Theory of Interest and Life Contingencies, with Pension Applications: A Problem-solving Approach [3 ed.] 1566983339, 1098765432

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ISBN 1-56698-333-9

P heory of Interest and Life

Contingencies,

with Pension Applications:

A Problem-Solving Approach Third Edition

\

Michael M.Parnnenter

Digitized by the Internet Archive in

2012

http://archive.org/details/theoryofinterestOOparni

THEORY OF INTEREST AND LIFE CONTINGENCIES WITH PENSION APPLICATIONS

A Problem-Solving Approach Third Edition

Michael M. Parmenter ASA, Ph.D.

ACTEX

Publications Winsted, Connecticut

To my Mother and

the

memory of my

Father

Copyright© 1988, 1999 by ACTEX Publications

No

portion of this

means without

book may be reproduced

in

any form or by any

the prior written permission of the copyright owner.

Requests for permission should be addressed to

ACTEX Publications P.O.

Box 974

CT 06098

Winsted,

Manufactured

in the

United States of America

1098765432 Cover design by

MUF

Library of Congress Cataloging-in-Publication Data

Parmenter, Michael

M.

Theory of interest and

life

contingencies, with pension

applications: a problem-solving approach

/

Michael M. Parmenter.

cm.

p.

Includes index.

ISBN 1-56698-333-9 1.

2.

Insurance, Life~Mathematics~Problems, exercises, etc.

Interest-Problems, exercises,

exercises, etc.

HG8781.P29

I.

ets.

3.

Annuities-Problems,

Title

1999

368.2'2'011076-dc 19

ISBN: 1-56698-333-9

88-38947

TABLE OF CONTENTS

CHAPTER ONE THEORY

INTEREST: THE BASIC 1.1

Accumulation Function

1

.2

Simple Interest

1

.3

Compound

1

.4

Present Value and Discount

4 6

Interest

1.5

Nominal Rate of Interest

1.6

Force of Interest Exercises

1

1

9

13

16

22

CHAPTER TWO INTEREST: BASIC APPLICATIONS 29 2.1

Equation of Value

2.2

Unknown Rate of Interest

2.3

Time-Weighted Rate of Return Exercises

29 33

34

36

CHAPTER THREE ANNUITIES

41

3.1

Arithmetic and Geometric Sequences

3.2

Basic Results

3.3

Perpetuities

3.4

Unknown Time and Unknown Rate of Interest

3.5

Continuous Annuities

3.6

Varying Annuities Exercises

41

44 52

66

58

57

53

Table of Contents

iv

CHAPTER FOUR AMORTIZATION AND SINKING FUNDS 4.1

Amortization

4.2

Amortization Schedules

4.3

Sinking Funds

4.4

Yield Rates

75

75

77

81

83

Exercises 85

CHAPTER FIVE BONDS

93

Bond 93

5.1

Price of a

5.2 5.3

Book Value 96 Bond Amortization Schedules 98

5.4

Other Topics Exercises

101

105

CHAPTER SIX PREPARA TION FOR LIFE CONTINGENCIES 6.1

6.2

Probability and Expectation

6.3

Contingent Payments Exercises

1 1

113

Introduction

1

14

119

123

CHAPTER SEVEN LIFE TABLES AND POPULATION PROBLEMS 127 127

7.1

Introduction

7.2

Life Tables

7.3

7.4

Analytic Formulae for ^x 133 The Stationary Population 137

7.5

Expectation of Life

7.6

Multiple Decrements Exercises

128

141

143

148

CHAPTER EIGHT LIFE ANNUITIES

155

8.1

Basic Concepts

8.2

Commutation Functions

8.3

Annuities Payable m^^^y

8.4

Varying Life Annuities 173 Annual Premiums and Reserves

8.5

Exercises

1

8

155 161 1

66 178

Table of Contents

CHAPTER NEVE LIFE INSURANCE 189 9.1

Basic Concepts

9.2

Commutation Functions and Basic

9.3

Insurance Payable at the

9.4

Varying Insurance

9.5

Annual Premiums and Reserves Exercises 210

189

199

202

CHAPTER TEN STATISTICAL CONSIDERATIONS 223 10.1

Mean Variance 223

10.2

Normal

10.3

Central Limit

10.4

Loss-at-Issue 235

Distribution

230

Theorem 233

Exercises 239

CHAPTER ELEVEN MULTI-LIFE THEORY 243 11.1

Joint-Life Actuarial Functions 243

.2

Last-Survivor Problems

25

11.3

Reversionary Annuities

254

1 1

Exercises 256

CHAPTER TWELVE PENSION APPLICATIONS Exercises

263

269

ANSWERS TO THE EXERCISES INDEX

297

Identities

Moment of Death

21

192 196

PREFACE

It is

in

impossible to escape the practical implications of compound interest

our

modem

The consumer

society.

is

faced with a bewildering choice

interest, and wishes to choose on her savings. A home-buyer is offered various mortgage plans by different companies, and wishes to An investor seeks to choose the one most advantageous to him. purchase a bond which pays coupons on a regular basis and is redeemable at some future date; again, there are a wide variety of

of bank accounts offering various rates of the one

which

will give the best return

choices available.

Comparing possibilities becomes even more difficult when the payments involved are dependent on the individual's survival. For example, an employee is offered a variety of different pension plans and must decide which one to choose. Also, most people purchase life insurance at

some point

in their lives,

and a bewildering number of

different plans are offered.

The informed consumer must be able in situations like

whenever

make an

those described above. In addition,

possible, she be able to

make

intelligent choice

it is

important

of mortgage payments

will, in fact,

that,

the appropriate calculations

For example, she should understand

herself in such cases. series

to

why

a given

pay off a certain loan over a

certain period of time.

She should also be able to decide which portion

of a given payment

paying off the balance of the loan, and which

portion

is

simply paying interest on the outstanding loan balance.

is

The

first

goal of this text

is

to give the reader

enough information

make an intelligent choice between options in a financial and can verify that bank balances, loan payments, bond

so that he can situation,

coupons,

how

etc. are correct.

Too few people

in today's society

understand

these calculations are carried out. In addition,

however,

we

are concerned that the student, besides

why they work. It how to apply it; you

being able to carry out these calculations, understands is

not enough to memorize a formula and learn

should understand

why

the formula

is

correct.

We

also wish to present

Preface

viii

the material in a proper mathematical setting, so the student will see the theory of interest

Let

appears

me

is

explain

in the title

of

why

the phrase "Problem-Solving

We

this text.

will prove a very small

formulae can be applied to a wide variety of problems.

needed to take the data presented

many

it

texts,

large

these

Skill will

a particular problem and see

in

so the formulae can be used.

where a

Approach" number of

how

formulae and then concentrate our attention on showing

rearrange

how

interrelated with other branches of mathematics.

how

be to

This approach differs from

number of formulae

and the

are presented,

student tries to memorize which problems can be solved by direct

of

application

a

particular

We

formula.

wish

emphasize

to

understanding, not rote memorization.

A

working knowledge of elementary calculus is essential for a all the material. However, a large portion of this book can be read by those without such a background by omitting the sections dependent on calculus. Other required background material such as geometric sequences, probability and expectation, is reviewed thorough understanding of

when

it is

required.

Each chapter exercises.

It

should be obvious that

to learn the material

number of examples and

in this text includes a large

is

for her to

Finally, let us stress that

the most

work it

calculator (with a y^ button) and

is

all

way

for a student

the exercises.

assumed

knows how

our ability to use a calculator that

efficient

that every student has a to use

it.

It is

because of

many formulae mentioned

in older

on the subject are now unnecessary. This book is naturally divided into two parts. Chapters 1-5 are concerned solely with the Theory of Interest, and Life Contingencies is

texts

introduced in Chapters 6-11. In

Chapter

1

we

present the basic theory concerning the study of

is to give a mathematical background for this and to develop the basic formulae which will be needed in the rest of the book. Students with a weak calculus background may wish to omit Section 1 .6 on the force of interest, as it is of more theoretical than practical importance. In Chapter 2 we show how the theory in Chapter 1 can be applied to practical problems. The important concept of equation

interest.

Our goal here

area,

of value

is

introduced,

problems are presented. concept of annuities.

emphasis

in this

and many worked examples of numerical Chapter 3 discusses the extremely important

After developing a few basic formulae, our main

chapter

is

on practical problems, seeing

such problems can be substituted

in the basic

formulae.

how It

data for

is

in this

Preface

ix

we have

section especially that

left

out

many of the formulae

presented

in other texts, preferring to concentrate on problem-solving techniques

rather than

rote memorization.

applications

of

material

the

Chapters 4 and 5 deal with further in

Chapters

through

1

namely

3,

amortization, sinking funds and bonds.

Chapter 6 begins with a review of the important concepts of probability and expectation, and then illustrates

combined with the theory of tables, discussing

Chapter 8

how

interest.

life

how

probability can be

Chapter 7

they are constructed and

concerned with

is

In

how

annuities, that

we

introduce

life

they can be applied. is

annuities

whose

payment are conditional on survival, and Chapter 9 discusses life insurance. These ideas are generalized to multi-life situations in Chapter 10.

Finally, Chapter

1 1

demonstrates

how many of these

concepts are

applied in the extremely important area of pension plans.

Chapters

1

through 6 have been used for several years as the text

material for a one semester undergraduate course in the Theory of Interest,

and

I

would

in earlier drafts.

Wanda Heath

who

thank those students I

am

pointed out errors

deeply indebted to Brenda Crewe and

job of typing the manuscript, and to my Narayanaswami, for his invaluable technical

for an excellent

Dr.

colleague.

like to

In addition,

P. P.

assistance.

Chuck Vinsonhaler, University of Connecticut, was supportive of this project, and introduced Publications, for

which

I

owe him

me

to the people at

strongly

ACTEX

Dick London did the

a great deal.

technical content editing, Marilyn Baleshiski provided the electronic typesetting, like

to

and Marlene Lundbeck designed the

thank them for taking such care

manuscript into what

I

hope

is

in

text's cover.

I

would

turning a very rough

a reasonably comprehensive yet friendly

and readable text book for actuarial students.

St.

John's,

Newfoundland

December, 1988

Michael M. Parmenter

PREFACE TO THE REVISED EDITION months since the original edition of this text was published, number of comments have been received from teachers and students

In the fifteen

a

regarding that edition.

We

most of the comments have been quite and we are making no substantial modifica-

are pleased to note that

complimentary to the

text,

tions at this time.

A edition

given

significant, is

in

rectified

that time

and thoroughly

justified, criticism

diagrams were not used to

of the original

illustrate the

examples

the second half of the text, and that deficiency has been

by the inclusion of thirty-five additional figures

in the

Revised

Edition.

Thus

it

is

fair to

say that there are no

new

topics contained in the

Revised Edition, but rather that the pedagogy has been strengthened.

For

this

reason

we

prefer to call the

new

printing a Revised Edition,

Second Edition. In addition we have corrected the errata in the original edition. would like to thank all those who took the time to bring the various

rather than a

We

errata to our attention.

February, 1990

M.M.P.

PREFACE TO THE THIRD EDITION It

now more

is

textbook.

than ten years since the original publication of this

In that time,

several very significant developments have

occurred to suggest that a

new

edition of the text

is

now

needed, and

those developments are reflected in the modifications and additions

made

in this

Third Edition.

First,

improvements

calculator

in

approaches to reach numerical

now

technology

In particular,

results.

us

give

many

better

calculators

include iteration algorithms to permit direct calculation of unknown

annuity interest rates and bond yield rates.

Accordingly, the older

approximate methods using interpolation have been deleted from the text.

Second, with the discontinued publication of the classic textbook Life Contingencies by C.W. Jordan, our text has become the only one

published in North America which provides the traditional presentation

of contingency theory. To serve the needs of those traditional approach, including the use

deterministic

life

table model,

who

still

prefer this

of commutation functions and a

we have chosen

to include various topics

contained in Jordan's text but not contained in our earlier editions.

These include insurances payable life

contingent

accumulation

at the

moment of death

functions

uniform seniority concept for use with

(Section

8.2),

(Section 9.3), the

table

of

Makeham and Gompertz annuity

values (Section 11.1), simple contingent insurance functions (Section 11.1),

and an expansion of the material regarding multiple-decrement

theory (Section 7.6). Third, actuaries today are interested in various concepts of finance

beyond those included introduced

the

ideas

in traditional interest theory.

of real

modified duration, and so on,

rates

in this

of return,

Third Edition.

To

that

end

we have

investment duration,

Preface

12

new

edition provides a gentle introduction to the more view of contingency theory, in the completely new modem stochastic Chapter 10, to supplement the traditional presentation. In connection with the expansion of topics, the new edition

Fourth, the

As

contains over forty additional exercises and examples.

numerical answers to the exercises have been errata in the previous edition

thank everyone

who

made more

have been corrected.

We

well, the

precise and the

would

like to

brought such errata to our attention.

