
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
(4)
(d) d20 when l = 0=065>
(e) v̈20 when l(4) = 0=1=
6. An amount of $100 is invested now and accumulated for two years. The
interest rate is 8% p.a. payable half yearly. Calculate
(a) the eective annual rate earned,
(b) the nominal rate p.a. payable monthly,
(c) the nominal rate p.a. payable quarterly,
(d) the accumulation at the end of the two years,
(e) the total interest earned on the investment.
7. Under an eective rate of interest of 10% p.a., how long does it take
an investment to
(a) double in value?
(b) treble in value?
8. Under an eective rate of interest of 7.1% p.a., calculate the accumulated value at time 25 years of payments of $300 at times 0> 3> 6> 9> = = = > 18
years.
9. An annuity provides payments annually in advance for 25 years. The
payment at the start of the wth year is $1> 000 (1=07w )=
(a) Derive an expression for the present value of this annuity at an
eective rate of interest l per annum, l 6= 0=07=
(b) Find the present value of this annuity when l = 0=05.
10. An annuity is payable annually in arrear for 25 years. The annuity
payment at time w years is 26 w for w = 1> 2> 3> = = = > 25.
Let [ denote the present value of the payments at eective rate l p.a.
(a) Write down an expression for [ in terms of y.
(b) By writing down an expression for (1 + l)[, or otherwise, nd an
expression for [=
(c) Calculate [ when l = 0=07=
11. At an eective rate of interest of 9% p.a., nd the present value of an
annuity under which the payments at times w = 1> 2> = = = > 15 years are
$1,000 and the payments at times w = 16> 17> = = = > 30 years are $2,000.
2.4. EXERCISES
47
12. An annuity of $1,000 is payable annually in arrear for 30 years. Find
the present value of this annuity if interest rates are as follows:
• an eective rate of 10% p.a. for the rst ten years,
• a nominal rate of 10% p.a. convertible quarterly for the next ten
years,
• a nominal rate of 10% p.a. convertible monthly thereafter.
13. An annuity consists of payments of $100 annually in arrear for ve
years, then $50 monthly in arrear for ve years. Interest rates are
5% p.a. eective for three years, then 1% per month eective for ve
years, and then 5% per half year eective for the remaining two years.
Calculate
(a) the present value of the series of payments,
(b) the value at time ve years of the series of payments,
(c) the accumulated value of the series of payments at the end of ten
years.
14. The same series of payments as in the previous exercise is made over a
ten year period. Interest rates over the period are the reverse of those
in the previous exercise, that is, 5% per half year eective for two years,
then 1% per month eective for ve years, and then 5% p.a. eective
for the last three years. Calculate
(a) the present value of the series of payments,
(b) the value at time ve years of the series of payments,
(c) the accumulated value of the series of payments at the end of ten
years.
Compare your answer to part (a) with part (a) in the previous exercise,
and explain the relationship between the two values.
15. An electrical store oers customers the opportunity to purchase goods
by paying one third of the purchase price at the date of purchase, one
third one year after the purchase date and the nal one third two years
after the purchase date. Additionally, a customer must pay 5% of the
purchase price at the same time as the third payment. Consider a
purchase price of $900. Find the eective rate of interest, l p.a., on
this transaction.
48
CHAPTER 2. VALUATION OF FINANCIAL TRANSACTIONS
16. Consider the previous exercise and suppose that the terms were changed
so that the second payment was 6 months after the purchase date, with
no other changes.
(a) Write down the equation of value, again for a purchase price of
$900.
(b) Solve this equation using linear interpolation.
17. Calculate the value of dl10 for values of l of 2%, 5%, 10% and 15%.
Observe how the values change as the interest rate increases. In each
case use 10y as a rough check on your answer, and observe how the
accuracy of your check changes as the interest rate increases.
18. A government is about to issue a bond with the following characteristics: term fteen years, coupon 12% payable half-yearly, redemption
at par. What price should an investor pay at the issue date to obtain
a yield of 9% p.a. eective?
19. Calculate the price on 1 September 1999 of a bond that is redeemable
at 110% on 1 September 2015, with dividends payable annually at 8%,
if the purchaser requires a yield of 8.5% p.a. eective.
20. On 1 July 2006, an investor bought a bond redeemable at par on 1
July 2018, with coupons payable quarterly at 10%. The investor paid
$96. What yield will the investor obtain if the bond is held until the
redemption date?
21. A local government is about to issue two bonds, as follows.
• Bond A has a face value of $100 and a fteen year term, with
coupons payable annually in arrear at 6.5%.
• Bond B has a face value of $100 and a twenty year term, with
coupons payable quarterly in arrear at 7%; the redemption price
per $100 of face value is $103.
Calculate the price of each of these bonds at the issue date if the yield
on each bond is 8% p.a. eective.
22. John borrows $100,000 at a rate of 12% eective p.a., to be repaid
by level annual instalments of capital and interest over twenty years.
Calculate the following
(a) the amount of the annual repayment,
2.4. EXERCISES
49
(b) the amount of capital outstanding immediately after the 10th repayment,
(c) the amount of interest paid in the 3rd year,
(d) the amount of capital repaid in the 11th year.
23. Janet borrows $50,000 from her bank, to be repaid by level annual
instalments of capital and interest over fteen years. The bank charges
interest at 10% p.a. eective.
(a) Calculate the amount of the annual repayment.
(b) At the time the 10th repayment is due, Janet makes an additional
capital payment of 50% of the capital outstanding after ten repayments under the original loan schedule. Calculate the revised
term of the loan if Janet continues to make the same annual repayments, and nd the amount of the nal payment.
24. A loan of $500,000 is issued with equal annual repayments which are
calculated using an eective rate of interest of 8% p.a. Under the
loan schedule, the amount of interest in the nal annual repayment is
$3,469.58. What is the term of the loan?
25. Consider a bank loan of $100,000, with a twenty year term and an
eective rate of interest of 7.5% p.a. Consider bank customers Bruce
and Sheila. Sheila will make fortnightly repayments, whilst Bruce will
make monthly repayments. Over the twenty year period, how much
more will Bruce pay in interest? You should assume that there are 26
fortnights in a year.
Chapter 3
Demography
3.1
Introduction
Demography is the study of populations. It deals with the composition of
a population, how it grows and declines, how it develops with the passage
of time. Actuarial work is concerned with demographic issues because it
frequently involves estimating the behaviour of a population, or of individuals
in a population. In addition to this, actuarial work is often concerned with
estimating the cost or value of such behaviour.
For example, the traditional actuarial area of life insurance involves the
provision of a benet under certain conditions which are associated with the
death or survival of an individual or groups of individuals. In order to cost
the provision of this benet, it is necessary not only to be able to put a
value on the payment of the benet at some time in the future, but also to
assess the likelihood of this benet being paid at any particular time. One
of the uses of demography is to collect information, and construct tools for
interpreting information, which enable defensible estimates to be made of the
survival of an individual from a given population.
The raw material of demography is the collection of data from actual
populations. These data are used to construct models describing the survival
experience of populations of a certain type, and these models are then used
to estimate the experience of other comparable populations, or individuals
within them.
While in general we will talk about human populations, the theory and
arguments hold good for other types of population. The dening characteristics of a ‘population’, for our purposes, are that it consists of a group of
individual members which is in some sense homogeneous, these individuals
may reproduce, and these individuals will eventually die. If we can dene
what we mean by ‘ceasing to survive’ in connection with an individual or
50
3.2. CHARACTERISTICS OF A POPULATION
51
object, in a sensible way, then we can apply the principles of demography to
a population of these individuals. For example, we can apply demographic
understanding to the survival behaviour of the number of members of a profession, an insect, or a car.
A demographic model describes the way the population, whatever it is, is
made up of dierent ages and classes of individuals (for example, by gender),
and describes also the means, rate and time, by which individuals enter and
leave the population. It thus constructs a picture of the whole population,
and can be used to deduce patterns of experience associated with individuals in the population. The following sections discuss the characteristics of
populations with special reference to national populations.
3.2
Characteristics of a Population
3.2.1
Sources of information
In order to construct a mathematical model which is representative of human
survival experience, it is necessary rst to examine the actual experience of
a group of individuals, from which to derive the model. There are several
sources of information available which help build a model of the population
of a country.
• Census data. In Australia, there is a major population census conducted every ve years, which collects information about every individual in the country on the census date. This is a massive undertaking, even for a relatively small population, providing a wealth of data,
which takes years to analyse and process. Other countries conduct a
full census at regular intervals, at periods of about ve to ten years.
Remark 3.1 Census data provide a wealth of detail regarding households
and individuals, their ages, history, occupations, beliefs, and education. It
is nevertheless subject to errors in the data, deliberate or otherwise. People
sometimes lie about their age, or their marital or economic status, for reasons of their own. They may wish, for example, to represent themselves as
older than they are in order to receive an age pension, or to join the armed
forces. Sometimes people are missed or counted twice, particularly if they are
travelling on the census date. The process is extremely expensive in time and
resources, and relies on the honesty and co-operation of the population, despite the use of trained data collectors. In some countries a full and accurate
census is simply impossible, due to conditions of war or poverty or inadequate
administrative and organisational resources. A census in a suburban area is
52
CHAPTER 3. DEMOGRAPHY
fairly straightforward by comparison to one attempted, say, in rural India, or
remote areas of Africa and Asia, or war torn parts of central Europe.
• Population Surveys. These are sample surveys, providing a limited
amount of information, based on a section of the population. These
are smaller, and can be conducted more frequently, than a full census.
The surveys can supply indications of how the general population may
be changing between census dates.
Remark 3.2 These surveys are much more economical to conduct, and can
therefore be conducted more often. They have, however, the disadvantage
that they do only question a sample of the general population, and can only
be used to construct comparative results in inter-census periods. They indicate
changes or trends rather than absolute experience.
• Registration data. Most countries conduct a system of registration of
births, deaths and marriages and have records of population changes
due to migration. These data are generally the most nearly complete
record of the make-up of a population.
Remark 3.3 In some countries registration data are not rigorously collected
or preserved, or may be lost or destroyed through acts of war or by some
other means. Migrants to countries whose records are accurate may have no
birth or other certicates and may provide inaccurate information on arrival.
Generally, such information is endorsed by some professionally accredited
person, such as a doctor or a registrar, and may be supposed to be accurate
when it exists.
Other sources of data for specic types of population are available, to
provide models of special populations, for example, pensioners, or annuitants,
or car owners.
• The records of social security registers, regarding the experience and
numbers of age pensioners.
• The records of life insurance companies, regarding their policyholders.
For example, the numbers and mortality experience of holders of life
insurance policies.
• The rolls of state schools can give comparative information about the
numbers and changes in geographical and age distribution of children.
• The records of pension funds regarding the mortality of their members.
• The records of general insurance companies, regarding the incidence of
car accidents, and thefts.
3.2. CHARACTERISTICS OF A POPULATION
3.2.2
53
Classication of data
Once the data are collected, they can be classied in numerous ways. The
object of classication is to divide the population into sub-groups which are
homogeneous in some respect, so that types of experience can be isolated,
and patterns described. For the purposes of studying human mortality experience, and for many other purposes also, the major classications are by
year of age and by sex. These classications may be grouped, or aggregated
for age groups, rather than single years of age, and it is common for population data to be presented in quinquennial age groups. Data may also be
classied by gender only.
There are a number of single summary gures which are used to describe
the population as a whole. We will rst discuss the summary population
statistics commonly used before looking at the experience of the individuals
in the population. In the sections which follow, discussion is of human and
national populations in particular.
3.2.3
Summary statistics
States
There are a number of summary statistics describing a population, some
describe the state of the population, and some describe the way in which the
population changes. First we dene some statistics which describe the state
of a population at a particular time.
Denition 3.1 Sex ratio: The sex ratio of a population is the number of
males per 100 females of the population, dened for a particular age or age
range. It is calculated as
100 ×
# males
at this age, or in this age group
# females
where the symbol # means ‘number of’.
Interestingly, the sex ratio at birth is about 105 irrespective of the nationality of the population. That is, for any country in the world, the number of
boys born is about 105 for every 100 girls born. As we will discuss later, the
mortality experiences of males and females have dierent characteristics, and
the sex ratio of a population depends on its age structure. Males generally
die younger than females, and the sex ratio becomes less than 100 for higher
ages. In the latest Australian population statistics, at ages in excess of 35
the age specic sex ratio is less than 100.
54
CHAPTER 3. DEMOGRAPHY
While the sex ratio at birth is fairly constant worldwide, the other age
specic sex ratios are dependent upon local circumstances. Where infant
mortality is high, male babies are more prone to death than female ones,
reducing the ratio. If male youths are exposed to war time conditions, the
sex ratio, again, is generally reduced by this. Other factors which aect the
sex ratio in adolescence and youth are suicide and accident, the main causes
of death of young people, both of which are more likely to occur amongst the
males in the population.
Denition 3.2 Child-woman ratio: This is dened as the number of children
(both male and female) in a population under a given age, typically 5, per
100 females in the population who are of reproductive age, typically taken as
ages 15 to 49. With these ages, the ratio would be calculated as
100 ×
# children aged 0 to 4
=
# females aged 15 to 49
This measure is simple to calculate from census statistics and can be used
as a rough measure of fertility in situations when more detailed data about
the population are unavailable.
Denition 3.3 Dependency ratio: There are several expressions, more or
less detailed, of this ratio. This ratio, in brief, is the number of individuals
who are not economically independent, usually by virtue of their age, divided
by the number of individuals who are qualied by age to be economically
independent. This ratio has traditionally been taken to be
100 ×
# aged 0 to 14 + # aged 65 and over
=
# aged 15 to 64
Whether this denition is useful or appropriate is to some extent dependent upon the socio-economic condition of the population, and on the
available population data. In an economically developed country, it is common for an extended period of formal education to take place, and so an
average age of nancial dependence might be more sensibly taken to be 18 or
even 20. In most countries, age retirement occurs between 60 and 70, though
the general trend is for this to be increasing gradually. The age dependency
ratio is
# aged 65 and over
100 ×
# aged 15 to 64
and the youth dependency ratio is
100 ×
# aged 0 to 14
=
# aged 15 to 64
3.2. CHARACTERISTICS OF A POPULATION
55
The age dependency ratio would give, for example, an indication of the
burden upon the working members of a population of supporting the costs
of the aged, largely related to health, and pensions. The costs associated
with youth are generally lower per capita, and associated with the health
costs of ante-natal and post-natal care, childhood health costs, and education
costs. Though these summary statistics are very simple and crude, they can
together give an insight into the experience of a population, and its age
distribution.
Age dependency ratios are tending to increase worldwide. This is due to
several factors, but in general people are surviving to older ages, and fertility
rates are dropping, increasing the proportion of the population at older ages.
Denition 3.4 Labour force participation rate: This rate represents the proportion of a given population who are employed, or actively seeking employment.
Labour force participation rates may be expressed specic to gender and
age group. A single aggregate rate will give an indication of the employment
rate in the population, but is not equivalent to it. There are several dierent
ways of classifying labour force participation, and in this case those who have
no job, but who are actively looking for employment, are said to be part of
the labour force.
Patterns of employment are changing. In the past, in an economically
developed country, the pattern of employment was inclined to show male
participation rates as being high and fairly constant throughout the working
lifetime, much higher than female rates. However, this pattern is changing.
Rates of unemployment now tend to be high for the young (15 to 25), and
as people live a longer and more active life, they are more inclined to seek
employment at higher ages. Female participation rates have been increasing
in recent years, but this trend shows signs of slowing, or even reversing.
In general, the male and female patterns of labour force participation are
becoming more like one another. However, the tendency remains for women
to have more breaks from full-time employment, (usually to take care of
family members), and to be more inclined to work part-time, than men.
Rates of change: crude rates
The statistics described so far give information about the shape and form of
the population. Others are used to describe the way in which the population
changes. Let Sw denote the population size at time w. Then we can describe
the changes in the population over a year in the following way:
Sw+1 = Sw + Eluwkvw@w+1 Ghdwkvw@w+1
+ Lppljudqwvw@w+1 Hpljudqwvw@w+1
56
CHAPTER 3. DEMOGRAPHY
where the subscript w@w+1 indicates the period from time w to w+1= That is, the
population increases by the number of births and the number of immigrants
into the population during the period, and is decreased by those leaving
the population by means of emigration or death. These means of increase
and decrease of the population may change the characteristics of the age
distribution and size of the population as time passes.
Denition 3.5 The number of births minus the number of deaths is the
natural increase of the population.
Denition 3.6 The number of immigrants minus the number of emigrants
is the net migration to the population.
There are a number of summary statistics which describe the natural
increase of populations.
Denition 3.7 A single gure statistic, based upon the number of events per
1,000 of population, is called a crude rate.
Denition 3.8 A gure based upon the number of events per 1,000 of a
specic section of the population, and relating only to that section, is called
a specic rate. Often the specied section is an age group, leading to age
specic rates.
Denition 3.9 The crude birth rate is the number of births in a period, per
1,000 of the average population in the period.
Denition 3.10 The crude death rate is the number of deaths in a period,
per 1,000 of the average population in the period.
These two crude rates give an indication of the rate of natural increase
of a population, but do not, in themselves, tell us very much about the
experience of individuals in the population. The advantages of crude rates are
that they are easily calculated, and give a single gure comparison between
populations. However, both rates depend very much on the age distribution
of the population to which they refer, and so should be interpreted in light
of this. For example, crude birth rates may be low in a population if women
have few children, or if the population has a high proportion of women too
old or too young to bear children. Similarly, a high death rate may indicate
that the population is relatively old, or relatively likely to die from other
causes, such as ill health, famine or acts of war. These other causes of death
are as likely to aect the very young, as the very old.
Crude rates are sometimes used as an indicator of the experience of a
country, because they are simple and easy to compare. However, it is important to remember that crude rates need careful interpretation, in the light of
the characteristics of a population and circumstances aecting it.
3.2. CHARACTERISTICS OF A POPULATION
57
Rates of change: specic rates
In general, a specic rate is the number of ‘events’ (for example, deaths)
occurring during a particular period, per 1,000 of the specied average population which are exposed to risk of that event occurring during the period.
The rate is specic in the sense that it counts only those events which occur
to those who belong to this specied sub-population.
For example, the age specic fertility rate per 1,000 is dened as 1,000
times the number of births to females in a given age range in a given year,
divided by the average number of females in that age range during the year.
Here the rate is age specic, and gender specic. The population ‘at risk’ is
the average number of women (only) belonging to the particular age range.
The total fertility rate (discussed in more detail later) is not age specic,
and refers to the whole of the population of women of childbearing age. The
total fertility rate is the sum of the age specic fertility rate for each age. It
thus gives an indication of how many children a woman of the population
bears over her reproductive lifetime. It is not the same as this, however, since
the age specic rates over a single year include the experience of women from
dierent generations, aged between (approximately) 15 and 50, in that year,
and does not give information about the complete childbearing history of any
one woman.
The age specic mortality rate per 1,000 is dened as 1,000 times the
number of deaths during the year of those in the specied age group, divided
by the average population in that age group during the year. As we will
discuss later, there is a strong link between mortality rates and age, though
there are general trends in mortality rates observable in many national populations.
A comparison of age specic rates for dierent populations will give information which is independent of the age distribution of the populations,
unlike the crude rates. Used together, the two types of rate can give a much
fuller description of the populations to which they refer. It is possible to
use age specic rates for a population, and to standardise these according
to a given population age distribution, and to construct a crude rate which
would apply had the population under consideration been of the ‘standard’
age distribution. In this way, crude rates can be constructed to enable direct
comparisons of a number of populations with very dierent age distributions.
Such standardised rates can be very useful one-gure summary measures of
comparison.
Example 3.1 (Calculation of crude, age specic and standardised or adjusted mortality rates.) The table below shows the following information for
two populations, denoted A and B: number in the population and number of
deaths, each in a given year.
58
CHAPTER 3. DEMOGRAPHY
(a) Calculate the crude death rate for each population.
(b) Calculate age specic death rates for each population.
(c) Calculate a standardised crude death rate for population B, using
population A as the standard population.
(d) Comment on these results.
Population A
Population B
Age Population No. of Population No. of
group
size
deaths
size
deaths
1
2,000
2
5,000
6
2
3,000
3
4,000
4
3
4,000
5
3,000
4
4
5,000
10
2,000
5
Remark 3.4 The two populations have dierent age structures, with population B having a relatively ‘younger’ population than population A. About
35% of population A is in the oldest age group, whereas about 15% of population B is in the oldest age group. In total, each population has the same
size.
Solution 3.1
(a) The crude death rate for population A is
20 × 1,000
= 1=4286
14,000
per thousand of population.
The crude death rate for population B is
19 × 1,000
= 1=3571
14,000
per thousand of population.
(b) The age specic death rates are as follows:
Population A
Population B
Age group
Age specic rate
Age specic rate
1
1,000 × 2@2,000 = 1
1,000 × 6@5,000 = 1=2
2
1,000 × 3@3,000 = 1
1,000 × 4@4,000 = 1
3
1,000 × 5@4,000 = 1=25 1,000 × 4@3,000 = 1=33
4
1,000 × 10@5,000 = 2
1,000 × 5@2,000 = 2=5
The rate for each age group is the number of deaths per thousand
of population in the age group.
3.2. CHARACTERISTICS OF A POPULATION
59
(c) Taking the standard age distribution as that of population A, we
apply the age specic rates obtained for population B, as shown
below.
RateE ×
Age group population sizeD
1
1=2 × 2,000
2
1 × 3,000
3
1=33 × 4,000
4
2=5 × 5,000
Number of
deaths per 1,000
2=4
3
5=33
12=5
Thus, the adjusted or standardised crude death rate for population
B, when standardised to the population age distribution of A, is
23=2333 × 1,000
= 1=6595
14,000
per 1,000 of population, compared to 1.4286 for population A.
(d) The crude rate for population B is lower than that for population
A. Does this mean that mortality rates are lower for population B,
and that individuals survive longer, or does this re ect the dierence in the age distributions? The age specic rates are lower for
population A than for population B for all age groups. Thus, the
dierence in the crude rates re ects the dierence in age structure
of the populations, and distorts the experience which is specic to
the age groups. The standardised crude death rate for population
B shows that mortality rates are about 22% higher for population
B than for population A, though the crude rate for population B
was actually lower than that for population A.
3.2.4
Rates of change: the population growth rate
The crude death rate and crude birth rate describe the rate of change in
the population due to specic causes, that is, mortality and fertility. The
population growth rate is a measure of the size of the population from time
to time, and is not specic as to cause of net change. Thus, if S (0) is the
number of individuals in the population at time 0> and S (q) is the number
of individuals in the population at time q years later, then u is the annual
growth rate of the population where
S (q) = S (0) huq =
That is, u is based on an assumption of exponential growth of the population
size over the period from 0 to q= This model, as we will see later in Section
60
CHAPTER 3. DEMOGRAPHY
3.5, has limited applicability. For example, if the exponential growth model
was assumed to hold over indenite periods of time, then populations which
were getting larger (u A 0) would eventually become innitely large, and
populations which were in decline (u ? 0) would eventually become extinct.
In general, this model is too simplistic to use over the long term.
This leads us to consider the larger view of how national populations
develop and grow, and a consideration of the factors involved. Such a view
is described by the Theory of Demographic Transition.
3.2.5
Demographic transition
The theory of demographic transition is an interpretation of the demographic
development of a national population having regard to changing socioeconomic conditions. It describes the cycle of change through which a country is inclined to develop as it becomes more economically ‘sophisticated’.
The model of the cycle of development is described thus. As a country
evolves from ‘third world’ conditions of tribal nomadic or agrarian culture,
into an industrial and urban society, one of the early changes which it makes is
to increase and improve access to health services, and to improve general nutritional and sanitary standards. The provision of easy access to clean water
can have a dramatic eect on infant mortality, in itself. These improvements
have a positive impact on ante-natal care, and have an immediate impact on
the infant mortality rate, reducing it. Though of course the whole population
benets from the changes, it is the increased number of infant survivals that
increases the size of the population very quickly. While older members of the
population may survive longer and be healthier for longer, these eects are
much slower to manifest. Thus, in the early stages of demographic transition,
the population grows and fertility increases.
As the eects of improved commercial conditions are felt, and people
become more economically wealthy with increased incomes and industrialisation, the fertility rate is inclined to drop again. A number of reasons
conspire to produce this eect. Individuals move to urban centres in search
of work, and the family support structure breaks down. Instead of perceiving a large number of ospring as some kind of insurance against poverty,
children become more expensive to support. In addition, the expectation of
each child’s survival increases, and the means to reliably control the number
of pregnancies becomes available. People begin to change their behaviour to
provide a higher standard of living, and education, for a smaller family than
previously. Thus, after a while, fertility rates decline.
After a longer period the improvement in conditions and standard of living
manifests as increased longevity, and the proportion of older individuals in
the population increases.
3.2. CHARACTERISTICS OF A POPULATION
61
To some extent therefore, it is possible to deduce the state of economic
development of a country by a study of the demographic characteristics of
the population. Typically third world countries, relative to the economically
established, have a high proportion in the very young ages, high fertility
and infant mortality rates, high mortality rates over all ages, and a low
expectation of life.
A graphical representation of a national (or other) population provides a
historical and current commentary on that population. A population pyramid shows the age and gender distribution of a country at a particular date.
In this picture we can trace both historical events and socio-economic conditions.
Figure 3.1 shows the population pyramid for Ethiopia in 2009. Population
pyramids for other third world countries generally have this shape. The
typical characteristics of such a population are:
• a high proportion of the population in young age groups;
• high infant and child mortality, with a low proportion surviving to
adult ages;
• generally high mortality rates;
• high fertility rates;
• low life expectancy.
Figure 3.2 shows the population pyramid for Australia in 2009. This
pyramid is typical of a country that has come through a period of economic
growth. The typical characteristics of such a population, compared with the
previous illustration, are:
• a higher proportion of the population in the middle and older age
ranges;
• lower fertility rates;
• higher life expectancy;
• a lower youth dependency ratio.
Remember that Figure 3.2 is a historical picture. It shows the eects of
an active migration policy (which increases numbers in the young and middle
age groups) and also of the ‘baby boom’ of the 1950s, although this eect
is masked somewhat by the migration eect. It also shows a recent increase
62
CHAPTER 3. DEMOGRAPHY
85+
80-84
75-79
70-74
65-69
60-64
55-59
50-54
45-49
40-44
35-39
30-34
Female
Male
25-29
20-24
15-19
10-14
5-9
0-4
8
6
4
2
0
2
Population in millions
4
6
8
Figure 3.1: Population pyramid for Ethiopia in 2009
100+
95-99
Female
Male
90-94
85-89
80-84
75-79
70-74
65-69
60-64
55-59
50-54
45-49
40-44
35-39
30-34
25-29
20-24
15-19
10-14
5-9
0-4
1000000
750000
500000
250000
0
250000
500000
750000
1000000
Population size
Figure 3.2: Population pyramid for Australia in 2009
3.2. CHARACTERISTICS OF A POPULATION
63
100+
95-99
90-94
Female
Male
85-89
80-84
75-79
70-74
65-69
60-64
55-59
50-54
45-49
40-44
35-39
30-34
25-29
20-24
15-19
10-14
5-9
0-4
2500000
1500000
500000
500000
1500000
2500000
Population size
Figure 3.3: Population pyramid for France in 2010
in fertility, in part due to government incentives to increase the number of
births in the country.
Figure 3.3 shows the population pyramid for France in the year 2010,
and illustrates the ‘pillar’ shape associated with a long established economy. France was part of the Industrial Revolution of the nineteenth century,
and demographic transition eects are well developed. At the highest age
groups are survivors of the two World Wars, with much higher numbers of
females than males in these groups. The country has experienced growth in
population after the Second World War, by birth rather than immigration,
which can be seen in the sudden increase in numbers in the age group 60—
64 compared with the 65—69 age group who were born during the war. By
comparison with the previous examples, the main features of the pillar shape
are:
• low fertility rates;
• low mortality at all ages;
• fairly even distribution of population between age groups;
• a resulting high proportion of population in the highest age groups,
that is, a high age dependency ratio.
Table 3.1 contains summary statistics for the three dierent countries,
and the contrasts and salient features can easily be seen from these.
64
CHAPTER 3. DEMOGRAPHY
Proportion of population aged under 15
Proportion of population aged 15—64
Proportion of population aged 65 and over
Child/woman ratio
Dependency ratio
Ethiopia Australia France
46.09%
19.06% 18.51%
51.25%
67.67% 64.85%
2.66%
13.27% 16.64%
80.85%
26.19% 27.19%
95.13%
47.77% 54.20%
Table 3.1: Population summary statistics
3.2.6
Development of world demographics
In the 1960s, following the increase in fertility in the industrialised world during the post-war period, there was much concern about the apparent ‘population explosion’. This expectation was based largely on the assumption that
the increase in fertility rates would continue indenitely. The prognostication was unfounded, and the general trend in many countries is that fertility
rates are falling and populations ageing relatively.
However, there is continuing concern about the depletion of the world’s
resources and increasing levels of pollution. As more countries become industrialised, the levels of pollution which they produce, and the rate at which
they destroy their own natural resources usually increase. The conservationist arguments sometimes appear insubstantial and too long term to a
government which is struggling to establish itself economically and industrially. Rich areas of unique and priceless natural value may be devastated in
order to provide short term gains of industrial establishment.
Part of the panic of the 1960s and 1970s about world population was the
fear that the world could simply not feed the expected numbers of inhabitants. In fact it is more generally accepted now that the capacity of the world
to produce enough to sustain the population is much more robust, and that
the major di!culty is with the distribution of the resources, rather than the
absolute quantity of them.
The move continues towards global co-operation and management of
global resources, and attempts are continuing to be made to support development of countries in an economically and ecologically sustainable way, and
to elicit responsible behaviour and accountability from established industrial
powers.
The matters which aect the world’s demography are complex and many
issues are involved. As we have discussed, aecting infant mortality is one
of the fastest evolutionary ways to change a population structure. As health
and nutrition improve globally, the fertility rate becomes more a matter of
management than before. The number of children surviving can increase by
improving infant mortality experience, but may also decrease as a result of
3.2. CHARACTERISTICS OF A POPULATION
65
choices made by the childbearing population.
Large scale natural disasters and catastrophe, or acts of war and genocide,
may have a sudden impact on a nation’s population. In general, these have
less impact than one might expect, though they seem large scale and shocking
when they happen. Also their effect is generally more local than global.
The slower effects of improving mortality and increased economic wealth are
inclined to contribute more to population changes in the long term. For
example, while the First World War took a terrible toll on, in particular,
young men in the period 1914—1918, this war was followed by an epidemic
of influenza which killed many more people, of all ages, than the war had.
While the capacity for human destruction is enormous given the scope of
modern weaponry, ‘modern’ warfare is inclined to have a low toll in human
life. In recent times, high numbers of casualties are more often associated
with ‘low tech’ warfare, such as the murder of thousands of civilians in intertribal African and European conflicts. The capacity of nuclear power for mass
destruction of life has been restrained since its only wartime use in 1945. As
more nations demonstrate the capacity for such weaponry, the likelihood of
the restraint continuing depletes.
Again, the impression is usually that epidemic and illness have a large
scale impact on world demographics. The AIDS virus is a global disease,
affecting nearly 50% of the adult population in some African countries. It
is not clear yet what the long term impact of this will be. However, bearing in mind that a population may be affected most quickly by changes in
fertility and infant survival rates, and considering that AIDS affects adults
of childbearing age and is passed on to any offspring, the virus may have a
considerable long term effect on some national populations. It is not unimaginable that if it continues to grow largely unchecked for some time, it will
decimate the emergence of the young generation of whole nations.
Attempts have been made to construct models of the future of the global
population, and these consider the influence of the above kinds of issue.
However, one factor which may dominate such models is that China is now
an emerging industrial nation which dominates (by number) the global population, and may display the pattern of development characterised by the
theory of demographic transition. Whether or not it does, as it emerges into
a more open association with the global economy and becomes more industrialised, by sheer weight of numbers it is likely to dominate the future global
demography and experience.
66
CHAPTER 3. DEMOGRAPHY
3.3
Individual Characteristics: Mortality
3.3.1
The survival function
We investigate the characteristics of mortality experience of an individual
by examining the experience of a group. We do not know what an individual’s experience will be. We can only infer, from our understanding of the
experience of the population from which the individual is drawn, what we
may expect that individual’s experience to be. We, essentially, deal only in
‘odds’ which we derive from past experience and a knowledge of current conditions and assumptions of future developments. This in fact characterises
much of the actuary’s work. Actuaries habitually deal with the valuation or
evaluation of events whose occurrence or timing are uncertain. In the case
of mortality, the uncertainty is associated with the length of time that an
individual will survive.
We can construct a mathematical model for this.
Denition 3.11 ({) denotes a person (or thing) which is aged { exactly.
Denition 3.12 W ({) is the length of the future lifetime of the individual
now aged exactly {=
We can construct a model of survival based on W ({), assuming it has a certain distribution of values, each with an associated probability. If the distribution is mathematically accessible, then the model is analytically tractable,
and we can construct general rules and formulae dealing with the behaviour
of this variable. In general, W ({) is a continuous random variable.
Consider a very simple example: W (0) represents the total lifetime of an
individual in this population, being the numbers of future years of life for a
newborn (0).
Example 3.2 Suppose our population is a pile of magazines. The lifetime
of each magazine is measured in weeks. You want to keep only the magazines
which are less than 4 weeks old. You will throw others out (they will cease to
‘survive’).
Solution 3.2 We immediately know something about the probability distribution of W ({) for this population.
W (0)
W (1)
W (2)
W (3)
W ({)
=
=
=
=
=
4 with probability 1,
3 with probability 1,
2 with probability 1,
1 with probability 1,
0 ;{ 4 with probability 1.
3.3. INDIVIDUAL CHARACTERISTICS: MORTALITY
67
A point to note here is that we do not admit the possibility of ‘recycling’ of
these magazines. That is, for our purposes, once survival has ceased, it does
not recommence. (Also, we have ignored the possibility of other forms of
decrement, e.g. all the magazines could be destroyed if a re destroyed your
home.)
We can construct functions to describe W ({)> and to express the probabilities associated with various events possible to the individual ({)=
Denition 3.13 v({) represents the probability that a newborn individual
will survive for at least { years. We call v({) the survival function.
Denition 3.14 I ({) represents the probability that a newborn individual
will not survive for { years, that is, will die before reaching age {=
We make certain assumptions about what ‘survival’ means, and these
assumptions impose conditions upon the functions dened here. Firstly, as
mentioned above, we consider that ‘dying’ is a monotonic process, in that
once survival has ceased, it does not resume.
From the above denitions it is clear that
v({) = 1 I ({)
and
0 v({) 1
since v({) is a probability. The same limits apply to I ({)> for the same
reason.
We assume that I (0) = 0> that is, that at the moment of ‘birth’ survival
is certain. It then follows that
v(0) = 1 I (0) = 1.
We assume that survival ceases at some time, and that the probability of
survival as time passes decreases monotonically to 0, i.e.
v(w) $ 0 as w $ 4
(‘you cannot live forever’) and, equivalently,
I (w) $ 1 as w $ 4
(‘you will die sometime’).
This condition we have dened in terms of 4> and clearly for some populations survival is limited to fairly short timespans. For humans, for example,
68
CHAPTER 3. DEMOGRAPHY
it is not likely that we would consider (at present at least) survival beyond
120 to 130 years. Such a lifespan is very rare indeed. We therefore dene a
value $ which stands for the upper age limit of the population. We restate
the two previous conditions as
v(w) $ 0 as w $ $,
I (w) $ 1 as w $ $.
For various species of animal, for example, $ might vary between very much
less than one year up to over one hundred years. The value put upon $ is a
characteristic of the population.
We can relate these functions to the random variable W ({)= The probability that a newborn will survive to at least age { is v({), and is therefore
equivalent to the probability that W (0) {, that is, the probability that the
total future lifetime of a newborn will exceed { years.
Similarly we can express I ({) in terms of W (0) = The probability that
a newborn dies in the next { years is I ({), that is, the probability that
W (0) ? {=
So far, we have considered the number of years survived from birth. We
can also use these functions to calculate probabilities associated with survival
between ages. If v({) is the probability that a newborn will survive to age
{> then the probability that a newborn will die between the ages of | and }
(i.e. will attain | but not }) is given by the following expression:
Pr(| W (0) ? }) = v(|) v(}).
Suppose now that we consider the case of a life aged (|) = We know that
the total lifetime is at least | years from birth. Having already reached age
|> what is the probability that this life will not survive to age }? We can
describe this by the following expression:
Pr(| W (0) ? }|W (0) A |) =
v(|) v(})
.
v(|)
Note that this is just an application of conditional probability. In general,
the probability of an event D occurring given that an event E has occurred
is denoted Pr(D|E) and calculated from
Pr(D|E) =
Pr(D and E)
.
Pr(E)
Thus, Pr(| W (0) ? }|W (0) A |) is the probability that the total lifetime of
the individual lies between | and }, given that the life has already survived
3.3. INDIVIDUAL CHARACTERISTICS: MORTALITY
69
for | years, or, the probability that a life (|) will die before age }= Note that
v(})
v(|) v(})
= 1
v(|)
v(|)
= 1 Pr((|) will survive to at least age }).
Example 3.3 The probability of survival for a member of a population is
described by the survival function v({) = 1 {@100 for 0 { 100=
(a) Conrm that this expression is suitable as a survival function.
(b) Calculate the probability that a newborn life from this population
will survive to age 80.
(c) Calculate the probability that a newborn life will die before age 30.
(d) Calculate the probability that a life now aged 30 will survive to age
80.
Solution 3.3
(a) In order to be suitable for a survival function v({) must satisfy
three criteria: v(0) = 1> v($) = 0> where in this case $ = 100, and
v({) must be non-increasing for 0 { $= This last condition
can be represented by v0 ({) 0 for 0 ? { ? $= It can be seen that
v({) = 1 {@100 for 0 { 100, satises all these conditions.
(b) The probability that a newborn will survive to age 80 is
v(80) = 1 80@100 = 0=2.
