Interest & Life Contingencies: Problem-Solving Approach

ISBN 1-56698-333-9
P heory of Interest and
Life
Contingencies,
with Pension Applications:
A Problem-Solving Approach
Third Edition
\
Michael M.Parnnenter
Digitized by the Internet Archive
in
2012
http://archive.org/details/theoryofinterestOOparni
THEORY OF INTEREST
AND LIFE CONTINGENCIES
WITH PENSION APPLICATIONS
A Problem-Solving Approach
Third Edition
Michael M. Parmenter
ASA, Ph.D.
ACTEX
Publications
Winsted, Connecticut
To my Mother and
the
memory of my
Father
Copyright© 1988, 1999
by ACTEX Publications
No
portion of this
means without
book may be reproduced
in
any form or by any
the prior written permission of the copyright owner.
Requests for permission should be addressed to
ACTEX Publications
P.O.
Box 974
CT 06098
Winsted,
Manufactured
in the
United States of America
1098765432
Cover design by
MUF
Library of Congress Cataloging-in-Publication Data
Parmenter, Michael
M.
Theory of interest and
life
contingencies, with pension
applications: a problem-solving approach
/
Michael M. Parmenter.
cm.
p.
Includes index.
ISBN 1-56698-333-9
1.
2.
Insurance, Life~Mathematics~Problems, exercises, etc.
Interest-Problems, exercises,
exercises, etc.
HG8781.P29
I.
ets.
3.
Annuities-Problems,
Title
1999
368.2'2'011076-dc 19
ISBN: 1-56698-333-9
88-38947
TABLE OF CONTENTS
CHAPTER ONE
THEORY
INTEREST: THE BASIC
1.1
Accumulation Function
1
.2
Simple Interest
1
.3
Compound
1
.4
Present Value and Discount
4
6
Interest
1.5
Nominal Rate of Interest
1.6
Force of Interest
Exercises
1
1
9
13
16
22
CHAPTER TWO
INTEREST: BASIC APPLICATIONS 29
2.1
Equation of Value
2.2
Unknown Rate of Interest
2.3
Time-Weighted Rate of Return
Exercises
29
33
34
36
CHAPTER THREE
ANNUITIES
41
3.1
Arithmetic and Geometric Sequences
3.2
Basic Results
3.3
Perpetuities
3.4
Unknown Time and Unknown Rate of Interest
3.5
Continuous Annuities
3.6
Varying Annuities
Exercises
41
44
52
66
58
57
53
Table of Contents
iv
CHAPTER FOUR
AMORTIZATION AND SINKING FUNDS
4.1
Amortization
4.2
Amortization Schedules
4.3
Sinking Funds
4.4
Yield Rates
75
75
77
81
83
Exercises 85
CHAPTER FIVE
BONDS
93
Bond 93
5.1
Price of a
5.2
5.3
Book Value 96
Bond Amortization Schedules 98
5.4
Other Topics
Exercises
101
105
CHAPTER SIX
PREPARA TION FOR LIFE CONTINGENCIES
6.1
6.2
Probability and Expectation
6.3
Contingent Payments
Exercises
1 1
113
Introduction
1
14
119
123
CHAPTER SEVEN
LIFE TABLES AND POPULATION PROBLEMS 127
127
7.1
Introduction
7.2
Life Tables
7.3
7.4
Analytic Formulae for ^x 133
The Stationary Population 137
7.5
Expectation of Life
7.6
Multiple Decrements
Exercises
128
141
143
148
CHAPTER EIGHT
LIFE ANNUITIES
155
8.1
Basic Concepts
8.2
Commutation Functions
8.3
Annuities Payable m^^^y
8.4
Varying Life Annuities 173
Annual Premiums and Reserves
8.5
Exercises
1
8
155
161
1
66
178
Table of Contents
CHAPTER NEVE
LIFE INSURANCE 189
9.1
Basic Concepts
9.2
Commutation Functions and Basic
9.3
Insurance Payable at the
9.4
Varying Insurance
9.5
Annual Premiums and Reserves
Exercises 210
189
199
202
CHAPTER TEN
STATISTICAL CONSIDERATIONS 223
10.1
Mean Variance 223
10.2
Normal
10.3
Central Limit
10.4
Loss-at-Issue 235
Distribution
230
Theorem 233
Exercises 239
CHAPTER ELEVEN
MULTI-LIFE THEORY 243
11.1
Joint-Life Actuarial Functions 243
.2
Last-Survivor Problems
25
11.3
Reversionary Annuities
254
1 1
Exercises 256
CHAPTER TWELVE
PENSION APPLICATIONS
Exercises
263
269
ANSWERS TO THE EXERCISES
INDEX
297
Identities
Moment of Death
21
192
196
PREFACE
It is
in
impossible to escape the practical implications of compound interest
our
modem
The consumer
society.
is
faced with a bewildering choice
interest, and wishes to choose
on her savings. A home-buyer is
offered various mortgage plans by different companies, and wishes to
An investor seeks to
choose the one most advantageous to him.
purchase a bond which pays coupons on a regular basis and is
redeemable at some future date; again, there are a wide variety of
of bank accounts offering various rates of
the one
which
will give the best return
choices available.
Comparing possibilities becomes even more difficult when the
payments involved are dependent on the individual's survival. For
example, an employee is offered a variety of different pension plans and
must decide which one to choose. Also, most people purchase life
insurance at
some point
in their lives,
and a bewildering number of
different plans are offered.
The informed consumer must be able
in situations like
whenever
make an
those described above. In addition,
possible, she be able to
make
intelligent choice
it is
important
of mortgage payments
will, in fact,
that,
the appropriate calculations
For example, she should understand
herself in such cases.
series
to
why
a given
pay off a certain loan over a
certain period of time.
She should also be able to decide which portion
of a given payment
paying off the balance of the loan, and which
portion
is
simply paying interest on the outstanding loan balance.
is
The
first
goal of this text
is
to give the reader
enough information
make an intelligent choice between options in a financial
and can verify that bank balances, loan payments, bond
so that he can
situation,
coupons,
how
etc. are correct.
Too few people
in today's society
understand
these calculations are carried out.
In addition,
however,
we
are concerned that the student, besides
why they work. It
how to apply it; you
being able to carry out these calculations, understands
is
not enough to memorize a formula and learn
should understand
why
the formula
is
correct.
We
also wish to present
Preface
viii
the material in a proper mathematical setting, so the student will see
the theory of interest
Let
appears
me
is
explain
in the title
of
why
the phrase "Problem-Solving
We
this text.
will prove a very small
formulae can be applied to a wide variety of problems.
needed to take the data presented
many
it
texts,
large
these
Skill will
a particular problem and see
in
so the formulae can be used.
where a
Approach"
number of
how
formulae and then concentrate our attention on showing
rearrange
how
interrelated with other branches of mathematics.
how
be
to
This approach differs from
number of formulae
and the
are presented,
student tries to memorize which problems can be solved by direct
of
application
a
particular
We
formula.
wish
emphasize
to
understanding, not rote memorization.
A
working knowledge of elementary calculus is essential for a
all the material. However, a large portion of
this book can be read by those without such a background by omitting
the sections dependent on calculus. Other required background material
such as geometric sequences, probability and expectation, is reviewed
thorough understanding of
when
it is
required.
Each chapter
exercises.
It
should be obvious that
to learn the material
number of examples and
in this text includes a large
is
for her to
Finally, let us stress that
the most
work
it
calculator (with a y^ button) and
is
all
way
for a student
the exercises.
assumed
knows how
our ability to use a calculator that
efficient
that every student has a
to use
it.
It is
because of
many formulae mentioned
in older
on the subject are now unnecessary.
This book is naturally divided into two parts. Chapters 1-5 are
concerned solely with the Theory of Interest, and Life Contingencies is
texts
introduced in Chapters 6-11.
In
Chapter
1
we
present the basic theory concerning the study of
is to give a mathematical background for this
and to develop the basic formulae which will be needed in the rest
of the book. Students with a weak calculus background may wish to
omit Section 1 .6 on the force of interest, as it is of more theoretical than
practical importance. In Chapter 2 we show how the theory in Chapter 1
can be applied to practical problems. The important concept of equation
interest.
Our goal here
area,
of value
is
introduced,
problems are presented.
concept of annuities.
emphasis
in this
and many worked examples of numerical
Chapter 3 discusses the extremely important
After developing a few basic formulae, our main
chapter
is
on practical problems, seeing
such problems can be substituted
in the basic
formulae.
how
It
data for
is
in this
Preface
ix
we have
section especially that
left
out
many of the formulae
presented
in other texts, preferring to concentrate on problem-solving techniques
rather than
rote memorization.
applications
of
material
the
Chapters 4 and 5 deal with further
in
Chapters
through
1
namely
3,
amortization, sinking funds and bonds.
Chapter 6 begins with a review of the important concepts of
probability and expectation, and then illustrates
combined with the theory of
tables, discussing
Chapter 8
how
interest.
life
how
probability can be
Chapter 7
they are constructed and
concerned with
is
In
how
annuities, that
we
introduce
life
they can be applied.
is
annuities
whose
payment are conditional on survival, and Chapter 9 discusses life
insurance. These ideas are generalized to multi-life situations in Chapter
10.
Finally, Chapter
1 1
demonstrates
how many of these
concepts are
applied in the extremely important area of pension plans.
Chapters
1
through 6 have been used for several years as the text
material for a one semester undergraduate course in the Theory of
Interest,
and
I
would
in earlier drafts.
Wanda Heath
who
thank those students
I
am
pointed out errors
deeply indebted to Brenda Crewe and
job of typing the manuscript, and to my
Narayanaswami, for his invaluable technical
for an excellent
Dr.
colleague.
like to
In addition,
P. P.
assistance.
Chuck Vinsonhaler, University of Connecticut, was
supportive of this project, and introduced
Publications, for
which
I
owe him
me
to the people at
strongly
ACTEX
Dick London did the
a great deal.
technical content editing, Marilyn Baleshiski provided the electronic
typesetting,
like
to
and Marlene Lundbeck designed the
thank them for taking such care
manuscript into what
I
hope
is
in
text's cover.
I
would
turning a very rough
a reasonably comprehensive yet friendly
and readable text book for actuarial students.
St.
John's,
Newfoundland
December, 1988
Michael M. Parmenter
PREFACE TO THE
REVISED EDITION
months since the original edition of this text was published,
number of comments have been received from teachers and students
In the fifteen
a
regarding that edition.
We
most of the comments have been quite
and we are making no substantial modifica-
are pleased to note that
complimentary to the
text,
tions at this time.
A
edition
given
significant,
is
in
rectified
that time
and thoroughly
justified, criticism
diagrams were not used to
of the original
illustrate the
examples
the second half of the text, and that deficiency has been
by the inclusion of thirty-five additional figures
in the
Revised
Edition.
Thus
it
is
fair to
say that there are no
new
topics contained in the
Revised Edition, but rather that the pedagogy has been strengthened.
For
this
reason
we
prefer to call the
new
printing a Revised Edition,
Second Edition.
In addition we have corrected the errata in the original edition.
would like to thank all those who took the time to bring the various
rather than a
We
errata to our attention.
February, 1990
M.M.P.
PREFACE TO THE
THIRD EDITION
It
now more
is
textbook.
than ten years since the original publication of this
In that time,
several very significant developments have
occurred to suggest that a
new
edition of the text
is
now
needed, and
those developments are reflected in the modifications and additions
made
in this
Third Edition.
First,
improvements
calculator
in
approaches to reach numerical
now
technology
In particular,
results.
us
give
many
better
calculators
include iteration algorithms to permit direct calculation of unknown
annuity interest rates and bond yield rates.
Accordingly, the older
approximate methods using interpolation have been deleted from the
text.
Second, with the discontinued publication of the classic textbook
Life Contingencies by C.W. Jordan, our text has become the only one
published in North America which provides the traditional presentation
of contingency theory. To serve the needs of those
traditional approach, including the use
deterministic
life
table model,
who
still
prefer this
of commutation functions and a
we have chosen
to include various topics
contained in Jordan's text but not contained in our earlier editions.
These include insurances payable
life
contingent
accumulation
at the
moment of death
functions
uniform seniority concept for use with
(Section
8.2),
(Section 9.3),
the
table
of
Makeham and Gompertz annuity
values (Section 11.1), simple contingent insurance functions (Section
11.1),
and an expansion of the material regarding multiple-decrement
theory (Section 7.6).
Third, actuaries today are interested in various concepts of finance
beyond those included
introduced
the
ideas
in traditional interest theory.
of real
modified duration, and so on,
rates
in this
of return,
Third Edition.
To
that
end
we have
investment duration,
Preface
12
new
edition provides a gentle introduction to the more
view
of contingency theory, in the completely new
modem stochastic
Chapter 10, to supplement the traditional presentation.
In connection with the expansion of topics, the new edition
Fourth, the
As
contains over forty additional exercises and examples.
numerical answers to the exercises have been
errata in the previous edition
thank everyone
who
made more
have been corrected.
We
well, the
precise and the
would
like to
brought such errata to our attention.
With the considerable modifications made in the new edition, we
is now appropriate for two major audiences: pension
actuaries, who wish to understand the use of commutation functions and
deterministic contingency theory in pension mathematics, and university
believe this text
students,
who
seek to understand basic contingency theory at an intro-
ductory level before undertaking a study of the more mathematically
sophisticated stochastic contingency theory.
As with
the original edition of this text, the staff at
Publications has been invaluable in the development of this
Specifically
I
would
like
to
new
ACTEX
edition.
thank Denise Rosengrant for her text
composition and typesetting work, and Dick London, FSA, for his
technical content editing.
February, 1999
M.M.P
CHAPTER ONE
INTEREST: THE BASIC THEORY
ACCUMULATION FUNCTION
1.1
The simplest of all financial transactions is one in which an amount of
money is invested for a period of time. The amount of money initially
invested is called the principal and the amount it has grow^n to after the
time period
This
accumulated value
is
called the
is
a situation which can easily be described by functional
at that time.
which the principal has been
invested, then the amount of money at that time will be denoted by A{t).
This is called the amount function.
For the moment we will only
consider values ^ > 0, and we will assume that / is measured in years.
We remark that the initial value ^(0) is just the principal itself.
In order to compare various possible amount functions, it is
convenient mathematically to define the accumulation function from the
notation.
If
the length of time for
is
/
amount function
as a{t)
=
^rj^
just a constant multiple of a(t)
is
We
.
,
note that a{0)
namely A(t)
=
k
•
=
1
a{t)
and that A(t)
where k
is
= A{0)
the principal.
What
any function
functions are possible accumulation functions?
a(t)
=
which money
we would
increasing. Should a(t) be continuous? That depends
if a(t) represents the amount owing on a loan t years
with a{0)
1
could represent the
accumulates with the passage of time.
hope
that a(t)
on the
after
is
situation;
it
way
In theory,
has been taken out, then a(t)
in
Certainly, however,
may
be continuous
continues to accumulate for non-integer values of
t.
if interest
However,
if a{t)
amount of money in your bank account t years after the
deposit (assuming no deposits or withdrawals in the meantime),
represents the
initial
jump
The graph of such an a{t)
We will normally assume in this text that a{t) is
make allowances for other situations when they
then a{t) will stay constant for periods of time, but will take a
whenever
interest
is
paid into the account.
will be a step function.
continuous;
turn up.
it
is
easy to
Chapter
In Figure
1
.
1
we have drawn
accumulation functions which occur
1
graphs of three different types of
in practice:
a{t)
a{t)
a{t)
(0,1)
(0,1)
(0,1)
(b)
{a)
(c)
FIGURE
Graph
of
where the amount of
(a) represents the case
constant over each year.
interest
earned
On
we would hope
earned also increases;
where
to be in a situation
earned
is
amount
This makes more
increasing as the years go on.
is
interest
the other hand, in cases like (b), the
sense in most situations, since
larger, the interest
1.1
in
that as the principal gets
other words,
"interest earns interest".
we would like
many
There are
which look roughly like the graph in
the one which will be of greatest interest
different accumulation functions
(b),
but the exponential curve
is
to us.
We
interest
money
is
is
remarked
withdrawn between these time periods.
interest paid
the
same
earlier that a situation like (c)
If the
is
height.
However,
all
amount of interest increases
then we would expect the steps
if
paid
the
be of
as the
to get
and larger as time goes on.
have used the term interest several times now, so perhaps
We
time to define
if
amount of
constant per time period, then the "steps" will
is
accumulated value increases,
larger
can arise whenever
paid out at fixed periods of time, but no interest
it is
it!
Interest
This definition
is
= Accumulated
Value
not very helpful
—
Principal
in practical situations,
since
we
are generally interested in comparing different financial situations to
most profitable. What we require is a standardized
and we do this by defining the effective rate of
interest i (per year) to be the interest earned on a principal of amount 1
over a period of one year. That is,
determine which
measure for
is
interest,
i
=
a{\)-\.
(1.1)
The Basic Theory
Interest:
We
a{t), if
3
can easily calculate
we
recall that A(t)
=
using the amount function A{t) instead of
/
k
Thus
a(t).
Verbally, the effective rate of interest per year
earned
interest
one year divided by the principal
in
the year.
There
definition.
We
is
the
at the
amount of
beginning of
nothing sacred about the term "year" in this
is
can calculate an effective rate of interest over any time
period by simply taking the numerator of the above fraction as being the
earned over that period.
interest
More
we
generally,
define the effective rate of interest in the n^^
year by
- a{n-\)
a{n-\)
_ A{n)-A{n-\) _
~
A{n-\)
a{n)
^"~
Note
that
i\,
calculated by (1.3),
is
same
the
as
/
,,
^.
^^'^^
'
defined by either (1.1)
or (1.2).
Example
1.1
Consider the function
a{t)
=
(a)
Verify that
(b)
Show that a(t)
(c)
Is
(d)
Find the effective
(e)
Find
a{t)
(3(0)
is
=
/^
+ +
r
1.
1.
increasing for
all
/
>
0.
continuous?
rate
of interest
+
=
/
for a{t).
/„.
[Solution]
(a)
(b)
(c)
a(0)
-
(0)2 -f (0)
1
1.
Note that a'(/) = 2t-\-\ > for alU > 0, so a{t) is increasing.
The easiest way to solve this is to observe that the graph of a{t)
a parabola, and hence a{t)
is
that all polynomial functions are continuous).
(d)
i
.
= a{\)- 1=3-1=2.
^ a{n)-a{n-\) ^ n" ^
a{n-\)
«^
-
/7
+
1
is
continuous (or recall from calculus
n
+
-[{n-\f + {n-\)
(n-\f + (n-\)-\-\
\
-^ \}
Chapter
SIMPLE INTEREST
1.2
There are two special cases of the accumulation function
will
1
examine
The
closely.
sionally, primarily
first
of these, simple
between integer
second of these, compound
interest,
it
by
is
accumulation function and will be discussed
mind
some
that in both
of these cases
interest, is
interest periods, but will
mainly for historical purposes and because
is
a{t) that
used occa-
be discussed
easy to describe.
the
far
in the
a{t) is continuous,
we
The
most important
next section.
Keep
in
and also that there are
where modifications must be made.
Simple interest is the case where the graph oi a(t) is a straight line.
Since a(0) =1, the equation must therefore be of the general form
However, the effective rate of interest / is
a{t) = \ -\- bt for some b.
practical settings
given by
i
—
a{\)
—
1
=
Z?,
so the formula
•^f)=rt-f/r,
is
r>0.
(1.4)
\-\-it
(0,1)
[FIGURE
1.21
case (a) graphed in Figure
1.1.
Remarks
1.
This
is
of interest earned each year
is
In this situation, the
constant.
original principal earns interest
amount
In other words, only the
from year to year, and
interest
accumulated in any given year does not earn interest in future
years.
2.
The formula
^(0)
=
a(0)
equal to
k,
=
a{t)
=
the
1
.
\
+
it
More
amount
at
applies to the case
where the principal
generally, if the principal at time
time
/
will be A(t)
:=z
k(\
-\- it).
is
is
Interest:
The Basic Theory
We
rate
noted above that the "/"
of interest for this
+
\
in
=
in a{t)
\
-\-
-
+
\\
also the effective
it is
function. Note however
that
i{n-\)\
\+i(ri-\)
+
Observe
that
not constant.
is
z„
(1.5)
/(«-l)
In fact,
decreases as n gets
/„
which should not surprise us. If the amount of
accumulated value increases, then
clearly the effective rate of interest is going down.
Clearly a{t) — 1 + /Y is a formula which works equally well for all
values of r, integral or otherwise. However, problems can develop
in practice, as illustrated by the following example.
larger, a fact
interest stays constant as the
Example
1.2
Assume Jack borrows 1000 from
15%
the bank on January
How much
simple interest per year.
does he
1996
1,
at
a rate of
owe on January
17,
1996?
Solution
The general formula
A{t)
=
amount owing
for the
1000(1 +.15/), but the problem
should be substituted into this formula.
is
An
at
time
general
in
t
is
what value of
to decide
obvious approach
is
t
to take
number of days which have passed since the loan was taken out and
number of days in the year, but should we count the
number of days as 16 or 17? Getting really picky, should we worry
about the time of day when the loan was taken out, or the time of day
the
divide by the
when we wish
to find the value
of the loan? Obviously, any value of/
only a convenient approximation; the important thing
consistent rule to be used in practice.
(a)
The
first
method
is
first.
owes 1000
+
In our case this
<")(3i)l
it is
it
we
use
,,
^.
(^•^)
•
365
is
have a
common:
number of days
counting the number of days
but not the
are
to
called exact simple interest, and with
t=
When
Two techniques
is
usual to count the last day,
would lead
1006.58.
Xo
t
=
^^P^ so Jack
Chapter
(b)
The second method is
Banker 's Rule), and with
1
called ordinary simple interest (or the
it
we
use
number of days
(1.7)
360
The same procedure
used for calculating the number of
= J^
360
/
1+(.15)(3^) -
1000
practice in
ordinary simple interest
is
we would have
days. In our case,
The common
as above
Canada
is
used
is
so the debt
is
1006.67.
to use exact simple interest,
in the
United States and
whereas
in international
D
markets.
1.3
COMPOUND INTEREST
The most important special case of the accumulation function a{t) is the
case of compound interest. Intuitively speaking, this is the situation
where money earns interest at a fixed effective rate; in this setting, the
one year earns interest itself in future years.
If / is the effective rate of interest, we know that a( 1 ) = 1 + /, so 1
becomes 1 + / after the first year. What happens in the second year?
Consider the 1 -h /as consisting of two parts, the initial principal 1 and
interest
earned
the interest
/
in
earned
in the first year.
The
the second year and will accumulate to
earn interest in the second year and will
amount
after
reasoning,
two years
we
is
1
principal
+
1
grow
+ + /(H- /) =
/
see that the formula for a(t)
«(/)
=
will earn interest in
+
+
z)^.
interest
will also
/)•
Hence the
By
continuing this
(l+/y. ^>o.
(h 1+0
[FIGURE
/
total
is
a(t)
(0,1)
to /(I
(1
1
The
/.
L3I
= (\+iy
(1.8)
The Basic Theory
Interest:
Remarks
1.
This
is
graph
2.
an example of the type of function shown
in
The formula
is
=
A{{))
equal to
3.
Figure
k,
Observe
More
a{t)
=
a{0)
the
=
(1 -f ij applies to the
More
1.
amount
time
at
case where the principal
generally, if the principal at time
t
will be A(t)
that the "/" in (1 -f ty
is
case
in this
same
the
is
/„
shouldn't be surprised, since this
pound
=
k(\
+
is
iy.
the effective rate of interest.
substituted into a(t)
it
was
fifty
—
(1
-\-
ty
.
we
years ago;
t,
We
for all positive integers n.
fits
with our idea
of interest
interest, the effective rate
Mathematically, any value of
than
of the
generally,
Hence,
4.
in part (b)
1.1.
is
constant.
whether integral or
This
is
com-
that, in
can be
not,
an easier task for us today
just have to press the appropriate
buttons on our calculators! Again, there are problems determining
what value of t should be used, but we can deal with them as we
did in the last section.
In practical situations,
sometimes used
however, a very different solution
case of compound interest. To find the
example) when Ms a fraction, first find the
integral values of / immediately before and
in the
amount of a loan
amounts for the
(for
immediately after the fractional value
interpolation
is
in question.
between the two computed amounts
Then use
linear
to calculate the
required answer.
This
is
equivalent to saying that
for integral values
values.
of r, and simple
In Figure
1.4,
whereas the dotted
interpolation
is
compound
interest
is
interest
the solid line represents a{t)
lines
indicate
used.
a{t)
FIGURE
1.4
the
is
used
used between integral
graph of
—
a{t)
{\
if
+
//,
linear
Chapter
As we
common
will see later, this
procedure benefits the lender
a financial transaction, and (consequently)
if
is
1
in
detrimental to the borrower
she has to repay the loan at a duration between integral values.
Example
1.3
Jack borrows 1000
15% compound
at
interest.
How much does he owe after 2 years?
How much does he owe after 57
(a)
(b)
between
interest
How much
(c)
How much
(d)
In
assuming compound
after
year and 57 days, under the same
1
in part (b)?
does he owe after
between
interpolation
(e)
owe
does he
assumption as
days,
integral durations?
how many years
1
year and 57 days, assuming linear
integral durations?
have accumulated to 2000?
will his principal
[Solution]
(a)
1000(1.15)2
(b)
The most
=
suitable value for
1000(1. 15)fi
1000(1. 15)^S
(c)
We
(d)
S7
/ is
^rzrc,
values
which
^
=
1000(1.15)2
is
1322.50.
^(2)— ^4(1)=
will
and the accumulated value
is
= 1022.07.
= 1175.38
=
must interpolate between ^(1)
A{2)=
f
1322.50.
accumulate
j(172.50)
=
172.50.
=
1150.00 and
57 days, assuming simple
in
26.94.
(1000)(1.15)
The difference between these
The portion of this difference
Thus the accumulated value
interest,
after
1
is
year
Observe that the
and 57 days is 1150.00 + 26.94= 1176.94.
borrower owes more money in this case than he does in part (c).
We
(e)
seek
/
such that 1000(1.15)^
=
2000, or that (1.15)^
=
2.
Using logs we obtain
'
To
close this section,
=
we
I^U3
will
=
D
'^9595 years.
compare simple
interest
and compound
graphs for
which gives the better return.
both simple interest and compound interest are drawn on the same set of
interest to see
axes.
In Figure 1.5,
The Basic Theory
Interest:
FIGURE
We know
1.5
+ ij is always concave
+ iy[ln{\ + /)]^, which is greater
that the exponential function (1
up (because the second derivative
than zero), whereas
+
1
// is
is (1
a straight line.
These
facts tell us that the
only points of intersection of these graphs are the obvious ones, namely
(0,1)
and
(1,
+
1
They
/).
also give us the
(l+/y<l+/Y,
two important
for
relationships
0</<l
(1.10)
/>1.
(1.11)
and
+ /y>l+//,
(l
Hence we conclude
first
that
compound
interest yields a higher return than
whereas simple interest yields more if
of these statements does not surprise us, since for
simple interest
The
>
for
if /
1,
r
< r < 1.
> 1, we
have interest as well as principal earning interest in the (1 + ij case.
The second statement reminds us that, for periods of less than a year,
simple interest is more beneficial to the lender than compound interest, a
fact
which was
1.4
illustrated in
Example
PRESENT VALUE AND DISCOUNT
In Section
1
.
1
we
defined accumulated value at time
the principal accumulates to over
value
1.3.
t
/
years.
years in the past as the amount of
the principal over
of that which
/
we have been
PRESENT
VALUE
money
In other words, this
years.
t
We now
as the
amount
that will accumulate to
is
the reverse procedure
discussing up to now.
ACCUMULATED
VALUE
PRINCIPAL
[FIGURE
1.6]
that
define the present
Chapter
10
For example,
much money
+
1
/
Ho\f
over one
over a single year.
needed, at the present time, to accumulate to
1
We will denote this amount by v| and, recalling that v accumulates
Therefore
+ we have v( + /) =
year?
to v(
is
accumulates to
1
1
i),
1
1
1
.
(1.12)
These two accumulations are shown
Figure
in
1.7.
1
1+/
+/
1
-1
1
FIGURE
From now
that
we
are in a
on, unless explicitly stated otherwise,
compound
this case, the present value
We
summarize
(l+zy
1.7
this
where
interest situation,
of
1, r
years
in the past, will
on the time diagram shown
TT7
t
^"
in
^^
Figure
t
we
a{t)
will
=
be
assume
+
=
v^
In
ij.
(1
^^
.
1.8.
AMOUNT
'^'
TIME
FIGURE
Observe
expresses
Hence
(1
or future.
all
+
that,
since
v^
=
(1
+
1.8
i)~\
the
function a(t)
=
(1
+
these values, for both positive and negative values of
ij gives the value of one unit (at time 0) at
The graph
is
shown
in
Figure
1
.9.
<m
(1
IFIGURE
1.9l
+
0'
any time
/,
//
t.
past
The Basic Theory
Interest:
Example
1
1.4
The Kelly family buys a new house for 93,500 on May 1, 1996. How
much was this house worth on May 1, 1992 if real estate prices have
risen at a compound rate of 8% per year during that period?
Solution
We
seek the present value, at time
93,500(^y^)
interest is
is
-4, of 93,500
at
time
0.
This
of present values
calculation
the
to
now
=
a{t)
is
1
+
is
D
assumed instead of compound
years in the past
value
=
=68,725.29.
What happens
function
t
simple
The accumulation
interest?
Hence, the present value of one unit
it.
given by
if
x,
where x{\
+ //) =
1.
/
Thus the present
is
The time diagram
1
^
+
for this case
is
shown
in
Figure 1.10.
1
AMOUNT
^'
^'^
^Y
-^-10
^
I
I
you are asked
V^
TIME
t
1
FIGURE
In Exercise 15,
^
1.10
to sketch the graph
of this
situation.
Unlike the compound interest case, this graph changes dramatically as
passes through the point
it
(0, 1).
We now turn our attention to the concept of discount. For the
moment we will not assume compound interest, since any accumulation
function will be satisfactory.
Imagine that
accumulated to 112.
figure,"
100
is
We
invested,
and that one year
later
it
has
have been viewing the 100 as the "starting
and have imagined that
interest
of 12
is
added to
it
at the
end of
However, we could also view 112 as the basic figure, and
is deducted from that value at the start of the year. From
the latter point of view, the 12 is considered an amount oi discount.
Students sometimes get confused about the difference between
interest and discount, but the important thing to remember is that the
the year.
imagine that 12
only difference
is in
the point of view, not in the underlying financial
Chapter
12
transaction.
we have
In both situations
nothing can change
1
100 accumulating to 112, and
that.
Since discount focuses on the total at the end of the year,
it
is
natural to define the effective rate of discount, d, as
^=^^.
In other words, standardization
of a(0), as was done
More
is
(1.14)
achieved by dividing by a{\) instead
define the effective rate of interest
in (1 .2) to
/.
generally, the effective rate of discount in the n^^ year
is
given by
(Compare
this
with the definition of /„, given by (1.3).)
Now we
will derive
some basic
identity follows immediately
'^~
Since
1
+ >
/
^(1)-1 _
-
a{\)
this tells us that
1,
identities relating
this
identity
The
exact analogy of
1
is
d<
we
accumulating to
1
+
/
(11^)
-rt7--=^
we would
to
1
expect from the
over one year
/,
we
(118)
will be asked to derive other identities in the exercises
and to give verbal arguments
will be
in
support of them.
We
note that
any accumulation function. For the
assumed that a{t) = (1 + ij
identities derived so far hold for
it
the
obtain
i=\^^
of this section,
is
over the same period.
Solving either of the above identities for
The reader
j^.
obtain
— J accumulates
1
.J
i.
exactly what
fact that
One
/.
/
i-^=i-Ti/ =
definition of J.
to
(1+0-1 __
1+/ -.Uk;ti
Immediately from the above
Actually,
d
from the definition oi d, namely,
all
rest
The Basic Theory
Interest:
In Section
years
we learned that to find the accumulated
we multiply by (1 + //, whereas to find the
1.3
in the future
value
years in the past
/
13
(1.17) tells us that
we
—d=
\
.
multiply by
f_
•
.
{
Hence,
if
present and accumulated value are reversed:
by multiplying by
d
is
However,
.
value
/
present
identity
involved, the rules for
present value
is
obtained
—d)', and accumulated value by multiplying by
(1
1
Example
1000
is
1.5
to be
accumulated by January
1995, at a
1,
compound
rate
of
discount of 9% per year.
(a)
Find the present value on January
(b)
Find the value of / corresponding to
1992.
1,
d.
[Solution
753.57.
M
^= ^=0989.
^-^1-3^7
d ~ .91
(b)
Example
compound
August
1,
Solution
Some
D
1.6
Jane deposits 1000
I
=
1000(1 -.09)^
(a)
in a
interest is
bank account on August
7%
1,
1996.
per year, find the value of this deposit on
I
students think that the answer to this question should be 0,
cal sense,
we know
the correct answer
We
of
1994.
because the money hasn't been deposited yet! However,
1.5
If the rate
is
that
money has
1000
value
(y^j^ =
in
a mathemati-
at all times, past or future,
so
D
873.44.
NOMINAL RATE OF INTEREST
will
assume
otherwise, in
Example
all
a(t)
=
(1
+
// throughout this section and, unless stated
remaining sections of the book.
1.7
A man borrows 1000 at an effective
How much does he owe after 3 years?
rate
of interest of
2%
per month.
14
Chapter
1
Solution
I
What we want
is
amount of the debt
the
effective interest rate
is
Thus the answer
periods.
The
given per month, three years
is
1000(1.02)^^
point of the above example
is
=
Since the
after three years.
36
is
interest
D
2039.89.
to illustrate that effective rates
of interest need not be given per year, but can be defined with respect to
any period of time. To apply the formulae developed to this point, we
must be sure
that
is
t
the
number of
effective interest
periods
in
any
particular problem.
many
In
real-life situations, the effective interest
year, but rather
fact hidden, as
you want
12%
some
it
period
might be to his benefit to do
When
per year.
so! For example, suppose
mortgage on a house and you discover a rate of
you dig a little, however, what you find out is that
which means that
the same thing? Not at
it
effective per half-year.
all.
what happens
=
Is that
to an investment
one
After
1.06.
(1.06)^
not a
to take out a
this rate is "convertible semiannually",
to
is
shorter period. Perhaps the lender tries to keep this
of
1
After half a year
.
(two
year
interest
So, over a one-year period, the
1.1236.
it
periods)
is
really
6%
Consider
has accumulated
it
has
amount of
become
interest
which means the effective rate of interest per year is
actually 12.36%. Although it may not be clear from the advertising,
many mortgage loans are convertible semiannually, so the effective rate
gained
.1236,
is
of interest
As
charges
rate
is
higher than the rate quoted.
another example, consider a well-known credit card which
1
8%
per year convertible monthly.
of interest
year,
1
is
R
jy =
1
This means that the actual
.015 effective per month.
will accumulate to (1.015)^^
interest per year is actually
=
Over the course of a
1.1956, so the effective rate of
19.56%.
The 18% in the last example is called a nominal rate of interest,
which means that it is convertible over a period other than one year. In
general,
we
convertible
of
^
use the notation
m
/^'"^
nominal rate of
to denote a
interest
times per year, which implies an effective rate of interest
per m^^ of a year. If
/
is
the effective rate of interest per year,
it
follows that
1+/-
jim)
1+V
ml
jt
(1.19)
The Basic Theory
Interest:
Example
1
5
1.8
Find the accumulated value of
000
1
after three years at a rate
of interest
of 24% per year convertible monthly.
SolutionI
This
2% effective per month, so the answer
namely 1000(1.02)^^ = 2039.89.
really
is
Example
1.7,
is
the
same
as
D
Remark
An
alternative
rate
of
method of solving Example
1
1.8 is to find
and then proceed as
interest per year,
+ -^ j -
1
would be 1000 (1.26824)^
= (l+.02)'2 = 2039.88.
Notice the difference of
.01 in the
1
in
=
Section
Of
in.
course, if
two answers. This
you use the memory
in
/,
time
preferable;
still
We
would
is
because
and some error
your calculator
unlikely that this type of error will occur. Nevertheless, the
is
the effective
.26824, and the answer
not enough decimal places were kept in the value of
crept
/,
1.3.
first
is
it
solution
on unnecessary calculations can be
spent
significant in examination situations.
It
will be extremely important in later sections
of the text to be
able to convert from one nominal rate of interest to another
convertible frequency
Example
If
z^^)
=
is
different.
Here
is
whose
an example of this.
1.9
.15, find the equivalent
nominal
rate
of interest convertible
semiannuallv.
[Solution
We have
(l
+
m
\ 2
V)
^
i}
^
'~^)
,
so
z^^)
^
2[(1.025)^
-
1]
=
.15378.
D
In the
same way
that
we
defined a nominal rate of interest,
we
could also define a nominal rate of discount, S"^\ as meaning an
Am)
effective rate of discount of ^^^^
identity (1.19),
it is
per m^" of a year.
Analogous
to
easy to see that
\^,d
m
(1.20)
Chapter
16
Since
1
—d
1+/ we
,
M
^
1
for all positive integers
Example
conclude that
=
+ ^m
1
\Jri
=
1_^
n
{\-d)-
(1.21)
^
m and n.
1.10
Find the nominal rate of discount convertible semiannually which
12%
equivalent to a nominal rate of interest of
is
per year convertible
monthly.
ISolutionI
12
d^^^
+
1
which we find
is
=
d^^'>
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.
<12)\12
/
=
=
)
1
+ =
/
P^^
=
\2{e^^'^^
-
\)
the result
=
D
.14082.
Remark
we tried to solve part (a) by first obtaining / = .15027 (as in
and then calculating 3000(1. 15027)^ we would get 6041.16, an
answer differing from our first answer by .10. There is nothing wrong
Note
that if
part (b)),
with this second method, except that not enough decimal places were
carried in the value of
/
an earlier admonition:
to guarantee an accurate answer.
Let us repeat
always wise to do as few calculations as
it
is
+
/T]
necessary.
Observe
that D[{\
the derivative with respect to
,
Let us see
=
why this
/.
=
{\
+
In {\
ij
Hence we see
-\- /),
where
happens to be
the derivative that Z)[a(0]
= Um
^(^
h-*Q
true.
-n
ha(,t)
(1.26)
Recall from the definition of
+ ^)-^W
h
^
3^
a{t+h)
a(0
stands for
that
;„(,+0=^^ = ^.
fact
D
-n
h
-
a{t)
^^-^'^
Interest:
The Basic Theory
-^^
The term
-rx
——
19
ajt^h)
over a very small time period
rate
the effective rate of interest
in (1.27) is just
a{t)
^
so
h,
-
corresponding to that effective
is
the nominal annual
which agrees with our
rate,
earlier
definition of ^.
The above analysis does more than
how
that,
however.
also indicates
It
the force of interest should be defined for arbitrary accumulation
functions.
First,
However,
to a constant
/„.
ln{\
-{-
is
i)
of compound
independent of
interest,
For arbitrary accumulation functions,
force of interest at time
.
we would normally
expect
For certain functions,
/.
corresponding
we
define the
by
/, (5/,
^-_
since
—
observe that 6
us
let
this is a special property
6t
^(^)
to
(1-28)
,
depend on
t.
more convenient
is
it
D[a{t)]
to use the equivalent
definition
6,
We also remark that,
=
since A{t)
D{ln{a{t))\
=
k
a(t)
,
it
(1.29)
follows that
6.= ^^=D[lr,(A(t))].
Example
Find
(1.30)
1.12
6t in
the case of simple interest.
ISolutionl
.
^t
—
D(\+it)
it
^it
1
l+/r
I
We now
have a method for finding the force of
any accumulation function
wish to derive a{t) from it?
To
start with, let
different variable,
/,
we
obtain
What
if
we
namely
r.
6r
=
interest, 6t,
are given
us write our definition of
derivative with respect to
to
a(t).
^^
D[ln{a{r)y\, where
6t
given
instead,
and
from (1.29) using a
D
now means
the
Integrating both sides of this equation from
20
Chapter
= [
6rdr
I
Jo
=
D[ln(a(r))]dr
1
ln{a{r))
Jo
= ln(a(t))-ln(am
=
=
since a(0)
and
1
/«
=
1
Then taking
0.
a(/)
Example
some
the antilog
(1.31)
we have
Jo'^^^^
(1.32)
1.13
Prove that
for
=
ln{a{t)\
if
a constant
is
(5
(i.e.,
independent of
then a{t)
r),
=
(1
+
ij
/.
Solution
\i6r
=
result
c,
is
hand side of (1.32)
the right
proved with
Example
=
i
e^
—
is
Jo^^^
z.z
=
e''
(ej. Hence the
D
\.
1.14
Prove that
A{t)
/
dt
6t
— A(n) - A(0) for any amount function A{t).
Jo
I
Solution!
The
left
["Ait)
Jo
hand side
dt
6,
=
is
["aH)
Jo
dt =
[^^1
A'}
L
J
rD[A(t)] dt
Jo
= A(t)
=
A(n)
— A{0)
as required.
D
The
identity
time
t
for the
represents the
in
amount of
number which
=
We now
(1 + /y.
6t
above example has an interesting verbal
dt represents the effective rate of interest at
"period of time"
infinitesimal
represents the total
a{t)
the
The term
interpretation.
is
amount of
interest
clearly equal to A(n)
return to the
It
is
dt.
interest earned in this period,
Hence A{t)6tdt
and
j^ A(t) St dt
earned over the entire period, a
— A(0).
compound
interesting to write
interest case
some of
where we have
the formulae already
For example S
developed as power series expansions.