With the considerable modifications made in the new edition, we is now appropriate for two major audiences: pension actuaries, who wish to understand the use of commutation functions and deterministic contingency theory in pension mathematics, and university believe this text

students,

who

seek to understand basic contingency theory at an intro-

ductory level before undertaking a study of the more mathematically sophisticated stochastic contingency theory.

As with

the original edition of this text, the staff at

Publications has been invaluable in the development of this Specifically

I

would

like

to

new

ACTEX edition.

thank Denise Rosengrant for her text

composition and typesetting work, and Dick London, FSA, for his technical content editing.

February, 1999

M.M.P

CHAPTER ONE INTEREST: THE BASIC THEORY ACCUMULATION FUNCTION

1.1

The simplest of all financial transactions is one in which an amount of money is invested for a period of time. The amount of money initially invested is called the principal and the amount it has grow^n to after the time period This

accumulated value

is

called the

is

a situation which can easily be described by functional

at that time.

which the principal has been invested, then the amount of money at that time will be denoted by A{t). This is called the amount function. For the moment we will only consider values ^ > 0, and we will assume that / is measured in years. We remark that the initial value ^(0) is just the principal itself. In order to compare various possible amount functions, it is convenient mathematically to define the accumulation function from the notation.

If

the length of time for

is

/

amount function

as a{t)

=

^rj^

just a constant multiple of a(t) is

We

.

,

note that a{0)

namely A(t)

=

k

•

=

1

a{t)

and that A(t)

where k

is

= A{0)

the principal.

What any function

functions are possible accumulation functions? a(t)

=

which money we would increasing. Should a(t) be continuous? That depends if a(t) represents the amount owing on a loan t years with a{0)

1

could represent the

accumulates with the passage of time.

hope

that a(t)

on the after

is

situation;

it

way

In theory,

has been taken out, then a(t)

in

Certainly, however,

may

be continuous

continues to accumulate for non-integer values of

t.

if interest

However,

if a{t)

amount of money in your bank account t years after the deposit (assuming no deposits or withdrawals in the meantime),

represents the initial

jump The graph of such an a{t) We will normally assume in this text that a{t) is make allowances for other situations when they

then a{t) will stay constant for periods of time, but will take a

whenever

interest

is

paid into the account.

will be a step function.

continuous; turn up.

it

is

easy to

Chapter

In Figure

1

.

1

we have drawn

accumulation functions which occur

1

graphs of three different types of in practice: a{t)

a{t)

a{t)

(0,1)

(0,1)

(0,1)

(b)

{a)

(c)

FIGURE Graph of

where the amount of

(a) represents the case

constant over each year. interest

earned

On

we would hope

earned also increases;

where

to be in a situation

earned

is

amount This makes more

increasing as the years go on.

is

interest

the other hand, in cases like (b), the

sense in most situations, since larger, the interest

1.1

in

that as the principal gets

other words,

"interest earns interest".

we would like many

There are

which look roughly like the graph in the one which will be of greatest interest

different accumulation functions (b),

but the exponential curve

is

to us.

We interest

money

is

is

remarked

withdrawn between these time periods.

interest paid

the

same

earlier that a situation like (c)

If the

is

height.

However,

all

amount of interest increases then we would expect the steps

if

paid

the

be of as the to get

and larger as time goes on. have used the term interest several times now, so perhaps

We

time to define

if

amount of

constant per time period, then the "steps" will

is

accumulated value increases, larger

can arise whenever

paid out at fixed periods of time, but no interest

it is

it!

Interest

This definition

is

= Accumulated

Value

not very helpful

—

Principal

in practical situations,

since

we

are generally interested in comparing different financial situations to

most profitable. What we require is a standardized and we do this by defining the effective rate of interest i (per year) to be the interest earned on a principal of amount 1 over a period of one year. That is,

determine which

measure for

is

interest,

i

=

a{\)-\.

(1.1)

The Basic Theory

Interest:

We a{t), if

3

can easily calculate

we

recall that A(t)

=

using the amount function A{t) instead of

/

k

Thus

a(t).

Verbally, the effective rate of interest per year

earned

interest

one year divided by the principal

in

the year.

There

definition.

We

is

the

at the

amount of

beginning of

nothing sacred about the term "year" in this

is

can calculate an effective rate of interest over any time

period by simply taking the numerator of the above fraction as being the

earned over that period.

interest

More

we

generally,

define the effective rate of interest in the n^^

year by

- a{n-\) a{n-\)

_ A{n)-A{n-\) _ ~ A{n-\)

a{n)

^"~ Note

that

i\,

calculated by (1.3),

is

same

the

as

/

,,

^.

^^'^^ '

defined by either (1.1)

or (1.2).

Example

1.1

Consider the function

a{t)

=

(a)

Verify that

(b)

Show that a(t)

(c)

Is

(d)

Find the effective

(e)

Find

a{t)

(3(0) is

=

/^

+ + r

1.

1.

increasing for

all

/

>

0.

continuous? rate

of interest

+

=

/

for a{t).

/„.

[Solution] (a)

(b) (c)

a(0)

-

(0)2 -f (0)

1

1.

Note that a'(/) = 2t-\-\ > for alU > 0, so a{t) is increasing. The easiest way to solve this is to observe that the graph of a{t) a parabola, and hence a{t)

is

that all polynomial functions are continuous). (d)

i

.

= a{\)- 1=3-1=2. ^ a{n)-a{n-\) ^ n" ^ a{n-\) «^

-

/7

+

1

is

continuous (or recall from calculus

n

+

-[{n-\f + {n-\) (n-\f + (n-\)-\-\ \

-^ \}

Chapter

SIMPLE INTEREST

1.2

There are two special cases of the accumulation function will

1

examine

The

closely.

sionally, primarily

first

of these, simple

between integer

second of these, compound

interest,

it

by

is

accumulation function and will be discussed

mind some

that in both

of these cases

interest, is

interest periods, but will

mainly for historical purposes and because

is

a{t) that

used occa-

be discussed

easy to describe. the

far

in the

a{t) is continuous,

we

The

most important

next section.

Keep

in

and also that there are

where modifications must be made. Simple interest is the case where the graph oi a(t) is a straight line. Since a(0) =1, the equation must therefore be of the general form However, the effective rate of interest / is a{t) = \ -\- bt for some b. practical settings

given by

i

—

a{\)

—

1

=

Z?,

so the formula

•^f)=rt-f/r,

is

r>0.

(1.4)

\-\-it

(0,1)

[FIGURE

1.21

case (a) graphed in Figure

1.1.

Remarks 1.

This

is

of interest earned each year

is

In this situation, the

constant.

original principal earns interest

amount

In other words, only the

from year to year, and

interest

accumulated in any given year does not earn interest in future years. 2.

The formula ^(0)

=

a(0)

equal to

k,

=

a{t)

=

the

1

.

\

+

it

More

amount

at

applies to the case

where the principal

generally, if the principal at time

time

/

will be A(t)

:=z

k(\

-\- it).

is is

Interest:

The Basic Theory

We rate

noted above that the "/"

of interest for this

+

\

in

=

in a{t)

\

-\-

-

+

\\

also the effective

it is

function. Note however

that

i{n-\)\

\+i(ri-\)

+ Observe

that

not constant.

is

z„

(1.5)

/(«-l) In fact,

decreases as n gets

/„

which should not surprise us. If the amount of accumulated value increases, then clearly the effective rate of interest is going down. Clearly a{t) — 1 + /Y is a formula which works equally well for all values of r, integral or otherwise. However, problems can develop in practice, as illustrated by the following example. larger, a fact

interest stays constant as the

Example

1.2

Assume Jack borrows 1000 from

15%

the bank on January

How much

simple interest per year.

does he

1996

1,

at

a rate of

owe on January

17,

1996? Solution

The general formula A{t)

=

amount owing

for the

1000(1 +.15/), but the problem

should be substituted into this formula.

is

An

at

time

general

in

t

is

what value of

to decide

obvious approach

is

t

to take

number of days which have passed since the loan was taken out and number of days in the year, but should we count the number of days as 16 or 17? Getting really picky, should we worry about the time of day when the loan was taken out, or the time of day

the

divide by the

when we wish

to find the value

of the loan? Obviously, any value of/

only a convenient approximation; the important thing consistent rule to be used in practice. (a)

The

first

method

is

first.

owes 1000

+

In our case this

SO

\2 2(1

-

1

-

.942045)

^=

=

.

(l.Oir^

=

D

1591.

1

somewhat theoretical, and Anyone wishing to proceed

note before starting this section that

independent of the rest of the

directly to particular,

it is

text.

more practical problems can safely omit this more background knowledge is required

understanding here than

is

Assume jim)

we want

^

nominal rates

to find

_

calculate these values

which are shown

TABLE m fm)

We

equivalent to

f'"^

which comes from

l]^

in

is /.

Table

full

=

.12,

and

The formula is

used to

1.1.

1.1

2

5

10

50

.12

.1166

.1146

.1140

.1135

f"^"*

a

In

reading.

i

identity (1.19),

1

observe that

for

first

that the effective annual rate of interest

^[(i_f/)i/^

material.

required for any other section; students with

only a sketchy knowledge of calculus might omit this on

that

.942045, from

FORCE OF INTEREST

1.6

We

^

decreases as

m

gets larger, a fact

will be able to prove later in this section.

We

which we

also observe that the

The Basic Theory

Interest:

are decreasing very slowly as

/^^^

values of

17

along; in the language of calculus,

This see

is,

what

in fact,

what the

derivation, so

limit

is

There

is.

=

is

further and further

we can

"

O'^'"

limit.

use L'Hopital's rule to

no need to assume

with arbitrary

Urn w[(l 4-

we go

seems to be approaching a

happening, and

we proceed

lim f^^

/^^'^

i

—

.\2

our

in

/.

l]

=

"^ ^i^ ^^

"^

~

^

(1.22)

^^i

m Since (1.22)

we

of the form ^,

is

take derivatives top and bottom,

cancel, and obtain

lim

w—>oo since lim (\

+

/« =

=

iY^""

This limit

lim

w—>cx:

[(1

+

ly^""

ln(\-\-i)]=ln(\+

1.

called the force of interest and

is

(1.23)

/),

is

denoted by

6,

we

so

have

S-^TiTTTlf In our example, 6

=

with the entries

Table

in

Intuitively,

6

/«(1.12)

=

1333.

.1

(1.24)

The reader should compare

this

1.1.

represents

a

nominal rate of interest which

is

more theoretical than practical However, 6 can be a very good approximation for f"^^

convertible continuously, a notion of

importance.

when

m

is

large (for example, a nominal rate convertible daily),

and has

the advantage of being very easy to calculate.

We note that identity (1 .24) can be rewritten as e^

\+i.

(1.25)

form is shown in the next example. Again we the importance of being able to convert a rate of interest with a

The usefulness of stress

=

given conversion

this

frequency to an equivalent rate with a different

conversion frequency.

18

Chapter

Example

1

1.11

A

loan of 3000 is taken out on June 23, 1997. 14%, find each of the following: (a) The value of the loan on June 23, 2002. The value of (b)

If the force

of interest

is

/.

(c)

The value of /^^^\

Solution (a)

The value

5 years later is

obtain 3000(e^^)^

= e'4-

(b)

/

(c)

f

1

1

+

^ry

3000(1

+ if

e'^

=

6041.26.

e^"^,

SO

we havc

3000

Using

.

e^

—

\

-\- /,

we

=.15027.

We

6.

note in passing that this series

(5.

us expand the expression

/

=

_

.

=

i

d{\

— d)~\

which

becomes

/-41 +^+^^+

(i^4- •••)

= ^ + ^^+

Again this shows us very clearly that / > d. have |(i| < 1 for this series to converge. yields an

amusing

the left

result:

also note that

(1.35)

we must d =2

In fact, trying to put is

i

=

.

_

^

=

—2, whereas

becomes 2 + all of which are positive + + Thus we have "proven" that —2 is a positive number! Next let us expand f^^ as a function of /. From (1.19) we have 2^

the right hand side

terms.

hand side

We

c/^4- •••.

li^)

=

/^>

= m

m[(\

i

+/)^^^-

w +

1],

'

are

we

•

,

so

2!

3!

'

2!

3!

Again, this converges for

Why

2^

|

/

1




They also give us means of calculating some of these functions, since often only the first few terms of the series are necessary for a high degree of accuracy. If you ask your calculator to do this work for you instead, it will oblige, but the program used for the calculation will often be a /

6 (although they certainly aren't needed for that).

a quick

variation of one of those described above.

22

Chapter

As

a final example,

let

us expand

^w

terms of 6.

in

d^"^"^

+ 0- ^ 1

(1

1

We have

g-«.

(1.37)

so

= m

From

-('-(-fe) 6i

^

= m m

iW

=

+

this

(

6

2\m

we



where

0,

i\

and

Then linear interpolation will be used to approxiii mate a value /q such that/(/o) = 0. To find i\ and ii, we use trial and error, aided by the fact that/(/) is an increasing function. We eventually obtain/(.ll) = -33.58 and/(. 12) = 187.88. are close together.