(c) The probability that a newborn will die before age 30 is
I (30) = 1 v(30) = 30@100 = 0=3.
(d) The probability that a life aged 30 will survive to age 80 is
v(80)@v(30) =
1 80@100
= 0=2857=
1 30@100
Let us dene another function which is used to describe the rate of death
at any moment, the force of mortality at age {, denoted by { . Consider the
probability of ({) dying in the next, innitesimally small, moment:
Pr({ W (0) ? { + {{|W (0) A {) =
I ({ + {{) I ({)
=
1 I ({)
This is the probability an individual will die before age { + {{, having
attained age {= If we take the limit of this function divided by {{, as {{ $ 0>
we have an expression for the instantaneous rate of death, known as the force
of mortality.
70
CHAPTER 3. DEMOGRAPHY
Denition 3.15 The force of mortality, { > is dened as
1
Pr({ W (0) ? { + {{|W (0) A {)=
{{<0 {{
{ = lim
That is,
1 I ({ + {{) I ({)
{{<0 {{
1 I ({)
1
gI ({)
=
1 I ({) g{
v0 ({)
.
=
v({)
{ =
lim
Example 3.4 The survival function for a certain population is v({) = 1 {@100 for 0 { 100= Derive the force of mortality associated with this
survival function, and evaluate it for age 25=
Solution 3.4 The force of mortality is { = v0 ({)@v({)> so
{ =
1@100
for 0 ? { ? 100
1 {@100
and
25 =
0=01
= 0=0133.
1 0=25
Given the characteristics of v({) we can see that the only condition which
must apply to { is that it cannot be negative. From the relationship { =
v0 ({)@v({) we can derive the following expression for v ({) in terms of { :
¸
Z {
| g| =
v({) = exp 0
This follows since
g
v0 (|)
log v(|) =
= | .
g|
v(|)
Integrating this identity over (0> {) gives
¶
Z {
Z {μ
g
log v(|) g| =
| g|
g|
0
0
and as the left hand side is just the integral of a derivative, we get
Z {
(log v({) log v(0)) =
| g|
0
which gives us v({) (noting that v(0) = 1).
3.3. INDIVIDUAL CHARACTERISTICS: MORTALITY
71
Example 3.5 The mortality experience of a population is described by the
survival function v({) = 1 {2 @900 for a range of ages {, 0 { $.
(a) For what range of values of { is v({) suitable as a survival function?
(b) What is the expression for the associated force of mortality?
(c) By what age is there a 50% probability that a newborn life has
already died?
(d) Calculate the force of mortality at age 25.
(e) Calculate the probability that a life aged 10 will die before reaching
age 25.
Solution 3.5
(a) A survival function must satisfy a number of criteria on the age
range to which it applies: v(0) = 1> v($) = 0> and v0 ({) 0= In
this case, where v({) = 1{2 @900> v(0) = 1> and v ($) = 0 implies
that
v($) = 1 $ 2 @900 = 0
, $ 2 = 900,
so that the survival function is suitable for values of { such that
0 { $ = 30= In addition, the survival function must be nonincreasing. We have v0 ({) = {@450 which is less than 0, and so
the condition is satised for all { in the range.
(b) { = v0 ({)@v({) for all { such that 0 ? { ? $ = 30. Hence
{ =
{@450
.
1 {2 @900
This expression is positive, as we require, if 1 {2 @900 A 0> which
is true for the range of values of { we are considering, namely
0 ? { ? $ = 30=
(c) We need to calculate the age at which the probability of survival to
that age becomes less than 0=5= If we call this age |> then we need
to nd | such that v(|) = 0=5= Thus
v(|) = 1 | 2 @900 = 0=5
s
, | = 450 = 21=21.
72
CHAPTER 3. DEMOGRAPHY
(d) The force of mortality at age 25 is given by
25 = (25@450)@ (1 625@900) = 0=1818=
(e) The probability that a life aged 10 will die before age 25 is
Pr(10 W (0) ? 25|W (0) 10) =
=
=
=
(v(10) v(25))@v(10)
1 v(25)@v(10)
1 0=3056@0=8889
0=65625.
Note that the probability of a newborn reaching, say, age D + E> is not
the same as the probability of an individual who has already reached age D>
surviving an additional E years to reach age D+E= The intuitive explanation
is that (D) has already survived the risks attached to the rst D years, and
is more likely to attain age D + E than a newborn.
3.3.2
The life table
The survival function leads to the probability distribution of the length of life
of members of a population, and acknowledges the fact that each individual
has a unique survival experience. In practice, it may be di!cult to construct
suitable functions, and, as we discuss later, it may be necessary to use different functions to describe dierent age ranges in the life cycle. In actuarial
work there is a tool constructed from population mortality to facilitate the
calculation of probabilities of death or survival. It is called the life table. A
typical life table has the following form:
{
o{
g{
t{
0 100,000 2,454 0=02454
1 97,546
156 0=00160
2 97,390
97
0=00100
3 97,293
===
===
The life table may be interpreted in various ways. It may be taken to represent a record of the expected number of lives who survive to each age, given a
starting population of births, and a probability of survival at each age. If we
interpret the life table in this way, the elements of this life table are dened
as follows.
Denition 3.16 o{ is the expected number of lives attaining age { from o0
newborn lives. We call o0 the radix, and it is commonly set to equal 100,000.
3.3. INDIVIDUAL CHARACTERISTICS: MORTALITY
73
100,000
80,000
60,000
40,000
20,000
0
0
10
20
30
40
50
60
70
80
90
100
Age
Figure 3.4: Female mortality table — o{
Denition 3.17 g{ = o{ o{+1 . We interpret g{ as the expected number of
deaths between ages { and { + 1 exact out of o{ lives at age { exact.
Figures 3.4 and 3.5 show the functions o{ and g{ from the Female Mortality Table (see Appendix 2). We observe that o{ is a decreasing function of {,
decreasing gradually up to about age 50. In contrast, g{ initially decreases
with {, before increasing to a peak at age 83 (in this mortality table), then
decreasing. Obviously, as the number of survivors from a given cohort of
births decreases, the number of lives who are exposed to the risk of death
decreases with increasing age, and hence we expect g{ to eventually decrease
with {, as illustrated in Figure 3.5.
Denition 3.18 t{ is the probability that ({) will die before reaching age
{ + 1= It is called the mortality rate. More generally, w t{ is the probability
that ({) does not survive to age { + w= In terms of the survival function, we
have
v({) v({ + w)
.
w t{ =
v({)
For survival probabilities, we have the following denition.
Denition 3.19 s{ is the probability that ({) will survive to age { +1= w s{ is
the probability that ({) will survive for w years, attaining age { + w. In terms
74
CHAPTER 3. DEMOGRAPHY
3,500
3,000
2,500
2,000
1,500
1,000
500
0
0
10
20
30
40
50
60
70
80
90
100
Age
Figure 3.5: Female mortality table — g{
of the survival function, we have
w s{ =
v({ + w)
.
v({)
There are many published life tables, and they are generally named according to the population whose mortality experience they describe, and the
date(s) at which the information was gathered. For example, ALT 2000—02
stands for Australian Life Tables, arising from the census information describing the general population of 2000—02. Other commonly used tables,
other than those describing the population of a country, are similarly named
after the population they describe. Some tables are specic to other significant subgroups, for example tables describing insured lives are also given
for smokers and non-smokers. For example, AMS00 is a table describing
assured male smokers, based on data from UK insurance companies from
1999 to 2002. There is often a male and a female table for any population —
AFS00 is the female counterpart of AMS00. In principle, a life table can be
constructed based on the survival function of any homogeneous group.
A life table is constructed given a series of age specic mortality rates t{
for a range of values of {= A value for the radix, o0 , is chosen, usually 100,000.
3.3. INDIVIDUAL CHARACTERISTICS: MORTALITY
75
From this and the mortality rates, the entire table is constructed. We have
g0 = t0 o0 >
s0 = 1 t0 >
o1 = o0 g0 = o0 s0 .
Thus, o1 is the expected number surviving to age 1 of the o0 newborn lives.
Note that we are using the word ‘expected’ in the statistical sense. The
values of g0 and o1 need not be integers, but they are frequently expressed
this way in published life tables. If the radix is chosen to be suitably large,
the degree of accuracy maintained is accepted as adequate.
In general, the probability of surviving to age { + w> having attained age
{, is given by
w s{ = o{+w @o{
and the probability that a life aged { will not survive to age { + w> is given
by
w t{
= 1 w s{
= 1 o{+w @o{ .
The prex w is generally omitted when its value is 1=
Successive values may be derived using the general relationships below:
g{ = o{ t{ >
o{+1 = o{ g{ = o{ (1 t{ )=
Example 3.6 Given the following values for mortality rates at ages 0—4 inclusive, construct a life table for this age range, showing values for o{ > g{ and
s{ =
{
0
1
2
3
4
t{ 0=02449 0=00157 0=00099 0=00069 0=00062
Solution 3.6 Choose a radix of 100> 000= Using the relationships dened
above, the following table is constructed.
{
o{
0 100> 000
1 97,551
2 97,398
3 97,302
4 97,235
t{
g{
s{
0=02449 2,449 0=97551
0=00157 153 0=99843
0=00099
96
0=99901
0=00069
67
0=99931
0=00062
60
0=99938
76
CHAPTER 3. DEMOGRAPHY
We can use the life table to calculate probabilities associated with circumstances of an individual belonging to the population.
Example 3.7 On the basis of the life table in Solution 3.6, calculate the
following:
(a) The probability that a life aged 1 dies before reaching age 3.
(b) The probability that a newborn dies within one year.
(c) The probability that a 1 year old survives to age 4.
(d) The expected number of 4 year olds who attain age 5.
Solution 3.7
(a) This is 1 S u(a 1 year old survives to age 3 exact). Hence
2 t1
= 1 o3 @o1
= 1 0=99745
= 0=00255.
Alternatively, 2 t1 = (g1 + g2 )@o1 .
(b) The probability that death occurs in the rst year is
g0 @o0 = (o0 o1 ) @o0 = t0 = 0=02449.
(c) The probability that a 1 year old survives to age 4 is
3 s1 = o4 @o1 = 97,235@97,551 = 0=99676.
(d) This is o5 , calculated as o4 g4 = 97> 235 60 = 97> 175.
Note that the life table gives integral values for o{ and g{ for the range of
ages shown here. Non-integral values are usually given only at higher ages
where the size of the population becomes relatively small. The Australian
Life Tables 2000—02 show only integer values up to a maximum tabulated
age of 109.
In actuarial work, it is common to assume that lives are independent with
respect to mortality. On this assumption, the probability that ({) and (|)
survive for w years is the product of w s{ and w s| . Similarly, the probability
that ({) is alive at time w and (|) is not, is w s{ (1 w s| ).
Example 3.8 On the basis of the Female Mortality Table, calculate the probability that of two lives now aged 50 and 55,
(a) both survive to age 65,
(b) exactly one of them survives to age 65.
3.3. INDIVIDUAL CHARACTERISTICS: MORTALITY
77
Solution 3.8
(a) The probability that both survive to age 65 is
15 s50 10 s55 =
o65 o65
= 0=7906=
o50 o55
(b) The probability that (50) survives to age 65 and (55) does not is
15 s50 (1 10 s55 ) = 0=0875>
and the probability that (55) survives to age 65 and (50) does not
is
10 s55 (1 15 s50 ) = 0=1098=
Thus, the probability that exactly one of (50) and (55) survives to
age 65 is
0=0875 + 0=1098 = 0=1973=
We nish this section with two more denitions.
Denition 3.20 We dene the function W{ by
W{ =
Z $3{
o{+w gw.
0
r
Denition 3.21 h{ is the complete expectation of life for ({). That is, it is
the expected future lifetime of an individual aged {= It can be shown that we
can calculate this from W{ as
W{
r
h{ = .
o{
Thus, we can interpret W{ as the expected future lifetime of a group of
r
individuals all aged {, since W{ = o{ h{ . (Again, this is just an interpretation
since o{ may not be an integer.)
r
Note that the total expectation of life for a newborn, h0 , does not equal
r
{ + h{ , which is the total expectation of life for an individual who has already
survived to age {=
78
CHAPTER 3. DEMOGRAPHY
Stationary populations
Generally populations of individuals ebb and ow and change in structure
over time. There are, however, cases where we can consider a population to
be in a stable condition, in terms of age and gender distribution, and it is
straightforward to construct a mathematical model to describe this. Having
constructed a model which is tractable we may use it, for example, to estimate
the size of some subgroup of the population some time in the future. In this
section we discuss this kind of situation.
Consider a population into which o0 lives are born each year. Assume that
the mortality experience is also unchanged each year, and that, for all { A 0,
exactly o{ lives survive to age { from o0 births. This describes a deterministic
model for mortality within this population, and gives rise to a special type
of population structure called a stationary population.
A stationary population is one which suers no net change in age distribution or population size from year to year. That is, the numbers and age
characteristics of those leaving and entering the population each year do not
vary. For simplicity we will assume that there is only one means of entry
into the population, and that is by being born. That is, the population is
increased by o0 each year. We will also assume that the only means of leaving,
‘mode of decrement’, is mortality. There will be a number of deaths each
year — g{ deaths at age {. Thus, for the population as a whole,
o0 =
"
X
g{
{=0
since the total number entering the population is the same as the total number leaving it. The total size of the population and the age distribution of
the population do not vary with time.
The life table may be interpreted as a stationary population, since it has
a population structure which ts this model. Let us dene two functions
which are elements of a life table and are used in the analysis of stationary
populations.
Denition 3.22 In a stationary population supported by o0 births p.a., O{
is the number of individuals who are aged {> that is, between ages { and { + 1
exact, at any particular time. We dene
Z 1
O{ =
o{+w gw.
0
Note that O{ ? o{ in general, since some who attain age { will die during
the following year. At any given time then, O{ equals o{ less those who have
3.3. INDIVIDUAL CHARACTERISTICS: MORTALITY
79
died since their {th birthday. If we assume that deaths are evenly distributed
throughout the year of age, then at any time, O{ will represent the group who
are aged { + 1@2 on average, and half of those who will die between ages {
and { + 1 have already died. We may thus use o{ g{ @2 as an approximation
to O{ .
Denition 3.23 In a stationary population supported by o0 births p.a., W{
is the total number of individuals who are aged { or over. That is
W{ =
$3{
X
w=0
O{+w =
Z $3{
o{+w gw.
0
We may use the o{ values from a life table to represent the mortality
experience of a stationary population by simply applying a scaling factor to
adjust for the relative sizes of the two populations. Consider a stationary
population covering the range of ages represented by the Male Mortality
Table (see Appendix 1), and whose mortality experience may be represented
by the o{ values in this table. Suppose, however, that the number of births
per year in this population is 10,000. We calculate a scaling factor such
that
10,000 = o0 , = 0=1.
We can now apply this scaling factor to any element of the standard table,
the Male Mortality Table in our case, to derive the value which applies to
the special population under consideration.
For example, in the Male Mortality Table we have W0 = 6> 909> 573, and so
the total size of the special population is given by W0 = W0W = 690> 957. We
can use this scaling technique to derive values for any stationary population,
of whatever age range, which experiences the mortality of a given life table.
Example 3.9 The membership of a university college alumni is a stationary
population. Members join the alumni at age 21, and enjoy life membership.
There are 100 new male members each year, and 120 new female members.
Their mortality can be represented by the Male and Female Mortality Tables
respectively.
(a) What is the size of the total membership?
(b) How many female members are there over age 60?
(c) How many male members are between the ages of 40 and 60?
(d) How many members die each year?
80
CHAPTER 3. DEMOGRAPHY
Solution 3.9
(a) If * denotes the experience of the alumni membership, there are a
iW
pW
+ W21
members, where p and i denote the gender.
total of W21
We can use the information regarding the number of new entrants
to calculate p and i , the scaling factors for the male and female
population size. We have
p
100 = p o21
, p = 100@97,877,
i
, i = 120@98,625.
120 = i o21
Thus, the total male population size is
pW
p
W21
= p W21
= 100
4,837,525
= 4,942=
97,877
The total female population size is
iW
i
W21
= i W21
= 120
5,595,545
= 6,808.
98,625
iW
i
W60
= i W60
= 120
1,851,109
= 2,252
98,625
(b) There are
female members aged over 60.
(c) This is the number of male members who are over age 40, but who
are not over age 60. That is,
pW
pW
p
p
W40
W60
= p (W40
W60
)
3,004,424 1,224,159
= 100
97,877
= 1,819.
(d) This is a stationary population, and so the total number leaving
the population each year is the same as the total number entering
it. The only mode of decrement is death, and the only means of
entering the population is joining at age 21. The number entering
at age 21 is 100+120, which must be the same as the total number
dying, 220=
3.3. INDIVIDUAL CHARACTERISTICS: MORTALITY
3.3.3
81
Characteristics, causes and trends of mortality
experience
We have discussed a model for mortality, and dened the mortality rate t{ >
enabling us to perform quantitative tasks related to population experience.
Let us now look in more detail at the qualitative aspects of mortality, and
some of the characteristics and trends observed in individual and population
experience.
Mortality risks for a given individual, and population, depend upon many
factors, some of which are apparently much more important than others. It is
generally true that mortality rates increase with age. As we have mentioned
above, this is not strictly true, as there are age ranges and circumstances
under which increases in age can reduce the risk of death. Infant mortality is
generally much higher than for older ages up to about 10 years old. The risk
of death in the rst year of life is often as high as it may be for an individual
in late middle age. Mortality rates decrease steadily until about 10 years of
age, then begin to increase with age again. Around the late teen years and
early twenties, for males especially, there is another range of high risk ages.
There has in the recent past been what is called an ‘accident hump’ in the
graph of mortality rates, associated with the peak in mortality rates for adolescents, associated largely with accidents arising from high risk behaviour.
For example, in Australia this is an age at which people are prone to fatal
car accidents, and also at risk from high suicide rates. This peak around age
20 is currently developing into a plateau-like feature which extends the high
mortality range out to the early thirties. Thereafter, mortality rates increase
steadily with age.
To illustrate this, the ALT 2000—02 (Males) table shows t0 = 0=00567>
a minimum value of t9 = 0=00012> and a ‘plateau’ of values of t{ for ages
{ between 20 and 35. It is not until age 57 that rates are again as high as
they are in the rst year of life, and they rise steadily thereafter. Figure
3.6 shows the mortality rates from the Male and Female Mortality Tables in
the appendices. These are plotted on a logarithmic scale so that the main
features can easily be seen. The pattern of mortality in this graph is typical
of that in a developed country.
Mortality experience also shows a strong link to gender: mortality rates
are almost always lower for females, than for males, and females generally
live longer than males. The dierential varies at dierent ages, being most
marked around the adolescent years. In recent years there has been observed
a new feature at very high ages of over 100, that male mortality rates are
lower than those for females at these ages. This feature has been observed in
successive Australian census data collections, and though data are scant at
these high ages, it is thought to be a ‘real’ development, and not an anomaly.
82
CHAPTER 3. DEMOGRAPHY
1.0000
0
10
20
30
40
50
60
70
80
90
100
Age
Mortality rate
0.1000
0.0100
0.0010
0.0001
Figure 3.6: Male and female (dotted line) mortality rates
The major causes of death for the Australian population include heart
disease and cancers, which between them account for about 50% of deaths.
Other major causes include stroke and respiratory ailments. Australia has a
very high incidence of asthma. Amongst the young, suicide and accident are
major causes of death, and degenerative diseases aect only the older age
groups. Accidents are associated with cars and also with leisure activities
(e.g. drowning). Young males are approximately twice as much at risk from
such causes as young females.
The importance of particular causes of death varies somewhat between
nations and races, and within populations. Some types of diet and lifestyle
are associated especially with some causes of death — for example smoking
with cancers and respiratory ailments, and high fat diets and high stress
lifestyles with heart disease. To a large extent these factors are ruled by
behaviour, which may be linked to social conditions and local customs. In
Japan, for example, where there is raw sh and little fat in the diet, the
incidence of death by heart disease is much lower than, for example, in the
United States. In Australia there is a large immigrant population, and it
is seen that to some degree and for some time after arrival, the benets or
drawbacks of the experience associated with the home country linger. The
eects, however, are not clearly understood.
Infant mortality is often taken to be a strong indicator of the health and
3.3. INDIVIDUAL CHARACTERISTICS: MORTALITY
83
socio-economic prosperity of a population. It is also a factor which quickly
impacts on the population as a whole. In situations where there is a risk
to the health of the general population, such as famine, outbreak of disease,
inadequate standards of sanitation and nutrition, it is the very young and the
very old who are at most risk. In countries where clean water and adequate
food are temporarily or habitually lacking, the rates of infant mortality are
high. Improvements in these basic living conditions have considerable impact
on infant mortality rates. Improvements in ante-natal and post-natal health
care subsequently lower infant mortality rates further.
Mortality rates for all ages have in general been ‘improving’, i.e. falling,
consistently as time passes, and longevity has been increasing. However,
infant mortality has improved to a greater extent than at other ages. Even
in a relatively short period infant mortality rates in Australia, for example,
have shown considerable improvement. In the general Australian population,
infant mortality has fallen from 18.2 per 1,000 live births in 1966 to 4.3 per
1,000 live births in 2009. The rates vary enormously from country to country.
For example, the rate in Singapore in 2008 was 2.1 per 1,000, whilst in many
African countries the rate is between 100 and 200 per 1,000.
Within Australia even, the experience of dierent ethnic communities
is divergent. In 2009, the infant mortality rate for the whole Australian
population was 4.3 per 1,000 live births. However, in the Northern Territory,
where there is a large aboriginal population, the rate was 7.1 per 1,000,
almost twice the national average.
Many factors aect the mortality experience of an individual or population. Clearly for an individual, age and gender are important factors, and it
is generally true that rates increase with increases in age (except in the age
ranges discussed above), and male rates are almost invariably higher than
female rates. To some extent there are links to race and ethnic origin, but
it is not entirely clear how this link works. It may be that much of this
is behavioural rather than genetic. Some life threatening diseases are more
prevalent in, or conned to, particular races and countries. For example,
sickle cell anaemia is almost exclusively conned to people of sub-Saharan
African descent. Many factors linked to increased mortality risk are clearly
behavioural. Among these are accidents associated with ‘risky’ behaviour,
e.g. hang gliding, fast driving, ballooning, ying small aircraft. Some ‘social’
or lifestyle behaviours are associated with high mortality risks. Smoking,
drug use and sexual practices which increase exposure to the risk of AIDS
infections, are obvious and well known examples. Occupations may be identied as risky — mining of any kind, working with chemicals and in high
risk environments such as quarries, on oil rigs, or with dangerous machinery.
Even apparently unrisky occupations have identiably higher risk to health
and life than others. For example, alcoholism and suicide are relatively high
84
CHAPTER 3. DEMOGRAPHY
in the dental and medical professions.
Studies would also indicate that socio-economic status is correlated to
mortality rates. The rich live longer, in general, than the poor. It is not
clear why, whether increased wealth improves access to health services which
prolong life, whether stress is higher in those living in poverty, whether the
wealthier individual simply eats and rests better. There is some link to level
of education achieved and standard of living achieved, and level of education
is to some extent an inherited trait — perhaps the lower mortality rates are
another part of the inheritance. It seems to be a complex of causes.
Interestingly there is an apparent link between longevity and marital
status. Married men generally live longer than men who have never married.
There is also a link for women, but this is not as marked as for males.
In general though, those who have never married have shorter lives than
those who have married. Again, there is no simple explanation for this, but
it should be borne in mind that those who have seriously impaired health
frequently do not marry, and this aects the statistics.
Scientic research is conrming that genetic factors are linked with a
predisposition to some life threatening diseases. Amongst women, for example, breast cancer is a signicant cause of death, especially amongst younger
women, and a genetic link has been identied in this case.
Many however defy genetic predisposition, and there is a complex interaction between genetic, behavioural and environmental factors aecting mortality and longevity. While we may safely identify correlations between some
characteristics or behaviours, it would not be justiable to deduce causes and
eects from these correlations.
3.3.4
Mathematical models of mortality
We require to be able to construct a mathematical model of mortality to
enable analytical representations of mortality rates. The curve of the force
of mortality, { , for humans cannot however be adequately modelled by a
single mathematical function, because it shows dierent characteristics over
dierent age ranges.
There are two well known mathematical models which have been suggested as being appropriate for use over limited age ranges:
Gompertz law: { = Ef{
and
Makeham’s law: { = D + Ef{
where D, E and f are suitably chosen constants. The behaviour of the force
of mortality as a function may roughly be split into three types for three age
ranges:
3.4. INDIVIDUAL CHARACTERISTICS: FERTILITY
85
(i) The age range of approximately 0 to 10 years is a period of rapidly
declining mortality rates.
(ii) In the age range from 10 to 30 years, { is an increasing function,
levelling o to a plateau between 20 and 30.
(iii) The remaining years show an exponential increase in the force of mortality.
These patterns correspond to a negative Gompertz curve ({ = Ef3{ ) for
the rst age range, then a bell-shaped curve as the force of mortality rises
to the plateau in the twenties and thirties, and then a Gompertz type curve
for the remaining years. Various renements are used in order to construct
smooth tabulations of rates using raw census data. The process of producing
tables of smoothed rates is called graduation. Graduation is a formal process
of constructing a smooth curve given the ragged raw data. The reason for
smoothing the raw data, rather than producing tabulations of raw data, is
that the tabulations of rates are used in practical situations which require a
smooth graduation from one age to the next. For example, in the calculation
of premium rates for life insurance products, undesirable consumer eects
may arise if mortality experience from one year of age to the next is modelled as other than gradual. A life table is therefore a representation of a
population’s mortality, and not a literal record of it.
3.4
Individual Characteristics: Fertility
3.4.1
Denition of fertility
Unlike mortality, the event which we are identifying when measuring incidences of fertility requires some careful denition. It is easy to recognise
when a death has occurred, but what kind of event are we counting when
considering fertility? In general, we dene fertility rates in terms of the
number of live births which occur.
There are other events, such as the number of conceptions, miscarriages
and abortions which take place, which are not counted as events associated
with fertility.
Fertility rates dier in their characteristics from mortality rates in several
respects. There is the di!culty of recognising when an event has taken
place. (Is a still birth an incidence of fertility? If a baby is born dead
some months prematurely, is this an incidence of fertility?) Also how do we
dene how many events have taken place? (In the case of multiple births,
are we interested in how many babies are born, or only in the fact that a
86
CHAPTER 3. DEMOGRAPHY
connement has taken place?). The denition of what we mean by fertility
is not a straightforward matter!
Apart from the di!culties involved in actually recognising an incidence of
fertility, there are several other dierences between fertility rates and experience and mortality. An individual may experience fertility more than once,
or not at all, whereas death occurs once and only once to each individual.
Mortality is a risk which exists at all ages and to all individuals, whereas
fertility is a risk to which only women in the childbearing years are directly
exposed. To some extent the exposure to the risk of fertility is a matter of
choice, given that it is entirely a behavioural risk, and even for those who
choose the risk bearing behaviour, contraception is widely available to most.
There are also a number who are barred from fertility for health reasons,
or because they have recently conceived. Generally, no attempt is made to
account specically for these individuals.
Having decided how an event is recognised, in order to construct rates
associated with fertility, we must also decide which population is exposed to
risk, and over what period of time the rates are measured. For example, we
may express fertility rates as the number of children born to a woman in her
reproductive lifetime, or the number born per year, or the number of live
births per number of population per year, or in some other way. In fact rates
are calculated for all of these situations and for others, including the number
of children born into a family (completed family size). Given the changes
which are occurring socially and in family formation, these statistics present
a rich picture of modern life.
We have dened the crude birth rate as the number of births in a period,
per 1,000 of the average population in the period. The crude birth rate,
as with the crude death rate, has the advantages that it is a single gure,
easily calculated, and easy to use in comparing populations. However, it has
the drawbacks that it is dependent upon the age and gender distribution of
the population and therefore must be used with care in comparisons. For
example, a population with a large proportion of women in the childbearing
age groups would clearly be expected to have a relatively high crude birth
rate.
3.4.2
Measures of fertility
Let us dene a number of other measures related to fertility.
Denition 3.24 The general fertility rate is the number of live births per
1,000 women of reproductive age, generally taken to be 15 to 49, per period.
Typically, the period is a calendar year.
3.4. INDIVIDUAL CHARACTERISTICS: FERTILITY
87
Note that this rate uses a population exposed to risk which is more narrowly dened than for the crude rate. Given both the crude and general rate,
we might deduce something about the characteristics of the population.
Denition 3.25 The age specic fertility rate is the number of live births
per 1,000 women in a specic age group, per period.
This rate renes the population exposed to risk even further. Because the
general fertility rate is the number of births per 1,000 women of reproductive
age, it is less dependent on the age and gender distribution of the whole
population. The age specic fertility rates give us information which is also
independent of the age distribution of the females in the population.
Denition 3.26 The total fertility rate (TFR) is the number of live births
per 1,000 women over their reproductive lifetimes, assuming no mortality.
Denition 3.27 The gross reproduction rate (GRR) is the number of live
female births per 1,000 women over their reproductive lifetimes, assuming no
mortality.
Suppose DVI U{ represents the age specic fertility rate between ages {
and { + 1, and DVI U{i represents the age specic fertility rate for female
births between ages { and { + 1= Then the total fertility rate is calculated as
X
DVI U{ =
WIU =
{
Intuitively, if the total fertility rate falls below 2 per woman, a population
should decline, and if it exceeds 2, a population should grow. However, this
is a simplied view of a population, taking account only of births. The gross
reproduction rate is the sum of DVI U{i over the reproductive age range, so
X
DVI U{i =
JUU =
{
The rate usually has a value close to 1 per woman in developed economies,
unless fertility rates are very high. This rate gives a measure of how the
female population is being replaced, at birth. Another reproduction rate is
also used, which allows for the mortality experience of the ospring.
The net reproduction rate (NRR) makes allowance for the age of the
mother, and for the survival rates of the female ospring in the population.
It measures the rate at which female ospring are born and also survive
to the age their mothers were when they were born. This rate not only
measures the rate at which babies are born into a population, but also the
88
CHAPTER 3. DEMOGRAPHY
mortality experience of these children, and the childbearing patterns of the
current population. The net reproduction rate for this population is the sum
(over the reproductive age range) of the product of age specic fertility rates
and survival probabilities, which are calculated on an appropriate female
mortality table. Assuming women giving birth at age { are, on average, {+ 12 ,
the probability that a newborn survives to her mother’s age at the time of her
i
birth is o{+1@2
@o0i where the superscript i indicates female mortality. Thus
QUU =
X
{
oi
i {+1@2
DVI U{ i .
o0
The net reproduction rate for a population will always be lower than the gross
reproduction rate, because some of the female births will not survive to the
age of their mother at their birth. To some extent, the dierence between the
gross and net rates will indicate the pattern of childbearing by age, since the
allowance for mortality will be greater for higher ages of mother. However,
the eect is complex and rates should be interpreted with care.
Example 3.10 The following table gives information regarding the births to
a population of women.
Average number of No. of female No. of male
Age group women in age group
births
births
20 29
1,000
45
49
30 39
1,200
63
65
40 49
1,200
26
28
In this example, we assume that the average age of the women in the ten year
age group from exact age {, is exactly { + 5. Calculate the following:
(a) the general fertility rate for this population;
(b) the gross reproduction rate for this population; and
(c) the net reproduction rate, assuming that the women experience the
mortality of the Female Mortality Table.
Solution 3.10
(a) The general fertility rate is the number of births to the population
in the period, per 1,000 of the average size of the population in the
period. In this case (276@3,400) × 1,000 = 81=18.
3.4. INDIVIDUAL CHARACTERISTICS: FERTILITY
89
(b) Assume that the age specic fertility rate is 45 per 1,000 for each
age in the 20 29 age group, with a similar assumption for the
other two age groups. The gross reproduction rate is the sum of
the age specic fertility rates, i.e.
μ
¶
45
63
26
10
+
+
1,000 = 1,191=7 per 1,000.
1,000 1,200 1,200
Thus, a woman in this population bears, on average, a total of
1.192 female children in her reproductive lifetime.
(c) The net reproduction rate incorporates a survival factor. It measures the rate at which the female reproductive population is reproducing a female population of the same age distribution.
The survival factor is chosen with reference to the original population. In this case we have information regarding the population
in terms of 10 year age groups, and the average age of the age
group { to { + 10. Let us assume that the population is evenly distributed over the age range, with an average age of { + 5. In this
case, therefore, it is appropriate to choose the survival factor for
i
each age range as o{+5
@o0i . This is the probability that a newborn
female child will survive to the age of its mother at the time of
the child’s birth. The female gross reproduction rates and survival
probabilities from birth to the middle age in each age group are
Age group
{ to { + 10 DVI U{i
20 29
45
30 39
52=5
40 49
21=67
Survival
i
probability, o{+5
@o0i
0=98407
0=97716
0=96181
For each age { in the age group 20 29 we multiply DVI U{i
by the survival probability, giving 44=283 at each age. The same
calculation for the next age group gives 51=301 at each age, and
20=839 for each age in the nal age group. Summing over all ages
from 20 to 49 we get the net reproduction rate as 1,164=2 per 1,000
females, i.e. 1=16 per female.
Remark 3.5 Note that in other circumstances a dierent survival factor
would be chosen. For example, suppose the age groups were dened as from
age { exact to less than age { + 5 exact. The average age for this group is
{ + 2=5 years. We might therefore use a survival factor of o{+2=5 @o0 . If we
suppose a uniform distribution of deaths in each year of age, we can say that
90
CHAPTER 3. DEMOGRAPHY
O{+2 is a good approximation to the value of o{+2=5 , and use the survival factor
O{+2 @o0 = We choose the survival factor which matches the characteristics of
the population information which is provided.
In some countries high fertility rates exist alongside high infant mortality
rates, and so the net reproduction rate might actually be relatively low, even
though the gross rate might be high. In populations enjoying high socioeconomic standards of living, both fertility rates and mortality rates tend to
be low. In such populations, the high survival rates of the female children
will lessen the eect of the low fertility rate on the growth of population. In
Australia, the trend is currently for women to have fewer children, and to
have these children later in their life. (The average age for a woman to have
her rst child is currently about 30 years.) Thus, the net reproduction rate
will be decreased by the decrease in the number of births, will be increased
by the generally improving mortality rates, especially the infant mortality
rates, and will be decreased because the generation gap is increasing. That
is, the children must survive longer to attain the age of the mother at their
birth. The interaction of these various eects is complex.
We have discussed rates which are specic to gender and to age, but it
is true that subgroups within a population may experience vastly dierent
fertility and mortality. This dierence is inclined to be related to socioeconomic status, which in turn may relate to behavioural and genetic risks.
The inter-relationship between these various features is complex and not
denitively understood. Within the Australian population, for example, the
indigenous peoples suer much higher rates of infant mortality, some two to
three times the rate of the population at large in some states, and their life
expectancy is much lower.
3.4.3
Factors aecting fertility
There are many factors which aect fertility, and many of these are matters of
choice. The standard of nutrition, sanitation and maternal and child health
care, however, are factors which tend to depend on the stability and economic
condition of a country. These factors are crucial in ensuring the survival of
infants. In emerging economies where such care and standards are put into
place, one of the earliest eects which can be observed is the increase in
fertility rates, and the increase in the net reproduction rate.
There are also genetic factors which may be beyond behavioural control
of the individual, at least to a degree. In Africa, for example, the incidence
of AIDS in the community is extremely high. It is estimated that there are
around 22 million cases of HIV or AIDS in Africa, though many cases have
not actually been reported. It is possible only to guess the impact of this on
the coming generations.
3.4. INDIVIDUAL CHARACTERISTICS: FERTILITY
91
In general, increased prosperity is inclined to lower fertility rates in the
long term, and other factors associated with economic development have
the same eect. As economic expectations increase, so do requirements for
education, and so individuals and couples are inclined to delay the birth of
their children when they can, and also to limit the family size.
For some, philosophical and religious convictions will dictate family size.
Some countries are dominated by a religion which forbids the use of contraception, while in other countries, for example India, governments are active in
promoting the use of the eective contraceptive methods available now. Such
methods are widely available to those who wish to determine the number and
timing of the children they bear. Equally, there are those who consider that
the world is overcrowded already, and so choose not to add to the problem.
In the recent past it was considered by many concerned with global overpopulation and the depletion of world resources to be irresponsible to have
large families. In modern China, family size is restricted by law, and the
state pursues the application of this law to the extent of imposing compulsory sterilisation on individual women. In some societies, large families are
seen to confer status and prosperity. In economies which are not urbanised
or industrial, children provide labour which contributes to the survival of the
whole family or group. In these societies, infant and child mortality rates are
often high. In this case, though fertility rates may be high, net reproduction
rates are not.
3.4.4
Modern trends in family formation
In western society family life is changing. In the middle of the twentieth century, one might have characterised nuclear family life as being one revolving
around a couple who married in their early twenties, or late teens, and with
two or three children. The children would be born while the parents were in
their twenties. The father would typically be in full time employment, and
maintain a steady pattern of work, and change his employer seldom, if ever.
The mother would typically take care of her children and not work, at least
until they were old enough to have left home. She would be on average about
three or four years younger than her husband, and have a lower standard of
education than him, and expect to live some seven to ten years longer than
him. Access to contraception and abortion would be limited, and perhaps
impossible, and births outside marriage were considered to be socially unacceptable. Divorce and separation carried a social stigma, and mostly could
not be obtained unless one party could be shown to have committed some
‘fault.’
Fifty years later, attitudes and behaviours in Australia have changed,
and are still changing. In the 1990s about 25% of births were outside of
92
CHAPTER 3. DEMOGRAPHY
marriage. Couples are quite likely to live in a relationship which is openly
‘de facto’, and not part of a formal marriage. No-fault divorce is possible,
and marital breakdown is a common occurrence, which is not condemned by
society. People who do marry are inclined to do so, for the rst time, at
higher ages. Couples have fewer children, and have them later in their lives.