=
ln(\-\-i)
becomes
6^i
L.
2
+
1
i_
3
_
+
L. 4_
4
(1.33)
Interest:
The Basic Theory
21
Convergence is a concern here, but as long as
the case, the above series does converge.
Another important formula was
- ^+f^ +
/•
Since
all
—
i
f^
—
e^
+
|
/
<
1
1,
1,
which
is
usually
which becomes
---.
(1.34)
terms on the right hand side are positive, this allows us to
conclude immediately that
converges for
Next
all
let
/
>
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
<
interested in
1.
power
series expansions? Well,
we have
already seen that they sometimes allow us to easily conclude facts like
>
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
<5\2
-
-)
(
<5\3
- -y
2!
3!
6'
V.rr?
(1.38)
2\m^
lim
easily see that
J^'"^
=
In other words, there
S.
need to define a force of discount, because
it
will turn out to
is
no
be the same
as the force of interest already defined.
EXERCISES
1.1
Accumulation Function;
1.3
Compound
1-1.
Alphonse has 14,000
(a)
1.2
Assuming
in
an account on January
simple
interest
accumulated value on January
(b)
Assuming compound
(d)
at
1,
interest
accumulated value on January
(c)
Simple Interest;
Interest
1,
8%
1,
per
1995.
year,
find
the
2001.
at
8%
per year,
find
the
2001.
Assuming exact simple interest at 8% per year, find the
accumulated value on March 8, 1995.
Assuming compound interest at 8% per year, but linear
interpolation between integral durations, find the accumula-
ted value on February 17, 1997.
Interest:
1-2.
The Basic Theory
23
Mary has 14,000 in an account on January 1, 1995.
Assuming compound interest at 11% per
(a)
(b)
accumulated value on January 1, 2000.
Assuming ordinary simple interest at 11% per
accumulated value on April
(c)
Assuming compound
7,
find the
year,
year, find the
2000.
\\%
interest at
per year, but linear
interpolation between integral durations, find the accumula-
ted value on April 7, 2000.
1-3.
For the a{t) function given
all
1-4.
Consider the function
(a)
(b)
(c)
1-6.
+
yj\
{P
<
/„+i
/„
for
>
i
0,
t
>
0.
/.
1
0.
\
-\- it
for
1.
Show
<
that a(t)
+
is
Let a{t) be a function such that a(0)
=
=
(1
(a)
Prove that
(b)
Can you conclude
a{t)
(1
+
i)'
sufficiently large.
1
and
that ait)
=
(1
+ iy
constant for
/„ is
iY for all integers
Let A{t) be an amount function.
(a)
prove that
-\-2i)t^,
1
1
define /„
1-7.
=
a(r)
ift
(d)
1.1,
Show that a{0) = and ^(1) = +
Show that a(t) is increasing and continuous for / >
Show that a(t) < + /Y for < / < 1, but a(t) >
t>
1-5.
Example
in
positive integers n.
/
>
for all
all n.
0.
t
>
0?
For every positive integer
n,
= A(n) — A{n—\).
Explain verbally what
/„
represents
- A(0) = h + h +
(b)
Prove that A(n)
(c)
Explain verbally the result
(d)
Is
(a)
In
it
true that a{n)
how many
—
^(0)
=
---¥ h-
in part (b).
i^
+
ii
+
-
-\-
in? Explain.
years will 1000 accumulate to 1400 at
12%
simple interest?
(b)
At what
in
(c)
rate
of simple
interest will
1000 accumulate to 1500
6 years?
Repeat parts
of simple
(a)
and (b) assuming compound
interest.
interest instead
24
1-8.
Chapter
At a
1
certain rate of simple interest, 1000 will accumulate to 1300
after a certain period
at a rate
of time. Find the accumulated value of 500
of simple interest # as great over twice as long a period of
time.
1-9.
Find the accumulated value of 6000 invested for ten years,
compound
1
1%
interest rate
is
7%
per year for the
first
if
the
four years and
per year for the last six.
1-10.
Annual compound interest rates are 13% in 1994, 11% in 1995
and 15% in 1996. Find the effective rate of compound interest per
year which yields an equivalent return over the three-year period.
1-11.
At a
certain rate of compound interest, it is found that 1 grows to 2
X years, 2 grows to 3
years, and 1 grows to 5 in z years.
Prove that 6 grows to 1 in z — x —y years.
my
in
grows to ^ in X periods at compound rate / per period and 1
1
grows \o K'my periods at compound rate 2/ per period, which one
of the following is always true? Prove your answer.
X <2y
(a)
(b)
x = 2y
1-12. If
(c)
(d)
(e)
x>2y
y= vx
y>2x
1.4
Present Value and Discount
1-13.
Henry has an investment of 1000 on January
(a)
(b)
(c)
1-14.
present value on January
1,
at
a
1,
year,
find the
1989.
Assuming compound discount
present value on January
(c)
1998
rate
Mary has 14,000 in an account on January 1, 1995.
Assuming compound interest at 12%) per
(a)
(b)
1,
of discount d = .12.
Find the value of his investment on January 1, 1995.
Find the value of / corresponding to d.
Using your answer to part (b), rework part (a) using / instead
ofd. Do you get the same answer?
compound annual
at
12%
per year, find the
1989.
Explain the relative magnitude of your answers to parts (a)
and
(b).
Interest:
The Basic Theory
25
Sketch a graph oi a{t) with
1-15. (a)
extension to present value in
its
the case of simple interest,
why — it is not
past, when simple
Explain, both mathematically and verbally,
(b)
the correct present value
interest
t
years in the
1
assumed.
is
constant in the case of compound interest.
1-16.
Prove that dn
1-17.
Prove each of the following identities mathematically.
and
(a), (b)
is
give a verbal explanation of
(c),
For parts
how you can
see that
they are correct.
d=iv
(d)
^-}
(b)
d=\-v
(e)
d(\
+ {)=i(\-^)
(f)
iyj\
-d =
i-d=
(c)
1-18.
=
(a)
id
1
dyJX
+/
Four of the following five expressions have the same value (for
/
> 0). Which one is the exception?
ruAi-^'
d'
(^)
(TT^ (^)T^
(c)(/-cOa'
3
J
(d);^-/V
___
,-3
/^\ ;2.
(e)i'cl
1-19.
The interest on L for one year is 216. The equivalent discount on
L for one year is 200. What is Z?
1.5
Nominal Rate of Interest
1-20.
Acme
Trust offers three different savings accounts to an investor.
Account A:
compound
interest
at
12%
per year convertible
quarterly.
Account B:
compound
interest at
1 1
.97% per year convertible
5
times per year.
Account C:
compound discount
at
11.8% per year convertible
10 times per year.
Which account
account
is
most advantageous to the investor?
most advantageous to Acme Trust?
is
Which
26
Chapter
1
16% per year
How much does she owe after 21
takes out a loan of 3000 at a rate of
1-21. Phyllis
convertible 4 times a year.
months?
1-22.
The Bank of Newfoundland
offers a
12% mortgage
convertible
semiannually. Find each of the following:
(a)
(d)
1-23. 100
(a)
(b)
(b)^(^>
/•
The equivalent
grows to 107
The effective
P^
(c)/<^2)
effective rate of interest per month.
in
6 months. Find each of the following:
rate
of interest per half-year.
(d) J(3)
(c)
/
+
^
,(6)
1-24.
Find n such that
1
«
^
""
.
1
1-25. Express J^^^ as a function oi
1-26.
1-27.
Show
Prove that
1-28. (a)
(b)
1.6
that
vM +
/<^)c/(»>
Prove that
^^
>
f"^^
)
=
(
/<8)
+
1
^
P\
^
+
^
(
)
^
-
^
j
\/l
-d.
f^'>d^^\
-
d^"^^
=
^
^
Provethat^-^ =
.
i.
Force of Interest
1-29. Find the equivalent value
(a)
/=.13
(b)
d=A3
(c)
/<^)
(d)
J(5)
= .13
= .13
of (5
in
each of the following cases.
Interest:
The Basic Theory
1-30. In Section 1.3,
Show
1-31.
that
Assume
(a)
+
1
it
was shown
—
//
27
{\-\-iy is
Show
maximized
of interest
that the force
<
that for
is
at
<
/
=
/
+
/y
—
In b].
1, (1
4[/«
/
<
1
+
it.
doubled.
that the effective annual interest rate
more than
is
doubled.
Show
(b)
that the effective annual discount rate
is
less than
doubled.
1-32.
Show
^^ -
.50.
=
+0"^-
that //w
1-33. Finda(/) if^,
.04(1
^(0
—
kd^^
1-34.
Obtain an expression for
1-35.
Using mathematical induction, prove that for
-^(v^'^S)
=
with respect to
v.
n,
1-36. Express v as a
1-37. Express
J as
a
1-38.
Prove that
/<")
1-39.
Prove that
d<
1-40.
Show
1-41.
Show
1-42.
Which
\)\,
where
all
positive integers
denotes derivative
-j-
series
power
series expansion in terms of/.
<
/<^> if
d^"^
n
>
<6 <
=
Jf^i
expansion
in
terms of 6.
m.
/<">
-
<
/
«
>
1.
where
D
is
for all
(Sif,
the derivative with
/.
that 6
is
+ i)(n —
b' c^'
power
that D{6()
respect to
—(\
6/ if
=
^ + ^^^ + ^^^^ +
larger,
i
—
6 or 6
^
—
•••
.
d? Prove your answer.
Chapter
28
1-43. (a)
1
Write a computer program which will take a given value of
/
and output values oi f"^^ for a succession of values of w.
(b)
Extend the program
using b
1-44. Write a
—
in part (a) to also give
you a value
for
(5,
Urn f'^\
computer program which
will take a given value
of ^ and
output the equivalent value of /. (Use the power series expansion.)
CHAPTER TWO
INTEREST: BASIC APPLICATIONS
EQUATION OF VALUE
2.1
In
every
terms,
simplest
its
interest
problem
involves
only
four
the principal originally invested, the accumulated value at
quantities:
the end of the period of investment, the period of investment, and the
rate
of
interest.
others are
Any one of these
In this section
the
four quantities can be calculated
if
the
known.
determination
we
will present a
of principal,
number of examples
accumulated value,
illustrating
and period of
investment; determining the rate of interest will be explored in Sections
2.2 and 2.3.
More complicated
situations involving several "principals"
invested at different times will arise in practice, and we will examine
some of these as well.
The most important tool in dealing with such problems is the time
diagram, which we encountered in chapter one, and the first step in any
solution should be to draw such a diagram. After that, all entries on the
diagram should be "brought" to the same point in time, in order that they
can be compared. Then an equation of value is set up at that point in
time, and a solution is obtained by algebraic means. The student should
carefully study the examples in this section to see
how
these steps are
carried out in practice.
We
remark that before calculators came into general use, the
some of these problems were quite difficult, and
it was necessary to develop a collection of techniques to deal with them.
Interest tables and log tables were in frequent use, and values which did
calculations involved in
not appear in the interest tables were handled by interpolation or other
approximate methods.
in the
For example, the power series expansions given
previous chapter could be used for calculation, since the
first
terms often give a good approximation to the correct answer.
few
We,
however, will use our calculators freely and will generally not need to
employ
the older techniques.
That does not mean that every question
30
Chapter 2
can be solved by pushing the appropriate button, however;
in particular
we
(e.g.,
see
will
interpolation)
necessary to
a
way such
where some approximate method
cases
required to obtain an answer.
is
first
upon
After
all,
your calculator
is
computation. The person with the problem
14%
it
often
such
in
to assist in solving
only an aid to mechanical
still
has to solve
it!
2.
Find the accumulated value of 500 after 173 months
of
linear
is
it
analyze the data very careftilly, and organize
that the calculator can then be called
the problem.
Example
In addition
convertible quarterly, assuming
compound
at a rate
of interest
interest throughout.
Solution
The
effective rate of interest
total
of 57| periods. Hence the answer
is
.035 per 3
is
month
period, and there are a
500(1.035)^^^^^
=
3635.22.
D
Remarks
1.
It
is
quite
common
assume compound interest over integral
between integral durations. Under
the answer to this example would be
to
durations, but simple interest
that
assumption,
500(1.035)^^11
answer
is
=
+(.035)('f)
larger than the
one
Observe
3635.69.
in the
that
example, agreeing with our
earlier observation that simple interest gives a higher return
the period
In
is
when
less than a year.
pre-calculator days the calculation of 500(1.035)'^^^^
require
this
some work.
Log
would
tables, if available, could give the
an-
only interest tables were available, you might
have to write the product as 500(1. 035)^^1. 035)^(1. 035)^^^ The
swer quickly but
if
values of (1.035)^° and (1.035)^ could be found in the interest
tables, in particular in the
n
—
SO and «
=
7 rows of the
/
= 3^%
no « = 57 row of most interest tables, which is
why (1.035)^^ would have to be broken up into two parts. The
table.
There
is
term (1.035)^^^ presents a special problem. Usually only integral
values of n are given in the interest tables, along with
fractional values such as ^, ^
and
-j^,
but not
|.
common
One could work
by observing that (1.035)^/^ = [(1.035)*^^2]^ but otherwise
log tables or a power series expansion would be required.
this out
Interest: Basic Applications
Example
31
2.2
Company
Alice borrows 5000 from The Friendly Finance
interest
of
18%
pays the company 3000.
How much does
2000.
Two
per year convertible semiannually.
Three years
owe seven
she
after that she
at a rate
of
years later she
pays the company
years after the loan
is
taken out?
Solutio n!
We will use a time diagram to aid
5000
\
\
\
\
3
4
5
6
\
\
our solution:
X
12
\
in
2000
3000
FIGURE
Let
X be the amount
still
owing.
2.1
of problem, our goal
In this type
obtain an equation of value which will yield the solution.
entries
h
7
To do
on the time diagram should be brought to the same point
so an equation can be found.
Any
is
to
that, all
in
time
point in time can be chosen, but the
most convenient one in this example is / = 7. The amount owing will
equal the accumulated value at time 7 of the loan, minus the
accumulated value at time 7 of the payments already made. Since the
actual rate of interest is .09 effective per half-year, we have
X= 5000(1.09)^^ - 3000(1.09)^0 _ 2000(1.09)^
Example
=
D
6783.38.
2.3
Eric deposits
8000
in
an account on January
1,
On
he deposits an additional 6000 in the account.
withdraws 12,000 from the account.
On
1995.
January
Assuming no
withdrawals are made, find the amount
in Eric's
January
1,
1,
1997,
2001, he
further deposits or
account on January
1,
2004, if/ =.05.
Solutionl
In this example,
we
see that withdrawals can be viewed as "negative
deposits" in an equation of value.
8000
6000
\
\
1995
-12,000
\
2001
1997
FIGURE
2.2
\
2004
32
Chapter 2
The
resulting balance
X=
is
+
8000(1.05)^
Example
6000(1.05)^
-
12,000(1.05)^
=
2.4
Two
John borrows 3000 from The Friendly Finance Company.
Two
he borrows another 4000.
later
D
6961.73.
At what point
additional 5000.
equivalent
if
=
/
in
years
years after that he borrows an
time would a single loan of 12,000 be
.18?
[Solution!
4000
3000
\
5000
\
\
\
_(,_
12,000
2
1
FIGURE
We
v'
=
"*"
-yj
.
3000
+ 4000v2 +
Inv
We
problems
remark that there
like
5000v^ where v
Taking logs of both sides of
,= M3±4v!+5}:V^Ll2 ^2
Example
it
this
= y^.
equation
Then
we
find
D
11789.
is
an
approximate method of solving
2.4, called the
not need to examine
will
2.3
at which a single
would be equivalent, and form the equation of value at
as 12,000v'
=
4
3
be the number of years after the 3000 loan
let /
loan of 12,000
time
1
1
method of equated
time, but
we
here since there are no difficulties in
obtaining an exact solution.
To conclude
rate
of interest
Example
is
this section,
we
give a very simple example where the
the unknown.
2.5
Find the rate of interest such that an amount of
money
will triple itself
over 15 years.
[Solution]
Let
/
that/
be the required effective rate of interest.
=
3'/'^
-
1
=
.07599.
We have (1 +
i)^^
=
3,
so
D
33
Interest: Basic Applications
UNKNOWN RATE OF INTEREST
2.2
When the
of interest
rate
To
cations often arise.
Example
the
is
unknown
illustrate this,
consider the following example.
2.6
Joan deposits 2000
1, 1995, and then
no other deposits or
the account on January 1 2000
her bank account on January
in
deposits 3000 on January
1998.
1,
If there are
withdrawals and the amount of money
is
an equation of value, compli-
in
in
,
7100, find the effective rate of interest she earns.
[Solution
2000
3000
1995
1998
[FIGURE
2000
7100
2.41
+ 3000(1 +/)2 = 7100 is the equation of value on January 1,
Now we have a problem. This equation is a fifth degree poly-
2000(1+0^
2000.
nomial
in
/,
and there
is
no exact formula for finding
its
Most
which will
solution.
students will have a subroutine available on their calculators
enable them to approximate the answer with a high degree of accuracy,
and
we encourage
To show how these approximations are
work out this example numerically.
2000(1+0^ + 3000(1+ /)2 - 7100. We wish to find
actually obtained,
=
Let /(/)
two values
for
/,
this approach.
we
will
and
i\
ii,
such that/(/i)
<
and/(/2)
>
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
<
the answer,
>
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
<
for withdrawals.
Then
'•
^
^/-l
+
1
(2.1)
^r-
represents the rate of interest earned in the time period between balances
Bt-\ and
Bf The time-weighted
/
=
rate
of return
is
then defined by
(l+/i)(l+/2)- ••(!+/.) -
1-
(2.2)
35
Interest: Basic Applications
Example
On
2.8
January
1,
1999, Graham's stock portfolio
On
worth 500,000.
is
At that
Six months
30, 1999, the value has increased to 525,000.
point,
adds 50,000 worth of stock to his portfolio.
later,
Graham
has dropped to 560,000, and
December
weighted
31, 1999, the portfolio
rate
is
April
Graham
the value
On
100,000 worth of stock.
sells
again worth 500,000. Find the time-
of return for Graham's portfolio during 1999.
ISolutionI
The accumulation
1
-f
z'l
from January
rate
= ^00000 ~
from
May
1
Immediately
^•^^'
purchase, the portfolio
is
Hence
=
.97391.
is
^^^^qqq
time-weighted rate of return for the year
Note
in
Example
30
April
the
stock
the accumulation rate
Finally, the
—
accumu-
1.08696.
The
is
.^q'qq^
is
found from the interval
accumulation factors as (1.05)(.97391)( 1.08696)-
=
1
D
.11153.
2.8 that the value of the portfolio decreased
May
during the period from
given by the factor
is
after
two months of the year
lation rate in the last
30
worth 575,000.
October 31
to
to April
1
1
to
October 3
interest is clearly not operating here.
1
,
so
we
Nevertheless,
see that
is still
it
compound
possible to
by considering only deposits
and withdrawals, and ignoring intermediate balances. Setting up the
equation of value by accumulating all quantities to December 31, 1999,
calculate a dollar-weighted rate of return
we
obtain
500,000(1+0
+
-
50,000(1 +/f^^
This could be solved by linear interpolation, as
an alternative approach to this type of problem
interest for periods less than a year.
500,000(1+0
+
Since this equation
is
50,000(1
We would then
+1
linear in
/,
-
=
100,000(1+0'^^
is
in
Section 2.2, but
to
assume simple
obtain
100,000(1 + ^0
the result
/
=
i
500,000.
=
500,000.
ino 000
~
09677
is
easily obtained.
In
Chapter 12
we
will see
how
the theories of dollar-weighted and
time-weighted rates of investment return are applied to pension funds.
Chapter 2
36
EXERCISES
2.1
Equation of Value
2-1.
Brenda deposits 7000 in a bank account. Three years later, she
withdraws 5000.
Two years after that, she withdraws an
One year
additional 3000.
Assuming
4000.
2-2.
=
after that, she deposits
and
.06,
how much
drawals are made,
the initial deposit
/
is
Eileen borrows 2000 on January
Assuming
/
=
.13,
Brenda' s account ten years after
made?
borrows an additional 3000.
2-3.
is in
On
how much
1,
On
1997.
January
1,
January
does she owe on January
prepared to pay as follows: 200 after
and 700
after
1,
year,
1
4 years. If
/
=
1998, she
2001, she repays 4000.
Boswell wishes to borrow a sum of money.
after 3 years
an additional
no other deposits or with-
that
2005?
In return,
500
.12,
1,
he
after 2 years,
is
500
how much can
he
borrow?
2-4.
Payments of 800, 500 and 700 are made at the ends of years 2, 3
and 6 respectively. Assuming i = .13, find the point at which a
single payment of 2100 would be equivalent.
2-5.
A vendor has two offers for a house: (i) 40,000 now and 40,000
two years hence, or (ii) 28,750 now, 23,750 in one year, and
27,500 two years hence. He makes the remark that one offer is
"just as good" as the other. Find the two possible rates of interest
which would make his remark correct.
2-6.
(a)
The present value of 2 payments of 1000 each,
the end oi n years and n
-\-
A years,
is
1250. If
to be
/
=
made
at
.08, find
n.
(b)
Repeat part
(a) if the
payments are made
at the
end of n
years and An years.
2-7.
In return for
payments of 400
end of 8 years, a
IX at the
woman
at the
end of
agrees to pay
end of 6 years. Find X'xii
—
3 years
and 700
at the
X at the end of 4 years and
.14.
37
Interest: Basic Applications
2-8.
How
that
it
long should 1000 be
may amount
deposited at the same time at
2-9.
accumulate at /= .12 in order
accumulated value of another 1 000
left to
to twice the
8% effective?
Fund A accumulates at 9% effective and Fund B at 8% effective.
At the end of 10 years, the total of the two funds is 52,000. At the
end of 8 years, the amount in Fund B is three times that in Fund A.
How much is in Fund A after 5 years?
1
Henry 500 every March 15 from 1996 to 2000 inclusive.
15 from 1998 to 2001
inclusive. Assuming f^"^ =17, find the value of these payments
on (a) March 15, 2005; (b) March 15, 1999; (c) March 15, 1995.
2-10. John pays
He
2.2
2-11.
also
pays Henry 300 every June
Unknown Rate
A
of Interest
consumer purchasing a
refrigerator
is
two payment
offered
plans:
250
Plan A: 150 down, 200 after
1
Plan B: 87 down, 425 after
year, 50 after 2 years
Determine the range of
the consumer.
1
year,
interest rates for
after 2 years
which Plan
A
is
better for
2-12. Find the effective rate of interest if payments of 300 at the present,
at the end of one year, and 100 at the end of two years
accumulate to 800 at the end of three years.
200
2-13.
Bemie borrows 5000 on January
January
1,
1998.
He
finishes repaying his loans
What
1,
1995, and another 5000 on
repays 3000 on January
1,
1997, and then
by paying 10,000 on January 1, 2000.
is Bemie being charged?
effective annual rate of interest
TV for 600 from Jean. John agrees to pay for the TV
by making a cash down payment of 50, then paying 100 every four
months for one year (i.e. three payments of 100), and finally
making a single payment 16 months after the purchase (i.e. four
months after the last payment of 100).
(a)
Find the amount of the final payment if John is charged
2-14. John buys a
interest at
(b)
Find
an effective rate of
the
payment
effective
is
350.
annual
1
2%
per year.
interest
rate
if
John's
final
38
2-15.
Chapter 2
A
company pays 7% effective on deposits at the end of each
At the end of every four years, a 5% bonus is paid on the
balance at that time. Find the effective rate of interest earned by
an investor if he leaves his money on deposit for (a) 3 years; (b) 4
trust
year.
years; (c) 5 years.
2-16.
The present value of a
years forever
of payments of 1 at the end of every 3
2S
equal to -^. Find the effective rate of interest
series
1
is
per year.
2.3
Time-Weighted Rate of Return
on January 1, 1997. On
withdrawn from the fund, and immediate-
2-17. Emily's trust fund has a value of 100,000
April
1997, 10,000
1,
January
(a)
1,
is
withdrawal the fund has a value of 95,000.
ly after this
1998, the fund's value
is
1
On
15,000.
Find the time-weighted rate of investment return for this
fund during 1997.
(b)
Find the dollar- weighted annual rate of investment return for
Emily's fund, assuming simple
(c)
interest.
Find the rate of return for Emily's fund using simple
interest,
and assuming a uniform distribution throughout the
year of all deposits and withdrawals.
2-18.
Assume
in
Question 17
also a
that, in addition to the
5000 deposit
to the fund
on July
information given,
1997.
there
is
(a)
Find the dollar-weighted annual rate of investment return for
1,
the fund, assuming simple interest.
(b)
Find the rate of return for Emily's fund using simple interest
and assuming a uniform distribution throughout the year of
all
(c)
Is
deposits and withdrawals.
it
possible to calculate the time-weighted rate of return?
If not,
why
not?
39
Interest: Basic Applications
2-19. Let
A
be the balance
in a
fund on January
1,
1999,
B
the balance
on June 30, 1999, and Cthe balance on December 31, 1999.
If there are no deposits or withdrawals, show that the dollar(a)
weighted and time-weighted rates of return for 1999 are both
equal to
(b)
If there
~^
was a
single deposit of
W immediately after the June
30 balance was calculated, find expressions for the dollarweighted and time-weighted rates of return for 1999.
(c)
(Assume simple interest for periods of less than a year.)
If there was a single deposit of
immediately before the
June 30 balance was calculated, find expressions for the
W
dollar-weighted and time-weighted rates of return for 1999.
(Assume simple
(d)
interest for periods of less than a year.)
Give a verbal explanation for the fact that the dollarweighted rates of return in parts (b) and (c) are equal.
(e)
Show
that the time-weighted rate
of return
in part (b)
larger than the time-weighted rate of return in part (c).
is
CHAPTER THREE
ANNUITIES
3.1
ARITHMETIC AND GEOMETRIC SEQUENCES
we
Before beginning the study of annuities,
will briefly
basic facts about arithmetic and geometric sequences.
review some
The formulae we
develop for geometric sequences will be used a number of times
later in
the chapter.
Recall that an arithmetic sequence
is
a sequence of
X\, X2,... where the difference between consecutive terms
For example, the sequence
10,
4, 7,
13,
..
.
is
is
numbers
constant.
an arithmetic sequence
with constant difference 3 (assuming the apparent pattern continues).
The sequence
5,
1,
—3, —7,
...
is
also arithmetic, having
common
difference —4.
Any arithmetic sequence is determined by its first term and its
common difference; if the first term is a and the common difference d,
the sequence
is
a,
a
+ d,
a
+ 2d,
a-\-3d, •••.
(3.1)
There are two important formulae about arithmetic sequences which
would
[Theorem
3T
Consider an arithmetic sequence with
ence d.
(a)
we
like to develop.
The
«^^
term of this sequence
first
term a and
common
differ-
is
a-\-(n-\)d.
(b)
The sum of
the
first
n terms of
this
formula
^[2a
+
(rj-\)d].
sequence
is
given by the
42
(a)
Chapter 3
Informally,
we
that the n'^
term will be a
see from the pattern a, a
+
(n—\)d.
using mathematical induction.
We
A
+ d,
a
+ 2d,
a
-\-
3d,
•
•
•
formal proof can be given
will leave this as an exercise
for the reader.
(b)
Let Sn denote the
(a),
sum of the
first
n terms. Using the result of part
we have
S„
=
a-\-{a-{-d)^(a
+
2d)-h----\-[a-\- {n-\)d].
Writing the terms on the right hand side of (3.2a)
order,
Adding together
+
[a-\-(n-2)d]
+
•••
+ (a+^ + a
and (3.2b) by combining
(3.2a)
together, then second terms together, and so on,
=
2S„
--.
+
[a
first
terms
obtain
+ («-l)J+a].
There are n equal terms on the right side of
n[2a
we
(3.2b)
[a-\-a-\-{n-\)d]-\-[{a-{-d)-^a-\-(n-2)d]
+
=
in the reverse
we have
Sn^[a-{-{n-\)d]
2S„
(3.2a)
(3.2c), so
(3.2c)
we have
+ {n-\)d\, or
S„
As an
as required.
=
^[2a-^(n-\)d],
(3 .2d)
exercise, the reader should give an alternate
V
derivation of (3 .2d) using mathematical induction.
Example
3
Find the 32"^ term and the
sequence
I
sum of
the first 18 terms of the arithmetic
5, 9, 13, 17,
Solution!
We have a=5andd=
Using
(3. 2d) the
Now we
sum of
4.
Hence, the 32"^ term
18 terms
is
^[10 +
is
(17)4]
will consider geometric sequences.
metric sequence
consecutive term
is
is
+ (3 1)(4) =
= 702.
5
129.
D
Recall that a geo-
a sequence of numbers X\, X2,
...
where each
obtained from the previous term by multiplying by a
43
Annuities
For example, the sequence
fixed number.
with
common
ratio 3,
with
common
ratio
a and common
and the sequence
— A The
4,
— ^,
...
geometric
is
geometric
first
term
be
ar, ar^, ar^,
a,
Theorem
— ^,
is
general geometric sequence with
.
ratio r will
5,
2, 6, 18, 54, ...
(3.3)
3.2
Consider a geometric sequence with
(a)
The
(b)
The sum of the
n^^
term of the sequence
first
first
term a and
common
ar"~^
is
n terms of the sequence
—
^.
is
_
ratio r.
-.
fPi^^of]
(a)
As
is
before, the pattern a,ar,ar^,
induction.
(b)
leads us to believe that ar^~^
...
the correct term, but a formal proof requires mathematical
We will
Let Sn be the
leave this proof as an exercise for the reader.
sum of the
first
n terms. Using the result of part
(a),
we have
Srr
=
+ ar^
a
Multiplying both sides by
rSrt^ar
r,
h
we
+ ar'^
ar^-y
obtain
-\-----\- ar".
Subtracting the second line from the
terms cancel, and
Sn
—
rSn
we
—
(3.4a)
first,
(3.4b)
we
observe that
ar
a-\-
—
ar
-\-
ar-^
—
ar'^
+
•
+ ar"-^ - ar"-^ - ar" =
Thus we have
*S'„(1
—
r)
S.
as required.
many
obtain
=
a(\
—
•
•
a
-
ar".
(3.4c)
r"), or
= '-^\^,
(3.4d)
V
44
Chapter 3
Example
Find the
3.2
13'^
term and the sum of the
sequence 48, —24,
12,
—6,
3,
first
9 terms of the geometric
— ^,
Solution
We
have a
=
48 and ^
that the 13'^ term
is
= -4-
48(-^)^2
=
Using part
^.
(a)
Using
we
find
sum of the
first
of Theorem
(3.4d), the
3.2,
1\9
9 terms
3.2
is
48
'-(-i)
i-(-^)
48
513
3
'-'
•
16
2
BASIC RESULTS
John borrows 1500 from a finance company and wishes to pay it back
with equal annual payments at the end of each of the next ten years. If
/ = .17, what should his annual payment be?
Jacinta buys a house and takes out a 50,000 mortgage.
If the
mortgage rate is 13% convertible semiannually, what should her monthly
payment be to pay off the mortgage in 20 years?
Eileen deposits 2000 in a bank account every year for 1 1 years. If
/ = .06, how much has she accumulated at the time of the last deposit?
All of these questions have one thing in common: they involve a
series of payments made at regular intervals. Such a series of payments
is called an annuity.
In the three cases above, the payments are of equal
amount, and that will be the case with all annuities studied in this
section. Later, however, we will study more general annuities.
Annuities turn up in many different types of financial transactions.
From the point of view of practical applications, a complete understanding of annuities is an absolute must!
We shall start by considering an annuity under which payments of
Sometimes a period
1 are made at the end of each period for n periods.
will be one year, as with John's loan above, but other periods are
certainly possible. It will be assumed throughout that, as with John's
loan, the interest period and the payment period are equal. When this is
not the case, as with Jacinta' s mortgage, for example, we will first find
the equivalent rate of interest per payment period and then proceed with
our solution.
Level payments of an amount other than
multiplying by the amount of the payment, as
examples.
1
can be handled by
we
shall
see in the
45
Annuities
^
i
1
2
3
[FIGURE 3.n
A
time diagram showing n payments of
present value of this
accumulated value of this annuity
We
now
shall
in turn,
'^\
This
is
the
at
v
we
denoted by
v(l
1
s-,.
is
crucial,
time
h v".
H
(3.5)
we
=
=
v and r
v.
obtain
-V
/•
'^
1
—
-v"
(3.6)
at
-v^)
+ -
1
o;^.
Taking the value
Section 3.1,
in
The
The
Figure 3.1.
sum ofn terms of a geometric sequence with a
^
Formula
is
a-,.
in
denoted by
obtain
+ v^ + v'
Using Formula (3.4d) developed
of the
given
is
time n
derive a formula for
of each of the payments
is
1
annuity at time
[^{l^'J
(3.6)
and will be used frequently throughout the
rest
text.
It is
easy
now
same annuity n years
to get a formula for s-y
after a-^ has
«;n(i
1
(1
tl
Since
been calculated,
s-^ is
it
the value of the
follows that
+ 'T
-v"
+
(1
0"
+o"-v''(i +0"
+ 0" i
1
u
•
(3.7)
46
Chapter 3
Let us immediately proceed to some practical examples.
Example
3.3
XX
Find John's payment
section.
-\
in the
problem stated
\
paragraph of this
in the first
X
\
\
10
FIGURE
3.2
Solution!
Let the payment be X.
ajQ,,
Since the present value of 10 payments of
the present value of 10 payments of ^^ will
1500
= X- a^^. Then;^^ i|M ^ 1500 ^
m
lOi
Example
be ^-
qjq^.
is
Thus we have
=
.^»^
i-(n7)
1
1
\10
321 98.
3.4
Find the accumulated value
in
Eileen's bank account in the problem
stated in the third paragraph of this section.
2000
2000
2000
11
X
FIGURE
I
3.3
Solution!
Since each deposit
by
X= 20005
Example
IJI
is
2000, the accumulated value will be given directly
-2UUU
29,943.29.
_Q^
3.5
Find Jacinta's mortgage payment
XX
in the
problem stated
in the
paragraph of this section.
H
X
240
50,000
FIGURE
3.4
1
second
47
Annuities
ISolutionI
As mentioned
earlier,
interest equivalent to
we first have to find the effective monthly rate of
13% convertible semiannually. This is because our
formulae for o^ and
are based
s-,
on the assumption
Letting this monthly rate bey,
period and payment period are the same.
we have
Note
\
-\-j
—
r
1
that there are
mortgage, so
Example
4-
n
^
that the interest
^^^
1
Now we
.
the mortgage
let
240 monthly payments
we have Jf qjm = 50,000
in the
20-year term of the
——
andX=
•
payment be X,
-
jao
—
573.77.
D
3.6
No
Elroy takes out a loan of $5000 to buy a car.
payments are due for
must
=
make 60 equal monthly payments. If /
.18, find (a) the amount of
each payment; (b) the amount of each payment if there is no paymentfree period, (i.e., if the first payment is due in one month and the
remaining 59 are made on a monthly basis thereafter).
the first 8 months, but beginning with the end of the 9^^ month, he
ISolutionl
(a)
We
(1
first
+y)i2
note that a monthly rate of interest y
=
we
\\s,
obtainy
=
(1.18)^^^2
-
is
1.
required.
Since
Let the amount of
each payment be X.
X
X
\
\
\
\
7
8
9
5000
i
H
[FIGURE
We now
month
8,
X
a-^^ will
3.51
is
into the standard annuity
fit
give us the value of the payments at
one month before the
loan at time 8
h68
10
observe that this does not
pattern, since
X
first
5,000(l4-yy, since
it
payment.
The value of
will accrue interest for eight
months, even though no payments are required. Thus
equation of value
X
a^^,
601
=
5000(1+/^,
_ so
.
Evaluating«^,X=50MH^
=
'601'"^
_^60
1
the
that
-
137.76.
X=
we have
the
^^MLtZL,
„^Q|
48
Chapter 3
we
In this case,
(b)
—
X=
— T—60 =
years, the total
payment
for 8
amount of extra money paid
months will be 863.40!
a-^,
=
—
1
—
(1+/)"^^
/"^
that,
days
solve for
for postponing the first
D
evaluation
the
we
over the next 5
beautifully demonstrates the
pre-calculator
In
5000, which
This shows
123.37.
The previous example
calculators.
X-a^^ =
just have
would have caused serious
power of our
of
term
the
Even
difficulties.
would not have helped, because the interest ratey is not
we do now, however, is press
few buttons (in the right order) and the answer appears!
interest tables
one for which tables were constructed. All
a
Example
3.7
v^
(a)
Prove the identity 'T^= ia^
(b)
Give a verbal interpretation of this
-f-
identity.
ISolutionI
=
—— v"
-1
j-^^, so ia-,
(a)
a-^
(b)
The term
ia-,
=
1
—
v",
and
1
=
ia-> 4-
v", as required.
can be thought of as the present value of an annuity
with level payment
3.6). The term v"
/
is
end of each year for n years (see Figure
the present value of 1 at year n.
at the
—
1
/
H
\
/
/
\
\
1
T
FIGURE 3.6
Imagine investing
interest
is
1
time
at
0.
At the end of the
first year,
the
separated off from the original investment, and the
amount of the investment
is
back to
1
.
This procedure continues
end of n years and the annuity of /
which was removed each year. The present value of these terms is
for n years, leaving
v" and
ia-^,
1
at the
There are two other symbols
namely
a-,
and
D
respectively.
5-1.
in
common
usage with annuities,
49
Annuities
a-, is
of the
last
first
the present value of the annuity described earlier at the time
payment, and
the accumulated value one year after the
s-, is
payment has been made. Our four functions,
a^,, a-,, s-^
and
are
s-,
illustrated in Figure 3.7.
1
\
1
1
1
\
h-
\
2
1
n+\
,^
FIGURE 3.7
Mathematically, there
s-^
=
s-^(\
-\-
i).
are analogous to
These relationships lead
Formulae
We
nothing very exciting going on here.
is
can see immediately from Figure 3.7 that
to
d-^
a-A\
formulae for
and (3.7) for
(3.6)
=
a-,
and
-\-
/),
and
a-,
s-,,
and that
s-,
that
respectively.
We have
I
=
where d
is
^'
17
1
"*
(3-8)
the effective rate of discount defined in Chapter
the reader should
show
1
.
Similarly,
that
^;^
=
^^^^^
(3-9)
Observe tRaf"^^pcafT also be described as tl^^feseht value of
payments of
1
made
at the
beginning of each period for
r?
periods, and
s-^
can be described as the accumulated value of the same payments at the
end of the last period.' Since the payments are at the beginnings of the
periods,
it
follows that
the last payment.
s-, is
their
accumulated value a
full
period after
50
Chapter 3
There are many
we have
identities relating the four quantities
we
introduced. In addition to the ones mentioned earlier,
note that
^r^-^Trrrr
o.io)
and
l=c/a;^
+ v",
(3.11)
both of which have nice verbal interpretations.
Other relationships will
be presented as exercises.
Let us do an example. Imagine that Henry takes out a loan of 1000
and repays
it
of the loan.
with
1
equal yearly payments, the
In this case,
X
if
appropriate equation of value
is
the
first
one due
at the
time
amount of each payment, an
would be
X-
0701
=
1000.
(3.12a)
X
X
+
2
1
10
9
1000
[FIGURE
We
use
djQ,
3:8]
here because the annuity symbol
qjq^
assumes
we
the present value one year before the first payment, and that
case with Henry's loan.
An
lOOOv,
(3.12b)
because lOOOv would be the value of the loan one year
first
not the
is
equally good equation of value would be
X^ioi^
allows us to use our
are taking
annuity symbol.
Once
earlier,
which
the rate of interest
/
is
known, we can find the amount of Henry's loan payment.
Let us observe as well that a third acceptable equation of value for
Henry's problem would be
X{\
Example
Using
all
+a^^=
1000.
(3.12c)
3.8
three equations of value, find Henry's loan
payment
if
z
== .16.
Annuities
5
ISolution
we
First
(a)
^
^
=
T-1-1
/
= t^1.16
X^
d=
(b)
consider Equation (3.12a), which
-h
Equation
^QQQ^o
1
-
(3.12b)
1000
X=^
To
-
is
Jf ajQ^
•
use this approach
=
Then
1000.
we need
to find
v'""
This leads to the answer
that
states
(n6)ci6)
\' \io
=
X-
(^fTT^r
=
X=
lOOOv.
178.36.
Then we have
178.36.
1
(c)
Finally, consider
Equation (3.12c), which gives us the equation
;^=_lMQ_^_im^^
1
+ o^i
1
+ "4^
D
178.36.
The moral of Example 3.8 is that there is more than one way to
work out this kind of problem. However, we reiterate the importance of
many decimal places as possible during your calculations.
many textbooks, the term annuity-immediate is used for the case"
where payments are made at the end of the period, and annuity-due is
used when payments are made at the beginning of the period. As we
keeping as
In
have just
illustrated,
however, the same techniques can be used
in
both
cases.
The following example
ways of analyzing annuities.
Example
illustrates that there are
m
possible
3.9
Consider an annuity which pays
-\-
many
n years.
Explain verbally
1
why
at the
beginning of each year for
each of the following expressions
gives the current value of this annuity at the end of year m. (See Figure
3.9 on the following page.)