Linear interpolation assumes that the function

between between

.11 i

=

and

.12.

.W

and

amount of this change /(/o) =:

is

-

The

is

187.88

that occurs

between

(-33.58)

=

occurring between .11 and fraction of the distance

the conclusion that

/o

to five decimal places.

change

total

/=.12

=

is

between .11

-

a straight line

value of the function

(-33.58)

=

The

221.46.

such that and a value Hence the fraction of the change

33.58. oo ro

i^

in the

is

221 46 .11

and

.11

~ .12.

/'o

15163, and

/q

must be

that

This reasoning leads us to

+ (.15163)(.01) =

.1 1

15163, or

/q

=

.11152

D

34

Chapter 2

Example

2.7

Obtain a more exact answer to Example

2.6.

Solution

To improve on




we

will start with values /land

ii

such that

where i\ and 12 are closer together than they were in the solution to Example 2.6. For instance, using i\ =.111 and Using these 72 =.112, we find /(.111) = -11.71 and/(.112)= 10.22.

f{i\)

values,

and/(/2)

we

obtain

/o

We

0,

=

11.71

.in

10.22-(-11.71)

(.001)

=

.11153.

remark that standard calculator techniques give

zq

=

.1 1

153 as

the correct answer (to five decimal places).

TIME-WEIGHTED RATE OF RETURN

2.3

The

rate

of interest calculated

weighted

rate

Section 2.2

in

of investment return.

is

often called the dollar-

A very different procedure

is

calculate the time-weighted rate of investment return, and that

we

used to is

what

We

remark before starting that in this section the compound interest assumption is no longer being made. To calculate the time-weighted rate of return, it is necessary to know the accumulated value of an investment fund just before each will consider here.

deposit or withdrawal occurs. Let ^o be the

initial

balance

in a fund,

B^

the final balance, B\, ..., B^-i the intermediate values just preceding

deposits or withdrawals, and Wi, ..., W„_\ the or withdrawal, where Wj

Let Wo

=

0.

>

amount of each deposit

for deposits and Wj


^.

be repaid by annual payments

each year for the next 20 years.

During the

payments are k per year; during the second

2k per year; during the third the fourth 5 years, 4k per year. If are

3-35.

Given

3-36.

Given

a-^

=

\2 and

=

9.370 and

d-^

a^;^^

=

5 years, /

=

first

at the

5

end of

years the

payments 3k per year; and during 5 years the

.12, find k.

21, find ct^y

a—n =

9.499, find the effective rate of

interest.

3-37.

An

Workers Compensation claim. It is payments of 20,000 a for the next 10 years and equal annual indemnity payments for the next 20 years. The medical payments will begin immediately, and the indemnity payments will begin in one year's time. The insurance company has established a fund of 680,000 to support these payments. Find the amount of each annual indemnity payment assuming / = .07. injured worker submits a

decided that she

3.3

entitled^to annual medical

is

Perpetuities

3-38. Prove identities (3.14), (3.15) and (3.16).

3-39.

Given

/

=

.15, find the present

continuing forever the first

payment

if (a)

is

value of an annuity of 100 per year

the first

payment

due immediately;

is

due

one year; (b) payment is due

in

(c) the first

in 5 years.

3-40.

A

perpetuity of 500 per year, with the

hence,

is

worth 2500. Find

first

payment due one year

/.

3-41. Deposits of 1000 are placed into a fund at the end of each year for

Five years after the next 25 years. payments commence and continue forever. amount of each payment.

the

deposit,

last

If

/

=

annual

.09, find the

Annuities

3-42.

A

11

loan of 5000

the

first

repaid by annual payments continuing forever,

is

one due one year

payments are X, IX, X, IX, 3-43.

.

loan

after the .

.

and

=

/

is

taken out.

If the

.16, find X.

At what effective rate of interest is the present value of a series of payments of 1 at the end of every two years, forever, equal to 10?

3-44. Albert Glover has just signed a contract with the Blue Jays will

pay him 3,000,000

five years.

To

at the

which

beginning of each year for the next

finance his retirement, the player decides to put a

same amount each year) into a fund which will pay him, or his estate, 400,000 a year forever, the first payment coming one year after his last salary cheque. If / = .08, how much salary does the player have left each year? part of each year's salary (the

3-45. Wilbur leaves an inheritance to four charities. A, B,

C

and D. The

of level payments

at the end of each 20 years, A, B and C share each payment equally. All payments after 20 years revert to D. If the present value of the shares of A, B, C and D are all equal, find /.

total inheritance is a series

During the

year forever.

3-46.

A

scholarship fund

is

first

accumulated by deposits of 400

The fund is to be used of 2000 in perpetuity, with the

at the end of pay out one annual scholar-

each year.

to

ship

first

one year

Assume / = .08. minimum number of deposits which must be made

after the last deposit.

(a)

Find the

(b)

Assume 25

in

scholarship being paid out

order to support such a fund. deposits are made.

Show

that

is is

possible to

pay out one scholarship as described above, but not possible pay out two such scholarships.

to (c)

Despite the result in

made,

it

is

(b),

and again assuming 25 deposits are

desired to pay out a second scholarship of 2000

on a regular basis as often as possible. integer value of

paid out every

/

/

Find the

minimum

such that a second scholarship could be

years, starting

/

years after the last deposit.

Chapter 3

72

3.4

Unknown Time and Unknown Rate 6000 from her

3-47. Joan takes out a loan of

pay

it

of Interest She wishes

local bank.

to

back by means of yearly payments of amount 800 for as

long as necessary, with a smaller payment one year

later.

If the

payment of 800 is due in one year and /= .11, find the number of payments required and the amount of the smaller first

payment. 3-48.

Do

Question 47 if the payments are 70 monthly, with the first payment due in one month, and / is still 1 1% per year. Assume the smaller payment is to be made one month after the last regular

payment. 3-49.

Do

Question 47

the loan

3-50.

A

is

if

the

first

payment

isn't

due

until

two years

after

taken out.

fund of 5000

is

accumulated by n annual payments of 50

to be

followed by another n annual payments of 100, plus a final payment, as small as possible,

payment. If/ 3-51.

At what

=

.08, find

effective

the end of every

monthly

month

made one year

after the last regular

n and the amount of the rate

final

payment.

of interest will payments of 200

at

for the next 3 years be sufficient to repay a

loan of 6500?

3-52. Write a computer

program which

cessive approximation.

by sucwhich are correct to 3

will solve Question 51

Print out answers

decimal places, then to 4 decimal places, then to 3-53. Write a general computer program like

3-54.

A

decimal places.

will solve

any problem

Question 51 to any required degree of accuracy.

fund of 25,000

is

to be

annual payments of 500 3-55.

which

5

accumulated

at the

end of 20 years by

at the end of each year. Find

A

/.

fund of 2200 is to be accumulated at the end of 10 years, with payments of 100 at the end of each of the first 5 years and 200 at the end of each of the second 5 years. Find the effective rate of interest earned

by the fund.

73

Annuities

3.5

Continuous Annuities

3-56. Prove each of the following identities:

a-^=

(b)

'-' 6

-

e""^

Show

3-57. (a)

1

that j^S-^

+

1

(5

-S-y

Verbally interpret the result obtained in part

(b)

3-58.

Redo Question

3.6

Varying Annuities

3-59.

=

Rank

3-6(a),

assuming the annuity

is

(a).

continuous.

the following in increasing order of magnitude, and give a

verbal explanation for your ranking,

3a^

(a)

3-60.

A man

(b)(/a)^

+ (Z)«)^

20 annual payments, the

X each

first

increasing by 100 each year. after the last installment is

A

loan

is

He

6

repays the loan with

If the first

payment

given out, and if/

is

is

1, 2, 1, 2,

.

=

due one year

.132, find X.

taken out.

is

first

that the

if

the payments

if

the payments

...

your answer to part

(b), find

limA^. Have you seen n—*oc

before?

Assume

.

Find a formula for the amount of the loan

this

is

/.

are 1,2, ...,«, 1,2, ...,«, \i An

in 5

100 and the payments

to

Find a formula for the amount of the loan are

(c)

time.

one equal

after the loan

effective rate of interest

(b)

(e)

repaid by annual payments continuing forever, the

one due one year (a)

(d) loj,

borrows money from a bank. He receives the money

annual installments, taking

3-61.

(c) 2(/a)^

Where?

74

3-62.

Chapter 3

Under an annuity, the first payment oin is made after one year, the second payment of « — 1 after two years, and so forth, until a payment of/? is made, after which payments cease. Show that the present value of this annuity

given by

is

3-63. Find the present value of a perpetuity under

100

made

is

after

one year, 200

payment of 1500

is

per year forever.

Assume

made,

after /

3-64. Find the present value at

=

which a payment of

after 2 years, increasing until a

which payments are

level at

1500

.075.

11%

effective of an annuity lasting

20

first payment of 1,000 is due immediately, and which each successive payment is 10% more than the payment

years in which the in

for the preceeding year.

which the payment is due six years from now, and in which the payments follow the pattern n, n—\,n—2, 2, 1,2, .,.,n—\,n.

3-65. Find an expression for the present value of an annuity in first

.

9%

3-66. Find the present value at

the

first

3-67. (a) (b)

1,

4, 9, 16, ...

Show that Find 4^

f.a-^

a-^

=

,

.

,

effective of a 20-year annuity, with

payment due immediately,

the pattern

.

in

which the payments follow

400.

-v{Ia)-y

evaluated at

/

=

0.

two perpetuities. The first has level payments of/? at the end of each year. The second is increasing such that the payments are q, 2q, 3q, Find the rate of interest which will make the

3-68. There are

difference in the present values of these perpetuities (a) zero; (b) a

maximum.

CHAPTER FOUR AMORTIZATION AND SINIGNG FUNDS 4.1

AMORTIZATION

To pay back

method

a loan by the amortization

is

means of installment payments at periodic intervals. saw how to calculate the amount of such a payment. will see

how

to find the outstanding principal

point in time, and in the next section into their principal

tion schedules for First,

This

principal. early, or

know

let

and

we

In

Chapter

3

In this section

on a loan

at

we we

any given

how to divide payments and how to construct amortiza-

will see

interest portions

repayment of loans.

us consider the problem of finding the outstanding is

if you want to pay off a loan any way at all, it is important to

of crucial importance, for

change your loan payments

in

amount of the outstanding

the

loan by

to repay the

loan.

Mortgage statements

typically give the outstanding principal at the time of the statement.

There are two approaches which can be used, and one preferable to the other depending on the situation.

may

According

prospective method, the outstanding principal at any point in time equal to the present value at that date of

According

to the retrospective

to the original principal

all

be

to the is

remaining payments.

method, the outstanding principal

accumulated to that point

in time,

is

equal

minus the

accumulated value of all payments previously made.

Some examples

will

illustrate

demonstrate when one method

Example

A

loan

is

is

the

two methods, and

will also

preferable to the other.

4.

being paid off with payments of 500

the next 10 years. ately after the

If

/

payment

=

.14, find the

at the

at the

end of each year for

outstanding principal, P, immedi-

end of year

6.

Chapter 4

76

Solution

500

500

500

500

1

1

1

1

1

1

1

T

T

L

P

FIGURE Here, the prospective method

is

principle

P = 500a^ =

is

amount of the loan L

then have

P=

Example

-

=

5005^.

P=

approach also produces

4.11

come

to

still

1456.86. Retrospectively,

to find the

1(1.14)^

10

although both methods will work.

easier,

Prospectively, there are 4 payments

A

1

6

2

1

500

C,

P — C being the amount of the premium.

with

5.2

corporation decides to issue 15-year bonds, redeemable at par, with

amount of 1000 each.

face

10%

payments are

If interest

to be

made

at the rate

happy with a yield of 8% convertible semiannually, what should he pay for one of these bonds?

of

convertible semiannually, and

if

George

is

Solution

H

\

FIGURE

A

r

=

.05

and

/

=

.04.

D

5.3

9%

100 par-va ue 15-year bond with coupon rate

annually

30

5.2

For these bonds we have F = C = 1000, n = 30. ThenP = 50^^.4 + 1000(1.04)-^^ = 1172.92. ^30j.04

Example

K

\

29

2

1

1000 50

50

50

50

is

convertible semi-

selling for 94. Find the yield rate.

4.50

H

4.50

4.50

\

\

29

2

1

FIGURE

100 4.50

H

30

5.3

Solution

We =

have 94

—

+

4.50a3Qj

lOOv^^, and our calculators give us the value

.04885 effective per half-year.

/

It is

P= i(P

+ Cv",

-Cv") =

form /,

Fra^^

=

we

use.

Fr(l

We

04^^1(7/

=

so Pi

30

-

v").

start

=

Fr(l-v")

/q

=

In this case

+ Cv"/, and

Therefore

with

how to we have

interesting to see as well

solve this problem by successive approximation.

.05

z

= ^^p\'}j^n\

in the right

hand

and

this is the

side,

and obtain

04881. Then using .04881 on the right hand side.

96

we

Chapter 5

get

/2

are done.