About 20% of children under fteen do not live with their natural parents. It
is common for a child to be living in a ‘blended family’ situation, where the
children in the household do not share the same parentage, though they may
have one parent in common. Approximately one in three marriages ends in
divorce or separation, and the ‘family unit’ is now likely to consist of a couple
who have re-partnered and both have children from previous partnerships,
who may or may not live with them. It is likely that the woman in the
household will go to work while her children are still at school, at least on a
part-time basis. It is also likely that her partner is nearer her own age, than
the partner of her mother and her generation. The male in the household is
unlikely to experience a stable working life of unbroken full time employment.
He may change his employer several times during his working life, and may
change his career once or twice also. He is also fairly likely to suer periods
of enforced ‘retirement’ or redundancy.
3.4.5
Trends in fertility rates, and consequences
The changing trends in social behaviour aect the demographics of our population, and the adequacy of the socio-economic structure of our society.
We can summarise the trends observed in fertility rates as typied by the
Australian population, as follows:
• Women are having fewer children over their reproductive lifetime.
• Women are choosing to have their rst child later in their life.
• Completed family size is falling.
The consequence is that the generation gap, the number of years of age
between one generation and the next, is increasing. Parents are now typically
30—40 years older than their children, rather than 20—30 years older as was
the case in the middle of the twentieth century. This, coupled with improved
mortality at all ages, results in an ageing of the population. The proportion
of the population in higher age groups is increasing.
This very simple description of changes in behaviour and experience leads
to a ramication of consequences. For example, as the population ages, a
relatively lower proportion of a nation’s resources is required for the education
of the young, and a relatively higher, and increasing, proportion is required
3.5. POPULATION PROJECTIONS
93
for the care of the elderly. In fact, statistics show that the elderly are very
much more expensive to care for than the young, largely due to their increased
medical costs. This, and the changes in family formation patterns make an
impact on the urban and suburban environment. The need for sheltered
housing, and housing for individuals living alone, increases, changing the
patterns of housing development. There is a changed need for schools and
those who teach and service the young and the old. Changing work patterns
render conventional and old established patterns of providing for retirement
inappropriate. Women can no longer rely on their husbands’ superannuation
to provide for them in their old age, and couples may no longer rely on one
another for physical care. With periods of broken and partial employment,
few can adequately provide for their own retirement, and with family support
fragmented also, governments in many countries are seeking solutions to deal
with the new shape of demographic and social structure. While individuals
may look forward to a longer, healthier, more active old age, nations are
struggling to adjust to the emerging conditions.
3.5
Population Projections
3.5.1
Context
There are many dierent reasons why an estimate of the size and distribution
of a national (or other) population might be required. The method chosen
to construct the estimate would depend on the reason and the time scale of
the requirement. We may be interested in local or widespread features of a
population, over the long or short term.
On the national scale, we may wish to estimate the size and age distribution of a national population in order to plan for the provision of necessary
facilities in the future — housing, education, health care, income in retirement, for example. These may involve short or long term estimates. We
may wish to estimate the impact of migration on the current population, in
order to determine adequate levels of immigration to maintain a youthful
population or to maintain, or otherwise control, population growth.
We may wish to calculate some detail of the constituent of a population
at some time in the past, or at some time in the near future. For example,
we may be considering the need for kindergarten and pre-school facilities in
a certain area. There are several things we would ideally like to know in
order to make some estimate of the emerging need. We would need to know
about the adequacy of the current facilities to meet the current population
of very young in the area, and also have some estimate of how many children
might be entering the pre-school years in the next and subsequent years. We
94
CHAPTER 3. DEMOGRAPHY
could look at recent trends in numbers of children enrolled in such facilities,
and establish whether this number was rising or falling. The pattern of
change could be established by looking at recent enrolment data, and be
used to extrapolate over the short term. In addition, information regarding
the number of new babies and toddlers in the local population could be taken
as an indication of how many places would be needed in the very short term,
amended to allow for our expectation of migration in and out of the area.
For example, in areas where new family housing estates are being built, an
increase in the number of young children might be expected as a result of
young families ‘migrating’ into the area. This kind of question requires a
short term horizon, and we can use recent information to give indications of
emerging patterns.
It may be that we wish to investigate a situation which obtains between
census dates, either for its own sake, or in order to give insight into what we
might expect in the future. In this case we can use census data for dierent
years, and interpolate, or otherwise approximate, between the two dates.
In other cases, however, short term estimates based on recent information
may not be at all appropriate. Many of the matters which concern actuaries
are very long term, and require long term projections. Projections involve
assumptions about future experience, and may attempt to model over a very
long time scale. Essentially, projections require extrapolation of some sort or
another, based on a more or less complicated set of assumptions. Projections
may be required of population size, or of the relative size of a constituent
of a population. An example of this which is used much in the context of
superannuation, is the estimated age dependency ratio. Population projections are used to calculate this ratio over very long time scales of fty years
and more. It must be said that in the past some published projections of
national population size have been wildly inaccurate. In the 1940s and 1950s
for example, many industrialised countries experienced periods of prosperity
and growth, and fertility rates rose, and there was much concern that populations would ‘explode’ in size. It was then assumed that high fertility rates
would continue unchecked. In these same countries, it has been found that
this has not eventuated, and in fact some of these nations’ populations would
decline if it were not for growth due to immigration.
Projections indicate that it is populations in the currently emerging nations, evincing the characteristics of demographic transition (particularly
China, India and South America), that are in danger of becoming overwhelming in size, and dominating world population growth. Many countries
in established economic conditions are ageing, and would decline if immigration were not used to support the younger age groups.
3.5. POPULATION PROJECTIONS
3.5.2
95
Projection models
Let Sw represent the size of a population at time w> and suppose that we wish
to construct a population projection to give an estimated value for Sw , given
information about the starting population, i.e. the population at time 0.
There are some common methods which might be used to do this, depending
on the amount of information already known, the detail required, and the
time scale of the projection, or estimate. The dierent methods described
here are appropriate under certain circumstances, and each has its advantages
and disadvantages.
(i) Polynomial model. We start with the most straightforward polynomial
model which is a linear model. Suppose we have information about the
population size at time 0 and time q years, that is we know the values
of S0 and Sq . We assume that the population size changes in a linear
fashion between these two dates, that is Sw = S0 + ew, where 0 w q,
and e is some constant. Since Sq = S0 + eq, e = (Sq S0 )@q, and if w
is such that 0 w q> we have the following equation for Sw :
w
Sw = S0 + (Sq S0 ).
q
This is an estimate based on linear interpolation. It is appropriate
when the time span is short, and when we have information regarding
the population size at two dates which enclose the date in which we
are interested. It has the advantage that it is simple and very easy
to calculate. It would be useful for calculations regarding inter-census
periods.
If w were such that 0 q w, the linear model might be used to
extrapolate a value for Sw , provided that the time w occurred soon after
time q= This model would not be appropriate for long term projections.
More generally, we could model Sw as a polynomial. If data are available
for a number, say p + 1, of dierent dates, then a curve may be tted
to these data. That is, a polynomial of degree p can be constructed
which ts the available data, and population size at some other date
may be estimated from this polynomial. Again, this method may have
use in projections over the short term particularly. Also, it is more
sensitive than the linear model, since it uses more information.
(ii) Geometric model. Suppose again that values are known for S0 and
Sq . The geometric model assumes that the population size follows a
geometric curve. A value can be found for u, the rate of growth per
period, dened by the following equation:
Sq = S0 (1 + u)q .
96
CHAPTER 3. DEMOGRAPHY
This ratio, u, derived from known population data, can then be used
to estimate the population size at time w:
Sw = S0 (1 + u)w .
This estimate is not appropriate for projections into the future over the
long term, since, if u is positive, the population size will ‘explode’ in the
long term. That is, Sw will go to innity as w becomes large. However,
over the short term, or for interpolation between dates, this may be an
appropriate, simple, model.
(iii) Logistic curve. The logistic curve describes the population size at time
w in the following way:
Sw =
1
,
D + Eh3uw
where the constants D> E and u are derived from known data. The
logistic model has some useful features which make it appropriate for
long term projections. Note that the value at time w = 0 is given by
S0 =
1
,
D+E
and the limiting size of the population as w $ 4 is given by
lim Sw =
w<"
1
.
D
This model allows for the size of the population to approach some
limiting value over the very long term, and this makes it suitable for
use in long term projections.
(iv) Component method. The component method is a detailed method,
which does not rely on a specic mathematical model of population
growth, but instead relies on assumptions being made about the experience of the population in terms of its modes of natural, and other,
growth. The population at time w is dened as
Sw = S0 + E0>w G0>w + L0>w H0>w
where:
E0>w is the number of births which occur between time 0 and time w,
G0>w is the number of deaths which occur between time 0 and time w,
L0>w is the number of immigrants into the population between time 0
and time w,
3.5. POPULATION PROJECTIONS
97
H0>w is the number of emigrants from the population between time 0
and time w.
The component method can be much more detailed than the previous
models, and relies upon assumptions about future mortality, fertility
and migration rates. The starting population is commonly analysed by
gender, and by age (quinquennial age groups are frequently used). Let
us consider the case when the grouping is in ve year age groups, by
gender, and that the size of the population is calculated every ve years.
p
Suppose that S{>w
represents the number of males in the quinquennial
age group aged between { exact and less than { + 5 exact, at time w=
Then we can apply a survival factor to this number to calculate the
number surviving for 5 years to enter the next age group, and adjust
this number by the expected number of migrants over the next 5 years:
p
p
S{+5>w+5
= S{>w
p
5 O{+5
+ qhw pljudwlrq,
p
5 O{
p
p
where q Op
{ = W{ W{+q . The estimate of net migration is often expressed as a percentage of the starting population. The survival factor
is the number aged between { + 5 and { + 10 divided by the number
aged between { and { + 5 according to the mortality experience assumed.
This method is adequate for calculating survivors of all age groups except the rst group, i.e. those aged between 0 and 5. The numbers
entering this age group are the male and female babies born during the
ve years, who also survive to the end of the ve years. The number
born during the ve years is calculated by applying an age specic fertility factor to the number of women in the childbearing age groups.
The assumed sex ratio at birth is then applied to this to give the number of male and female births. An average, age specic, survival factor
is then applied to calculate the number of male and female children
surviving to give the size of the population of age less than 5 years
at the end of the ve year projection period. In this case, the survival
factor used is of the form 5 O0 @5o0 . The total number born is o0W per year,
(where denotes the special population), and their survivors represent
the population aged between 0 and 5 years old at the end of the ve
year period. So the total number of births is 5o0W and the population
surviving is 5 OW0 = W0W W5W =
The process is repeated for each ve year projection, with the survivors
of each age group in year w becoming the population in the next ve
year age group at time w + 5=
98
CHAPTER 3. DEMOGRAPHY
Clearly the component method can incorporate specic assumptions about
the components contributing to the change in population size, in ways the
other methods do not. The other methods make an assumption about the aggregate eect of components of population growth. In this sense, the component method may appear to be more accurate, being more detailed. Modern
computation capabilities make it possible to construct component method
type calculations and projections which may be done quickly and relatively
easily. This means it is possible to examine the long term eects of changes
in any of the components assumed. We can see what might happen to population size if fertility rates fall dramatically, or rise signicantly, for example,
or investigate the eect on population demography of changes in the rate
of migration. This enables a study and understanding of the interrelationship between the components of growth, and the sensitivity of the overall
population distribution to the various factors.
The shortcomings of the component method are that the accuracy of the
projections depends upon the accuracy of the assumptions made as to the
long term experience of mortality, fertility and migration of the population.
Typically ‘long term’ may mean fty years. While one can justify various
assumptions reasonably, this is clearly a very long time span over which to
predict such factors. In some respects, fertility may seem to be the hardest
factor to predict, since it appears to involve the largest element of choice.
By and large, couples can exert a fair amount of control over the number
and timing of births to which they give rise. While it is generally agreed
that mortality rates are falling, and it is still true that these fall most in the
infant years, to what extent can this improvement in mortality be expected to
continue? Infant mortality rates are about one third of what they were only
fty years ago, and life expectancies have increased by about seven years over
the same period. What might the next fty years bring in terms of changing
these features? How would one account for the spread of AIDS in an African
or Asian country, where the rate of infection may be as high as 50% of the
adult population?
Figure 3.7 shows Australia’s population (in millions) from 1901 to 2009.
We can see that the rate of population growth has increased in recent years,
and that a model that might have been appropriate for Australia’s population
at the start of the 20th century would not be appropriate at the end of that
century.
Example 3.11 The population of a certain country was 10 million in 1980,
and 12.2 million in 1990. Calculate the size of the population in 2000 using
each of the following models of population growth:
(a) linear growth,
(b) geometric growth.
3.5. POPULATION PROJECTIONS
99
25
Population in millions
20
15
10
5
0
1900 1910 1920 1930 1940 1950 1960 1970 1980 1990 2000 2010
Year
Figure 3.7: Australia’s population from 1901 to 2009
Solution 3.11
(a) Let 1980 stand for year 0, and express the population size in millions. The assumption of linear growth gives a value for the population size in 2000 dened by the following extrapolation:
20
S20 = S0 + (S10 S0 )
10
= 10 + 2(2=2)
= 14=4 million.
(b) Using the same terminology, the geometric model gives a rate of
growth u derived from the following:
S10 = S0 (1 + u)10
, 12=2 = 10(1 + u)10
, (1 + u)10 = 1=22
, u = 0=020084.
We use this to calculate S20 as
S20 = S0 (1 + u)20
= 10(1=22)2
= 14=884 million.
100
CHAPTER 3. DEMOGRAPHY
Example 3.12 The population of a certain country was 20 million in 1985,
23 million in 1990, and 25=5 million in 1995. Fit a logistic curve to these
values and use this to estimate this country’s population in 2000.
Solution 3.12 Let 1985 be time 0, and express the population size in millions. Then we have
1
>
D+E
1
S5 = 23 =
>
D + Eh35u
1
S10 = 25=5 =
,
D + Eh310u
S0 = 20 =
giving
D + E = 1@20 = 0=05>
D + Eh35u = 1@23 = 0=04348>
D + Eh310u = 1@25=5 = 0=03922.
(3.1)
(3.2)
(3.3)
Then, subtracting (3.2) from (3.1) we have
E(1 h35u ) = 0=00652
and subtracting (3.3) from (3.2) we have
Eh35u (1 h35u ) = 0=00426.
Hence
0=00426
,
0=00652
giving u = 0=08505. We then nd that E = 0=01883, and hence D = 0=03117.
Thus, our estimate of the population at time 15 based on this model is
¡
¢31
S15 = 0=03117 + 0=01883 h31=2758
= 27=45 million.
h35u =
3.6
Exercises
1. In a certain country, the population size at the end of a year was 30.43
million. During the year there were 0.74 million births in the population, 0.68 million deaths, 0.18 million immigrants and 0.05 million
emigrants.
3.6. EXERCISES
101
(a) Calculate the natural increase in the population during the year.
(b) Calculate the net migration in the population during the year.
(c) Estimate the crude birth rate for the population.
(d) Estimate the crude death rate for the population.
2. The table below gives information about the populations of two towns
in a certain state.
Raintown
Sunshine
Number in Number of Number in Number of
Age group population
deaths
population
deaths
0—14
15,000
12
16,000
10
15—39
24,500
31
26,300
25
40—64
22,700
15
24,900
17
65+
12,400
80
15,200
120
The number in the population is an estimate based on a recent census
and the number of deaths is the number in the last calendar year.
(a) Calculate the dependency ratio, the youth dependency ratio and
the age dependency ratio for Sunshine.
(b) Calculate the crude death rate for each town.
(c) Calculate age specic death rates for each town.
(d) For this state, the population size is as follows:
Number in
Age group population
0—14
159,000
15—39
260,000
40—64
218,000
65+
68,000
Calculate the standardised crude death rate for each town using
the state’s population as the standard basis.
3. Figure 3.8 shows the population pyramid for Japan in 2009.
(a) Assuming no change in recent trends in mortality and fertility
rates, what will the population pyramid look like in 2049?
102
CHAPTER 3. DEMOGRAPHY
100+
95-99
90-94
Female
Male
85-89
80-84
75-79
70-74
65-69
60-64
55-59
50-54
45-49
40-44
35-39
30-34
25-29
20-24
15-19
10-14
5-9
0-4
6000000
4000000
2000000
0
2000000
4000000
6000000
Population size
Figure 3.8: Population pyramid for Japan in 2009
(b) Assuming fertility rates remain at the 2009 level, but no change
from recent trends in mortality rates, what will the population
pyramid look like in 2049?
(c) Assuming mortality rates remain at the 2009 level, but no change
from recent trends in fertility rates, what will the population pyramid look like in 2049?
4. The following statistics relate to a particular country in 2008:
Youth dependency ratio 23.7
Crude birth rate
10.8 per 1,000
r
h0 for males
78
State with reasons whether the country is most likely to be Cambodia,
Canada or Colombia.
5. On the basis of the survival function
v({) = (1 {@110)3@4
for 0 { 110, calculate
(a) the probability that a newborn life will survive to age 10,
3.6. EXERCISES
103
(b) the probability that a newborn life will die between ages 60 and
70,
(c) the probability that a life now aged 20 will survive to age 40,
(d) the probability that two lives, each aged 60, will both be dead by
age 80,
(e) the force of mortality at age 65.
6. Conrm that the function j({) = exp{{ }, where > A 0, is suitable as a survival function. Find and given that on the basis of
this survival function,
(i) 40 = 410 , and
(ii) the force of mortality at age 25 is 0.001.
7. (a) Consider the function
s
121 {
v({) =
11
for 0 { 121= Verify that this function satises the conditions
to be a survival function.
(b) On the basis of the survival function in part (a), calculate
i. 10 s20 ,
ii. the probability that only one of two lives now aged 35 and 40
survives to age 60,
iii. the probability that a life aged 30 dies between ages 60 and
70,
iv. 35 .
8. On the basis of a constant force of mortality of 0=004, calculate
(a) 10 s30 ,
(b) 5 t15 ,
(c) the age { such that a newborn life has a 90% probability of surviving to {.
9. (a) Complete the missing values in the following extract from a life
table.
{
o{
g{
t{
50 96,251 279
51
285
52
0.00351
104
CHAPTER 3. DEMOGRAPHY
(b) Assume that o50+w = d + ew + fw2 for 0 w 2= Find the values of
d, e and f, and hence calculate O50 .
10. The following is an extract from a life table.
{
o{
g{
t{
85 24,906
0.0930
86
2,240
87
2,152 0.1057
88 18,197 2,051 0.1127
(a) Calculate the missing values.
(b) Calculate the probability that (85) dies aged 87 last birthday.
(c) Assuming that o{ is a linear function of { for 85 { 86, calculate
85=5 .
11. Using the Male Mortality Table, calculate
(a) 10 s40 ,
(b) the expected number of survivors to age 3 from 10,000 newborn,
(c) the probability of a 50 year old dying before age 54,
(d) the probability of a 60 year old surviving for 15 years,
(e) the age at which a life has the highest chance of survival for the
next year.
12. (a) Given that
½ Z {
¾
v({) = exp | g| >
0
show that
½ Z {+w
¾
| g| =
w s{ = exp {
(b) Suppose that { = 0=01 + 0=001{ for 70 { 90. Calculate the
probability that a 75 year old dies between ages 80 and 85.
13. (a) Show that when Gompertz law ({ = E f{ ) applies
¾
½
E {
(f 1) >
v({) = exp
log f
and hence write down an expression for w s{ . (Remember that
f{ = exp{{ log f}.)
3.6. EXERCISES
105
(b) A life table has been constructed using Gompertz law. Given that
10 s70 = 0=61334 and 10 s80 = 0=38228, nd the parameters E and f
and hence calculate 10 s85 .
14. (a) Let v({) = 1 {@$ for 0 { $. Show that
r
h{ =
${
=
2
(b) You are given values of o| for | = {> { + 1> { + 2> = = = . Assume that
o|+w = (1 w)o| + w o|+1 for integer values of |. Show that under
this assumption
o| + o|+1
O| =
>
2
and hence for integer {,
"
1 X
h{ = +
w s{ =
2 w=1
r
15. The membership of a certain society has been stationary for a large
number of years. It is supported by 2,000 new entrants each year at
age 35. These individuals are subject to a constant force of mortality
throughout their lifetimes of 0.02.
(a) Show that o35+w = o35 exp{0=02w}.
(b) Calculate the total membership of the society.
(c) Calculate the number of members between ages 40 and 60.
(d) Calculate the age { such that one half of the membership is aged
less than {.
16. The population of a retirement village has reached a stationary condition, and experiences the mortality of the Male and Female Life Tables.
Entrants to the village are accepted only at age 70 exact. There are
20 female entrants per year, and 15 male entrants. At exact age 90, all
individuals leave the retirement village and move to a sheltered housing
facility. Calculate
(a) the number of men living in the retirement village,
(b) the number of female residents who die each year,
(c) the number of people in the village aged between ages 85 and 90,
106
CHAPTER 3. DEMOGRAPHY
(d) the number of people who move into the sheltered housing facility
each year.
17. The population of a certain country is in a stationary condition and the
survival function is given by v({) = h3{@70 for { 0= There are 350,000
births per annum into the population, and there is no migration. In
this country
• all adults aged over 20 and under 60 contribute $[ a year to a
social insurance scheme (with contributions being made continuously throughout each year of age),
• a funeral benet of $3,000 is paid on the death of any person over
age 60,
• an annuity of $20,000 p.a. is paid continuously to each person in
the population aged over 60.
The contribution income each year equals the benet outgo. Calculate
[.
18. A western European country produces population life tables for males
and females, each with a radix o0 = 100,000= State with reasons which
of the following pairs you would expect to be larger:
(a) t0 for males and t20 for males,
(b) o70 for females and o70 for males,
(c) g70 for males and g90 for males.
19. The table below shows data for the last calendar year for a large city.
Age Number of Number of
Number of
group
females
male births female births
15—19
44,300
451
425
20—29
92,000
6,345
6,213
30—39
87,500
2,866
2,794
40—49
83,200
333
321
(a) Calculate the age specic fertility rate for each age group.
(b) Calculate the total fertility rate for a female from this city.
(c) Calculate the gross reproduction rate for a female from this city.
3.6. EXERCISES
107
(d) Assuming that female mortality is such that the probability that
a newborn female survives w years is 0=998w , calculate the net reproduction rate for a female from this city.
20. A student has been asked to analyse fertility data for New Zealand
from 2008. The student has calculated the following gures:
Gross reproduction rate 3,017 per 1,000,
Net reproduction rate
2,988 per 1,000.
State with reasons whether or not you think the student’s calculations
are correct.
21. The following data relate to Australia in 2009.
Female
Age group ASFR population
15—19
16.7
727,168
20—24
53.8
782,583
25—29
102.2
791,698
30—34
123.9
751,566
35—39
68.7
814,971
40—44
14.2
769,345
45—49
0.7
793,905
(a) Estimate the gross reproduction rate, stating any assumptions
that you make.
(b) Estimate the number of female births in Australia in 2009, stating
any further assumptions that you make.
22. The population of a certain country was 5.2 million on the census date
1992. On the basis of linear population growth at 2% p.a., estimate
(a) the country’s population on the census date in 2002, and
(b) on which anniversary of the 1992 census date will the country’s
population exceed 7 million.
23. Repeat the previous exercise assuming population growth at a compound rate of 2% p.a.
24. In a certain country, censuses are held every ten years. The estimates
of the country’s population on the three most recent census dates are
25 million, 26.2 million and 27.1 million respectively. Use a logistic
curve to estimate the country’s population at its next census.
108
CHAPTER 3. DEMOGRAPHY
25. A remote island is inhabited by a certain type of animal. The table below shows the female population aged { (last birthday) and age specic
fertility rates (female), denoted DVI U{i .
{ Population DVI U{i
0
2,000
0
1
1,500
0.49
2
1,400
0.52
3
1,200
0.48
4
1,000
0
5
600
0
6
200
0
Use the component method to project the population of female animals
at times 1> 2> = = = > 5 years from the present assuming v({) = 1 {@7 for
0 { 7 and fertility rates are as in the table above. State all
assumptions that you make. You will nd it easiest to answer this
exercise using a spreadsheet calculation.
Chapter 4
Actuarial Practice
4.1
Introduction
In this chapter we describe some of the areas in which actuaries work. Historically, the actuarial profession grew out of the life insurance industry, but
nowadays actuaries are involved in much more than life insurance.
Nevertheless, insurance companies remain a major employer of actuaries
and actuarial skills are required in the pricing and marketing of insurance
products, in the design of new policies, and in safeguarding the interests of
an insurance company’s policyholders.
We start by discussing some basic principles of insurance. Then we describe some life insurance products, and consider some actuarial aspects of
life insurance. We also discuss other areas in which actuaries contribute, and
explain the actuary’s role in these areas.
4.2
Principles of Insurance
The basic idea underlying insurance is that it provides nancial protection
to individuals. Individuals transfer their nancial risks to insurers. Insurers
manage these risks by pooling them. That is, insurers manage the risks of
large groups of people seeking protection against similar risks.
As a very simple example, consider a group of one hundred individuals
who each face the possibility of a loss of $1,000 in the coming year. Suppose
that, for each individual, the probability of the loss occurring is 0.01, so that
each individual’s expected amount of loss is $10, i.e. 1% of $1,000. The
individuals could act in one of two ways. Each individual could absorb their
own loss and pay the $1,000 if necessary. On average, we would expect that
only one of these individuals would actually have a loss, since the probability
109
110
CHAPTER 4. ACTUARIAL PRACTICE
of a loss occurring is one in one hundred. Alternatively, the individuals
could pool the risks. They could each pay $10 (each individual’s expected
loss) into a fund, giving a total of $1,000. This provides enough funds to meet
the expected loss of the group. By paying a small amount, each individual
is protected against the possibility of a substantial loss.
The basic principle of insurance is pooling of risks. An insurance company
sells insurance policies which allow individuals to share the risks they face
with other individuals facing similar risks. The pooling of risks in this way
means that rather than face an uncertain loss amount (which could be very
large) an individual simply pays the cost of insurance, which is usually based
on the average loss amount. Insurance is thus attractive to those who prefer
to pay a certain, but relatively small, amount rather than run the risk of a
large (perhaps catastrophic) loss.
Insurance companies make their business out of risk. Much of the responsibility of an insurance company actuary is managing risk. Basic tasks for
an actuary are therefore assessing and pricing risks.
4.3
Life Insurance
The idea behind a life insurance policy is a simple one. An individual enters
into a contract with an insurance company under which the insurance company agrees to pay a sum of money to the individual’s estate on the death of
the individual. In return for this benet, the individual makes a one-o payment, or a series of payments at regular intervals, to the insurance company.
This is a very simple example of a life insurance contract. Before discussing
this contract, and variations of it, let us introduce some terminology.
The contract is referred to as an insurance (or assurance) policy, or simply
a policy, and the individual becomes a policyholder once the contract is
eected. The benet that is paid under a life insurance policy is known as
the sum insured. If the policyholder secures the benet by a one-o payment,
the payment is known as a single premium. If the benet is secured by a
series of regular payments, these payments are known as premiums.
Nowadays, there are two distinct types of life-insurance policy — traditional and unitised policies. We start by giving a description of the three
most important traditional life insurance policies — a whole life insurance
policy, a term insurance policy, and an endowment insurance policy.
4.3.1
Whole life insurance policy
Under a whole life insurance policy, the benet is payable on the death of the
policyholder, whenever this may be. In practice, the benet is actually paid
4.3. LIFE INSURANCE
111
a short time after a policyholder’s death, the delay being due to legalities,
e.g. the insurance company would want to see a death certicate. Typically,
whole life insurance policies are secured by regular premiums, and these
premiums may be payable monthly, quarterly or annually in advance, or
at some other frequency. The premiums may be payable throughout the
whole of the individual’s life. Alternatively, they may be payable only for a
maximum number of years. For example, an individual might choose to pay
premiums throughout their working life.
A key point about premiums, applicable to all types of life insurance
policy, is that they are payable in advance. To see why this is so, consider
what would happen if all insurance companies collected premiums annually
in arrear. A policyholder could then eect a policy without paying a premium
for a year, i.e. have insurance cover without paying for it. At the end of the
rst year, the policyholder could simply leave the insurance company, eect
another policy with a dierent company, and enjoy a second year of insurance
cover without paying a premium. In such a situation, a policyholder could
be insured for life without paying any premiums! We mention in passing that
the conditions of a life insurance policy are usually such that it is unattractive
for a policyholder to withdraw from the policy.
Since a whole life insurance policy provides a benet only on the death
of a policyholder, the market for such policies includes individuals who have
dependants. For example, a parent might eect such a policy to provide
the means to pay o a mortgage and provide support for children until they
turned 21, should the parent die. A whole life policy may be used to protect
someone or some organisation (for example a small business partnership)
against losses arising from the policyholder’s death.
The historical roots of life insurance lie not just in insurance companies,
but also in friendly societies. Modern friendly societies operate very much
like insurance companies, but in nineteenth century industrial Britain, where
these societies started, these were usually societies of people with a common
interest, and one common interest was to have su!cient funds at death to
avoid the ignominy of a pauper’s grave. Today, providing the cost of a funeral
remains a reason why people eect whole life insurance policies.
4.3.2
Term insurance policy
A term, or temporary, insurance policy is very similar to a whole life insurance
policy, the dierence being that the sum insured is payable on death only
if it occurs during a specied period, known as the term of the policy. For
example, if an individual aged exactly 40 eected a term insurance policy
with a term of 20 years, then if the individual died at age 65, the sum insured
would not be payable because the term of the policy stipulates that death
112
CHAPTER 4. ACTUARIAL PRACTICE
must occur before age 60 if the sum insured is to be payable. The contract
ceases at the end of the term. Thus, unlike under a whole life insurance
policy, it is not certain at the issue of a term insurance policy that the sum
insured will be paid.
Regular premiums can be paid with any frequency. For this type of
business, single premium policies are not unusual. There are two reasons for
this. First, the term of the policy can be quite short — often less than one
year. Second, if a policy is eected at a fairly young age, and the term is
not very long, the single premium may not be very high. This is because the
probability of the death benet being paid is quite small.
Reasons for eecting term insurance policies are varied. A common reason
for an individual eecting such a policy is to provide funds to pay o a
mortgage, or to protect dependant children until they are old enough to
support themselves. Related to the former reason is a variation on a term
insurance policy known as a decreasing term insurance, under which the
sum insured decreases from year to year in a fashion that broadly mirrors
the decrease in capital outstanding under a loan. Term insurances with a
short term are a popular way for the relatively young to obtain low cost life
insurance cover, since mortality rates are relatively low for ages under 40.
4.3.3
Endowment insurance policy
An endowment insurance policy is dierent from the two types of policy
mentioned above in one crucial respect — the policyholder’s death is not the
only event that triggers payment of the sum insured.
An endowment insurance is a policy with a xed term. The benet under
such a policy is paid either if the policyholder dies before the end of the
policy’s term or if the policyholder survives to the end of the policy’s term.
The benet is paid at the time of the policyholder’s death, if this occurs
during the policy term, or at the end of the term if the policyholder survives
until then. This illustrates a key property of life insurance policies — the
contingencies under which benets are payable can be death or survival.
(Other contingencies include severe illness — see Section 4.3.5 on trauma
insurance). The benet under an endowment insurance is certain to be paid
because the policyholder must either die before the end of the policy’s term
or survive until that time.
As with policies discussed previously, regular premiums can be paid with
any frequency. Single premium endowment insurance policies are very unusual because the term is normally quite long, say 20 years, and a single
premium would be very expensive. Notice that we can deconstruct an endowment insurance into two policies — one providing a death benet on death
within the policy term (i.e. a term insurance), and the other providing a ben-
4.3. LIFE INSURANCE
113
et on survival to the end of the policy term. This latter type of policy, which
pays no benet on death, is known as a pure endowment.
Endowment insurances may therefore be viewed as a policy with two components — an insurance component (death benet) and a savings component
(survival benet). These policies are therefore eected by individuals who
expect to survive the policy’s term. (If this were not the case, individuals
would eect term insurances at a much smaller cost!) There are many reasons why individuals would want to save. For example, a person might want
to save towards retirement, or to pay o a housing loan.
4.3.4
Unitised insurances
Unitised insurances have the same ingredients as traditional policies. Policyholders pay premiums to the insurance company and the company pays
benets as specied under the policy’s conditions. The major dierence from
traditional insurance products is that premiums are allocated to specied assets, and the amount of benet paid under a policy depends directly on
the investment performance of these assets. That is, the sum insured varies
depending upon the value of the units purchased by the premiums.
Unit linked policies
Under a unit linked insurance policy, premiums buy ‘units’ in a unit fund.
When a policyholder dies, the death benet is simply the value of the units
owned by the policyholder (and similarly when a policy matures). A policy
may have a guaranteed minimum death benet, so that if the policyholder
dies very soon after the policy is issued (or if the investment performance of
the insurance company is extremely bad) the death benet would be at least
the guaranteed amount.
The unit fund is an internal fund, created by the insurance company
for this particular line of business. The insurance company is responsible
for investing the assets of such a fund. Typically, an insurance company
has many such funds, and policyholders have a choice of where they want
their premiums invested. For example, funds may be geographical in nature,
investing only in shares of companies in a given region, e.g. a European fund,
or a fund may invest solely in a particular type of asset, e.g. xed interest
bonds.
To give an idea of how such a policy works, suppose that the price of
units in a fund is as follows:
Date 1/6/06 1/6/07 1/6/08
Price $2.00
$2.50
$3.00
114
CHAPTER 4. ACTUARIAL PRACTICE
Suppose that the policy has an annual premium of $1,000 and has a guaranteed minimum death benet, payable at the end of the year of death, of
$5,000. Let us suppose that the insurance company allows for expenses by
deducting a certain percentage of premium paid (we give a brief description
of how these can arise in Section 4.3.8) and invests 60% of the rst premium
in units, and 98% of subsequent premiums in units. Then, on 1/6/06, there
is $600 to be invested, purchasing 300 units. Similarly, on 1/6/07, there is
$980 to be invested, purchasing a further 392 units. Now suppose that the
policyholder died between 1/6/07 and 1/6/08. Then, there are 692 units
attaching to this policy at 1/6/08, worth a total of $2,076. Since this is less
than $5,000, the death benet paid would not be the value of the units, but
would be $5,000.
With unit linked policies, the insurance company passes the investment
risk on to the policyholder. If the company has a good investment performance, and the unit values show good growth, this is passed directly on to
the policyholders. Similarly, a poor investment performance by the company means that policyholders’ benets may be lower than they anticipated.
Typically, unit values vary fairly frequently.
Investment accounts
Such policies are similar in operation to unit linked policies, but instead of
the benet depending on the performance of units, the benet is equal to
the accumulation of the premiums. The rate of accumulation credited to an
account in any year may not be the same as the rate of return earned by
particular assets in that year. Instead, the return credited to the accounts
is often smoothed, so that it is relatively stable. For example, if returns in
successive years were 10.5%, 8% and 11%, premiums paid at the start of
these years might be accumulated at 9% each year. This approach leads to a
smooth progression of the accumulation of premiums. Such a policy is very
much like a savings account with an element of life insurance.
Unitised insurance policyholders
Each of the products described above has an element of life insurance. Other
than that, each is very similar to other nancial vehicles — we may compare
unit linked policies to unit trust investments (i.e. investing in shares through
a company specialising only in investments), and investment accounts to bank
accounts. The policies may then be viewed as a combination of life insurance
and another product. Why do individuals not purchase these separately?
There are at least two reasons for this. First, the expense involved in dealing
with an insurance product may well be less than the expense of buying two
4.3. LIFE INSURANCE
115
separate products. Second, in some countries, there are tax incentives associated with insurance policies that do not apply to other investment vehicles.
It may therefore be desirable on tax grounds for individuals to purchase insurance company products, even if the element of insurance in the product
is very small.
Example 4.1 An Australian insurance company sells an investment bond.
This bond provides a means for policyholders to save, and to have life insurance cover.
The bond has a term of at least 10 years. During this period, premiums are
invested in a unit fund selected by the policyholder. The investment earnings
of the fund are taxed, the tax being payable by the insurer. At the end of
the 10 year period (or later), the policyholder is entitled to the accumulation
of the premiums and is not required to pay tax on the dierence between the
amounts invested and the accumulation of these amounts.
The premium under this policy is eectively the amount the policyholder
wishes to invest. The policyholder has the option of paying a single premium,
or of making regular premium payments, subject to the constraint that the
amount paid in premiums in any one year cannot be more than 125% of the
premiums paid in the previous year. If this constraint is violated, for tax
purposes the 10 year period is taken as starting anew.
Premiums are used to buy units in a fund, and the insurance benet under
the bond — payable on death, disablement, or withdrawal (e.g. at the end of
10 years) — is the accumulation of the premiums less charges imposed by
the insurer. These charges include a management charge (e.g. to cover the
insurer’s expenses) and an investment charge (to cover the cost of dealing in
bonds and/or shares).
Policyholders can choose to have their premiums allocated to one of three
funds. Essentially these funds can be described as low risk, medium risk and
high risk. The low risk fund is designed to provide steady, but unspectacular
returns. The main investments of this fund are xed interest loans, such
as bonds and mortgages. The high risk fund is designed to provide high
returns, but there is much greater volatility in unit prices in this fund as the
prices of the underlying assets, particularly shares, can uctuate considerably.
The medium risk fund is designed to steer a course between the other two.
Policyholders may switch between funds. For example, a policyholder who has
been saving towards retirement via the high risk fund might choose to switch
to the low risk fund a few years before retirement. Since low risk funds tend to
produce stable returns, such a strategy simply allows a policyholder to protect
what they have accumulated over the last few years of investment.
Because this is a savings product, the insurer requires very little in the way
of underwriting. Prospective policyholders will eect this type of policy only
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CHAPTER 4. ACTUARIAL PRACTICE
if they expect to survive a reasonable period. The policy carries withdrawal
penalties if the contract is terminated within its rst few years, and there is
also a tax liability for the policyholder in these circumstances. The policy is
therefore unattractive to anyone who does not anticipate being a policyholder
for at least 10 years.