(a)
a^,{\ + ir^'
(b)
a^^{\+ir
(c)
s^^v"-'
(d)
s—^^"
(e)
5;;;^
(0
^H
(g)
i+^H + «;rTi
+ a;^
+^
h-^
;
^—
•-r;
•
/Vf/[-
/
^^
52
Chapter 3
w+1
^
w4-«-l
m-\-n
T
FIGURE
3.9
Solution
diagram for
In the time
t
<^^^q^ is the
value at year
years into the future, so
(b)
i^
^;;rM
denoted by
which we want the value of the annuity.
the point at
(a)
we have
this annuity in Figure 3.9,
—
1
To
.
we have
get to
<3f^^(l 4-
+
So <^;;^(1
^^^ value at year 0.
]
,
we must move m +
0^^'
i)"^
1
•
moves us
m
years
into the future.
(c)
is
5^;qp;^
the value at year
m
-\-
n
-
Hence we move back n
1.
—
1
years.
m
(d)
5^q:;;T is
the value at year
(e)
5—n
the value of the first
is
(f)
is
the value of the last n
s—, is the
value of the
—
1
first
last n payments
Here the single payment of 1
value of the
(g)
-\-
in part (f), leaving
a—
n, so
we move back n
m+
1
payments
at
years.
time m, and
payments at time m.
m payments at time w, and
at the
at
same
time
m
a—n
d-^ is
the
time.
is
separated off from the o;^
D
f|.
The above example should illustrate how careful we must be when
working with these functions, but also that we have considerable flexibility in using them to express an annuity value at some point of time.
3.3
PERPETUITIES
A perpetuity
diagram
is
is
an annuity whose payments continue forever.
shown
in
Figure 3.10 below.
1
1
1
[FIGURE
3.101
The time
53
Annuities
The value of
this annuity
one year before the
first
payment
is
a—,.
We
have
lim
n^oD
oo|
lim
n—^oo
^
1
-v"
-
•
/
1
(3.13)
7'
=
since lim v"
We
0, as
long as
can see verbally
/
>
0.
why Formula
(3.13) should be true:
=
if
a
can be
principal of
y
removed
at
the end of each year, leaving the original principal intact
in
Section 3.2, the symbol d—^ represents the value of a
is
invested at rate
/,
then the interest
(
y
j
/
1
forever.
As
perpetuity at the time of the
left
first
payment. The following identities are
as exercises for the reader:
^^ = ^^(1+0,
(3.14)
and
^
3.4
i
(3.16)
UNKNOWN TIME AND UNKNOWN
RATE OF INTEREST
We
examples involving annuities where the
of interest involved is the unknown.
will consider here several
length of time or the rate
Example 3.10
A
fund of 5000
is
used to award scholarships of amount 500, one per
year, at the end of each year for as long as possible.
number of scholarships which can be awarded, and
fund one year after the
last
If
the
/
=
.09, find the
amount
scholarship has been awarded.
left in
the
54
Chapter 3
ISolutionI
-^
500
500
\
\
500
\-
\
n+\
«
2
1
5000
,
,
[FIGURE
If
n
number of
the
is
so that
a-^< 10
<
v"
=
reduces
to
scholarships, then 500
a—
1
3.711
Putting
q.
—
so
10/,
o;^
=
«
=
10,
a-^
•
we
——^
=
logv
year
5000(1.09)2^
Example
A
trust
the
after
scholarship
last
_ 5005— =
<
5000
obtain
^
^;^,
•
This
10.
we
Hence,
26.7.
been
has
500
~^ =
The amount
conclude that 26 scholarships can be awarded.
one
<
fund
in the
awarded
be
will
D
363.84.
3.11
fund
is
by means of deposits of amount 5000
to be built
at the
at the end
monthly pay-
end of each year, with a terminal deposit, as small as possible,
of the
The purpose of
final year.
ments of amount 300
month
fund
is
to establish
into perpetuity, the first
of
If the rate
after the final deposit.
convertible quarterly, find the
of the
this
payment coming one
interest
is
12%
number of deposits required and
per year
the size
final deposit.
ISolution!
We
require effective yearly and monthly rates equivalent to
The yearly
y
=
(1.03)'^^
payment
is
ty 50005;^
5-
=
6.06,
rate
—
1.
is
/
—=
^2^
30,298.03
we
(1.03)"^
1,
whereas
monthly
the
<
obtain
30,298.03.
50005;;:^, so that
(1
+
/f
=
1.6758,
We
start
<
6.06
s-^
so
n
=
=
f"^^
The value of the perpetuity one month before
^=
<
=
—
.12.
rate
is
the first
with the inequali-
< 5;^.
Solving
Hence four
4A.
deposits of 5000 each are required.
H
12
\
5000
\
5000
300 per month
\
\
1
3
4
5
5000
5000
FIGURE
3.12
X
ll
illl lll
l
l
l
6---
55
Annuities
The
of the
size
equation
^-
X=
given by X, and
final deposit is
5000 ^^^-^
=
(1+/)
is
it
found from the
D
3184.30.
Example 3.12
At what effective yearly rate of interest is the present value of 300 paid
at the end of every month, for the next 5 years, equal to 15,000?
300
300
\
\
1
300
1
60
2
1
15,000
IFIGURE
Let
/
rate
be the monthly rate of
Hence
/.
we would
lator
=
a-^^
then
interest;
we have 300
we must now
50 and
solve for
/.
•
a-^,
As
—
in
15,000
at
Chapter
2,
expect students to solve this equation using standard calcu-
subroutines,
50/(1+/)^^
3.131
-
equation 50
since the
{\-¥if^ +1=0.
two numerical methods
However, we
is
equivalent to
will also briefly outline
for obtaining approximate solutions, partly for
and partly to show
historical reasons
— a^ =
how
the calculator subroutines
actually work.
[Solution Onel
Define /(/)
=
50
— a^. We
will use linear interpolation to estimate the
=
which /(/)
0.
Observe that /(.006) = - .26213, and
/(.007) = 1.14413. Using linear interpolation between these values we
26213
=
obtain / = .006
value of
/
for
+
is
The
.00619.
(.001)
effective yearly rate
therefore (1.006
[Solution
As an
Twol
alternative to linear interpolation,
we
could use an interesting
method called successive approximation. As before, we have
SO -
—Y-— =
trial
value of
1
_
^60
obtained
is
1
50, or
/
in
/
=
_
;-/
The
idea
now
is
to substitute
the right hand side of the above equation.
then used as the next
the right hand side.
degree of accuracy
is
trial
value, and
This procedure
obtained.
a-^.
=
50,
60
is
is
some
The value
substituted back into
continued until the required
56
Chapter 3
For example, let us try /q
1 -(1.007)-6Q
— V•60
50
50
=
i\
.00684
Continuing
this procedure,
=
.00661
/5
=z
.00647
=
=;
=
=
=
=
=
=
=
=
=
=
=
i\\
in
i\5
hi
i\9
hi
'23
hs
hi
h9
hi
subroutines
We
being
we
show
/
remark
iterative,
.00684.
Next
we
=
-v
substitute
,60
1
obtain
'
=
50
.00671.
obtain the following values:
.00653
.00638
k
h
.00642
.00631
ho
.00629
.00627
.00625
.00624
hi
iu
.00622
he
.00621
.006205
h%
.006201
.006198
ho
.006195
.006193
hi
.006191
.0061898
hA
.0061887
.0061878
he
.0061870
.0061860
hs
.0061858
.0061854
ho
.0061850
.00634
.00623
.0061847
/
=
.006185, to 6 decimal places.
-
(1.006185)'^
=
/2
We
estimate.
l4
obtain
therefore
is
we
^3
h
h
rate
initial
the right hand side, yielding
in
In this solution
.007 as an
1
=
(Standard
.0768.
The yearly
calculator
D
.006183 to 6 decimal places.)
that successive
and hence
it
is
approximation has the advantage of
easy to write a computer program which
you any degree of accuracy you require. However, care is
needed since this procedure will not converge for all functions. In some
will give
cases convergence never occurs, whereas in others convergence will
only
occur
if
the
starting
Furthermore, convergence
For example when
A:
is
s-, is
value
more
close
may depend on
to
/
=
the
actual
answer.
the form of the function used.
involved, the expression
a constant, diverges, whereas
A
is
(1+/^)^^"
—
(1+0"
-1
/
=
1
will converge.
,
where
detailed discussion of successive approximation will not
be given here, but the interested reader should consult any standard text
in
numerical analysis.
57
Annuities
CONTINUOUS ANNUITIES
3.5
A
type of annuity which
commonly
arises
payments, but a quoted interest rate which
is
practice has
two sections we saw
In the last
that this situation could easily be handled
converting the interest rate to a monthly
approach
monthly
effective yearly or half-
Mortgages are a good example of this type of annuity.
yearly.
found
in
in interest tables.
For
rate.
Older textbooks found that
because the
inconvenient, primarily
this reason,
by
symbols
new
like
rate
could not be
a-
and s-
were
developed, where a^Vepresents the present value of an annuity with
amount
^
paid at the end of every m^^ of an interest period for the next
The basic idea
n periods.
now been
is
that the usual
m parts of ^
divided into
The formulae developed
our
modem
payment of
1
per period has
each.
for these
new symbols
are not required in
approach to annuities and will not be stated here. However,
the limiting case
is
of interest.
We define
a-,= Urn a^\
a-^
can be considered as an annuity lasting n periods, where the periodic
payment of 1
sum becomes
is
paid continuously throughout the period.
integral
and
/
where the term
v^ dt in the integral
v'dt,
arise in practice,
it
payment period
is
evaluate,
(3.18)
can be thought of as the present value
of an infinitesimally small amount dt payable
Such an annuity
To
we have
^= JoJO
the
(3.17)
at
time
t.
called a continuous annuity.
While
it
may
not
serves as a good approximation to annuities where
is
very small
We easily obtain
(e.g., daily).
58
Chapter 3
dt
'^
Jo
n
[lnv\
v" ln\
1
1
-v"
+/)
ln{\
1-v^
(3.19)
b
Formula (3.19) compares very nicely with Formulas (3.6) and
and a-i, respectively, derived earlier in this chapter.
(3.8) for
a-^
Similarly
J;^
=
(l
+
'To;^
(l+zT-l
(3.20)
6
Formula
[c.f.
(3.7).]
We also have
---[
(1
+
which has a similar verbal interpretation
Formula
to that given a^^ as defined
by
(3.18).
VARYING ANNUITIES
3.6
To
(3.21)
i)'dt.
this point, all the annuities considered
We now remove this
payments.
restriction
have had a
level series
of
and consider annuities with a
varying series of payments.
Any
type of annuity can be evaluated by taking the present value
or accumulated value of each payment separately and adding the results.
There
are,
several types of varying annuities for which
compact expressions are possible. The only general
study is the type where payments vary in arithmetic
however,
relatively simple
type
we
will
59
Annuities
Specific examples of other types will be considered as
progression.
well.
First let us
assume
payments vary in arithmetic progression.
payment is P and payments increase by Q
that
In other words, the first
thereafter, continuing for
n years.
P + {n-\)Q
P+Q
FIGURE
3.14
The value of this annuity one year before the
^ ^ Pv +
(P
+
ey
Multiplying (3.22) by
(1
+
-f
+
1
(/>
/,
+ 2Qy +
we
•
•
•
payment
first
+
[P
is
given by
+ {n-\)Q]v".
obtain
^
= P + (P + 0v -h (P + 20v2
+ [P + {n-\)Q]v"-\
/>4
Subtracting (3.22) from (3.23),
(3.22)
we
(3.23)
obtain
= P+Q(v-\-\- ^i-v"-^)-Pv" -(n-\)Qv"
iA
=
P(l-v")
+
e(v
+ v^+
•••
+
v"-^
4-
v")-nQv\
(3.24)
Hence
A = P [1 -v"
+^ f'^
-nv"-
.
I
^-"v"!
^nl
+Q
(3.25)
/'
The accumulated value of these payments
equal to^(l
+
/)",
P^nl
Two
at
time n
is,
of course,
and hence equals
+Q
(3.26)
special cases often occur in practice.
increasing annuity where
P=
1
and
Q=
\.
The
first
of these
is
the
Chapter 3
60
2
1
?
[FIGURE
The value of
this annuity at
time
3.T5]
is
denoted by
{Id)-^,
following formula, derived from (3.25) by substituting
-
V^">'/7|
""1
and has the
P= Q=
1.
/
'
I
dj^^- „v"
(3.27)
The value
at
time n
(^^);;i
The second
is
therefore given by
=
(/a);^(l+/r
_
^'^^-^
special case
(3.28)
the decreasing annuity
is
where
P—
rj
and2=-l.
—
n
n
\
—
H
\
1
\
2
1
?
(^^)^
(^^);^
[FIGURE
The present value of this annuity
by
at
3.T6]
time
0,
derived from (3.25),
is
given
Annuities
61
(Da)-,
-
a-.-nv"
1(1)
na^^
I
Kl-v")-a;;|4-A7v"
/
(3.29)
i
The accumulated value
(Ds);;,
=
time n
at
is
(Da);^(l+0"
_n{\^i)"-s-,
(3.30)
I
Finally, let us consider an increasing perpetuity as
shown
in
Figure
3.17.
12
2
1
T
IFIGURE
An
3
3
3.171
appropriate symbol for this would be {Ici)—\, and
(/or)—
I
=
Urn
(la)-,.
Since lim
ci-,
1
+
=
1
+
easily seen that
it is
7, and lim nv"
=
0,
we
obtain
!
(^^)^
= j-^i'
Alternatively,
we
(3.31)
could obtain Formula (3.31) from
first
principles
as follows:
(la)—^
=
v
=
(v
+ 2v2 +
+ v2+
3v^
v^-f
+
•••
•••)
+ (v^+ 2v^+
= a^ + v(v + 2v2+ 3v^+
•••)
•••)
62
Chapter 3
Solving for (la)—^,
we
obtain (7^)^ (1
v) ==
y so
,
that
1
~"
(la]
-
/(1-v)
1
1
=
=
i-^'
(3.31)
as before.
Example 3.13
Find the value, one year before the
200, 500, 800,
...
if
=
/
.08
first
payment, of a series of payments
and the payments continue for 19 years.
ISolutionI
Using
P=
200(^191
+
200,
^^^
Q=
300 and «
-^^^
19v'^
=
=
19,
we
1920.72
obtain,
+
from Formula (3.25),
19,504.01
=
21,424.73.
D
Example 3.14
Find the present value of an increasing perpetuity which pays
end of the fourth year, 2
at the
twelfth year, and so on, if/
=
end of the eighth year, 3
at the
1
at the
end of the
.06.
[Solution]
This could be done either by
(3.31).
rate
of
Hence
If the
formula
is
first
to be used,
interest per four-year period.
the answer
is
.26248
principles or
we must
first
This equals
=
by applying Formula
compute the effective
(1.06)"^
—
1
=
.26248.
18.32.
(.26248)^
Example 3.15
payment of an annuity where
payments start at 1 increase by annual amounts of 1 to a payment of n,
and then decrease by annual amounts of 1 to a final payment of 1
Find the value one year before the
first
,
.
Annuities
63
Solution
n-
1
n-
n-
1
n-\-\
[FIGURE
1
time diagram for this is shown in Figure 3.18. Note
payment occurs at time 2/7—1. The present value is
+ v\Da)j^, =
a-,-nv"
1
1
\
that the final
+ {n-\)v"-v"a—
-I- rt
'2n-
3.18]
A
(lah
1
v"
— v"n
-v^
(3.32)
""nl-^nl-
n
Example 3.16
Show
both algebraically and verbally that (Da)-,
=
{n-\-\)a-,
Algebraic Solution
{n^\)a-^
- (lah =
(«+l)
l_vn
\^n-f^^
n+\ -nv" -v"
n+\ -v"
-cti
n-(d-,+v"-\)
"-^
(D^)ny
-d-^-}- nv"
-
(la)-.
64
Chapter 3
[Verbal SolutionI
Here
is
a diagram for {n
+
Ifc:
«+
n+\
A7+
1
1
h-
\
1
T
IFIGURE
Here
is
3.191
a diagram for {Id)-y
2
1
1
T
[FIGURE
3.201
Subtracting the second diagram from the
IFIGURE
This
is
first,
we
obtain the following:
3.211
(Da)-,, as required.
Example 3.17
An
annuity provides for
payments
is
preceeding
the final
15
annual payments.
200, and each subsequent payment
it.
is
The
5%
if
/
=
of these
less than the one
Find the accumulated value of this annuity
payment
first
at the
time of
.09.
[Solution]
-^
200
\
200(.95)
\
200(.95)2
\
200(.95)^^
h
15
T
FIGURE 3.22^
65
Annuities
We
want
earlier
S
=
a
=
is
in
this annuity at
just the
200(.95)^'^
Observe
time 15.
that the
geometric, rather than arithmetic sequence, so our
From
formulae cannot be applied.
200(.95)'^
This
of
to find the value
payments are
+ 200(.95)'^(1.09) +
sum of the
=
and r
first
principles the value
first
200(.95)'2(i q9)2 _^
.
.
.
15 terms of a geometric sequence with
The formula developed
-M^.
.95
is
+ 200(1. 09)'^
Theorem
in
1.091 '^
3.2(b) tells us that S^s
=
200(.95)'^
^
L
1
^
_
.95
J
109
=
4541.70.
.95
Example 3.18
Frank to make 8
The first payment of 10,000 is made
payments are to increase according to an
In settlement of a lawsuit, the provincial court ordered
annual
payments to Fred.
immediately, and future
assumed rate of inflation of
payments assuming / = .07.
Find the present value of these
.04 per year.
SolutionI
10,000(1.04)
10,000
-H
10,000(1.04)'
10,000(1.04)2
\
h
\
1
FIGURE 3.23
The required present value
P=
is
10,000+10,000(^1^) +•••
+
10,000(^1^)
\
=
,
0,000
fi^l#
=
Note
that an alternative
+ = TM'
/
D
72,574.66.
that the expression for
1
1.07
^
y
^^^^
^^
P
is
way of
solving
Example 3.18
is
just 10,000^gj at a rate of interest
/
to realize
such that
often called the real rate of interest, and can be
calculated in any problem
where
inflation
is
involved.
66
Chapter 3
EXERCISES
3.1
Arithmetic and Geometric Sequences
3-1.
Find the
term and the sum of the
17^^
first
terms of each of the
1 1
following:
(a)
(b)
(c)
The arithmetic sequence 2, 7, 12, 17, ...
The arithmetic sequence with a = 7\ and d = —3.
The arithmetic sequence whose 5^^ term is 19 and whose
term
9'^
47.
(d)
The geometric sequence
(e)
The geometric sequence
(f)
The geometric sequence whose
term
3-2.
is
5, 15,
3,
45,
— 4,
-f^,
5^^
— A,
term
4 and whose
is
S'''
^.
is
Prove Theorems 3.1(a), 3.1(b), 3.2(a) and 3.2(b) using mathematical induction.
3.2
Basic Results
3-3.
Henrietta borrows 6500 in order to buy furniture. She wishes to
pay the loan back by means of 1 2 annual payments, the first to be
made one year after the loan is taken out. If / = .13, find the
amount of each payment.
3-4.
Answer Question
payments, the
3-5.
be paid back with 144 monthly
one due one month after the loan is taken out.
3 if the loan is to
first
Alphonse deposits 450 in a bank account at the beginning of each
and continuing for 20 years. If/ = .08, find
the amount in his account at the end of 1996.
year, starting in 1977
3-6.
An
annuity pays 1000 a year for 8 years.
If
/
=
.08, find
each of
the following:
(a)
(b)
(c)
(d)
(e)
The
The
The
The
value of the annuity one year before the
that
its
payment.
payment.
value of the annuity at the time of the fifth payment.
number of years the annuity would have to run in order
value of the annuity one year after the
current present value be doubled.
The number of years
that
its
first
last
the annuity
would have
current present value be tripled.
to run in order
67
Annuities
3-7.
Prove each of the following
identities:
= «h + """"ni
= «;^ - ^%
(a)
«^t;^
(b)
a-z;^
(d)
^;^ =
-(!+'• r^H
^;;^
3-8.
Give verbal interpretations for each of the
3-9.
Prove that
^+
=
;|r
identities in
Question
7.
/.
3-10. Prove each of the following identities:
(a)
^=
(b)
^-^
=
1
+ ^;r:n
^^TT|
-
1
3-11. Give verbal interpretations for the identities in Question
3-12.
Rank
n, a-^
and
s-^ in
increasing order of magnitude.
conditions will equality hold for
all
10.
Under what
nl
end of 25
end of each of the
the end of each of the last 1 5 years,
3-13. Harriet wishes to accumulate 85,000 in a fund at the
years.
If she deposits
first
years,
1
and
1
1000
000
find X if the fund earns
3-14.
Showthat^ +
4-
x
7%
in the
at
fund
at the
effective.
^-^ =
l.
3-15. Prove each of the following identities:
+ X-v"
(a)
a^^^a-^
(b)
s-^^s-^-X+iX+i)"
3-16. Give verbal interpretations for the identities in Question
b
3-17.
Show
that
^(^^ -
s-^
=
5^ -^ai' (^+1-^)-
15.
Chapter 3
68
3-18.
An
annuity runs for 25 years as follows:
first
ten years 500
years 300
is
years before the
3-19.
is
paid.
Edward buys
a
paid,
If
first
/
=
and then
at the
at the
.08, find the
end of each of the
end of each of the
last 15
value of this annuity three
payment.
new house and
To
takes out a mortgage of 60,000.
pay off the mortgage, he will make monthly payments with the
payment due in one month. Given p-"^ = .12, find the amount
of his payment if (a) the payments will continue for the next 25
years; (b) the payments will continue for the next 20 years; (c)
the payments will continue for the next 10 years.
first
3-20.
Rework Question 19
semiannually
3-21.
A man
16%
is
if
the nominal rate of interest convertible
instead of 12%.
wishes to accumulate a small pension by depositing 2500
the beginning of each year for 25 years.
year in which the final deposit
3-22.
is
made, he
will
make 20 annual
withdrawals.
Find the amount of each withdrawal,
during the
25 years and
A
series
first
of «
+
1
/
=
payments are made as follows:
\.
Show
3-23. Give a verbal explanation of
if
=
/
.07
.11 thereafter.
the first year, 2 at the end of each of the next n
the end of year n-\-
at
Starting at the end of the
1
at the
end of
years, and
1
at
of these payments
at
the formula in Question 22
is
that the value
why
—
1
correct.
3-24.
An
annuity consists of « payments of
1,
the first to be
end of 7 years and the other payments to be made
intervals thereafter.
Show
made
at the
at three
year
that the present value of the annuity
is
69
Annuities
3-25. Albert Glover, star third
baseman with the Blue
Jays,
is
given a
choice of contracts:
3,200,000 per year for the next five years, payable
(a)
at the
end
of each year.
3,000,000 per year for the next five years, payable
(b)
at the
beginning of each year.
1,800,000 per year for the next ten years, payable at the end
(c)
of each year.
If
/
=
value of each of these contracts at the begin-
.04, find the
ning of the
first
Repeat for
year.
=
/
3-26. Find the range of interest rates for
.06.
which each of the contracts
in
Question 25 has a higher present value then the other two.
where ? payments of
3-27. Consider an annuity
occurring k years from
now
1
are made, the first
with the payments continuing at A:-year
intervals thereafter, until a period
ci
that the present value
3-28.
Show
that the
Prove
of n years has passed.
of these payments
is
—
equal to -^.
accumulated value of the annuity
in
Question 27
s-\
immediately after the
last
payment
is
-p
s-i
3-29. Give verbal interpretations for the formulae in Question
27 and
Question 28.
3-30. Prove that the present value of an annuity
of each
m
of a year for the next n years
present value
is
which pays
is
equal to
^
at the
j^
.
the time of the last
is
This
denoted by a'^K
3-31. Prove that the accumulated value of the annuity in Question
value
end
payment
(\Jr-iY
—
1
^^
is
-/^^
fm)
.
30
at
This accumulated
denoted by s'^K
3-32. Derive an expression for the present value of an annuity under
which payments are
next 25 years.
2,
1,2,
1,
.
.
.
at the
end of every year for the
70
Chapter 3
3-33. Ifo-^
3-34.
A
—
X and
a^;;-^
loan of 25,000
— y,
to
is
express (^as a function of x and>^.
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<^. Once
so the outstanding
we would
that
is
The reader should
done,
have
first
we would
verify that this
D
1456.86.
4.2
7000 loan
is
being paid off with payments of 1000
at the
end of each
year for as long as necessary, plus a smaller payment one year after the
last regular
payment.
after the loan
is
If
taken out,
/
=
payment
due one year
find the outstanding principal, P, immediately
.11
and the
first
is
payment.
after the ninth
[Solution]
1000
1000
1000
-^
T
T
7000
P
FIGURE
In this case the retrospective
the necessity of finding the
smaller payment.
By
in
«
\
77+
1
4.2
method is more convenient, since it avoids
number of payments and the amount of the
We obtain P -
this point
?
\
9
2
1
1000
\
\
\
7000(1.
1
1)^
the text, the reader
=
-
lOOO^^,
is
certainly
3742.29.
D
aware of the
importance of being able to convert a given interest rate to an equivalent
rate with a different conversion frequency.
trates this point
The following example
with respect to a mortgage loan.
illus-
Amortization and Sinking Funds
Example
11
4.3
John takes out a 50,000 mortgage on a home
He
annually.
the first one due one
after his 60^^
12|%
at
convertible semi-
pays off the mortgage with monthly payments for 20 years,
month
mortgage
after the
is
taken out. Immediately
He
payment, John renegotiates the loan.
agrees to repay
by making an immediate cash payment of
10,000 and repaying the balance by means of monthly payments for ten
years at 11% convertible semiannually. Find the amount of his new
the remainder of the mortgage
payment.
Solution
First
we have
rate
of interest j equivalent to the nominal rate
(1
+y)6
=
1
payment X. To do
to find the old
+
X= ^M^ =
J25^
3Q
^j^^^
we
Next,
557.05.
J
^
(1 0625)^/^
we need
that,
-
/^^^
=
Then
1.
a monthly
.125.
Thus
we have
will find the outstanding principal P,
"2401
immediately after the 60^^ payment.
prospective method
P=
557.05ajgQj
=
is
Either
Since
easier.
180
Now we just
45,954.19.
method will work, but the
payments remain, we have
consider this as a
new
loan
Actually, the amount to be repaid is 10,000 less, namely
The new monthly rate is j = (1.055)^^^ - 1. Note that the
new payment is made for only 120 months, so it is equal to
of amount P.
35,954.19.
35,954.19
-
490.32.
1201
4.2
AMORTIZATION SCHEDULES
To begin
interest
at
how
with, let us study
and principal
a certain time
First,
t.
principal, P, at time
t
—
\,
a
payment can be divided up into its
say a payment of X is made
the amount of outstanding
As an example,
we must know
parts.
one period before the time we are interested
in.
X
\
t-
\
t
1
T
p
FIGURE
4.3
Chapter 4
78
The amount of
iP,
so that
on the loan during
interest earned
the interest portion of
is
X
this period is equal to
Hence the other
portion,
X— iP,
goes toward repayment of principal.
Example 4.4
A
1000 loan
payment.
repaid by annual payments of 150, plus a smaller final
is
If
/
=
.11,
and the
first
payment
is
made one year
time of the loan, find the amount of principal and
after the
interest contained in
the third payment.
Solution
150
150
150
1
1
1
1
2
1
3
T
1000
IFIGURE
= 2 is, by the retrospective method, equal
=
915.60. Hence the interest portion at = 3 is
15052J
= 3 is
(915.60)(.l 1) = 100.72, and the principal portion at
50 - 00.72 = 49.28.
D
The outstanding
to 1000(1.1 1)^
equal to
equal to
4.4
principal at
t
-
/
/
1
1
For certain types of problems,
it
is
convenient to use some easily
developed formulae for dividing payments into principal and interest
portions.
The reader should be
careful,
however, to use these formulae
only in situations where they are applicable.
Consider a loan which
1
for
is
being repaid by equal annual payments of
n years.
1
1
1
T
FIGURE
Such a loan has a value of
a-,
4.5
one year before the
outstanding principal at time
t is
fore the interest portion of the
(/
+
first
payment.
(prospectively) equal to a-^^.
1)^^
payment
'a;^=l-v"-',
is
The
There-
equal to
(4.1)
Amortization and Sinking Funds
and the principal part
is
1
More
79
-(l-v"-0 =
generally, if a loan
is
v"-'
(4.2)
being repaid by equal payments
payment
for n years, then the interest part of the k'^
,n-k+\
Xi
of^
will be
(4.3)
),
and the principal part will be
Xv'
Now
us see
let
how
-k+\
(4.4)
to construct an amortization schedule for
repayment
of a loan, such a schedule being simply a table which shows
payment
is
divided into principal and interest.
how
can be very useful pictorial tools, as they visibly demonstrate
interest portion decreases
goes on. They also show
as
each
Amortization schedules
how
the
and the principal portion increases as time
how
the outstanding balance decreases to zero
payments are made.
As an example, consider
amount
that the
X=
X of
1387.05. Here
is
5000 at 12% per year to be
due one year hence. We know
a loan of
repaid by 5 annual payments, the
first
each payment
given by Xaj,
is
=
5000, so that
an amortization schedule for this loan.
TABLE 4.1
Payment
Interest
1
1387.05
600.00
2
1387.05
3
1387.05
4
1387.05
5
1387.05
Duration
Principal
Outstanding
Repaid
Principle
5000.00
It
is
easy to see
is
4212.95
505.55
881.50
3331.45
399.77
987.28
2344.17
281.30
1105.75
1238.42
148.61
1238.44
a table
was
constructed.
The outstanding
is
5000, the amount of the loan. The amount of each
listed in
Column 2 opposite the appropriate duration. The
Column 3 is obtained by multiplying 5000 by
principal at time
payment
how such
787.05
entry of 600 at the top of
.12.
We
then subtract 600 from 1387.05 to obtain 787.05, and then
Chapter 4
80
ISl. 05
is
4212.95.
subtracted from 5000 to obtain the
The procedure
is
now
new
repeated (using 4212.95 as the out-
standing loan balance) to obtain the next row.
outstanding principal
is
outstanding principal,
zero; notice in our
This continues
example
second decimal place has led to a small imbalance
In general the rules to obtain
until the
that rounding off the
at the last stage.
an amortization schedule are as
follows:
Take the entry from Column 5 of the previous row, multiply
it by /, and enter the result in Column 3.
I.
2
— Column
5
of previous row
III.
Column
Column
IV.
Continue.
II.
Example
3
= Column 4.
— Column
4
=
Column
5.
4.5
Construct an amortization schedule for the loan described in Example 4.4.
I
Solution]
One way
would be to first find the duration and amount of
payment and to proceed as above. Alternatively, we
can just start forming the amortization schedule and worry about the
final payment later! Here are the first 12 rows of this schedule.
to solve this
the smaller final
TABLE
Duration
Payment
Interest
4.2
Principal
Outstanding
Repaid
Principle
1000.00
1
150
110.00
40.00
960.00
2
150
105.60
44.40
915.60
3
150
100.72
49.28
866.32
4
150
95.30
54.70
811.62
5
150
89.38
60.72
750.90
6
150
82.60
67.40
683.50
7
150
75.19
74.81
608.69
8
150
66.96
83.04
525.65
9
150
57.82
92.18
433.47
10
150
47.68
102.32
331.15
11
150
36.43
113.57
217.58
12
150
23.93
126.07
91.51
Amortization and Sinking Funds
Note
8
that at this point the outstanding principal
is
=
91.51, so the interest
The
required in the next payment
is
(91.51)(.l 1)
13'^
payment
is
91.51, in order that the balance be
Hence
the final
required in the
brought to
row
0.
13 of the schedule
I
Of course,
as
is
10.07
+ 91.51 ^
principal
101.58, and
would be
101.58
13
I
payment
10.07.
91.51
10.07
I
we have
D
.
|
|
I
seen earlier,
it
is
not necessary to construct
the entire amortization schedule if one simply wants to divide any one
payment into principal and interest portions.
should follow the method demonstrated in Example 4.4.
In that case
particular
4.3
one
SINKING FUNDS
Instead of amortization, an alternate
way of repaying
a loan
is
for the
borrower to pay the interest on the loan as it comes due, keeping the
amount of the loan constant, and then repaying the principal by a single
lump-sum payment
at
some point
strated in Figure 4.6, for a loan
in the future. This method is demonof amount L with interest at rate /.
L
iL
\
\
1
iL
iL
^-
\
n
2
T
L
[FIGURE
4:6]
This lump-sum payment of principle
is
often accumulated by
periodic deposits into a separate fund, called the sinking fund.
Usually
the sinking fund will accumulate at a rate of interest different
(usually smaller than) that charged by the lender.
From
point of view, the total outlay at any point of time
is
from
the borrower's
the
sum of
the
payment, made directly to the lender, and the sinking fund
payment. Observe, however, that the lender is oblivious to the sinking
interest
fund; she only cares that she receives her regular interest payments and
her principal back at the end of the period.
If the
borrower
is
able to obtain a higher rate of interest on the
sinking fund than that charged on the loan, then his total annual outlay
will be less than
method.
On
if
he were repaying
this
loan by the amortization
the other hand, a lower rate of interest in the sinking fund
implies a higher annual outlay for the borrower.
82
Chapter 4
Let us consider some examples.
Example
4.6
John borrows 15,000 from a trust company at 17% effective annually.
He agrees to pay the interest annually, and to build up a sinking fund
which will repay the loan at the end of 15 years. If the sinking fund
accumulates
12%
at
annually, find (a) the annual interest payment;
(b)
the annual sinking fund payment; (c) his total annual outlay; (d) the
annual amortization payment which would pay off this loan
in 15 years.
ISolutionI
(a)
(b)
The annual interest payment is 15,000(.17) = 2550.
The annual sinking fund payment X is given by
^^T5|.i2
(c)
(d)
His
=
total
15,000,
soX^
annual outlay
is,
the equation
402.36.
therefore,
T=
2952.36.
The amortization payment K would be given by the equation
Kaj^^^^= 15,000, so that a: -2817.33.
K<
which
Observe
that
loan
17%, whereas the sinking fund only earns
is
T,
is
not surprising since the interest rate on the
12%
D
annually.
Example
interest at
4.7
Helen wishes to borrow 7000 to buy a car. One lender offers a loan in
which the principal is to be repaid at the end of 5 years. In the meantime, interest at 1 1 % effective is to be paid on the loan, and the borrower
is to accumulate her principal by means of annual payments into a
sinking fund earning 8% effective. Another lender offers a loan for 5
years in which the amortization method will be used to repay the loan,
with the first of the annual payments due in one year. Find the rate of
interest, /, that this second lender can charge in order that Helen finds
the two offers equally attractive.
ISolutionI
Consider the terms offered by the
first lender.
Helen's annual interest
=
payment is 7000(.l 1)
770, and her annual sinking fund payment Xis
Hence Helen's total
given by Xsj^q^ =^ 7000, so that
1193.20.
X=
1963.20.
She would find the two offers equally
were also her amortization payment to the second
Hence the rate of interest we want is the rate / for which
1 963.20(337.
Using our calculator, as described earlier in the
annual outlay
attractive
lender.
7000
=
text,
we
is
if this
find
/
=
.124425.
D
Amortization and Sinking Funds
We
the
first
is
higher than either rate involved
in
This reflects the fact that the sinking fund
is
note that this interest rate
lender's offer.
earning interest
outlay
83
at a rate less
greater than
than that of the loan, so Helen's annual
would be
it
by the amortization method.
paying is greater than 1 1%.
is
if
she were paying the
first
lender back
Hence, the actual rate of interest she
is
YIELD RATES
4.4
Consider an investor
who makes
time, and receives other
a
number of outlays
payments
in return.
which the value of his expenditures
interest rate for
of the payments he receives, when
all
at
There
various points in
is
(at least)
one
will equal the value
are considered at the
same point
in
on his investment.
We have seen a number of examples of yield rates. If a loan
company lends Harry 10,000 which he pays back by the amortization
method at 15% per year, then the yield rate earned by the company is
15%. Exercises 26(c) and 31(b) in Section 4.3 are examples of yield
time. This rate
is
rates in reverse;
that a
borrower
called the yield rate he earns
we
is
calculate, in those situations, the real rate of interest
paying on a loan. Also,
in
Exercise 2-15,
we
calcula-
ted an investor's yield rate, without using that term.
Here are several examples of problems involving yield rates. The
remember is that only the payments made directly to, or
crucial thing to
directly by, the person involved should be considered in evaluating that
person's yield rate.
Example
4.8
Herman borrows 5000 from George and agrees
annual installments
at
1
1%, with
George sells his right to future
yield Ruth 12% effective.
Find the price Ruth pays.
years,
will
(a)
first
(b)
Fin d George's overall yield
to repay
payment due in one
payments to Ruth, at
it
in
year.
10 equal
After 4
a price
which
rate.
[Solution]
First,
we need
to find the
PaToi,, =5000, so
that
amount of each installment P. This
P=
|^ =
849.01.
is
given by
Chapter 4
84
(a)
before, we should concentrate attention on the
payments involving Ruth. She pays a certain amount of money, X,
As mentioned
and
due
X=
(b)
in return she receives the final
one year.
in
If
=
849.01^6112
she
6 payments of 849.01, the
to
is
first
12%, we must have
earn
3490.63.
George pays out 5000 at time 0. In return, he receives 4 payments
of 849.01, and, at the same time as the last payment, he receives an
additional 3490.63 from Ruth.
849.01
1
1
1
1
1
1
3490.63
849.01
849.01
849.01
1
1
1
2
3
4
T
5000
FIGURE
His yield rate
is
4.7
the rate of interest at which these payments
+ 3490.63vl This
We shouldn't
balance, which implies that 5000 -: 849.01^4,
equation
is
solved by calculator giving
be surprised that the answer
is
less
/
than
=
1
.10526.
1%; the
fact that
Ruth
is
earning a higher rate of interest than George on the annuity means
that
George has
Herman
is
something by
to be losing
oblivious to
all
paying 11%, pure and simple,
Who is George Herman Ruth?
Example
selling.
Of
course,
the financial wheeling and dealing; he
until the loan is repaid.
is
Query:
D
4.9
At what yield
rate are
payments of 500 now and 600
equivalent to a payment of 1098 at the end of
1
at the
end of 2 years
year?
Solution
We
wish
to solve
500
+
600v2
=
1098v, or 300v^
Using the quadratic formula, we find v
V
=
^^^
or
^^^.
answer to a question
Hence
like this
/
=
=
-^k
.023 or .173.
need not be unique!
- 549 v + 250 =
We
,
0.
which gives
observe that the
D
Amortization and Sinking Funds
85
Example 4.10
Henri buys a 15-year annuity with a present value of 5000
price
which
at
9%
at a
him to accumulate a 15-year sinking fund to
7%, and will produce an overall yield rate of 10%.
will allow
replace his capital at
Find the purchase price of the annuity.
Solution
We
must be very clear what
this annuity.
fmd
is
going on here. Henri pays a price, P, for
^^KioQ
i^l.uy
=
We
he gets payments for 15 years.
In return,
the amount, X, of each payment.
X==
5000, so that
^^ =
must
first
This comes from the equation
620.29.
We know
that the price,
"15|.09
7%
P, must be replaced by a
fund deposit
7^— =
is
10%, we need. IP
+
sinking fund in
.0397946247P.
.0397946247P
=
1
5 years, so the sinking
In order that the yield rate be
620.29. Finally,
P=
D
4437.15.
EXERCISES
4.1
4-1
Amortization;
A
4.2
loan of 50,000
years at
six
13%
months
is
Amortization Schedules
being repaid with semiannual payments for 10
convertible semiannually.
first
payment
is
due
after the loan is taken out.
Find the outstanding loan balance
(a)
The
at the
end of the sixth
year.
Divide the
(b)
4-2.
4-3.
13^^
payment
into principal
and
A
interest.
loan is being repaid by 20 annual payments. The
ments are 300 each, the next 8 are 400 each, and the
=
first 5 install-
last
7 are 600
each.
Assume
(a)
Find the loan balance immediately after the tenth payment.
(b)
Divide the
A
loan
is
/
1
.14.
\'^
payment
into principal
and
interest.
being repaid by monthly payments of 200, the
first
due
one month after the loan is taken out, along with a smaller final
payment. If / = .11 and the loan balance at the end of 1 8 months
is 5,000, find the amount of the original loan.
86
4-4.
Chapter 4
A
loan of 1000
first
due
is
being repaid by annual installments of 200, the
one year, and a smaller
in
payment.
after the last regular
If
final
i=
payment made one year
.16, find the
outstanding
principal immediately after the 4^^ payment.
4-5.
A
loan
is
repaid by level payments at the end of each month.
principal outstanding on
was R; on May
1,
1998,
May
was
S;
The
was Q\ on May 1, 1997,
on May 1, 1999, was T. Determine
1,
1996,
whether or not each of the following
(b)
(c)
4-6.
true:
+ RXS-\-T) = {R + Sf
{Q-R){S-T) = {R-Sf
{Q
Garfield
is
repaying a debt with 20 annual payments of 1000 each.
makes an extra payment of 2000.
remaining payment period by two years, and
At the end of the fourth
He
is
Q+T<R-^S
(a)
then shortens his
makes
level
year, he
payments over
that time. If
/
=
.12, find the revised
annual payment.
4-7.