=

.04885. Next

we

use .04885, and obtain

Observe how quickly

this particular

ii,

=

.04885, so

example converges

we

to an

D

answer.

BOOK VALUE

5.2 In the

same sense

time,

we

that a loan has an outstanding balance at

can talk about the book value of a bond

Prospectively the definition

any point

in

any time

t.

the same, namely that the

book value is payments. If we assign the usual meanings

is

the present value of all future

at

symbols F, C, r, / and n for a bond, and if we are at a point in time where the t^^ coupon has just been paid, then the book value at time t is the value of the remaining payments: n — t coupons and a payment of C at time n. Hence the book value is to the

=

Bt

{Fr)a^^

Cv-^.

(5.6)

book value is P, and when t = n the book value is C. For values of t between and /?, the book value lies between P and C, and represents a reasonable value to assign to the bond on that date. Book values are often used by investors when preparing financial statements, and are also important in constructing bond Observe that when

/

amortization schedules, as

Example

=

+

the

we

shall see in the next section.

5.4

Find the book value immediately after the payment of the

14^^

a 10-year 1,000 par-value bond with semiannual coupons,

if r

the yield rate

is

12%

coupon of .05 and

=

convertible semiannually.

[Solution]

50 \

\

50

We

have

are 6

F=C =

coupons

left,

1000, n

=

20, r

so the book value

= is

\

20

19

14

FIGURE

1000 50

\

\

\

2

1

50

50

5.4

.05

and

/

=

.06.

At

/

=

14 there

97

Bonds

(1000)(.05)a^ that this

is

+

1000(1.06)-^

950.83.

Observe, as mentioned above,

P—

885.30, but smaller than the

larger than the price

C=

redemption value

Example

=

D

1000.

5.5

Let Bt and Bt^\ be the book values just after the f^ and

Show

are paid.

that 5/+i

—

Bt(\-\-i)

—

(t

+

\y^ coupons

Fr.

Solution

We know that B^ = B,(\-hi)-Fr

{Fr)a-;;zji

+

Cv"'^.

\-v'

(Fr)

Hence

(l+/)

+ Cv^-^(l+/)-Fr

I

Fr

_

Fry

i

We

/

=

n-t-\ Fr 1-v

=

(Fr)a;^^^

=

B t+\-

+ Cv n-t-\

now seen how to find the book value at the time a coupon what do we do between coupon payment dates? The answer we assume simple interest at rate / per period between adjacent have

is

paid, but

is

that

coupon payments, just

Example

as

we

did with loans in Chapter 4.

5.6

Find the book value of the bond the 14'^

coupon

is

in

Example

5.4 exactly 2

months

after

paid.

[Solution]

We know

ft-om

Hence

answer

the

Example is

5.4 that the

950.83

book value

(i)C06)

at

969.85,

/

=

14

is

950.83.

D

Chapter 5

98

Observe coupon,

that if

we would

we

extend Example 5.6 to 6 months after the

obtain a book value of 950.83(1 4-

This

/)•

is,

14^^

in fact,

book value yw^/ before the next coupon is paid. After the coupon is paid, however, the value goes down by the amount of the coupon, and becomes 950.83(1 + /) — Fr. This agrees with our result in Example 5.5. The book value calculated in this way is called X\\q flat price of a bond. If instead of allowing the book value to increase from 950.83 to 950.83(1 + /), and then to drop sharply as the coupon is paid, we simply interpolate linearly between successive coupon date book values, we obtain what is called the market price (or sometimes the amortized

the

This procedure has the advantage of giving us a

value) of the bond.

smooth progression of book values from

bond

In practice, the

P=

(flat) price is

However,

price plus accrued interest.

it

Bq

to

C=

8^.

usually quoted as the market is

also

common bookkeeping

procedure for the book value of a bond to be considered equal to the

market

If the latter

price.

procedure

is

used, then any accrued interest in

the financial statements must be handled separately.

Example

5.7

Find the market price of the bond after the 14^^ I

Solution

coupon

is

in

Example

5.4 exactly

two months

paid.

I

We know that the book value at = 14 is 950.83, and the book value at /=15 equals 950.83(1.06) - 50 = 957.88. We interpolate between these values to obtain 950.83 + |(957.88-950.83) = 953.18. D r

Observe

Example

5.6,

that the

answer to Example 5.7

is

less

demonstrating that the market price

is

than the answer to less than the flat

price.

5.3

We

BOND AMORTIZATION SCHEDULES saw

how

to find the book value of a bond at any saw that this corresponds to the outstanding balance of a loan. In the same way that amortization schedules were constructed for loans in Section 4.2, we can now construct bond amortization schedules in which the final column gives the book value of the bond. Furthermore, the bond amortization schedule will show us in the last section

point in time, and

we

also

99

Bonds

book value changes over time from P to C, just as a loan amortization schedule shows us how the outstanding principal decreases

how to

the

over time.

The basic idea is familiar. The book value is Bt at time r, and the amount of the coupon at time / + 1 is Fr. Since the investor is earning a yield rate of /, the amount of interest contained in this coupon is Bti. The difference, Fr — Bti, is therefore the change in the book value of the This gives us row / + 1 of the bond bond between these dates.

we

amortization schedule, and

Example

Before presenting a

continue on.

5.8

Find the amount of interest and change 15^^

full

an example.

table, let us give

coupon of the bond discussed

in

in book value contained Example 5.4.

in the

Solution!

The book value

/=

was seen

amount of The amount of interest is greater than the amount (50) of the coupon itself! However, this is no problem; it just means that the book value is increasing instead of decreasing, and to get the new book value we add 7.05 to obtain B\s = 957.88, as we saw Example 5.7. If the amount of interest were less than the coupon, that would tell us that P > C and the book value was decreasing. In particular, the excess of the coupon over the amount of interest would be the size of decrease in the book value. D interest in the

We schedule.

time

will

at

15'^

is

now proceed

As was

to be 950.83, so the

=

57.05.

a

complete amortization

this

can be constructed from

(950. 83)(.06)

to

true in the case

construct

of loans,

Horrors!

without knowledge of intermediate book values.

only part of a schedule that

14

coupon

of Example

Example

is

required, the

method

However,

if

to be followed should be

5.8.

5.9

Construct a bond amortization schedule for a 1000 par value two-year

bond which pays yield rate of

interest at

6% convertible

8%

convertible semiannually, and has a

semiannually.

00

Chapter 5

Solution

40

40 1

1

1

1

2

3

IFIGURE

We

have

P=

F= C =

1000(.04)a4j

1000, «

+

=

4, r

1000(1.03)-^

= -

1000 40

40

1

1

4

5.5

.04 and

=

/

Hence

.03.

the price

is

This means that over the

1037.17.

two-year period the book value of this bond will decrease from 1037.17 to 1000.00.

=

book value is 1037.17. When / = 1 (measured in half-years), the first coupon of (1000)(.04) = 40 is paid. From the investor's point of view, the amount of interest in this coupon is Hence the amount of principal adjustment (1037. 17)(. 03) = 31.12. (change in book value) is 40 — 31.12 = 8.88. Noting that book values are decreasing, the new book value will be 1037.17 - 8.88 = 1028.29. The first two columns of the bond amortization schedule are as At

/

0, the

follows:

ITABLE5.1I Principal

Book

Time

Coupon

Interest

Adjustment

Value

1

40

31.12

8.88

1037.17

This procedure the

t

=

2 row.

is

now

continued, using the

The complete schedule

is

1028.29

new book

shown

in

value to construct

Table 5.2; the student

should verify the entries in this table.

TABLE

5.2

Principal

Book

Time

Coupon

Interest

Adjustment

Value

1

40

31.12

8.88

1028.29

2

40

30.85

9.15

1019.14

3

40

30.57

9.43

1009.71

4

40

30.29

9.71

1000.00

1037.17

Observe how nicely time goes on.

this tells us

what happens

to the value

of the bond as

D

Bonds

101

bond

If the

Example

in

5.9

were bought

at

a discount instead of at

a premium, exactly the same procedure would be followed, except that

Column

the entries in

would to

Column

5.4

would be

3

Column

larger than those in

This

2.

us that the differences between the columns should be added

tell

so that the

5,

book values would increase

as time goes on.

OTHER TOPICS we

In this section

will deal with a

number of

problems related

different

to bonds.

Example 5.10 Find the price of a 1000 par-value 10-year bond which has quarterly 2% coupons and is bought to yield 9% per year convertible semiannually. Solution]

I

20

20

H

\

FIGURE In this problem, the

not coincide, so

We =

r

have .02,

/

+ =

1

first

(1.045)^/2

_

1^

„

rate conversion period

do

find the equivalent quarterly yield rate

Now we

(1.045)'/l

/

-

40

5.6

coupon period and yield

we must

H

\

39

2

1

1000 20

20

^

proceed with

F=C=

/.

1000,

40, and obtain 40

20(3401

Example

+

=

1000

D

940.75.

(1.045)^^^

5.

Find the price of a 1000 par-value 10-year bond which has semiannual

coupons of

1

last half-y ear, I

the

first half-year,

bought to yield

9%

20 the second half-year,

.

. .

effective per year.

Solution!

10 \

\

1

20

180

\

\

2

18

FIGURE

5.7

190 \

19

1000 200

h— 20

,

200 the

Chapter 5

102

The The

yield rate price

is

effective per half-year,

/,

is

given by

+ =

1

(1.09)'/'^,

/

then given by

P=

10(/a)2oj

=

1614.14.

+

-20

201

=

1000(1+/)-"^

20v

20

+

10

lOOOv^^

Example 5.12 Find the price of a 1000 par-value 10-year bond with coupons convertible semiannually, and for which the yield rate

year for the

first 5

years and

6%

per half-year for the

5%

is

11%

at

per half-

last 5 years.

Solution

1000 55

55

55

55

1

t

1

1

1

1

1

1

1

1

....

2

1

10

first

10 coupons,

The value of the

we have

of 55a 10|.05

is

(SSq-yq^^^XX .05)'^^

=

The present value of

the redemption

amount

1000(1.05)-^^(1.06)-^^

=

price equals the

three present values,

Let us

now

those studied so

A

342.81.

which

is

The

424.70

1

20

19

5.8

=

r

1

1

a value at time

10 coupons at

last

55

1

11

FIGURE For the

55

+ 248.51 +

at

/

342.81

is

=

424.70.

=

248.51.

given by

sum of

these

1016.02.

D

consider a type of bond which differs somewhat from

far.

callable

redeem the bond

bond at

is

one for which the borrower has the right to

any of several time points, the

being named the call date and the date of maturity of the bond.

earliest possible date

latest possible date

Once

the

bond

is

being the usual

redeemed, no more

coupons will be paid. Possible Redemption

Purchase Date

Call Date

FIGURE

5.9

Maturity Date

Bonds

103

Calculating prices and yield rates gets tricky here because of the

Complicating the matter

uncertainty concerning the date of redemption. is

the fact that sometimes the redemption values will differ, depending

Two

on the date chosen.

Example

examples should help

clarify the situation.

5.13

5%

Consider a 1000 par-value 10-year bond with semiannual

Assume

at

par at any of the

last

4 coupon

Find the price which will guarantee an investor a yield rate of

dates. (a)

bond can be redeemed

this

coupons.

6%

4%

per half-year; (b)

per half-year.

Solution

000 50

50

50

H

2

1

50

50

\

\

\

\

17

18

19

Call

Date

FIGURE (a)

Since the yield rate

is

buying

known

to be the last 50(32oj

this

+

5.10

greater than the

1000(1.06)-^^

=

coupon

rate, the investor

885.30. If the redemption date

is

any

is

uncertain at the time of purchase,

earlier,

1

20 Maturity Date

bond at a discount. If the redemption date were coupon date, the investor's price would be

will be

equal to

50

he should pay more, but, since the redemption date if

he pays 885.30, he will be

guaranteed of earning at least 6%. (b)

Here the yield

rate is less than the

greater than 1000.

coupon dates 50ap7| later,

+

coupon

rate,

so the price will be

redemption date were known to be 4

If the

earlier than the maturity date, the price

1000(1.04)-'^

=

1

121.66.

If the

would be is any

redemption date

then he should pay more, so this price will guarantee him a

return of at least

D

4%.

Example 5.14 Consider a 1000 par-value 10-year bond with This bond can be redeemed for

1050

at the

coupon.

of (a)

6%

time of the

What

19'^

1

100

at the

5%

semiannual coupons.

time of the

coupon, or for 1000

at the

18'^

coupon, for time of the 20'^

price should an investor pay to be guaranteed a yield rate

per half-year? (b)

4% per half-year?