The policy is available to anyone aged over 10, although children aged
10 to 16 require written parental consent to purchase a bond. Parents or
relatives may purchase a bond for children under 10. The bond is aimed at
policyholders who wish to save, particularly individuals who pay income tax
at the highest rates.
4.3.5
Trauma insurance
Trauma insurance is a product which was introduced to insurance markets in
the early 1990s. Trauma insurance provides a policyholder with cover against
some specied event severely aecting health, for example the policyholder
has a stroke. The benet under such a policy is normally a lump sum payment
at the occurrence of a specied event. The insurance is designed to oer
protection against the event of a severe illness which involves considerable
medical (and other) expense and loss of earnings or capacity.
Typically, insurance premiums are payable on an annual basis. However,
this product is not normally sold by level annual premiums. Insurance companies usually guarantee cover to an individual, but do not guarantee the
premium rate at the issue of a policy. Thus, if the insurance company perceives that the risk of payment under a policy increases as a policyholder
ages, the insurer will increase the premium from year to year.
Example 4.2 A leading Australian insurer oers trauma insurance which
provides a lump sum benet to the insured life if the insured life survives for
fourteen days following the diagnosis or occurrence of one of a specied list
of conditions. These include:
• Heart disorders, for example a heart attack or open heart surgery;
• Nervous system disorders, for example Parkinson’s disease or motor
neurone disease;
• Body organ disorders, for example loss of speech or chronic liver failure;
• Blood disorders, for example occupationally or medically acquired HIV;
• Other events, for example being unable to perform any two activities
of daily living (these include dressing and feeding oneself, being able
to go to the toilet independently, being able to control one’s bowel and
bladder, being able to get into a chair or a bed independently).
4.3. LIFE INSURANCE
117
The policy conditions give a very strict denition of what is meant by each
medical condition, and the lump sum benet will be paid only if the denition
in the policy conditions is satised.
Premiums for this policy can be paid monthly, quarterly, half-yearly or
annually. A prospective policyholder must provide the same sort of information to the insurer as would be required in a proposal for life insurance.
The main factors which aect the premium are a policyholder’s age, sex and
health condition at the date of the proposal for insurance. The insurer accepts
proposals from lives aged 18 to 59. Future premium rates are not guaranteed
by the insurer, and may increase from time to time. The insurer will not,
however, cancel a policy simply because a policyholder’s circumstances (e.g.
occupation or state of health) have changed.
The policyholder selects the sum insured, up to a maximum of $1,000,000.
The insurer, however, requires a minimum level of premium each year. In
addition to the premium, the policyholder must pay a policy fee each time a
premium is payable (e.g. $5 per monthly premium, $50 per annual premium)
and there is an extra charge (to cover administration costs) if premiums are
payable more frequently than annually.
Insurance cover under this policy is available until the policy anniversary
preceding the insured life’s 70th birthday. The insurer oers this policy as
a stand alone policy, but it can also be purchased in conjunction with life
insurance.
As with most forms of life insurance, the insurer will not pay a benet if a
claim occurs due to self-in icted injury. There are additional circumstances
under which the insurer will not pay a benet, for example if an insured
person is diagnosed with cancer within three months of eecting a policy.
4.3.6
Disability insurance
Disability insurance is designed to provide an income to individuals who are
unable to earn wages or salary during periods of disability. For example,
a self-employed dentist would be unable to work with a broken arm. A
disability insurance policy could provide such an individual with an income
until a return to work was possible. It is available only to those who have
an earned income, and does not cover those involved in unpaid work, such as
care of your own children. It is dierent from trauma insurance, in that it is
designed particularly to provide income to compensate for loss of earnings,
and not an immediate lump sum payment.
The policyholder pays premiums throughout the term of the policy at an
agreed frequency, e.g. monthly. The benet under the policy is an income
during ‘disability’, and the maximum amount of the benet is usually 75% of
income (from wages or salary) foregone due to disablement. The policy will
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CHAPTER 4. ACTUARIAL PRACTICE
state in detail exactly what disability means, and the denition of disability
may vary from one insurer to the next. Usually, it is required that an individual is healthy when a policy is issued. If a policyholder becomes disabled
(according to the policy’s denition of disability) during the policy term, the
benet becomes payable for a specied maximum term, e.g. for two years or
until the policyholder reaches age 65.
A feature of these policies is that they are guaranteed renewable. This
means that when the term of the policy has elapsed, the policyholder can
renew the policy. Renewal is usually at a higher premium, depending upon
age, since policyholders are more prone to claim as they get older.
Disability insurance policies have what is known as a waiting period. If a
policyholder becomes disabled the policyholder may not make a claim until
the end of the waiting period and will receive no income for disability during
the waiting period. For example, if the waiting period was three months
(which is a typical waiting period in Australia), and a policyholder fell sick
on 1 June, the policyholder could start to receive benet on 1 September if
they were still disabled then. Since sickness or disablement for short periods
is more common than for long periods, the longer the waiting period is,
the cheaper the policy is to provide, and the lower the premiums will be.
Administering claims for this type of policy is expensive, given the di!culties
with identifying, proving and monitoring the condition of disability.
An important distinction between life and disability insurance is that,
in the former, there is no doubt whether a claim has occurred — a person
is either alive or dead at a given point in time. By contrast, even with a
tightly worded denition of disability, it can be a matter of debate as to
whether a person is entitled to claim at a given point in time. Thus, under
disability insurance, an insurer is subject to what is known as moral hazard.
The opportunity exists for a dishonest policyholder to make a fraudulent
claim. It is too expensive for insurers to monitor the condition of all their
policyholders who are claiming.
Example 4.3 A leading Australian insurer oers disability insurance. The
maximum disability benet payable under this policy is 75% of the insured
person’s salary, where salary is dened as pre-tax salary over the last twelve
months before claiming. The minimum amount of monthly benet that is
available is $1,500. For self-employed individuals, there is also the possibility
of increasing cover to include expenses associated with business.
The insured person has a choice of seven dierent waiting periods, ranging
from two weeks up to two years. As the waiting period increases, the premium
decreases. Premiums can be paid monthly, quarterly, half-yearly or annually.
As with other forms of life insurance, the key determinants of the premium
are a person’s age, sex and health at the date of proposing for insurance. The
4.3. LIFE INSURANCE
119
insurer accepts proposals from people who are aged between 18 and 55 and
who are in full time employment.
Cover can continue up to the policy anniversary preceding the insured
person’s 55th, 60th or 65th birthday. Insurance cover is guaranteed during
this period, even if the insured person’s circumstances change, but the level
of premium will not be aected by such changes. Premiums may, however,
increase as the insured life’s age increases.
The policy oers ve dierent maximum benet periods, from one year
up to lifetime. The shorter the benet period is, the cheaper the premium will
be.
This cover can be obtained as a stand alone policy, or in conjunction
with life insurance. The insured life need not be the person buying the insurance policy. For example, a professional person might take out insurance to
provide protection should a business partner become disabled, providing the
policyholder with an income to employ a temporary replacement.
Remark 4.1 It is possible for individuals to eect other types of insurance
(including life insurance) on another person’s health or life, if they can show
‘insurable interest’. In general, an insurable interest means that the person
who is eecting the policy depends nancially in some way upon the condition
of the person they wish to insure. However, it is not generally permissible to
insure the life of a child.
4.3.7
Reverse mortgages
A reverse mortgage is an example of a more general type of product known
as an equity release product. The basic idea of these products, which are
issued by banks and insurance companies, is to provide funds to individuals
or couples who are around retirement age and who possess a home, but do
not have much in the way of savings or potential income. Such individuals
are popularly referred to as ‘asset rich, income poor’.
The idea of a reverse mortgage is that the individual (or couple) borrows
a sum of money based on the value of their home. The amount borrowed is
a percentage of the value of the home and might range from 15% to 40%.
Generally, the lower the borrower’s age, the lower the percentage that can
be borrowed. Once the loan is in place, no repayments are required until the
borrower leaves the home, either through death — in the case of a couple on
the second death — or due to incapacity. Meantime, the loan accumulates
with interest, and the loan must be repaid once the borrower leaves the home.
The idea is that the borrower has access to the equity in the home during
the borrower’s lifetime, and the loan will be repaid from the sale proceeds
of the home. From the borrower’s point of view it has the advantage of
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providing funds for retirement. A potential disadvantage for the borrower, if
the borrower wishes to leave an inheritance to family members, is that the
borrower’s estate will be signicantly reduced.
From the borrower’s point of view, the risk associated with entering into
such a contract is that the sale proceeds of the home could be less than
the accumulated amount of the loan, leaving the borrower’s estate with the
problem of repaying part of the loan. To prevent such a situation, many
reverse mortgage contracts include a no negative equity guarantee (NNEG).
This guarantees that the borrower’s debt to the lender is capped at the sale
price of the home. Thus, for example, if the accumulated amount of a loan
was $300,000 and on the borrower’s death the home sold for $280,000, there
would be a shortfall to the lender of $20,000, whereas if the home sold for
$350,000, the loan would be repaid in full and the borrower’s estate would
get $50,000 from the sale of the home.
The NNEG makes a reverse mortgage more attractive to borrowers, and
hence allows lenders to conduct a greater volume of business, but it creates a
risk for the lender. One way in which the lender can compensate for this risk
is to charge a higher rate of interest on a reverse mortgage than it would on
a standard loan. In determining the terms of a contract, the lender needs to
model the mortality of the borrower as well as the likelihood of the contract
ending because the borrower exits the contract for other reasons, in particular
due to disability. (Repayment of loans is possible, but does not represent a
risk to the lender.) House prices, both current and potential, must also be
taken into consideration. The techniques commonly used to put a cost on
the NNEG are well beyond the scope of this book.
4.3.8
The role of the actuary in life insurance
Underwriting
Underwriting is the process by which an insurer decides which risks (i.e. lives
in life insurance) it is prepared to cover, and under what conditions. Insured
lives are subdivided into homogeneous groups and a premium is calculated
for each group. Subdivision usually occurs on the basis of age, sex and other
rating factors. For example, most insurance companies oer life insurance
at lower premiums to non-smokers than to smokers, because the mortality
experience of their policyholders indicates higher mortality rates for smokers
than for non-smokers.
Insurance companies are under no obligation to accept proposals for insurance, be it life insurance or any other type. The rst task for an insurer
in considering a proposal for insurance cover is to check whether there is
anything in the proposal which suggests that it is an unusual risk. For ex-
4.3. LIFE INSURANCE
121
ample, in a proposal for life insurance, prospective policyholders might be
asked questions about, say, any history of heart disease in their family, or
whether they indulge in high risk activities such as hang gliding. Proposals
which indicate that the proposer may be at risk to higher mortality are not
necessarily rejected. Life insurance has existed for many years, and the extra
mortality risk posed by a condition such as obesity may simply be re ected
in a higher premium.
An issue facing life insurers is selection — both self-selection and adverse
selection. Selection occurs when there is some feature or characteristic of
a policyholder which means that they are in some way dierent from the
general population, so that they belong in a select group. For example, selfselection would occur when an individual wishes to purchase a life annuity
by a single premium. (A life annuity is a contract under which a policyholder pays a single premium to an insurer in return for regular payments
to the policyholder throughout their lifetime.) A rational individual would
not make a substantial outlay to an insurance company for an annuity if the
individual’s own assessment of their survival prospects was other than good.
The self-selection eect in this case would be increased longevity relative to
the general population. Adverse selection (that is selection not in the o!ce’s
best interest) would occur when an unhealthy individual, anticipating a high
likelihood of early death, wishes to take out life insurance. Clearly it is not
in an insurance company’s interests to issue a life insurance policy to an individual who seems likely to pay only a few premiums. Proposal forms for
insurance by prospective policyholders contain questions which attempt to
identify the danger of adverse selection.
It should be noted that proposers for insurance are normally legally bound
to disclose all relevant information to an insurer. If a proposal is accepted,
and it is subsequently found that the policyholder has lied in the proposal,
the insurer may not be legally bound to pay the sum insured.
Reinsurance companies often provide assistance or advice on underwriting. Reinsurance companies are simply companies which eectively share
risks with an insurer. For example, if an individual wished to purchase a
one-year term insurance with a sum insured of $1 million, an insurance company may issue the policy, but could protect itself from paying all the sum
insured by eecting reinsurance with one or more reinsurers. The o!ce issuing the policy would pay a premium to the reinsurer(s). Reinsurers generally
accept many kinds of insurance risks and thus have built up a store of information about risks which are considered to be non-standard.
Example 4.4 A leading Australian insurance company oers a term insurance policy which pays a benet either on a policyholder’s death or if a policyholder is diagnosed as terminally ill with less than 12 months to live. The
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company does not require a medical examination of proposers. However, it
does ask proposers questions on the following topics, requiring details in some
cases.
1. Smoking: have you smoked any substance (not just tobacco!) in the
past year? If so, what, and in what quantities?
2. Drinking: do you have more than 14 standard alcoholic drinks in a
week, and if so, what is your average daily consumption?
3. Sex life: do you or your sexual partner(s) have HIV infection or AIDS,
or are you at risk of exposure to these?
4. Hazardous pursuits: do you engage in hobbies such as diving, motor
racing, ying, or other such pursuits?
5. Health: have you ever been advised about an adverse health situation,
e.g. cancer, heart condition, mental illness?
6. Medication: are you taking any medication or shortly due to have a
medical procedure or test?
7. Insurance history: have you ever been refused insurance or oered insurance at special rates?
Thus, although the insurer does not require a medical examination, it does
require a considerable amount of information from a proposer.
Pricing
The pricing of insurance contracts is fundamental to actuarial work. Actuaries build mathematical models to help them set premiums for insurance
policies. The two most important elements to be modelled in the pricing of
life insurance policies are
(i) the mortality experience of the policyholders, and
(ii) the return that the insurance company can achieve on its investments
(including investment of premiums to be received in the future).
Another important factor to be taken into account in pricing life insurance
policies is expenses. Life insurance companies may sell their products through
agents (e.g. insurance brokers who advise clients and who are not in any way
connected to the insurance company), or through their own sales force, i.e.
company employees. In each case, it is normal for commission to be paid at
4.3. LIFE INSURANCE
123
the time a policy is issued, and this commission is often a high proportion of
the rst premium. Thus, insurers incur what are known as initial expenses,
i.e. expenses associated with establishing a policy (and they normally include
more than commission). Once a policy has been sold, the insurer is then faced
with what are known as renewal expenses, such as ongoing commission, the
expense of collecting premiums, the expense involved in maintaining data
about policyholders, or the expense of notifying the policyholder about the
status of the policy (e.g. for a unit linked policy, there would be an annual
statement showing the current price of units).
The pricing of traditional life insurance policies has until recent years
largely been by using the principle of equivalence, which is discussed in detail in the next chapter. Nowadays, another method, called prot testing,
has found much favour, particularly in the pricing of unit linked policies.
The approach here is to consider the expected cash ows in each year of the
policy, and to look at the impact of dierent premium levels on the values of
some specied measures. For example, the issue of a policy may involve the
insurance company making some capital available to cover initial expenses,
and an important measure in determining the premium for such a policy
would be the rate of return achieved on this capital.
Reserving
In most countries it is a statutory requirement for insurance companies to
carry out a valuation at regular intervals, e.g. annually. The main purpose of
a statutory valuation is to assess whether or not an insurance company has
su!cient funds to meet its emerging liabilities. Thus, a valuation provides
information about the solvency of an insurance company. In simple terms,
the actuary has to quantify the expected present value (formally dened in
Chapter 5) of the dierence between the future outgo (benets and expenses)
and income (premiums) from insurance policies. The means for doing this is
a reserve calculation. A reserve is the amount of money held by an insurer
for a group of policies (e.g. whole life insurance policies). The reserve should
be such that the reserve plus the expected present value of income exceeds
the expected present value of outgo. Broadly speaking, if this situation holds
for all types of business (i.e. whole life insurance, term insurance, etc) we
say that the insurer is solvent. Reserves are built up by the accumulation of
premium income (and hence investment income).
The interest rate and mortality table that are used for the valuation
may be specied in law, to ensure that the valuation is a robust test of the
company’s solvency.
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Investment
Insurance companies manage billions of dollars of assets. Central to any
insurance company is its investment department. The role of the investment
department is not simply to obtain the best return possible on an insurance
company’s assets. It must invest in assets appropriate to the company’s
liabilities.
Many actuaries are employed in investment departments in insurance
companies. The unique skill that they bring to these departments is an
understanding of the company’s liabilities. The nature of life insurance business is generally long term. Many policies will be on an insurance company’s
books for twenty years or more, and it is important that premium income
is invested with this in mind. In the context of life insurance, appropriate
investments are shares and bonds with a long term until redemption.
Analysis of surplus
One way of viewing premium calculation is to say that actuaries use their
best estimate of what will happen in the future, and will set the premium
accordingly. However, the actual experience of an insurance fund, i.e. a
collection of similar policies, is not usually exactly as expected. There are
several ways in which the actual experience may dier from the assumptions.
For example:
• Fewer policyholders than expected may die in a given period meaning
that the insurance company collects more in premium income and pays
out less in benets. In the short term at least, such conditions give rise
to prot.
• Adverse economic conditions may mean that an insurance company
earns less on its invested funds than expected. A loss will occur if the
insurance company is unable to earn the rate of return it has assumed
in its premium calculation.
• The insurer may control its expenses more tightly than anticipated,
resulting in less of the premium being used for expenses, and so a
prot arises.
An important task for the actuary is to analyse the position of insurance
funds from year to year and to determine how the actual experience of a fund
compares with that expected. In particular, there is a task called analysis
of surplus which attributes the gain (or loss) in a particular period to the
major sources mentioned above — mortality, investment and expenses.
4.4. PRIVATE HEALTH INSURANCE
125
Prots in life insurance funds can be distributed in two ways. If the insurance company is a mutual company, it is owned by the policyholders, and
all the prots belong to them. Actuaries distribute bonus to traditional policies in two ways. The rst is a reversionary bonus. Each year the insurance
company declares a reversionary bonus as a percentage of the sum insured
and any previously declared bonuses (i.e. it is a compound bonus). When
the sum insured is payable, the insurance company may declare a terminal
bonus. This is a one-o bonus payment, the amount of which usually re ects
the conditions that a policy has experienced throughout the term of the contract. Bonus distribution is a di!cult topic. We emphasise that the actuary’s
role in this task is to ensure an equitable distribution of bonus, so that the
bonuses that attach to a particular policy broadly re ect the contribution
that policy has made to the overall prot of the insurance fund.
If an insurance company is a proprietary company, then it is owned by its
shareholders, and not the policyholders. In this case the shareholders must
be included in the division of prots in the same way as shareholders of other
proprietary companies are, and prots are re ected in rises in value of the
shares in the market, and the amount of declared dividends.
4.4
Private Health Insurance
In Australia, private health insurance is the term that is applied to a policy
that provides a policyholder with cover for hospital and other health care
related expenses. These expenses can range from a visit to the dentist to a
lengthy stay in hospital, and may be for traditional or alternative therapies.
Indeed these days the scope of health insurance is widening, and some funds
pay benets associated with preventative therapies, for example costs related
to ceasing smoking.
Individuals, couples and families can obtain private health insurance by
joining a private health insurance fund. By legislation, health funds in Australia charge contributions on a basis known as community rating (although
there are some exceptions). Under this scheme, the contribution paid by an
individual is the same as for all other individuals with the same benets,
regardless of age, sex and health status, although the cost can vary from one
state to another. Contributions are calculated without subdividing the lives
into homogeneous groups.
For example, the contribution for a specied level of cover for a 20 year
old man is the same as for a 60 year old man, and family (parents and
dependant children) membership costs the same whatever the number or
ages of family members. It is usual in insurance to identify classes of risk
in more detail than this, and the community rating system gives rise to
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CHAPTER 4. ACTUARIAL PRACTICE
various undesirable eects. For example, the young generally incur much
lower health costs than the elderly — and so health insurance is relatively
unattractive to them, and they do not eect policies. This means that the
experience of the policyholders re ects an older membership, and the costs
of insurance are inclined to rise. This creates an upward spiral of costs, and
government intervention in the form of tax incentives and penalties has been
used to encourage membership of private health funds.
On the assumption that health costs increase with age, we would say that
young people are subsidising the cost of cover for older people. Of course,
when these younger people grow old, they would in turn receive a subsidy
from the younger members. Unfortunately, the current situation in Australia
is that young people are not entering health insurance funds, presumably
because they think that their chances of using the cover are small and the
premiums are high. Also, like all members of the population, young people
have access to a public health scheme which provides a level of basic necessary
health care and emergency treatment. Private health insurers oer dierent
levels of cover, targeted at people at dierent ages, from low cost policies with
limited benets aimed at younger people, to policies oering comprehensive
cover which are targeted at older people.
At present, there is much scope for actuarial involvement in health insurance, but the extent to which actuaries can contribute may depend on
the government. Health is a political issue, and politicians may choose not
to change the way in which health insurance operates, even if the basis of
operation is demonstrated to be unsound. In Australia, the government is
trying to encourage membership of private health funds by various methods
in the hope of relieving the burden on the public health system. One such
method is a tax rebate on premiums. Another is a system called ‘lifetime
rating’ under which the amount of premium paid by individual members of
private health funds depends on the age at which they rst eect insurance,
provided this membership is continued, relatively, unbroken. Individuals who
eect hospital cover after age 30 pay a higher premium than those of the same
age who have held continuous hospital cover from before age 30. This is an
example of community rating not applying.
As Australia’s population ages, the cost to the public health care system
will increase considerably from current levels, but this burden on the public
purse can be reduced if more individuals eect private health insurance.
Example 4.5 Most Australian health funds oer a range of packages tailored
to the requirements of couples, families and single people. The following is
a description of benets available under a saver’s package for single people
oered by a leading health fund. Such a package is aimed at young people in
good health, not likely to require major surgery or spend a prolonged period
4.5. SUPERANNUATION
127
in hospital. People buying this package self-select, i.e. they anticipate that
they are unlikely to require a greater level of cover.
The following cover is provided for a period of one year. This cover is
secured either by monthly premiums or by an annual premium equal to twelve
times the monthly premium.
Hospital cover: Members receive 100% cover (i.e. they pay nothing) for
treatment and a stay in hospital following an accident. The same level of
cover is also available for some specied surgical procedures, including the
removal of an appendix, tonsils or wisdom teeth. For other types of hospital
treatment, there is full cover for hospital stays in public hospitals, a maximum daily allowance for hospital stays in a private hospital, and a maximum
allowance for an out-patient visit to any hospital.
A feature of hospital cover is that there is a waiting period. For example,
new members cannot claim hospital cover within one year of joining the health
fund to obtain treatment of a condition which existed at the time the member
joined.
Everything listed below, is considered an ‘extra’. There is normally an
upper limit to the amount a member can claim in a year for extras.
Dental cover: In the rst year of membership, members can claim up to
$500 for certain specied dental work. This limit increases by $100 a year
over ve years to a maximum of $1,000. Specied work includes dentures,
crowns and bridgework.
Optical cover: Members may claim up to $180 for optical costs, for example
the cost of a pair of contact lenses.
Miscellaneous: Members may claim for visits to chiropractors, physiotherapists, osteopaths or naturopaths. There is a maximum amount that may be
claimed in a year, and there are prescribed payments for visits.
4.5
Superannuation
Superannuation is concerned with savings for retirement. Provision for retirement is generally from three possible sources: the government age pension
designed to relieve poverty, an occupational benet in association with an
employer, and any additional savings made by the individual. In Australia,
it is most common for these superannuation savings to be accumulated until
a person’s retirement age, and for the savings to be paid in the form of a
lump sum benet. In many other countries, the payment is in the form of
an annuity, referred to as a pension, payable from the date of retirement
throughout a person’s remaining lifetime.
Occupational superannuation can be arranged for individuals in association with their employers. In Australia, all employers must contribute a leg-
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islated minimum percentage of an employee’s earnings (when these exceed a
specied minimum amount) to an approved superannuation fund. Sometimes
employees eect additional superannuation to top up the expected benets
from membership of an employer’s superannuation fund. Additionally, in
some countries the government may provide a pension in old age, as part of
their social welfare provisions. In Australia this government age pension is
means tested, i.e. you receive an old age pension only if your income and
assets in retirement are below a certain specied level. In other countries,
the age pension may be provided to all those who full the age requirement.
This payment without means testing is called a universal pension. In general
there is some restriction preventing individuals from getting access to their
retirement benets before retirement age. This practice of earmarking benets for use in retirement only is referred to as preservation, and preservation
rules vary in their strictness from country to country.
The main purposes of superannuation are to provide income in old age,
and to allow individuals to maintain a standard of living in retirement commensurate with that while employed. It is generally agreed that the benet
should provide some income throughout retirement, though the emphasis on
benets being taken as income varies amongst dierent countries. There is
debate over what an ‘adequate’ standard of living, or level of income, is. The
basic age pension is generally described in terms of the average wage (or similar measure), and the proportion granted is about 25%, though this again
varies considerably from country to country. Occupational superannuation
schemes usually aim to provide a total retirement income of between one
half and two-thirds of a person’s income immediately prior to retirement for
individuals retiring after about 40 years of full time employment.
In this section we consider two major types of occupational and personal
superannuation schemes — dened benet and dened contribution schemes.
Dened benet schemes are largely employer related. Dened contribution
schemes need not be. Indeed, many nancial institutions oer superannuation products which individuals can purchase, independent of their employer.
4.5.1
Dened benet schemes
As the name suggests, in a dened benet superannuation scheme, the benets that are payable are dened when a person joins the scheme. Usually
they are dened in terms of a ‘nal salary’ prior to exit from the scheme, the
period of membership of the scheme and a rate of accrual of benet per year.
The term ‘nal salary’ means dierent things from scheme to scheme. For
example, it could be dened as the total salary earned in the twelve months
preceding retirement, or it could be the average annual salary over the three
years preceding retirement. The years of eligible membership may also be
4.5. SUPERANNUATION
129
dened in dierent ways for dierent schemes.
Most dened benet schemes are employer organised, and employees join
when they begin employment. The benets from the scheme are generally
funded by contributions from both the employer and the employees. Typically, they each pay a percentage of the employees’ salary, and the total
level of contributions is reviewed regularly to ensure that the accumulated
assets will be su!cient to meet the costs of the benets over the long term.
This is known as controlled funding. Normally, the employer will contribute
at a greater rate than the employee, with the employer’s contribution rate
perhaps being of the order of twice as much as the employee’s.
This practice of accumulating contributions in anticipation of the payment of the long term benets is called advance funding.
Dened benet schemes can oer a variety of benets, dened in a number
of ways. The most common benets oered by dened benet schemes are:
• An age retirement pension, which would be payable to the employee
from normal retirement age until death. The pension would typically
be calculated as a xed percentage of nal salary for each year of service.
For example, if the xed percentage was 1.5%, and an employee was
about to retire with 35 years of service and a nal salary of $100,000,
the annual amount of pension would be
0=015 × 35 × 100> 000 = 52,500.
• A lump sum retirement benet, which would be payable when an employee retired. The amount of the lump sum would be calculated using
a similar formula, with a higher percentage. For example, the percentage might be 15%. The dierence in percentage arises from the fact
that the lump sum is a notional cash equivalent of a lifetime annuity.
(As a very general statement, we might explain the dierence between
this percentage and that under the age retirement pension in the following way. Suppose that around retirement age, the value (i.e. the
purchase price) of an annuity of $1 per annum to a male retiring at
65 is about $10. Hence, a lump sum benet of $525,000 ‘is equivalent
in value’ to a pension of $52,500. That is, if the cash were applied to
purchase an income, it would buy $52,500 p.a. The relative value, or
commutation rate, depends upon many factors.)
• A death benet, providing a lump sum payment should an employee die
before normal retirement age. The amount of this benet could be
calculated in a number of ways, for example as a benet depending on
length of membership of the scheme, or as a specied amount such as
twice the employee’s annual salary in the year preceding death.
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• A spouse’s pension, payable to an employee’s spouse following the employee’s death in retirement. Note that such a benet is payable only if
the employee has died after retiring. The amount of pension is normally
between 50% and 75% of the employee’s pension.
From an employee’s point of view, the actual amount of pension or lump
sum payable becomes known only in the last few years of the employee’s
working life, although the benet level relative to current salary is known.
Any salary increases (or decreases) in this period can have a dramatic eect
on the amount of benet payable.
Retirement benets cannot normally be taken before normal retirement
age, but special provisions usually exist in cases of sickness or hardship after
age 55.
Example 4.6 Members of a large Australian superannuation scheme contribute 7% of their salary to the scheme, and employers contribute 14% of
each member’s salary. Thus, the annual contribution for a member is 21% of
that member’s salary. For most members, the normal means of contribution
is by a fortnightly deduction from their pay. These contributions are invested
on behalf of the members, with employers’ contributions rst being subject to
a 15% contributions tax.
The scheme currently comprises two funds — a dened benet fund and
an accumulation fund — and members must choose the fund to which they
wish to contribute. The benets under the dened benet plan include the
following.
Retirement benet: This is available when members retire, at any age from
55 onwards. They have a choice of taking a lump sum, an indexed pension,
a combination of these, or deferring payment of the benet. The lump sum
is expressed as a multiple of the member’s benet salary. This benet salary
is the average of a member’s salary over the last three years of employment,
adjusted for in ation to the time of retirement. For members who have always
been employed full time, the multiple is the product of two terms — number of
years of scheme membership and a lump sum factor. The lump sum factor
varies with retirement age, increasing from 21% for retirement at age 55 to
23% for retirement at age 65. Thus, a member retiring at age 65 with 35
years of full time membership and a benet salary of $80,000 would be entitled
to a lump sum of
80,000 × 0=23 × 35 = 644,000.
If a member selects to receive an indexed pension instead, the annual
amount of the pension is calculated as a percentage of the member’s benet
salary. For members who have always been employed full time, the percentage
is the product of two terms — number of years of scheme membership and
4.5. SUPERANNUATION
131
a pension factor. The pension factor varies with retirement age, increasing
from 1.30% for retirement at age 55 to 1.70% for retirement at age 65. Thus,
a member retiring at age 65 with 35 years of full time membership and a
benet salary of $80,000 would be entitled to an annual pension of
80,000 × 0=017 × 35 = 47,600.
Note that the member would receive this amount each year only if in ation
— as measured by the Consumer Price Index (CPI) — was 0% p.a. each year
in the future. The annual amount of pension is increased each January to
re ect changes in the CPI. The pension benet is payable monthly.
Resignation benet: This is available to members who leave the scheme
before age 55. They have a choice of taking a lump sum, or deferring payment
of their benet. The lump sum is expressed as a multiple of the member’s
benet salary. For members who have always been employed full time, the
multiple is the product of two terms — number of years of scheme membership
and a lump sum factor. The lump sum factor varies with resignation age.
For example, it is 18% for resignation at age 30 and 20% for resignation
at age 50. Thus, a member resigning at age 50 with 20 years of full time
membership and a benet salary of $60,000 would be entitled to a lump sum
of
60,000 × 0=2 × 20 = 240,000.
Preservation rules require that benets be set aside until at least age 55. Options available to members in these circumstances include depositing funds in
a personal deposit account held by the scheme, or transferring the benet into
another superannuation fund (if a member has moved to a new employer).
Death benet: This is payable to a member’s estate on a member’s death.
The basic death benet is the value of the resignation or retirement benet
as at the member’s date of death. Additionally, if death occurs before age 60,
there is a top up benet equal to a multiple of the member’s benet salary,
the multiple being 21% of the dierence between 60 and the member’s age at
death.
Disability benets: The scheme also pays benets if a member ceases employment permanently due to illness or disability. This benet is in the form
of a monthly pension, but a lump sum payment is also possible. If a member
ceases employment temporarily due to illness or disability, a monthly income
benet is payable for a maximum of two years. In either case, the annual
amount of benet for a member who has always been employed full time is
60% of the member’s benet salary.
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CHAPTER 4. ACTUARIAL PRACTICE
Dened contribution schemes
Under these accumulation schemes, it is the amount, or rate, of contribution
that is paid into the scheme that is specied. For example, an employee
may pay 6% of salary into a scheme throughout their working life. These
contributions, together with contributions by an employer, accumulate until
the employee retires, at which time a lump sum benet and/or a pension is
payable. Thus, in many respects we can think of such a scheme as operating
like a bank account where the return on the deposits made to the fund will
vary from year to year depending on the investment performance of the fund.
There are important dierences between dened contribution and dened
benet schemes. In a dened contribution scheme, the member knows the
value of the accumulated contributions represents the value of their benets,
and the member primarily bears the risk associated with the investment
performance of the fund. In a dened benet scheme, the member does not
in general know the value of the benet they will receive, but does know
what relative standard of living they might expect to realise at retirement,
and the investment performance risks are borne by the employer. It is likely
that most members would underestimate the amount of cash required to
purchase a particular level of income, however, so in some respects the dened
contribution plan may be more di!cult for the member to interpret in terms
of their long term needs.
Example 4.7 In the scheme described in the previous example, members
have a choice of benets. We consider here the Investment Choice Plan of
this scheme.
Contributions by both members and employers are as described in the
previous example. Benets are payable under the contingencies described for
members of the Dened Benet Plan. It is only the way in which the benet
amount is calculated that diers. We therefore consider only the retirement
benet, as the principles that apply to it also apply to other benets.
The retirement benet is available when members retire, at any age from
55 onwards. They have a choice of taking a lump sum, an indexed pension, or
a combination of these. The lump sum benet is the balance in the member’s
accumulation account. Basically, a member’s accumulation account contains
the accumulation of the member’s and the employer’s contributions, subject to
certain deductions (e.g. an administration charge and a tax on the employer’s
contributions). Each member has a choice of how the contributions to their
account are invested. The member must specify which of twelve investment
strategies should be followed for their account. These strategies range from a
cautious approach providing fairly low, but stable, returns to the most risky
approach which involves investment in shares, both in Australia and overseas.
4.5. SUPERANNUATION
133
Thus, two members with identical proles with respect to age, salary and
scheme membership, would almost certainly receive dierent lump sum benets on retirement if they had chosen dierent investment strategies for their
contributions.
4.5.3
The role of the actuary in superannuation
The role of the actuary in dened benet schemes is, in many respects, similar to the role of the actuary in life insurance. Broadly speaking, setting
the contribution rate is a similar task to pricing insurance contracts. From
a modelling point of view, if benets are payable on leaving the scheme by
reasons other than retirement, we need to construct a model which allows
for such possibilities. Just as in life insurance, the actuary must make assumptions about the future. In the context of a dened benet scheme,
assumptions are required on investment earnings, on future salary rates of
employees, on mortality rates, and on other rates of decrement (e.g. withdrawing from employment).
Actuaries will normally carry out regular (possibly annual) valuations of
dened benet schemes. The aims of these valuations will be to ensure solvency (again, as in life insurance, to ensure that existing funds plus expected
future income are su!cient to provide the expected future outgo), and to
check that the current contribution rate to the scheme is at an appropriate level. (If it is not, recommendations will be made for an increase or a
decrease.)
The actuary’s role in dened contribution schemes is much less. The
reason for this is simply that a person’s contributions accumulate, and accumulation from one year to the next is a straightforward task.
Actuaries may be involved in other aspects of either type of scheme. For
example, they may give investment advice to the scheme. Another area in
which actuaries may advise is in regard to the associated life insurance benets. Many schemes (either dened benet or dened contribution) do not pay
out death benets from the schemes’ funds, but may arrange such cover for
their members with a life insurance company as a group life contract. Under
a group life contract, a premium is paid for insurance cover for each member
of the group. Such cover is available only for large schemes. The actuary can
advise on an appropriate source of insurance cover and an appropriate level
of insurance benets.
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4.6
CHAPTER 4. ACTUARIAL PRACTICE
General Insurance
Broadly speaking, the term general insurance refers to insurance which is
not insurance on a life. There are many dierent types of general insurance
policy. Some examples are as follows.
• A comprehensive motor insurance policy provides the policyholder with
insurance cover against any loss that may occur as a result of damage
to, or theft from, a specied motor vehicle during a given period.
• A travel insurance policy provides a policyholder with cover against
losses incurred on a trip. The cover normally includes items such as
theft of belongings and medical costs. It may also provide some life
insurance cover for the duration of the policy.
• A household contents policy provides a policyholder with protection of
their belongings against various acts such as theft or re during a xed
period of time.
• A re insurance policy provides the cost of rebuilding or restoring a
property destroyed by re during a xed period of time. The property
could be a home or a business.
• A satellite insurance policy provides cover against a loss incurred as a
result of an accident to a satellite in space, for example, if there was
an explosion shortly after launch.
• A workers’ compensation policy provides benets to individuals who
sustain illness or injuries in the workplace during a xed time period.
This cover is not only for injuries or illnesses that are reported during
the period of insurance cover. Many illnesses, such as asbestosis, do
not manifest themselves for many years.
General insurance policies are very dierent in nature to life insurance
policies. Typically, general insurance policies have the following characteristics:
(i) The insurance cover for the policy is for a limited term. In many lines
of business, e.g. motor insurance, the typical period of cover is one
year.
(ii) The premium is often a single premium, although monthly premiums
can be payable under a policy with a term of one year.
4.6. GENERAL INSURANCE
135
(iii) There is no guarantee from the insurer that a policyholder can renew
a policy. If a policy is renewed, the premium may not be at the same
level as for the previous period. However, in some lines of business,
insurers oer a no claims discount, giving a policyholder a reduction in
the standard premium if the policyholder has not claimed in previous
years.
(iv) There is a moral hazard under many general insurance policies. For
example, under a travel insurance policy, it is often di!cult to ascertain
whether an individual making a claim for theft of possessions actually
had the possessions in the rst place, and also lost them.
Example 4.8 A major Australian travel company oers travel insurance to
customers travelling overseas. Although the insurance is purchased through
the travel company, it is actually underwritten by a separate insurance company. The policy provides travellers with insurance cover for the following.
Medical expenses: these include visiting a doctor, hospital costs, ambulance charges and emergency dental treatment. Additionally, in the event of
death, the policy provides up to $15,000 for a funeral overseas or to transport
remains to Australia.
Cancellation costs: these cover a variety of circumstances. The most obvious event is the cancellation of the trip, in which case the policy covers any
losses the policyholder incurs due to cancellation. Other events such as expenses due to travel delays (e.g. accommodation) also fall into this category.