A
90,000 mortgage
is
repaid by payments at the end of each
for the next 25 years.
The
rate
of interest
ii
11^%
2
is
month
convertible
semiannually.
(a)
(b)
first payment into principal and interest.
Find the outstanding principal immediately after the 75^^
Divide the
payment.
(c)
Divide the 76^^ payment into principal and
(d)
Find the
total
amount of
interest.
interest paid during the life
of the
mortgage.
(e)
The borrower
is
temporarily unable to
through 94 inclusive.
ments so
4-8.
A
He
make payments 76
then wishes to increase his pay-
mortgage will
still
be paid off
scheduled time. Find the amount of the
new payment.
loan of 1000
years at
5%
the loan
is
installment
that the
is
being repaid with annual installments for 20
effective, with the first installment
taken out.
is
30.98.
at the
Show
that the
amount of
due one year
interest in the
after
1
V^
Amortization and Sinking Funds
4-9.
A loan
= .17.
being repaid with 30 equal annual installments at
what installment are the principal and interest portions
is
In
/
most nearly equal
4-10.
A
loan
to
each other?
repaid by 20 equal annual payments at
amount of
the
4-11.
is
87
principal in the 4^^
is
effective.
150, find the
in the 12'^
A
being repaid with 30 equal annual installments.
loan
is
payment.
payment
is
is
352.87. Find
A
is
being repaid with 20 annual installments of
loan
/
The
247.13, and the interest
portion
at effective rate
If
amount
of interest
principal portion of the 11^^
4-12.
payment
1%
1
/.
for the first
1
years,
and j for the
1.
last
Interest
1
is
years.
Find an expression for each of the following:
The amount of principal repaid in the eighth installment.
The amount of interest paid in the last installment.
(a)
(b)
4-13. In order to pay off a loan of ^,
end of each year.
rate
/,
and
balance
4-14.
The
at the
original
on the
first
on the excess
is at
Interest
interest
Herman makes payments of Xat
end of the
r^^
year,
B
of the unpaid balance
rate/
assuming
it
amount of an inheritance was
the
is at
Find the outstanding
to be
more than B.
just sufficient at
8%
end of each year for 10 years. The
payments of 5000 were made for the first 5 years even though the
fund actually earned 10% effective. How much excess interest
effective to
was
in the
4-15. George
pay 5000
fund
at the
at the
end of 5 years?
was making annual payments of Xon a 16% 10-year
loan.
After making 4 payments, he renegotiates to pay off the debt in 3
more years with the lender being
4-16. Harriet
is
repaying a loan with payments of 3000
every two years.
/
=
.13.
amount of interest
amount of principal
If the
2,982.31, find the
Assume
14% over
new payment.
satisfied with
entire period. Find an expression for the
in the
in
the
at the
5^^
8'^
the
end of
installment
is
installment.
Chapter 4
88
4-17.
Henry borrows 5000
14%
at
per year, and wishes to pay
with 6 equal annual payments, the
first
due
in
it
back
one year. Construct
an amortization schedule for this loan.
4-18. Construct an amortization schedule for the loan in Question
1.
4-19. Construct an amortization schedule for the loan in Question 2.
4-20. Construct an amortization schedule for the loan in Question 4.
4-21. Consider a loan
of
1,
(a)
the
first
which
is
being repaid by n equal annual payments
due one year
Assuming «
>
5,
after the loan
construct the
is
taken out.
first 5
rows of the amortiza-
tion schedule.
(b)
Assuming n >
(c)
Construct the
20, construct the 20^^
row of the amortization
schedule.
(assume n
4-22.
Assuming
4-23.
Use
>
last
3
rows of the amortization schedule
3).
a loan is to be repaid by equal payments, write a
computer program which will output an amortization schedule for
such a loan. Test your program by redoing Questions 17 and 18.
the program of Question 22 to produce an amortization
schedule for Question
7.
4-24.
Modify the program of Question 22 so that it can handle problems
like Question 19. Test your program by redoing Question 19.
4-25.
Modify the program of Question 22 so that it can handle problems
like Question 20. Test your program by redoing Question 20.
Amortization and Sinking Funds
4.3
4-26.
89
Sinking Funds
A
loan of 10,000
of
tive rate
is
taken out on
of
interest
8%
March
per year.
1,
1995, at an effec-
Interest
is
paid annually,
is established to repay the principal on March
Payments are made annually into the fund beginning on
March 1, 1996, and the fund earns interest at 9% per year.
Find the amount of each payment made to the sinking fund.
(a)
Find the total amount the borrower must pay each year.
(b)
From the point of view of the borrower, what rate of interest
(c)
and a sinking fund
1,
2002.
she really paying each year?
is
4-27.
A
Repayis taken out at 1 1% per year effective.
by the amortization method, with equal payments at the
end of each year for the next 20 years. Immediately after the 10^^
payment, the borrower renegotiates with the lender, and they agree
that the remainder of the loan will be repaid by the sinking fund
method, where the interest rate on the loan increases to 12% per
year and the sinking fund earns interest at 14% per year. Interest
payments and sinking fund deposits will still be annual and
continue for 10 more years. Find the sinking fund deposit, and
compare the borrower's new total payment with his old one.
loan of 50,000
ment
4-28.
A
is
loan of 1000
The
years.
and
/',
is
rates
to be repaid
by the sinking fund method over 10
of interest on the loan and the sinking fund are
The borrower's
respectively.
total
payment each year
/
is
125.
=
(a)
If/
(b)
What
4-29. Kelly has
/',fmd/.
is
the
maximum
possible value for /?
borrowed 1000 on which she
effective per year.
She
is
is
paying interest
At the end of the eighth
borrower makes a total payment of 149.06.
(a)
How much of the 149.06 pays interest on the loan?
effective to repay the loan.
(b)
How much
(c)
What
is
In
11^%
at
9%
year, the
of the 149.06 goes into the sinking fund?
the sinking fund balance at the end of the eighth
year?
(d)
at
accumulating a sinking fund
which year
will the principal be paid off?
90
Chapter 4
9%
4-30. Walter borrows 5000 for 20 years at
11%
years and
interest yearly
annual payment
If his total
much he
(b)
He wishes
to
pay
and to repay the principal by annual payments into
a sinking fund earning interest at rate
(a)
effective for the first 10
effective for the last 10 years.
is
/.
is
570 and
/
=
how
.09, find
short of repaying the loan at the end of 20 years.
If his total annual
payment
is
570, find
/
such that the loan
is
exactly repaid after 20 years.
(c)
If
/
=
.10
and
amount of
his total annual
that
payment
in
payment
is
constant, find the
order that the loan be exactly
repaid after 20 years.
(d)
If
/
=
and the sinking fund deposit
.10
amount of
is
constant, find the
that deposit in order that the loan
is
completely
repaid after 20 years.
4-31. Ashley borrows
interest yearly,
3000 for 10 years at 13% effective. She pays the
and the principal by means of two sinking ftands.
is repaid by a sinking fund earning 16%)
and the other two-thirds by a sinking fund earning 1 1%)
One-third of the principal
effective,
effective.
(a)
Find Ashley's
(b)
Determine the
total
rate
annual payment.
of
interest she is really
paying on her
loan.
(c)
Redo
(a) if
Ashley puts one-third of her
4.4
Yield Rates
4-32.
A
borrows
1000 from B, and agrees to repay
installments at
18%
B sells her right to
C 19% effective.
price which C pays to B.
year. After 3 payments,
a price which yields
(a)
Find the
(b)
Find the overall yield rate to B.
(c)
Find the overall yield rate to C.
it
11%
in
fund.
8
payment due
future payments
effective, with the first
fund
total sinking
deposit in the 16%) fund and two-thirds into the
equal
in
one
to
C at
Amortization and Sinking Funds
91
Assume, in
more payments, C sells his right to ftiture
at a price which yields B 20% effective on
4-33. Consider the transactions described in Question 32.
addition, that after 3
payments back to
5
those remaining payments.
Find the price which
(a)
B now pays to
Find the overall yield rate to
(b)
C.
C
on the
entire set
of transac-
B
on the
entire set
of transac-
tions.
Find the overall yield rate to
(c)
tions.
In
(d)
each of parts (b) and
earned
if
(c),
the rate of inflation
fmd
is
the real rate of interest
5%
per year.
(See end of
Section 3.6.)
4-34. Ellen
is
repaying Friendly Trust a 10,000 loan with 8 equal annual
payments of principal, the
In addition, she
out.
first
due one year
pays interest
standing principal each year.
An
payments from Friendly Trust
11^%
effective.
at
11%
after the loan
effective
is
taken
on the out-
investor wishes to purchase these
at
a price which will yield her
Find the price.
which should be paid for an annuity of 500 per year
is to be 11% and if the
principal can be replaced by a sinking fund earning 8% per year
for the next 6 years and 7%) per year for the following 4 years after
4-35. Find the price
for the next 10 years, if the yield rate
that.
made 6 annual payments of 500 each on
3000 loan
was
taken out, the borrower decides to repay the balance of the loan
over the next 5 years by equal annual payments of principal in
addition to the annual interest due on the unpaid balance. If the
lender insists on a yield rate of 14% over the final 5-year period,
4-36. After having
at
1 1
%
effective, with first
payment one year
a
after the loan
find the total payment, principal plus interest, for the tenth year.
Chapter 4
92
4-37. Seven years ago, Jean took out a 20-year 30,000 loan at
8%
on which she was making annual payments, with the
first
effective
payment due one year
after the loan
was taken
out.
She
now
wishes to make a lump-sum payment of 6000, and then pay off the
loan in 5
more
years.
Find the revised annual payment under each
of the following situations:
(a)
(b)
The lender
The lender
is
satisfied with earning
is
satisfied with
but insists on an
(c)
The lender
loan.
11%
insists
8%
8%
effective.
effective for the past 7 years,
yield for the next 5 years.
on an 11% yield
for the entire life of the
CHAPTER FIVE
BONDS
PRICE OF A BOND
5.1
When
money, it issues bonds
bond is a certificate in
which, in return for receiving an initial sum of money from the investor,
the borrower agrees to pay interest at a specified rate (the coupon rate)
until a specified date (the maturity date), and, at that time, to pay a fixed
sum (the redemption value). The coupon rate is customarily quoted as a
nominal rate convertible semiannually, and is applied to the/ace (or par)
value, which is stated on the front of the bond. Usually the face and
and
a corporation or
sells
them
government needs to
to a large
number of
redemption values are equal, but
may
This
this is not
appear complicated
nothing fancy or
new going on
raise
investors.
A
always the case.
at first glance,
here.
It's really
but there
is,
in fact,
a familiar story:
in
some money, the investor obtains regular
interest payments and a final lump sum payment at the end of the term.
For example, consider a bond of face amount 500, redeemable at its par
value in 10 years with semiannual coupons at rate 11%, compounded
semiannually. When an investor buys this bond he receives, in return,
20 half-yearly payments of (.055)(500) — 27.50 interest, and a lump sum
payment of 500 at the end of the 10 years.
The investor's price for this bond may differ from 500. If she is
able to buy it for less than 500, then we know from Section 4.4 that her
return for lending the borrower
yield rate (the rate of return on her investment) will be greater than
If she
pays more than 500, her yield rate will be
less
than
1
11%.
1%.
Let us introduce the notation to be used in this section, and also
take this opportunity to review
F =
=
r
some
basic terminology:
the face value or par value of the bond.
the
coupon
rate
rate per interest period.
Throughout
this section, the
quoted will usually be a nominal rate 2r convertible semi-
Observe that the amount of each semiannual
payment (coupon) is Fr.
annually.
interest
94
Chapter 5
C =
the redemption value of the bond.
"redeemable
—
=
P =
/
n
Often
C=
F, and the phrase
at par" describes this situation.
the yield rate per interest period.
number of interest periods until the redemption date.
bond to obtain yield rate /.
the
the purchase price of the
Our knowledge of
annuities easily gives us a formula for P, in
terms of the other quantities described above.
C
Fr
Fr
Fr
1
1
n
2
1
T
p
FIGURE
In return for paying out
P
at
time
receives n coupons of
0, the investor
value Fr each and a fmal payment of
yield rate
5.1
C
at
time
n.
Hence, to obtain a
/,
P=
The student
{Fr)a-^^
+ C{\^-ir\
will be asked to derive a variation
(5.1)
of the above formula
in
the exercises.
Example
A
5.
13%
per
year convertible semiannually. Find the price to yield an investor (a)
8%
bond of 500, redeemable
effective per half-year; (b)
1
par after 5 years, pays interest at
at
6%
effective per year.
[Solution]
(a)
We
=
We
/
(b)
Formula
use
.08,
and obtain
32.
the
5a^.
In each part
less than the
is
+
yield
rate
500(1+0"^^
of the
last
F = C = 500, w =: 10, r = .065,
+ 500(1. 08)- ^^ - 449.67.
F=C= 500, w = 10, r =- .065 and
= (1.16)'/^ — 1. Then we have
= 459.08.
D
with
'H.Sa^yQ^
use Formula (5.1) with
calculate
P-
(5.1)
P=
/
example, the investor
redemption value
(i.e., at
is
buying the bond for
a discount), because the yield rate
higher than the coupon rate. If the coupon rate were higher, she would
Bonds
95
have to buy the bond
Example
A
premium.
at a
In that case
we would have P >
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)
<
have
Pr{X)
<
1,
and
never occurs, whereas
if
We now give several examples.
6.1
Find the probability that one card drawn from a deck of 52
is
a Jack.
ISolutionI
Since there are 52 cards, 4 of which are Jacks, the answer
Example
is
A = tt- D
6.2
Find the probability that two cards drawn from a deck are both Jacks.
[Solution Onel
Imagine drawing the cards one
card
is
a Jack
is
yW.
at
a time.
The
probability that the first
After that, there are 51 cards
left
of which 3 are
T
Jacks.
Hence
The answer
lities to
is
the probability that the second card
pr x yU
=
jjr, where
we
is
a Jack
is :?y
=
1
-rW.
multiply the separate probabi-
obtain the probability of both events occurring.
Preparation for L ife
Solution
Con tingene ies
115
Twd
There are
2
(
)
^^ys of choosing 2
choosing 2 Jacks out of the 4
in the
cards, and there are
Hence
deck.
f
2
answer
the
)
ways of
is
given by
(2)
V 2 j
ways of choosing
Example
A fair die
r items out
D
of «.
6.3
is
thrown. Find the probability of getting a
5.
Solution
There are six possible outcomes, so clearly the answer
Example
D
i
is
6.4
A
fair die is thrown twice.
numbers obtained is 9.
Find the probability that the
total
of the
[Solution!
This
is
more
the answer
false
interesting.
is
The beginner might be tempted
jj, since there are 11 possible
because the different totals are not
all
to think that
However
totals.
equally likely.
this is
Instead
should concentrate on the 36 different possible pairs of numbers
Of these,
could obtain from the two throws.
4
+ 5, 5 + 4, 6 + 3).
Example
Hence
the answer
4 pairs add up to 9 (3
+
^ = ^.
is
we
we
6,
D
6.5
A
fair die is thrown twice.
Find the probability
numbers obtained is 4 or larger.
that the total
of the
[Solution]
We
could just
Observe
list all
that the
the possibilities, but there
is
an easier approach.
probability that the total equals 2
probability that the total equals 3
probability that the total equals
bility that the total is
4 or
more
is
^
(1 H-
2or3is^ +
1
is
equal to
1
2 or 2
?
is
+
Jz, and the
1).
^ = jW, and
— tU =
Hence
the
1
^4.
the proba-
D
Chapter 6
116
Example
A
6.6
bag contains 4 white,
black and 3 green
5
balls.
Determine the
probability of each of the following:
Drawing
Drawing
Drawing
Drawing
Drawing
Drawing
(a)
(b)
(c)
(d)
(e)
(f)
a green ball
a white or a black ball.
two white
two white
balls (no replacement).
balls (with replacement).
a green ball followed by a black ball (no replacement).
a green ball, then a black ball (with replacement).
ISolutionl
(a)
The answer
is
(b)
The answer
is
(c)
The
is
^=
^•
=
\-
y2
probability that the
first ball is
no replacement, the second
white. Hence, the answer
(d)
The answer
(e)
We obtain^
(f)
The answer
Example
A
^s
=
i x i
\s
yj
—
k
ball has probability
^ x
Since there
Y[
of being
^ = tV.
i
H ^•
i x ^ = ^
x
""
D
6.7
insurance
life
is
^s
white
company determines
that the probability
of surviving
and 0.4 for persons aged 40, 50, 60 and
70, respectively. Determine each of the following:
(a)
The probability of dying in the next ten years for each group.
for ten
(b)
(c)
years
The
The
is
0.9, 0.8, 0.6
probability that a 40-year-old lives to age 80.
probability that a 50-year-old dies between ages 70 and 80.
[Solution]
(a)
We
use Pr{X)
=
1
—
Pr{X) to obtain the following
results:
= - 0.9 = 0.1.
— 0.8 = 0.2.
Pr( 5 0-year-old dies in next 10 years) =
— 0.6 = 0.4.
Pr(60-year-old dies in next 10 years) =
—
=
0.4 = 0.6.
Pr(70-year-old dies in next 10 years)
Pr(40-y ear-old dies
in
next 10 years)
1
1
1
1
(b)
In order for a 40-year-old to reach age 80, he
each
of the
intervals.
(0.9)(0.8)(0.6)(0.4)
=
Hence
0.1728.
the
must
required
live
through
probability
is
Preparation for Life Contingencies
In order that a 50-year-old dies
(c)
between age 70 and 80, she must
The answer is
first live to
70 and then die between 70 and 80.
(0.8)(0.6)(1
-0.4)
Another topic
that
=
several
D
0.288.
we need
expectation or expected value.
the
1 1
to explore at this time
If a particular
game
is
mathematical
or experiment has
outcomes x\, jc2, ..., and the corresponding
then we define
each of these outcomes \^f(x\),f{x2),
possible
probability for
.
.
.
,
the expected value to be
E=
That
is,
Xi/(X,
+ X2f{X2) + X3/(X3) +
)
the expected value
sum of terms
the
is
possible outcome and /(x,)
is
there will be only finitely
many
its
probability.
•
•
•
Xif(Xi),
In all
(6.3)
.
where
x, is
one
of our examples,
possible outcomes, so the expected
n
value
is
equal to
^
Expected value should be thought of as the
Xif{xj).
average value to be obtained
Example
in
a particular
game
or experiment.
6.8
you throw a single die, what is the expected value of the number of
dots which appears on the die's upward face?
If
ISolutionI
There are six possible values which could appear, and each has
Making use of Formula
probability i.
we
(6.3)
find the answer to be
(l)(g)
+ (2)(^) + (3)(i) + (4)(1) + (5)(i) + (6)(i) =
this
just an average value;
3
is
A dots cannot
Example
we
3^.
Note
shouldn't be bothered by the
that
fact that
D
actually appear in one throw!
6.9
Henry enters a betting game involving a
a row, he wins 15 dollars.
fair coin.
Otherwise, he loses
If
1
he can get 4
dollar.
tails in
What
is
his
expected value?
[Solution
The
]
probability of getting 4 tails in a
row
is
(
bility
of losing
15(-iV) -h
is
1
(— 1)(t^)
~ TZ —
=
0.
f^-
^
)
=
-iV,
so the proba-
Thus we have an expected value of
Therefore, in the long run, Henry should
expect to neither win nor lose.
D
Chapters
118
Example 6.10
Consider the data
lity
estimate,
we
will
Example
in
that a person
assume
also
between the endpoints of an
What
is
Assume,
6.7.
in addition, that the probabi-
aged 80 will survive ten years
the expected
is
0.
As
a rough
that all deaths occur exactly halfway
interval
(i.e., at
age 45, 55, 65, 75 or 85).
span of a 40-year-old?
life
Solution
We have the following for a 40-year-old:
Pr(dying at age 45) = 0.1
Pr(dying at age 55) = (0.9)(0.2) = 0.18
Pr(dying at age 65) = (0.9)(0.8)(0.4) = 0.288
Pr(dying at age 75) = (0.9)(0.8)(0.6)(0.6) = 0.2592
PKdying at age 85) - (0.9)(0.8)(0.6)(0.4) = 0.1728
Then
the expected value, using Formula (6.3),
55(0.18)
65(0.288) 4- 75(0.2592)
45(0.1)
+
+
Example
is
+
85(0.1728)
=
67.248.
D
6.
In a certain lottery, 5 million
1
-dollar tickets are sold each week.
The
following prizes are awarded:
5 prizes
of 100,000 each
20 prizes of 25,000 each
45 prizes of 5,000 each
100 prizes of 1,000 each
1950 prizes of 100 each
24,500 prizes of 10 each
Find the expected value of a single
ticket.
[Solution]
Since 5 million tickets are sold, the probability of winning 100,000
5
000 000
~
1
000 000
'
Similarly, the probabilities of winning the
other prizes are easily calculated.
'oO'OOo(5;ooo:ooo)
is
Hence
the expected value
is
+25.000(5^0^) +5000(5^0^)
+ 'OOo(5;oSoo) + '00(5W00) + 10(5^^0) =
This means that the expected return on a
1
dollar ticket
is
35 cents.
-353.
D
Preparation for Life Contingencies
19
1
Example 6.12
Take the data from Example 6.1 1 and assume, in addition, that 1,250,000
prizes of 1 free ticket on the next draw are awarded. Find the expected
value in this case.
Solution
This means
that,
tion as the ticket
E=
(previous
so that
J5E =
on any draw, 1,250,000
you purchase. So,
expectation)
and
.353,
4- £"(
E=
tickets
E
if
^'ooo'ooo
have the same expecta-
we
the expectation,
is
)
^~
^^^^
"
All. The expected return
is
obtain
+ .25£',
-^^^
now 47
cents.
D
CONTINGENT PAYMENTS
6.3
Henry borrows 1000 from Amicable Trust and agrees to repay the loan
in one year. If repayment were certain, the trust company would charge
13%
is
a
interest.
5%
From
prior experience, however,
Amicable Trust ask Henry
determined that there
This situation
is
What should
at all.
to repay?
company
value to be (.95)X-{- (.05)(0), where
X
first
is
the
determines their expected
amount
to be repaid.
done
be repaid and a
just a very special case of the type of calculation
section, since there
is
To
a good example of a contingent payment.
solve the problem, the finance
is
it is
chance that Henry will not repay any money
a .95 chance that
X will
company will receive nothing.
The present value of this expected value
This
in the last
.05
chance
that the
.95X(1.13)-', so
X=
1000(1. 13)(.95)-^
the trust
the
we have
13%
Trust
whom
is
=
.95(X)(1.13)-^
1
company charges Henry
originally given.
making a
We
189.47.
is
=
at the
time of the loan
1000, from which
we
is
find
observe that the rate of interest
18.947%, considerably higher than
This does not mean, however, that Amicable
fortune; if they
were
to lend
money
only 95 actually repay, then the company would
to
1
00 people of
make 13% on
its
investments.
Let us look at
some more examples.
Example 6.13
The All-Mighty Bank lends 50,000,000
to a small Central
country, with the loan to be repaid in one year.
It is
felt that
American
there
is
a
120
Chapter 6
20%
chance that a revolution will occur and that no money will be
repaid, a
30%
chance that due to soaring inflation only half the loan will
50%
be repaid, and a
repayment were
chance that the entire loan will be repaid.
certain, the
interest should the
What
bank would charge 9%.
If
of
rate
bank charge?
Solution
Let Jf be the amount to be repaid. Then the expected value
(.5)Z+
{.2>){\X)
which gives
+ (.2)(0) =
X= 83,846,153.85.
~
50 QQO 000
and
.65^,
The
rate
is
=
.65X(1.09)-'
50,000,000,
of interest charged
^^^' ^^ ^^^ country must pay 67.7% on
its
is
therefore
D
loan!
Example 6.14
Mrs. Rogers receives 1000
The
probability
years,
30%
is
80%
at the
she will survive one year,
50%
is alive.
she will survive 2
she will survive 3 years, and negligible that she will survive
longer than 3 years.
If the yield rate is
Rogers place on these payments
I
end of each year as long as she
15%, what value should Mrs.
now?
Solution!
We
solve this by finding the expected value of Mrs. Rogers' payments
for each year,
and then bring them back
1000(.8)(1.15)-'
+
1000(.5)(1.15)-2
+
to the present.
1000(.3)(1.15)-^
The answer
=
is
D
1,270.98.
Example 6.15
An
insurance
company
which pays 50,000 at the end of
two years. The
live for one year is .99936, and the
issues a policy
the year of death, if death should occur during the next
probability that a 25-year-old will
is
.99858.
What should
a policyholder to earn
1% on
investment?
probability he will live for
company charge such
two years
1
its
the
[Solution
We have to
occur
be a
little
more
in the first
year
is
careful here.
The
probability that death will
.00064, so the expected value of that payment
is
(.00064)(50,000). To find the expected value of the second payment, we
must calculate the probability that death occurs during the second year.
The probability that death occurs within the first two years is .00142.
Hence the probability we want is .00142 - .00064 = .00078, and the
expected value is (.00078)(50,000). Thus the company should charge
(.00064)(50,000)(1.11)-' +(.00078)(50,000)(l.ll)-2
=
60.48.
D
Preparation for L ife Contingencies
1
2
Example 6.16
borrow some money from Friendly Trust.
He
at the end of each year for the next 10 years, but
there is a 5% chance of default in any year. Assume that once default
occurs, no further payments will be received. How much can Friendly
Trust lend Alphonse if it wishes to earn 9% on its investment?
Alphonse wishes
to
promises to repay 500
Solution
500
500
500
10
FIGURE
6.1
Note that unlike examples given in earlier chapters, here the payments
of 500 each are not all guaranteed. We must find the expected value of
each payment and then discount these values to the point where the loan
is taken out.
The fact that there is a 5% chance of default in any year
means that there is a 95% probability of receiving the first payment,
90.25%
Hence
(
the
(.95)^
i.e.,
answer
500(.95)(1.09)-'
Summing this
is
of receiving the second payment, and so on.
j
given by
+500(.95)2(1.09r2+
series
we
obtain 500
•••
(^j^]
+ 500(.95)^0(1.09)-'o
~
"^s
(
)
=
2534.76.
Let us note one thing about the previous example. We solved the
problem by using the formula for the sum of n terms of a geometric
sequence, developed in Chapter Three. Alternatively we could have
noted that our expression
is
a-^
could
now
is
this
method
rate
of interest which Alphonse pays on his loan.
is
=
q^ A ^q — 1. The
be used to solve the problem. We note that
of 500 per year, where the rate of interest
formula for
of a 10-year annuity
just the value at time
j
quite illuminating, since y (roughly .147)
The reader should
is,
in fact, the
also be aware that the fact of no further
payments once default occurs
is
necessary for our calculations to work.
Chapter 6
122
Example 6.17
Redo Example
further
6.16, without the restriction that once default occurs,
no
payments will be received.
Solution!
In this case, there
is
simply a
95%
chance that each payment
is
made.
Then
500(.95)(1.09)-'
+
500(.95)(1.09)
-2
500(.95)(1.09)-*o
+
=
(500)(.95)aioio9
=
3048.39
Example 6.18
A
20-year 1000 face value bond has coupons
annually and
is
redeemable
given half-year, the coupon
Assume
at par.
is
at
a
14%
2%
convertible semi-
chance
that, in
any
not issued, and that once default occurs,
no further payments are made. Assume as well that a bond can be
redeemed only if all coupons have been paid. Find the purchase price to
yield on investor 16% convertible semiannually.
ISolutionl
70
70
FIGURE
The same considerations
70[(.98)(1.08)-'
+
as in
...
1000
70
39
40
6.2
Example 6.16
(.98)2(1. 08)-2+
70
tell
us that the price
+
(.98)^^1.08)-^°]
+
1000(.98r(1.08)
is
•40
40,
40
="(ta)
v^^)
.98
1.08
40
+ iooo(y^)
692.44.
D
Preparation fori ife
Con tingene ies
1
23
EXERCISES
Introduction;
6.1
6-1
.
6.2 Probability
and Expectation
Given a normal deck of 52 cards, determine the following probabilities:
(a)
(b)
(c)
(d)
(e)
6-2.
Drawing the 2 of hearts.
Drawing a heart.
Drawing a 2 or a heart.
Drawing the 2 of hearts in 2 draws (without replacement).
Drawing the 2 of hearts in 2 draws (with replacement).
Given a fair die, determine the following probabilities:
(a)
Throwing a 3.
Throwing a number 3 or larger.
(b)
(c)
In
(d)
In
(e)
In
two
two
two
rolls,
throwing a
rolls,
throwing a number smaller than 3 each time.
rolls,
throwing a
total
total
of 8.
of
1
5.
6-3.
Given a fair coin, determine the following probabilities:
(a)
Throwing 2 consecutive heads.
Throwing 2 heads and then a tail.
(b)
Throwing exactly 2 heads in 3 tosses.
(c)
(d)
Throwing at least 2 heads in 3 tosses.
6-4.
The probability of a 45-year-old surviving
80
is
4.
The
probability of a 45-year-old dying between 60 and 80
is
#.
Find
to age
the probability of a 45-year-old surviving to age 60.
6-5.
A
single die is thrown. If a 1, 2 or 3 turns up, player A wins that
amount of money (1, 2 or 3). If a 4, 5 or 6 turns up, player B wins
the amount of money showing. Find the expected value for each
player.
Chapter 6
124
6-6.
A
box contains
You
bills.
3
lO-doUar
6 5-dollar
bills,
are allowed to pull
two
bills
bills,
and 4
1
-dollar
(without replacement)
from the box. If both bills are of the same denomination you can
keep them. Otherwise, you lose 3 dollars. What is your expected
value?
6-7.
For a
1
-dollar ticket, a lottery offers the following prizes:
1 prize of 25,000;
20 prizes of 1,000 each;
50 prizes of 100 each;
100 prizes of 25 each;
1500 prizes of 5 free
If
6-8.
A
6-9.
A
tickets.
100,000 tickets are sold, find the expectation per
ticket.
current quiz program gives the contestants 5 true/false statements and awards 5 for each correct answer. If all 5 are answered
correctly, the contestant gets 1 ,000 extra. Find the expected value
for someone who guesses at each answer.
9.
perishable product
is
purchased by a
Based on past experience,
items can be sold,
it is
k of the time
There
time 12 items are sold.
is
retailer for 5
and sold for
estimated that 4 of the time 10
1 1
items are sold, and
also a
A
A of
the
probability that a strike
and no items can be sold at all. If an item is not sold,
5.
Find the number of items he should
purchase in order to maximize profit.
will occur
the retailer will lose
6.3
Contingent Payments
6-10.
The Trustworthy Trust Company would
16% on
their loans.
are not paid.
What
rate
of
of all loans
like to obtain a yield
Past experience indicates that
5%
of interest should Trustworthy Trust
charge?
He will repay the loan with a
end of one year. The lending agency has a
of interest of 13%, but estimates there is an 8%
6-11. Mr. Hill wishes to borrow 5000.
single
payment
"risk-free" rate
at the
chance that Mr. Hill will not repay the loan.
they ask Mr. Hill to repay?
How much
should
125
Preparation for Life Contingencies
borrow some money. She wishes to repay the
loan with a single payment of 3000 in two years' time. It is fek
that there is a 5% chance she will not repay the loan. How much
will the lending agency lend to Mrs. Kelly if they wish a yield of
6-12. Mrs. Kelly wants to
14%?
repays the loan in
If she
full,
what yield
rate
was
actually
realized?
6-13. Friendly Finance
22%
year.
Company wishes
a yield rate of 15%, but charges
on loans repayable with a single payment
What default rate is being assumed?
6-14. Mrs. Trudeau
is
interested in ensuring that her
have sufficient funds for higher education.
award a 5000 scholarship
The
university.
probability that a
if
scheme
6-15.
to
end of one
newborn son
A
will
certain plan will
her son survives to age 18 and enters a
cost for this plan at time of birth
newborn male survives
Mrs. Trudeau feels she can make
probability
at the
must she assign
to age 18
17% on
The
is
200.
is
.9821.
If
her money, what
to her son entering university for this
be worthwhile?
How much
would you lend a person today
if
he promised to repay
end of each year for the next 1 years? Assume there
is a 3% chance of default in any year and you wish to earn 1 \% on
your money. Also assume that once there is a default, no further
2000
at the
payments are forthcoming.
6-16.
The All-Mighty Bank wishes to lend 100,000,000 to a South
American country and would like to earn \2% on its investment.
Repayment of the loan will be by two equal annual payments, the
first
is
a
40%
due
20%
in
one year.
It is
estimated that, in any given year, there
chance that no payment will be made.
There
is
also a
chance that half the payment will be made, and a 40%) chance
no payment is made the first year, it is
no payment the second year. It is
considered possible, however, that there might be partial payment
the first year and full, partial or no payment the second year. How
much will the bank ask the country to repay each year?
of payment
assumed
in full.
If
that there will be
1
26
6-17.
Chapter 6
An
insurance
company
probability of surviving
an
sells
1
year
annuity to
a
and whose probability of surviving 3 years
is
negligible.
annuity pays 3000 at the end of each year, and
wishes to earn 14%, what
6-18.
Do
A
t
years
is
The coupon
rate is
purchase price to yield
the
If the
company
12%
if
the probability of
^
estimated to be
20-year 100 face value bond, redeemable
sale.
if
is .45,
a fair price for this annuity?
Question 17 for a perpetuity of 3000
surviving
6-19.
is
person whose
of surviving 2 years
is .65,
at par, is offered for
convertible semiannually.
8%
convertible
probability of default in any 6-month period
Find the
semiannually,
is
if
the
1%. Assume
that
once default occurs, no further payments are received.
6-20.
The Happy Finance Company experiences a 10% default rate on
one-year loans. The Super Finance Company experiences a 7%
If the Super Finance Company
default rate on one-year loans.
charges 16%) on loans, what should the Happy Finance Company
charge to obtain the same return?
6-21.
Agatha pays 770 for a 1000 face value bond paying interest at 1 1%)
convertible semiannually, and redeemable at par in 20 years. If
her desired yield was 12% convertible semiannually, what rate of
default did she expect?
Assume that once default occurs, no
further payments are made.
6-22. Charles buys a 1000 face value 20-year
12%
with semiannual coupons at
bond redeemable
determines his purchase price to yield
14%
who
Exactly 7 years
later,
Charles
sells the
determines her purchase price to yield
par
He
convertible semi-
annually, and to allow for a default probability of
year.
at
convertible semiannually.
bond
1%
per half-
to Elizabeth,
10%
convertible
semiannually, and to allow for a 2%o half-yearly probability of
default.
Assume
that the probability
of default in any given year is
in any other year. Find
independent of the probability of default
each of the following:
(a)
Charles' purchase price.
(b)
Elizabeth's purchase price.
(c)
The
yield rate to Charles, assuming no default occurred
during the
first
seven years.
CHAPTER SEVEN
LIFE TABLES
AND
POPULATION PROBLEMS
7.1
In
INTRODUCTION
we saw how
Chapter 6
combine the theory of
to
interest with
elementary probability theory to obtain the present value of contingent
payments.
In practical situations the
following question
is
how
crucial:
do we determine the appropriate probabilities to be used
in
these
calculations?
The answer is that we must have data
what percentage of borrowers do not repay
able to
money
to guide us.
We
must know
and we must be
identify high-risk borrowers and either refuse to lend them
at all, or lend
them money
at
their loans,
we
use
remainder of
this
higher rates of interest than
for low-risk customers.
In almost all
of the examples
text, the probabilities
we
study
in the
required are those of surviving to certain ages or of
dying before certain ages. Data required to calculate these probabilities
is
collected empirically and
is
published
in life tables.
In this chapter
will introduce the basic notation underlying life tables,
calculate required probabilities.
and see
Section 7.2 will consider the
how
life
we
will see
how
the
same
to
tables
as presenting survival data for a given fixed initial population.
Section 7.4
we
In
table can also be interpreted as
giving information about a stationary population.
Historically,
Makeham and
attempts
describe survival from age
some of these
Finally,
were
made by de Moivre, Gompertz,
would
examine
others to find an analytic function S{x) which
to
age
x.
In Section 7.3
we
will
possibilities.
we
briefly
examine
of multiple decrement theory.
in Section 7.6 a
few of the basic ideas
Chapter 7
128
LIFE TABLES
7.2
TABLE
Age
4
^.
1,000,000
1580
1.58
998,420
680
68
2
997,740
485
.49
3
997,255
435
.44
presented an excerpt from a typical
such a table the column
For
4
make
x.
population
In our case £q
As we
sufficed.
this to
shall see,
individual numbers,
which
life table. In
denotes the number of lives which have
survived to age
^o-
1000^;,
1
we have
In Table 7.1
7.1
it
=
is
we have
sense,
to
assume a starting
would have
1,000,000, but any value
the ratio of entries in the table, not the
important. In particular, the ratio
is
— ''^'^io
from
sents the probability of surviving
function of x,
Note
have died
it is
called the survival function and
that in
Table
in the first
jc
+
1
.
7.1, £i
year of
denotes the number of
age
birth to age x.
lives,
=
life,
998,420.
and
this
is
is
As
a general
denoted S{x).
This means that 1580 lives
the entry
cIq.
In general,
dj^
out of those aged x, which do not survive to
Thus
i.
(7.1)
^x+r
qx denotes the probability that a life aged x will not survive to age
X
+
1.
Thus
a
In our case, the final
column
-^
tells
us that ^2
(7.2)
49
1000
.00049, and
485
.00049 as well.
997,740
Using a life table we can compute numerous probabilities concerning survival. In Example 7.1, we use Table 7.1 to assist us. In Examples
7.2 and 7.3
we
which could be converted
a complete life table.
will determine expressions
numerical answers
if
we had
access to
to
Life Tables
Example
and Population Problems
1
7.1
Use Table 7.1 to find each of the following:
The probability that a newborn will live to age 3.
(a)
The probability that a newborn will die between age
(b)
Solution
I
29
1
and age
3.
I
|=
j^^ =
997255.
(a)
This equals
(b)
The number of deaths between ages
probability
Example
is
^^^-=-^
=
amm(\
1,000,000
.
1
=
and 3
is £i
—
£3.
Thus the
D
001 165.
7.2
Find an expression for each of the following:
(a)
(b)
(c)
The
The
The
probability that an
1
8-year-old lives to age 65.
probability that a 25-year-old dies between ages 40 and 45.
probability that a 25-year-old does not die between ages 40
and 45.
(d)
The
probability that a 30-year-old dies before age 60.
[Solution!
(a)
This equals
(b)
Since the number of people dying between age 40 and age 45
£40
(c)
-^45,
This
is
-f^.
•^18
this is
the
^—^.
J
is
^^V^-
complement
of
Alternatively,
(b),
we
so the answer
is
given by
could obtain this as the
sum of
•''25
——^
-^^
the probability of dying before age 40, and -p-, the
probability of dying after age 45.
(d)
This
is
^30
- 4o
^
-^30
Example
7.3
There are four persons,
now aged
40, 50, 60 and 70.
for the probability that both the 40-year-old
Find an expression
and the 50-year-old will die
within the five-year period starting ten years from now, but neither the
60-year-old nor the 70-year-old will die during that five-year period.
30
Chapter 7
Solution
Working out each probability separately and multiplying the
<Hi^)rii^)(
obtain
This
constructed in the
answer
a
is
that qx
life
table
The numbers 4 and d^ do not represent
people, so where did they come from?
The
place?
first
numbers of
actual
we
'70
and good, but how was such a
well
all
is
)(
60
results,
real
was estimated from observations of mortality data from
study sample, and these values of q^, together with the
4, determine the whole table.
suitable
arbitrary starting value
This
Now we
is
done as follows.
can find
on. In general,
we
£\
=
io
—
Start with £q.
cIq.
Then
We know that ^o^o =
= ^\ — d\, and
q\i\ =z ci\, £2
<^o-
so
continue with the basic identities
gx^x
=
(7.3)
^x
and
4+1 =£x-cix
Example
A
q\
7.4
of Golden-Winged Warblers.
scientist studies the mortality patterns
She
=
(7.4)
establishes
.20,
construct a
q2
—
following probabilities of deaths:
the
.30,
q^
=
.70
and ^4
=
1.
Starting
qo
with £0
= .40,
— 100,
life table.
Solution
Age
Before continuing
4
d.
qx
100
40
.40
1
60
12
.20
2
48
14
.30
3
34
24
.70
4
10
10
1.0
let
us introduce a bit
more
notation.
We
let/?;^
represent the probability that an individual just turning age x will survive
to
agex
+
1.
Hence
p.= ^-^='\-q.
(7.5)
Life Tables
More
nPx
1
3
generally,
=
the probability that an individual just turning age
to
nqx
and Population Problems
=
age X
x
will survive
n.
-{-
the probability that an individual just turning age x will not
survive to age x
-i-
n.
Thus
The reader should
new
of x
In the special case
notation.
= \-nq.
rewrite the answers to
(7.6)
Examples
=
0,
7.2
and 7.3 using
we have
„po
=
this
S(n), the
on page 128.
survival function defined
Example
h^
4
=
nP.
7.5
Explain both mathematically and verbally
why
each of the following
is
true.