Chapters

04

Solution

50

1100 or 1050 or 1000 50 50 50

50

18

20

19

FIGURE5.il (a)

is greater than the coupon rate, the investor is bond at a discount. Hence the worst scenario for him is if the bond is redeemed at / == 20, and he should assume this case in fixing his price in order to guarantee the 6% return. Hence the answer is the same as in Example 5.13(a), namely 885.30. Here we have a trickier situation. In Example 5.13, assuming an

Since the yield rate

buying

(b)

this

earliest possible

redemption date gave the answer, but the different

possible values for

here

is

work out

C cause trouble

in this

the prices. If redemption occurs at 50(3fY8|+ 1100(1.04)-'^

t=

19, the price

=

at

/

=

In

+

should be

18, the price

If

redemption

occurs

at

=

1155.07.

If

1050(1.04)"'^

:..

1135.90.

135.90 to be sure of earning 4%. Then,

earlier, his yield will turn

where a

can do

20, the maturity date, then the price

should be 50^201+ 1000(1.04r20 1

t=

1175.96.

should be SOa^^

redemption occurs

pay

we

example. All

the three cases separately, and pick the lowest of

if

Hence he should redemption occurs

D

out to be larger.

Example 2.4 of Section 2.1, we saw how to find a point in time single payment was equivalent to a series of payments made at

different times.

We

remarked that there

is

an approximate method,

method of equated time, also available for such problems. These approaches work just as well for bonds as they did in the earlier examples. In addition, two other concepts are introduced here with particular application to bonds. As with the Chapter 2 concepts, they are generally used to provide indices of the average length of an called the

investment.

Let R\, Ri, tn.

•, Rn

be a series of payments made

The duration of the investment, denoted d,

d =

'-^

^v'JRj

,

is

at

times

t\, t2, ...,

defined by

(5.7)

Bonds

105

and the modified duration, denoted

v, is

defined by

(5.8)

The term

volatility is

sometimes used for modified duration

Example 5.15 Find

d and

v for the

bond of Example

5.10.

[Solution 40

Y^jvJ{2Q) +40v40(1000)

d^

^=' 40

£v>(20) +v^^(1000)

^

(/g)^ ^4^

by evaluating

27.48,

at effective rate

V

Note

in this

+ 2000v^Q + 50v^'

example

/

=

—

(1.045)^^-^

= j^. =

1

per quarter. Then

26.88.

that times are given in quarters

of a year.

D

EXERCISES 5.1

Price of a

5-1.

A at

Bond

10-year 1000 face value bond, redeemable at par, earns interest

9%

investor

convertible

8%

semiannually.

Find the price to yield an

convertible semiannually.

Chapters

106

5-2.

A

1000 par value bond, with r

January

bond

is

and July

1

1,

and

—

coupons payable on redeemed July 1, 2001. The

.055, has

will be

bought January

1,

1999,

to

12%

yield

convertible

semiannually. Find the price.

5-3.

Derive the alternate price formula

P^C^(Fr-Ci)a-^. 5-4.

One bond of r

—

A

.025 costs 15.1A.

and r

=

.04 costs

1

5-5.

bond with semiannual coupons Both are redeemable at par in n years

similar

12.13.

and have the same yield

rate

/.

(a)

Find

/

(b)

Find

n.

Two

1000 face value bonds, redeemable

same

period, are bought to yield

One bond

costs

12%

at

par at the end of the

convertible semiannually.

879.58 and pays coupons

half-year. Find the price

5-7.

is,

same

yield.

price of a similar

same

5-8.

A r

redeemable

.035

is

r

\,

=

at par,

have

Prove that the price of a 10-year

9% semi+ .\0A2.

and

1 1

yield rate, equal to ^1 'J

ji and the price

is

\

-\-p.

Find the

which r

=

4

Assume both bonds

/.

are

at par.

100 par-value

=

per

bond with the same number of coupons and the

yield rate, but for b

redeemable

same

at the

For a bond of face value

7%

priced at^i and A2, res-

1 00 face value bond, with a redemption value of

annual coupons,

per year at

of the second bond.

Two 10-year 100 face value bonds, each 8% and 10% semiannual coupons and are pectively, to give the

10%

at

The other bond pays coupons

convertible semiannually.

5-6.

semiannual coupons and

100 with

value

face

(5.2)

10-year bond with

selling for 103

.

semiannual coupons and

Find the yield

rate.

Bonds

5-9.

107

A

100 par-value bond with semiannual coupons

is redeemable at At a purchase price of 105.91, the yield rate exactly 1% less than the coupon rate per half-year.

the end of 4 years.

per half-year

is

Find the yield 5-10.

A

6% coupons and sells them at a price yielding 4% effective. It is proposed to replace them with 5% bonds having annual coupons. How long must the new bonds run so that investors will still realize a yield which is at least 4%? corporation issues par-value bonds with annual

maturing

5-11.

rate.

A 6%

in

5

100 face value bond with annual coupons, redeemable

end of n years year.

years,

105, sells at 93.04 to yield

at

5%

Find the price of a

coupons, redeemable

at the

l\%

at the

effective per

100 face value bond with annual

end of In years

at 104, to yield

7^%

effective per year.

5-12. In addition to the notation already introduced in this section, let

k= ^^^andg- ^. (a)

Derive Makeham's Formula, which

is

P = Cv" + ?(c-Cv"V

=g-

^

(b)

Show

(c)

Show that

^ - ^ l+^.-f^/^

(d)

Using the

first

tion

(e)

that

/

/~g-

two terms of

(d) that

i

(f)

part (c), derive the approxima-

^^^1

I 14-

Conclude from

(5.3)

^

Approximating ^ 2rj

g--

m /

"

\ 1+ ""

^^ 1

(5.4)

•

"

^" ^^^'

^^^^^" ^^^

^^"^

sales-

man's approximation

(g)

Apply Formula

(5.5) to find the yield rate in Question 8.

Chapter 5

08

5.2

Book Value

5-13. For the

bond

Question

in

book value

find the

1,

at

each of the

following times: (a)

Just after the 7^^

(b)

4 months after the

7^^

(c)

Just before the

coupon has been

5-14. For the

bond

coupon has been

S^''

Question

in

paid.

coupon has been

2, find the

paid.

paid.

book and market values on

each of the following dates: (a)

June 30, 1999

(b)

July 1,2000(12:01 a.m.).

(1

(c)

March 1,2001.

(d)

June 23, 2001.

1:59 p.m.).

5-15. Give a verbal argument for the result

5.3

in

Example

5.5.

Bond Amortization Schedules

5-16. Construct a

face r

5-17.

shown

=

Do

bond amortization schedule

amount, redeemable .035 and

/=

Question 16

5-18. Construct the for the

/

at

for a 3-year

bond of 1000

par with semiannual coupons,

if

.025.

if r

=

bond given

8

=

.035 and

and

in

t

=

/

=

.04.

\1 rows of the amortization schedule

Question

1

program which will construct bond amortization If it works, schedules. Test your program on Question 16. construct the entire bond amortization schedule for Question 1

5-19. Write a computer

5-20.

A

5000 par-value bond with semiannual coupons and

=

.03 has a

5%, convertible semiannually. Find the book value of bond one year before the redemption date.

yield rate of this

r

Bonds

5-21.

109

A

10-year bond of 1000 face

redeemable

at par, is

bought

at

amount with semiannual coupons, a discount to yield

12%

convertible

semiannually. If the book value six months before the redemption date

is

985.85, find the total amount of discount in the original

purchase price.

5-22.

A

1000 par-value 10-year bond with semiannual coupons

convertible semiannually

is

bought to yield

annually. Find the total of the interest

9%

column

at

8%

convertible semi-

in the

bond amorti-

zation schedule.

5-23.

A

10,000 par-value 20-year bond with semiannual coupons

is

at a premium to yield 12% convertible semiannually. If the amount of principle adjustment in the 18^^ coupon is 36, find the amount of principle adjustment in the 29^^ coupon.

bought

5.4

Other Topics

5-24. Find the price of a 1000 par-value 10-year

bond with coupons

10% convertible semiannually if the buyer wishes (a) 12% per year; (b) 1% per month.

Find the duration and modified duration for the bond

5-25. (a)

at

a yield rate of

in

Question 5.24(a).

Using the same notation as

(b)

in the text, the

method of equated

time gives a single-time value of

1='4

Find 5-26.

A

1

for the

bond

in

(5.9)

.

Question 5.24(a).

1000 par-value 10-year bond has semiannual coupons of

the

first 5

years and

7%

for the last 5 years.

investor should pay if she wishes to earn (a) (b)

14%

per year.

6%

for

Find the price an

7%

per half-year;

Chapters

110

5-21

.

A

10-year par-value bond of 1000 face amount has annual coupons

which

200 and decrease by 20 each year

start at

to a final

coupon

of 20.

5-28.

(a)

Find the price to yield

(b)

Find the yield rate

A

2%

1

per year.

bond

if the

is

purchased

at its face value.

1000 par-value 15-year bond has semiannual coupons of 60

each.

This bond

is

callable at any of the last 10

coupon

dates.

Find the price an investor should pay to guarantee a semiannual yield rate of (a)

5-29. In

Example

7%;

5.13(a),

(b)

5%;

we saw

(c)

6%.

that an investor should

pay 885.30 to

6% on the bond described. What would the investor actually earn if this bond were

guarantee himself a return of yield rate

redeemed

at

5-30. Consider a

date from t

=

18 instead of at the last possible date?

/

15-year bond with

100 par-value

Assume

coupons.

=

/

=

that this

10 to

29 inclusive, and

/

=

at

bond

is

20 inclusive, 100

at the

semiannual

2%

any coupon 104.50 from / = 21 to

callable at 109 at at

time of the final coupon.

What

price should the investor pay to guarantee himself a yield of (a)

5-31.

2^%

semiannually? (b)

1

^%

semiannually?

Ten 1000 par-value bonds with semiannual coupons of 50 are issued on January 1, 1992. One bond is redeemed on January 1, 2003, another on January

1,

2004, and so on until the

last

one

is

redeemed on January 1, 2012. What price should an investor pay for all ten bonds on January 1, 1992, in order to earn 11% convertible semiannually? [Bonds like these are called serial bonds. The student will find that Formula (5.3), Makeham's Formula given in Question 5-12, is more convenient than Formula (5.1) for finding the price of a serial bond.]

Bonds

5-32.

1 1

A

bond

preferred stock can be thought of as a

which the is no

in

coupons (dividends) continue forever and for which there redemption date. (a)

Find the price of a preferred stock which pays semiannual dividends of

3, if

12%

the purchaser wishes to earn

per year

convertible semiannually. (b)

State a general formula for the price of a preferred stock

paying a yearly dividend of X,

if

the desired yield rate

is /

per year. (c)

State a general formula for the price of a preferred stock

paying a quarterly dividend oiX,

if

the desired yield rate

is

/

per year.

5-33.

Common

stock differs from preferred stock in that the

the dividend paid a

common

is

not constant.

In theory,

amount of

however, the price of

stock should be equal to the present value of

dividends, and one would try what these dividends are likely

vary widely

in the

to settle to be.

Deepwater Oil

on a price by estimating however, prices

market because of the influence of investors

However,

Inc.

future

In practice,

buying and selling various stocks, and this into account.

all

let

we have

us try one problem in this area.

has a policy of paying out

as quarterly dividends.

not taken any of

25%

of

its

earnings

estimated that Deepwater will earn 2

It is

per share during the next quarter, and that earnings will increase at a rate of

2%

per quarter thereafter.

Find the theoretical price an

10%

per year convertible quarterly;

investor should pay to earn (a) (b)

6% per year convertible quarterly.

5-34. Let/(/)

=

'^v^J Rj denote the denominator of Equation (5.7). 7=1

= -4r^. /(O

(a)

Show

(b)

Another concept sometimes encountered

that V

f"(

defined by c

= r^J

.

/(O _

respect to

/

is

is

the convexity,

'\

equal to v^

Show

—

c.

that the derivative

of v with

CHAPTER SIX PREPARATION FOR LIFE CONTINGENCIES 6.1

INTRODUCTION

Mary

takes out a loan of 5000 from Friendly Trust and agrees to pay

back by the amortization method.

Unfortunately,

Mary

it

runs short of

is not able to make all her payments. The All-Mighty Bank lends a large sum of money to a small Central American country. Due to an extremely high rate of inflation,

cash and

the country defaults on

The above

its

loan.

situations are

common

in real life,

and any financial

them into account when lending money. However, these possibilities were ignored in the first five chapters of this book; we always assumed that all payments were made. It is the uncertainty of events in the real world which forces interest rates on most loans to be higher than the prime rate, and also forces rates on loans to high-risk borrowers to be higher than those on loans to others. We are continually reading in the news how a certain country has an AAA credit rating, institution has to take

whereas another country might have an AA,

A

or a

B

rating.

Lower

ratings indicate a higher risk and, consequently, a higher rate of interest will

have to be paid.

The mathematical

which deals with uncertainty is probability and statistics. In this and subsequent chapters, we will see how elementary probability theory can be combined with the theory of interest

to

produce the

discipline

important

contingencies.