Death benet: the policy provides a xed sum should the policyholder die
during the trip, or die within twelve months of the trip as a result of an
injury that happens during the trip. As with life insurance policies, there are
restrictions on the payment of the benet. For example, the benet would not
be payable if the policyholder died from a self-in icted injury.
Luggage and personal eects: the policy covers the policyholder against
loss or theft of personal items. This covers situations like fraudulent use of
a policyholder’s stolen credit card, or replacement (temporary or otherwise)
of essential goods, e.g. if a policyholder arrived at a ski resort only to nd
their skis had been lost in transit. There is an onus on the policyholder to
provide evidence in such cases, including making a report to the local police,
and keeping receipts for replacement items.
Personal liability: the policyholder is covered if they accidentally cause
damage to life or property and are required to pay legal expenses and/or
compensation. There are exclusions here, including any damage that occurs
if the policyholder acts in a malicious or unlawful way.
Legal expenses: the policy provides cover should the policyholder wish to
pursue damages or compensation as a result of an injury to the policyholder
during the trip. The cover also applies should the policyholder be killed.
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CHAPTER 4. ACTUARIAL PRACTICE
The premium depends on the duration of the cover (a maximum of twelve
months), and would be a single payment prior to the trip. Family cover
(i.e. at least a couple) is available at slightly less than twice the cost of
cover for a single person. There is eectively no underwriting. The onus
is on the individual to provide all relevant information. Policy exclusions
provide a level of protection to the insurer. For example, proposers for cover
must disclose all information about existing medical conditions, and the death
benet is not payable if death is caused by an existing medical condition.
For each benet listed above, there is a prescribed maximum amount
payable. For example, the death benet is $25,000 and the maximum amount
of legal expenses is $12,000. With the highest level of cover this company
oers, there is no limit on medical expenses.
Individuals generally purchase travel insurance for any type of overseas
trip, be it business or pleasure. Premiums are relatively small, around $125
for a fortnight’s standard cover, and the majority of claims are small. However, one only needs to think of the cost of, say, a two month hospital stay
in America to realise that claims can be very large.
Example 4.9 The insurance arm of a leading Australian bank oers Home
Insurance. Policyholders are covered against a range of events — re, explosions, lightning strikes, storms, rainwater damage, vandalism, theft. Additionally, legal liability cover is available. This provides the policyholder with
protection against damages or legal fees arising out of, for example, death or
injury caused by an accident at the insured home.
Such insurance is often purchased when a policyholder eects a mortgage
with the bank. There is generally little in the way of underwriting. The
policyholder will advise the insurer of the value of the home and contents, and
the premium will re ect these values. The insurer would assess factors such
as a property’s age, location and condition, as well as a proposer’s claims’
record. The proposer is legally bound to disclose all relevant information that
the insurer requires, and failure to do so may result in the insurer declining
to pay a claim.
The main reason for people purchasing such insurance is simply that their
home is their greatest asset and they want to protect it.
Contracts of this nature are tightly worded. Although the policy oers protection against storm damage, a claim would not be paid for interior damage
caused, for example, by the policyholder leaving a window or door open. There
are many such exclusions, often designed to make the policyholder act in a
responsible fashion.
Premiums are payable monthly or annually. If a policyholder eected
the policy with a mortgage, the premiums could be payable by debit from a
bank account at the same time as mortgage repayments are due. To reduce
4.6. GENERAL INSURANCE
137
the number of small claims, policies have what is called a policy excess. A
standard level of policy excess is $100. This simply means that if a loss is,
say, $500, the policyholder is liable for the rst $100 and will receive $400
from the insurer. Thus, the insurer pays the amount of the loss that is ‘in
excess of’ $100.
Perhaps the most important distinction between life and general insurance
policies from an actuarial point of view is that the modelling of claims is
quite dierent. Consider a life insurance policy. At the start of a given year,
the insurer knows the sum insured under a given policy, can estimate the
probability of that benet being paid, and can be sure that the sum insured
would only be paid, at most, once, and that the policy would cease after
a claim was paid. In contrast, consider a comprehensive motor insurance
policy, providing cover for one year. The insurer does not know how many
claims the policyholder will make in the year. The insurer therefore needs a
model to describe the number of claims a policyholder might have. If a claim
is made, the amount payable under that claim can be anything from a few
dollars to the cost of replacing the motor vehicle. Thus, the insurer needs
another model to describe the amount of the payments that could be made
under the policy. Payment of a claim does not result in the policy ceasing,
and the policyholder may renew a policy when its term is over.
Statistical modelling is therefore very important in general insurance.
We will not discuss statistical modelling here, but simply restrict ourselves
to saying that the premium for most risks would normally exceed the average
amount that the insurer expects to pay out under the policy.
4.6.1
The role of the actuary in general insurance
Pricing
As in life insurance, a fundamental task for actuaries is setting an appropriate
level of premium for a policy. The collection and analysis of statistics is vital
in this task. Since most general insurance policies have a short term, premiums are reviewed on a regular basis. For many types of general insurance
policy, for example motor vehicle insurance, the market is very competitive
and the actuary must be aware of both the insurer’s aims in the marketplace
and the strategies and prices of rivals.
The actuary will also be expected to assess the level of expense associated
with a policy, and to consider what level of prot is desired from a policy.
Of course, market conditions may dictate what level of prot an insurer can
actually achieve.
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Reserving
In general insurance, perhaps the most important task for an actuary is estimation of outstanding claims reserves. For many types of general insurance
policy, claims are reported and settled very quickly. For example, it would
not take an insurer long to settle a claim under a home contents insurance
policy following a burglary. However, for other types of policy it can take
years to fully settle a claim. For example, compensation payments to someone incapacitated by a workplace injury could be paid over the person’s
remaining lifetime.
The actuary’s task is to estimate an insurer’s liability in respect of claims
that will be attributable to a given period’s premium income. In doing so,
the actuary must take account of factors such as the time until the claim
will be settled nally and in ation over this period. There is a range of
mathematical models available to help the actuary in this task, but often it
is su!cient to perform case estimates, that is to consider all the circumstances
of a claim and to make a judgement on the likely outcome, and hence the
reserve required.
Reinsurance
Many general insurers eect reinsurance, that is they eectively share a risk
with a reinsurance company. There are a number of dierent types of reinsurance arrangement available, the most common being sharing claims in
an agreed proportion, or the insurer paying claims only up to some agreed
limit, with the reinsurer paying the remaining amount. The tasks for an
insurance company’s actuary include
• assessing what type of reinsurance is appropriate for a given risk;
• assessing what level of reinsurance is appropriate, for example what
is the maximum amount the insurer would be prepared to pay out on
claims generated by a given event such as a windstorm or hailstorm;
• assessing the cost of reinsurance protection.
4.7
National Insurance
In many countries, governments provide welfare benets. Examples of such
benets are an old age pension, a disability pension, free health care and
unemployment benets. Crucial to the costing of such benets is a knowledge
of the current demographic and economic structure of a country’s population,
and the ability to estimate the nancial impact of changes in that structure.
4.7. NATIONAL INSURANCE
139
Broadly speaking, we can label the provision of such benets as national
insurance. In some countries, advice to governments on the provision of
certain benets is provided by a government department employing actuaries.
In Australia, there is a Government Actuary who provides advice to the
government on a range of issues. In other countries, such as the UK, there
is a Government Actuary’s Department. Typically, the number of actuaries
employed in the public sector is not large, and their role is rather specialised.
National insurance diers from conventional insurance in a number of
ways. The most obvious of these is that in most cases national insurance
is not optional. Benets are generally paid out of government levies (i.e.
taxes), although in some countries there is a tax (which could be called a
levy or a contribution) specically associated with a particular benet. For
example, in Australia there is the Medicare levy, which goes towards the cost
of funding health care. In other countries, employees may pay a mandatory
percentage of their earnings into a national insurance fund, and this may
entitle them to a certain level of benets. There are great variations in detail
from country to country.
A second dierence is that, as the ‘insurer’, a government does not have
to meet the same statutory requirements as an insurance company. In most
countries, insurance companies have to demonstrate that they hold su!cient
assets to meet their emerging liabilities. In contrast, a national insurance
fund can be a notional fund, i.e. a fund that does not actually exist. Such
a fund operates on a pay-as-you-go basis, with the notional income of the
fund (i.e. that amount raised in taxes and levies that would go into the fund
if the fund existed) being used to pay benets. If this income is insu!cient
to pay benets, the government can use funds raised by other means to pay
benets. The power of a government to raise money is substantially greater
than that of an insurance company, and so is its ability to change the terms
of the ‘contract’. Governments may bow to political expediency, and change
the level and nature of benets without regard to equity. Generally, there
is no choice of type or amount of benets available, and the level of benet
may not be very high.
The role of the actuary in national insurance is therefore substantially
concerned with demographic issues. Specialist knowledge is required in the
sense that actuaries operating in this eld have to take account of factors such
as unemployment, and workforce participation rates, which would not normally enter into calculations for, say, life insurance and superannuation. The
reason for taking account of unemployment is that taxes (or contributions)
are normally levied only from the workforce, not the unemployed. Given a
population structure, the actuary can estimate the current and future costs
of benets, allowing for items such as changing demographics (e.g. declining
fertility rates) and increases in the level of insurance benets (e.g. indexing
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old age pensions to increase at a rate linked to the rate of in ation).
4.8
Exercises
1. In Australia, some life insurance companies oer a whole life insurance
policy designed purely to cover funeral costs without any underwriting,
provided that the proposer for insurance is under age 80.
(a) Do you think a lack of underwriting is wise?
(b) To what risks might such an insurance company be exposed?
(c) How might the insurance company manage the risks in part (b)?
2. State which is the larger in each of the following pairs:
(a) the single premium for a 15 year term insurance for a male aged 45
and the single premium for a 15 year term insurance for a female
aged 45,
(b) the annual amount of annuity that can be bought at age 65 by
a woman paying a single premium of $500,000 and the annual
amount of annuity that can be bought at age 65 by a man paying
a single premium of $500,000,
(c) the reserve required after 15 years for a 25 year term insurance
issued to a male aged 25 with sum insured $100,000 and the reserve
required after 15 years for a whole life insurance issued to a male
aged 25 with sum insured $100,000.
You should assume that this exercise refers to a developed economy.
3. What features do an endowment insurance policy and a term insurance
policy have in common, and what features dier?
4. A large insurance company oers both traditional life insurance and
general insurance, with a wide variety of products under each line of
business. An actuarial student has suggested that this insurance company could follow the same investment strategy for each line of business.
State with reasons whether or not you agree with the student.
5. Under a reverse mortgage with a no negative equity guarantee, is the
lender exposed to a moral hazard? If so, what steps might the lender
take to reduce this risk?
4.8. EXERCISES
141
6. You are the actuary to a large dened benet superannuation scheme
which provides a full range of benets including an index-linked pension
on retirement. Describe the sort of modelling that would be required
to conduct a valuation of the scheme.
7. The following statements have been made by actuarial students. State
with reasons whether or not you agree with them.
(a) Under dened benet superannuation the investment risk is borne
by the member.
(b) Under dened contribution superannuation the retirement benet
is closely related to the member’s salary at retirement.
(c) The investment strategy of a dened contribution superannuation
fund should be similar to that of a life insurance company selling
traditional business.
8. What features do a one year term insurance policy and a one year motor
vehicle insurance policy have in common, and what features dier?
9. Do you think that the setting of assumptions to calculate a premium
is more important in general insurance than in life insurance? Justify
your answer.
10. At the time of writing (2010), there has been much discussion in western European countries about raising the age at which people become
entitled to a state provided age pension. With reference to demographics, suggest why governments might wish to adopt such a change, and
brie y discuss alternative measures that governments might take.
Chapter 5
Valuation of Contingent
Payments
5.1
Introduction
In situations described previously, we have always assumed that payments
would be made. For example, in valuing an annuity payable annually in
arrear for q years, we did not consider the possibility that one or more of the
annuity payments would not be made.
In many practical situations, we are required to place a value on a series
of payments when either
(i) we may be uncertain about the number of payments to be made, or
(ii) we may know the number of payments that should be made, but we
are uncertain of their amounts.
These are not the only types of uncertainty that can arise when valuing a
series of payments, but we shall concentrate on these types.
An example of the rst type of uncertainty is as follows. Suppose you
wish to enter into a contract with an insurance company under which it will
pay you $10,000 annually until you die, the rst payment being one year
from the date on which the contract is agreed upon. The insurance company
must decide upon a price for this contract. A problem that it faces is that it
does not know how many payments it will make under the contract.
An example of the second type of uncertainty would be the situation
where a borrower defaults on loan repayments. In such a situation, the lender
may not receive payments promised on a given date. Thus, the number and
timing of repayments may be specied in a contract, but the lender may
not receive all the amounts stipulated at the specied times. (Of course the
142
5.2. DISCRETE RANDOM VARIABLES
143
lender may have recourse in the law, but this may not help if the borrower
is bankrupt). This type of uncertainty arises in most types of investment or
lending.
There is a major dierence between the two types of risk mentioned above,
namely that in the rst instance, the insurance company will have collected
a great deal of information about people wishing to enter such contracts, and
it can use this information to help it establish a fair price for the contract.
By contrast, there may be only a limited amount of information available
about an investment opportunity, and the opportunity may be unique.
In each of the above examples, the common feature is that the amount
of a payment at a given time is uncertain. In the following we establish a
framework which allows us to place a value on such a series of payments.
First, we give a very short review of some elementary probability.
5.2
Discrete Random Variables
A discrete random variable can take only one of a countable set of values.
For example, under a motor insurance policy, the number of claims that a
policyholder can make in a year is one of 0> 1> 2> = = = . Under a one-year term
insurance policy, the number of claims a policyholder can make in a year is
either 0 or 1. In each of these situations, we would model the number of
claims that a policyholder makes as a random variable because we do not
know at the start of the year how many claims will be made in the year.
We use capital letters to denote random variables. In the case of the motor
insurance policy mentioned above, we might use the letter Q to denote the
number of claims that a policyholder makes in a year. We would write, for
example, Pr(Q = 1) to mean ‘the probability that the number of claims in a
year is 1’.
Denition 5.1 The probability function of a discrete random variable [ is
represented by Pr([ = {), where { is a possible value of the random variable.
For example, an insurer might assess that for a particular policyholder,
the probability function of Q, the number of claims that a policyholder can
make, is as follows:
Pr(Q = 0)
Pr(Q = 1)
Pr(Q = 2)
Pr(Q = 3)
=
=
=
=
0=6>
0=3>
0=05>
0=05.
(5.1)
144
CHAPTER 5. VALUATION OF CONTINGENT PAYMENTS
Notice that the sum of the Pr(Q = q) values over all possible values of q
(i.e. 0> 1> 2 and 3) is one. This is always true of a probability function. In
general, for a discrete random variable [>
X
Pr([ = {) = 1>
{
where summation is over all possible values that the random variable [ can
take.
An important characteristic of a random variable is its mean or expected
value.
Denition 5.2 We denote the expected value of a random variable [ by
H([). For a discrete random variable [, we dene
X
H([) =
{ Pr([ = {)
(5.2)
{
where summation is over all possible values that the random variable can
take.
Example 5.1 What is the mean of the random variable whose probability
function is given by (5.1)?
Solution 5.1 From formula (5.2) we have that
H(Q) = (0 × 0=6) + (1 × 0=3) + (2 × 0=05) + (3 × 0=05) = 0=55.
Remark 5.1 We can interpret the above statement H(Q) = 0=55 as saying
that the average number of accidents the policyholder will have in a year is
0.55. Loosely, we could interpret this as saying that we would expect about
one claim every second year from this policyholder.
Example 5.2 The random variable N({) represents the number of whole
years lived in the future by a life now aged {. Show that
H(N({)) =
"
X
w s{ =
w=1
Solution 5.2 First, note that the possible values for N({) are 0> 1> 2> = = = .
To nd Pr(N({) = w) consider rst specic values for w. First, N({) = 0 if
({) does not survive to age { + 1. Second, N({) = 3 if ({) survives to age
{ + 3 but does not survive to age { + 4. Applying this reasoning generally,
5.3. VALUATION OF A SINGLE CONTINGENT PAYMENT
145
we see that N({) = w if ({) survives to age { + w but not to age { + w + 1.
Thus N({) = w if w W ({) ? w + 1, and so
Pr(N({) = w) = Pr(w W ({) ? w + 1) = w s{ w+1 s{ =
By formula (5.2),
H(N({)) =
"
X
w Pr(N({) = w)
w=0
=
=
"
X
w=0
"
X
w=0
w ( w s{ w+1 s{ )
w w s{ "
X
w w+1 s{ =
w=0
Note that the rst term of the rst sum is 0, so that the lower limit of summation can be written as w = 1, and in the second summand we can write
w = (w + 1) 1, giving
H(N({)) =
"
X
w=1
w w s{ "
X
(w + 1) w+1 s{ +
w=0
"
X
w+1 s{ >
w=0
and as the rst two sums are identical,
H(N({)) =
"
X
w s{ =
w=1
5.3
Valuation of a Single Contingent Payment
Let us start by considering a one year term insurance policy. Suppose that
under the terms of the policy, the policyholder pays the insurance company
an amount S now, and if the policyholder dies within the policy term, the
insurance company will pay the policyholder’s estate $100,000 at the end of
the policy term. How does the insurance company calculate S ?
Before addressing the question of how to calculate S , we introduce some
terminology.
Denition 5.3 If a payment under an insurance policy, or any other nancial transaction, depends on the occurrence of a certain event (e.g. a person
dying within a given period, or a person surviving a given period), we call
such an event a contingent event. The corresponding payment is referred to
as a contingent payment.
146
CHAPTER 5. VALUATION OF CONTINGENT PAYMENTS
Thus, for our one year term insurance policy, the contingent event is the
policyholder’s death within the policy term.
In order to calculate S the insurance company has to make some assumptions. Let us suppose it assumes the following:
(i) it can invest its funds to earn interest at 8% p.a. eective, and
(ii) the probability of the policyholder surviving for one year is 0.98.
At the end of the contract, the insurance company needs either $100,000 (if
the policyholder dies during the year) or 0. Thus, at the start of the contract
it needs either 100,000@1=08 or 0. We say that at the start of the contract
the insurance company requires
100,000@1=08 with probability 0.02,
0
with probability 0.98.
Thus, the amount the insurance company requires at the start of the contract
is a random variable, which has a two point distribution. We calculate S by
invoking the principle of equivalence. In this situation, this principle simply
says that the fair value of S is the expected value of the above probability
distribution. Thus
S = 0=02 × 100,000@1=08 = 1,851=85=
The above idea applies more generally. Suppose we are required to place
a value on a payment of amount D which may be due in w years’ time, using
an eective interest rate of l p.a. Let us assume that
(i) the payment will be either D or 0, and
(ii) the probability of receiving D is s (and hence the probability of receiving 0 is 1 s).
The present value of the payment at time w is a random variable. The present
value is
0
with probability 1 s>
w
Dy with probability s>
and the mean of this distribution, which we call the expected present value,
is
0(1 s) + Dyw s = Dyw s.
This is the value we place on the payment at time 0. This representation is
simple and appealing. The expected present value in this case is the product
of three things: the amount of the payment (if it is made) multiplied by the
discount factor multiplied by the probability that the payment is made.
5.3. VALUATION OF A SINGLE CONTINGENT PAYMENT
147
Example 5.3 Assuming an eective rate of interest of 5% p.a. and that the
probability that you survive w years from now is 0=99w , calculate the expected
present value of a payment of $5,000 ten years from now if you are alive
then.
Solution 5.3 The expected present value is
5,000 × 0=9910 × 1=05310 = 2,776=06=
Example 5.4 You have been oered a choice between (i) $40 now, and (ii)
$100 exactly one year from now provided it rains on that day. The meteorological o!ce advises you that the probability of rain on that particular day is
s. Assuming you can earn interest at 6% p.a. eective, for what values of s
is option (i) more attractive?
Solution 5.4 Under option (ii), the expected present value of the payment
in one year’s time is 100s@1=06. This is less than 40 if
40 × 1=06 A 100s,
i.e. if 0 ? s ? 0=424.
In the above discussion, there were two possibilities for the payment at
any given time — the payment was either zero or greater than zero. The ideas
discussed above extend to situations when this is not the case. Again, let us
start with a simple insurance contract as an example.
Suppose that a couple enter into a ve year contract with an insurance
company. Under the terms of the contract, the couple pay the insurance
company at the start of the contract. At the end of the contract the
insurance company will pay $20,000 to the couple if both are alive, or $10,000
to the survivor if only one is alive. What is a fair value for ?
Let us again make some assumptions. Suppose the insurance company
can earn 8% p.a. eective and assesses that the probability that both of
the couple survive for ve years is 0.9, and the probability that exactly one
of them survives for ve years is 0.06. Then at the end of the contract the
insurance company needs
20,000 with probability 0.9,
10,000 with probability 0.06,
0
with probability 0.04.
Equivalently, at the start of the contract it needs
20,000 y5
10,000 y5
0
with probability 0.9,
with probability 0.06,
with probability 0.04,
148
CHAPTER 5. VALUATION OF CONTINGENT PAYMENTS
where y = 1=0831 . Notice that we have set things up in exactly the same way
as in the previous insurance illustration. The only dierence here is that the
present value of the payment at the end of the contract is a random variable
with a three point distribution. The fair premium, , is the expected value
of this distribution:
= 0=9 × 20,000 y 5 + 0=06 × 10,000 y5 = 12,658=85=
More generally, suppose that w years from now you could receive
D1
D2
D3
with probability s1 ,
with probability s2 ,
with probability s3 = 1 s1 s2 .
Then, at an eective rate of interest of l p.a., the expected present value of
the payment at time w (years) is
D1 s1 yw + D2 s2 yw + D3 s3 y w ,
where y = 1@(1 + l). Notice that the same appealing form has been retained.
We are summing terms of the form: amount of payment at time w multiplied
by the probability of that payment being made multiplied by the discount
factor. This approach generalises to any number of possible payments at
time w, but we shall not consider such examples.
Example 5.5 Exactly six months before the English F.A. Cup Final you are
oered the choice of (i) an immediate payment of $200 or (ii) a payment of
$250 on the day of the nal if Manchester United are in the nal, with an
additional $250 if they win the nal. Your local bookmaker assesses that the
chance of Manchester United being the losing nalist is 0.3 and the chance of
them being the winning nalist is 0.4. Under this assessment, nd the range
of values for l, the eective annual rate of interest, which makes option (i)
more attractive.
Solution 5.5 Taking 6 months as our unit of time, and m to be the eective
rate of interest for a six month period, the expected present value of the
payment on F.A. Cup Final day is
250 × 0=3 y + 500 × 0=4 y = 275y
where y = 1@(1 + m). This is less than 200 if
200
1
Ay=
275
1+m
5.4. VALUATION OF A SERIES OF CONTINGENT PAYMENTS
149
giving
m A 0=375=
Hence option (i) is more attractive only if m A 0=375, implying an eective
annual rate of l = 89=1%.
The principle of equivalence is fundamental in actuarial work. In the
above insurance examples, we have set the expected present value (at the
issue of a policy) of the benet(s) under the policy to be equal to the (single)
premium. In a more general context, for example when a policy is issued by
annual premiums rather than a single premium, or when the benet under
the policy is a series of payments (e.g. an annuity) rather than a single
payment, the principle of equivalence applies as follows.
Denition 5.4 If a premium is calculated according to the principle of equivalence, then the expected present value of premium income equals the expected
present value of the benets under the policy.
Remark 5.2 This denition is set in the context of an insurance policy.
However, the principle equally applies to other nancial transactions. The
crucial point is that if any quantity is calculated by this principle, the expected
present value of payments received by an individual or institution must equal
the expected present value of the payments to be made by that individual or
institution.
Remark 5.3 Throughout this text, we ignore the expenses of issuing an insurance policy in premium calculation. If we were to include expenses, we
would extend the above denition by replacing ‘benets’ by ‘benets and expenses’. Basically, the insurer would then be calculating a premium by matching the expected present values of income and outgo under the policy.
5.4
Valuation of a Series of Contingent Payments
In practice, evaluation of a series of contingent payments is a much more common problem than evaluation of a single contingent payment. However, the
ideas underlying the evaluation of a single contingent payment are important
because we can generalise them.
Let us again start with an insurance example to illustrate ideas. Suppose
that on 1 June an individual aged { entered into a contract with an insurance
company under which the insurance company will pay the individual $10,000
150
CHAPTER 5. VALUATION OF CONTINGENT PAYMENTS
on each subsequent 1 June provided the individual is alive. What is a fair
price for the individual to pay to enter into this contract?
Let us suppose that at the outset of the contract the insurance company
assumed that it would earn interest at 7% p.a. eective. Then, if we measure
time in years from the issue of the contract, the present value of the payment
at time w, w = 1> 2> 3> = = = > is
10,000 yw
0
with probability w s{ ,
with probability 1 w s{ ,
where y = 1=0731 . Hence, the expected present value of the payment at time
w is 10,000 yw w s{ . The question we now have to address is whether or not we
can sum these expected present values to get the expected present value of
the series of payments, so that the fair price for the contract is
"
X
10,000 yw w s{ .
w=1
To see that this is the fair price we have to consider the possible times of death
for ({) and the associated present values. If ({) dies before time w = 1> then
no payments are made, and the present value of these payments is trivially
0. If ({) dies between times w and w + 1, for w = 1> 2> 3> = = =, then annuity
payments will be made at times 1> 2> = = = > w and the present value of these
payments is 10,000 dw . The probability that ({) dies between times w and
w + 1 is w s{ w+1 s{ . Thus, if the random variable S Y represents the present
value of the payments, then S Y takes the value 10,000 dw with probability
w s{ w+1 s{ for w = 1> 2> 3> = = = and hence
"
X
H(S Y ) =
10,000 dw (w s{ w+1 s{ ) =
Recalling that dw =
Pw
w=1
m
m=1 y , we can write
H(S Y ) = 10,000
" X
w
X
w=1 m=1
ym (w s{ w+1 s{ ) >
and changing the order of summation we get
" X
"
X
H(S Y ) = 10,000
ym (w s{ w+1 s{ )
= 10,000
= 10,000
m=1 w=m
"
"
X
X
m
y
m=1
"
X
m=1
w=m
ym m s{ =
(w s{ w+1 s{ )
5.4. VALUATION OF A SERIES OF CONTINGENT PAYMENTS
151
More generally, if at time w from the present, we could receive Dw with
probability sw , or 0 with probability 1 sw , then the expected present value
of this series of payments is calculated as
X
Dw sw yw .
w
Example 5.6 A corporation issues bonds with term ten years, redeemable at
par and with coupons payable annually in arrear at 15%. At the issue date,
you assess that the probability that the interest payment at the end of year w
will be made is 1 0=01w for w = 1> 2> ===> 10. What price would you pay per
$100 nominal of a bond at the issue date if you wished to obtain a return on
your investment of 15% p.a. eective?
Solution 5.6 The price you would pay would be the expected present value
of the payments. Per $100 nominal, the interest payments are $15, so the
expected present value is
10
X
w=1
15(1 0=01w) yw + 100 y10
where y = 1@1=15. One straightforward way to calculate this is to use a
spreadsheet. Using Excel, we could do this as follows.
In Row 1, set up column headings in columns A to D as Time, Discount
factor, Expected payment and Expected present value.
In cell A2, insert 1, then insert = D2 + 1 in cell A3, highlight cells A3 to
A11 and use the Edit Fill Down command which will put the values 3 to 10
in the empty highlighted cells.
In cell B2, insert = 1=15ˆ(D2). Highlight cells B2 to B11 and again apply
the Edit Fill Down command to get the discount factors y 2 > y 3 > = = = > y 10 .
In cell C2, insert the expected payment at time 1 as = 15 (1 0=01 D2).
Highlight cells C2 to C10 and again apply the Edit Fill Down command to
get the expected payments at the end of years 1 to 9. In cell C11, insert
= 15 0=9 + 100, allowing for the redemption money at the end of the 10th
year.
We have the discount factors and the expected payments in columns B and
C. All that remains is to multiply these together, then add them. In cell D2,
insert = E2 F2, then highlight cells D2 to D11 and again apply the Edit
Fill Down command. Finally, in cell D13, insert = VXP(G2 : G11), giving
the expected present value of all the proceeds of the bond.
The values are set out in Table 5.1.
The above ideas extend to the situation where there is a three point
distribution for the payment at a given time. A classical actuarial problem
152
CHAPTER 5. VALUATION OF CONTINGENT PAYMENTS
A
B
C
D
Expected
Discount Expected present
1 Time
factor
payment
value
2
1
0=8696
14=85
12=91
3
2
0=7561
14=70
11=12
4
3
0=6575
14=55
9=57
5
4
0=5718
14=40
8=23
6
5
0=4972
14=25
7=08
7
6
0=4323
14=10
6=10
8
7
0=3759
13=95
5=24
9
8
0=3269
13=80
4=51
10
9
0=2843
13=65
3=88
11
10
0=2472
113=50
28=06
12
13
96=70
Table 5.1: Spreadsheet output for Example 5.6
illustrating this is the valuation of an annuity payable at a given rate to a
married man whilst he is alive, with a reduced rate of payment to his widow
following his death. We now illustrate such a calculation.
Example 5.7 On retirement, a man receives a lump sum payment of
$100,000 from his employer’s superannuation fund. He decides to buy an
annuity from an insurance company, which will pay $[ monthly in arrear
while he is alive. Under the terms of the contract, the annuity will continue
to be paid to his wife following his death, but the level of the payment will be
cut to $0=6[. The insurance company assumes that it can earn an eective
rate of interest of 0.75% per month. It further assumes that the probability
that the annuity payment will be $[ w months from the date the contract is
eected is exp{0=01w} and the probability that the annuity payment will be
$0=6[ at that time is (1 exp{0=01w}) exp{0=008w}. On this basis, nd
[.
Solution 5.7 At time w (months), the payment under the contract will be
[
0=6[
0
with probability exp{0=01w},
with probability (1 exp{0=01w}) exp{0=008w},
with probability 1 exp{0=01w} (1 exp{0=01w}) exp{0=008w}.
5.4. VALUATION OF A SERIES OF CONTINGENT PAYMENTS
153
The present value of the payment at time w is thus
[ yw
0=6[ yw
0
with probability exp{0=01w},
with probability (1 exp{0=01w}) exp{0=008w},
with probability 1 exp{0=01w} (1 exp{0=01w}) exp{0=008w},
and the expected present value of the payment at time w is
[ yw exp{0=01w} + 0=6[ yw (1 exp{0=01w}) exp{0=008w},
where y = 1@1=0075. Setting the expected present value of all the payments
equal to 100,000 we get
100,000 =
"
X
[ y w exp{0=01w}
w=1
+
"
X
w=1
= [
"
X
0=6[ yw (1 exp{0=01w}) exp{0=008w}
y w exp{0=01w}
w=1
+0=6[
"
X
w=1
We have
"
X
y w exp{0=008w} 0=6[
yw exp{0=01w} =
w=1
"
X
y w exp{0=018w}.
w=1
y exp{0=01}
= 56=74>
1 y exp{0=01}
and similar formulae for the other summations give
100,000 = 56=74[ + 0=6[(64=13) 0=6[(38=76)
= 71=96[.
Thus [ = 1,389=67.
Remark 5.4 You may be wondering why the upper limit of summation in
the above example is innity. Clearly the individuals are not going to live
forever. We have used a simple model of mortality in this example, and a
defect of this model is that there is a non-zero probability of each person
being alive at any time in the future. Of course, as w gets large, terms like
exp{0=01w} get very small, and so make virtually no contribution to the
relevant summation. In practice, we normally base our assessment of the
probabilities of the lives being alive at a given time on a mortality table, in
which there would be a xed upper age beyond which we would assume that
survival is not possible.
154
CHAPTER 5. VALUATION OF CONTINGENT PAYMENTS
5.5
Premium Calculation
5.5.1
Whole life insurance
Let us consider premium calculation for a whole life insurance policy as
described in the previous chapter. To illustrate ideas we use a very simple
model of mortality. We assume that the force of mortality is constant, i.e.
{ = for all {. Although this is not a realistic model for mortality at all
ages, its simplicity allows us to demonstrate points that are generally true
about premium calculation with a more realistic model of mortality. Let us
suppose that a policy is issued to an individual aged {, with premiums of S{
payable annually in advance throughout the individual’s lifetime. Then the
expected present value of these premiums at an eective rate of interest of l
p.a. is
"
X
S{ yw w s{
w=0
where y = 1@(1 + l), and the probability of a premium being paid at time
w is w s{ , i.e. it is the probability that the individual, now aged {, is alive at
time w. To simplify our presentation, it is convenient to work in terms of the
force of interest, rather than the eective rate of interest.
Under our model of mortality,
½ Z w
¾
½ Z w
¾
{+u gu = exp gu = exp{w}.
w s{ = exp 0
0
Hence, the expected present value of the premiums is
"
X
S{ h3w h3w =
w=0
S{
.
1 h3(+)
Denition 5.5 The symbol d̈{ — called “d due {” — denotes the expected
present value of a payment of 1 annually in advance payable as long as a life
now aged { survives.
Remark 5.5 It is usually clear from the context what the valuation rate of
interest is. However, if there is any possible confusion, we could write, for
example, d̈6%
{ , to indicate an eective rate of interest of 6% per period.
Thus, under our model of mortality,
d̈{ =
"
X
w=0
h3w w s{ =
1
.
1 h3(+)
(5.3)
5.5. PREMIUM CALCULATION
155
Suppose now that the sum insured is V, and it is payable at the end of
the year of death. (Again, this is a simplifying assumption for ease of presentation.) The probability that the benet is paid at time w, w = 1> 2> 3> = = =, is
the probability that the policyholder survives to time w1, at which time the
policyholder is aged { + w 1, then dies during the following year, between
ages { + w 1 and { + w. The probability of this is
¡
¢
3(w31)
1 h3
w31 s{ t{+w31 = h
since t{+w31 = 1 s{+w31 = 1 h3 . Hence, the expected present value of
the benet is
"
X
3w 3(w31)
Vh
h
w=1
"
¡
¢
¡
¢X
3
3
h3w h3(w31)
1h
= V 1h
w=1
=
V (1 h3 ) h3
.
1 h3(+)
Denition 5.6 The symbol D{ denotes the expected present value of a payment of 1 at the end of the year of death of a life now aged {.
Remark 5.6 We could write Dl{ to emphasise that the valuation rate of interest is l p.a. eective.
In general,
D{ =
"
X
3w
h
w31 s{ t{+w31 =
w=1
"
X
h3w
w=1
g{+w31
>
o{
and so under our model of mortality,
D{ =
"
X
¡
¢ h3 (1 h3 )
h3w h3(w31) 1 h3 =
.
3(+)
1
h
w=1
(5.4)
Using the principle of equivalence, we set the expected present value of the
premiums equal to the expected present value of the benet. Thus
S{ d̈{ = V D{
,
S{
V h3 (1 h3 )
=
>
1 h3(+)
1 h3(+)
giving
S{ = V h3 (1 h3 )=
156
CHAPTER 5. VALUATION OF CONTINGENT PAYMENTS
Thus S{ is a function of V, and . To see how S{ behaves as a function
of these parameters, we simply need to dierentiate S{ with respect to each
parameter. First,
C
S{ = h3 (1 h3 ) A 0>
CV
so that S{ is an increasing function of V. This should be obvious — the larger
the sum insured, the greater the premium. Second,
C
S{ = V h3(+) A 0
C
so that S{ is an increasing function of . What this says is that if we have
two individuals aged {, whose forces of mortality are 1 and 2 respectively,
with 1 A 2 , to insure the individual with force of mortality 1 requires
payment of a higher premium. The intuitive explanation of this is that the
insurer expects to pay the death benet sooner to the life with the heavier
force of mortality, and consequently the insurer will receive fewer premiums
from that individual. In life insurance practice, the most obvious example
of this feature is that, in general, a woman would pay a smaller premium
than a man of the same age because female mortality is lighter than male
mortality. Finally,
C
S{ = Vh3 (1 h3 ) ? 0
C
so that S{ is a decreasing function of . Thus, the lower the rate of interest
that the insurer can earn (and hence the lower the value of ), the higher the
premium must be.
We stress that the above mortality model is a very simple one, which
leads to premiums which are independent of age. In life insurance practice,
premiums increase with age because the mortality risk increases with age.
The above calculations take no account of the expenses of issuing and
running an insurance policy. In insurance practice, the premium calculation
would also allow for the expenses of running the business. We have ignored
this in the above, for simplicity.
Example 5.8 A life insurance company uses the following assumptions to
calculate the premium, payable annually in advance, for a whole life insurance
policy under which the sum insured is payable at the end of the year of death:
Interest: a force of interest of p.a.,
Mortality: a constant force of mortality of .
(a) Calculate the annual premium for a policy with sum insured
$50,000 when = 0=02 and = 0=05.
5.5. PREMIUM CALCULATION
157
(b) Calculate the annual premium for a policy with sum insured
$50,000 when = 0=03 and = 0=05.
(c) Calculate the annual premium for a policy with sum insured
$60,000 when = 0=02 and = 0=05.
(d) Calculate the annual premium for a policy with sum insured
$50,000 when = 0=02 and = 0=06.
Solution 5.8 In each case we denote by S the annual premium. The basic
equation that gives S is
S d̈{ = V D{
where { is the policyholder’s age, V is the sum insured, and d̈{ and D{ are
given by (5.3) and (5.4) respectively. As we have seen above, with a constant
force of mortality, the values of d̈{ and D{ are independent of age.
(a) With = 0=02 and = 0=05, we have
d̈{ =
and
1
1
=
= 14=792
3(+)
1h
1 h30=07
h3 (1 h3 )
1 h3(+)
h30=05 (1 h30=02 )
= 50,000
1 h30=07
= 13,930=39=
50,000 D{ = 50,000
Setting
14=792 S = 13,930=39
gives
S = $941=78.
(b) With = 0=03 and = 0=05, we have
d̈{ = 13=007
and
50,000 D{ = 18,282=87
giving S = $1,405=65.
