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English Pages 313 [324] Year 1999
ISBN 1-56698-333-9
P heory of Interest and Life
Contingencies,
with Pension Applications:
A Problem-Solving Approach Third Edition
\
Michael M.Parnnenter
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2012
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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-