In Section 6.2

then

in

indicate,

Section

contingencies problems.

we

6.3,

area

of mathematics called

lay the foundation for later

the

basic

life

work and

approach to solving

life

114

Chapter 6

PROBABILITY AND EXPECTATION

6.2

Let us briefly highlight some of the basic concepts of probability which

needed

will be

in the rest

on the

subject, since this section

Assume fail

is

that a certain event

happen

to

in

probability that

may need

is

to consult a standard text

intended primarily as a review.)

X can

b different ways,

X occurs

(A reader with no previous

of the book.

experience with probability theory

all

happen

in

a different ways and

of which are equally

The

likely.

defined by

^K^=^.

(6.1)

Formula (6.1) says that the probability equals the number of ways in which X can occur, divided by the total number of possibilities. We stress that, in order for this formula to be valid, all the possibilities have to be equally likely. Later examples will show how crucial this requirement

The

is.

probability that

X does not occur

is

equal to

Pr(X)=X-^, = ^,. In general,

Pr{X) = Pr(X) =

1

1,

Example

'\i

X

is

any event,

- Pr(X). then

X

is

If

we

Pr(X)

=

will always 0,

certain to occur.

then

X

(6.2)




=

^

^/),

(7.26)

k=\

and

we

also have

=

4^) -J

2.

= I000(p65+05)V+^

it

is

is

only

unusual.

Hence

in

the

first

In general,

the net single

year that

we

premium

see that is

(1000)0765+. 05),_,j^66V^

t=2 PC

PC

=

1000;765V

-

1000/765V

+

^(1000);765 t-iPeev'

4-

^

(1000),/?65 v'

+

/=2

1000^65

+

50v

+

^

50 r^ipee v^

00

PC

+

50v

50v

+ 50vJ2 /=1

+

50va66

«»«(S;)+T^("-»

iPee v'

165

Life Annuities

Now A^66 =

^67

+

—

^66

•000(^) +

Hence our

1710.

price

is

S(. + ^)= 6001.69.

The previous example could be a

D more general

special case of a

where a "select" group of the population has a different life tables. Sometimes, as in example, a group of people will have a higher than average

situation

mortality experience from that given in the the last

perhaps because they are

probability of survival,

in excellent health.

In

other cases, the select group might have a higher than usual mortality rate,

perhaps because of employment

The notation

p[x\

and

q[x]

probabilities if a person age

x

dangerous surroundings.

in

often used to denote these differing

is

is in

the

year of being

first

in the select

group. Probabilities in subsequent years of being in the select group are

denoted

and so on.

/?[;,]+,,;?[;,] +2,

For

Other notation involving select groups follows naturally. example, first

^[30]

payment

member of the

A

life

would be the net

single

premium

for a life annuity of

one year, to a person aged 30

in

in his first

1,

year as a

select group.

table

which involves a

group

select

is

often called a select-

and-ultimate table.

Example

A

8.8

two years. Select by the relationships

select-and-ultimate table has a select period of

probabilities are related to ultimate probabilities

= ^60 =

P[x]

select

\J^)P^ ^^^ /^W+1

=

1900, De\

1500,

temporary annuity

= \%)P^+^and d^^.^^ =

^" 11,

ultimate

when

/

=

shows

table

Find the

.08.

^[501 20]

ISolutionI 19

We

will proceed

where

from

first principles.

tp^Q is the probability

the select period at age 60. 2?60 =;^[60iP[60]+i

=

is

have

of survival for

t

^r^oi 20]

—

In general,

^

+

X^ ^^?60,

years of a person entering

Now we knowp^Q — P[60] —

(^) iPeo-

since the select period

We

t%Q

(tA)/^60 and

= (^^j

,peo,

t

only 2 years. Therefore the annuity value

> is

2,

66

Chapter 8

«(60).20j

=

= We

1

+ (to)

+ (So)

^^60

__

,

are given that ^^0201

VL

_3J

200

(Si)''.60.201

"^^PeO

v/?6o

=

^6cr/^

"

""

t|-

^^"^

^

life

tables

is

illustrated

8.9

The following values A^38

" S^

^^^^'

Another variation on the idea of changing by the following example.

Example

,9;>60

^P^^-

(200;^^^)" 200 ~ V200JvT9J

^[60].20j=

+ (Sq) v"

••

vn

200

=11. Also

+

=

5600,

A^39

are based on a unisex life table:

^

5350,

-

iV40

5105,

7V41

=

4865,

A^42

=

4625

assumed that this table needs to be set forward one year for males and set back two years for females. If Michael and Brenda are both age 40, find the net single premium that each should pay for a life annuity of 1000 per year, if the first payment occurs immediately. It is

I

Solution]

Michael should be treated as if he were age 41, so his premium will be 1000iV41 1000(4865) 20,270.83. - ^r.^nr.o. 240 Z)4i Brenda should be treated as if she were 38, so her premium will be

i»

8.3

= lOOgOO) ^

ANNUITIES PAYABLE

In practice life annuities are often

year, with

D

22,400.00.

monthly being a very

m'^^y

payable more frequently than once a

common

frequency.

situation arose with interest-only annuities in

Chapter

Exactly the same 3, as

with mort-

gages, for example, where the payments were usually monthly.

encountered no difficulty

in

dealing with this in Chapter

3,

We

where we

simply converted the given interest rate to an equivalent rate for the

payment period and proceeded as

usual.

Life Annuities

167

However

life

We

annuities do present a problem.

are totally

dependent on commutation functions for calculating values of

commutation functions are only tabulated for standard

of

rates

a^,

and

interest,

and {ox yearly probabilities of survival. To apply commutation functions to life annuities payable monthly, for

which don't

We to a

denote by Ux

spaced payments of similar notation

of Chapter

3.

tables

the present value of an immediate life annuity

aged x where each yearly payment of

life

we would need

example,

Another method must be found.

exist.

^

each, the

was introduced

first

due

at

1

m evenly-

divided into

is

age x

+

^.

Recall that a

for interest-only annuities in Exercise

The difference between

a^ and Qx

is

30

illustrated in Figure

8.7.

1

1

ar

1

1

1

1

1

^

x+2

x+1

1

1

1

1

1

1

1

x+3

x+A

FIGURE

8.7a

1

jc+5

1

m

1

1

1

m

1

1

1

;C+2

x-\-^

••

1

m

m

Im)

1

1

1

X

FIGURE

1 1

m-\ ^

x+\

8.7b

We now If,

for the

oiy, then

proceed to derive a good approximate formula for Jx^\ moment, we allow ourselves to use Dy for non-integral values

we have

PC

m

'=0

7=1

(8.22)

mDx .

Using

linear interpolation

Dx+,+j/m

between successive Dy

~ Ax+/ +

4i(Dx+,+i

for integer >^,

-Dx^d

we

obtain

(8.23)

Chapter 8

168

Substitution of (8.23) into (8.22) yields

,{m)

mDx Y^ Y, (a+, +

^(i).+,+i

- A+,))

_/=0 j=\

"I

Y^

+ mD^

mD^ 7=1

'

^^+'

i=\

m 1

+a

= ar+\ -

=

ax

m

^

m{m-\-\)

In?

+ m-2m

(8.24)

Formula (8.24) for a^ is very important, and in this and subsequent sections.

will be required

numerous

times

Example 8.10 Linda, aged 47, purchases a

of 1000 each, the

premium I

Solution

first

life

annuity consisting of monthly payments

payment due in one month. Find the net = 850 and A^48 = 6000.

single

for this annuity if 1)47 I

000 47

47-L ^' 12

000

47^

FIGURE Since the total yearly payment

is

12,000, the answer

approximation (8.24) gives us a^^

12,000(^^

8.8

^

=

+ ^) = 12,000(^^ +

^347

+

2^)

44.

^

i

is

12)

12,000^4^

Hence

90,205.88

the .

.

Our

answer

is

D

169

Life Annuities

What payment

if

to be

our m^^^y

made

at

(m)

_

annuity

life

age x

is

deferred for n years, with the

first

+ « + ^? We then have

(D^±n\ Am) 'x+w

I

m-\ 2m

I

(%•)(

\

r ^K.1^^^ V^ \^

^'*m)(^}

(8.25)

Using (8.24) and (8.25) together, it is easy to obtain a formula for a temporary life annuity payable m times a year for n years. We have (m)

for all X

8-40.

A man

is

fix

=

05

50.

offered the choice of a continuous

annuity paying

life

20,000 per year or a continuous 5-year annuity with certain

X

payments at per year, followed by a continuous life annuity paying per year. Assuming S = .05 and /i^ = 06 for all x, find

X

^such 8.4

that the

two

annuities are equivalent.

Varying Life Annuities

8-41. Pauline purchases a life annuity

which

will

pay 2000

time, with annual payments that will increase by thereafter.

Find the present value of this annuity

36, given N31

=

24,000,

8-42. Repeat Question 41 if a

Assume

5*47

=

5*37

=

300,000 and D^e

maximum

160,000 and N47

=

Pauline

if

=

is

aged

1500.

of 10 payments

9,500.

in one year's 400 per year

is

to be

made.

L ife A nnuities

8-43.

1

Repeat Question 41

if

payments increase

and then remain constant

thereafter.

to a

Make

the

87

maximum

of 5600 same assumptions

as in Question 42.

8-44. Repeat Question 41

if,

instead of increasing, the payments de-

crease by 400 per year until reaching zero.

and7V42

=

Assume

5*42

=

185,000

15,000.

8-45. Prove each of the following identities. n

{Da\-^

(a)

=

^a

=

ri-N,^,-(S,^2-S.^n+2)

t^\

(b)

{Da\-^

(c)

{Id%

=

^

commutation symbols the present value at age .x of a commences with a payment of 10 at age x, increases annually by 1 for 5 years to a maximum of 15, and then decreases annually by 1 until it reaches zero.

8-46. Express in life

annuity which

commutation

8-47. Find formulae for n\(J^)x and (Da)^-, in terms of

symbols.

8-48.

As with

level

life

annuities,

(Ia)x

annuity where each yearly payment

payments paid

at

intervals

approximate formulae:

8-49.

^

represents is

divided

of length ^.

an

increasing

into

m

equal

Derive the following

(a)

(Iat>

=

(la),

+

«.

(8- 46a)

(b)

(Idt^

=

(Id),

-^a,

(8.46b)

A

40-year-old purchases a

which

will

commence

in

life

annuity with annual payments

exactly 10 years.

1000 and payments will increase by

show

that the net single

premium

(^^^)(:o;'4o)(l+e5o).

8%

The

first

per year.

for this annuity

is

payment If

/

=

is

.08,

Chapters

188

8-50. Let {Ia)x denote the present value of a continuous life annuity (to

a person age

jc)

which pays

at the rate

of

1

per year during the

first

year, at the rate of 2 per year during the second year, and so on.

Explain verbally

why {la^

Oa)x denote the net

8-51. Let

annuity (to a person age

approximately equal to {Ia)x

is

premium

single

which pays

jc)

kdx.

H-

for a continuous life

at the rate

of

/

per year at

moment of attaining age x-\-t.

the

roc

Explain

(a)

why

=

(7^);r

/

tv^ tp^dt.

Jo

Show that

(b)

8.5

(la)^

is

approximately equal to (Ia)x

+

tW

.

Annual Premiums and Premium Reserves Example 8.18

8-52. Repeat

if

Arabella

is

paying for the annuity with

semiannual instead of annual premiums. 8-53. Eric, aged 40, purchases a deferred life annuity

monthly payments of 300

commencing

at

age 60.

at

the

Eric pays a

each year for the next 20 years. His

Deo 8-54.

120, andA^60

=

month

premium at the beginning of premium is 4000, and the

X if 1)40 =

500, A^4o

=

8600,

1000.

Again consider Eric's purchase of the deferred Question 53.

will provide

of each

first

remainder are each equal to X. Find

=

which

beginning

This time,

if

life

annuity in

Eric dies before reaching age 60, net

premiums paid prior to death are refunded with interest. Using the same data as in Question 53, and assuming / = .08, find X. 8-55. Repeat Question 53 if 1000 out of the first issue expenses, and

20%

premium

is

required for

of each subsequent premium of

X

is

required for administrative upkeep.

8-56. If

30%

of each gross premium

is

required for loading (when

paying for a deferred annuity), what

premium

to the net

premium?

is

the ratio of the gross

CHAPTER NINE LIFE INSURANCE BASIC CONCEPTS

9.1

In the previous chapter

annuities,

see

we saw how

techniques from the theory of

can be combined with elementary probability theory to study

interest

how

life

which are annuities contingent upon survival. Now we will same ideas can be used to study life insurance, where the

the

contingency of interest

is

that of

dying

at certain

saw one example of this type of problem

Example

in

times in the future.

Section 6.3. Here

is

We

another.

9.1

Rose is 38 years old. She wishes to purchase a life insurance policy which will pay her estate 50,000 at the end of the year of her death. If 7

=

.12,

find an expression for the (actuarial) present value of this

benefit.