(c) With = 0=02 and = 0=05, we have d̈{ = 14=792, and with a
sum insured of $60,000, the expected present value of the benet
is
60,000 D{ = 16,716=46
giving S = $1,130=14.
158
CHAPTER 5. VALUATION OF CONTINGENT PAYMENTS
(d) With = 0=02 and = 0=06, we have
d̈{ = 13=007
and
50,000 D{ = 12,127=54
giving S = $932=41.
Remark 5.7 If we take the situation in part (a) above as being a benchmark
— sum insured 50,000, = 0=02 and = 0=05 — then we can see that we have
conrmed numerically the points made above about how the premium behaves
as parameters change. We see that
(i) increasing the force of mortality, , by 0.01 results in a premium increase from $941.78 to $1,405.65,
(ii) increasing the sum insured from $50,000 to $60,000 results in a premium increase from $941.78 to $1,130.14, and
(iii) increasing the force of interest, , by 0.01 results in a premium reduction from $941.78 to $932.41.
Remark 5.8 Note that in Parts (b) and (d), the total of + is 0.08,
resulting in the same value for d̈{ in each part. However, we do not get the
same value for D{ in each case. This illustrates a general point, namely that
in calculating the expected present value of an annuity, dierent assumptions
on mortality and interest can lead to the same expected present value. It is
the combination of and that determines the value of d̈{ .
5.5.2
Endowment insurance
Let us consider premium calculation for an endowment insurance, using exactly the same model as in Section 5.5.1. We assume that the term of the
policy is q years, and that the sum insured, V, is payable either at the end of
the policy term or at the end of the year of the policyholder’s death (provided
that death occurs before time q). We assume that premiums are payable annually in advance for at most q years.
Denition 5.7 The symbol d̈{:q denotes the expected present value of an
annuity of 1 payable annually in advance for at most q years as long as a
life now aged { survives.
5.5. PREMIUM CALCULATION
159
Let S{ (q) denote the annual premium. Then the expected present value
of the premium income is
S{ (q) d̈{:q
= S{ (q)
= S{ (q)
q31
X
w=0
q31
X
h3w w s{
h3w h3w
w=0
= S{ (q)
1 h3(+)q
.
1 h3(+)
Now note that for w = 1> 2> = = = > q, the probability that the sum insured is
payable at time w because the policyholder dies between times w 1 and w is
¡
¢
3(w31)
1 h3 ,
w31 s{ t{+w31 = h
just as it was under the whole life insurance policy, and the probability that
the sum insured is payable at time q because the policyholder is alive then
is h3q .
Denition 5.8 The symbol D{:q denotes the expected present value of a benet of 1 payable at the end of the year of death of a life aged { if the life dies
within q years, or payable at the end of q years if the life survives that term.
The expected present value of the benet is
!
à q
X
¡
¢ 3w
3(w31)
3
3q 3q
h
1h
h +h h
V D{:q = V
= V
à w=1
h3
q
X
w=1
h3(w31) h3(w31) q
X
w=1
h3w h3w + h3(+)q
!
μ
¶
3(+)q
3(+)q
3 1 h
3(+) 1 h
3(+)q
= V h
h
+h
1 h3(+)
1 h3(+)
¶
μ
3(+)q
h3(+) h3(+)q
3 1 h
.
= V h
1 h3(+)
1 h3(+)
Using the principle of equivalence we get
S{ (q) d̈{:q = VD{:q
¡
¡
¢
£
¢¤
, S{ (q) 1 h3(+)q = V h3 (1 h3(+)q ) h3(+) h3(+)q
160
giving
CHAPTER 5. VALUATION OF CONTINGENT PAYMENTS
¶
μ
h3(+) h3(+)q
3
.
S{ (q) = V h 1 h3(+)q
Thus, S{ (q) is a function of the sum insured, V, the force of mortality,
, the force of interest, , and the term, q. It is left as an exercise for you to
verify that if we x three of these parameters and vary the other, then
(i) S{ (q) is an increasing function of V,
(ii) S{ (q) is an increasing function of ,
(iii) S{ (q) is a decreasing function of , and
(iv) S{ (q) is a decreasing function of q.
Remark 5.9 The above comments do not apply to all types of insurance.
For example, it is possible for the premium for term insurance to decrease
with the policy term. Such a situation could arise if insurance is eected at
an age at which mortality rates are decreasing. For example, in the Male
Mortality Table, mortality rates decrease from age 21. On this mortality
table, the (single) premium for a one year term insurance for a life aged 21
could be less than the (annual) premium for a ten year term insurance. This
feature is being driven by mortality rates during a longer policy term being
lower on average than the mortality rates during the shorter policy term.
Remember that our model is a simplied one, where the force of mortality
is assumed constant throughout the term of the contract, and no allowance
has been made for factors other than interest and mortality. In fact, the
expenses incurred by the company in conducting their business are a major factor to be taken into account, and may aect the protability of the
business signicantly if the actual expenses incurred vary much from the
assumptions made in premium calculations.
Example 5.9 A life insurance company uses the following assumptions to
calculate the premium, payable annually in advance, for an endowment insurance policy under which the sum insured is payable at the end of the year
of death:
Interest: a force of interest of p.a.,
Mortality: a constant force of mortality of .
(a) Calculate the annual premium for a policy with sum insured
$50,000 and term ten years when = 0=02 and = 0=05.
5.5. PREMIUM CALCULATION
161
(b) Calculate the annual premium for a policy with sum insured
$50,000 and term ten years when = 0=03 and = 0=05.
(c) Calculate the annual premium for a policy with sum insured
$60,000 and term ten years when = 0=02 and = 0=05.
(d) Calculate the annual premium for a policy with sum insured
$50,000 and term ten years when = 0=02 and = 0=06.
(e) Calculate the annual premium for a policy with sum insured
$50,000 and term fteen years when = 0=02 and = 0=05.
Solution 5.9 As in Example 5.8, we denote by S the annual premium in
each case. The basic equation that gives S is
S d̈{:q = VD{:q
where { is the policyholder’s age, and V is the sum insured. Then
d̈{:q =
and
1 h3(+)q
1 h3(+)
1 h3(+)q h3(+) h3(+)q
=
1 h3(+)
1 h3(+)
As we have seen above, and as in Example 5.8, with a constant force of
mortality the values of d̈{:q and D{:q are independent of age.
D{:q = h3
(a) With V = 50,000, q = 10, = 0=02 and = 0=05, we have
d̈{:10 =
and
D{:10 = h30=05
giving
1 h30=7
= 7=446
1 h30=07
h30=07 h30=7
1 h30=7
= 0=63684
1 h30=07
1 h30=07
0=63684
= 4,276=23.
7=446
(b) With V = 50,000, q = 10, = 0=03 and = 0=05, we have
S = 50,000
d̈{:10 = 7=162
and
D{:10 = 0=65069
giving
S = 50,000
0=65069
= 4,542=38.
7=162
162
CHAPTER 5. VALUATION OF CONTINGENT PAYMENTS
(c) With V = 60,000, q = 10, = 0=02 and = 0=05, we have
S = 60,000
0=63684
= 5,131=48.
7=446
(d) With V = 50,000, q = 10, = 0=02 and = 0=06, we have
d̈{:10 = 7=162
and
D{:10 = 0=58289
giving
S = 50,000
0=58289
= 4,069=13.
7=162
(e) With V = 50,000, q = 15, = 0=02 and = 0=05, we have
d̈{:15 = 9=615
and
D{:15 = 0=53105
giving
S = 50,000
0=53105
= 2,761=45.
9=615
Remark 5.10 If, as in Example 5.8, we take the situation in part (a) above
as being a benchmark — sum insured 50,000, term ten years, = 0=02 and
= 0=05 — then we can see that we have conrmed numerically the points
made above about how the premium behaves as parameters vary. We see that
(i) Increasing by 0.01 results in a premium increase from $4,276.23 to
$4,542.38.
(ii) Increasing the sum insured from $50,000 to $60,000 results in a premium increase from $4,276.23 to $5,131.48.
(iii) Increasing by 0.01 results in a premium reduction from $4,276.23 to
$4,069.13.
(iv) Increasing q from 10 to 15 results in a premium reduction from
$4,276.23 to $2,761.45.
Remark 5.11 Note that Remark 5.8 applies to an annuity payable for at
most q years — the values of d̈{:10 are identical in parts (b) and (d) above.
5.5. PREMIUM CALCULATION
163
Example 5.10 On the basis of the Male Mortality Table and interest at 5%
p.a. eective, calculate the premium, payable annually in advance for a ten
year endowment insurance with sum insured $25,000 issued to a life aged 40.
Solution 5.10 Let us solve for the annual premium, S , using a spreadsheet
calculation. We have
!
à 10
9
X
X
S
y w w s40 = 25,000
yw w31 s40 t40+w31 + y 10 10 s40
w=0
or
w=1
!
à 10
X
o
g
o
40+w
40+w31
50
.
yw
= 25,000
yw
+ y 10
S
o
o
o
40
40
40
w=0
w=1
9
X
Using Excel, we start by setting up values of { and o{ in columns A and B
of our spreadsheet. Set these as headings in cells A1 and B1, then set the
values 40> 41> = = = > 50 in cells A2 to A12, and the values o40 > o41 > = = = > o50 in cells
B2 to B12. P
To calculate 9w=0 y w w s40 set up column headings w, y w , w s40 and Product in
cells C1 to F1. In cells C2 to C12, enter the values 0> 1> = = = > 10. Next, in
cell D2, enter the expression = 1=05ˆ F2, then copy this formula into cells
D3 to D12. In cell E2, enter = E2@E$2, then copy this formula into cells
E3 to E11. Note that the denominator in the entry in cell E2 is E$2 so that
when we copy this formula, we retain the value of o40 as the denominator in
w s40 . In cell F2, enter = G2 × H2, then copy this formula
P into cells F3 to
F11. Finally, in cell F13 enter = VXP(I 2 : I 11) to get 9w=0 y w w s40 . Thus,
P9
w
w=0 y w s40 = 7=9723.
We now want to calculate the expected present value of the insurance benet.
For the death benet, insert the heading g40+w31 @o40 in cell G1, then in cell
G2 enter = (E2 E3)@E$2. Copy this formula into cells G3 to G11. Enter
the heading Product in cell H1, then enter = G3 J2 in cell H2. Copy this
formula into cells H3 to H11. Summing the entries in cells H2 to H11 in cell
H13 gives
10
X
g40+w31
yw
= 0=0367.
o40
w=1
Finally, enter the expected present value of the survival benet in cell H15
(with an appropriate reference in cell G15) as = G12 E12@E2.
The premium, which we calculate in cell H17 is given by the formula =
25,000 (K13 + K15)@I 13, giving S = $1,945=39.
The full spreadsheet is shown as Table 5.2.
164
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
CHAPTER 5. VALUATION OF CONTINGENT PAYMENTS
A
{
40
41
42
43
44
45
46
47
48
49
50
B
C
o{
w
94,718 0
94,417 1
94,088 2
93,727 3
93,332 4
92,900 5
92,427 6
91,910 7
91,345 8
90,728 9
90,054 10
D
yw
1.0000
0.9524
0.9070
0.8638
0.8227
0.7835
0.7462
0.7107
0.6768
0.6446
0.6139
E
F
G
s
Product
g
w 40
40+w31 @o40
1.0000 1.0000
0.0032
0.9968 0.9494
0.0035
0.9933 0.9010
0.0038
0.9895 0.8548
0.0042
0.9854 0.8107
0.0046
0.9808 0.7685
0.0050
0.9758 0.7282
0.0055
0.9704 0.6896
0.0060
0.9644 0.6527
0.0065
0.9579 0.6175
0.0071
7.9723
H
Product
0.0030
0.0032
0.0033
0.0034
0.0036
0.0037
0.0039
0.0040
0.0042
0.0044
0.0367
Survival
0.5837
Premium
1,945.39
Table 5.2: Spreadsheet output for Example 5.10
5.6
Relationships between Actuarial Functions
In Chapter 2 we saw that relationships exist between present values of annuities. For example, we saw that
d̈q = 1 + dq31 =
Similar relationships exist between expected present values of annuities under
which payments are contingent on survival. Consider an annuity of 1 p.a.
payable in advance to a life aged ({) as long as ({) is alive. The expected
present value of this annuity at eective rate l p.a. is
d̈{ =
"
X
w
y w s{ =
w=0
"
X
w=0
yw
o{+w
=
o{
Separating out the rst term in the sum we see that
d̈{ = 1 +
= 1+
"
X
w=1
"
X
u=0
yw
o{+w
o{
y u+1
o{+u+1
o{
5.6. RELATIONSHIPS BETWEEN ACTUARIAL FUNCTIONS
o{+1 X u o{+1+u
y
= 1+y
o{ u=0
o{+1
165
"
= 1 + y s{ d̈{+1 =
(5.5)
This is a useful relationship which allows us to evaluate the expected present
value of annuities recursively, given survival probabilities and an interest
rate. We can interpret this result as follows. The expected present value of
the rst payment is just 1 — the payment is certain to be made. The term
y s{ d̈{+1 is the expected present value at age { of payments from age { + 1,
which we can think of as the expected present value at age { + 1 of payments
from age { + 1 (i.e. d̈{+1 ), ‘discounted’ for one year allowing for both interest
and survival.
A similar argument (see Exercise 13) gives
D{ = y t{ + y s{ D{+1 .
(5.6)
Example 5.11 On the basis of a certain mortality table and interest at 4%
p.a. eective, d̈50 = 15=998= Given that s50 = 0=9952 and s51 = 0=9944
calculate d̈52 =
Solution 5.11 We rst calculate d̈51 from
d̈50 = 1 + y s50 d̈51
giving
d̈51 = 1=04
d̈50 1
= 15=673=
s50
d̈52 = 1=04
d̈51 1
= 15=346=
s51
Similarly,
Formulae (5.5) and (5.6) relate a function at one age to its value at the
next age. We now show that there is a link between d̈{ and D{ . Before doing
so, we introduce a piece of notation. Let g = l@(1+l), so that g is the present
value at time 0 of l at time 1. With this notation we have
d̈w = (1 + l) dw = (1 + l)
1 yw
1 yw
=
>
l
g
and we use this expression to obtain our relationship between d̈{ and D{ .
Recall that d̈{ is the expected present value of payments of 1 annually
in advance as long as ({) is alive. The present value of these payments is a
166
CHAPTER 5. VALUATION OF CONTINGENT PAYMENTS
random variable, and the possible values that this variable can take are as
follows:
if ({) dies before age { + 1>
if ({) dies between ages { + 1 and { + 2>
if ({) dies between ages { + 2 and { + 3>
===
if ({) dies between ages { + w 1 and { + w>
===
1 = d̈1
1 + y = d̈2
1 + y + y 2 = d̈3
===
d̈w
===
The expected value of the present value random variable is found by multiplying together the possible values the random variable can take with the
probability the random variable takes that value, then adding these together,
as in formula (5.2). As the probability that ({) dies between ages { + w 1
and { + w is w31 s{ w s{ , we have
d̈{ =
"
X
w=1
=
d̈w (w31 s{ w s{ )
"
X
1 yw
g
w=1
Now
P"
(w31 s{ w s{ ) =
w=1 (w31 s{ w s{ ) = 1 and
w31 s{ w s{ =
o{+w31 o{+w
g{+w31
=
>
o{
o{
so
d̈{
1
=
g
=
Ã
"
X
g{+w31
1
yw
o{
w=1
!
1
(1 D{ ) =
g
Alternatively we can write
1 = g d̈{ + D{ =
(5.7)
To interpret this equation, consider a loan of 1, repayable after q years, with
interest (at l p.a. eective) payable annually in advance. The equation of
value for this transaction is
1 = g d̈q + y q
5.7. PARAMETER VARIABILITY
167
where the right hand side represents the present value of the interest payments (g at the start of a year is equivalent in value to l at the end of that
year) plus the present value of the repayment of 1. If we now consider a loan
of 1 to ({), repayable at the end of the year of death of ({), with interest
(at l p.a. eective) payable annually in advance, then the expected present
value of the interest payments is g d̈{ and the expected present value of the
repayment of 1 is D{ .
Formula (5.7) is extremely useful as it can reduce the amount of calculation required when we calculate the premium for a whole life insurance
policy.
Example 5.12 As in the previous example, suppose that d̈50 = 15=998 on
the basis of a particular mortality table and interest at 4% p.a. eective.
Calculate the annual premium for a whole life insurance policy issued to (50)
with sum insured $100,000 payable at the end of the year of death.
Solution 5.12 Let S denote the annual premium. Then the principle of
equivalence gives
S d̈50 = 100,000 D50 = 100,000 (1 g d̈50 ) =
Hence
5.7
μ
¶
1
S = 100,000
g = $2,404=63=
d̈50
Parameter Variability
In Chapter 2 we considered the situation when rates of interest change. The
techniques for nding the present value of a certain payment when interest
rates change equally apply to nding the expected present value of a contingent payment. Consider rst a payment of 1 at time w to a life ({) provided
that ({) is alive at time w. At an eective rate of interest l p.a. the expected
present value of this contingent payment is y w w s{ = Suppose now that interest
rates change over (0> w) and that y(w) represents the present value of a (certain) payment of 1 at time w. The present value of a payment of 1 at time w
to ({), provided that ({) is alive, is a random variable which takes the value
y(w) with probability w s{ and takes the value 0 with probability w t{ . Hence,
the expected present value of this payment is y(w) w s{ .
Example 5.13 Find the expected present value of a payment of $100,000 ten
years from now to a life now aged 50 provided that the life is alive then. The
life is subject to the Male Mortality Table, and the eective rate of interest
is 5% p.a. for 5 years and 4.5% p.a. thereafter.
168
CHAPTER 5. VALUATION OF CONTINGENT PAYMENTS
Solution 5.13 The present value of a certain payment of 1 ten years from
now is
1=0535 1=04535 = 0=62874>
and as 10 s50 = 0=87794, the expected present value of the payment is
100,000 × 0=62874 × 0=87794 = 55,199=72=
The valuation of a series of contingent payments under varying interest
rates presents no new challenges. The expected present value of a series of
contingent payments is equal to the sum of the expected present values of
the individual payments.
Example 5.14 Find the expected present value of payments of $1,000 annually in advance for 10 years to a life subject to a constant force of mortality
of 0.001. Assume an eective rate of interest of 6% p.a. for the rst 5 years
and 5% p.a. thereafter.
Solution 5.14 The probability that the life survives w years is h30=001w . Thus,
the expected present value of the payment at time w is
1=063w h30=001w
for w = 0> 1> 2> 3 and 4, and
1=0635 1=053(w35) h30=001w
for w = 5> 6> 7> 8 and 9. Hence the expected present value of the annuity is
!
à 4
9
X
X
1=063w h30=001w +
1=0635 1=053(w35) h30=001w
1,000
μ
w=0
w=5
1 (h30=001 @1=06)5 h30=005 1 (h30=001 @1=05)5
+
= 1,000
1 h30=001 @1=06
1=065 1 h30=001 @1=05
¶
= 7,830=34=
In some situations we might assume that individuals are subject to different levels of mortality throughout their lifetimes. For example, people
whose occupation exposes them to a higher risk of accidental death might
be deemed to be subject to a higher level of mortality pre-retirement than
post-retirement. We can value life contingent payments in such a situation
by noting that for positive p and q,
p+q s{ = p s{+q q s{
and hence we can nd the expected present value of a payment to ({) at
time p + q if ({) is subject to one mortality model from age { to { + q and
another mortality model thereafter.
5.7. PARAMETER VARIABILITY
169
Example 5.15 A man aged 30 is subject to an extra risk between ages 30
and 40 such that his force of mortality is 10% greater than that of the Male
Mortality Table for all ages {, 30 { 40. From age 40 he is subject to
the Male Mortality Table. Using an eective rate of interest of 5% p.a., nd
the expected present value of a payment of $100,000 to this man at age 50,
provided he is alive then.
Solution 5.15 The probability that the man survives to age 40 is
½ Z 10
¾
exp 1=1 30+w gw = (10 s30 )1=1
0
since
½ Z 10
¾
30+w gw
10 s30 = exp 0
e
and exp{d e} = (exp{d}) . Hence the probability that the man survives to
age 50 is
(10 s30 )1=1 10 s40 = 0=93004>
and the expected present value of the payment is
100,000 y 20 (10 s30 )1=1 10 s40 = 35,052=37=
Example 5.16 A woman aged 40 is subject to a force of mortality
{ + 0=009479 for 40 { 60 where { is according to the Female Mortality
Table, and is subject to the Female Mortality Table thereafter. On the basis
of an eective rate of interest of 5% p.a., nd an expression in terms of actuarial functions for the expected present value of an annuity of $20,000 p.a.
payable in advance to this woman.
Solution 5.16 Let w sW40 denote the probability of survival to age 40 + w, allowing for the addition to the force of mortality up to age 60, and let w s40
denote the probability of survival to age 40 +w on the Female Mortality Table.
For 0 w 20, the probability that the woman survives to time w is
½ Z w
¾
W
(40+v + 0=009479) gv
w s40 = exp 0
½ Z w
¾
= exp 40+v gv exp{0=009479 w}
0
=
w s40 exp{0=009479 w}=
Hence the expected present value of the payment at time w, for w = 0> 1> 2> = = = > 19,
is
20,000 × 1=053w w s40 exp{0=009479w} = 20,000 × 1=063w w s40
170
CHAPTER 5. VALUATION OF CONTINGENT PAYMENTS
since exp{0=009479}@1=05 = 1@1=06= Thus, the expected present value of the
payments at times 0> 1> 2> = = = > 19 is
6%
=
20,000 d̈40:20
For w = 20> 21> 22> = = = > the expected present value of the payment at time w is
20,000 × 1=053w w sW40
3w
= 20,000 × 1=05
½ Z 20
¾
Z w
exp (40+v + 0=009479) gv 40+v gv
0
20
= 20,000 × 1=053w exp {20 × 0=009479} w s40 >
and so the expected present value of the payments at times 20> 21> 22> = = = is
20,000 exp {20 × 0=009479}
= 20,000 exp {20 × 0=009479}
= 20,000 × 1=06320 20 s40
"
X
"
X
1=053w w s40
w=20
"
X
1=053(u+20) u+20 s40
u=0
1=053u u s60
u=0
since u+20 s40 = 20 s40 × u s60 and exp {20 × 0=009479} × 1=05320 = 1=06320 .
Hence, the expected present value of the annuity is
´
³
6%
20
5%
20,000 d̈40:20 + y6% 20 s40 d̈60 =
5.8
Exercises
1. The random variable [ has the following probability function:
Pr([ = 10) = 0=05
Pr([ = 12) = 0=12
Pr([ = 15) = 0=36
Pr([ = 20) = 0=27
Pr([ = 25) = 0=13
Pr([ = 32) = s
Calculate s and hence calculate H([).
2. Consider the random variable N({) from Example 5.2 which denotes
the number of whole years a life now aged { lives in the future. Under the assumption of a constant force of mortality, , show that the
probability function for N({) is
Pr(N({) = n) = h3n (1 h3 )
5.8. EXERCISES
171
for n = 0> 1> 2> = = = . Hence show that
H(N({)) =
1
h 1
and verify that H(N({)) is a decreasing function of .
Hint: You can use
"
"
" X
X
X
n
n{ =
{m .
n=1
n=1 m=n
3. A one year term insurance provides a death benet of $10,000, payable
at the end of the policy year. On the basis of interest at 6% p.a.
eective and the Male Mortality Table, calculate the premium for a
life aged (a) 40, (b) 45, and (c) 50. Comment on the ordering of these
premiums.
4. An insurance policy provides $20,000 if a man now aged exactly 45
survives ten years. On the basis of an eective rate of interest of l p.a.
and the Male Mortality Table, the single premium, payable at the issue
of this policy, is $11,190. Find l.
5. An insurance policy provides $20,000 20 years from the present if two
lives are alive, but only $10,000 if only one of the lives is alive. On the
basis of interest at 6% p.a. eective and a constant force of mortality
of 0.004 for each life, calculate the single premium payable now for this
policy. You may assume that the lives are independent with respect to
mortality.
6. Smith is subject to a constant force of mortality of 0.002, and Jones
is independently subject to a constant force of mortality of 0=002 + n.
They eect an insurance policy which provides $50,000 if exactly one
of them is alive 10 years after the policy is issued. On the basis of
this mortality and interest at an eective rate of 7% p.a., the single
premium, payable at the issue of the policy, is $1,224.75. Calculate n.
7. On the basis of the survival function v({) = 1 {@100 for 0 { 100>
calculate d̈55:20 when l = 0=07=
8. On the basis of the survival function v({) = h3{ for { 0 and interest
at l p.a. eective, nd an expression for D{ .
9. An industrial organisation is about to issue a bond which has a term of
25 years, a coupon of 12% payable annually in arrear, and is redeemable
at 120. You assess that the probability that the payment due at the
172
CHAPTER 5. VALUATION OF CONTINGENT PAYMENTS
end of the mth year is 0=995m , for m = 1> 2> ===> 25. What price would
you pay for this bond at its issue date to obtain a yield of 14% p.a.
eective?
10. The probability that a man is alive exactly w years from the present
is 0=95w for w = 1> 2> 3> = = =, and the probability that his wife is alive
exactly w years from the present is 0=97w for w = 1> 2> 3> = = =. Assuming
the couple are independent with respect to mortality, calculate the
expected present value at 8% p.a. eective of a pension of $20,000 p.a.,
payable annually in arrear and starting one year from now, provided
the man is alive, reducing to $12,000 p.a. if only his wife is alive.
11. A man aged 25 has just purchased an endowment insurance policy
with sum insured $100,000 and term 30 years. The insurance company
assumes that the man is subject to a constant force of mortality of
0.001 and that it can earn interest at 7% p.a. eective for 10 years,
and at 6% p.a. eective thereafter. Calculate the annual premium for
this policy.
12. A woman aged exactly 45 has just purchased a deferred annuity from
an insurance company. The annuity commences at exact age 60 (if she
is alive then) and is payable annually thereafter provided that she is
still alive. Each annuity payment is $50,000. The insurance company
assumes the woman is subject to a constant force of mortality of 0.025
up to age 80 and thereafter to a constant force of mortality of 0.03.
Calculate the single premium for this deferred annuity using an eective
rate of interest of 6% p.a.
13. Show that D{ = y t{ + y s{ D{+1 =
14. Show that 1 = g d̈{:q + D{:q .
15. A life eects an endowment insurance policy with term 20 years and
annual premiums of $2,000, payable in advance. The premium has
been calculated using an eective interest rate of 7% p.a. The expected
present value of a unit benet, payable at the end of the year of death
or at the maturity date, is 0.2965. What is the sum insured under this
policy?
Use a spreadsheet to solve the following problems.
16. Solve Exercise 9 using a spreadsheet calculation.
17. Solve Exercise 10 using a spreadsheet calculation.
5.8. EXERCISES
173
18. On the basis of the Male Mortality Table and interest at 8% p.a. effective, calculate d̈40:10 and d̈50:10 . Explain why the former exceeds the
latter.
19. On the basis of the Male Mortality Table and interest at 7% p.a. eective, calculate the annual premium for an endowment insurance policy
with sum insured $50,000 (payable at the end of the year of death or
on maturity) and term 15 years, issued to a life aged exactly 45.
20. By how much would the premium in Exercise 19 increase if the policy
conditions changed so that the benet on survival to age 60 was $60,000
rather than $50,000?
21. Repeat Exercise 11, but now assume that the man is subject to the
Male Mortality Table.
22. Repeat Exercise 12, but now assume that the woman is subject to the
Female Mortality Table up to age 80, and that after age 80 her force
of mortality is 50% greater than under the Female Mortality Table.
23. On the basis of the survival function v({) = 1 {@100 for 0 { 100
calculate D{ for { = 80> 81> 82> = = = > 99 when l = 0=06 using the result in
Exercise 13.
Further Reading and
Acknowledgements
The topics introduced in Chapter 2 are described in great detail by
McCutcheon and Scott (1986). This textbook also provides an in-depth
discussion of many other problems involving compound interest.
There are many textbooks available on demography. At an introductory
level, Pollard et al. (1990) describe demographic techniques and illustrate
many of these using Australian data. Hinde (1998) gives a slightly more
advanced treatment of the subject. Cox (1975), although less up-to-date,
provides a very broad discussion of the subject, ranging from the practical to
the theoretical. United Nations Demographic Yearbooks are a useful source
of information. Nowadays, very up-to-date demographic data can be found
on the internet. Data from websites of the Australian Bureau of Statistics,
the Institut national d’études démographiques (France) and the U.S. Census
Bureau were used to construct tables and graphs in Chapter 3.
The information on insurance products in Chapter 4 is based largely on
customer information brochures produced by Australian companies. Nowadays most insurance companies provide detailed information brochures, and
many companies also publish these on the internet. Indeed the internet is
a very useful resource for nding information about an insurance company
and its products. Much information about superannuation funds can also be
found on the internet. A very accessible collection of essays on the material
covered in Chapter 4 can be found in Renn (ed.) (1998).
In Chapter 5 we provided an introduction to the valuation of contingent
payments. The ideas in this chapter are expanded on by Dickson et al.
(2009). See also Bowers et al. (1997) who not only describe the pricing of
life insurance policies, but also discuss some mathematical models used in
the pricing of general insurance products.
The mortality tables included in the appendices are hypothetical ones.
They have been constructed to re ect the mortality experience of a developed country such as Australia and are constructed from formulae given in
Benjamin and Pollard (1980).
174
175
The population pyramids in Chapter 3 were produced in Excel. There
are a number of internet sites which explain how to do this. The spreadsheet
solutions provided in this text are described in terms of Excel commands.
This is the software that has been used in teaching at Melbourne. However,
any spreadsheet package can be used and the authors’ use of Excel should
not be taken as a recommendation.