(a)
(b)
I
4 - 4+^7 =
m+nPx — mPx
Solution
(a)
+
<^x
+
<^x+\
•
•
+
•
Cix+n~\-
nPx+m-
'
I
Mathematically,
^x
+ ^^+1 +
•
•
•
+ ^x+n-i =
+
Verbally,
at
age x
4 — 4+«
-\-
•
(4 - 4+1 )
•
•
+ (4+M-i
+ (4+1 - 4+2)
~ 4+«) — ^x ~ 4+«-
number of people alive at age x but dead
number of people who die between age x
But dx-\- •• 4- ^x+«-i is just the sum of the
is
the
n, (i.e., the
and age x-{-n).
numbers of people dying at various ages between age x and age
X ^- n — \ (inclusive), which is the same as above.
(b)
Mathematically,
„
m+nyx
Verbally, m+nPx
years.
X
+ w,
To do
live
probabilities
-x+m+n
—
is
9~
f
—
^x+m
f
"-x+m+n
'
f
—
„
"^^^
'
the probability that a person aged
this,
he has to
n more years.
first live
m
„
nfx+m-
jc
lives
m
years and then, at
n
age
-\-
Hence m^nPx is the product of the
of these two events, namely n^p^ nPx+m^
32
Chapter 7
Example
30%
7.6
who
of those
between ages 25 and 75 die before age 50. The
is 20%. Find isPso-
die
probability of a person aged 25 dying before age 50
I
Solution!
We
want
^^^
and that ^^^^~
£25
—
=
to find isPs^)
-
^75)
=
are given
The second
.20.
When
1.25£5o.
.30(1.25^50
=
We
-p--
substituted
l-25£50-
in
.3 0(^25
^75)
relation says that
.
=
80^25
hs — 4o,
=
4o, or
the first expression, this gives
Thus .12540
^50-
—
=
.30£75,
and
finally
hi^A^^ 4167.
t50
D
-3
Finally, let us
been defined thus
Say, for example,
other cases?
aged 20
that a person
able directly from a
by assuming
we would
remark that the expressions ^Px and nq^ have only
of n. What should we do in
far for integral values
lives to
life table,
we want
to find
i/4/?20,
the probability
age 204. This information
but
we can
is
obtain a good approximation
that deaths occur uniformly over a given year.
expect that 4
year, leaving £20
— i
not obtain-
In that case
J20 individuals die during the first
•
Hence an approximate value
4 of the
for
i/4/?20
used linear interpolation between £20 and
£21 in
alive.
<^20
is
—
= ho J^^J
1/4/^20
•
<^20
ho-\{ho-h\)
ho
we have
In other words,
the
life table.
It is
possible to use
formulae, but linear interpolation
'
more sophisticated finite difference
seems to be sufficiently accurate for
most purposes.
Example
7.7
Using Table
7.1
and assuming a uniform distribution of deaths over each
year, find each of the following:
(a)
4/3^1
(b)
The
the
probability that a
first
two months
newborn
thereafter.
will survive the first year, but die in
and Population Problems
Life Tables
Solution
I
(a)
By
133
I
linear interpolation
-
4iip\
we
obtain
-
-
^^
^^
-.yvyi^/.
^^
(b)
The number dying
answer
7.3
is
period described
in the
^^ =
is
k-d\.
Hence, the
D
.00011333.
ANALYTIC FORMULAE FOR iX
Calculations of the type described in the previous section are often
straightforward
if
we assume
a simple analytic formula for 4-
Here
is
an example.
Example
Given
7.8
4=
lOOOf
(b)
—
1
r^ ), determine each of the following:
45
(a)
£o
(f)
The probability that a 30-year-old dies between ages 55 and 60.
The probability that a 30-year-old dies after age 70.
The probability that a 1 5-year-old reaches age 1 10.
The probability that, given a 20-year-old and a 30-year-old, one
(g)
(h)
(i)
(d)
(c) ^20
\spis
(e) 15^25
but not both of these individuals reaches age 70.
[Solution]
j^) =
(a)
4=
(b)
45^1000(1-^) =667
-
1000(^1
(note the answer
must be an
^^^V^ =
=-
(c)
^20
(d)
MPis
(e)
15<725=
= 1^ =
1
1000.
1
integer).
-•^^^24-.01176.
.78571.
-15^25=
1-^=. 1875.
^25
(f)
(g)
^^^
^^Q
7
This
is
which
-
.0667.
equal to the probability that a 30-year-old reaches age 70,
is 40/730
=
TT^
=
.4667.
Chapter 7
134
(h)
Observe that
£105
=
required probability
so no one
0,
Note
is 0.
We
x
<
Hence
alive at age 105.
that our
<
used for values of x satisfying
(i)
is left
formula for
4
the
can only be
105.
must consider separately the two possible cases and add the
results.
The
probability that the 20-year-old reaches age 70 but the
30-year-old does not
^f\-ho\ ^
is
2\96.
soPio
The
•
—
(1
40P30),
which
probability that the 20-year-old does
not reach age 70 but the 30-year-old does
which
is
.2745.
Of course
there
Hence
is
the answer
no reason
reliable formula for £x, but
it
evaluated as
is
is
+
.2196
is
(1
.2745
—
50/^20)
=
to expect that there will be
may
well be true that
40/^30
D
.4941.
any
totally
some function
will
give a very good approximation to the observed values of 4, and this
making calculations and
function could then be used in practice for
predictions about the future state of the population.
What
properties should a formula for
will certainly be decreasing.
We
would
4
possess?
also expect
4
First
of
all,
to decrease
it
more
rapidly for very small x and for x around 65 or 70 than for middle values,
since
individuals in those age groups are subject to higher probabilities
of dying. Figure
7.1 gives a typical
curve for
4
derived from empirical
data.
FIGURE
Note
that there
is
some age
u; at
7.1
which no survivors remain,
cc
is
called
the terminal age of the population.
As
a
first
early 1700's that
approximation,
4
Abraham de Moivre suggested in the
line. In Example 7.8, we
be represented by a straight
Life Tables
saw a
and Population Problems
135
The general formula
particular case of de Moivre's idea.
in this
type of setting will be
= «(l-5).
where a
=
i^
the starting population.
\s
to a final value
=
of £^
so
0,
uj is
(7.7)
Note
that
4
decreases steadily
the terminal age for this group of lives.
Although very rough, de Moivre's approximation
is
reasonable in the
middle range of ages.
Example
Let
7.9
4 = \/lOO-x
(a)
Calculate
(b)
Henry and Henrietta are both 19 years old. Find the probability
that Henry lives at least 17 years, Henrietta lives at most 45 years,
and at least one of them survives for 32 years.
2%Pi>6-
[Solution
(a)
,,;,3,
(b)
We
= |i =
=
6
,75.
work out several cases and then add the results. The
probability that Henry lives for 32 years and Henrietta does not
will
'Kfe)
(>
-
= (5) (9) =
tt)
ST-
Note
that the additional
conditions on the problem do not affect this case, but they do
affect the case
This
where Henrietta
probability
Finally
we
is
lives
32 years and Henry does
(^M__^) (^il^) .
This
is
(|t)(^^)
the answer
is|j
+
^ + ^ = |j.
In Section
1
.6
we
is
is
called force of mortality.
denoted by
/x.
4 we
density of deaths.
= (|)(|) = ^.
Hence
The
The force of
introduced the concept of force of interest.
corresponding notion here
Multiplying by
= ^.
calculate the probability that both persons survive for
32 years.
mortality at age x
(^) (|)
not.
have
ix^,
=
and
is
defined by
-^.
— -D(4) —
4M;c>
(7.8a)
which
is
called the expected
Chapter?
136
Example 7.10
Find
for the
/i;r
4
given
Example
in
7.8.
Solution
1000
ZjH
-
000 en „
__L
-105-.-
\
105(1--)
000
""''^'^"'"^'"'^""--(l-Tfe)"
n
Example
7.1
4
Find the force of mortality for the
expected density of deaths
at
given
in
Example
and the
7.9,
age 36.
ISolutionl
i)(4)
=
1(100
_
is
//;r
=
J
•
£35^36
We
_ V The
9/100
\/l00-36
r
is
so the force of mortality
,
1
=
o^
age 36
^J——
- x)-^'^{-\) =
expected density of deaths
°
1
= 2(100-36) ^16-
observe
that, as
with force of
r-,
interest, force
of mortality
Note
instantaneous measure of the rate at which death occurs.
both examples
^^^
becomes
at
infinitely large as
is
an
that in
x approaches the terminal
age of the population, reflecting the certainty of death as survivors
The minus
approach that age.
convenient
that
4
is
sign in the definition of ^^
way of guaranteeing
that all
a decreasing function, so
As with
its
we
force of interest,
is
simply a
answers will be positive. (Recall
derivative
is
negative.)
can express the formula for ^^
in
terms of logarithms as
^i,
Then
it
follows
that
=
J^ firdr
-D{ln£,y
=
—ln£x
-\-
(7.8b)
Inio,
which
leads
to
the
relationship
e-Io^rdr^k=^p^ =
We
will
now
briefly consider
(7.8c)
S(x).
two other formulae
for
4
which
have been used over the years. In the early 1800's, Benjamin Gompertz
and Population Problems
Life Tables
investigated the case
we
integrating,
constant
A
where
see that this
or, in
Z)( yjis
1
=
)
-/?
same
the
•
yp
as /«(
for
y^
j
some constant
= —hx + A
for
for suitable constants
by
B
(7.8c),
and
we
k, g,
constants
mate the
Using
c.
(7.9)
this definition
=
middle 1800's,
/^^
and the
(7.10)
—
c appropriately, this formula can
curve very well, especially
In the
of
kg'\
c are suitable constants. Note that ^o
g and
4
some
obtain Gompertz' formula
i.
where
By
other words,
lJi.-Bc^
relations given
h.
37
in the
By choosing the
be made to approxi^g-
middle age range.
W. M. Makeham
took a more sophisticated
approach, assuming that
fi,=A + Bc\
After integration,
we
see that this gives
£.
where
k, s, g,
formula
is
The
=
ks^g'\
c are suitable constants.
(7.12)
Note
that Iq
=
kg.
Makeham'
more accurate than Gompertz', but both approximations have
been extremely useful
7.4
(7.11)
in the historical
construction of
life tables.
THE STATIONARY POPULATION
life
table introduced in Section 7.2 traced the future survival of a
newborn individuals. The symbol 4 represented
number of those individuals who survive to age x and d-^ represented
number who die between age x and age x -\- \. Other symbols were
particular group of £o
the
the
introduced as needed.
Imagine
now
a population in
which
births occur at
even intervals
throughout the year, and the level of mortalit\ remains the same from
year to year. Let us assume that there are Iq annual births.
Px and qx will retain their original meanings.
It
The symbols
may be shown under
Chapter 7
138
these conditions that, after a period of time, the total population will
remain stationary and the age distribution will remain constant. Our
table can
now
be used to study
some of
this "stationary population", but
introduced earlier take on a different meaning
the symbols
life
this
in
context.
As mentioned above,
£o will
be the number of births occurring
any given one-year period. This period could be January
31, but
May
it
is
equally true that
4
16 of the following year.
births will occur
to
1
between
in
December
May
17 and
Because of the stationary population, the
number of deaths must equal the number of births in any given year, so
£o is also the total number of deaths occurring in any twelve month
period.
More
generally,
4
the
number of those
number of people who reach
will represent the
any given year.
their x^^ birthday during
individuals
who
—
Since d^
qx^x, dx represents
die before reaching age x-\-
But
\.
the assumption of constant mortality over time allows us to conclude that
dx
also equal to the
is
between ages x and
stationary
for
conclude that
between age
Now
jc
let
1
This
.
4 — £x-^n
and age x
the
is
die during any given year
latter interpretation is the
population problems.
number of annual deaths
we take a census of the population at some fixed
how many people aged x last birthday will we discover?
Lx.
However, some of these
will achieve age x.
Assuming
Lx
could expect
is
^
•
dx deaths to occur.
individuals
we
is
see that,
Hence an approximate value
for
given by
precise definition of Z^
\{£x
+ 4+i)
.
The number of
is
(7.
13a)
is
Lx= I Utdt.
time
4
will die before the census
uniform distribution of births and deaths,
Lx^£x-\-dx =
The
occur
If
In the year immediately preceding the census date,
we
that
+ n.
We will denote this number by the symbol
taken.
important one
Similar reasoning allows us to
us deal with an important kind of problem which arises in
population study.
point in time,
number of people who
;c -I-
individuals aged
given by the symbol 4, where
jc
and over
(7.13b)
at
any fixed point
in
Life Tables
and Population Problems
139
oc
= ^!,+,.
7;
(7.14)
/=0
Under the uniform
distribution assumption
we have
CXj
OO
1=0
/=1
Theprecisedefinitionofrj, using (7.14) and (7.13b),
/•OO
•OO
r,=
/
\
Utdt ^
("^0 cX^
is
i'y^t
^
(7.15b)
7o
Before giving an example,
4
let
us review the basic difference
on the one hand, and L^ and T^ on the other hand, in
the stationary population model. L^ and Tx represent numbers of individuals at a fixed point in time; Lx, for example, is the number of people
aged X last birthday at that instant.
4 and dx, however, represent
between
and
d^,
numbers of individuals achieving
their x^^ birthday or
dying
at
age x
last
birthday in a given calendar year.
Example 7.12
An
organization has a constant total membership.
members
join at exact age 20.
10% of those remaining
(b)
(c)
(d)
Twenty per cent leave
after 10 years,
leave after 20 years, and the rest retire at age 65.
Express each of the following
(a)
Each year 500 new
in
terms of
life
table functions:
The number who leave at age 40 each year.
The size of the membership.
The number of retired people alive at any given
The number of members who die each year.
time.
[Solution]
(a)
number of new entrants were 4o? we would argue as
At age 30, .804o of the original entrants are still
members. By age 40 this has become .SO^o, and the number who
leave is (.10)(.804o) = .OS^o- Since the number of new entrants
If the
follows:
is
500, the correct answer is4P^(.084o)
= ^7^.
140
Chapter 7
This
(b)
given by the total number of individuals aged 20 and over,
is
and would be simply 720 if ho individuals started and no one left
except by death. Taking into account the various ways of leaving
the population,
we
obtain
This expression says
^
(7^20
that, since
-
20%
.20730
-
.08740
-
.72765).
leave the population at age
30, .20730 of the expected individuals are not present (for each age
20%
over 30,
of those people have
to the other terms,
This
(c)
is
just
and over
^(.72765)
in the total
^(^20
"^20
the
-
number of new
who do
A
-20^30
is
and
Similar reasoning applies
who
-
all retire at
age 65.
number of people aged 65
did not leave at an earlier age.
.7245).
is
given by the expression
Note
entrants each year,
that the
above
risk
is
is
just
minus the number of people
not die but leave the membership for other reasons.
measure of the
is
the
stationary, this
.08£40
with stationary populations
rrix
= ^^^,
population
Since the population
(d)
left).
and the remaining .llTss
of dying which
is
D
often useful in dealing
the central death rate.
This
is
denoted by
defined as
m,
=
(7.16)
j^.
Numerous
identities relating nix to other
to derive.
The reader should
symbols are available and easy
which assume a uni-
verify the following,
form distribution of deaths:
h- \-dx
\my
i-qx
(7.18)
m.
and
\
my
(7.19)
2
My
Life Tables
and Population Problems
we
In this section
Section 7.4.
population be
continue with our study of a population, as introduced
The
results presented here
We
stationary.
are
do not require
interested
first
expression for the average future lifetime, denoted
X.
4
EXPECTATION OF LIFE
7.5
in
1
This
is
e^,
just the expected value of the future
in
that the
obtaining an
of a person
lifetime,
at
age
and our
discussion in Chapter 6 indicates that the answer should be given by
^x
we
In other words,
=
^
titPx
-
(7.20)
t+\Px)-
are multiplying each possible
the probability that an individual will live that
more. The above expression
is
number of years
many
full
t
by
years but no
called the curtate expectation and can be
simplified as follows:
ex^
=
{^)iPx-lPx)-^{^){2Px-3>Px)+'--
Px-^2Px
+ 3Px^
Curtate expectation counts only full years of future
by
(7.20).
denoted by
A
Cx,
more accurate measure
life,
as
shown
is
the complete expectation,
tPxdt.
(7.22)
and defined by
%=
r
JO
A
special case of work
when
done
Section 8.3) shows that
assumed we obtain the useful
later in this text (see
a uniform distribution of deaths
is
approximation
%^ex-]-\.
From Formulas
(7.22) and (7.15b)
we
(7.23)
easily obtain
142
Chapter 7
=
^.
= 4^x- Since Cx is
of 4 individuals, we
cross-multiply this identity,
we
obtain T^
the
average number of future years of
life
for each
see
If
we
(7.24)
J.
that Tx
the total
is
This interpretation
stationary,
number of years of future life for these individuals.
of T^ has meaning in any population, whether or not
and will be extremely important
in the exercises.
Example 7.13
If tP35
—
(-98)^ for all
Com-
find e^s and e^s without approximation.
/,
pare the latter with the approximate value of ^35 given by (7.23).
Solution
E
= ^,P35 =
e35
=1^=49-
-98'
(=1
i=\
^ =
(.98)'
Approximately,
=
e^s
^35
+
A
=
49.498.
D
49.50.
Example 7.14
Interpret verbally the expression T^
—
Tx+„
—
n
-
4+w
[Solution]
Tx
the total future lifetime for a group of
is
age
X,
age X
and Tx+n
-\-
n.
Therefore Tx
next n years for
4
individuals after they turn
the total future lifetime for the
is
4
—
individuals after they turn age x.
Now
n
4+n
•
is
-\-
individuals
the total
after
number of years between age x and age x n lived by the
who survive to age x -\- n. Hence Tx — Tx+n — ^ 4+« is
number of years to be lived by those of the 4 individuals who
exactly the
4+n
same individuals
Tx+n represents the total future lifetime in the
will die in the next
•
D
n years.
The complete expectation
ex
is
problems concerning the average age
represents expected future lifetime,
of a person currently aged
jc
is
we
given by
especially helpful
at death.
when
solving
In particular, since ex
see that the average age at death
Life Tables
and Population Problems
x
Putting X
—
death for
all
^
+ ?.=x+|^.
we have
expression,
in this
143
(7.25)
e^
=
y-, the average age at
'0
deaths
among
the £q individuals.
A
more intricate problem involving average age
illustrated by the following example.
death
at
is
Example 7.15
Find the average age
X
at
who
death of those
die between age x and age
+ n.
Solution
In
Example 7.14 we saw
by
this
are
4 — 4+„
time
is
that the total
group of individuals
+
^^
—
~
—
!^+^
/'^^
7
^
number of future years
we
^^^^.
p
'
—
given by T^
individuals in the group,
given by
equal to x
is
Tx+n
—
n
•
£x+n-
see that the average future
Hence the average age
life-
death
is
D
Examples 7.14 and 7.15 most commonly arise
In Example 7.15, for example, the
at
at
^^+\
population setting.
average age
to be lived
Since there
in
a stationary
result gives the
death for all those in the stationary population
who
die
between ages x and x+n, not just those who are the survivors from one
particular group of £o newborns.
Similarly, eo
is
the average age at death
for all deaths in the stationary population.
7.6
MULTIPLE DECREMENTS
Up to now we
have been assuming
that,
except for retirement
at a certain
on a body of lives.
Consider the case of an employer, however, where disability or
withdrawal would be other reasons for terminating employment. The
insurance company covering claims for the employer would want to treat
death and disability separately, since the amounts of the claims would be
different. It might also be true that, in some situations, different causes
age, death
is
the only cause of decrement acting
of death should be analyzed separately.
Multiple decrement theory
with these kinds of problems;
is
it
the area of mathematics
which deals
allows us to study each kind of
Chapter 7
144
decrement individually, and to draw conclusions from the
section
we
notation.
results.
In this
will examine very briefly some of the basic terminology and
Our goal
is
merely to introduce these ideas to the student,
in
case they are required later on.
we have
Let us assume
dently on a body of lives.
total
number of lives
initial
group of
causes of decrement,
Our
original
life
we
all
4
Note
will
that
now be written tx\ the
we have now returned to
table as presenting the survivorship data for a
Now
£o individuals.
denote by
cause {k) between ages x and x
decrements from
causes of decrement acting indepen-
attaining age x.
our original view of a
fixed
m
+
if
the
(T^
.
.
.
,
{m) are the
m
number of decrements from
Also,
\.
(1), (2),
(Tx
is
the total
number of
we have
causes, so
4^>
=
^
^/),
(7.26)
k=\
and
we
also have
=
4^) -J<^)
qx
is
the probability that a
because of cause
(A:),
so
life
£<;».
(7.27)
aged x will leave within one year
we have
^x
qx
2ir\dpx
are analagous to our old qx and/7;f, so
we have
m
= > y^
<l'p
(7.29)
k=\
and
PV-=
1
-,
"
(7.30)
4^^
Similarly,
nP^'
and
-
(7.31)
Life Tables
and Population Problems
nq^I^
It is
how
easy to see
add to get
=
-
(7.32)
nP^:\
a multiple-decrement table can be constructed
^q\ we obtain
cr^'
1
There will be a separate column for each
from sample data.
Starting with
=
145
X] <^
•
=
(^^^
^[f^4^^
Next we obtain
for each value of
=
£,
£q
—
cTq
A:,
cVx
'
and then
and proceed
inductively as before.
We
can also talk about a central rate of decrement for each cause,
was done for death alone in Section 7.4. The central rate of decrement from all
causes at age x, assuming uniform distribution of
as
decrements,
is
""
i4W-
=
The
central rate
As
before,
of decrement from cause
several
{k) is
'""
given by
We
follow immediately.
identities
example here and leave others
will
do one
to the exercises.
Example 7.16
Show
mV
that q^^^
1
1
+
1
-mr"
Solution
4*'
-f
i
+
^-r
- qx
1,1/
2U)-i.4^
cf;^
'
We may
M
//'-i-4^>
'
also
define the force of decrement from
combined. Analagous to Formula (7.8a)
.1"
=
all
causes
we have
-^
(7.35,
146
Chapter 7
The force of decrement from cause k
.f =
where
=
4^^
which leads
4^^
+
^^,
+
is
defmed by
-^,
Then
•••
.
it
(7.36)
follows that 4^^
= X^ff
to
/^r
= Em^.
(7.37)
Example 7.17
For a person age
x,
=
//^^^
.03
and
/i^^^
=
.02 for
/
>
0,
where
indicates death due to illness and (2) indicates death due to
all
(1)
other
causes. Find each of the following.
(a)
(b)
The
The
probability that (x) will die within the next 8 years.
probability that {x) will die within the next 8 years due to
illness.
[Solution!
(a)
We know that y^^l, =
(-)
tP.
= e-rvr^. =
,q^;^^\-s/;^
(b)
Our approach
.03
^-
+
05/_
.02
.05,
so
Then
= \-e-'' =
this
=
3291.
time uses the multiple-decrement analogue to
the single-decrement formula derived in Exercise 7-26(b).
We
have
Jo
=
.03
-.05/
/
e-""'' dt
Jo
.03'
^
.05
-.1978.
D
Life Tables
and Population Problems
Finally
we remark
147
that the absolute
annual rate of decrement,
\ is a concept which the student may encounter. These rates arise
from consideration of the family of single decrement tables associated
The theory behind this is more
with any multiple decrement table.
q'x
advanced than we wish
to
to
go
into,
but there
is
a formula which allows us
approximate the absolute rates of decrement when the probabilities
^i*^
are given. This
is
c^'^^^z
W^
—rf^^^To
#'^^i'Ml + ^(^i"-^i")
K^"^-^^'))
(7-38)
I-
Another important relationship
is
= X[{\-qf\
P^P
(7.39)
k=\
which
is
derived from (1 31) and can be verified approximately by use of
(7.38).
Some
tricky and
appear
questions involving absolute rates of decrement can be
Here
require careful handling.
is
one example, and others
in the exercises.
Example 7.18
A
company
is
affected by
and termination
rate
of de ath
is
two preretirement decrements, mortality (w)
£30 = 5000 and £31 — 4800. If the absolute
Assume
(/).
.003, find
^^3'^.
ISolutionl
We are given that
given that
[^30
r -
,,
^
the
^^3^^
.003
^
=
2.04^3Q^
+
two answers
ru
^
A and
A ^^
(between
1)
q'il
=
= g'^ - q\t =
g'^f
-
=
^^3'^^
[l
+
1
-
-
(^^3^^
^^3^^^)
We are also
+
^ (.04-
= 0. Using the quadratic
2.04 ± x/(2.04)2-4(.006)
^,
^
^
formula,
.04,
so .003
q%^[\
.006
^
•
is
M=
qy^'
.03105.
2
^"^
04)2 -.024
2.04- ^(2.
^^-^^2
=
or
^^3;'^)],
we
obtain
.,
^
feasible
solution
.00295.
Finally,
,
,
.
D
Chapter 7
148
EXERCISES
7.1
Introduction;
7-1.
For a certain type of
qi
=
7.2 Life
.40 and ^3
=
Tables
we
insect,
Starting with £0
1.0.
=
.30,
1000, construct a
life
find that
=
=
^q
-70, ^i
table.
7-2.
Write expressions for each of the following:
(a)
(b)
(c)
The
The
The
probability that a 20-year-old lives 25 years.
probability that a 20-year-old reaches age 25.
probability that a 20-year-old dies between ages 25 and
26.
(d)
(e)
The
The
probability that a 20-year-old lives for at least
40
years.
probability that a pair of 20-year-olds do not both
survive to age 60.
7-3.
80% of people age 25 survive to age 60. 40% of people who die
between age 25 and age 60 do so before age 45.
Find the
probability that a 45-year-old will die before reaching age 60.
7-4.
Four persons are all aged 30.
Find an expression for the probability
(a)
that
any three of them
will survive to age 60, with the other dying
between age 50
and age 55.
(b)
Find an expression for the probability that
persons will survive for
7-5.
at
Explain, both mathematically and verbally,
true.
(a)
(b)
\,^)
(d)
4=
^x 4- <i,+i
4+«
=4
m+nPx
qx
+
^^
+
^x+2
+
Px 'Px+\
nPx
'
Px- qx+]
•
•
•
Px+n-\
mPx+n
+ iPx
•
qx+2
at least
2 of the
most 30 years.
+
•
•
•
=1
why
the following are
Life Tables
7-6.
and Population Problems
Complete the missing
entries in the following table.
4
X
149
^.
1000
Px
qx
100
1
750
2
.80
3
300
4
.60
5
6
7-7.
If/7;c
=
-95 for all
(a)
/?20
(c)
The
jc,
(b)
find each of the following:
2^30
probability that a 20-year-old dies at age 50 last birth-
day.
(d)
The
probability that a 20-year-old dies between age 50 and
age 55.
7-8.
Find an expression for the probability that a 30-year-old will die
in
the second half of the year following her 35th birthday.
7-9.
</<
Derive each of the following approximations, where
(a)
(b)
(c)
1.
= £x - i- dx
= (1-0 -4 +^-4+]
ip^= -tq^
4+/
4+/
\
7-10. Let nlm^x denote the probability that a person aged x will die
between ages x
-\-
n and x
-'r
n
+ m. (When w =
1
it
is
omitted
in
this notation.)
(a)
Show
that
,U^,
(b)
Show
that
„|;„^;f
= ^-^n-Un^m
=
np^
—
n+mPx-
Explain this identity verbal-
ly.
(c)
Show
that n\mqx
verbally.
=
nPx(^
-
mPx+n)-
Explain this identity
Chapter?
150
7-1
1.
You
are given the following probabilities:
That two persons age 35 and 45 will both
(i)
live for 10 years
equals .80.
That a person age 60 will die within 5 years, whereas
(ii)
another person age 55 will live for 5 years
That a person age 35 will
(iii)
live
30 years
is .05.
is .60.
Find the probability that a person age 35 will die between ages 55
and 60.
7-12.
The
probability that a person age 10 will survive to age 30
is .80.
Sixty per cent of the deaths between ages 10 and 40 occur after
age 30. The probability that three lives aged 30, 50 and 70 will
survive for 20 years
is .20.
Find sqPaq-
7.3
Analytic Formulae for ix
7-13.
Given
4=
—
1
(b) ^120
(a)
£o
(f)
The
(g)
lOOOf
jJtv ),
determine each of the following:
(d) 20/?30
(c) <^33
most 25 years.
at
The
probability that three 25-year-olds
4
qx and ^x-
(e) 30^20
probability that a 25-year-old lives for at least 20 years
and
7-14. For the
all
survive to age 80.
given in Question 13, calculate general formulae for px,
Then sketch graphs of all four
functions.
7-15. Prove that in the general case of de Moivre's formula, given
(7.7),
all
we have
^ix
=
u^-
7-16. Obtain an expression for
Makeham's second
by
if
/jLx
4=
ks^w^ g^.
(This
=
gc'(c-i)
is
formula.)
7-17.
Show that Gompertz' formula
7-18.
Show that Makeham's formula
for £x implies /7;f
for
4
implies px
=
sg^^^~^\
called
Life Tables
and Population Problems
5
Gompertz' formula 4 = ^S^\ verify that the
force of mortality is yi^ = Bc^ for suitable B.
Starting with Makeham's formula 4 = ks^g^\ verify that
Starting with
7-19. (a)
(b)
/i;r
7-20.
1
Show
= ^ + Bc^
that,
for suitable
A and
B.
under de Moivre's formula, n\qx
independent of
is
«.
(See Question 7-10 for the definition of this notation.)
4=
7-21. If
100,000
f^^]
and
=
£35
44,000,
each
find
of the
following:
The value of c.
The terminal age in the life table.
The probability of surviving from birth
The probability that a person aged 5
(a)
(b)
(c)
(d)
1
40 and age
7-22. If
4=
lopo
/X70
(c)
The terminal age of the population.
//;c
=
Show
.0017 for 20
(b) sPio
that if
<
jc
(c)
<
qi2>
30, find each of the following:
(d) 5^23
(e) 41^23
tPxI^x+t
7-25. During the first 12
=
<
^^ for all
months of
life,
/
<
4|3^23
=
fix
-^
\_
,
where
Show that ^p^
Use
(b)
=
e'L^^"^^"^'
=
e'l^^^'"^^.
(7.8a) to derive the formula ^q^
=
/
tPxfJ^x+t dt.
Jo
7-27. If X
is
is
jc
is
months. Calculate the probability that a newborn will
survive for 4 months but not for 7 months.
7-26. (a)
jc
1
infants in a developing country
are subject to a force of mortality given by
in
(0
uniform distribution of deaths over year of age
assumed, then
measured
between age
each of the following:
(b)
(a) P20
7-24.
.80jc)'/\ find
(a)
7-23. If
will die
50.
-
250(64
to age 50.
fixed and
/
is
a variable,
show
that
fi.,+[
=
^y^^.
Chapter 7
152
lA The
7-28.
A
Stationary Population
membership of
service club has a constant
1,000.
X
new
entrants are added each year at exact age 35.
Withdrawals are
either by death or by retirement. 40% of those who reach age 50
retire at that time and all others retire at age;65. Express each of
the following in terms of population functions: ^^
The number of annual entrants at age 35.
(a)
The number of members who retire each year at age 50.
(b)
(c)
The number of members who die each year.
7-29. Estimate
m^
if
7-30. Prove that rux
7-31. Verify that W;,
4=
>
3825 and 4+i
q^ for all ages x.
=
/i;c+i/2
=
5713.
When
for the function
does equality occur?
4=
lOOOf^l
-
^V
was maintained
by 10,000 annual births. 40% of the population was under age 15.
Beginning in 1980, annual births increased to 12,000. Assuming
mortality rates did not change at any age, what was the total
7-32. Before 1980, a stationary population of 500,000
population on January
7.5
1,
1995?
Expectation of Life
7-33. If
4=
lOOOf
1
—
TTiri), calculate ^90
and
ego (both exactly
and
approximately).
7-34.
Using integration by
7-35.
You
parts,
show
that
t
ej^
•
tPxl^x^t dt.
are given the following values for a stationary population:
4
X
10,000
69.0
55
8,250
19.5
70
5,380
10.3
Find the average age
die before age 70.
at
death for those
who
survive to age 55 but
and Population Problems
Life Tables
7-36.
We
eio
=
are given the values £50
=
1
o
8200, ^50
=
20, £70
=
53
3000, and
10.
death for those surviving to age 50.
(a)
Find the average age
(b)
Find the average age at death for those surviving to age 70.
(c)
Find the average age
at
at
death for those
who
survive to age
50 but die before age 70.
If £0
(d)
=
10,000 and the average age at death of the entire
population (assumed to be stationary)
age
at
who do
death for those
is
65, find the average
not survive to age 50.
number of deaths is
2000 annually. The complete expectation of life for a 40-year-old
is 30 years. 60% of the population is under age 40.
7-37. In a stationary population of 120,000 lives, the
What
(a)
the average age at death for an individual in this
is
population?
What
(b)
the average age at death for an individual
is
who
dies
before age 40?
7-38.
An army
new
of mercenaries has a constant size of 1000. Each year
all
entrants are at exact age 25, and any soldier reaching age 55
must
No
retire.
one can leave the army except by death or
There are seventy deaths each year among
retirement (at age 55).
soldiers, the average age at death
soldiers enter each year? (b)
being 35.
How many
(a)
How many new
soldiers retire each year?
7-39. Find an expression for the expected age at death of a person
who
survives to age 40 and either dies before age 50 or dies after age
75.
7-40. In a small country with a stationary population, a special system
is
On January 1, each person whose
between 20 and 65 contributes 100 to a pool. On the same
payments of A^ are made from the pool to all persons aged 65
used for supporting the elderly.
age
is
date,
and over, with the entire pool being allocated.
(a)
Derive an expression for K.
(b)
If A^
=
life at
of
500, £20
age 65
life at
is
age 20?
=
2
•
45, and the complete expectation of
what is the complete expectation
15 years,
Chapter?
154
7.6
Multiple Decrements
7-41. Consider a population in
which three decrements are
mmQ-
acting,
\y{a\{b\{c).
Write an expression for the probability that a 40-year-old
(i)
will leave the population because
of decrement {b)
at
age 47
last birthday,
Write an expression for the probability that a 40-year-old
(ii)
will leave the population because
of decrement (b) or
(c)
between ages 44 and 48.
Write formulae for q^^ and W3Q
(iii)
If it is discovered that q^^^ + ^1^^ = 1 - p^P
what can you conclude about decrement (c)?
(iv)
7-42.
Show
S^)
that p^
—
1
'
~
1
+
^
•
m^'^
•
wi
2
f
i:f),
for all
x
>
50,
assuming uniform distribution of
decrements.
7-43. In the first year following training for soldiers, the central death
and withdrawal rates are
What
still
7-44.
A
is
is
affected by
b,
respectively.
and disability
(/).
two preretirement decrements, mortality
Assume 4o
=
the absolute rate of disability at age 50
7-45.
=
a and mx
be a soldier one year later?
company
(d)
—
rrix
the probability that a soldier just finishing training will
Under a multiple decrement
absolute rate of death
is
2%
table, at
l^'^OO and
is .06,
find
is
ct^q
9300.
If
.
each age from 50 to 70 the
and the absolute
for all causes other than death
4V =
rate
of termination
4%. Find a good approximation
to each of the following probabilities:
(a)
(b)
That an employee, aged 55, will still be employed at age 60
That an employee, aged 55, will work for at least 5 years and
most 10 years with the company.
employee, aged 55, will die while employed
between his 58^'' and 59'^ birthdays.
at
(c)
That an
CHAPTER EIGHT
ANNUITIES
LIFE
8.1
BASIC CONCEPTS
we saw how
In Section 6.3
payments, and
in
of contingent
to calculate the present value
we
Chapter 7
how
learned
probabilities concerning
survival can be calculated from life tables (or, occasionally,
to
from an
combine these ideas
are contingent on either
two chapters we
solve problems involving payments which
analytic formula).
In the next
will
survival or death.
Example
8.
Yuanlin
is 38 years old.
If he reaches age 65, he
payment of 50,000. If / — 12, find an expression
payment to Yuanlin today.
will receive a single
for the value of this
.
[Solution]
The
probability of survival to age 65
is
Hence
27/?38-
the answer
is
50,000(27/?38)(1.12)-2^.
To
tables.
27/^38
=
obtain a numerical answer to
If,
for example, i^s
0397
—
Example
8.1 that
4=
=
^0
On
1523.60.
1
—
=
X
jkc
105
,
we
8.1
541
1,
could consult
be
the other hand,
then 27P3S
=
45
f
'38
=
life
we would have
then
would
value
the
value would be 50,000('|2Vl.l2)-27
Example
Example
8327 and ^s
so
.64981,
50,000(.64981)(1.12)-2'7
in
=
if
equal
to
we assumed
40
_
= ^^
67
and the
1399.81.
an illustration of what
pure endowment,
that a unit of
money is to be paid t years from now to an individual currently aged x, if
the individual survives to that time. The value of this payment at the
8.1 is
and a formula for the general case
present time
is
is
is
called
easy to find.
3.
Assume
equal to
tE,
=
{tP,){\-^i)-^
=
v',p,.
(8.1)
Chapters
156
This important expression will be used to
one time point to another
pure
endowment
is
in the rest
of
move payment
this text.
also called the net single
values from
The present value of a
premium for the pure
endowment.
A more common type of situation is called a life annuity. Example
6.14 was one illustration of a life annuity, and here is another.
Example
Aretha
is
8.2
27 years
old.
Beginning one year from today, she will receive
10,000 annually for as long as she
is
alive.
Find an expression for the
present value of this series of payments assuming
/
=
.09.
Solution!
10,000
10,000
28
29
27
FIGURE
8.1
We can view this as a series of pure endowments of the type described
the previous example.
10,000(;727)(1.09)->
Thus the answer
+
1
in
is
0,000(2/727X1.
09)-^
+
...
PC
=
^(10,000)ap27)(1.09)-^
k=\
Although
this
since kPii
=
appears to be an infinite sum,
life
table for each k,
a serious problem.
it
appears that there
formed a geometric sequence, we would here have
To
obtained
If
kPu
is
no nice way of
is
where the terms
to resort to
adding
at a time.
get around this difficulty
life
tables have
these terms have already been added together.
life
will be finite
D
calculating this sum. Unlike our examples in Chapter 3,
terms one
it
eventually.
Now, however, we have
from a
in practice
tables
columns
constructed in conjunction with a number of
encountered rates of
interest,
possible to look up such
and for various values of x and
sums
in the tables.
in
which
In other words, there are
commonly
/
The name given
it
will be
to these
Life Annuities
sums
is
1
commutation functions, and we will study them carefully
5
in the
next section.
A
general formula for a
with interest-only annuities,
is easily written down.
As
assume constant payments of 1 per
alive, with the first payment due at
annuity
life
we
will
year for as long as the individual
is
the end if the year, as illustrated in Figure 8.2.
1
1
\
\
^
\
jc+2
jc+1
FIGURE
The symbol
for the present value of these
and the formula
Again we remark
value
is
payments
to a life
aged
;c
is Ux,
is
=
premium
8.2
Jv',ft.
(8.2)
that the present value
for the annuity.
(In
also used; in this text
some
we
is
also called the net single
texts the phrase actuarial present
will continue to use the simpltr present
value.)
If
above
we have
a formula for
or p^,
we might
be able to sum the
series algebraically.
Example
8.3
Consider Aretha's
premium
=
(b)4 =
(a)Px
4
life
annuity
for this annuity in each
.95 for each x
4[i-^
in
Example
8.2.
Find the net single
of the following cases:
Chapter 8
158
I
Solution
Since kp^
(a)
=
(-95/ for
the required value
all k,
is
PC
= ^(10,000)(.95/(1.09r*
10,000^27
^=1
J2
ik
.95
(10,000)
1.09
k=\
io^oo(t§)
67,857.14.
.95
1.09
(b)
In this case,
0,000
,,p,
=
^
F
78)<i-«9r'
+
since payments will be
haveiM
Now
let
^ 105-x-A:
105 -X
(^y'm-'+]
after
17-ajj^
=
.09
us introduce a bit
so
for all k,
we have
- i2iP(c.fei„.
77 years. From Formula (3.29),
D
93,879.59.
more
we
notation.
Perhaps a
life
annuity
has payments which will end after a certain period.
1
1
1
jc+1
x-{-2
x^n
FIGURE
A
temporary
years
is
life
denoted
8.3
annuity which will only continue for a
a^-^,
and the formula
EC
a^
maximum of n
is
'.p.
n
(8.3)
t=\
An
«-year deferred
to a person
life
now aged x does
annuity
is
one
in
which the
not occur until age x-\-
n+
1.
first
payment
159
Life Annuities
\
Is
x+n x+n+\
Jc+2
denoted hy
\
\
\
[FIGURE
This
1
1
Hx+\
„\a,
8.4]
we have
and
n\cix,
x-\-n-^2
=
=
£
(1
+
/r-^+r/^.
^v',/;,.
(8,4)
t=n-^\
We
note that
Gx, so
n\cix
can be thought of as omitting the
first
n payments from
we have
(8.5)
'x:n\
As with
due
interest-only annuities in Chapter 3, there are annuities-
in life contingencies as
analogous:
dx
denotes
well as annuities-immediate. The notation
a
life
whose
annuity
first
payment occurs
immediately.
Time diagrams
'^X
H
^
\
jc+l
for dx, d^-^,
and
n\^x are
shown below.
h-
x+2
IFIGURE
8.4al
1
a: .77
X
jc+l
x+n
x+n-\
JcH-2
FIGURE
8.4b
1
^
1
+
n\'^x
JC+1
x+w-1
[FIGURE
8.4cl
X+n
is
jcH-«+l
Chapter 8
60
Hence we have
d^=\-^
=
^xM
1
a^.