~>-

Solution

50,000 +

38

Death

[FIGURE 9T1 The expected value of this benefit payable / + 1 years hence is the probability that Rose dies at age last birthday 38 + / multiplied by 50,000, which is (,/?38)(^38+/)(50,000). The present value of the entire policy is the sum of the present values of these terms, which is oc

50,000^Gp38)(^38+/)(112)~'~'.

ment

is at

To life

the end of the

t^^

Note

that v'+^

is

required since pay-

D

year.

Example 9.1, we could consult would be laborious to add all the Section 9.2 that commutation functions

obtain a numerical answer to

tables but, as with life annuities,

terms together.

We

can be used to aid

will see in

it

in the calculation.

On

the other hand, if a simple

90

Chapter 9

formula for/?^ example, ^38+/

=

\i

is

assumed,

=

p^

may be

sum can be

that this

calculated. For

then tP3s — (-94)^ from which 06, so our present value is

.94 for

— P3s+t —

1

it

all x,

=

50,OOoJ(.94)'(.06)(l.l2)-'-'

we

find

J (^)'

50,OOo(-ffi^)

/=0

/=0 (=0

1

50'000(t^)(-

.94

1.12

= Example

9.1 is

16,666.67.

an illustration of a whole

a fixed amount, thQ face value,

life

policy, a policy

where

paid to the insured's beneficiary at the

is

end of the year of death, whenever that

may

be.

The

policy with face value of

aged

x, is

given by the symbol

Ay.

The formula

1,

to an insured

price of such a

is

(9.1)

tP.q.^tv'^'^

J2 /=0

Note that eventually tPx = 0, so this sum is actually finite. It will be assumed for the rest of Sections 9.1 and 9.2, as well as in the exercises for these sections, that insurances are payable at the end of the year of death.

Example Michael

9.2 is

50 years old and purchases a whole

value 100,000.

4=

If

1000

A

- j^]

and

/

policy with face

life

=

.08, find the price

of

this policy.

[Solution] 54

The required

price

Note

sum terminates

that our

We

values. ^50+/

=

1

-pso+t

100,000^50

is

have

=

,^30

=

=

100,000^;

tPso

because

ipso

54

at

/

-

7^ -

^%5"-^50

qso+t (1.08)-'-^

= ^

for all larger

= ^33^-

- IosIsqI/ Hence the premium

1

-

^"^

is

54 1

00,000^(5^)(33L)(1.08)^

/=o

'

'

'

/

.

\ 55

22,397.48.

-

(m)

D

Life Insurance

191

some cases

In

company

a

that the face value

is

period. If the period

is

paid only

death occurs within a prescribed

if

n years and the insured

denoted ^' - (for a payment of

which means

will sell term insurance,

1),

is

aged

and the formula

then the price

x,

is

is

/=0

Example

9.3

Calculate the price of Rose's insurance in

Example 9.2 if both For Rose, assume />;f =

Example

9.1

and Michael's

insurance in

policies are in force for a term of only

30 years.

.94 for

all x.

[Solution In Rose's case, the price

is

29

50,000

Y,

29

;/^38^38+/(1.12r-^

=

50,000

^

(.94y(.06)(1.12r

/=0

/=0

30 1

-

50,000(jfi^) 1

_

^

= In Michael's case,

we

Jl. 1.12

16,579.74.

obtain

29

00,000^

,;75o

^50+/

(l.OSr-^

/=0

29

- 100,000^(5^) (33L_)(1.08)'='

=

We

^oo,ooo ^/^ V

30 \

\(

)\\m)\

55

^

v»08J

20,468.70.

D

could also talk about ^lA^, the price for deferred insurance,

where the policy of face amount 1 is purchased at age x but does not come into force until age x -\- n. This is not as important as the other two cases, so

we

will not stress

it

here, but

it

should be noted that

92

Chapter 9

= Al- + „\A.-

A,

Finally, there

value

paid

is

policyholder

if

Section

is

8.

1

«-year

endowment

the

if

The price for this benefit, with face value 1, is denoted sum of /7-year term insurance and a pure endowment (see

the at

)

age x

In this context the

Hence we have

-^ n.

symbol

A x:n-

0.

Why?

CHAPTER TEN STATISTICAL CONSIDERATIONS 10.1

MEAN AND VARIANCE

The concept of expected value was introduced in Section 6.2, and that idea was then applied in almost all of the calculations in subsequent DC

=

For example, A^

chapters.

of which

is

Yl

tPx(]x+t^''^^

is

a

sum of terms, each one

the product of the present value of one dollar paid at the end

of a given year

times the probability of death occurring during Such an expected value is called the mean of a random variable Z (in this case Z = v^"^^ ), and is denoted E[Z]. Two important general properties of the mean, to be referred to (v^+^),

that year (tp^qx+t)-

later,

are

£[Zi+Z2]-£[Zi]+£[Z2]

(10.1)

and E[rZ] for

any number

=

r

E[Z],

(10.2)

r.

Before proceeding,

we

should acknowledge that

we

are being less

We

than rigorous in our presentation of concepts in this chapter.

however, that students

this

somewhat informal approach

whose background

One measure of

in statistics is

— E[Z\f].

In practice,

we

Var(Z)

will

^

random is

variable

Z

is

the

defined to be equal to

always use the formula

E[Z^]

-

(E[Z])\

which can be derived using Properties (10.1) and example.

hope,

be helpful to

not particularly strong.

the dispersion of a

concept of variance, denoted Var(Z), which

E[{Z

will

(10.3)

(10.2).

Here

is

an

224

Chapter 10

Example Let

Z

10.1

denote the present value random variable,

whole

life

policy with a death benefit of

of death, purchased by a 20-year-old. and

tP2o

=

(.97y for

all

1

payable

at

policy issue, for a

at the

end of the year

Find E[Z\ and Var{Z)

if

i

=

.07

/.

ISolutionI

E[Z]

is

the

same as^20> which

is

=

^(.97)'(.03)(i^)

•01 1.07

/=0

To

calculate Var(Z),

probabilities as Z,

we

first

1

^

1

^

- -^ 1.07

We

find £'[2^].

=

.30.

see that Z^ has the

but the present value of the death benefit

is

same

now

2t+2

(rin)"'

-{ 1.07 j

Hence

2t+2

^^1.07, ;=0

-

=

03

.03

'

.97 V/,(1.07)2

/

^

1

1.1449 \^i-

1.1449

/

.17153.

Then we obtain Var(Z) == E[Z'] =

-

{E[Z\f

.17153 -(.30)2

= .08153.

The concepts of mean and variance ble at the

moment of

also apply to insurances paya-

death; in fact, the formula A^

which was derived in Section 9.3, for the random variable Z — v^

is

—

J^v'

ip^fix+t^f,

just the expected value calculation

225

Statistical Considerations

|Example 10.2

Z

Let

denote the present value random variable,

whole

life

policy with a death benefit of

6

=

.07.

policy issue, for a

at

payable

whom

by an individual for

death, purchased

1

^^

at the

—

moment of

04 for

all

x and

Find E[Z] and Var(Z).

ISolutionI

£[2]

is

just

^;t,

which

is

roc

»oo

/

Jo

Jo

I

DC

=

.04/

dt

Jo

As

in the

.36364.

previous example, to get E[Z^]

E[Z^]=

-

we just

square v^ so

e-^^'e-'^^\M)dt

/ Jo

.04/ e-'^^dt Jo 22.

Then VariZ)

£[Z^]

is

called the

-

.22

-

(.36364)^

=

D

.08999.

second moment of the random variable Z. Note works out just like E[Z\ but with the

that in the last example, E[Z^]

force of interest doubled.

Because of that, the notation

'^A^

can be used

to represent E[Z^] in cases like this.

Unlike the mean, the variance does not usually satisfy the linearity properties given

by (10.1) and

(10.2).

However,

it

does satisfy the

properties

Var{rZ)

=

p-

•

Var{Z)

(10.4)

and

Var{r+Z)=

Var(Z),

(10.5)

226

Chapter 10

any number

for

application to

Example

all

Property (10.4) has an immediate and important

r.

of our work.

10.3

Redo Example

1000 payable

10.1 if the death benefit is

end of the

at the

year of death. Solution!

We know value

from Chapter 9 (or by using Property (10.2)) that the expected

now

is

=

1000^^20

the variance will

now

more decimals

in

1000(.30)

=

Property (10.4)

300.

be equal to (1000)2(.08153)

Example

10.1

gives

the

=

tells

us that

(Keeping

81,530.

more precise answer

D

81,526.587.)

When we

turn our attention to life annuities, there

a problem

is

oc

The expression

that needs to be addressed at the outset.

a^

=

^V

tPx,

t=o

while intuitively appealing,

Ax

=

X^v'"*"'

which ax,

The

of a conceptually different nature from

latter consists

of a sum of terms, each one of

the product of a possible value of the present value

is

variable

tPxqx+f

is

Z

random

and the probability associated with that value. In the case of

however, the possible values of the present value of all payments are

sums of

the

terms (depending on

v'

how

long the individual survives).

Nevertheless, the linearity properties given by (10.1) and (10.2) assure DC

US that ax

=

Yl^' tPx

variance (which

in

still

correctly calculates the mean.

general

is

not linear)

is

However, the

another matter entirely. The

next example illustrates these problems and shows

how

to

go about

solving them.

Example

10.4

In a very

ill

2/790

=

we know that /?9o = .80, A 90-year-old wishes to purchase a

population of elderly people,

.50, 3/?9o

=

.25

and

4/790

=

0.

payment in one year. If / .08, find the mean and variance of the random variable for the present value of payments under this annuity. life

=

annuity of 10,000 per year with the

first

221

Statistical Considerations

Solution

I

The mean

is

just

10,000^90

=

10,000

=

13,678.68.

The variance

is

+(.50)(y^)V.25(^)

(^(.80)(^)

more complicated; we

show two methods of finding

will

it.

Method One To start, we assume

the

annuity

possibilities for the present value

survives

1

is

of future payments.

year but not 2 (the probability of which

the value will be r-jyo-

probability .50

—

.25

=

There are three

per year.

1

is

impossible.)

+

value will be

.25), the

Hence /

2

-r4o

(

J

+

the second .

/

—

.50

person

=

.30),

person survives 2 years but not 3 (with

If the

+

-r-ljo

(

y4yg

the person might survive 3 years (with probability .25) in

value will be -r4yg

If the

.80

is

.

jXr^ (

j

.

j

.

Finally,

which case the

(Note that surviving 4 years

moment E[Z^]

is

x2-2

+-25(T:k+(Lk)

•3o(i:k)

2

+ •25(T:k+(T:M)'+(T:k)y = Then

the variance for an annuity of

1

per year

2.712567155 -(1.367868211)2

is

=

.841503712.

Using Property (10.4), the answer to our question (10,000)2(.841503712)

=

2.712567155.

is

84,150,371.20.

Method Two Let

and Ax. ax

Y be the present value random variable whose expected value is ax, let Z be the present value random variable whose expected value is

We saw in V —A -j-^. = —

Y=

^ ~j

(10.5) then

,

Chapter 9 that A^

\

—

ddx

=

v

—

da^, and consequently

(This actually follows from the more general relation

which tell

=

is

us that

derived in the same way.)

Properties (10.4) and

Chapter 10

228

=

Var{Y)

Then we can solve

=

Var(^^

- |) = Jj

by finding Var{Z).

this

Var{Z).

•

First note that

.8246023548

and E[Z']

=

%o = =

Hence

the

d— T^,

.6845863477.

of

Z

get Var(Y)

=

variance

we

is

^^ 90

Method Two

Example

is

-

=

(^90)^

0046173042.

in

a

the only reasonable

annuity

life

way

Since

In the usual situation

.8415037, as before.

where the number of future payments large,

25{j^y + Himy

-20(1^)' + .30(1^)' +

is

potentially quite

D

to proceed.

10.5

Determine the variance of the present value random variable for a continuous 6

=

annuity of

life

1

per year assuming

fix

—

-03 for all

x and

.05.

Solution In Chapter 9,

have

Z—

1

—

we ^

•

noted the identity A^ F,

=

\

—

ba^.

More

generally

and consequently

VariY) In the insurance case,

= Var(^-^) =

j^

Var(Z).

we know that

E[Z]=

/

e-°5'e-°\03)c//=|

J

and E[Z^]

Hence Var(Z)

—

j^

= /"e-^^V^^^(.03)J/= ^.

~ (i)

"^

^^'

^"^ ^^^ required variance

^«K>0=;^(^)=

36.06.

is

we

229

Statistical Considerations

Another example of an expected value seen of the complete expectation of that the formula for this was

encountered

life

earlier

in

was

the notion

Section 7.5.

Recall

/•OO

=

^r

tPx dt.

/

Jo

This

is

the expected value of the future lifetime T{x) of a person aged x.

Although annuity

where

/

=

calculating the variance will not recall

Exercise

premium

e^ is just the net single

the case

in

7-33

for a continuous life

the approach just demonstrated for

0,

work

here.

from Section

7.5,

The

crucial observation

is

to

which gave the alternative

formula ttPxfJ^x+tdt.