Appendix 1
Male Mortality Table
176
177
Age, {
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
o{
100,000
99,014
98,948
98,908
98,878
98,852
98,828
98,805
98,783
98,760
98,737
98,713
98,687
98,658
98,622
98,574
98,509
98,422
98,312
98,180
98,033
97,877
97,718
97,563
97,414
97,273
97,139
97,011
96,886
96,761
96,634
96,502
96,362
96,212
96,050
95,874
95,682
95,473
95,244
94,993
g{
986
66
40
30
26
24
23
22
23
23
24
26
29
36
48
65
87
110
132
147
156
159
155
149
141
134
128
125
125
127
132
140
150
162
176
192
209
229
251
275
s{
0.99014
0.99933
0.99960
0.99970
0.99974
0.99976
0.99977
0.99978
0.99977
0.99977
0.99976
0.99974
0.99971
0.99964
0.99951
0.99934
0.99912
0.99888
0.99866
0.99850
0.99841
0.99838
0.99841
0.99847
0.99855
0.99862
0.99868
0.99871
0.99871
0.99869
0.99863
0.99855
0.99844
0.99832
0.99817
0.99800
0.99782
0.99760
0.99736
0.99711
t{
0.00986
0.00067
0.00040
0.00030
0.00026
0.00024
0.00023
0.00022
0.00023
0.00023
0.00024
0.00026
0.00029
0.00036
0.00049
0.00066
0.00088
0.00112
0.00134
0.00150
0.00159
0.00162
0.00159
0.00153
0.00145
0.00138
0.00132
0.00129
0.00129
0.00131
0.00137
0.00145
0.00156
0.00168
0.00183
0.00200
0.00218
0.00240
0.00264
0.00289
{
0.01520
0.00529
0.00054
0.00035
0.00028
0.00025
0.00024
0.00023
0.00023
0.00023
0.00024
0.00025
0.00028
0.00033
0.00043
0.00057
0.00077
0.00100
0.00123
0.00142
0.00155
0.00161
0.00161
0.00156
0.00149
0.00141
0.00135
0.00130
0.00129
0.00130
0.00134
0.00141
0.00150
0.00162
0.00176
0.00192
0.00210
0.00229
0.00252
0.00277
r
h{
69.10
68.78
67.83
66.85
65.87
64.89
63.91
62.92
61.93
60.95
59.96
58.98
57.99
57.01
56.03
55.06
54.09
53.14
52.20
51.27
50.35
49.42
48.50
47.58
46.65
45.72
44.78
43.84
42.90
41.95
41.01
40.06
39.12
38.18
37.24
36.31
35.38
34.46
33.54
32.63
O{
99,424
98,942
98,927
98,892
98,865
98,840
98,816
98,794
98,772
98,749
98,725
98,700
98,673
98,641
98,599
98,543
98,467
98,369
98,248
98,107
97,955
97,797
97,640
97,488
97,343
97,205
97,075
96,948
96,824
96,698
96,569
96,433
96,288
96,132
95,963
95,779
95,579
95,360
95,120
94,858
W{
6,909,573
6,810,149
6,711,208
6,612,281
6,513,389
6,414,524
6,315,684
6,216,868
6,118,074
6,019,302
5,920,554
5,821,828
5,723,128
5,624,455
5,525,814
5,427,215
5,328,672
5,230,205
5,131,836
5,033,588
4,935,481
4,837,525
4,739,728
4,642,088
4,544,600
4,447,257
4,350,052
4,252,977
4,156,029
4,059,205
3,962,507
3,865,939
3,769,506
3,673,218
3,577,086
3,481,123
3,385,343
3,289,764
3,194,404
3,099,284
178
Age, {
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
APPENDIX 1. MALE MORTALITY TABLE
o{
94,718
94,417
94,088
93,727
93,332
92,900
92,427
91,910
91,345
90,728
90,054
89,319
88,518
87,645
86,696
85,664
84,544
83,330
82,016
80,595
79,062
77,411
75,637
73,735
71,700
69,529
67,221
64,774
62,190
59,472
56,625
53,658
50,581
47,409
44,159
40,852
37,512
34,165
30,842
27,574
g{
301
329
361
395
432
473
517
565
617
674
735
801
873
949
1,032
1,120
1,214
1,314
1,421
1,533
1,651
1,774
1,902
2,035
2,171
2,308
2,447
2,584
2,718
2,847
2,967
3,077
3,172
3,250
3,307
3,340
3,347
3,323
3,268
3,180
s{
0.99682
0.99652
0.99616
0.99579
0.99537
0.99491
0.99441
0.99385
0.99325
0.99257
0.99184
0.99103
0.99014
0.98917
0.98810
0.98693
0.98564
0.98423
0.98267
0.98098
0.97912
0.97708
0.97485
0.97240
0.96972
0.96681
0.96360
0.96011
0.95630
0.95213
0.94760
0.94266
0.93729
0.93145
0.92511
0.91824
0.91078
0.90274
0.89404
0.88467
t{
0.00318
0.00348
0.00384
0.00421
0.00463
0.00509
0.00559
0.00615
0.00675
0.00743
0.00816
0.00897
0.00986
0.01083
0.01190
0.01307
0.01436
0.01577
0.01733
0.01902
0.02088
0.02292
0.02515
0.02760
0.03028
0.03319
0.03640
0.03989
0.04370
0.04787
0.05240
0.05734
0.06271
0.06855
0.07489
0.08176
0.08922
0.09726
0.10596
0.11533
{
0.00304
0.00334
0.00367
0.00403
0.00443
0.00487
0.00536
0.00589
0.00647
0.00712
0.00783
0.00860
0.00946
0.01040
0.01143
0.01257
0.01381
0.01518
0.01669
0.01834
0.02015
0.02214
0.02433
0.02673
0.02937
0.03225
0.03542
0.03890
0.04270
0.04687
0.05144
0.05644
0.06191
0.06789
0.07443
0.08157
0.08938
0.09789
0.10716
0.11727
r
h{
31.72
30.82
29.93
29.04
28.16
27.29
26.42
25.57
24.73
23.89
23.07
22.25
21.45
20.66
19.88
19.11
18.36
17.62
16.89
16.18
15.48
14.80
14.14
13.49
12.86
12.24
11.65
11.07
10.51
9.96
9.44
8.93
8.45
7.98
7.53
7.10
6.69
6.29
5.92
5.56
O{
94,570
94,255
93,910
93,532
93,119
92,667
92,172
91,632
91,041
90,396
89,692
88,924
88,088
87,177
86,187
85,112
83,945
82,682
81,315
79,838
78,247
76,534
74,697
72,729
70,626
68,387
66,009
63,493
60,842
58,059
55,151
52,128
49,002
45,790
42,509
39,184
35,838
32,500
29,202
25,975
W{
3,004,426
2,909,856
2,815,601
2,721,691
2,628,159
2,535,039
2,442,372
2,350,200
2,258,568
2,167,527
2,077,131
1,987,439
1,898,515
1,810,428
1,723,250
1,637,063
1,551,952
1,468,007
1,385,325
1,304,010
1,224,172
1,145,926
1,069,391
994,694
921,966
851,340
782,953
716,944
653,451
592,609
534,550
479,399
427,271
378,268
332,479
289,969
250,786
214,948
182,448
153,246
179
Age, {
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
o{
24,394
21,334
18,427
15,702
13,185
10,897
8,852
7,058
5,516
4,219
3,152
2,297
1,630
1,124
751
486
303
182
105
58
30
g{
3,060
2,907
2,725
2,517
2,288
2,045
1,794
1,542
1,297
1,067
855
667
506
373
265
183
121
77
47
28
15
s{
0.87456
0.86374
0.85212
0.83970
0.82647
0.81233
0.79733
0.78152
0.76487
0.74710
0.72874
0.70962
0.68957
0.66815
0.64714
0.62346
0.60066
0.57692
0.55238
0.51724
0.50000
t{
0.12544
0.13626
0.14788
0.16030
0.17353
0.18767
0.20267
0.21848
0.23513
0.25290
0.27126
0.29038
0.31043
0.33185
0.35286
0.37654
0.39934
0.42308
0.44762
0.48276
0.50000
{
0.12829
0.14026
0.15326
0.16737
0.18265
0.19922
0.21716
0.23650
0.25728
0.27981
0.30400
0.32973
0.35736
0.38746
0.41922
0.45384
0.49110
0.52989
0.57178
0.62638
0.67620
r
h{
5.22
4.89
4.59
4.30
4.03
3.77
3.52
3.30
3.08
2.88
2.69
2.52
2.35
2.19
2.05
1.91
1.78
1.66
1.54
1.41
1.31
O{
W{
22,853 127,270
19,866 104,418
17,048 84,551
14,425 67,503
12,021 53,078
9,854 41,057
7,934 31,203
6,266 23,269
4,848 17,003
3,667 12,155
2,708
8,488
1,949
5,780
1,365
3,831
927
2,467
611
1,539
389
929
238
540
140
302
80
161
43
82
22
39
Appendix 2
Female Mortality Table
180
181
Age, {
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
o{
100,000
99,155
99,099
99,068
99,046
99,028
99,013
98,999
98,986
98,973
98,960
98,946
98,931
98,914
98,894
98,870
98,841
98,807
98,768
98,724
98,676
98,625
98,572
98,518
98,463
98,407
98,351
98,293
98,234
98,173
98,109
98,041
97,969
97,891
97,807
97,716
97,617
97,509
97,391
97,262
g{
845
56
31
22
18
15
14
13
13
13
14
15
17
20
24
29
34
39
44
48
51
53
54
55
56
56
58
59
61
64
68
72
78
84
91
99
108
118
129
142
s{
0.99155
0.99944
0.99969
0.99978
0.99982
0.99985
0.99986
0.99987
0.99987
0.99987
0.99986
0.99985
0.99983
0.99980
0.99976
0.99971
0.99966
0.99961
0.99955
0.99951
0.99948
0.99946
0.99945
0.99944
0.99943
0.99943
0.99941
0.99940
0.99938
0.99935
0.99931
0.99927
0.99920
0.99914
0.99907
0.99899
0.99889
0.99879
0.99868
0.99854
t{
0.00845
0.00056
0.00031
0.00022
0.00018
0.00015
0.00014
0.00013
0.00013
0.00013
0.00014
0.00015
0.00017
0.00020
0.00024
0.00029
0.00034
0.00039
0.00045
0.00049
0.00052
0.00054
0.00055
0.00056
0.00057
0.00057
0.00059
0.00060
0.00062
0.00065
0.00069
0.00073
0.00080
0.00086
0.00093
0.00101
0.00111
0.00121
0.00132
0.00146
{
0.01301
0.00453
0.00044
0.00027
0.00020
0.00017
0.00015
0.00014
0.00013
0.00013
0.00014
0.00015
0.00016
0.00019
0.00022
0.00027
0.00032
0.00037
0.00042
0.00047
0.00050
0.00053
0.00054
0.00055
0.00056
0.00057
0.00058
0.00060
0.00061
0.00064
0.00067
0.00071
0.00077
0.00083
0.00089
0.00097
0.00106
0.00116
0.00127
0.00139
r
h{
76.73
76.39
75.43
74.45
73.47
72.48
71.49
70.50
69.51
68.52
67.53
66.54
65.55
64.56
63.57
62.59
61.61
60.63
59.65
58.68
57.71
56.74
55.77
54.80
53.83
52.86
51.89
50.92
49.95
48.98
48.01
47.04
46.08
45.11
44.15
43.19
42.24
41.28
40.33
39.38
O{
99,473
99,126
99,083
99,057
99,037
99,020
99,006
98,993
98,980
98,967
98,953
98,939
98,923
98,904
98,882
98,856
98,824
98,788
98,746
98,700
98,651
98,599
98,545
98,491
98,435
98,379
98,322
98,264
98,204
98,141
98,075
98,006
97,931
97,850
97,762
97,667
97,564
97,451
97,328
97,192
W{
7,673,452
7,573,979
7,474,853
7,375,770
7,276,714
7,177,677
7,078,656
6,979,650
6,880,658
6,781,678
6,682,712
6,583,759
6,484,820
6,385,897
6,286,993
6,188,110
6,089,254
5,990,430
5,891,642
5,792,896
5,694,196
5,595,545
5,496,946
5,398,401
5,299,911
5,201,476
5,103,097
5,004,774
4,906,511
4,808,307
4,710,166
4,612,090
4,514,085
4,416,154
4,318,305
4,220,542
4,122,875
4,025,311
3,927,860
3,830,533
182
Age, {
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
APPENDIX 2. FEMALE MORTALITY TABLE
o{
97,120
96,965
96,795
96,609
96,405
96,181
95,935
95,666
95,371
95,047
94,692
94,304
93,879
93,414
92,905
92,349
91,741
91,077
90,353
89,563
88,703
87,767
86,749
85,644
84,445
83,147
81,743
80,228
78,595
76,839
74,955
72,938
70,785
68,494
66,063
63,493
60,786
57,948
54,986
51,910
g{
155
170
186
204
224
246
269
295
324
355
388
425
465
509
556
608
664
724
790
860
936
1,018
1,105
1,199
1,298
1,404
1,515
1,633
1,756
1,884
2,017
2,153
2,291
2,431
2,570
2,707
2,838
2,962
3,076
3,177
s{
0.99840
0.99825
0.99808
0.99789
0.99768
0.99744
0.99720
0.99692
0.99660
0.99627
0.99590
0.99549
0.99505
0.99455
0.99402
0.99342
0.99276
0.99205
0.99126
0.99040
0.98945
0.98840
0.98726
0.98600
0.98463
0.98311
0.98147
0.97965
0.97766
0.97548
0.97309
0.97048
0.96763
0.96451
0.96110
0.95737
0.95331
0.94889
0.94406
0.93880
t{
0.00160
0.00175
0.00192
0.00211
0.00232
0.00256
0.00280
0.00308
0.00340
0.00373
0.00410
0.00451
0.00495
0.00545
0.00598
0.00658
0.00724
0.00795
0.00874
0.00960
0.01055
0.01160
0.01274
0.01400
0.01537
0.01689
0.01853
0.02035
0.02234
0.02452
0.02691
0.02952
0.03237
0.03549
0.03890
0.04263
0.04669
0.05111
0.05594
0.06120
{
0.00153
0.00168
0.00184
0.00202
0.00222
0.00244
0.00268
0.00295
0.00325
0.00357
0.00392
0.00431
0.00474
0.00521
0.00573
0.00630
0.00693
0.00762
0.00838
0.00922
0.01013
0.01114
0.01224
0.01346
0.01479
0.01626
0.01787
0.01964
0.02158
0.02371
0.02605
0.02862
0.03143
0.03452
0.03791
0.04162
0.04569
0.05014
0.05502
0.06036
r
h{
38.44
37.50
36.57
35.64
34.71
33.79
32.87
31.97
31.06
30.17
29.28
28.40
27.52
26.66
25.80
24.95
24.12
23.29
22.47
21.66
20.87
20.09
19.32
18.56
17.81
17.08
16.37
15.67
14.98
14.31
13.66
13.03
12.41
11.80
11.22
10.65
10.11
9.58
9.07
8.57
O{
97,044
96,881
96,704
96,509
96,295
96,060
95,803
95,521
95,212
94,872
94,501
94,095
93,650
93,164
92,632
92,050
91,414
90,721
89,964
89,140
88,242
87,266
86,205
85,053
83,805
82,455
80,996
79,423
77,729
75,909
73,959
71,874
69,652
67,291
64,791
62,152
59,379
56,478
53,458
50,330
W{
3,733,340
3,636,297
3,539,415
3,442,712
3,346,203
3,249,908
3,153,848
3,058,045
2,962,524
2,867,312
2,772,440
2,677,939
2,583,844
2,490,193
2,397,030
2,304,398
2,212,348
2,120,934
2,030,213
1,940,249
1,851,109
1,762,867
1,675,601
1,589,396
1,504,343
1,420,538
1,338,083
1,257,087
1,177,664
1,099,936
1,024,027
950,068
878,194
808,542
741,250
676,460
614,308
554,929
498,451
444,994
183
Age, {
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
o{
48,733
45,472
42,147
38,782
35,403
32,039
28,722
25,485
22,360
19,381
16,578
13,978
11,604
9,472
7,592
5,967
4,591
3,453
2,534
1,811
1,258
g{
3,261
3,325
3,365
3,379
3,364
3,317
3,237
3,125
2,979
2,803
2,600
2,374
2,132
1,880
1,625
1,376
1,138
919
723
553
410
s{
0.93308
0.92688
0.92016
0.91287
0.90498
0.89647
0.88730
0.87738
0.86677
0.85537
0.84317
0.83016
0.81627
0.80152
0.78596
0.76940
0.75212
0.73385
0.71468
0.69464
0.67409
t{
0.06692
0.07312
0.07984
0.08713
0.09502
0.10353
0.11270
0.12262
0.13323
0.14463
0.15683
0.16984
0.18373
0.19848
0.21404
0.23060
0.24788
0.26615
0.28532
0.30536
0.32591
{
0.06621
0.07260
0.07957
0.08718
0.09550
0.10457
0.11443
0.12519
0.13690
0.14960
0.16340
0.17836
0.19457
0.21213
0.23105
0.25150
0.27350
0.29715
0.32268
0.35014
0.37938
r
h{
8.10
7.64
7.21
6.79
6.39
6.01
5.64
5.30
4.97
4.66
4.36
4.08
3.81
3.56
3.32
3.09
2.88
2.66
2.46
2.25
2.03
O{
47,109
43,814
40,467
37,092
33,718
30,375
27,095
23,911
20,856
17,962
15,258
12,769
10,515
8,509
6,756
5,257
4,001
2,975
2,156
1,520
1,041
W{
394,664
347,554
303,740
263,273
226,181
192,463
162,088
134,994
111,083
90,228
72,266
57,007
44,238
33,723
25,214
18,458
13,201
9,200
6,225
4,069
2,549
Appendix 3
Solutions to Exercises
Chapter 2
1. The equation to be used to complete the table is
S (1 + u w) = D=
The complete table is as follows, with the answers shown in bold font.
Principal, S
1,000
1,000
1,500
3,000
1,200
1,562.50
1,000
1,071.43
2,053.57
u
Years, w Accumulation, D
8% p.a.
10
1,800
4.5% p.a
5
1,225
7% p.a
5.714
2,100
4% p.a.
1.667
3,200
12.5% p.a.
2
1,500
6% p.a.
10
2,500
6.67% p.a.
3
1,200
8% p.a.
5
1,500
3% p.a.
4
2,300
2. Under simple discount at 8% p.a., an investment of $1,000 at the start
of a year grows to
1,000
= 1,086=96
1 0=08
at the end of a year. Thus, the equivalent rate of simple interest, u
p.a., satises
1,000 (1 + u) = 1,086=96
so that u = 8=7% p.a.
184
185
3. Under simple discount of 6% p.a., the accumulated amount of $1,000
at the end of ve years will be
1,000
= $1,428=57=
1 0=06 × 5
Under simple interest of 7% p.a., the accumulated amount of $1,000 at
the end of ve years will be
1,000 (1 + 0=07 × 5) = $1,350=00=
Under simple interest of 3% every half year, the accumulated amount
of $1,000 at the end of ve years will be
1,000 (1 + 0=03 × 10) = $1,300=00=
Hence, simple discount of 6% p.a. is the preferred choice as it results
in the greatest accumulation over ve years.
4. The accumulation of a unit investment to time q years under an eective rate of 5% p.a. is 1=05q , whilst the accumulation of a unit investment under 6% p.a. simple interest is 1+0=06q= Thus, the accumulation
under compound interest is greater when
1=05q A 1 + 0=06q=
There is no analytical solution to the problem of nding the least value
of q for which this inequality holds. We can solve easily in a spreadsheet
and nd that q = 9 years. Relevant calculated values are
1=058 (1 + 0=06 × 8) = 0=0025>
1=059 (1 + 0=06 × 9) = 0=0113=
5. (a) From
¶6
μ
l(6)
h = 1+
6
we get = 0=09918=
(b) From
we get l = 0=08243=
¶4
μ
l(4)
1+l= 1+
4
186
APPENDIX 3. SOLUTIONS TO EXERCISES
(c) We have
1 y 10
d̈10 = (1 + l) d10 = (1 + l)
l
where
y
10
310
= (1 + l)
and
¶320
μ
l(2)
= 1+
= 0=50257
2
¶2
μ
l(2)
l= 1+
1 = 0=071225>
2
giving d̈10 = 7=48142=
(d) We have
(4)
d20 =
where
1 y 20
l(4)
¶4
μ
l(4)
1=065 = 1 +
4
(4)
gives l(4) = 0=06347, and hence d20 = 11=28356=
(e) We have
(1 + l)20 1
v̈20 = (1 + l) v20 = (1 + l)
l
¡
¢4
(4)
and 1 + l = 1 + l @4 = 1=10381, giving v̈20 = 66=02457=
6. (a) The eective annual rate l satises
¶2
μ
l(2)
1+l= 1+
= 1=0816>
2
so l = 0=0816 or 8=16%=
(b) The nominal rate p.a. payable monthly, l(12) , satises
¶12
μ
l(12)
>
1+l= 1+
12
giving l(12) = 0=0787 or 7.87%.
(c) The nominal rate p.a. payable quarterly, l(4) , satises
¶4
μ
l(4)
>
1+l= 1+
4
giving l(4) = 0=0792 or 7.92%.
187
(d) The accumulation at the end of two years is
100 (1 + l)2 = $116=99=
(e) The total interest earned on the investment is the dierence between the accumulated amount and the original investment, i.e.
$16=99.
7. (a) An investment of $D doubles in value in q years if
D (1=1q ) = 2 D
giving q log 1=1 = log 2, and so q = 7=2725=
(b) An investment of $D trebles in value in q years if
D (1=1q ) = 3 D
giving q log 1=1 = log 3, and so q = 11=5267=
8. With l = 0=071, the accumulated value at time 25 years is
¢
¡
300 (1 + l)25 + (1 + l)22 + (1 + l)19 + = = = + (1 + l)7 =
The terms to be summed are in geometric progression, and as there are
7 terms in the sum, the accumulated value is
300(1 + l)25
1 y 21
= $6,839=16=
1 y3
9. (a) The present value is
1,000
24
X
y w (1=07w ) = 1,000
w=0
1 (1=07 y)25
=
1 1=07 y
(b) When l = 0=05, the present value is
1,000
1 (1=07@1=05)25
= $31,643=63=
1 1=07@1=05
10. (a) As the payments are 25> 24> 23> = = = > 2> 1 we have
[ = 25y + 24y2 + 23y 3 + = = = + 2y24 + y 25 =
188
APPENDIX 3. SOLUTIONS TO EXERCISES
(b) Multiplying [ by 1 + l gives
(1 + l) [ = 25 + 24y + 23y 2 + = = = + 2y23 + y 24
and subtracting [ gives
l [ = 25 y y2 y3 y25
= 25 d25 >
and so
[=
25 d25
=
l
(c) When l = 0=07>
d25 =
1 1=07325
= 11=65358
0=07
and so [ = 190=66=
11. There are two ways to value this annuity. First, we can value it as the
sum of an annuity in arrear of $1,000 for 15 years and a 15 year deferred
annuity of $2,000 for 15 years. Using this approach the present value
is
1> 000 d15 + 2> 000 15 |d15
= 1> 000
1 1=09315
1 1=09315
+ 2> 000 (1=09315 )
0=09
0=09
= 8,060=69 + 4,425=93
= 12,486=62=
The second approach is to value an annuity in arrear of $2,000 for 30
years, and to subtract from this the present value of an annuity in
arrear of $1,000 payable for 15 years. Using this approach the present
value is
2,000 d30 1,000 d15
= 2,000
1 1=09315
1 1=09330
1,000
0=09
0=09
= 20,547=31 8,060=69
= 12,486=62=
189
Note that each approach leads to the following expression for the present
value:
1 + 1=09315 2(1=09330 )
=
1,000
0=09
12. The present value is the sum of the present values of the payments in
each of years 1 to 10, 11 to 20 and 21 to 30. The present value of the
payments in years 21 to 30 is
¶340
μ
0=1
310
1,000 (1=1 ) 1 +
d10m
4
where m is such that 1 + m = (1 + 0=1@12)12 , so that m = 0=104713 and
d10m = 6=022104= Hence the present value of the payments in years 21
to 30 is
1,000 × 0=385543 × 0=372431 × 6=022104 = 864=70=
The present value of the payments in years 11 to 20 is
1,000 (1=1310 ) d10n
where n is such that 1 + n = (1 + 0=1@4)4 , so that n = 0=103813 and
d10n = 6=045197= Hence the present value of the payments in years 11
to 20 is
1,000 × 0=385543 × 6=045197 = 2,330=69=
Finally, the present value of payments in years 1 to 10 is
1,000
1 1=1310
= 6,144=57>
0=1
so that the present value of the annuity is
864=70 + 2,330=69 + 6,144=57 = 9,339=95=
13. To solve this problem (and the next) the idea is that we break the time
period from time 0 to time 10 years into dierent intervals, based on
when either the interest rate changes or the amount of the payment
changes. There are four distinct intervals, as follows:
• years 1—3, when the payments are $100 annually in arrear and the
interest rate is 5% p.a. eective,
• years 4—5, when the payments are $100 annually in arrear and the
interest rate is 1% per month eective,
190
APPENDIX 3. SOLUTIONS TO EXERCISES
• years 6—8, when the payments are $50 monthly in arrear and the
interest rate is 1% per month eective,
• years 9—10, when the payments are $50 monthly in arrear and the
interest rate is 5% per half year eective.
(a) To nd the present value of the payments, we can nd the present
value of payments in each of the above four time periods, then
add these together.
The present value of payments in years 1—3 is
100 d35% = 272=3248=
The present value at time 3 of payments in years 4—5 is
¡
¢
100 1=01312 + 1=01324
and so the present value (at time 0) of the payments in years 4—5
is
¡
¢
1=0533 × 100 1=01312 + 1=01324 = 144=6941=
The present value at time 5 of payments in years 6—8 is
1%
50 d36
= 1,505=3753
and so the present value (at time 0) of the payments in years 6—8
is
1%
= 1,024=1508=
1=0533 × 1=01324 × 50 d36
The present value at time 8 of payments in years 8—10 is
(12)
600 d2
at eective rate m p.a. where 1 + m = 1=052 > giving m = 0=1025 and
(12)
m (12) = 0=097978= Hence d2 = 1=809562, and so the present value
(at time 0) of the payments in years 9—10 is
(12)
1=0533 × 1=01360 × 600 d2
= 516=2670=
Hence the present value of all the payments is 1,957=44.
(b) The value at time 5 is the value at time 0, accumulated for 5 years.
Thus, the value is
1,957=44 × 1=053 × 1=0124 = 2,877=19=
191
(c) The value at time 10 is the value at time 0, accumulated for 10
years. Thus, the value is
1,957=44 × 1=053 × 1=0160 × 1=054 = 5,003=75=
14. We break the time period from time 0 to time 10 years into dierent
intervals, based on when either the interest rate changes or the amount
of the payment changes. There are four distinct intervals, as follows:
• years 1—2, when the payments are $100 annually in arrear and the
interest rate is 5% per half year eective,
• years 3—5, when the payments are $100 annually in arrear and the
interest rate is 1% per month eective,
• years 6—7, when the payments are $50 monthly in arrear and the
interest rate is 1% per month eective,
• years 8—10, when the payments are $50 monthly in arrear and the
interest rate is 5% p.a. eective.
(a) The present value of payments in years 1—2 is
¡
¢
100 1=0532 + 1=0534 = 172=9732=
The present value at time 2 of payments in years 3—5 is
¡
¢
100 1=01312 + 1=01324 + 1=01336 = 237=3940
and so the present value (at time 0) of the payments in years 3—5
is
¡
¢
1=0534 × 100 1=01312 + 1=01324 + 1=01336 = 195=3047=
The present value at time 5 of payments in years 6—7 is
1%
50 d24
= 1,062=1694
and so the present value (at time 0) of the payments in years 6—7
is
1%
1=0534 × 1=01336 × 50 d36
= 610=7551=
The present value at time 7 of payments in years 8—10 is
(12)
600 d3
(12)
at 5% p.a. Hence d3 = 2=78511, and so the present value (at
time 0) of the payments in years 8—10 is
(12)
1=0534 × 1=01360 × 600 d3
= 756=7516=
Hence the present value of all the payments is 1,735=78.
192
APPENDIX 3. SOLUTIONS TO EXERCISES
(b) The value at time 5 is the value at time 0, accumulated for 5 years.
Thus, the value is
1,735=78 × 1=054 × 1=0136 = 3,018=72=
(c) The value at time 10 is the value at time 0, accumulated for 10
years. Thus, the value is
1,735=78 × 1=054 × 1=0160 × 1=053 = 4,437=14
In the previous exercise, the interest rate in the rst 3 years is 5%
p.a. eective. In this exercise, the interest rates in the rst 3 years
are equivalent to 10.25% p.a. eective in the rst 2 years, and 12.68%
p.a. in the third year. When the order of the interest rates changes
from the previous exercise, the present values of the earlier payments
thus decrease, and a little maths shows that the present value of each
payment decreases. For example, the present value of the payment at
time 8 years is
¡
¢
50 1=0533 × 1=01360
in the previous exercise, whereas it is now
¢
¡
50 1=0534 × 1=01360 × 1=0531 =
15. As 5% of the purchase price is $45, the purchaser of a $900 item pays
$300 at the purchase date, $300 one year later, and $345 two years
later. Thus, the equation of value is
900 = 300 + 300y + 345y2
or
345y2 + 300y 600 = 0=
The positive solution of this quadratic equation is y = 0=953802, giving
l = 0=0484=
16. The only dierence from the previous exercise is the timing of the
second payment. Thus, the equation of value is now
900 = 300 + 300y 0=5 + 345y2
or, equivalently,
3=45y2 + 3y0=5 6 = 0=
193
Now let i (l) = 3=45y2 + 3y0=5 6= The return to the store is greater
than in the previous exercise as the second payment is earlier. This
means that a trial solution should exceed 0.0484. We nd that
i (0=057) = 0=0059316
and
i (0=058) = 0=0012823=
Assume that i(l) is linear in l over the interval (0=057> 0=058), say i(l) =
d + el. Then
i(0=057) = 0=0059316 = d + 0=057e>
i(0=058) = 0=0012823 = d + 0=058e>
giving 0=001e = 0=0072139= Hence e = 7=2139 and d = 0=4171= We
want l such that i(l) = 0, so l = d@e = 0=0578=
17. The values are shown in the table below.
l
2%
5%
10%
15%
d10
8=9826
7=7217
6=1446
5=0188
10y
9=8039
9=5238
9=0909
8=6957
We note that 10y is an upper bound for d10 > based on y q ? y for
q = 1> 2> = = = and l A 0= The lower the rate of interest, the closer y q is
to y> and hence the closer d10 is to 10y= For example, when l = 0=02>
y = 0=98039 and y 10 = 0=82035> and when l = 0=15> y = 0=86957 and
y10 = 0=24718=
18. The price of the bond per $100 nominal is
(2)
12 d15 + 100 y15
= 12
@ 9%
1 1=09315
+ 100 (1=09315 )
0=08806
= 126=31
since 0=09(2) = 2(1=091@2 1) = 0=08806=
19. The term of the investment is 16 years and so the price per $100 nominal
194
APPENDIX 3. SOLUTIONS TO EXERCISES
is
8 d16 + 110 y 16
= 8
@ 8.5%
1 1=085316
+ 110 (1=085316 )
0=085
= 98=42=
20. The term of the investment is 12 years. Let l denote the eective rate
of interest p.a. Then the equation of value per $100 nominal is
(4)
96 = 10 d12 + 100 y 12 =
Roughly, the return is
10 + (100 96)@12
= 0=1076=
96
In performing this calculation we have made no allowance for quarterly
coupons, so we are underestimating the return. Let
(4)
j(l) = 10 d12 + 100 y12 96
so that the return is the value of l such that j(l) = 0= Then
j(0=11) = 0=12753>
j(0=111) = 0=47705=
Assuming that j(l) = d + e l over the interval (0=11> 0=111)> we obtain
0=12753 = d + 0=11 e>
0=47705 = d + 0=111e>
giving e = 604=576 and d = 66=63087= Then j(l) = 0 gives
l = d@e = 11=02%=
21. For Bond A, the price per $100 nominal is
6=5 d15 + 100 y 15
and
d15 =
@ 8%
1 1=08315
= 8=55948>
0=08
195
giving a price of $87.16.
For Bond B, the price per $100 nominal is
(4)
7 d20 + 103 y20
@ 8%
and
1 1=08320
= 10=10797>
0=07771
since l(4) = 4(1=081@4 1) = 0=07771. Hence the price is $92.85.
(4)
d20 =
22. (a) Let [ denote the annual repayment. Then at 12% eective p.a.,
100,000 = [ d20
giving
[=
100,000 × 0=12
= 13,387=88=
1 1=12320
(b) The amount of capital outstanding after the 10th repayment is
[ d10 =
100,000 d10
100,000 (1 1=12310 )
=
= 75,644=50=
d20
1 1=12320
(c) The amount of capital outstanding after the 2nd repayment is
[ d18 =
100,000 (1 1=12318 )
= 97,057=70=
1 1=12320
Hence the amount of interest in the payment at the end of the 3rd
year is
0=12 × 97,057=70 = 11,646=92=
(d) From part (b), the amount of interest paid at the end of the 11th
year is
0=12 × 75,644=50 = 9,077=34=
Hence the amount of capital repaid is
[ 9,077=34 = 4,310=54=
23. (a) Let [ denote the annual repayment. Then at 10% p.a. eective,
50,000 = [ d15
giving
[=
50,000 × 0=1
= 6,573=69=
1 1=1315
196
APPENDIX 3. SOLUTIONS TO EXERCISES
(b) After 10 repayments, the amount outstanding under the original
schedule is
50,000 d5
50,000 (1 1=135 )
[ d5 =
=
= 24,919=45=
d15
1 1=1315
Janet makes a special repayment of half this amount at time 10
years, and continues to make repayments of [. Thus, if q is the
revised term, we have
0=5 × 24,919=45 = [ dq
which gives
dq = 1=8954=
As the annuity present value is small, the term q must also be
small, and without doing any mathematics we can easily compute
from rst principles that d2 = 1=7355 and d3 = 2=4869= Thus, the
revised outstanding term of the loan is 3 years, with payments of
[ at the end of the rst two years, and a reduced payment of \
at the end of the nal year. The equation of value that gives \ is
0=5 × 24,919=45 = [ d2 + \ y3 >
from which we calculate \ = $1,398=67=
24. Let the term of the loan be q years, and let the annual repayment be
[. In the nal year, interest is payable on the loan outstanding at time
q 1 years, after the repayment due then. Let this amount be Oq31 =
As the repayment in the nal year reduces the loan outstanding to 0,
the amount of the loan repaid at time q years is Oq31 and as the nal
repayment (of [) comprises capital and interest,
[ = Oq31 + 0=08 Oq31 = 1=08 Oq31 =
As 0=08 Oq31 = 3,469=58> we have [ = 46,839=33=
Thus,
500,000 = [ dq
which gives
500,000
1 1=083q
=
=
0=08
[
If we rearrange this identity we obtain
dq =
1=083q = 0=146017>
so that
q log 1=08 = log 0=146017
giving the term as q = 25 years.
197
25. Let [ denote the total repayment in a year by Sheila and let \ denote
the total repayment in a year by Bruce. Then, at 7.5% p.a. eective,
(26)
100,000 = [ d20 = [
1 1=075320
l(26)
where l(26) = 26(1=0751@26 1) = 0=07242= Thus [ = 9,471=96> and so
the total of Sheila’s repayments is
20 [ = 189,439=14>
of which $100,000 is capital, meaning that she pays a total of $89,439.14
in interest.
For Bruce,
1 1=075320
l(12)
where l(12) = 12(1=0751@12 1) = 0=07254= Thus \ = 9,487=35 and
the total of Bruce’s repayments is $189,746.99, meaning that he pays
a total of $89,746.99 in interest. Hence, Bruce pays $307.85 more in
interest.
(12)
100,000 = \ d20 = \
Chapter 3
1. Let S0 denote the population size (in millions) at the start of the year.
Then
S0 + 0=74 0=68 + 0=18 0=05 = 30=43>
giving S0 = 30=24=
(a) The natural increase in the population is the number of births
minus the number of deaths, i.e. 0.06 million.
(b) The net migration in the population is the number of immigrants
minus the number of emigrants, i.e. 0.13 million.
(c) We can estimate the average population size (in millions) as
1
(30=24 + 30=43) = 30=335, so our estimate of the crude birth
2
rate per 1,000 of population is
0=74 × 1,000
= 24=39=
30=335
(d) Our estimate of the crude death rate per 1,000 of population is
0=68 × 1,000
= 22=42=
30=335
198
APPENDIX 3. SOLUTIONS TO EXERCISES
2. (a) For Sunshine, the dependency ratio is
100 ×
16,000 + 15,200
= 60=94=
26,300 + 24,900
The youth dependency ratio is
100 ×
16,000
= 31=25
26,300 + 24,900
and the age dependency ratio is
100 ×
15,200
= 29=69=
26,300 + 24,900
(b) For Raintown, the total number of deaths is 138 and the total
population size is 74,600. Hence the crude death rate is
1,000 ×
138
= 1=85=
74,600
For Sunshine, the total number of deaths is 172 and the total
population size is 82,400. Hence the crude death rate is
1,000 ×
172
= 2=09=
82,400
(c) For Raintown, the age specic death rate for the age group 0—14
is
12
1,000 ×
= 0=8=
15,000
A similar calculation applies at each age group, and results are as
follows:
Age group Raintown Sunshine
0—14
0.8
0.63
15—39
1.27
0.95
40—64
0.66
0.68
65+
6.45
7.89
(d) To calculate the standardised crude death rate for Raintown we
apply the age specic death rates to the standard population to
obtain the number of deaths that would occur in the standard
population. As the age specic death rates in part (c) are per
1,000 of population, we can nd the number of deaths as
0=8 × 159 + 1=27 × 260 + 0=66 × 218 + 6=45 × 68 = 1,038=9=
199
As the total population of the standard population is 705,000, the
standardised crude death rate is
1,000 ×
1,038=9
= 1=474=
705,000
Similarly, the number of deaths expected in the standard population if we apply the age specic death rates for Sunshine is
0=63 × 159 + 0=95 × 260 + 0=68 × 218 + 7=89 × 68 = 1,032=2=
The standardised crude death rate is thus
1,000 ×
1,032=2
= 1=464=
705,000
3. (a) The pyramid for 2009 indicates that fertility rates have been falling
with the numbers in successive age groups increasing from the 0—4
age group up to the 35—39 age group. In 2049, those aged 0—39 in
2009 will be aged 40—79, and so if recent patterns of mortality and
fertility continue, the pyramid for 2049 will fan out from its base,
with increasing numbers in successive age groups, until around
the 75—79 age group, after which the numbers in successive age
groups will decline.
(b) The shape of the pyramid will not be very dierent from part (a).
The main dierence will be that the base of the pyramid will be
slightly wider, with more population in each of the age groups
from 0—4 up to 35—39.
(c) Again the shape will not dier greatly from that in part (a). A
reduction in mortality rates should cause an increase in the population in each age group, but as mortality rates are very low
already in Japan (a well developed economy), any reduction in
mortality rates is likely to be a small reduction.
4. The country is most likely to be Canada. The key piece of information
is expectation of life at birth, which we expect to be high for a well
developed economy. We would expect a much lower expectation of life
at birth in Cambodia, and a somewhat lower level than 78 in Colombia.
For these three countries we would expect fertility to be highest in
Cambodia and lowest in Canada, which would suggest a fairly low
crude birth rate for Canada. Comparison based on youth dependency
ratios is not an easy task.
200
APPENDIX 3. SOLUTIONS TO EXERCISES
5. (a) The probability that a newborn life will survive to age 10 is
v(10) = (1 10@110)3@4 = 0=9310=
(b) The probability that a newborn life will die between ages 60 and
70 is
v(60) v(70) = (1 60@110)3@4 (1 70@110)3@4 = 0=0853=
(c) The probability that a life now aged 20 will survive to age 40 is
v(40) (1 40@110)3@4
=
= 0=8282=
v(20) (1 20@110)3@4
(d) The probability that (60) will die before age 80 is
(1 80@110)3@4
v(80)
=1
= 0=3183=
v(60)
(1 60@110)3@4
1
Hence the probability that two lives aged 60 both die before age
80 is 0=31832 = 0=1013=
(e) The force of mortality at age { is { = v0 ({)@v({). As
¶
μ
{ ´31@4 1
3³
0
1
>
v ({) =
4
110
110
we have
{ =
giving 65 = 0=01667=
3
3
=
,
440(1 {@110)
440 4{
6. We must check three conditions: j(0) = 1, lim{<" j({) = 0 and j 0 ({) ?
0, so that the function is decreasing. The rst two conditions are easily
veried. To check the third,
j 0 ({) = {31 exp{{ }>
which is negative since > A 0, and so all three conditions are satised.
As { = j 0 ({)@j({), we have { = {31 = The condition 40 = 410
gives
4031 = 4 1031
so that
431 = 4=
Thus = 2, and as 25 = 0=001> we have
giving = 2 × 1035 .
2 × 25 = 0=001>
201
7. (a) The rst condition (v(0) = 1) is satised since
s
121
= 1=
v(0) =
11
Second, v({) decreases with { since
1
g
v({) =
(121 {)31@2 ? 0
g{
22
for 0 ? { ? 121. Third, it is straightforward to see that v(121) =
0.
(b)
i. We have
v(30)
=
10 s20 =
v(20)
r
91
= 0=94920=
101
ii. We require that either (35) survives to age 60 and (40) doesn’t,
or vice versa. The required probability is thus
μ
¶
μ
¶
v(60)
v(60)
v(60)
v(60)
1
+
1
v(35)
v(40)
v(40)
v(35)
=
v(60) v(60)
v(60) v(60)
+
2
v(35) v(40)
v(35) v(40)
= 0=24827=
iii. The required probability is
v(60) v(70)
= 0=07011=
v(30)
iv. As { = v0 ({)@v({) we can use the answer from part (a) to
write
1
(121 {)31@2
1
22
=
{ = 1
=
1@2
2(121 {)
(121 {)
11
Thus, 35 = 0=00581=
8. On the basis of a constant force of mortality of 0.004, we have w s{ =
exp{0=004w}> independent of {=
(a) 10 s30 = exp{0=04} = 0=96079=
(b) 5 t15 = 1 5 s15 = 1 exp{0=02} = 0=01980=
202
APPENDIX 3. SOLUTIONS TO EXERCISES
(c) We require { such that { s0 = 0=9= Thus
exp{0=004{} = 0=9
so that
{=
log 0=9
= 26=34=
0=004
9. (a) First,
t50 =
g50
279
= 0=00290=
=
o50
96,251
Next, o51 = o50 g50 = 95,972, leading to
t51 =
g51
285
= 0=00297=
=
o51
95,972
Finally, o52 = o51 g51 = 95,687, leading to
g52 = o52 t52 = 336
(where we have rounded to the nearest integer).
(b) We assume that o50+w = d + ew + fw2 for 0 w 2= Then by setting
w = 0, 1 and 2 we have
o50 = 96,251 = d>
o51 = 95,972 = d + e + f>
o52 = 95,687 = d + 2e + 4f=
Then
o52 2o51 = 2f d
giving f = 3, and from the equation for o51 we obtain e = 276=
Now
Z 1
Z 1
¡
¢
o50+w gw =
d + ew + fw2 gw
O50 =
0
= d+
0
e f
+
2 3
= 96,112=
203
10. (a) The missing values are calculated as follows:
g85 = o85 t85 = 2,316>
o86 = o85 g85 = 22,590>
t86 = g86 @o86 = 0=0992>
o87 = o86 g86 = 20,350=
(b) The probability that (85) dies aged 87 last birthday is
g87 @o85 = 0=0864=
(c) Assume that o85+w = d + ew for 0 w 1= Setting w = 0 gives
d = o85 and setting w = 1 gives e = o86 o85 = g85 , so that
o85+w = o85 w g85
and so
g
o85+w = g85 =
gw
Hence
85+w =
giving 85=5 = 0=0975=
1 g
g85
o85+w =
>
o85+w gw
o85 w g85
11. (a) 10 s40 = o50 @o40 = 90,054@94,718 = 0=95076=
(b) The expected number of survivors to age 3 from 10,000 newborn
is
o3
10,000 = 9,890=8=
o0
(c) The probability of a 50 year old dying before age 54 is
4 t50 = 1 o54
= 0=03729=
o50
(d) The probability of a 60 year old surviving for 15 years is
15 s60 =
o75
= 0=51671=
o60
(e) The age at which a life has the highest chance of survival for the
next year is found by reading the s{ column and nding the largest
value. This occurs when { = 7=
204
APPENDIX 3. SOLUTIONS TO EXERCISES
12. (a) We know that w s{ = v({ + w)@v({), so
½ Z {+w
¾
½ Z {
¾
| g| ÷ exp | g|
w s{ = exp 0
0
½ Z {+w
¾
= exp | g| =
{
(b) The probability that (75) dies between the ages of 80 and 85 is
v(80) v(85)
= 5 s75 10 s75 =
v(75)
Now for 0 ? w 20>
w s70
½ Z 70+w
¾
= exp { g{
70
½ Z 70+w
¾
= exp (0=01 + 0=001{) g{
70
( μ
¶¯70+w )
0=001{2 ¯¯
= exp 0=01{ +
¯
2
70
= exp{0=08w 0=0005w2 }=
We get the following values:
w
w s70
5
10
15
0.66199 0.42741 0.26915
As 10 s70 = 5 s70 5 s75> we get 5 s75 = 0=64565 and similarly we nd
that 10 s75 = 15 s70 @ 5 s70 = 0=40657, giving the required probability
as 0.23908.