(8.6)
+ ^x.;^^
(8.7)
n-\
(8.8)
and
=
n\(^x
k?x-
However, the reader should be careful! Because of the uncertainty
of payments in life annuities, other formulae do not carry over directly.
For instance, the identity a-, — {\ -\- i) a-^ does not extend to life
annuities d^-,
=
(1
+
O^xni-
^^
obtain the correct analogue,
we
note
that
n-\
I=\
n~]
=
(1
1+0^' +^(1+0"'-',/'.
+
1
I
t=\
Therefore
n-\
A-i^'x^
= (1+0
t=\
= (1+0
(1 4-
i)-^p,-i
+ X!
(1
+
0~'"'
/+iPx-i
n
= (1+0 Y.(\^ir^p'x-l
=
This
is
(l
+
0«,^,.;^-
usually written as
^x-l:n\
The reader should
=
(8.9)
''P^-^^x:n\-
try to give a verbal explanation for this identity.
Other relationships similar to (8.9) are presented
in the exercises.
Life Annuities
8.2
161
COMMUTATION FUNCTIONS
we
commutation functions which are
important when dealing with life annuities. After learning how to manipulate these functions, we will see some further examples of life annuity
In this section
will introduce those
problems.
Recall
first
from Section
n years hence to a
symbol nE^
is
life
8.1 that a
pure endowment of
1
to be paid
The
currently age x has present value v" „px.
used to denote such an endowment, so
,x-\-n
,E.
-
x+n
x+n
v"
L
The important expression v^4 which appears
denominator is given the symbol D^. That is
=
i)x
so
(8.10)
v'L
in
both numerator and
v^4,
(8.11)
^.
(8.12)
we have
„E,=
Now we
recall the
formula for a
^^
=
life
annuity,
5ZvS/7,
-^,£,.
t=\
We
can also write this as a^
(8.13)
t=\
=
^ -^ =
-X-^Dx+i
4- Z)x+2
+
•
•
)•
We define a new commutation function
PC
TV,
Then we have
=
^
/),+,.
(8.14)
the formula
a.
=
^.
(8.15)
Chapters
162
Clearly
Qx.
if
It is
we have
values of
A'^
tabulated, then
we
can easily calculate
here that commutation functions are very useful in
life
contin-
gencies.
Example
8.4
Marvin, aged 38, purchases a life annuity of 1000 per year. From tables,
learn that N^^ = 5600 and N^g — 5350. Find the net single premium
we
Marvin should pay for this annuity (a) if the first 1000 payment occurs
in one year; (b) if the first 1000 payment occurs now.
Solution
1000
1000
38
40
38
FIGURE
(a)
The
premium
net single
Observe
that
D^s
—
lOOOf-^j, from Formula
is
—
A^38
8.5
=
A^39
250, so the price
is
(8.15).
given by
1000(^^^) =21,400.
(b)
Since
cix
Example
=
^
+
ctx,
the price
is
D
22,400.
8.4(b) could also be obtained using the general formula
^^
=
(^-1^)
Wx-
Here are some other basic identities, all of which can be obtained
by suitable substitution of commutation functions into formulae derived
in Section 8.1. The student should verify that all of these identities are
correct.
^x«i
-l^-D^-
DT
/=1
^.
=
%^
^
^
(8.18)
'X
^.nl
-
^'
J!'^"
'X
(81^)
163
Life Annuities
Analogous
to the
interest-only annuity
we have
accumulated values of annuities,
sx.n\
-r
and
the
life
and
s-^
x.n\
meaning of nE^ defined by
(8.1).
s-^
for
contingent symbols
for the accumulated values of life annuities.
-7
5'
symbols
Recall the
Then we have
and
(8.21a)
^x:n\^nE,-S^-^.
Using (8.12), (8.17), and (8.19) for ^Ex, (^xn\^ and
have the commutation function expressions
'^JC.77|
d^-^, respectively,
we
~
^x+n
and
^.:7i
=
Nr-N.
'*X+A7
'
X
(8-21b)
^x+n
down and
Other commutation function expressions can be written
given
verbal interpretation, such as the one in the following example.
Example
8.5
State verbally the
—^•
meaning of —^^7^-
[Solution]
This
is
premium, for a life annuity
payment at age 35 and a maximum of
by Figure 8.6.
the present value, or net single
bought by a 20-year-old, with
20 payments
H20
first
in total, as illustrated
11
\
21
\
\
34
35
11
\
36
FIGURE
1
1
53
54
\
55
8.6
Here are further examples involving commutation
More can be found in the exercises.
Example
Given N,
functions.
8.6
=
5000, iV^+i
=
4900, A^^+2
=
4810 and
q,
=
.005, find
/.
Chapters
164
I
Solution
We
have D,
=
=
-7V.+,
iV.
^ = ^ = ^^T^ =
100,
""P^-
and
Z),+,
^"^
A=
=
1
7V,+i
-
- 7V,+2 =
=
.005
90,
Hence
.995.
/=^-l=.1056.
Example
so
D
8.7
life annuity which will provide annual
payments of 1000 commencing at age 66. For the next year only,
Margaret's probability of survival is higher than that predicted by the
life tables and, in fact, is equal to/?65+.05, where /765 is taken from the
standard life table. Based on that standard table, we have the values
Des = 300, Dee = 260 and Nei = 1450. If / = .09, find the net single
Margaret, aged 65, purchases a
premium
for this annuity.
Solution
We
will attack this
from
The general formula
first principles.
for the
DC
price of a 1000 life annuity
is
1000^65
^\000(tpe5)v^, where
==
the probability that the individual involved
survives
Now,
years.
/
tpes is
Margaret's case, these probabilities will not be the standard ones.
avoid confusion of symbols,
Margaret survives
let
us write (Pes for the probability that
net single premium is equal to
Hence our
years.
t
in
To
DC
^(\000)(iPes)^'-
N0WP55 = /?65+.05,
but the two-year probability 2^55
t=\
since
is Pes -Pee — (P65+05)/766,
Margaret's probability of survival
iPes
=
(P65+05)/_i/?66, for
1000^65
r
>
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)
<4 =
(m)
«i
ar
im)
I
-«i«^
+ w—
2m
a,-„\a,
=
Example
m—\
+ -i^
2m
I
+
x-\-n
(^)(%
X
l-^5s±a
D.
mci-,+
x:n\
2m {\-nE,)
(8.26)
8.11
Find Linda's premium
years.
"i-^
Assume
A^68
=
in
Example 8.10
1400 and
Z)67
=
if
the annuity
is
to last only
310.
ISolutionl
47
1000
1000
1000
000
47-
47^
66
67
12
FIGURE
8.9
12
20
170
Chapter 8
The answer
is
(12)_
12,000
2,000a'
"47;20|
47.-20I
^
24
Dai
A^48-A^68
12,000
11
D47
I
6000-1400
12,000
^.noy
24 V
,
850
"^
850A
11 f540\
24V850;j
D
68,435.29.
we can
Similarly,
first
payment of
^
as being just like ai \ but with the
define a;
occurring immediately. Hence
It
1
im)
dr'
we have
(8.27)
follows from (8.24) and (8.27) that
cfr'
ax
+ m-2m +
^ m
= d,-\ + m-\
2m ^ m
1
I
=
Next we wish
function,
dx
(m)
Ux
=
AQ
-fy—-
_
— Nx-
,
for ai
to
.
m-\
a
2m
define
corresponding
a
Analogous
to dx
=
However, we know from (8.28)
^^
Dx
.
-t^,
m^^^^
commutation
we would
that dx
=
^ jf ~ ^~
Nx-^Dx.
Im)'
Corresponding annuities n\d\
identities for these are
and
im)
cv - are
have
like to
We conclude that the appropriate definition
N^."'^
and
(8.28)
'
,
or
is
(8.29)
defined as expected,
given in the exercises.
We
will give
one
L ife A nnuities
171
example here showing how formulae for these related annuities can be
expressed in terms of commutation functions.
Example 8.12
Show
that „|ai
^
=
^^
^
Solution!
From
(8.25)
we have
„\a^
~
"l^^+
_
Nx±n±l,V^)^
2m )[ D,
{
)
D
Just as
we
did in Section 3.5,
we can now
define a continuous
life
annuity ax as
ax
An
integral
=
lim
m—>oc
(8.30)
ai'"\
representation for ax can be obtained by changing the
summation sign
in (8.2) to
an integral, obtaining
ax= f
This expression
is
useful if
we have
v^ tPxdt.
an analytic formula for
case, an approximate value for ax can be
approximate Formula (8.24) for
ai"^^
(8.31)
iPx.
In
any
found by taking the limit of the
derived earlier.
We
obtain
Chapters
172
Qx
=
^
lim a\
Urn
If we define a
a,
+
(8.32)
^.
continuous commutation function A^v to be
= 7V,-^.D„
7V,
it
^
commutation functions, we obtain the approximate formula
In terms of
then
+
-^
we can
follows from (8.33) that
5.
The reader
(8.34)
write
= §.
(8.35)
will be asked to derive formulae for the continuous life
annuities a^-^ and
ri\'^x
in the exercises.
Example 8.13
Find the value of a continuous
year-old
I
if
we assume
that 6
=
annuity of 2000 per year for a 50-
life
and
.08
fix
=
05 for
all x.
Solution!
^50+
From
From
Section 7.3,
Section
1
.6,
«50
we know
that tPso
we know that
=
2000/
e-
v
=
=
e~^^.
08,^-.
/r'05^
^^^n^
= -
Hence
the answer
50
- -
.
^
e
.05/
is
°''dt
JQ
-2000/
e- '"dt
Jo
-2000(13)
=
Note
that the
15,384.62.
above example
is
rather unrealistic, because the age
of the individual involved had no effect on the
final
answer!
Life Annuities
1
VARYING LIFE ANNUITIES
8.4
now we have assumed that all payments in a
same size. As with interest-only annuities, however,
Up
73
to
life
annuity are the
this
may
not be the
and the methods of handling varying payments are exactly the
case,
same here
as they
were
Section 3.6.
in
We will
develop formulae for the
where payments are increasing or decreasing in arithmetic
In other situations an argument from first principles will
special cases
progression.
be appropriate.
Example 8.14
Ernest, aged
50,
purchases a
5 years,
8000 each
that.
year, find an
pay for
expression for
3000
annuity which offers him annual
and
payment occurs in exactly one
the net single premium that Ernest should
payments of 5000 for
year after
life
for the five subsequent years,
If the first
this annuity.
[Solution]
5000 3000
5000
-H
\
50
1
8000
\
\
60
56
55
51
3000
\
[FIGURE
8000
\
61
62
8.10]
There are several ways to solve this problem. We could, for example,
view this as the sum of a 3000 life annuity, a 2000 temporary life
annuity with a term of 5 years, and a 5000 life annuity deferred for 10
years. Hence the premium is equal to
30007V51
We
+
2000(iV5i-A^56)
+
50007^61
_
note that there are other correct
payments.
We
8000A^5i
-
5 years.
1
+
50007V61
ways of analyzing
could view this as an 8000
subtracted for each of the next
of the next
5000A^5i
life
-
2OOO7V56
the above
annuity, with 5000
years and 2000 added back on for each
This gives
5000(A^5i -A^6i)
+
2000(7V5i
-
Nse)
D50
= 5000^51+5000^., - ^QQQ^^^
as before.
D
Chapters
174
Let
us
now
consider
arithmetical sequence.
life
annuity, first
Let {Ia)x denote the present value
payment
in
x^\
for {Ia)x
at
1,
in
age x of a
2, 3,
1
x+3
x-h2
[FIGURE
The general formula
increasing
3
\
\
^
one year, having payments of
2
1
\
of annuities
case
the
8.1TI
is
ex:
(/a),-^/vS/7,.
(8.36)
t=\
Note
that
we can
{Ia\
=
write the above as
^v' ip^-\-^v' ,p^-\-^v' iP^-\--/=3
t=2
t=\
oo
= Erl^..
(8.37)
/=o
In terms
of commutation functions,
(/«).
=
we have
J %^.
/=o
Defining the
new commutation
(8.38)
""
function
oo
S,
= J2 ^:^+"
(8-39)
/=0
we
then have the formula
(/a),
=
^.
(8.40)
175
Life Annuities
An
increasing
annuity with a term of n years
life
^
\
\
\
\
x^n
x+2
jc+1
denoted by
n
2
1
\
is
[FIGURE
x+n+\
8.121
We clearly have
A good way of thinking about this
is
to observe that
PC
k=\
=
(Ia)x
-V\p^
n^^
'
V^ kPx+n
-
V" nPx
=
(la):,
-V\p:c- n-
^x+1
_
^x+\
~
Dx^n
^x+n+\
a^+n
-
^x+n+
~
^
'
X^^V^ kPx'x+n
k=\
k=\
V" nPx {lo) x+n
_
1
^x+n
Sx+n+]
^x+n+'
(8.42)
Example 8.15
Georgina, aged 50, purchases a
one year, 5500
thereafter
in
two
annuity which will pay her 5000 in
years, continuing to increase
Find the price
.
life
if ^'51
=
5000,
A^5i
[Solution]
\
50
5000
5500
\
\
51
6000
52
FIGURE
\
53
8.13
=
by 500 per year
450, and ^50
=
60.
Chapter 8
176
To make
this
the pattern,
fit
level life annuity
500.
we must view
the annuity as the
of 4500 and an increasing
life
sum of
a
annuity which starts at
We have
4500^50
-f
500(/a)5o
=
(^^) + ^^^{d^)
4500
^4500(^)+500(^)=
75,416.67.
D
Example 8.16
Repeat Example 8.15
if
maximum
the payments reach a
and then remain constant for
life.
Assume
Ss^
=
level
of 8000,
2100.
Solution
5000
5500
8000
8000
50
52
51
57
FIGURE
is
58
59
8.14
There are several correct ways to solve
approach, but the reader
8000
h
H
this
problem; here
encouraged to find others.
is
one
Instead of the
payments increasing to 8500, 9000, 9500, and so on, they stay constant
So we can imagine subtracting an increasing life annuity of
at 8000.
500, 1000, ... from the original to obtain our answer. Since the 8500
payment occurs when Georgina
is
58, the subtracted annuity should start
with that payment. Hence the price
4500^50
+
500(/a)5o
-
500(/a)57
•
is
^=
- 500(^^')
75,416.67
-75,416.67-500(^2122^
D
57,916.67.
We
can also talk about (Da)^-,, a decreasing
annual payments of
n,
n—\, n—2,
..., 3, 2,
1,
life
the first
annuity with n
payment
year.
n
+
x+1
x+2
FIGURE
x-\-n
8.15)
x-\-n+
in
one
L ife A nn uities
The reader
1
will be asked to derive a formula for {Da)^-^ in terms
77
of
the exercises.
Here we simply note the
will not be surprised to learn
of the existence of symbols
commutation functions
in
important identity
The reader
Fortunately, these symbols
such as {Id\, {Dd)^-^, and so on.
exactly what
them
we would
expect from our earlier work, so
we
mean
will leave
for the exercises.
Example 8.17
Two
annuities are of equal value to Jim, aged 25.
The
first is
guaranteed
payment in 6
years. The second is a life annuity with the first payment oi X in one
year. Subsequent payments are annual, increasing by .0187% each year.
We are given that / = .09. We are also told that A^26 = 930 and
D25 = 30 in the 7% interest tables. FindX
and pays him 4000 per year for 10 years, with the
first
Solution
4000
4000
1
1
1
1
1
1
1
1
4000
1
1
26
25
32
31
•
FIGURE
X
25
Since the
first
=
X(1.0187)
1
1
1
26
27
annuity
is
16,684.15.
40
1
41
8.16a
1
IFIGURE
v^(4000)(7tqTqq
•Toj.09
1
••• ...
x{\m%if
1
1
28
8.16bl
guaranteed,
its
present value
is
equal to
The second annuity has present value
Chapters
178
;725(1.09)-'X+2j925(1.09)-2jr(1.0187)
4-
•
•
•
+
;/725(109r^X(1.0187y-*
+
•
•
•
V+
-
y(
Un n-0187 \ „ / 1.0187
^VT:0T87J[^25^-i;09"J+2/?25^^;09"j
=
^(roW) ^25(1.07)-' + 2^25(1. 07)-2 +
1
,
.
.
.]
^ ^(roTsrJv^o")But the two annuities are of equal value, so
^^(16,684.15X30X1.0187)^3^3^^
8.5
D
ANNUAL PREMIUMS AND PREMIUM RESERVES
way to pay for a deferred life annuity would be with a
premium payments rather than a net single premium. Most
commonly the premium payments would continue for the length of the
One
possible
series of
deferred period, but other situations are possible. Problems of this type
should be treated
text: the
in the
same way
as equations of value earlier in the
present values of the two annuities (premiums and benefits)
should be set equal to each other.
Example 8.18
Arabella, aged 25, purchases a deferred
with the
first
this annuity
each
benefit
coming
in
life
annuity of 500 per month,
exactly 20 years.
She intends
with a series of annual premiums paid
at the
year for the next 20 years, dependent on survival.
annual premium
if
D25
=
9000,
D/^s
=
5000, ^25
=
to
pay for
beginning of
Find her net
15 and ^45
=
1
1.5.
179
Life Annuities
Solution!
Let
P be
the net annual
premium.
500 500
P
P
P
FIGURE
The present value of her
value of future benefits
p^
is
future
60 00(^45
V
-
12
8.17
premiums
60002o|"25
ai5
A
45 45-L
44
26
25
^
"^25
•
^^^25
^"^ ^^^ present
2oi'
Hence we have
•
g
-^)
"7^
«45
is
D
^45
^^^
D
4,274. 19.
formula for problems of the above type
general
easily
is
obtained, but the reader should be cautioned against memorizing too
many
Most of these problems
formulae.
However,
first principles.
payable
life
at the
if
we
let
from
easily handled
tP{n\^x) be the annual premium,
beginning of each year for
t
annuity-due, then the equation of value
tPinWx)-
more
are
Ci^.J\
=
years, for an /7-year deferred
clearly
is
n\cix,
SO
we compare
If
net
premium
is
this notation
we
with Example 8.18,
denoted by the symbol 20^
lol^^s
)•
Next we consider the concept of reserves. This
concept of "outstanding principal" discussed
is
future premiums.
the reserve after
/
In this section
we
is
Chapter
in
just the present value of all future benefits
all
see that Arabella's
analogous to the
4, so the
reserve
minus the present value of
are specifically
years of an w-year deferred
life
concerned with
annuity of
1
which
is
being paid for by annual premiums for n years, of size „P(„|^x) each.
The symbol for the reserve
must be considered.
If
/
>
/7,
then
all
after
Therefore
years
premiums have been
reserve will just equal the value at age x
is dxjf-t.
t
-\- 1
of
is ^jV{n\dx).
paid.
all
Two
cases
In this case, the
future benefits,
which
Chapters
180
=
".ViM)
\it<n,
a,+,
^
=
,
if >
(8.45a)
«.
then the present value of all future benefits
premiums
the present value of future
is
is n-t\^x-\-t,
and
equal to Pd^^^.-^^, where
P^nP{n\ci,). Hence
lV{n\dx)
=
=
n-t\ Ci,+i
-
Pci^+,:;^\
^-^-^n-P(N.^.-N..n)
^
if,^„
(8 45b)
^x+t
Finally, let us realize that
when
calculating premiums for deferred
which we have not yet considered.
annuities, there are practical aspects
For example,
may be expenses
there
underwriting the
risk,
in
the
to
insurer
in
issuing the contract, and in the continuing
administration of the account.
in
involved
The premium which
is
actually charged
a business transaction, including expenses and other costs,
is
called
premium, and the amount by which the gross premium exceeds
the net premium is called loading.
Problems involving loading should be solved from first principles.
We present one example here; others can be found in the final section of
the gross
the next chapter.
Example 8.19
Let us return to Arabella's annuity
of her
first
10% of all
premium
is
in
required for
Example
initial
Assume
8.18.
that
50%
underwriting expenses, and
subsequent premiums are needed for administration. In addi-
100 must be paid for issue expenses.
premium.
tion,
Find Arabella's gross
ISolutionI
Let
G
be the gross premium.
We
again set up an equation of value,
noting that the portions needed for expenses do not contribute to paying
for the annuity, so
we have (Ag)
4-
(n)^)<^25
T9j
—
100
=
60002o|^'25
•
Thus
G= ^"°"jf^ +_'"" = 5,021.16.
D
Life Annuities
1
8
EXERCISES
8.1
Basic Concepts
8-1.
Find the net single premium for a 28-year pure endowment of
20,000 sold to a male aged 36,
(a)
(b)
(c)
4=
(e)
Henri,
years old, wins
1 1
.95 if
40
<
<
x
70
x/lOO-jc
first
can have 1,000,000 francs
=
FindXif/
8-3.
each of the following cases.
/
jc
(d)
8-2.
in
= .12.
^36 = 9618,44 -7100
p, = .96 for all X
< < 40,p, =
Pjc = .9S if
4 = 4[i-jfo
Assume
prize in the Parisian Lottery.
if alive at
age 21, or
X francs
He
today.
.13andio;?ii -= -975.
aged 30, purchases a contract which provides for three
payments of 2000 each at ages 40, 50 and 55, if she is alive.
Elaine,
Given
i^
=
1
10
—x
and
/
=
.09, find the net single
premium
for
this contract.
8-4.
Find the net single premium for a
with the
first
payment due
in
life annuity of 5000 per year,
one year, sold to a 30-year-old in
each of the following cases,
=
4=
(a)
px
(b)
1000
1
^
^
-
115
8-5.
Do
Question 4
if
the first
8-6.
Do
Question 4
if
the
8-7.
/
/
,
payment
is
deferred until age 40.
maximum number of payments
Find an expression for the present value of a
will
pay 500
annuity
8-8.
= .09.
and = .13.
.96 for each x, and
Do
is
at the
end of every two years
sold to a person aged 45.
Question 7 ifp,
=
.96 for
all x.
will be 40.
life
annuity which
if
=
/
.13
and the
Chapters
182
8-9.
Do
Question 7
if
4=
8-10. For a given population,
net single
premium
2^
1
£c
115
^
4=
120
—
jc.
Given
that
age 60 for a deferred
at
annual payments of 1000 commencing
at
/
=
age 70
.07, find the
annuity with
life
if at
most twenty
payments will be made.
8-11. Derive each of the following identities.
(a)
a^
=
vp^d^^\
(b)
d^
=
1
(d)
nl^x
(f)
-{vpxd^+]
=
y" nPx
=
-^X
/7;c-l
(1
Cix+n
+/)a;,_l
8-12. Give verbal explanations for each of the identities in Question
8-13. Julio's mortality for
1
Harold's mortality for
If
/
—
at the
.07,
fmd
<
1
/
<
<
r
4
<
is
5 is
the value at time
Harold
is still
Assume
by /A
(a)
in
=
governed by
—
3(4
tPx
—
—
/),
-25(5
1.
and
—
f).
of an annuity which pays 1000
is
alive.
paid as long as either Julio or
alive.
Question 13 that Noreen's mortality
is
also governed
.3(4-0.
at r =
of a life annuity paying 1000 at the
end of each year as long as at least two of Julio, Harold and
Find the value
Noreen
(b)
tp^
end of each year as long as both Julio and Harold are
8-14. Repeat Question 13 if the annuity
8-15.
governed by
1
Do
survive.
part (a) if the annuity
is
Harold and Noreen survive.
paid only
if a/
most two of Julio,
183
Life Annuities
8.2
Commutation Functions
2000 per year. From
and
2290
7^27
Die = 75.
single premium Andrea should pay for this
first payment occurs in one year.
single premium Andrea should pay for this
first payment occurs in two years.
single premium Andrea should pay for this
first payment occurs immediately.
8-16. Andrea, aged 25, purchases a life annuity of
we
tables,
(a)
find Nis
Find the net
annuity
(b)
if
the
Find the net
annuity
if
the
Find the net
(c)
annuity
(d)
Find
/
if
the
=
if/?25
=
=
2450,
-9353.
8-17. Brenda, aged 38, purchases a 20,000 life annuity with the first
payment
(a)
in
10 years.
Find Brenda's net single premium
in
terms of commutation
functions.
(b)
Find Brenda's net single premium
functions, if payments
terms of commutation
24 inclusive are to be
in
15 through
omitted.
(c)
Find Brenda's net single premium
functions
if
the
her estate, but
first 5
all
in terms of commutation
payments are guaranteed to Brenda or
subsequent payments are contingent upon
survival.
8-18. State verbally the
(,)
(b)
8-19.
meaning of each of the following.
fc^ +
10 -40^30
A^x+l
+
-g^+2+---
A^;,
=
2000,
Given
7V;,+
,
=
1900,
A^^+2
=
1820
and
/
=
.ll,
find qx.
8-20. Determine
which of the following expressions are equal
other.
'X
(c)
v"p,+„
(d)
„£.
to each
Chapters
184
8-21. Harold, aged 60, purchases a life annuity
annual payments of 1000 commencing
beginning
death,
Given
namely
7^60
A
=
+
.
10,
4650, Ne\
provide
For the year
age 61.
subject to a higher risk of
is
from the standard
3950, N^i
=
3350 and
=
/
life table.
.07, find the
for this annuity.
life
table has a select period of
two
Select probabilities are related to ultimate probabilities by
the rules py^^
Dis
at
will
is
where ^60
=
select-and-ultimate disabled
years.
8-23.
q^o
premium
net single
8-22.
age 60 only, Harold
at
which
=
^
\ Px and
•
2000, and D26
=
p^^^^x
1800, find
=
!•/?;,+ ,.
Given
dis
=
17,
a[25].
= .08 and that we are dealing with a four-year select
period. We know that ^po] = •40,/>[30]+i = .80, ^[30]+2 =10 and
^[30^_^3 = 10. Also D34 = 1000 and ^35 = 920. Find the probabiAssume
/
.
lity that
remain
a person entering the select group at age 30 (a)
in the
population for 5 years;
(b)
remain
will
will
in the
population for at most 3 years.
8.3
Annuities Payable m^^'^
8-24. (a)
(b)
(c)
(d)
How much money
must be invested to provide John, aged
50, with monthly payments of 400 for life if ^50 = 16.5?
The first payment will occur in exactly one month.
Repeat part (a) if the payment is 1200 every three months,
the first payment occurring in exactly three months.
Repeat part (a) if the first 400 payment occurs immediately.
Can you solve part (a) if the first payment occurs in exactly
13 months? If not, what additional information is required?
8-25. Derive each of the following approximate formulae:
(a)
„|4"')=
„|a,-(^)„£,
8-26. Explain verbally
why
the formula a -
= ^ + ^^~\
is
incorrect.
185
Life Annuities
8-27. Derive each of the following:
d,
=
(b)
n\dr^
=
—
{"j^y^
_
^
^
-
^-^^
(a)
D,
N^r'
-^
D^
-^
m
'X
8-28. Express
cr
^
4
Give two an-
terms of commutation functions.
in
x:n\
A^
swers, one involving
terms, and one using regular
Ny terms.
8-29. Marvin, aged 50, purchases an annuity of k per month, the first
payment to be made immediately. For the first 60 months, payments will be guaranteed (i.e., will be made independent of
Marvin's survival). After that, payments will continue for as long
as Marvin is alive. Marvin pays 50,000 for the entire package.
Let i = .07, Dso = 5200, Dss =4100, and Nss = 60,000. Find k.
8-30. Repeat Question
29
addition, that A^50
=
8-3
1
.
Jeannette, aged 65,
if
Assume,
no payments are guaranteed.
in
83,500.
is
about to
retire.
Her
salary
is
70,000 per year
and, because of her long service and senior position, she will
50%
of her final salary at the end of each year
as long as she survives. Find the present value of these benefits if
Nes = 700 and Des = 82.
receive a pension of
8-32.
Repeat Question 31 if the yearly payment is to be divided into
twelve equal monthly payments, the first occurring in one month.
8-33. Marilyn,
aged 45, works for the same company as Jeannette
(Question 31) and will have the same retirement benefit, beginning
at age 65.
Marilyn's current salary is 25,000 per year and her
7%
each year for the next 20 years. If
Marilyn will not die before age 65,
.08,
find the present value to Marilyn of her future retirement benefits.
salary will increase at
/
=
and
it
is
assumed
that
8-34. Repeat Question 33 if the yearly pension
into
twelve equal monthly payments, the
payment
first
is
to
be divided
occurring one month
after retirement.
8-35. Repeat Questions 33
and
we assume
that
and 34
20A5
=
if
•
death
^4.
is
possible before retirement,
Chapters
186
was
life annuity of 400 at age 65, but
payment was due. A death benefit,
equal in value to half the value of the annuity, was payable to his
beneficiary. Edgar's beneficiary is his girlfriend Linda, aged 22.
The benefit is a 48-month temporary life annuity, commencing
8-36. Edgar
monthly
entitled to a
died on the day before the
immediately, of
D22
=
X per
=
2500, N22
first
month.
Find
53,000, D26
=
X if D(,s =
=
100, Nes
=
1950, and 7V26
900,
43,500.
8-37. Derive each of the following approximate formulae:
(b)
(C)
^.:n\-^KM-^^xM)
~
nl^x
n|^;c
+
2 •n^;c
8-38. Express a^;^ and nt^^ in terms of commutation functions.
8-39. (a)
Find the present value of a continuous
per year for a 40-year-old
/jLx
(b)
=
04 for
Redo
if
life
we assume
annuity of 1000
that 6
=
and
.06
all x.
part (a) if the annuity
is
temporary, lasting for only 30
years.
(c)
Redo
part (b) if the force of mortality changes to
>
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-
<:^+''E^-
=
sometimes used
is
(9-4)
in
place of nE^
9.4
Example
Calculate the price of Rose's insurance in
insurance in Example 9.2
insurance. For Rose,
I
In this case the face
end of;? years, he receives the face value
alive at the
^^:nl
Example
insurance.
death occurs within a prescribed A7-year period or,
is still
at that time.
A^-., and
is
(9.3)
if
9.1
and Michael's
both policies are to be 30-year
assume /?;r
=
.94 for
endowment
all x.
Solution]
We
will use the results of
endowment
in
Example
9.3
and simply add a 30-year pure
+ 50,000(3o;?38)( 1.1 2)-^°,
= 16,840.52. For Michael,
each case. For Rose, 16,579.74
+
+ (100,000)(3o/?5o)(1.08)-^^ which
(100,000)(||)(1.08)-^o
= 24,985.85.
20,468.70 +
which
we
is
50,000(.94)^0(1. 12)-^^
16,579.74
obtain 20,468.70
The reader will show
in the
<7i
Note
that the
examples given
is
D
exercises that
<A^< K^V
in this section
(9-5)
support these inequalities.
COMMUTATION FUNCTIONS AND
9.2
BASIC IDENTITIES
We
begin this section by introducing two
which
will be helpful in solving
also develop
some
new commutation
functions,
problems involving insurance.
We
will
nice relationships between the insurance symbols
Ay^^
L ife Insurance
1
A^ -, and A^-^, and the annuity symbols discussed
these identities
life
The denominator
Using
8.
8.
of our insurance formulae involve sums of terms
first that all
like {tPx)qx+t"^^^^
Chapter
insurance problems can sometimes be solved by
using the commutation functions defined in Chapter
Note
in
93
which can be expressed as
,
our familiar D^.
in (9.6) is
The numerator
is
denoted
by Cx+t, so we have the new commutation function
C,
Analogous
to
A^;^,
we
=
v^+'ci,.
(9.7)
define
oo
M, = ^C,+,.
In terms
of these symbols
we
(9.8)
then obtain
oo
A,
= )
,^+1
\
tPx qx+t
t=0
—
(=0
=
term insurance
(9.9)
we have
=
^U
1=0
M, - M+„
Z).
Since A^-^
= A\-
-\-
„£;,,
we have
(9.10)
94
Chapter 9
<
Example
9.5
Find the net single premium for Rose's insurance
=
M38
2500 and D^^
=
in
Example
9.1
if
7200.
Solution
The premium
is
50,000(2^^ =
The student should
try
D
17,361.11.
to
obtain
"feeling"
a
commutation formulae work. After some practice
to write
down
it
should be possible
that this can
9.6
Juan, aged 40, purchases an insurance policy paying 50,000
occurs within the next 20 years, 100,000
and 70, and 30,000
premium
these
the price for complicated benefits directly in terms of
commutation functions. The following example also shows
be done in more than one way.
Example
how
for
if
if
if
death
death occurs between ages 60
death occurs after that.
Find the net single
for this policy in terms of commutation functions.
ISolutionI
50,000
or
—\
or
100,000
\
40
60
70
FIGURE
The answer
is
^Q.QQQ(Mo- Meo)
The reader can see how
benefits.
Simplifying,
Can you give a
the above
we
30,000
\
obtain
+
is
9.2
100,000(M6o- M70)
+
30,000^70
obtained as a sum of 3 different
+ 50,000A/.o - 70,OOOA/.o
^O-OOQ^^"
verbal explanation as to
why
this
second expression
is
correct?
Another approach would be to consider the policy as a whole
life
policy of 30,000, plus a 30-year term policy of 20,000, plus a 10-year
term
policy,
30,OOOM4o
+
deferred for 20 years, of 50,000.
20,000(M4o- M70) + 50,000(^60- Myp)
that this expression
is
equivalent to the other two.
This
^^
^^^..
gives
^^^
D
Life Insurance
Example
195
9.7
Describe the benefit whose net single premium
is
——
—
80
n'^^
40
^
SolutionI
—^^
Ao
—
represents a 20-year term insurance deferred for 10 years to
a person aged 40.
^80
jy^
fore this 40-year-old
and age 70.
is
is
a pure
endowment
and then 10 more years to reach
age 80, she will receive a pure endowment
There are a number of important
Two
A,
age 80. There-
insured for the 20-year period between age 50
If she survives that period,
annuity symbols.
to be paid at
D
at that time.
identities relating insurance
and
of the most useful are derived as follows:
= "^
tPx qx+t^''^^
t=0
PC
CX)
tPxPx+t
t=0
/+1
y
t=0
oo
vcix
- ^t+\PxV'^^
/-O
oo
X^rP.v^
= vd^-a^
(9.12a)
= \~dd,
(9.12b)
identities can be very helpful. If we are given appropriate values
of Njc and D^, together with the value of /, we can now compute A^. We
These
see therefore that the commutation symbols
lation rather than absolute necessities.
M^ and
Q are aids to calcu-
Chapter 9
196
Example
aged
Phyllis,
7V4o
=
9.8
5000,
purchases
40,
A^4i
=
4500 and
a
/
=
whole
of 50,000.
policy
life
.08, find the price
If
of this policy.
Solution
We have
Ao=
= l-(t^)(g^)
1-^^.0
-^ _
08 V5000A
Vio8A5oo;is 50,000^4o = 12,962.96.
(
1
Thus the
net single
premium
There are many other
and we
identities
which can be similarly derived,
will leave these for the exercises.
INSURANCE PAYABLE AT THE
9.3
D
MOMENT OF
DEATH
Until now,
we have always assumed
case and
death.
it is
more common
that insurance
is
payable
In practice, however, this
of the year following death.
is
at the
end
often not the
for insurance to be payable at the
moment of
We will now see how to deal with problems of this type.
In a similar
manner
to Section 8.3,
we
first
consider the case
'^
where a payment of
1
is
due
at the
death occurs. The net single
is
end of the
premium
1
^
part of a year in
which
for this insurance to a life aged
x
given by
^(-).vif^^Vv^fH^^V-4
M 4
;
Taking the
limit
obtain the net single
of
this
death. Denoting this by
^
m
approaches
for insurance payable at the
A x, we
Z, = lim/r^ = - /"v'
m-DC
expression as
premium
(9.13)
'
we
moment of
infinity,
have
•
^^ = - f
yg
This gives us the important formula
t^
Jq
v'
•
%ti
tx
^(4+,)
t.
197
Life Insurance
=
Expressions like
expected meanings.
v\p, fix^t
/
Jo
and so on,
^^.;;;i,^^;^,
We
(9.14)
dt.
all
exist
and have the
immediately obtain
nn
(9.15)
Jo
and
^xn\=
(9.16)
/ V^tPxfJ'X+tdt-^v" np^.
Jo
Example 9.9
Find the net single premium for a 100,000
at the
/•
=
moment of death, purchased by
.06
and
tp^o
=
(.98)' for all
a
life
life
insurance policy, payable
aged 30
if
it
is
assumed
that
t.
Solution!
We have A 30
iP30^30+tdt.
Jo
Since
/i3o+r
100,000
=
^n
~ — /«(.98), the answer is
1M)'"<-
9S)dt= -100,000/^.98)
"(ti)
=
'"'•''''yf' =
25.745.24.
O
"•(tl)
Example 9.10
Repeat Example 9.9
if
the 30-year-old
insurance instead of whole
life
is
insurance.
purchasing 20-year endowment
Chapter 9
198
SolutionI
Now we want
00,000^,0
701
30:201
=
100,000(Z 30:201
5^
'
+ v^SoPso)
•20
=
1
00,000 /
v\p3ofi30+t dt
+
1
00,000v20
20/^30
Jo
n 20
(tiy
100,000/a7(.98)
20
+ ,oo,ooo(j^)
'"(tS)
_
=
Example
Repeat
4=
I
100,000/A?(.98)
20
20
+ 100,000(^1)
D
41,202.36.
9.1
Examples 9.9
105
(M\
h
J
-x,0
<x<
and
9.10
if
we now assume
—
6
.06
and
105.
SolutionI
We now have ip^o =
n7
100,000^30
^
and
/130+t
=
100,000
~
100,000
75
=
75
_
.
for
<
/
<
75.
Then
/ e-^^'(^)clt
75
=
-.06
21,975.36.
For the endowment insurance,
00,0007^30
201
=
100,000 /
_
~
100,000
75
=
37,616.59
^~^^'{js)^^~^
.06
100,000e-^6(20)^55^
+ 100,000^-^2^^^
L ife
Ins urance
1
In Section 8.3, the
with
along
approximate formula a^
formula
corresponding
the
M^ = Nx - ^Dx.
In the case
for
=
a^
-\-
at the
death, the relationships are the very different looking
M, = JM,.
h was derived,
commutation
of insurance payable
99
functions
moment of
A^ = iAx and
(See Exercise 9-39.)
Using integration by parts on the expression for A^ given by
(9.14),
we
obtain the important identity
A=\-6a,.
Ax
=
9.4
The reader should note the similarity of
— ddx given by Formula (9.12b).
this
formula
to the
\
VARYING INSURANCE
In Section 3.6
we
and {Da)-^, and
{Ia)x,
(9.17)
in
studied interest-only varying annuities such as {la)-^
Section 8.4 the corresponding
were introduced.
Now we
life
annuities, such as
consider the analogous situation for
life
whole
is
insurance.
We
will denote increasing
life
insurance by
policy which provides a death benefit of
second year, and so on, increasing by
the year of death.
1
{lA)^.
in the first year,
1
This
a
2 in the
per year, payable at the end of
We note that
1=0
Cx+t
t=0
^{Cx + 2G+, +3C,+2 +
+ (G4-i + G+2 +•••)+
^^((G H-G+,
4-
••)
^{Mx +
+
Mx^2
Mx^x
•••)
+
•
•
),
•••)
(9.
18)
200
Chapter 9
using the commutation functions from Section
commutation function
/?,
we
= M, + M,+ +
9.2.
If
we
define a
••,
,
new
(9.19)
obtain
{IA\
= ^.
(9.20)
Example 9.12
Roger, aged 45, purchases a whole
each year of his age
premium
single
Z)45
=
at the
for
this
life
policy which will pay 2000 for
end of the year of
policy
given
Find the net
his death.
—
R^s
=
M45
500,
45
and
110.
Solution
2000x
X Death
45
FIGURE
The insurance
year,
life
benefit
and so on.
We
is
90,000
x-\-\
9.3
in the first year,
92,000
in the
second
can view this as the sum of an 88,000 level whole
policy and a 2,000 increasing whole
life
policy, so the
premium
is
given by
88,000^^45
+ 2,000(Z4)45 =
88,000
r^)
+ 2,000(122) =
45,090.90.
D
Increasing term insurance, {lA)^ -, refers to an increasing policy
with a final benefit of n
life
in the n^^ year.
This can be viewed as a whole
increasing policy, {IA\, with the later benefits of
deleted.
These
later benefits can, as in
Example
«+l, /?+2,
...
9.12, be considered the
sum of a constant benefit of « and an increasing benefit of 1,
They must be evaluated at age x -\- n, and then brought back
Thus we have
2, 3,
to
age
x.
Life Insurance
201
D^-^ ^ .^^
^
_ R,-Rx+n
n Myx^-n
(9.21)
D.
Finally,
we
(iX4)' -,
where the
will
derive
first
formula for decreasing term
a
payment
is
insurance,
n and payments decrease annually to
x:n\
a final benefit of
1
in the
th
n"
year.
Noting that
{DAY - + {lA^ - =
we
{n
+
\)A^ -,
(9.22)
obtain
{DAl-^
=
(n+\)[ ^^-j^^^" )
_
{n-h\)M,
_ n-M^
-
R,
- ^^-^^^n-n M,^„
- M,+, + R,+„
-Rx+\ -^Rx+n+\
(9.23)
Example 9.13
Again consider Roger, aged 45, of Example 9.12. Let us assume that his
policy is as stated up to and including age 65, but after that the benefit
will decrease
premium
that
Mes
by 13,000 per year
for this policy
=
10,
Res
=
if,
in
210 and
until
it
reaches zero. Find the net single
addition to earlier information,
R-je
=
1
we know
10.