Jo

We recognize that this the term inside

integral

is

(The

probability tPxl^x+t-

latter is called

proper statistics terminology.) the

same

direct

Example Given

up just like the integral for^;^, since amount / times the instantaneous

set

the product of an

is

way

as

we

3.

probability density function in

Hence we can

calculate the variance in

did for A^.

10.6

4=

lOOOf

- t^),

1


0, so a negative lower bound tells us nothing, and we also know that the largest value

Z can

in part (b)

obtained.

take

we

is

-p^

part

probability

of

Z

(^j^) =

.816




the interval

directly

obtained .87106,

the first

+ (.97)(.03)

two

=

are in the interval (whereas

So

what the

is:

since

Z

will only

years.

.0591,

so

The the

Chebyshev only

Chapter 10

232

Often what

is

of most interest

constant c such that Pr{Z

bound on how


;+/

=

and

both subject to the force

(y) are


3 =

7.2.

month for payment occurring immediately, with 50 per month continuing to her husband, aged 65, if she dies. If her husband Ethel, aged 60, purchases an annuity paying 80 per life,

first

dies before Ethel, the annuity to Ethel increases to

month. Find the net single premium for

C=10.C = 8.5and4'o'^3 = 11-37.

this annuity

100 per

given

6.5.

Wally, aged 65, purchases an annuity paying 400 per month as long as both he and his wife, aged 60, survive, with the

first

pay-

If either dies, the benefit is ment occurring immediately. reduced to 300 per month for the life of the survivor. Find the

net single

premium

for this annuity given

4o —

10,

45 —

8.5

Chapter 11

262

1

1-38.

Prove algebraically that ay\^

(a)

=

Y1^^

tPx {t-\\qy) ^x+t-

t=\

Give a verbal explanation for the identity

(b)

11-39.

Darryl and David are both 30 years old.

in part (a).

Annuity

A

pays

1

per

year as long as exactly one of the men is alive. Annuity B pays 3 per year while Darryl is alive, and 2 per year to David after Darryl dies.

Annuity

A

has present value 8.50 and Annuity

B

Find the present value of an annuity per year while both Darryl and David are alive.

has present value 32.25.

paying 11-40.

1

George has just retired at age 65, and his retirement benefit entitles him to 1000 per month for life (starting immediately), with 500 per month continuing to his surviving spouse if he should die. George is four years older than his wife. A unisex life

table gives us the values d^^

"57 66

~

8.17.

We

—

9.15,

il2)

hi

11.43 and

are told that the table should be set forward

one year for males and

set

back four years for females. Find the

present value of George's retirement benefit. 11-41.

A

reversionary annuity provides payments of 100 at the beginning of each month during the lifetime of Brown, aged 60, after the death of Smith, aged 65. You are given the following

values from a

life

table

which follows Makeham's law: 4'^^

X

You

.(12)

60

10.5

8.8

61

10.2

8.5

62 63 64 65

10.0

8.2

9.8

8.0

9.6

7.8

9.3

7.5

are also given the following values

from a table of uniform

seniority.

Difference in

Ages

Addition to

Younger Age

1

0.5

2

1.0

3

1.6

4

2.2

5

2.8

Find the net single premium for this annuity.

CHAPTER TWELVE PENSION APPLICATIONS One of

major areas of application for

the

annuities in particular,

is in

contributions for participants in a pension plan, and final chapter

contingencies,

life

and

the calculation of the values of benefits and

we have

devoted the

of this text to pension applications.

The reader

no new mathematics

will be pleased to learn that

The terminology

required here. earlier chapters,

and

this

should be quite temporary.

may

We

differs

initially

will

somewhat from that used cause some confusion, but

is

in it

spend the entire chapter presenting

number of worked examples, which, hopefully, will prepare the reader for most problems which can arise. This can be put to the test in a large

the exercises at the end of the chapter.

Example

12.

Francisco, aged 45, works for a large oil company. retire first

At age 65 he

and begin collecting a pension of 20,000 per year for

payment coming one year

after retirement.

of Francis co's future benefits given

7V66

=

life,

will

with the

Find the present value

950 and

D/^s

=

180.

[Solution!

20,000

20,000 \

\

\

45

65

66

FIGURE The present value

Example Repeat

is

12.1

20,000 (^1=20,000 (US')

=

105,555.56.

D

12.2

Examp

e 12.1

if

the

20,000 per year, with the ment.

\

67

Assume

D(,s

=

80.

payments are 5000 per quarter instead of payment coming 3 months after retire-

first

Chapter 12

264

Solution

5000 5000 5000 5000

FIGURE The answer

66

65

45

12.2

is

= 20,000(g^)(..3 +

20,000(g^)aS>

20,000

Das

20,000(950 180

Some pension life

applications

contingencies.

Example

may

i)

+

30)

108,888.89.

D

require just the theory of interest and not

The next two examples

illustrate this point.

12.3

At the end of each year, Wanda's employer will contribute 10% of Wanda's salary to a pension fund. Wanda's salary increases by 5% each year, and the contributions earn interest at a rate of 9% per year. If Wanda's current salary is 20,000 per year, how much will the fund contain after 15 contributions assuming Wanda remains alive and employed throughout the period? I

Solution!

The answer

is

2000[(1.09)'4 -f (1.05X1.09)^3

=

2000(1.09)^4

-

2000(1.09)^4

1

+ (1. 05)^(1. 09)'^ +

+

M 1.09

+

V'QV

+

•

•

•

+ (1.05)i^:

Vi-09;

=

78,177.71.

(l:o9J

In general,

any time a pension calculation assumes, for convenience, that is then done

the probabilities of mortality are negligible, the calculation at interest only.

265

Pension Applications

we saw how

In Section 2.3

of which is just assumed. Each of these

to calculate the time-weighted rate

investment return, as opposed to the dollar-weighted

our usual yield rate

when compound

interest is

measures of investment return are encountered

we

pension situations, as

in

see in the following example.

Example

On

rate,

12.4

January

March

1,

2000, a pension fund has a market value of 3,000,000.

31, 2000, a contribution of 160,000

made.

is

On

Immediately after

made, the market value of the assets is 3,300,000. On August 31, 2000, a lump sum distribution of 12,000 is made. Immediately after this distribution is made, the market value of the assets

this contribution is

is

On December

3,250,000.

31, 2000, the assets have a market value of

3,500,000. (a)

Find the dollar-weighted rate of return of the fund during 2000.

Assume simple (b)

interest for periods less than a year.

Find the time-weighted rate of return of the fund during 2000.

ISolutioril

(a)

In calculating dollar-weighted rates, the market values at interme-

no significance; we are only concerned with the amount of each deposit or withdrawal. The equation of value,

diate times are of

using simple interest for periods less than a year,

is

3,000,000(1+0+ 160,000(1 + 10- 12,000(1 + ^0 leading to 3,1 16,000/ (b)

In this

=

352,000, so

/

=

.

1

=

3,500,000,

1297.

case the market values at intermediate times are very

Without such information we simply could not calculate a time-weighted return. The fund value just before the March 31 deposit is 3,300,000 - 160,000 = 3,140,000, and the accumuimportant.

lation 1

+

/"i

from

rate

=

I'oOo'ooO

withdrawal

rate

'

^^^

3,250,000

is

fund actually

January

lost

+

to

1

f\Jind

=

the

~^

'^

~

and

is

equal to

rate

in

the

given

is

by

final

3,262,000.

We

see that the

summer, and the accumulation

between March 31 and August 31

accumulation

31

value just before the August 31

12,000

money during

March

is

1

four

+

/2

=

s'sqq'qqq

months

of

-

2000

^^^ is

^^^ dollar-weighted return for the year is ^Sso'oOO then obtained by subtracting from the product of these fractions, ^

"

1

.1

142.

D

Chapter 12

266

Example

12.5

Vivienne's retirement benefit

She

65.

is

assume

life

5%

You

future retirement benefits. to

of her fmal year's

salary, paid in

payment made at age Assuming currently age 20 and earns 20,000 per year.

Vivienne's salary increases at

you are

50%

is

equal monthly installments for

with the

first

per year, find the present value of are given

=

/

.09

—

and ^^5

11,

all

and

Vivienne does not die or leave the company

that

before age 65. Solution

Vivienne's salary on retirement will be 20,000(1.05)'^'^, and her annual pension

10,000(1.05)'^'^

is

value of these payments Its

value

which

at

retirement is

is

(85,571.50)45^^

Example

= 941,286.50. = 19,477.33,

(941, 286. 50)(1. 09)"'^^

calculated at interest only since survival to age 65

is

The present

85,571.50, payable monthly.

present time

the

at

=

is

assumed.

12.6

retirement at age 65, Bruce, now aged 55, will receive a pension of 500 per month, increasing by 20 per month for each completed year of

Upon

retirement for

Dss

=

life.

1400, Des

=

Find the present value of these benefits, given that 550,

^

A^65

4800, ^55

=

34000.

Solution!

500 500500 \

\

55

\

520 520

\

\

66

65

FIGURE The

benefit

is

500 per month

Dss (12)(480)4f

at

520 per month

age 55 (see Exercise 8-48)

+ (12)(20)(/4'5'^

{\2)(4S0)[Nes-^,-De5]^(\2){20)[Ses-^,-Nes] Dss

=

24,162.86.

540 h67

12.3

in the first year,

second and so on. The present value

Dei

\

in the is

267

Pension Applications

Example

12.7

Corinne, aged 60, will retire at age 65. If she should die before reaching retirement age, her estate

of 1000 for each

entitled to a benefit

is

completed year of service, payable

at the

end of the year of death. Find

the present value of this future death benefit if Corinne 45, and if M6o

Mes

=614

=

=

790, Me\

=

and Deo

=

765, M^i

729, Me3

=

was hired 693,

Me4

at

=

age

656,

2500.

Solution

The death

benefit

age 64.

present value

Its

15,000(g^)

_

+

15,000

is

is

at

age 60, 16,000

16,000(gl)4-.--

1000[15M60

+

age 61,

at

19,000

...,

at

equal to

M61

+

+ 19,000(gi) +

M62

M63

+

M64

-

19M65]

^60

=

1210.80.

n It

is

pension plan upon retirement, as the present value of the

all

in the earlier

Example

12.7,

and a

examples. In such a case

future benefits can be calculated

two benefits separately and adding the

Example

would

quite possible that an employee's benefit package

involve both a preretirement death benefit, as in

by working out

results.

12.8

Rosalind's defined benefit pension

monthly installments for

life

is

50%

of her

final salary,

payable

in

beginning when she turns 65, offset by the

annuity provided by the participant's defined contribution plan account balance.

10% of salary

the end of each year.

is

contributed to the defined contribution plan at

Salary increases are

contribution plan earns

8%

30,000 per year and has accumulated

6%

per year, and the defined

At age

per year. 1

45, Rosalind

plan account, just after the last contribution

is

made.

projected defined monthly benefit, assuming she will

employed

at the

time of retirement.

is

earning

1,000 in the defined contribution

Assume

d^^

=

9.

Find Rosalind's still

be alive and

Chapter 12

268

Solution

The defined

contribution plan balance in 20 years will be of amount 19

11,000(1.08)20+ ^(.l)(30,000)(1.06y(1.08) /=0

=

11,000(1.08)20

4-

19-/

m

(3,000X1.08)^9

20

269,343.78.

1.06

1.08

:.(12)

we see that the defined contribution plan Dividing the above by ^^5 will provide an annual pension of 29,927.09, payable in monthly ,

installments.

defined

The

benefit

salary

final

pension

is

contribution value, giving benefit

is

Example

is

30,000(1. 06)^^

90,767.99, so her

minus

annually

45,384

=

the

defined

Hence her projected monthly

15,456.91.

D

1,288.08.

12.9

A pension plan offers three equivalent options. pant 1000 per month for

life.

Option

II

Option

I

pays the partici-

pays the participant 800 per

month for life, and pays the participant's spouse a reversionary annuity of 600 per month for life. Option III pays the participant K per month for

life,

month I

and pays the participant's spouse a reversionary annuity of /C per

for

Find K.

life.

Solution!

If

jc

is

the age of the participant and

=

have 10004^^^

8004'^^

y

+ 6004[J\

is

the age of the spouse, then

so that 4^^^

= ^(4^^Uai|;>), SO (1000 us 3(1000 -K) = K,so4K= 3000 and K =

,(12)

3aAy

10004'^^

Likewise

'

(12)

Ka'x\y

we

This gives

D

750

Example 12.10 Agatha's retirement plan pays 200 per month during the joint lifetime of

Agatha and her husband, plus 300 per month during the lifetime of the survivor following the first death. The first payment comes at the time of retirement.

If

Agatha

retires at

age 65 and

is

4 years older than her

husband, find the present value of these benefits on the date of ment.

You

are given

.•(12)

1

2a'

61

15.

245^^= 105 and

12a^'^^ 61.-65

retire-

130.

269

Pension Applications

Solution

The present value 24004->3

is

+ 3600a at

6%

times

Answers

278

26.

^2-