13. (a) Under Gompertz law { = Ef{ , and so
¾
½ Z {
| g|
v({) = exp 0
½ Z {
¾
|
= exp Ef g|
0
½ Z {
¾
= exp E exp{| log f} g|
0
¯{ ¾
½
¯
E
exp{| log f}¯¯
= exp
log f
0
½
¾
E {
(f 1) =
= exp
log f
205
As w s{ = v({ + w)@v({) we have
½
¾
¢
E f{ ¡ w
f 1 =
w s{ = exp
log f
(b) Note that
log w s{ =
so that
¢
E f{ ¡ w
f 1 >
log f
¢
E f70 ¡ 10
f 1 >
log f
¢
E f80 ¡ 10
f 1 =
=
log f
log 10 s70 =
log 10 s80
Thus
f10 =
log 10 s80
= 1=9671
log 10 s70
giving f = 1=07= We can now solve for E using either the expression
for log 10 s70 or log 10 s80 , and we obtain E = 0=0003= Hence
½
¾
¢
E f85 ¡ 10
f 1 = 0=25958=
10 s85 = exp
log f
14. (a) With v({) = 1 {@$ we have
w s{ =
1 ({ + w)@$
${w
w
v({ + w)
=
=
=1
v({)
1 {@$
${
${
and hence
r
h{
Z $3{ μ
1
=
w
${
0
¯$3{
¯
w2
¯
= w
2($ {) ¯0
= ${
=
${
=
2
${
2
¶
gw
206
APPENDIX 3. SOLUTIONS TO EXERCISES
(b) Assuming that o|+w = (1 w)o| + w o|+1 , we have
Z 1
o|+w gw
O| =
0
Z 1
=
((1 w)o| + w o|+1 ) gw
0
Z 1
Z 1
= o|
(1 w) gw + o|+1
w gw=
0
As
Z 1
w gw =
0
we have O| = 12 (o| + o|+1 ).
0
Z 1
1
(1 w) gw = >
2
0
Next,
r
"
W{
1X
=
O|
o{
o{ |={
μ
¶
1 o{
+ o{+1 + o{+2 + = = =
=
o{ 2
"
1 X
+
=
w s{ =
2 w=1
h{ =
15. (a) We have
½ Z w
¾
½ Z w
¾
35+v gv = exp 0=02 gw = exp{0=02w}>
w s35 = exp 0
0
and as w s35 = o35+w @o35 we have
o35+w = o35 exp{0=02w}=
(b) The total membership of the society is W35 where o35 = 2,000=
Now
Z "
Z "
o35
W35 =
o35+w gw = o35
exp{0=02w} gw =
0=02
0
0
so that the total membership is
2,000
o35
= 100,000=
×
o35
0=02
207
(c) The number of members between ages 40 and 60 is
(W40 W60 ) =
As we have a constant force of mortality, o{+w = o{ exp{0=02w}
for any age {, so that
o{
=
W{ =
0=02
Thus,
¶
μ
o60
2,000 o40
(W40 W60 ) =
o35
0=02 0=02
μ
¶
o40 o60
= 100,000
o35 o35
= 100,000 (exp{0=02 × 5} exp{0=02 × 25})
= 29,831=
(d) We require age { such that the number of members aged between
35 and { is 50,000. Thus
50,000 = (W35 W{ ) =
We know from part (b) that W35 = 100,000> so
W{ = 50,000>
which gives
o{
2,000
= 50,000>
×
o35
0=02
or o{ @o35 = 0=5= As
o{
= {335 s35 = exp{0=02({ 35)}>
o35
we nd that { = 69=66=
p
p
p
16. (a) The number of men in the village is p (W70
W90
) where p o70
=
15 and the superscript p indicates males. Thus, the number is
15
(534,526 8,481) = 139=
56,625
208
APPENDIX 3. SOLUTIONS TO EXERCISES
i
=
(b) The number of women who enter the village each year is i o70
20> and the number who move to the sheltered housing facility
i
each year is i o90
. As the population of women in the village is
stationary, the number of entrants each year equals the number of
exits, and there are two modes of exit — death and moving to the
sheltered housing facility. Hence the number of deaths is
¶
μ
i
o90
16,578
i i
20 o90 = 20 20 i = 20 1 = 16=
74,955
o70
(c) The number of people in the village aged between ages 85 and 90
is
i
i
p
p
p (W85
W90
) + i (W85
W90
)
=
20
15
(41,040 8,481) +
(192,463 72,266)
56,625
74,955
= 41=
(d) The number of people who move into the sheltered housing facility
each year is
i
op
o90
i
p
p o90
+ i o90
= 15 90
+
20
p
i
o70
o70
16,578
3,152
+ 20
= 15
56,625
74,955
= 5=
17. As v({) = h3{@70 , we have
o{+w = o0 v({ + w) = o0 h3({+w)@70
which gives
W{ =
Z "
0
= o0
o{+w gw
Z "
h3({+w)@70 gw
0
= o0 70 h3{@70 =
209
The information about births gives o0 = 350,000= The total payable
in contributions each year is
¡
¢
[ (W20 W60 ) = [ × o0 × 70 h320@70 h360@70
¡
¢
= [ × 350,000 × 70 h320@70 h360@70
= 8=014059 [ × 106 =
The total amount of death benets each year is
3,000 (g60 + g61 + g62 + = = =) = 3,000 o60
= 3,000 o0 h360@70
= 445=5915 × 106 =
The total payable in annuities each year is
20,000 W60 = 20,000 × 350,000 × 70 h360@70
= 207,942=7 × 106 =
Equating the contributions each year with the outgo in annuities and
death benets gives
8=014059 [ = 445=5915 + 207,943=7
so that [ = $26,002=84=
18. (a) The value of t0 will be higher. Mortality rates initially decrease
with age, then increase and we would not expect the mortality
rate to return to the level of t0 until about age 50.
(b) The value for females would be higher. Females are subject to
lighter mortality rates than males, and so the expected number of
survivors to age 70 from 100,000 female births would be greater
than from 100,000 male births.
(c) The value of g70 would be higher. We expect g{ to increase with
age from around age 30 to around age 75, then to decrease. We
also expect a fairly low value for g90 as the expected number of
survivors to age 90 from 100,000 births will be small.
19. (a) The age specic fertility rate for age group 15 to 19 is
1,000
451 + 425
= 19=8=
44,300
Similar calculations for the other age groups give age specic fertility rates as 136.5, 64.7 and 7.9.
210
APPENDIX 3. SOLUTIONS TO EXERCISES
(b) The total fertility rate is the sum of the age specic fertility rates.
We calculate this as
5 × 19=8 + 10 × (136=5 + 64=7 + 7=9) = 2,189=3
since we are assuming that the rate that applies to a particular
age group applies at each age in the age group. Note that the rst
age group contains only 5 ages, whilst the other groups contain
10. Per woman, the total fertility rate is 2.19.
(c) We calculate the age specic fertility rate for female births for
each age group. For age group 15 to 19 this is
1,000
425
= 9=6=
44,300
Similar calculations give the following gures for the other three
age groups: 67.5, 31.9 and 3.9. The gross reproduction rate is
thus
5 × 9=6 + 10 × (67=5 + 31=9 + 3=9) = 1,081=2>
where we make exactly the same assumptions as in part (b). Per
woman, the gross reproduction rate is 1.08.
(d) The net reproduction rate is
5×9=6×0=99817=5 +10×(67=5×0=99825 +31=9×0=99835 +3=9×0=99845 )
which gives 1,021.6, or 1.02 per woman.
20. The ordering of the values of gross reproduction rate and net reproduction rate is correct. The net reproduction rate is a measure of women
replacing themselves, i.e. female births only. Taking into account a sex
ratio at birth around 1.05, the student’s calculations say that women
are (on average) having six children. This gure is much higher than
is currently observed in developed economies such as New Zealand,
so we would conclude that the student has performed the calculations
incorrectly.
21. (a) We assume that the female population aged 15 to 49 is the reproductive population. We further assume that the age specic
fertility rates for an age group apply at each individual age in that
age group. On the assumption that the sex ratio at birth is 1.05,
our estimate of the gross reproduction rate per 1,000 is
1 X
5
(16=7 + 53=8 + = = = + 0=7) = 927=3=
DVI U{ =
2=05 {=15
2=05
49
211
(b) We further assume that the female population at each age in an
age group is one fth of the total population for that age group.
The estimate of the number of female births if we had data at
individual ages would be
1 X
DVI U{ × Female population aged {,
2=05 {=15
49
so with our assumptions, the estimate becomes the sum of the
product of age specic fertility rate and population per group
rather than at individual ages. Hence the estimate of the number
of births is
1
(16=7 × 727=168 + = = = + 0=7 × 793=905) = 144,266=
2=05
22. (a) The estimate of the population (in millions) on the census date in
2002 is
5=2 (1 + 10 × 0=02) = 6=24.
(b) The population on the census date in 1992+w will exceed 7 million
if
5=2 (1 + 0=02w) A 7>
which gives w A 17=3. Hence the population rst exceeds 7 million
on the 18th anniversary of the 1992 census date.
23. (a) The estimate of the population (in millions) on the census date in
2002 is now
5=2 (1=0210 ) = 6=34.
(b) The population on the census date in 1992+w will exceed 7 million
if
5=2 (1=02w ) A 7>
which gives w log 1=02 A log(7@5=2), i.e. w A 15=01= Hence the
population rst exceeds 7 million on the 16th anniversary of the
1992 census date.
24. Under the logistic model of population size, the population size at time
w is
1
Sw =
=
D + Eh3uw
212
APPENDIX 3. SOLUTIONS TO EXERCISES
Working in units of 1 million, we have
1
= 25>
D+E
1
=
= 26=2>
D + Eh310u
1
=
= 27=1=
D + Eh320u
S0 =
S10
S20
We can re-write these equations as
D + E = 1@25>
D + Eh310u = 1@26=2>
D + Eh320u = 1@27=1=
Thus
Eh310u Eh320u =
and
¡
¢
1
1
= Eh310u 1 h310u = 0=001268=
26=2 27=1
(A.1)
1
1
= 0=001832=
25 26=2
Dividing equation (A.1) by equation (A.2) we obtain
E(1 h310u ) =
(A.2)
h310u = 0=691882>
and equation (A.2) gives E = 0=005946. Finally, as D + E = 1@25>
we have D = 0=034054. Thus the population at the next census is
estimated as
1
S30 =
= 27=76 million.
D + Eh330u
Note that it is su!cient to nd h310u and that the actual value of u is
not required.
25. We assume that the population ‘aged {’ are aged { + 12 on average.
Thus, if S{>w is the population aged { at time w> we have
¡
¢
v { + 1 12
¢
S{+1>w+1 = S{>w ¡
v { + 12
for { = 1> 2> = = = > 5= The population aged 0 at time w + 1 is made up of
the survivors of births in the last year. We nd the number of births
by applying the age specic fertility rates to the population at a given
213
age. Thus, for example, the projected number of births in the next
year is
(1,500 × 0=49 + 1,400 × 0=52 + 1,200 × 0=48) = 2,039>
and so the projected
¡ 1 ¢ number of survivors aged 0 one year from the
present is 2> 039 v 2 , where we assume that lives are, on average, born
half way through a year. Repeating this procedure over successive years
leads to the following population structure for times 1> 2> = = = > 5 from the
present (time 0).
Age\Time
0
1
2
3
4
5
6
0
2,000
1,500
1,400
1,200
1,000
600
200
1
2
3
4
5
1,893 1,848 1,823 1,824 1,774
1,692 1,602 1,564 1,543 1,544
1,227 1,385 1,311 1,279 1,262
1,089 955 1,077 1,020 995
857
778
682
769
728
600
514
467
409
462
200
200
171
156
136
Chapter 4
1. (a) In general, it is prudent for insurers to go through the process
of underwriting. If the insurer does not go through this process
it must try to anticipate the risks and their consequences. An
important point about such policies is that the sum insured is
not high — perhaps between $5,000 and $15,000 — so the insurer
is not as concerned about the consequences of not underwriting
compared with the situation of a person proposing for a whole life
insurance policy with a sum insured of, say, $1 million.
(b) The rst risk to which the insurance company is exposed is that
terminally ill people may wish to buy insurance. Such individuals would anticipate paying a premium signicantly less than the
benet they would receive. A second risk is that individuals who
buy such cover may be subject to higher levels of mortality than
individuals who are screened for insurance via the process of underwriting.
(c) To protect against the rst risk in part (b), the insurance company
could have an exclusion clause in the contract, so that it pays a
death benet in the rst year of the contract only on accidental
death. To protect against the second risk, the insurance company
214
APPENDIX 3. SOLUTIONS TO EXERCISES
would simply price the policies on an appropriate mortality table
which re ects the past mortality experience of insured lives.
2. (a) The single premium will be higher for the male as there is a higher
probability of the death benet being paid as male mortality rates
are higher.
(b) The amount bought by the man will be greater as the woman
is expected to live longer and consequently to receive a greater
number of annuity payments.
(c) The reserve will be much higher for the whole life insurance as the
death benet is certain to be paid whereas the probability of the
death benet being paid is small under the term insurance.
3. The common features are:
• underwriting would normally take place,
• each policy has a xed term,
• the benet is secured by the payment of premiums at regular intervals (unless the term is very short, in which case a single premium
would be appropriate),
• each policy has a death benet.
Dierences are:
• the endowment insurance has a survival benet, the term insurance does not,
• the benet is certain to be paid under the endowment insurance,
but not under the term insurance.
4. The key point about investments is that the investments held should
be appropriate to the liabilities. Most life insurance policies are long
term contracts, and the benets are commonly paid many years after
a policy has been issued. Thus, it is appropriate for a life insurance
company to hold assets such as shares or property which will not need
to be turned into cash soon after their purchase. By contrast, most
general insurance business is short term business, with many policies
having a term of one year, and hence if claims arise they are paid soon
after the issue of a policy. It is thus appropriate for a general insurance
company to invest in assets that can be easily turned into cash. Thus,
the student’s suggestion is inappropriate.
215
5. The no negative equity guarantee means that if the sale price of the
property is less than the borrower’s debt, the lender (insurance company or bank) has a shortfall at the date of sale. A problem that can
arise for the lender is that if the borrower knows that the debt exceeds
the property value whilst the borrower is still alive and living in the
property, there is no incentive for the borrower to maintain the property to a high standard because the borrower’s debt is capped under
the no negative equity guarantee. Property maintenance should benefit
the lender by making the property attractive to prospective purchasers.
Thus, the lender is exposed to a moral hazard.
The lender can only really protect itself at the issue of a loan.
The lender can assess a property’s value at the issue date and offer an
appropriate level of loan. A cautious approach would be to offer only
a small percentage of the property’s value as a loan, thus reducing the
chances of the accumulated debt exceeding the property value when
the property is sold to repay the debt.
6. Modelling is required for two categories of member: those who are
active and are contributing to the superannuation scheme, and those
who have retired and are in receipt of benefits. For active members we
need to model the probability of exit from the active status by modes
such as death, ill-health retirement, age retirement and withdrawal
from the scheme. We also need to model salaries for these members,
taking account of both increments in salaries (e.g. due to promotions)
and salary inflation. Thus, we need to model inflation. For those in
receipt of pensions, we need to model their mortality, and if a spouse’s
pension is payable on the death of a pensioner, we must model spouses’
mortality. As the valuation of liabilities would be done on a collective
rather than individual basis, we would also need to make assumptions
about the proportion of pensioners married at each age. As the scheme
pays index-linked pensions, inflation would need to be modelled, and
the model adopted here could be different from that used for salary
inflation.
Financial modelling is also required. Taking account of the type
of assets that the scheme holds, a rate of interest would need to be set
to value the future contributions and the liabilities.
7. (a) This statement is incorrect. Under defined benefit superannuation
the investment risk is borne by the superannuation fund (and to
some extent the employer). Members’ benefits are expressed in
terms of quantities like years of scheme membership and salary on
216
APPENDIX 3. SOLUTIONS TO EXERCISES
leaving the scheme, and are not determined by the fund’s investment performance.
(b) This statement is also incorrect. Under dened contribution superannuation the member’s benet is essentially the accumulation
of the member’s contributions, and as such there is no relationship
between the benet amount and the salary at exit.
(c) This statement is also incorrect. The investment strategy of a life
insurance company is dictated by the nature of its liabilities, which
are typically long term. Members of dened contribution superannuation schemes normally have a choice of investment strategy
(ranging from conservative to aggressive), and the assets that such
a scheme holds will depend on members’ investment choices.
8. Features in common are
• policyholders usually pay a single premium (although in general
insurance it is also common for policyholders to be oered the
option of paying monthly premiums),
• the policy has a xed term and claims can occur only if certain
events occur within that term.
Features that dier are
• the benet level is xed under the term insurance policy, but not
under the motor vehicle policy,
• the number of claims under the term insurance policy is 0 or 1,
but under the motor vehicle policy the number of claims could be
0> 1> 2> = = =, and in practice there is an upper limit to the number
of claims a person would make,
• there are incentives not to claim under the motor vehicle policy
— through no claims discount — but the death of an insured life
would certainly result in a claim,
• a claim under the motor vehicle policy will most likely mean a
payment for the policyholder through a policy excess, but a claim
under term insurance does not cause a payment by the policyholder’s estate.
9. The setting of assumptions for premium calculation is an important
task for any type of insurance. A key dierence between life insurance
and general insurance is that most life insurance policies are long term
contracts, whereas general insurance policies are short term contracts.
217
Thus, once a premium is set for, say, an endowment insurance policy,
the insurance company is locked into that premium for the duration of
the policy. For example, if the insurance company uses an interest rate
assumption that is too high, the insurer may find that the accumulation
of premiums to the maturity date does not provide the sum insured. By
contrast, the premium for, say, motor vehicle insurance can change after one year, so if inappropriate assumptions have been used in setting
the premium, these can be changed and a more appropriate premium
can be charged. However, policyholders are unlikely to be satisfied if
general insurance premiums fluctuate greatly from one year to the next
— what they expect is a small (inflationary) increase in general insurance premiums from one year to the next, and they are free to change
insurance companies if they can obtain more favourable terms from
another company. Thus, there are different reasons that make setting
assumptions important for both life insurance and general insurance.
10. A major issue facing many western European governments is an ageing
population, meaning that the proportion of the population entitled to
age pension benefits is increasing over time. The obvious reason why
governments want to raise the age at which people become entitled
to an age pension is that it reduces the cost of age pensions to the
national insurance scheme. However, it may also be argued that the
age at which people become entitled to a pension was set many years
ago at a time when expectation of life from retirement was lower than
it is today, and consequently it is appropriate to change the age of
entitlement to an age pension.
There are various alternative measures a government could take,
although, like increasing the age of entitlement, none may be particularly popular with the voting public. One option would be to freeze the
level of benefits, resulting in a lower level of benefit in real terms (i.e.
allowing for inflation) as time progresses. A second option would be
to increase the level of contributions towards an age pension. As the
proportion of the population contributing is decreasing, this is likely
to increase contributions by a considerable amount. A third option
would be to means test the pension, so that only those with a retirement income below a certain level would be eligible for an age pension.
This will not be a cost-less exercise — meaning that a government will
have to spend money to conduct means testing, when its objective is
to reduce spending — and means testing is likely to cause resentment
amongst those who have made contributions throughout their working lifetime. A fourth option is to do nothing specifically relating to
a national insurance scheme and simply fund the increased cost of age
218
APPENDIX 3. SOLUTIONS TO EXERCISES
pensions through general government revenue. As this is most likely
to come through taxation, this is similar to increasing contributions.
Some combinations of these options could also be implemented.
Chapter 5
1. The sum of the probabilities is 1, which means that s = 0=07= We
calculate H([) as
0=05 × 10 + 0=12 × 12 + = = = + 0=07 × 32 = 18=23=
2. Under the assumption of a constant force of mortality, w s{ = h3w > so
Pr(N({) = n) = n s{ n+1 s{
= h3n h3(n+1)
= h3n (1 h3 )=
We have
H(N({)) =
"
X
n Pr(N({) = n)
n=0
=
"
X
n=0
n h3n (1 h3 )
3
= (1 h
)
= (1 h3 )
"
X
n h3n
n=0
"
" X
X
h3m
n=1 m=n
where the nal step follows from the formula given in the question.
Now
"
X
h3
h3m =
1 h3
m=n
219
giving
3
H(N({)) = (1 h
=
"
X
"
X
h3
)
1 h3
n=1
h3
n=1
h3
1 h3
1
=
= h 1
=
We see that as increases, the denominator of H(N({)) increases, and
hence H(N({)) decreases as increases, so that H(N({)) is a decreasing
function of .
3. For a life aged { the premium is S{ = 10,000 t{ @1=06=
(a) As t40 = 0=00318, S40 = $30=00=
(b) As t45 = 0=00509, S45 = $48=02=
(c) As t50 = 0=00816, S50 = $76=98=
As the probability of death in a single year increases, so too does the
premium for one year term insurance.
4. The expected present value of the survival benet is
20,000 (1 + l)310
o55
85,664
=
= 20,000 (1 + l)310
o45
92,900
Setting this equal to the single premium gives
(1 + l)10 =
20,000 85,664
= 1=6481=
11,190 92,900
Thus 1 + l = 1=64811@10 = 1=0512, so l = 5=12%=
5. Under the assumption of a constant force of mortality,
½ Z w
¾
½ Z w
¾
{+v gv = exp 0=004 gw = exp{0=004w}>
w s{ = exp 0
0
independent of {. Let s denote the probability of survival for 20 years,
so that s = exp{0=08}. Then, the expected present value of the
benet (which is the single premium) is
¡
¢
20,000 s2 + 10,000 × 2 s (1 s) y 20 = 20,000 s y20
220
APPENDIX 3. SOLUTIONS TO EXERCISES
since the probability that both lives survive 20 years is s2 and the
probability that exactly one life survives is 2s(1 s). As l = 6% the
single premium is $5,756.64.
6. Let sV and sM denote the probabilities that Smith and Jones respectively survive for 10 years. Then
½ Z 10
¾
(0=002 + n) gw = exp{0=02 10n}>
sM = exp 0
and sV = exp{0=02}. Then
1,224=75 = 50,000 y10 (sV (1 sM ) + sM (1 sV ))
at 7%, giving
1,224=75 (1=0710 )
= sV + sM 2sV sM
50,000
and hence
μ
¶
1
1,224=75 (1=0710 )
sV = 0=970446=
sM =
2sV 1
50,000
Thus
giving n = 0=001=
exp{0=02 10n} = 0=970446
7. With v({) = 1 {@100, we have
w s{ =
100 { w
w
v({ + w)
=
=1
=
v({)
100 {
100 {
Under this survival function,
¶
μ
w
d̈55:20 =
=
y w s55 =
y 1
45
w=0
w=0
19
X
Now
19
X
y w = d̈20 = 1=07
w=0
and let
V=
w
19
X
w=0
19
X
w
1 1=07320
= 11=3356
0=07
w yw = y + 2y 2 + 3y 3 + = = = + 19y 19
221
so that
(1 + l)V = 1 + 2y + 3y 2 + = = = + 19y 18
giving
d̈19 19y 19
= 82=9347=
V=
l
Thus,
d̈55:20 = 11=3356 8. We have
D{ =
"
X
w=1
and
82=9347
= 9=4926=
45
yw
g{+w31
o{
g{+w31
o{+w31 o{+w
=
= w31 s{ w s{ = h3(w31) h3w =
o{
o{
Thus
D{ =
"
X
w=1
=
¡
¢
yw h3(w31) h3w
y
yh3
1 yh3 1 yh3
1 h3
=
=
1 + l h3
9. Consider $100 nominal. The coupon at the end of each year is $12, so
the expected present value of the coupon at the end of the wth year is
12 y w 0=995w
with y = 1@1=14 for w = 1> 2> 3> = = = > 25> and the expected present value
of the redemption payment is 120 y 25 0=99525 = Thus, the price to yield
14% p.a. eective is
25
X
12 y w 0=995w + 120 y 25 0=99525
w=1
= 12 y 0=995
1 (y 0=995)25
+ 120 y25 0=99525
1 y 0=995
= 79=5995 + 4=0007
= 83=60=
222
APPENDIX 3. SOLUTIONS TO EXERCISES
10. The payment at time w years is 20,000 if the man is alive and 12,000
if only the wife is alive. At an interest rate of 8% p.a. eective, the
expected present value of the payment at time w years is
¡
¢
yw 20,000 × 0=95w + 12,000 (1 0=95w ) 0=97w =
Thus, the expected present value of the annuity payments is
20,000
"
X
w
w
y 0=95 + 12,000
w=1
=
"
X
w=1
yw (1 0=95w ) 0=97w
20,000 y 0=95 12,000 y 0=97 12,000 y 0=95 × 0=97
+
1 y 0=95
1 y 097
1 y 0=95 × 0=97
= 146,153=85 + 105,818=18 69,766=56
= 182,205=47=
11. For w = 1> 2> 3> = = = > 30> the death benet is payable with probability
g25+w31 @o25 = w31 s25 w s25 = h30=001(w31) h30=001w =
Thus, the expected present value of the payment at time w for w =
1> 2> 3> = = = > 10 is
¡
¢
100,000 × 1=073w h30=001(w31) h30=001w =
Similarly, the expected present value of the payment at time w for w =
11> 12> 13> = = = > 29 is
¡
¢
100,000 × 1=07310 × 1=06103w h30=001(w31) h30=001w =
Finally, the expected present value of the payment at time 30, allowing
for payment of a death benet or a maturity benet, is
¡¡
¢
¢
100,000 × 1=07310 × 1=06320 h30=029 h30=03 + h30=03 =
Thus, the expected present value of the benet (on death or maturity)
is
100,000
10
X
w=1
¡
¢
1=073w h30=001(w31) h30=001w
310
+100,000 × 1=07
30
X
w=11
¡
¢
1=06103w h30=001(w31) h30=001w
+100,000 × 1=07310 × 1=06320 h30=03 =
223
We have
10
X
w=1
¡
¢
1=073w h30=001(w31) h30=001w
= 1=0731
¢
1 (h30=001 @1=07)10 ¡
1 h30=001
30=001
1h
@1=07
= 0=006992>
and
30
X
w=11
¡
¢
1=06103w h30=001(w31) h30=001w
= 1=0631 h30=01
= 0=011264>
¢
1 (h30=001 @1=06)20 ¡
1 h30=001
30=001
1h
@1=06
so that the expected present value of the benet (on death or maturity)
is
699=245 + 572=620 + 15,382=116 = 16,653=98=
Similarly, the expected present value of an annuity of 1 p.a. in advance
for at most 30 years is
9
X
3w 30=001w
1=07 h
w=0
=
+
29
X
1=07310 1=063w+10 h30=001w
w=10
30=001
1 (h30=001 @1=07)10
@1=06)20
310 30=01 1 (h
+
1=07
h
1 h30=001 @1=07
1 h30=001 @1=06
= 7=4857 + 6=0728
= 13=5585=
Thus, if S is the annual premium, 13=5585 S = 16,653=98> giving S =
$1,228=31=
12. The single premium is the expected present value of the annuity. For
w = 15> 16> 17> = = = > 35> the probability that the woman is alive at time
w years is h30=025w , and for w = 36> 37> 38> = = =, the probability that the
224
APPENDIX 3. SOLUTIONS TO EXERCISES
woman is alive at time w years is h30=025×35 h30=03(w335) . Thus, at 6% p.a.
eective, the expected present value of the annuity is
!
à 34
"
X
X
50,000
yw h30=025w +
y w h30=025×35 h30=03(w335)
w=15
w=35
¶
μ
30=025 20
y35 h30=025×35
)
15 30=025×15 1 (y h
+
= 50,000 y h
1 y h30=025
1 y h30=03
= 50,000 (2=9106 + 0=6420)
= 177,627=63=
13. We start from
D{ =
"
X
w=1
yw
g{+w31
o{
and isolate the rst term so that
"
g{ X w g{+w31
y
D{ = y +
o{
o{
w=2
= y t{ +
"
X
u=1
yu+1
g{+u
o{
o{+1 X u g({+1)+u31
= y t{ + y
y
o{ u=1
o{+1
"
= y t{ + y s{ D{+1 =
14. We follow the same line of argument that gives the relationship
1 = g d̈{ + D{ . Recall that d̈{:q is the expected present value of an
annuity of 1 p.a. payable in advance for at most q years. Now note
that the present value takes the value d̈w with probability w31 s{ w s{
for w = 1> 2> 3> = = = > q 1 and takes the value d̈q with probability q31 s{
since all q annuity payments will be made if ({) survives until (at least)
time q 1= Hence
d̈{:q
=
q31
X
w=1
=
d̈w (w31 s{ w s{ ) + d̈q q31 s{
q31
X
1 yw
w=1
g
(w31 s{ w s{ ) +
1 yq
q31 s{ =
g
225
Now
Pq31
w=1 (w31 s{ w s{ ) = 1 q31 s{ and
w31 s{ w s{ =
giving
d̈{:q
1
=
g
Ã
g{+w31
>
o{
q31
X
g{+w31
1 q31 s{ yw
+ (1 yq ) q31 s{
o{
w=1
!
Ã
q31
X
1
g{+w31
1
=
yw
y q q31 s{ =
g
o{
w=1
!
Now note that
q31 s{ =
o{+q31
o{+q31 o{+q + o{+q
g{+q31 + o{+q
=
=
o{
o{
o{
so that
1
d̈{:q =
g
Ã
q
X
g{+w31
1
yw
y q q s{
o
{
w=1
!
=
1
(1 D{:q ) >
g
or, equivalently, 1 = g d̈{:q + D{:q =
15. The equation from which the sum insured, denoted V, is calculated is
2,000 d̈{:20 = V D{:20 = 0=2965 V=
Now
D{:20 = 1 g d̈{:20
gives d̈{:20 = 10=7535> and so V = $72,536=26=
P
w
16. We want to calculate 25
w=1 Dw y sw where Dw = 12 for w = 1> 2> = = = > 24
and D25 = 132> y = 1@1=14 and sw = 0=995w . Part of a spreadsheet
calculation is shown below.
w
1
2
3
...
...
24
25
Dw
12
12
12
...
...
12
132
yw
0.8772
0.7695
0.6750
...
...
0.0431
0.0378
sw
Product
0.9950
10.47
0.9900
9.14
0.9851
7.98
...
...
...
...
0.8867
0.46
0.8822
4.40
The sum of the terms in the column headed ‘Product’ gives the answer,
$83.60.
226
APPENDIX 3. SOLUTIONS TO EXERCISES
17. From Solution 10, we know that the expected present value of the
annuity payments, with y = 1@1=08, is
20,000
"
X
w
w
y 0=95 + 12,000
w=1
"
X
w=1
yw (1 0=95w ) 0=97w =
We can set up a spreadsheet calculation with columns y w , 0=97w and
0=95w > then an EPV column calculated as
¡
¢
20 y w 0=95w + 12 yw 1 0=95w 0=97w =
The sum of values in this nal column, multiplied by 1,000, gives the
expected present value of the annuity. As the sums are innite, we must
truncate them to perform a spreadsheet calculation, and the authors
choose 170 as the upper limit of summation. The rst three lines of
the spreadsheet calculation are shown below.
w
1
2
3
...
...
yw
0=97w
0=95w EPV
0.9259 0.9700 0.9500 18.13
0.8573 0.9409 0.9025 16.42
0.7938 0.9127 0.8574 14.85
...
...
...
...
...
...
...
...
18. At 8% p.a. eective we calculate
d̈40:10 =
9
X
yw w s40 = 7=133
w=0
and
d̈50:10 =
9
X
yw w s50 = 6=960=
w=0
As the probability of (40) surviving w years is greater than the probability of (50) surviving w years for w = 1> 2> = = = > 9, payments are more likely
to be made to (40) and so the expected present value of the payments
to (40) is greater.
19. Let S denote the annual premium. Then
S d̈45:15 = 50,000 D45:15
227
at 7% p.a. eective. It su!ces to calculate
d̈45:15 =
14
X
yw w s45
w=0
as 1 = g d̈45:15 + D45:15 . We nd that d̈45:15 = 9=3257, giving S =
$2,090=50.
20. Let Ŝ denote the annual premium. Then
Ŝ d̈45:15 = 50,000 D45:15 + 10,000 y15 15 s45 =
From the previous solution we know that
50,000 D45:15 = 2,090=50 × 9=3257 = 19,495=37>
and as
10,000 y15 15 s45 = 3,084=58>
we nd that Ŝ = $2,421=26, an increase of $330.76.
21. Let S denote the annual premium. Then the expected present value of
the premium income is
29
X
y(w) w s25
S
w=0
where
y(w) =
½
1=073w
1=07310 1=063(w310)
for w = 0> 1> 2> = = = > 9>
for w = 10> 11> = = = > 30=
Similarly, the expected present value of the death benet is
100,000
30
X
w=1
y(w) (w31 s25 w s25 )
and the expected present value of the maturity benet is
100,000 y(30) 30 s25 =
The equation for S is
13=4040 S = 3,578=66 + 13,958=89>
giving S = $1,308=38=
228
APPENDIX 3. SOLUTIONS TO EXERCISES
22. For w = 15> 16> 17> = = = > 35> values of w s45 are calculated as o45+w @o45 . For
w = 1> 2> 3> = = = > we have
½ Z w
¾
W
1=5 80+v gv = (w s80 )1=5
w s80 = exp 0
where denotes the mortality experience from age 80, and so for
w = 36> 37> 38> = = = >
1=5
=
w s45 = 35 s45 (w335 s80 )
Thus, the single premium is calculated as
!
à 35
55
X
X
50,000
yw w s45 +
yw 35 s45 (w335 s80 )1=5
w=15
w=36
where y = 1@1=06= (As the nal tabulated age in the Female Mortality
Table is 100, the upper limit of summation is 55 in the second sum.)
This gives a single premium of $724,585=81=
23. From the solution to Exercise 7 we know that
w s{ =
100 { w
100 {
so s{ = (99 {)@(100 {) and t{ = 1@(100 {)= Then s99 = 0 and
t99 = 1, giving D99 = y = 1=0631 . We then use the equation from
Exercise 13, working backwards through D98 > then D97 , until we nd
that D80 = 0=57350.
References
Benjamin, B. and Pollard, J.H. (1980) The Analysis of Mortality and Other
Actuarial Statistics, 2nd edition. London: Heinemann.
Bowers, N.L., Gerber, H.U., Hickman, J.C., Jones, D.A. and Nesbitt, C.J.
(1997) Actuarial Mathematics, 2nd edition. Schaumburg, IL: Society of Actuaries.
Cox, P.R. (1975) Demography, 5th edition. Cambridge: Cambridge University Press.
Dickson, D.C.M., Hardy, M.R. and Waters, H.R. (2009) Actuarial Mathematics for Life Contingent Risks. Cambridge: Cambridge University Press.
Hinde, A. (1998) Demographic Methods. London: Arnold.
McCutcheon, J.J. and Scott, W.F. (1986) An Introduction to the Mathematics of Finance. London: Heinemann.
O!ce of the Australian Government Actuary (2004) Australian Life Tables
2000—02. Parkes, ACT: Australian Government Actuary.
Pollard, A.H., Yusuf, Y. and Pollard, G.N. (1990) Demographic Techniques,
3rd edition. Sydney: Pergamon Press.
Renn, D.F. (ed.) (1998) Life, Death and Money. Oxford: Blackwell.
United Nations (annual publication) Demographic Yearbook.
unstats.un.org/unsd/demographic/products/dyb/dyb2.htm
Internet Sites
Australian Bureau of Statistics: www.abs.com.au
Institut national d’études démographiques: www.ined.fr
U.S. Census Bureau: www.census.gov
229
Index
Accumulation, 8
Advance funding, 129
Age specic rates, 57
AIDS, 65, 83, 90
Analysis of surplus, 124
Annuity
accumulated value, 22
certain, 21
deferred, 20
in advance, 20
in arrear, 17
Bonds, 27
nominal amount, 28
redemption price (or value), 28
Bonus
reversionary, 125
terminal, 125
Child-woman ratio, 54
Community rating, 125
Commutation rate, 129
Contingent payment, 145, 149
Controlled funding, 129
Coupon rate, 28
Crude rate
crude birth rate, 56
crude death rate, 56
Demography, 50
Dependency ratio, 54
age dependency ratio, 54
youth dependency ratio, 54
Dividend date, 28
Equation of value, 25
Expectation of life, 77
Expected present value, 146, 148, 150
Expected value, 144
Expenses
initial, 123
renewal, 123
Fertility rate
age specic, 87
general fertility rate, 86
total fertility rate, 87
Final salary, 128
Friendly society, 111
Future lifetime, 66
Gompertz law, 84, 105, 204
Graduation, 85
Guaranteed minimum death benet,
113
Housing loan, 34
Insurable interest, 119
Insurance
disability, 117
endowment insurance, 112, 158
general, 134
group life, 133
life, 110
national, 138
principles of, 109
private health, 125
term insurance, 111
trauma, 116
unit linked, 113
whole life insurance, 1, 110, 154
230
INDEX
Interest
compound interest, 7
eective rate of interest, 7, 9
force of interest, 12
nominal rate of interest, 10
simple interest, 5
Investment, 124
Investment account, 114
Labour force participation rate, 55
Life table, 72
Lifetime rating, 126
Lump sum benet, 129
Makeham’s law, 84
Maturity date, 28
Means test, 128
Moral hazard, 118, 135, 140
Mortality rate
age specic, 57
standardised, 57
Mutual company, 125
231
Principal, 5
Principle of equivalence, 146, 149
Probability function, 143
Prot testing, 123
Proprietary company, 125
Prospective valuation, 41
Pure endowment, 113
Radix, 72
Reinsurance, 121, 138
Reproduction rates
gross reproduction rate, 87
net reproduction rate, 87
Reserves, 123, 138
Retrospective valuation, 41
Reverse mortgage, 119, 140
Selection
adverse selection, 121
self-selection, 121
Sex ratio, 53
Simple discount, 6
Solvency, 123
Stationary population, 78
No claims discount, 135
No negative equity guarantee, 120, 140 Sum insured, 110
Superannuation
Pension
dened benet, 128
age pension, 128
dened contribution, 132
universal pension, 128
Survival function, 66
Perpetuity, 19
Theory of demographic transition, 60
Policy excess, 137
Time line diagram, 14
Policyholder, 110
Population
Underwriting, 120
data sources, 51
natural increase, 56
Waiting period, 118, 127
net migration, 56
Population projection
component method, 96
models, 95
Population pyramid, 61
Premium, 1, 110
Present value, 8
Preservation, 128