[Solution]
The policy
benefit reaches a
maximum of
130,000
then decreases by 13,000 per year for 10 years.
premmm
is
in the 21^^ year.
Hence
It
m
the net single
'""Km + '-""X'Km + "•'>"»<™>i
15
Note that we must be careful to count the 130,000 benefit only once
the above sum; it is counted in the (^^)A.T7^j term. Our premium is
in
202
Chapter 9
^^^^^^^
88,000
(
)+ 2,000 ( ^^^~^g7^^^^0
+ 13,000(i^^^i^|^±^)(gi)
= 88,000(^^) 4-2,000(^) + 13,000(^^) =
30,818.18.
D
ANNUAL PREMIUMS AND RESERVES
9.5
In Section 8.5
we saw how
a deferred
life
annuity might sometimes be
paid for by a series of annual premiums rather than a single premium.
This method of payment
we
will investigate
it
is
extremely
common
with
life
insurance, and
in this section.
many
which can arise with
book is that the student
should think through problems rather than memorize formulae, and this
rule certainly applies to the current section. The basic idea is that the
present value of all future premium payments should equal the present
value of all future benefits. Since the first payment will almost always
Clearly there are
this
kind of problem.
occur
at the
The
time the policy
different situations
cardinal rule of this
is
taken out, our basic equation of value will
have the general form
Pd =
A.
(9.24)
Here A denotes some insurance symbol (whole life, term, varying,
and a denotes some annuity symbol (life, temporary, etc.). Here
etc.),
is
an
example.
Example 9.14
Marie, aged 30, wishes to purchase a whole
payable
at the
life
beginning of each year for as long as she survives.
the
premium in each of the following
= 2250 and A^30 = 120,000
cases:
(a)M3o
(b) A^30
=
policy of 10,000,
end of the year of death, with a series of premiums
120,000,
/
=
.07 and D^q
=
10,000
at the
Find the amount of
203
Life Insurance
ISolutionl
Let
P
Pd^Q
From Formula
be the amount of each premium.
=
(9.24)
we have
10,000^^30-
30
Death
31
T
10,000
IFIGURE
(a)
In this case,
(b)
Using the
^^^^^
P=
identity
y^;^
=
—
1
=
9.41
87. 50.
1
ddx, derived in Section 9.2,
10,000(1-^^30)
^30
_
10,000(Z)30
- ^^^30)
^30
_
1
0,000[ 1 0,000
- ( T%
) ( 1
20,000)]
120,000
We
in
will present a
few basic formulae
general problems should be solved from
^
^^ ^
-i/y.iJ.
u
^
for this type of question, but
first principles.
Occasionally
premium P to indicate the type of
insurance purchased. For example, if a whole life policy is to be paid for
we
will include subscripts with the
by a
life
annuity, the symbol
is
Px and
A=f
If «-year
term insurance
contingent on survival,
is
=
(9.25a)
f.
being purchased with n annual payments
we have
p\
_
x:n\
^.:n\- d^-^
If the
we have
number of premiums
is
_ My — Mx+n
- Nx-Nx+n-
limited to at most
employed. For «-year term insurance
we have
/q ')/;\
^^'^^^
/,
the symbol iP
is
204
Chapter 9
A bar over the P
in
any of these cases means that the premiums are
paid continuously. For example
^^
If insurance
net single
after the P.
=
t=f
payments are made
premium symbol
(^-^Sb)
•
moment of
at the
for the insurance
=
4f
=
^
denotes the annual premium for a whole
moment of death.
is
If the
(9-25C)
life
insurance paid at the
premiums are paid continuously as
well, then a
also used over the P, as in
^(^^)
An
death, then the
written in parentheses
For example,
PiA.)
bar
is
A2-year
=
t=f
term insurance, with benefit paid
premiums paid continuously
Other similar
for at
(9-25d)
•
at the
most t years,
t
<
moment of death and
n/\s denoted by
identities are given in the exercises.
However
the reader
should be alert to problems which do not
fit easily into any given
Here are two more examples. (Again, and for the rest of this
section, we assume death benefits paid at the end of the year of death,
formula.
unless specified otherwise.)
Example 9.15
Matilda, aged 40, purchases an insurance policy which pays 1000 in the
first
year, increases
by 1000 per year to a
maximum
of 10,000, and then
She wishes to pay for this policy with an initial payment of X and subsequent annual payments of 2X contingent on survival.
Find Jf given D40 = 390, A^40 = 7500, R40 = 5000 and Rso = 3500.
remains constant.
205
Life Insurance
ISolutionI
age 40 of
The value
at
1000(Z4)^|^
— + 10,000^50 (£^)-
payments
future
future benefits
all
Similarly, the value at age
+ Xd^\
given by Xd^Q
is
1000(/<T^+
^
_
given by the expression
is
i
-jy^
j
Equating,
.
10,000^50
we
40 of
all
obtain
(gg)
(1000)(7?40-^50-10A/5o)+ 10,OOOM5o
A^40+A'41
-
1000(1500)
7500
+
7110
_
-
^
^^^-^''
Example 9.16
Garth, aged 45, purchases a
life
insurance policy by means of annual
The death benefit at any age is the
sum of the net premiums paid to that time plus an additional 1000. Find
the premium if M45 = 1700, A^45 = 45,000 and R^s = 27,000.
payments contingent on survival.
Solution
If
P
is
the annual
premium, then the benefits are
The present value of
••.
The present value of
=
PC.,
,000.«
+P
.
all
all
future benefits
future
(/.,. so P
premiums
is
= .JfO^
is
P+
1000,
1000^45
Pd/^s-
IP
+P
+
•
1000,
(Z4)45.
Thus we have
= lOOOgO) ^
,, ,,
D
It is
made
also possible that
at intervals other
yearly on a whole
Px
and
life
payments
for an insurance policy could be
than yearly. If payments are to be
policy, the total annual
premium
made
is
m times
denoted
by
we have
/"i"*
=
^.
(9.29)
206
Chapter 9
Using Formula (8.28) gives the approximate formula
_
p{m)
Ax
m-
dx
\
^
K
(9.30)
(^h
Example 9.17
Repeat Example 9.14(a)
if
made monthly.
the payments are to be
addition to the given information, assume D^,^
=
In
10,000.
[Solution
^
p02)
^
10,OOOA/3o
payment
will be equal to
The concept of
J
^^^^
94 93
^(194.95)
^^^^^ ^^^^ monthly
D
16.25.
loading, discussed in the final section of the last
chapter, also applies in this setting.
and several more
=
^^-^^
^^^^
We
will present
two examples here
in the exercises.
Example 9.18
Kim
purchases a 50,000 whole
payable
include
50%
at the
(i)
20%
of the
life
of the
first
first
expenses.
=
year gross premium and
Find
annual
gross
the
10 0,000 and ^30
=
life.
year gross premium for issuing expenses;
thereafter for administrative expenses;
A^30
Premiums are
Expense provisions
policy at age 30.
beginning of each year for
10% of each
(iii)
gross
(ii)
premium
100 for additional start-up
premium
given
M30
=
1200,
4350.
ISolutionI
Let
G
be the gross annual premium.
In every year except the
contributes to the purchase of coverage. In the
contributes to the insurance.
first
first,
.9G
—
100
year only .3G
Then (.9G)a3o - .6G — 100
=
50,000^3o,
so
^
^-_
50,000^30
(.9)^30
+
100
-.6
_
-
50,000^30 + \OOD30
.97V30-.6D30
_
-
.q. c.
^^l-^^-
n
^
207
Life Insurance
Example 9.19
A
whole
1000
insurance policy
life
is
The death
issued at age 35.
benefit
premium is payable at the beginning of each year
The following expense provisions apply:
gross annual
is
The
year and increases by 1000 per year thereafter.
in the first
for the
next 30 years.
(i)
40%
(ii)
10% of the
(iii)
100
(iv)
2 at the beginning of each year of the policy.
Find
/?35
of the
year's gross annual premium.
premium
gross annual
after the first year.
at issue.
gross
the
^
first
premium given
annual
20 000 and N^s
=
1)35
=
3000,
A^35
=
70,000,
28,000.
ISolutionI
The
G
death
future
all
be the gross annual premium,
=
1000(Z4)35
is
of
value
present
Letting
(.
-
90)^35. 3o|
.3G
-
100
-
benefits
we have
^_
is
+
100
(-9)^35 3^
+
we
1000i?35
+
100Z)35+2A^35
.9(A^35-A^65)-.3Z)35
We obtain
_
"
ceo
Q-;
^^^•^^•
of premium
main importance now in connection with
than annuities. The reserve is defined to be the excess
briefly encountered the concept
This idea takes on
insurance rather
_
-
2^35
-.3
In Section 8.5
reserves.
of the policy.
life
the appropriate present value of this expense.
1000(Z4)35
1000(Z4)35.
The yearly expense of 2
2^35.
not restricted to 30 years, but continues for the
Hence 2^35
is
the equation of value
its
of the present value of future benefits over the present value of future
premiums.
For a whole
life
payable by net premiums
premium reserve / years
policy of
at the
later is
,V,
issued to a person aged x and
1,
beginning of each year, the net level
denoted
tVx,
= A,+, - P,
•
and
is
given by
a,+,.
(9.31)
The reserve given by (9.31) is called the policy year terminal reserve,
which means that / is assumed to be an integer, so that tV^ exists at the
end of a policy year.
Many
Formula (9.31) are encountered, depending on
premium payments and the type of insurance. For
example, if premiums are only paid for n years, then for t > n the second
variations of
the duration of the
term
in (9.31)
disappears, and
we
find that the reserve
is
simply equal to
208
Chapter 9
Here are a few examples. Other situations
arise in the exercises.
Example 9.20
Ralph purchases a 10,000 whole
policy with a series of
next twenty years. If
premiums
Mo =
life
policy at age 40.
for the
beginning of each year for the
at the
100, A^40
1
He pays
=
57,000 and N^o
=
28,000, find
the terminal reserve for this policy at each of the following ages:
(a)
(b)
(c)
(d)
Age
Age
Age
Age
= 820, 7V50 = 41,000 and D^q =
M70 = 290 and Djo = 750.
50, if M50
70, if
2010.
premium is paid.
premium is paid.
40, just before the
40, just after the
[Solution]
(a)
We
must
^^40.20j
calculate the amount, P, of the premium.
first
=
P=
10,000yl40, so
l^.QQQMo
present value of future benefits
of
premiums
future
(10,000X820)
(b)
-
is
la^Q.jQj.
=
_
-
(379.31X13,000)
2010
Since no further premiums are
10,000^70
10,000
("^^ =
—
379 31
10,000^^50,
is
379.3
^
We
At age 50 the
and the present value
Thus
... ^^
^^^^-^^
the
reserve
•
to be paid, the reserve
Since one premium payment has been made, the reserve
- ^'^40 191 =
terms together are
P=
Example
0,
^^4020]
10,000^40
from part
(c),
The
so the reserve
is
now
first
two
is
simply
D
9.21
in the first year,
maximum
pay for
- ^^40.2^ + P-
379.31.
Annette, aged 50, purchases a whole
to a
just
~^
(d)
10,000^40
is
3,866.67.
The reserve
10,000^4o
is
,
(c)
is
have
this
life
policy paying 50,000 for death
55,000 the second year, and increasing by 5000 annually
of 100,000,
with
an
at
which point
increasing
series
it
She will
payments of P
remains constant.
of annual
immediately and then 2P, 3P, and so on, for as long as she survives.
Find an expression,
reserve at age 55.
in
terms of commutation functions, for her terminal
209
Life Insurance
I
Solution
At age 55 the benefit
75,000, so the present value of Annette's future
is
+
+ 30,000^60^- The present
value of future premiums is (5P)as5 + P{Id)ss. The premium itself is
found from P{Id)sQ = 45,000^50 + 5000(7^).^^ - + 55,000^60
(^)'
45,000^50 + 5000(/?50 -^60-10^60) + 55,OOOM6o
=
^^^
so that P
benefits
70,000^55
is
5000(Z4)^'.-
SsQ
terminal reserve at age 55
J-(70,000M55+
_
therefore
50007^55 -5000i?60+5000M6o)
/ 5A^55+5'55
V 45,000M5Q+5000/^5O-5000/e60+5000M6o A
^50
A
)'
^55
I
is
D
There are a number of useful formulae which can be developed for
For example, using A^+t = 1 — dcix+t (see Formula (9.12b)),
reserves.
we
obtain
[
Using Px
=
-T^
Ux
Vx
—
—
Ax+t
=
\-{Px+d)dx^t.
PxCix+t
—-—^ = J—
=
Ux
we have
<:/,
Ux
tVx
(9.32)
=\-
%^.
~
^x+t
the very useful formula
(9.33)
We also have
/
Kx
—
Ax-i^t
Px
'
= A+,(l-;^),
since /',+,
=
^^±^.
Substituting ^;,+,
,F,
=
=
P^+i
(9.34)
d^+, into (9.34) gives
(/>,+,-/>,)«.+„
(9.35)
210
Chapter 9
which leads
to
Px^t
~
Px
= ^17+7'
trx'^^
since
1
-r—
=
P^+z
Sometimes
J.
(9-36)
Ar+/
+ ^•
important to
is
it
know
the connection between the
terminal reserves of two successive policy years.
can either be worked out from
Problems of this type
principles or by using the formula we
first
= A^^t — Px ^x+t- We note that
thought
of
the
value
can
be
as
of
the
first year's benefit, vq^+t, plus
Ax+t
the present value of the remaining benefits, which is given by
will
now
vpx+t
Again
derive.
Ax+t+\-
tVx
recall that ;F^
Similarly, a^+t
=
"^qx+t
=
vpx+t(Ax+t+\
=
+
vpx+i
=
Ax+t+\
n\Vx +
iyPx^-t)
+ ^Px+t
1
- Px
'
-
Px(\-^vpx+t
cix+t+\)
•
vqx^t
-
cix+t+\-
•
Px
Thus we have
•
cix+t+\)
+ vq^+t - Px
(9.37)
.
This formula can be used either to obtain tVx given /+i K^, or to go in the
other direction. A nicer symmetrical form is obtained by transposing Px
and multiplying by
(1
1
+
+
/,
obtaining
i){,Vx
+ Px) =
qx+t
+Px+iUi
(9.38)
Vx).
EXERCISES
9.1
Basic Concepts
9-1.
Find the price of whole
life
insurance with a face value of 100,000
sold to a person aged 40 in each of the following cases.
(a)
(b)
/?;,
=
ix=
.96forallxand/=-.09.
1000(^1
- -^\ and/ -.13.
9-2.
Repeat Question 1 if the payment of 100,000 is
end of the 5-year period in which death occurs.
9-3.
Repeat Question
1
if the
insurance
is
to be
made
at the
a term policy for 30 years.
211
Life Insurance
9-4.
Repeat Question
1
9-5.
Repeat Question
1
for 30-year
if
endowment
insurance.
the policy has a face value of 100,000 for the
30 years only. If the insured survives to age 70, he is paid
70,000 and the remaining 30,000 is retained as a whole life
first
benefit.
9-6.
Repeat Question
still
9-7.
Prove that the identity
-
yi'
•^
9-8.
each case,
in
1, if
=
/"^^^^
12, but the benefits are
.
paid yearly.
<
^^^
<
x:n\
A^-\
^-"1
ri\Ax<Ax\in>\. Give
Prove that
is
x and for
true for all
all
a verbal explanation for this
inequality.
9-9.
Julio's mortality for
law
by
—
tPx = -3(4
—
=
.25(5
fp^
/).
<
1
/
<
4
assumed
is
Harold's mortality for
If
t).
/
=
to
1
be governed by the
/ < 5 is governed
<
.07, find the price at
time
of an
insurance policy which will pay 100,000 at the end of the year in
which the second of Julio or Harold
9-10. Repeat Question 9 if the insurance
which the first of the two men
=
is
paid at the end of the year in
dies.
9-11. Repeat Question 9 if the policy
9-12. Prove the identity ^;f
dies.
is
for 2-year
endowment
insurance.
v^q^ -\-pxAx+]).
9-13. Give a verbal argument for the identity in Question 12.
9-14. Prove that
A = (\-A,+„)Al-4-A^.-,'A,+„.
x:n\
insurance policy which pays
occurs within that period, or
1
at
premium
at age x for an
end of 10 years if death
the end of the year of death if death
9-15. Obtain a formula for the net single
at the
occurs after 10 years.
9-16.
Assume
qx+n
+k
that
for
unchanged.
kv^^\px{\
a
single
rate
some constant
Show
that
- Ax^n^x).
A^
of mortality,
q-^+n,
is
increased to
All other values of qy remain
will be increased by the amount
k.
212
9-17.
Chapter 9
endowment of 10,000 issued at
premium is to be returned in the
event of death before age x-\-n. If the premium is not returned,
the net single premium is 7000. Find the net single premium for a
pure endowment of 10,000 issued at the same age and for the same
period if half of the net premium is to be returned in the event of
The
net single
premium
age X for n years
is
for a pure
8000
if
the
death.
9.2
9-18.
Commutation Functions and Basic
Identities
Herman, aged 45, purchases a whole life policy of 100,000. Find
premium if (a) M45 = 250 and D^s = 520;
=
A^45
(b)
8000, Z)45 = 520 and / = .04.
the net single
9-19. Repeat Question 18(a) for a 20-year term insurance policy given
Mes
=
35
.
premium
9-20. Express the net single
for the following policy, issued to
a person aged 30, in terms of commutation functions: 50,000 if
death occurs in the next 20 years,
20-year period after
50,000 after
that,
1 00,000 if death occurs in the
and 10-year endowment insurance of
that.
9-21. Prove each of the following identities:
= v-da^
(a)
A^
(d)
A^-^
(f)
M, = D,- dK
=
-
1
lax:t-\\
(g)
9-22. (a)
(b)
1+
/
da^-^
_ M^ - M^^t + D^+t
~
A
Find the rate of interest
Given M^
=
3000, M^+i
if
a;^
=
=
15.5 and^;^
2800 and
q,
=
=
.25.
.01, find
D^+i
L ife
21
Ins urance
9-23. Ronald, aged 40, purchases a
whole
life
policy paying 10,000
during the next 10 years and 20,000 during the ten years after that.
If
Ronald
M60
=
=
420, Z)4o
net single
200
alive at age 60, he will receive
is still
each month for the rest of his
premium
=
4500, Z)60
at the
end of
= 750, M50 = 600,
= 10,000, find the
Given M40
1100 and A^60
life.
for this policy.
9-24. Angela, aged 55, purchases a deferred life annuity of
year commencing at age 65. Before age 65 there
is
4000 per
a 10,000 death
benefit payable at the end of the year of death. Find the net single
premium
A^65
=
for
package
this
110, Nee
=
95 and
/
=
given
N^s
9-25. Repeat Question 24 if the life annuity
either
9.3
Angela or her
estate, for
Insurance Payable at the
9-26. (a)
is
Do
iPaq
is
=
/
.06
9-27. (a)
Do
and
40
if
it
t.
purchasing 30-year term
insurance.
is
purchasing 30-year endow-
insurance policy, payable at
moment of
Do
that 6
Do
life
=
.05
and ^x
=
life
part (a) if the 40-year-old
that the life
terminal age
is
1
is
is
purchasing 30-year term
insurance.
is
purchasing 30-year endowlife
continue to use
insurance.
(5
=
.05 but
now assume
subject to a de Moivre's survival function with
10.
Question 28(a)
20 years.
we
it
-04 for all x.
part (a) if the 40-year-old
9-28. Repeat Question 27 if
insurance.
death, bought by a life aged 40 if
ment insurance instead of whole
Do
(.97)' for all
Find the price of a 100,000
insurance instead of whole
9-29.
insurance
life
assumed
(c)
life
part (a) if the 40-year-old
life
to a life aged
ment insurance instead of whole
the
(b)
230,
guaranteed, payable to
is
moment of death,
part (a) if the 40-year-old
that
=
Moment of Death
—
assumed
A^56
50 years.
insurance instead of whole
(c)
250,
Find the net single premium for a 50,000
policy, payable at the
(b)
—
.03.
if
we assume
the insurance
is
to be deferred for
214
9-30.
Chapter 9
George is informed
whole life insurance
70,000.
=
/ijc
6
is
policy, payable at the
for a 200,000
moment of
death,
is
George is subject to a constant force of mortality
and \i 6 is also constant, find the net single premium for
a 5-year deferred
whole
Helen
premium
If
.03,
9-31. Helen
that the net single
informed that the net single premium for 100,000 of
is
life
is
policy of 200,000.
life
insurance, payable at the
moment of death,
subject to a constant force of mortality ^^
is
=
65,000. If
0275, and
also constant, find the actuarial present value of a
whole
if
life
annuity of 5000 per year payable continuously to Helen.
9-32.
Brenda purchases
life
insurance which will pay 100,000
during the next 5 years and 200,000
benefits are payable at the
if
moment of
death.
b — .08, and ^^ = 04 for the next 8 years and
Find the net single premium for this insurance.
9-33. (a)
If the force
of
fix
It
A^
she dies
=
known
is
The
that
.05 thereafter.
interest is increased, but the force
stays the same, does
if
she dies after that.
of mortality
Explain your
increase or decrease?
answer,
(b)
If the force
of mortality
stays the same, does
A^
is
increased, but the force of interest
Explain your
increase or decrease?
answer.
9-34.
Using integration by
9-35. (a)
(b)
9-36.
parts, derive the
= — Sa^-, —
— 6a^-y
A^-^ =
Derive the formula ^
Derive the formula
formula ^;r
1
^;;;;|
n such that A ^-^
1
—
Sa^.
v^nPx-
\
Assuming a constant force of mortality
interest 6, find
=
=
fi^
2A ^-,.
and a constant force of
L ife
215
Ins urance
9-37.
Herb purchase a 100,000 whole
life
moment of
benefit payable at the
insurance policy, with the
The policy pays an
death.
additional 50,000 if death occurs during the first 5 years due to
Assuming
and the force of decrement due
than {a)
9-38.
—
specified cause {a).
to cause {a)
A
is
.005,
is
6
.07, the force
premium
.045, find the net single
20-year-old purchases a 100,000 whole
payable
all X,
at the
moment of
to causes other
for Herb's insurance.
life
=
Given d
death.
of decrement due
policy with benefit
.05
and
=
.02 for
80%
certain
fix
find each of the following:
The net single premium for this policy.
A number X such that the insurance company
(a)
(b)
that the value at time
will be
less than
of purchase of their eventual payout
(Such a number
or equal to X.
80^^ percentile
sometimes called the
is
X
is
of the present value of
the benefit.)
Assuming uniform
9-39 (a)
of age
—
show that ^ ^^
Extend the result of part
(b)
9.4
jc,
distribution of deaths throughout the year
i
'^'tj-
(a) to
show
that
Ax
=
t
A^^jf-
Varying Insurance
(Note:
Unless stated otherwise,
assume death benefits payable
9-40. Tim, aged 50, purchases a
benefit
is
thereafter.
Rso
=
and
20,000,
Find
40
whole
net
=
220.
if
exercises in Sections 9.4 and 9.5
end of the year of death.)
life
benefits
the
2300 and Mso
9-41. Repeat Question
all
at the
single
policy.
increase
In the first year his
by
5000 per year
premium given Dso
=
500,
Tim's policy is only to last for 20 years.
assume M-jo = 60 and R-jo = 400.
In
addition to the above data,
40 if Tim's policy is to increase as stated up to
and including age 70, and then remain constant at 120,000 per year
thereafter. Use the data given in Questions 40 and 41.
9-42. Repeat Question
216
9-43.
Chapter 9
Do
Question 40
the insurance policy begins with an initial
if
6000 annually
benefit at 120,000, and then decreases by
Assume
reaches zero.
until
it
the data given in Questions 40 and 41.
9-44. Derive each of the following identities:
(b)
9-45.
A
= v{Id\ - {Ia\
= d,- d{Id\
{IA\
{IA\
(a)
newborn
2000
in the second year, and so on up to 10,000 in the tenth year. The
policy continues to provide coverage of 10,000 up to age 51. At
age 51 and over it provides 50,000. Write the net single premium
for this policy in terms of commutation functions.
whole
child.
9-46. Douglas
retire
insurance policy
life
The policy pays 1000
is
hired
when he
by
XYZ
sold to the family of a
is
if
death occurs in the
first
year,
Publishing Company at age 50, and will
his 62"^ birthday he becomes eligible
On
turns 65.
1000 for each year of service with XYZ.
The benefit will terminate when he retires. Find the present value
of this policy on his 62"^^ birthday, given M(,2 = 800, ^53 = 765,
for a death benefit paying
M64
Ne4
= 728,
Mes = 690,
Nei =
= 17,700 and 7V65 == 15,700.
9-47. Ellen, aged 60, purchases a
until
age 65, 25,000
thereafter until
it
9-48
=
=
Let (lA
19,800,
insurance policy paying 50,000
life
reaches 5000, at which point
=
500, Mes
=
A^63
age 65 and decreasing by 2500 per year
at
Find the present value
Meo
M74
22,000,
at
remains constant.
it
age 60 of Ellen's future benefits given
=
350, Deo
1200, Res
=
4000,
/?74
=
1600 and
150.
)x
denote the net single premium for whole
which pays
1
increasing by
in the first year,
1
payable
per year,
at
Similarly, \eX(IA)x denote the net single
insurance which pays
/
moment of
at the
life
insurance
2 in the second year, and so on,
the
moment of
premium
death
if
for
death.
whole
age x+t.
(a)
Show
(b)
Explain
that (lA )x
why
=
(7Z);,
d^
=
—
S(Id)x-
/
-'o
= 0^-
tv' tp^fi^+t dt.
_
(c)
Show
(d)
Justify verbally the approximation (I
that (I
A)^
6{I d)^.
A)x
=
life
death occurs at
(lA
)x
— ^A x.
217
L ife Ins urance
9.5
Annual Premiums and Reserves
whole
9-49. Roy, aged 50, purchases a
policy of 10,000 with a series
life
Find the amount of
of annual payments contingent on survival.
each payment
M5o=
M50 =
M50 =
(a)
(b)
(c)
each of the following cases:
in
150andiV50
150,
•=
/
=
2200
.04 and Dsq
=
150, A^5o
2200,
=
Z)5o
220
220 and the payments are
=
made monthly
9-50.
A
10,000 term insurance policy for 20 years
The
net annual
thereafter
M45
(a)
is
-
800,
A^45
Des
=
=
9-51. Derive each
rc.\
P -
/^\
(a)
D
Mes
each year
is
=
520, N^s
=
25,000,
^45
=
1200 and
5500.
A^65 =^
(b)
Find
issued at age 45.
for the first year
X and for
X in each of the following cases:
premium
2X.
is
D45
25,000,
=
1200,
/
=
=
A^65
.03,
5500
and
initial
death
400.
of the following formulae:
-
1^
- Mc - M^+n + Dx+n
x.n\
9-52.
A
tPx=
life
^X
-^ =
AJ
Mc
_\r
insurance policy
is
issued at age 30.
The
At age 40 and each year thereafter, the death
benefit increases by 5000 per year for ten years. No death benefit
is payable after age 50.
Annual premiums
are payable at the
benefit
is
10,000.
beginning of each year for 10 years.
M50
7^40
= 800, R40 =
= 68,000.
33,000,
R50
=
Find
X
X given
20,000
9-53. Repeat Question 52 if the death benefits
N30
=
do not stop
but increase indefinitely by 5000 per year, and if the
payable for as long as the purchaser survives.
=
1200,
90,000,
and
M^o
20 years,
premiums are
after
218
Chapter 9
9-54. Aloysius, aged 40, purchases a
death benefit
the
sum of the
the death benefit
later,
payable
7V60
is
=
at the
13,000,
is
life
net
20,000.
Before age 60, the
policy.
premiums
At age 60 and
Find the net annual premium,
paid.
beginning of each year for 20 years, given
Ne\
=
N^q
12,010,
=
61,000,
7^40
=
/
—
.05,
31,000 and
/^60- 11,500.
60
9-55. Repeat Question 54 if the death benefit before age
the net
9-56.
A
premiums paid plus
100,000 whole
payable
at the
life
are payable at the
20%
gross
annual premium
/
=
is
issued at age 25.
60%
premium
if A^25
=
the
sum of
of the
first
Premiums
are
and death benefits
life,
The following
end of the year of death.
of the gross premium
10% of the
and
policy
beginning of each year for
required for expenses:
is
interest.
are
year's gross premium,
in
years two through five inclusive, and
in
each subsequent year. Find the gross
120,000, Nie
=
112,000, N^o
=
96,000
.06.
9-57. Repeat Question 56 if an extra expense of 3
is
required at the
beginning of every year throughout the policy.
9-58.
A
whole
life
insurance policy
annual premium
life.
is
120 payable
is
Expense provisions include
The gross
issued at age 30.
at the
beginning of each year for
50%
of the
first
year's gross
annual premium, 10%) of each subsequent gross annual premium,
and 200
at issue.
amount of the
9-59.
Given A^o - .068 and a^o
=
15, find the face
policy.
A
deferred life annuity, issued at age 35, will provide monthly
payments of 1000 commencing at age 65. Annual premiums are
payable at the beginning of each year for 20 years. During the
deferred period, the death benefit is equal to the sum of the gross
premiums paid.
Gross premiums are equal to 110% of net
premiums. Find the gross annual premium given 7V65 = 5000,
Des
Rs5
= 500,
A^55 = 12,500,
= 7500 and Mes = 300.
7V35
=
38,000,
R35
=
20,000,
L ife
219
Ins urance
9-60.
A
life
insurance policy issued at age 50 provides a death benefit of
sum of
premiums which would
otherwise be paid after the date of death. Premiums are paid at the
beginning of each year for 30 years. The loading is 10% of each
net level annual premium. Find an expression for the gross annual
10,000, less the
the gross annual
premium.
commutation functions for the gross
9-61. Derive a formula in terms of
annual premium
age
at
insurance policy of
payable for n years, for a whole
x,
amount of h
amount
the expenses consist of a flat
1, if
life
per year, as long as payments continue, plus an additional
oi k
plusy percent of each premium paid, plus an
in the first year,
extra expense of r in the year the claim
9-62.
Morgan, aged 40, purchases a whole
made.
is
policy paying 50,000 in
life
case of death within the next 10 years and 100,000 thereafter.
Assume M40
=
740,
M50
=
580 and N^q
59,000.
Find the net annual premium for this policy.
(a)
Find the terminal reserve
(b)
A^45
=
42,000 and D45
=
Ds5
=
1150and7V55
=
at
at
are funded
by
his future
age 55, assuming M55
years and
3X thereafter.
=
475,
of Morgan's future benefits that
whole
life
policy paying 5000 the
year and increasing by 5000 per year thereafter.
at the
675,
premiums.
9-63. Rosalita, aged 20, purchases a
annual premiums
=
27,000.
In part (c), find the portion
(d)
age 45, assuming M45
1700.
Find the terminal reserve
(c)
first
=
She pays
beginning of each year of Jf for the
Express
in
first
10
commutation functions each of
the following:
(d)
The
The
The
The
Show
that u.-x-\ ^x
(a)
(b)
(c)
9-64.
value of J^
reserve just after the
first
premium payment.
terminal reserve at age 25.
terminal reserve at age 50.
population.
=
^
—
Px,
where
u; is
the terminal age of the
Chapter 9
220
9-65. Benjamin, aged 35, purchases a 30-year term insurance policy
Level premiums are payable
with a face value of 50,000.
Assume
beginning of each year for the next 20 years.
A^35
=
7300,
and Nee
(a)
(b)
=
=
iV36
=
7000, Nss
2400,
=
A^56
2265,
A^65
at the
= .03,
= 1050
/
960.
Find the amount of each premium.
Find the
A^45
=
10'^
year terminal reserve for this policy, assuming
4200 and
A^46
=
4000.
(c)
Find the 20'^ year terminal reserve for this policy.
(d)
Find the 30^^ year terminal reserve for
this policy.
9-66. Repeat Question 65 if the death benefit, instead of having a face
value of 50,000,
with the
first
is
a 20-year interest-only annuity of 5000 per year
payment
at the
end of the year of death.
9-67. Betty, aged 40, purchases a 100,000
premium of 36,000. At the same
100,000 whole life policy with a
payable
at the
whole
policy with a single
life
time, Betty also purchases a
net annual
beginning of each year for
premium of 2200
life.
Assume
/
is
the
for both policies. The net amount at risk (the excess of the
amount over the terminal reserve) for the tenth year on the
first policy is equal to K, and the net amount at risk for the tenth
year on the second policy is L. The net single premium at age 50
same
face
for a 100,000
whole
life
policy
is
48,600. Find ^.
9-68. Prove each of the following identities:
(^)
(b)
'^-
,V,
_ PAN, - K^,)
=
a.,
M, - M,^,
-
-t
= i^x+t ~
9-69. Find ,+|F, given
D
^x
A = .013, ,F^ = A3,q^+, = .004 and
= A+, (l +
^
-
9-70.
Show that
9-71.
Show that, F,= 1-(1-|F,)(1-|F,+ ,)
,
F,
P.)
^
/
P.-
•••(1-1 F,+,_|).
=
.03.
Life Insurance
9-72.
Assuming 6
221
=
.07
and ^^
=
.04 for all x, find each of the follow-
ing:
(a)
9-73.
P(A.)
(b)
P,
As with premiums,
(c)
P(Ax)
(d) P(aIj{)
where the payments are made
continuously are denoted by placing a bar over the V, and reserves
where the insurance is payable at the moment of death are denoted
by including the net single premium for the insurance in paranreserves
theses after the V.
(a)
(b)
Given a^ = 23, a^^s
Given a^.-^ = 16,
=
= .03,
= ^^
^;c+io;^^:To|
19 and 6
find 5V{Ax).
^"^
6
=
.03,
find
lo^a.;^)-
9-74.
Under "normal" circumstances, we would expect
/F;^
>
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
<
x
<
105, find the
mean and variance
of r(30), the future lifetime of a person currently aged 30.
Solution
The mean
is
just e^o
calculate the second
=
ip2odt=
/
moment
/
'^7
as
EIT\30)]= [ t',p,om+,dt
Jo
dt
_
=
Then the variance
1
f(^5f
1875.
is
1875
-
(37.5)^
=
468.75.
^
dt
=
37.5.
Next we
230
Chapter 10
To
close this section,
of median.
The median of
we would
like to briefly
a set of measurements
of the measurements are
that precisely half
mention the notion
is
Questions involving the median are best answered from
Here
is
m
number
a
such
m.
less than or equal to
first principles.
an easy example.
Example
10.7
Find the median of
71(30) in
Example
10.6.
Solution
To
solve this,
we
simply need to find a time
the time 71(30) will be less than
-nT
^
=
We
last
A. Solving,
obtain
/
=
such that precisely half of
this just
means
that tPzQ
=
is
^,
or
D
37.5.
remark that although the mean and median were equal
example, this
10.2
we
But
/.
t
in the
not the usual situation.
NORMAL DISTRIBUTION
The primary purpose of the previous
of variance.
In
practical
section
interested in the square root of the variance.
deviation and
is
denoted by
cr,
G
Although E[Z]
is
so
=
unlikely that any specific value
between E[Z]
—a
to introduce the notion
This
is
we
will
be more
called the standard
we have
y/Var(Z).
the expected value or
somewhat more
was
applications, however,
is
mean of a random
(10.6)
variable Z,
exactly equal to E[Z].
We
can be
confident, however, that a specific value might
and E[Z\
—
+ a.
This confidence increases
if
we
it is
lie
extend
2a), and continues to increase if we
from
the mean.
allow further and further deviation
To be more precise about the ideas discussed in the last paragraph,
we need to know the underlying distribution of Z. However, there is an
observation due to Chebyshev which can be used in all cases.
the interval to {E[Z\
2cr,
E[Z]
-\-
J
23
Statistical Considerations
Chebyshev's Rule
If
A:
>
0,
and E[Z\
the probability that a specific value lies
+ ka
is
greater than or equal to
—
1
— ka
between E[Z\
-j
Chebyshev's Rule can be restated as saying that
<ka)>\-^.
Pr{\Z-E[Z\\
For instance, setting
Example
In
A:
=
2
we
see that Pr{\Z
—
(10.7)
<
E[Z]\
2a)
>
.75.
10.8
Example
Chebyshev's Rule
10.1, use
to find
an interval for
Z
such
that the probability that a specific value lies in the interval is at least
(a).75;(b)|.
I
Solution]
We
=
have E[Z]
that Pr{\Z
-
.30
<
E[Z\\
(E[Z]
-
and a
2a)
>
= y/M\53 =
2a, E[Z]
we have Pr(\Z -
For part (b)
(E[Z]
-
3cr,
+ 2a) =
<
E[Z]\
E[Z]
+
is
(-.27106, .87106).
3o-)
3a)
we know
.28553. For part (a)
.75, so the interval
=
>
§, so the answer
is
D
(-.55659, 1.15659).
A
moment's thought, however, tells us that these intervals are not
helpful. We already know from practical considerations that Z > 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
<
probability
of
is
gives us .75).
(if
being
in
.87106 and
this
if
easily
we
the
(y^) =
is
1
.03
that
Z
calculate
interval
death occurs
occurring
.9409 that
death occurs in the
probability
(a),
be bigger than .87106
probability
.93458
know with
we can
actually
In
=:
in
873
first year).
is in
>
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
<
c)
large the payout
underlying distribution
in
problems of this type
is
to
fmd
a
relatively high, since this puts an upper
is
is
likely to be.
If
we assume
the
symmetric, then Chebyshev's Rule can be
is
we saw in the last example that it is
much stronger assumption would be to
applied to this sort of problem, but
A
not likely to be very useful.
assume
that the underlying distribution
tables can be found in
any
statistics
is
normal.
Normal
distribution
textbook, but the only values
we
will
need are the following:
Pr(Z
E[Z]
.842a)
H-
=
(10.8a)
.80
Pr(Z
<
E[Z]
+
1.282cr)
-
.90
(10.8b)
Pr(Z
<
E[Z]
+
1.645a)
=
.95
(10.8c)
.975
(10.8d)
.99
(10.8e)
Pr(Z
Pr{Z
Example
<
<
<
E[Z]
+
E[Z]
1.960cr)
=
+ 2.327a) =
10.9
Assuming a normal distribution in Example 10.1, fmd a constant c such
that Pr(Z < c) is equal to (a) .80; (b) .90; (c) .95; (d) .99.
ISolutionI
(a)
c
(b)
c
(c)
c
(d)
c
=
=
=
=
+ .842a = .30 + .842(.28553) =
+ 1.282(.28553) = .666
.30 + .645(.28553) = .769
.30 + 2.327(.28553) = .964
E[Z]
.54
.30
1
D
While the above answers seem more reasonable than those of
Example 10.8, some of them are still inaccurate. For example, (c)
indicates that
we
are
95% certain Z
Z is less than
is
The moral
is
that
we have
less than .769,
whereas an exact
.769 only 91 .3% of the time.
calculation reveals that
to
be very careful about drawing
conclusions based on an assumption that the underlying distribution
normal. Students with a background
about
last
why
the normal distribution
example.
is
in statistics are
is
encouraged to think
not particularly appropriate in this
Statistical
10.3
Cons iderations
233
CENTRAL LIMIT THEOREM
After examining the last section, the reader might feel that the normal
distribution assumption will be of
the case at
we
First
and
Var{Z\)
we
In this section,
all.
recall
little
from Section
are
how
it
this
is
not
can be applied.
10.1 that, in general,
Var{Z\
However, when Z\
+ Z2)
and Z2
VariZi)
independent random events, these expressions are equal.
represent
-\-
different.
extends to sums of more than two random
Moreover,
this observation
variables.
What do we mean by independent? Simply
random
different individuals.
that the events
The fundamental example
have no effect on each other.
the present value
to
But
use in our work.
will see
for us will be
variables for insurance (or annuities) sold to
Because the survival of one individual
is
assumed
have no effect on the survival of another, these random variables are
independent.
we have
(Actually,
assumed
tacitly
this in a
number of
places earlier in the text.)
Example 10.10
Example
assume 100 policies of the type described are sold to
and let Z denote the present value of aggregate
future death benefits for all those policies. Find E[Z\ and Var{7).
In
10.1,
different individuals
[Solution]
The expected value of aggregate death
E[Z\
=
benefits
100(.30)
By assuming independence among
=
is
30.
the different individuals, the variance
is
Var{Z)
Note
we were
=
100(.08153)
=
8.153.
D
that if instead of 100 unit policies sold to different people,
amount 100 sold to one
30 but Var{Z) would now be
equal to (100)^(. 08153) = 815.3. It makes sense that this answer should
be larger than before. Any deviation of a single individual's present
value of benefit from its expected value would be magnified if the face
amount of the policy were increased, whereas deviations among a group
talking about a single policy of face
individual, then
we would
still
have E[Z]
=
of 100 individuals would tend to cancel each other out.
234
Chapter 10
The important connection of all
this
with the normal distribution
given by the Central Limit Theorem, stated here
will apply
in the
way
is
which we
in
it.
Central Limit Theorem
If
independent random variables Z] Z2,
,
tion,
and
n
if
then
large,
is
the
.
.
.
,
Z„ have the same distribu-
sum Z
=
Zi
+ Z2
-h
•
+ Z„
is
approximately normally distributed.
Although
be
«
in
=
in general
it
is
not that easy to decide
order for the approximation to be very good,
how large n must
many authors use
30 as a rough yardstick. Here are some examples.
Example
10. 11
Example 10.10, find the value of c such
(b) Pr{Z <c)= .99.
In
that (a)
Pr{Z
<
c)
—
.95
and
ISolutionI
The Central Limit Theorem allows us
normal distribution here, with a
we have 30 +
30
(1.645)(2.855)
+ (2.327)(2.855) =
=
= y
to
more confidently assume a
8.153
34.7,
=
and
2.855.
for
Then
part
(b)
for part (a)
we have
D
36.6.
Example 10.12
50 independent
lives,
each age 40, contribute equal amounts to establish
pay 1000 at the moment of death of each of the
individuals. Given /ix = 02 for all x and 6 = .07, find the amount that
each must contribute to have 90% confidence that all claims will be
a fund which
will
paid.
ISolutionI
If Zj
is
the
random variable representing present value of the death
benefit for the
i^^
person,
we have
'DC
E[Z,]
=
1^000
/
Jo
_
~
2000
9
.02t-.01t
(.02)^-^^'^-^^'^^
235
Cons iderations
Statistical
and
=
Var{Zi)
/ {m)e- ^^'
(1000)2
e'
^^' dt
-
{E[Z,]f
Jo
= (1000)2(1)=
we
(^)
75,617.28.
Since the lives are independent,
values
(1000)2
if
Zis the sum of the individual present
then have
E[Z]
= 50(^^) =
11,111.11
and
=
Var(Z)
Hence a
=
amount
1944.44, and the total
+
confidence must be 11,111.11
=
to be contributed for
=
13,603.88.
90%
Each
D
272.08.
LOSS-AT-ISSUE
Consider a whole
insurance policy of
life
If death
were
to
occur
at
1,
payable
who
year of death, purchased by a person age x
Q.
3,780,864.
(1.282)( 1944.44)
person must contribute ^(13,603.88)
10.4
=
50(75,617.28)
at the
end of the
pays annual premiums of
age x-hk, the value
at
time of purchase of
would be v^+', whereas the value of the premiums paid
a-j—^^. The difference
the death benefit
would be
Q
L
is
= v'+'-Q-d^^
(10.9)
called the loss-at-issue for this policy.
This
new random
variable
L
is
quite important because
whether the issuer of the policy has
transaction.
In
this
section,
we
lost or profited
will
remember from before that £[v^^^] = A^.
Note that we are not assuming here that
Px. Instead, we are opening up the possibility
we
will
assume
that
Q=
Px-
measures
apply the ideas developed
previous sections to answer questions about L.
be different from that calculated earlier.
it
from the given
It
Q
is
in
will be important to
necessarily equal to
that the
premium
Q may
In our first example, however,
Chapter 10
236
Example 10.13
L
is
the loss-at-issue
payable
random variable
pays annual premiums of P30.
fmd E[L] and
for a
whole
life
end of the year of death, purchased by a
at the
Given
tp^Q
—
insurance of
life
(.91)' for all
/
age 30
and
Var(L).
Solution
First
we
calculate P^n
=
da^.
=
-jt^
<^30
^30
We
fmd
^^30
= E^S/^30 =
^(t^)
/=0
^=
-
/=0
1.07
SO that
P30
=
- (t^ )(iQ-7)
1
10.7
3
107
and
^30
= l-(y^)(10.7) =
Next we fmd
=
v^+i
-
v^+^
=
v^+'
-
3
,
107'h+i\
f\-v'^'
3
107
-
,^+1
[
^
\
1.07
3
.07
107
k+\
i\-v''^')
10^M\
3
7'
Then
E[L]
=
J^£[v^+i]-
—
lyj
10
^
A
7
-ion
-^- 7 uoyA_37 _Q
10
10.7,
/
=
1,
who
.07,
237
Statistical Considerations
Next we need
to find
=E
E[U-]
49
^
iOO^r
^
49"^
49
7A+2n_60M
49 ^Lv^
J
A
49^07 +
49
We have
E[v^''^^]
=
2k+2
Y.f^P^oq30+ky
SO that
^[^^]
18
00
= ^(.17153)-^
+
^=
6638.
and finally
Var(L)
In the
zero.
= .16638-0 2 _
above example,
it
In fact, the reader will
when
occurs precisely
premium
is
Q=
was no
16638.
surprise that E[L] turned out to be
be asked to prove
Px-
in the exercises that this
Sometimes the phrase "the annual
determined by the equivalence principle"
is
used to describe
this situation.
If a
whole
life
policy has a benefit of
1
payable
at the
death (rather than at the end of the year of death), and
are paid continuously, then the appropriate formula for
L
if
moment of
the
premiums
is
(10.10)
The next example
illustrates this situation.
Example 10.14
Let L be the random variable for fully continuous
(i.e.,
premiums paid continuously) whole life insurance
The total annual
Assuming fi^ = 02 for all jc and 6 = .08, find E[L] and
the time of death and
with face amount 1000 purchased by a 20-year-old.
premium
Var{L).
is
24.
benefit payable at
238
Chapter 10
Solution
We have
Z=
1000v^-24a^
= 1000v^-24(^i^)
=
1300v^-300.
Then
E[L\= 1300£[v']-300
=
,/?20/^20+y ^/
1300/
-
300
Jo
poo
=
1300(.02)/
=
-40.
e-^^'^/-300
We also find
E[L^]
=
(\300fE[v^']
=
(1300)2
/
-
+ 90,000
(2600)(300)£[v^]
^p20/^20+/V^^ dt
-
780,000(.20)
+ 90,000
Jo
Finally
=
(1300)2(.02)('^)
=
121,777.78.
-
66,000
we have
Var(L)
=
The next example
121,777.78
illustrates
-
(-40)2
how
^
120,177.78.
the normal distribution can play
a very important role in loss-at-issue calculations.
Example 10.15
In
Example
10.14, find the
minimum number of
described that an insurer must issue
total loss will
in
order to be
policies of the type
95%
certain that the
be negative.
[Solution]
We
must assume that the different policies are independent. Let the
number of policies issued be n. From the last example, we know that the
expected value of the
total loss is
—40n and
the variance
is
120,1 77. 78«,
[
239
Statistical Considerations
so that the standard deviation
mation,
y^>
we
must
14.258 and
-40«
have
>
A?
Using a normal approxi-
346.7 a/^.
is
+ (1.645)(346.7y^) <
Hence
0.
D
204.
EXERCISES
10.1
Mean and Variance
10-1.
Let
Z
for a
denote the present value random variable,
whole
policy with a death benefit of
life
at
1
policy issue,
payable at the
end of the year of death purchased by a 30-year-old. Find E[Z]
and Var{Z) if/
=
10-2.
Do
Question
1
if
the death benefit
10-3.
Do
Question
1
if
the policy
is
for 30-year term insurance.
10-4.
Do
Question
1
if
the policy
is
for 30-year
10-5.
Do
Question
1
if
the insurance
Do
Question
10-6.
1
.09
if
and
=
,;?30
is
(-97/ for
all
t.
1000 instead of
is
1.
endowment
insurance.
deferred for 30 years.
the death benefit
is
payable
at the
moment of
death.
10-7.
Let
Z
for a
denote the present value random variable, at policy issue,
whole
life
policy with a death benefit of
moment of death purchased by
for all
X and 6
=
.
1 1
.
Do
Question 7
10-9.
Do
Question 7 assuming
Let
Z
10-10.
payable
at the
—
.03
<
100.
Find E\Z\ and Var{Z).
10-8.
if
1
a 30-year-old, assuming ^^
the policy
(5
is
=
for 40-year term insurance.
.1 1
and
4=
100
-
jc,
<
x
denote the present value random variable for a whole
policy of
Var{Z)
if fi
1
payable
=
fix
at the
moment of
death.
and S are both constant and if6
life
Find E\Z\ and
=
2//.
240
10-11.
Chapter 10
Determine the mean and variance of the present value random
variable for a
year, given
10-12.
Do
10-13.
A
Question
.09
if
1 1
and
per year,
1
is
in
one
t.
continuous.
annuity will pay 2
life
payment
first
{-915y for all
at the
end of the
year, 5 at the end of the second year, and 4 at the end of the
Given
third year.
=
tPx
—
the annuity
three-year temporary
first
/
annuity of
life
=
/
= 98, ps\ = .978, ps2 = 975 and
variance of the present value random
that pso
mean and
.07, find the
variable for this annuity issued to a 50-year-old.
10-14.
10-15.
Given that p.^ is constant, 6 — .09, and
and -^a^.
(Recall that the superscript
doubling of the force of interest.)
Given
that
/x^
-^A^
to
=
i
.10, calculate ax
the
denotes a
left
and 6 are both constant, find expressions for each
of the following:
The mean, variance and median of the present value
random variable for a whole life insurance of 1 payable at
(a)
the
moment of death.
The mean, variance and median of the present value
random variable for a continuous life annuity of 1 per
(b)
year.
10-16.
Find the mean and variance of 7(40)
in
each of the following
cases:
(a)
£x=\00-x,0<x<
(b)
,;7,
=
10-17.
Find the median of 7(40)
10-18.
Given
aj^>
10-2
10-19.
4=
10,
100
(.97yforalU
1000(^1
where
in
- y^]
T represents
each of the cases
and
/
=
in
Question
16.
.07, find the probability that
the future lifetime of a
life
aged 50.
Normal Distribution
In
Question
1,
use Chebyshev's Rule to find an interval for
such that the probability that a specified value
is at
o
least (a) .75; (b) g.
Are these
lies in
intervals useful?
Z
the interval
24
Statistical Considerations
10-20.
What
value of k
10-21.
Assuming
-
(c) Pr(\Z -
Show
in
<
Chebyshev's Rule
ka)
>
<
c)
c)
=
Pr{Z
<
E[Z]\
-
Pr(Z <
Pr(Z < c)
.90; (b)
.90; (d)
is
Using the
order to
1,
find the constant
c)
=
=
.975.
.95;
where r
is
any positive
equal to r times the standard deviation of Z.
result
death benefit
of Question 22, redo Questions 19 and 21
is
1000 instead of
1.
Redo Question 21
if
if
the
(This could also be done
using Question 2 but the suggested method
10-24.
in
.95?
that the standard deviation of rZ,
number,
10-23.
E[Z\\
a normal distribution in Question
c such that (a)
10-22.
needed
is
guarantee that Pr{\Z
is
preferable.)
the death benefit of Question
1
is
payable
at
moment of death.
the
10-3
Central Limit Theorem
10-25.
assume 1000 policies of the type described are
let Z denote the present value of
aggregate future benefits for all these policies. Find £'[Z] and
In
Question
1,
sold to different individuals and
VariZ).
10-26.
Redo Question 25
if
a single policy with death benefit 1000
is
sold.
10-27.
Redo Question 25
if
10 policies with death benefit 100 each are
sold.
10-28.
100 independent
establish a fund
lives,
which
each age 20, contribute equal amounts to
pay 10,000 at the end of the year of
Given ^/?20 = (-98/ and / = .06, find
each must contribute to have 95% confidence
will
death of each individual.
the
amount
that
that all claims will be paid.
10-29.
Redo Question 28
if
there are 1000 lives instead of 100. Explain
the relative magnitude of your answers to the last
two
exercises.
242
10-30.
Chapter 10
Redo Question 28
the insurance
if
is
payable
at the
moment of
death.
10-4
Loss-at-Issue
10-31.
Let
L be the loss random variable for a fully discrete (i.e., annual
premiums and benefit payable at the end of the year of death)
whole life insurance of 1 purchased by a life age 25. Find £[1]
and Var{L)
(a)
d=
(b)
7%
=
/
in
each of the following cases:
.40, ^A25
=
.23,
and the annual premium
is
of the amount of insurance.
4=
.08,
premium
10-32.
—
A25
.08,
is
1
10
-X
<
for
jc
<
1
and the annual
10,
determined by the equivalence principle.
Let L be the loss random variable for a fully continuous whole
life
insurance of
Var(L)
(a)
6
(b)
/
in
1
purchased by a
life
Find E[L] and
age 40.
each of the following cases:
= .06, fix = .03 for all x, and the annual premium is .04.
= .06, iP4o = (.96y for all and the annual premium is
/,
.05.
10-33.
In a fully discrete setting,
(a)
E[L]
prove that
only
Q=
Px
if
and only
if
0.
In a fully continuous setting,
(b)
10-34.
=
=
if E[L]
prove that
Q = P(Ax)
if
and
0.
Let L be the loss random variable for a fully continuous whole
life
insurance of
net annual
but that
5%
well that
fi
1
purchased by a
premium
is
=
is
life
age
x.
Assume
that the
determined by the equivalence principle,
added to each premium for expenses. Assume as
fix and 6 are both constant and that 6 = 4fi. Find
E[L] and Var(L).
10-35.
In
Questions 31(a) and 32(a), find the
minimum number of
policies of the type described that an insurer
be
99%
must
certain that the total loss will be negative.
sell in
order to
CHAPTER ELEVEN
MULTI-LIFE THEORY
11.1
JOINT-LIFE ACTUARIAL FUNCTIONS
we have
In the previous chapters
relative to a single life:
individual or an insurance payable
this chapter
we
discussed annuities and insurances
an annuity payable during the lifetime of a given
will see
how
upon the death of that
individual.
to extend this theory to handle
In
problems
more than one individual. Clearly
types of questions which can arise, and rote
involving the death or survival of
many
there are
different
memorization of formulae will not be very helpful. As in previous chapters, we will stress the solving of these problems from first principles.
There
that will be
is
some
done
additional notation
first.
Then we
will
which must be introduced, and
discuss life annuities which are
dependent on the survival of a group of individuals, as well as life
insurance payable upon the death of the first member of this group;
these are cdiW^d joint-life functions.
In subsequent sections other types
of multi-life problems will be discussed.
We saw briefly in earlier chapters how probabilities involving
more than one person could be calculated. Here are two examples which
will recall these ideas.
Example
Herman
is
11.1
30 years old and Hermione
is
25.
Find expressions for each
of the following:
(a)
The
probability that both
Herman and Hermione
live for
30 more
years.
(b)
The
probability that at least one of them does not live for 30
more
years.
(c)
(d)
The
The
probability that
Herman and Hermione both reach age
60.
probability that one dies before age 45 and the other dies after
age 55.
(e)
The
45.
probability that at
most one of them dies
at
age
last
birthday
Chapter 11
244
ISolutionI
(a)
(30^30 )(30j^25)
(b)
1
(C)
(30/^30 )(35/?25)
(d)
There are two cases here since either person could be the
-
(30/?30)(30/?25)
The answer is (1 - i5P3o)(3oP25)
the complement of the case
death.
This
(e)
birthday 45, so
Example
Given
+
we have
1
-
-
{isPi>o){^
loPis)-
that both die at age
is
(15^30
-
leP^oXioPis
10/735
=
.8,
-
first
last
ixPis)-
.2
1 1
=
io/?20
-9,
—
xspio
75 and
find the probability that
George (aged 20) and Ellen (aged 30) will both reach age 35, whereas
Maurice (aged 20) and Bemadette (aged 30) will die between age 35 and
age 45.
[Solution]
Since
is given by {xsPioXsPioXxsPiQ - isPioXsPso - \5P3o)of these probabilities are among the given values we must solve
The answer
not
all
= (•9)(. 75) =
=
= «4375. Also 3P30 =
-^- =
Then we have (.84375)(.9375)(.16875)(.1875) = .025028.
for
We
=
them.
have
S ^
In several of the
several lives
notation
X],
JC2,-
•
all
=
we
above problems
surviving for a
tells
is
one of the
Example
it is
Show that
1 1
^
q^^
+
q^,,
-
~
=
.3
=
D
new
denoted nPx^xr-x^- Our work
q^^q^^.
nPx\X2-
(1
)•
n<3xiX2-Xr„^
lives dies during the next
not true that nqx^xr-x^
q^^^^
.9375.
independent lives of ages
x^= (nPx, ){nPx2 )" (nPx^
n^x\X2-Xm ^^
that
f=
In this case, a
years.
m
Other formulae follow naturally. For example,
Note
and
us that
nPx,X2-
that at least
.675,
calculated the probability of
number of
will all survive for n years
Xm
on probability
S
(io/?2o)(i5/?3o)
helpful: the probability that
is
•,
25/^20
n years,
•
1
the probability
is
given by
V^^-^)
Xm-
(W;c,)(/7^X2)-
1
'
inqxj-
245
Multi-Life Theory
ISolutionI
+
qx,
The
tion.
dies,
- Px, +
1
q^j
we
=
-Px^Px2
1
- Px^ - {X-p^X^-p^j)
1
Example
1
1 1
-Px.xj
=
next year,
is
qx^
two
lives,
the probability that the first
the probability that the other
is
n
qx,x2-
has a nice verbal interpreta-
.3
the probability that at least one of the
qx^x2 is
together
=
=
qx,qx2
relation given in
X2, dies in the
and
-
qx2
are counting twice the case
where both
jci
and
aged
;ci,
adding these
In
life dies.
aged
life,
lives die, so
we must
subtract ^^1^x2 to get a correct total.
We may now
joint-life situation.
long as
m
denoted by
lives,
define
A
life
life
life
annuity paying
ages x\, X2,
ax^x2.xn,^
annuity and
x^,
...,
insurance functions for a
at the
1
end of each year as
survive has a present value
all
^^ ha\Q
PC
~
^^lJf2--^/«
2-^ ^
tPxiX2-Xn,
t=\
PC
= E^V., )(/;'..)• ••Cp.J-
(11.3)
t^\
A
first
life
insurance policy paying
death occurs in a group of
value denoted Ax^x2
1
at the
lives,
end of the year
ages X], X2,
.
.
.
,
in
which the
x^, has present
We have
-x^-
^x.xr-xn,
w
=
^
^''^\tPx,X2-Xr„
"
(1 1-4)
t^\Px,xr xj'
If one of the lives subscripted in an insurance present value
symbol has a numeral over it, then the insurance pays at the end of the
year of that death provided the deaths occur in the order named. For
example, Al
of (x)
if {x)
after (y).
is
the present value of an insurance that pays on the death
dies before (y), and A^^y pays
Note
that
Ax
=
A\y
on the death of
{x) if
(jc)
dies
+ Aly.
Other joint-life functions can be defined analogously to their
single-life counterparts.
For example
X\X2--Xrr,
^tPx^Xr-Xrr.'
/-I
(11.5)
246
Chapter 11
A
few problems involving
joint-life annuities
earlier in the text (see Exercises 13-15
was required beyond
that
developed
solution to Exercise 8-13 using our
Example
1 1
now
Let us
at the time.
new
were actually given
8), since no theory
give a
notation.
.4
Julio's mortality for
mortality for
of Chapter
<
1
r
<
1
<
5 is
of a
the value at time
^
<
4
is
governed by
governed by
life
tPy
=
=
tPx
.25(5
-3(4
—
t).
—
If
/).
/
=
Harold's
.07, find
annuity which pays 1000 at the end of each
year as long as both Julio and Harold are alive.
ISolutionI
We
wish to calculate
= 1000^ ^\tPx)itPyX
1000a;c>'
in
our
new
notation.
t=\
Substituting the above formulae for (Px and tPy,
1000[(1.07r '(.9)(1)
we have
+ (1.07)-2(.6)(.75) + (1.07)-^(.3)(.5)] =
1356.61.
D
we
commutation
functions defined in Chapters 8 and 9.
First, however, we need an
analogue of 4- Since nPxyxr-x„, = (nPx^XnPxj)- inPxJ, we have
For future problems
will require analogues of the
•
nPXiXj-Xm
Thus
here
it
is
4
^xy+n^X2+n •••^x„+n
f f ...f
-^JCl ^X2
^Xm
=
'
{^^f^^
v^A"/
4^- The only problem
terms are often quite large, so this product might be
makes sense
that the
_
~
unmanageable. This
to define £xiX2-Xn,
is
dealt with
4, 42*
*
"
by defining
where k is some fixed constant of proportionality, usually a power of
10~^ so the size of the product is reasonable.
We then
have
r.P.,.r:.
=
^"
'
^p^"-^^"
.
(11.8)
^X\X2-Xm
We define
d'X\X2-Xm
—
4iJC2-Jfm
~
^X\
+ \.X2+\--Xr„+\-
V^^-")
247
Multi-Life Theory
Hence
^XyXj
-Xm
—
i
_
~
1
~ Px\X2
X„
+ \:X2+\--Xr„+l
^X]
f
^X\X2-Xm
=
7'^^--^"'
(11.10)
.
-X\X2-Xm
was previously defined by
Recall that the force of mortality
formula
fix
—
—D{ln£x).
If
we
think of jc as a "starting age" and use
the variable, this can be rewritten as fix+i
—
—4-Alrj£x+t)-
the
/
as
This leads us
to the joint-life analogue.
t^Xi+r.X2+t---x„+t
From Formula
(1 1.1 1)
fJ^Xi+t:X2+t--X„ + t
we
—
—
(11-11)
x„+t]
~'^A^^^Xi+t:x2+t
obtain the pleasing formula
~'^iy^
(^^Xi+/^X2+/
- - ^^ (In k
-i-
In £x,+,
'
'
+
'
^Xrr,^t)\
---{- In £x^+,)
= -|(/«4.+/) + ---+(-^(/^4.+.))
=
flx^+t -^
What about commutation
fJ'X2
+
t
-^
functions?
^
fJ^X^
+
(11-12)
t-
Well, recall from
(1 1.3)
and
(11.8) that
PC
<^;ClX2-X^
—
/
^
^ tPx\X2-Xm
t=\
= f^v'
l=\
^^'^p^'
'-"-'
.
^X\X2-X„
(11.13)
Chapter 11
248
a
It's
little
tricky to see
observe that
if
we
to
this into
fit
we obtam
,
m
'x^xr-xrr,
—
/_^
.xi
£.
^
+
+.X2
v-^
t=i
E
our Chapter 8 format, but
multiply both numerator and denominator of (11.13)
m
by V
how
X\+r.X2+t---Xm + t
•••+Jr;„
L'X\Xi-
m
y
^xi+r.x2+t---x„+t
{x\ -\-X2-\
\-Xm) f
^
V
t=\
--Xn
^XiX2-Xm
(11.14)
Hence we define
x^=V
^;c,X2
X]
+ X2
-\-
\-Xff,
^
4,;c2--;c„-
(H.IS)
Similar reasoning leads us to the definition
= y^'^^'^
I
C.,.r-^.
The joint-life functions
as sums of D and C
A^,
~T~
2^
_j_
I
"^^...-.„-
(11-16)
M, S and R are then defined in the usual way
The student should verify that the
functions.
following formulae are consequences of the definitions just given:
^XiX2
Xn,
and
^,.r-..
The case of 2
is an example
Example
lives,
aged x and
>^,
=
^(1
+ O^D^Dy.
(1118)
^^X\X2-Xm
is
often encountered in practice. Here
illustrating that setting.
11.5
Show that Ax>
= #^^'^^-
249
Multi-Life Theory
Solution
I
k(\
We
+ iy-^D.Dy =
saw
in
because
useful
k(\
+
O"^ v^£,vy£y
=
v'+y-'^ k£,£y
=
V
=
D,y.
2
£^y
n
Chapter 8 that the notation
A^i""^
led to the convenient formula
it
notation can also be extended to several lives, as
=
N^
dx"^
we
- ^^^ D^ was
•
=
-A—-
This
illustrate
with the
following example.
Example
1
1.6
John, aged 65, can collect
1000 a month for as long as he
Equivalently, John can elect a benefit
years guaranteed, and
X per month
aged 60, are
his wife, currently
beginning of the month. Find
=
Z)60:65
I
650and7V^^^,^3
=
paying^ per month
lives.
for the next 10
thereafter as long as both John and
Each payment
alive.
X given = .07, Des =
/
is
150,
made
=
N^f^
at the
1250,
1100.
Solution]
John's
the monthly
first benefit,
12,00045^^
= 2,000 ("i^^ =
1
life
annuity, has present value equal to
100,000.
We must compute the present
value of the other benefit in two parts. Since the monthly interest rate
j
=
(1.07)'/'^
xl}-zihp)
—
1,
—
is
the guaranteed portion has present value given by
(1
+y)
=
87.4456830X The other portion has present
J
value equal to 12A^-
^^Q^ = \2x(^^) = 20.3076923X
present value of the joint-life option
is
107.7533753J(^
the present value of the single-life option,
The next example
illustrates
we
obtain
one way
annuities can occur in joint-life situations.
in
Thus the
Equating
X = 928.05.
this to
D
which continuous
250
Chapter 11
Example
George,
11.7
who
has just retired, will receive a continuous annuity of 30,000
per year as long as both he and his wife are alive.
reduces to 20,000
exactly one of
if
continuing until the second death.
annuity assuming ^^
S
=
=
them
The annual
alive,
is
benefit
with payments
Find the net single premium for this
07 for George,
=
fiy
.06
wife,
his
for
and
.05.
ISolut ion
The required present value
is
30,000^;,^ 4- 20, 000(0:, -a:,y)
+ 20,000(^^ -a;,^)
=
+ 20,000a^ -
20,000a;,
10,000^;,^.
We have
12
'0
Jo
'CXj
ay
=
-.05r^-.06/
'0
Jo
and
a^y=
Then
e'^^^tPxydt^
/
the answer
o
o
-^- J8
c.
is
20,000 ('-|2 +
Note
/
in the
^)
-
10,000f
previous example that
if
example and then using
cix
=
ax
-\~
j
first
or Qx
we
could obtain a good
doing the calculation as
=
ax
—
j.
We
out proof) that the same approximations apply to axy and
We
close this
D
292,929.29.
the annuity were payable at
the beginning or at the end of the year,
approximation to the present value by
=
^)
section by mentioning that
if
data
is
in the
remark (with-
cixy.
obtained from a
which follows Makeham's law, then joint-life values can be
in which all lives are of the same age. It is beyond the
scope of this book to discuss how this is done in general, but in the
special case of two lives a Table of Uniform Seniority is often given
life
table
reduced to those
indicating
how
this
should be carried out.
For example,
if
the table
Multi-Life Theory
251
indicates that given an age difference of 4 then 2.4 should be
we
the younger age,
=
^60:64
^62.4:62.4
If the life table follows
simplification
~
^62:62
"
•4(^62:62— ^63:63)-
Gompertz' law instead, an even greater
possible in that
is
all
calculations can be reduced to the
Again, a Table of Uniform Seniority shows
single-life case.
added to
could conclude that
reduction can be carried out.
For example,
if
how
this
the table indicates that
given an age difference of 4 then 6.6 should be added to the older age,
we
obtain.
=
<^60:64
11.2
<370.6
~
^70
"
-^Cavo— «7l).
LAST-SURVIVOR PROBLEMS
The annuity and insurance problems considered
concerned either with a group of lives
death
in
a given group of lives.
all
This
is
insurance
may
Section 11.1 were
first
Other situations are possible.
example an annuity may continue as long as
or, similarly,
in
surviving or with the
at least
not be paid until
all
one person
is
For
alive
individuals have died.
called a last-survivor situation, and will be discussed in
more
detail later in this section.
The reader
will
have no trouble dreaming up other possible
For example an annuity
scenarios.
may be
payable as long as
at least
half the individuals survive. All of these problems should be approached
from
first principles.
Here are two worked examples.
Exercise 8-14)
is
later in the section after
will
do
it
Example
In
here from
1
Example
The
first
(seen previously as
actually a last-survivor situation and will be redone
appropriate theory has been developed, but
we
first principles.
1.8
1
1
.4
find the value at time
as either Julio or Harold
is
alive.
if
the annuity pays 1000 as long
Chapter 11
252
SolutionI
We
we
Next we
consider 3 cases separately and add the results. In Example 11
saw
that the value if both continue to be alive
consider the cases of only one being alive.
Harold
is
not
1356.61.
is
The value
.4
if Julio is alive
but
is
l,000[(1.07)-'(.9)(0)
+ (1.07)-2(.6)(.25) +
Similarly, the value if Julio
is
(1.07)-^(.3)(.5)]
dead but Harold
alive
is
is
=
253.46.
given by
+ (1.07)-2(.4)(.75)
+ (1.07)-^(.7)(.5) + (1.07)-4(1) (.25)] =
1000[(1.07)-i(.l)(l)
Hence the
Example
Three
value
total
1
831.92.
D
244 1 .99.
1.9
aged
lives are
premium
is
x,
y and
Find a nice expression for the net single
z.
for an annuity paying
least two of the
at the
1
end of each year as long as at
lives survive.
SolutionI
I
axy
is
the
premium
to
pay for the annuity while x and y survive, and
and ayz are similarly defined.
Thus
+ a^z + ayz
axy
is
annuity while two lives survive, but the case where
counted 3 times here.
answer
we must
The
new
do want
w
nPx,X2
We
To compare
nPx\X2-
="
nPxxxy xm
aged x\ jc2,
this
Xm
•
premium
three survive
ynPx\
i
nPxi
is
counted once, so to get the correct
+ axz + <^yz —
'^cixyz-
^
important enough to warrant some
is
x^, will survive for n years.
•,
We have
(1119)
-(^-nPx,)(^-nPx2y-<^-nPxJ-
with nPxxxr-Xm^
~
•
,
^
all
denote the probability that at least one of
let
lives,
xrr,
it
subtract off 2^;^^^, obtaining a^y
last-survivor situation
notation.
a group of
We
the
axz
for the
^^
+
\nPx\X2
+
y^ nPx, - X]
*
multiply out and obtain
'
*
'
•••
+
nPxm)
nPx\XT,
+
"^^'^;
i
'
'
"
i
nPxm-\Xm)
(-1)"'+'.;',•^1-^2
+
.
.
.
(
-
Xn
1)^+^ nPx,X2--
KJ
(11.20)
253
Multi-Life Theory
We
also have
nQx^xr
x^
=
1
-
nPx,x2-x^
,
where nqx.xj-xrr, denotes the
Hence
probability that all individuals die within n years.
ngx,X2
x^
In practice, the case
In this special case, the
=
{nqxXnqx^)
of two
lives,
'
'
"
(1 1-21)
(W;c^)-
aged x and
y, occurs quite often.
above formulae become
nPxy
=
nPx
+ nPy "
=
(.nqxXnqy)-
(1 1-22)
nfxy
and
nqxy
Other
last
(11.23)
survivor functions follow easily. For example,
PC
= ^^' tPx.xr
^x,xr-x„
x^
(11-24)
-
t=\
Using Formula
(1
.20) for tPxxxr-xm ^
we
^a,, -^^,,,^+
aX\X2-Xn
The
1
obtain
...
+ (_!)-+!
^^_^^...^^.
(11.25)
special case of two lives gives
a^ ^
The reader
is
a^
+ ay-a^y.
(11.26)
encouraged to give a verbal explanation for
In the case of insurance,
this identity.
we have
oc
"^x^xr-'X^
=
^^^'
2-/
(tPx^xr
x^
-
/=0
Example 11.10
Repeat Example
1 1
.8
using our
new
notation.
t+\PxiX2-xJ-
(1 1-27)
254
I
Chapter 11
Solution
This time
we
Example
1
use the formula lOOOo^
we had
1.4,
=
+
1000^;,
\Q00axy =1356.61.
If
lOOOa^.
lOOOa^,
=
=
1000[(1.07)-i(l)
+
+
Hence
11.3
A
lOOOo^^.
=
1610.07
XOOQa^y.
In
x refers to Julio we have
[(1.07r'(.9) + (1.07)-2(.6) + (1.07)-^(.3)]
Similarly with 3^ referring to Harold, we have
1000a;,
-
=
1,610.07.
(1.07)-2(.75)
+
(1.07r^(.5)
+ 2188.53 -
=
(1.07)-4(.25)]
1356.61
=
2,188.53.
D
2441.99.
REVERSIONARY ANNUITIES
reversionary annuity
is
an annuity which begins at the end of the year
of death of a given individual,
now aged
second individual,
now aged y. The payments
x, as
long as he
is
alive.
are
Note
made
that
to a
x must
be alive when y dies for payments to begin. The notation for this annuity
is Gy^x,
and a formula for
payment occurs
Hence
^y\^
at
time
^
^
t
its
present value
if,
and only
A/;^x)(l
-
if,
is
y
is
easily derived.
dead but x
Clearly a
is still alive.
tPy)
t=\
PC
=
'OO
a,-a,y.
(11.28)
Formula (1 1.28) makes sense. Our reversionary annuity is an annuity to
X, which has present value a^, but will not pay during the joint lifetime
of X and>^, so we must subtract a^y.
Example
In
11.11
Example
1 1
Julio paya ble
.4,
find the value at time
of a reversionary annuity to
upon the death of Harold.
[Solution]
We
can carry out a direct calculation, obtaining a present value of
1000[(1.07)-^(.9)(0)
+ (1.07)-2(.6)(.25) + (1.07)-^(.3)(.5)] =
253.46.
255
Mult i- Life Theory
Note that this is just one of the terms calculated
Example 1 1.8. Alternatively, we could use ay\x = ^x
a;,
=
(1.07)-'(.9)
us \000a,
^
+ (1.07)-2(.6)4- (1.07)-2(.3)=
1610.07.
Thus we have 1610.07
From Example
-
1356.61
More complicated
as
we
=
11.4
in the solution to
—
^xy
As
before,
1.61007, which gives
we have
1000^;,^
=
1356.61.
D
253.46.
reversionary annuities can also be worked out,
see in the following examples.
Example 11.12
State verbally the
life
meaning oi ajy\z, and then express
it
in
terms of joint-
functions.
ISolutionl
ajy\z is
an annuity payable to z upon the death of both x and
y.
Then
DC
t=\
00
= "^^ytPz-
tPzitPx+tPy-tPxyTj
/=1
=
Clz-
Clzx
The symbol a^?
m times
a year.
-
is
Clzy
used to denote a reversionary annuity payable
Analogous
to (1
1
=
a^;^^
If
D
+ Cl^xy
.28)
we have
dr^-d-:>.
approximate formulae are used for a\
^^^~
will cancel out,
and
we
(11.29)
and a\y\ the correction terms
obtain the approximation
a%^a,-a,y.
A
more
careful approximation,
this text, is
given by
whose
derivation
(11.30)
is
beyond the scope of
Chapter 11
256
Many
included
interesting problems involving reversionary annuities are
We
in the exercises.
problems are often not as
will close with an
difficult as they
might
example showing
first
that
appear.
Example 11.13
Siegfried has the option of choosing any of the following three annuities,
first
payment
(a)
(b)
A
A
life
one year:
in
annuity of 1000 per year.
annuity of 750 per year, together with a reversionary
life
annuity of 600 per year to his wife after his death.
(c)
A
life
annuity of
T per
year payable as long as either Siegfried or
his wife survives.
If all three annuities are equivalent, find T.
Solution
Let X be Siegfried's age and
y be
values of options (a) and (b)
his wife's age.
we have
XQQOa^
=
Equating the present
ISOa^
+
600(0^—^;^^),
= j^i^y ~ ^xy) — 2A(ay — a^y). The present
value of option (c) is Ta^, = Tax + Kay—a^y). Equating this to the value
of option (a) we have Tax + T{ay—axy) — 1000^;^, so that
from which we find a^
1000[(2.4)(a -....)]
{2A){ay
-
axy)
+ {ay -
2400,, 05.88.
axy)
3.4
D
EXERCISES
11.1 Joint-Life Actuarial Functions
11-1.
Mary
is
15 years old and Helen and Harry are 25-year-old twins.
Find expressions for each of the following:
(a)
(b)
(c)
(d)
The
The
The
The
probability that
1-2.
Show that
three live 40 years.
40 years.
probability that at most one does not live for 40 years.
probability that
of the twins
1
all
probability that at least one does not live for
lives for
^?g^ =
Mary
reaches age 50 and exactly one
20 years.
(px)(n-iPx+i,)-
257
Multi-Life Theory
11-3.
Show
two lives, one aged 30 and the
same age last birthday is equal to
that the probability that
other aged 40, will die at the
10/?30(l+^40:40)
"
2
•
1
l/'SoCl +^40:41 )
two lives are both governed by the formula
< X < 120, find /i45+/5o+/.
11-4.
If
11-5.
Prove each of the following
1
1-6.
+ 04o)(l l/'SoX 1+^4
=
4=
120
1:41 )•
— x,
for
identities:
- Dx+\y +\
vD^y
(a)
C^y
(b)
D,y^^{D,,){Dyy)
graduate at age 22, find expressions for each of
If 6 students all
the following:
The
(a)
be alive
The
(b)
more than
probability that not
at
3 of the students will
age 50.
probability that not less than 3 of the students will be
alive at age 50.
The
(c)
more than one of the students
probability that not
will
die at age 50 last birthday.
1
1-7.
Francis, aged 65, purchases a life annuity paying 1000 at the end
of each year provided she and her husband, aged 60, are both
Find
alive.
iV60:65
1
1-8.
11-9.
=
net
premium
single
Z)60:65
=
for
this
annuity
if
85.
payment occurs immediately.
Repeat Question 7
if
the first
Repeat Question 8
if
the first 3 annuity payments are guaranteed.
Assume
11-10
the
1500 and
The
A^63 68
=
1280 and
/
== .09.
curtate-future-lifetime of a given situation
is
said to be
the situation survives for n years but does not survive for
years.
If
^;c
=
05,
^;,+
i
=
.07,
qy
=
M
and
n
if
n+
1
^^^i^.H,
determine the probability that the curtate-future-lifetime of the
joint life status {xy)
11-11.
is
equal to
Given two independent
/i;^
=
^'
lives,
1
each subject to a force of mortality
determine the probability that the joint
(40, 50) fails during the next 5 years.
life
status
Chapter 11
258
11-12.
Repeat Question
and the 50-year-old
11-13.
the 40-year-old
if
1 1
is
subject to ^y
is
=
subject to ^^
-OOOSx
— M\y.
George and Sarah purchase a continuous annuity of 20,000 per
Assuming
year, payable as long as both of them survive.
11^ — .05 for George, iiy = .04 for Sarah, and 6 — .08, find the
present value of this annuity.
11.2
11-14.
Last-Survivor Problems
Do Example
1
1.8 if the annuity
of Julio or Harold
11-15.
A
life
pays 1000 as long as exactly one
is alive.
insurance policy
is
payable
at the
end of the year
occurs the second death of two lives, aged 50 and 60.
in
which
Show
that
the probability that the death benefit will be paid exactly
years later
1-16.
is
0/^50 (l-/?60)+ 10/?60(1-J^70)
"
10/?50:6o(l -/?60:7o)•
Express the answer to Question 15
in
terms of single-life proba-
1
l
1
bilities.
11-17.
An
annuity of
1
is
payable yearly to a person aged x as long as
aged y survive, and for n more years
(provided x is alive), except that no payments
both she and another
after the death
will be
1
1-18.
made
of jv
after
life
m years from the present time {m >
n).
Show
that the present value of this annuity
is
An
while both lives survive, 4
if
«-year temporary annuity pays
the first
survives.
life
If the
1
only survives, and 4
if
the second
ages are x and y, find the present value
life
in
only
terms
of annuity functions.
11-19.
Rank
^yx->
the following in increasing order of magnitude:
^^^xy•
ax, a^y,
259
Multi-Life Theory
1
1-20.
If
two independent
of mortality
yix+t
lives (x)
=
M>;+/
=
and
both subject to the force
(y) are
<
03,
/
<
10, find the probability
that the last survivor status (3cy) will survive for 6 years.
11-21.
George and Sarah purchase a continuous annuity of 20,000 per
year payable as long as at least one of them survives. Assuming
lix
=
.05 for
net single
11-22.
George,
premium
//^
—
.04 for Sarah,
and 6
—
.08, find the
for this annuity.
George and Sarah purchase 100,000 of whole life insurance.
Assuming ^^ — 05 for George, iiy — .04 for Sarah, and 6 = .08,
find the net single premium for this insurance in each of the
following cases.
The insurance
The insurance
(a)
(b)
1
1-23.
Smith
year
is
is
annuity
age 50 and
is
payable
at the
time of the
is
payable
at the
time of the second death.
Brown
is
is
age 51.
The
payable to Smith.
equal to the net single
A
life
net single
premium
first
death.
annuity of 100 per
premium
for this
for an annuity of
per year payable as long as either Smith or
Brown
is
K
alive.
Annuity payments are made at the end of each year. A life table
following Gompertz' law tells us that aso = 16.08, as] = 15.72,
as9 = 12.76, and a^o = 12.39. A table of uniform seniority is as
follows:
Difference
in
Ages
Addition to
Older Age
9.12
1
8.63
2
8.16
3
7.71
4
7.27
5
6.86
Find K.
11.3
1
1-24.
Reversionary Annuities
Paul, aged 65, purchases a reversionary annuity of 1000 per year
for his wife,
aged 60.
annuity given the values
andZ)60
65
=
370.
Find the net single premium for this
A^6i
=
1350, D(,o
=
300,
iV6i:66
=
620
Chapter 11
260
11-25.
Can you
calculate the present value of a reversionary annuity
payable to Paul upon his wife's death, from the data given in
Question 24? If not, what additional information
1
1-26.
11-27.
Show
that ax\y
—
ay\x
=
ay
—
a^.
Bertha, aged 40, wishes to purchase an annuity.
two equivalent plans:
500 per month for
(a)
required?
is
She
is
offered
life,
with 300 continuing to her husband
life,
with 150 continuing to her husband
after her death.
600 per month
(b)
for
after her death.
=
Given d\Q
will be
payment under
31, and that the first
made immediately, fmd
either plan
the present value of the benefits
payable to Bertha and her husband.
1
1-28.
In
Question 27, what
to Bertha's
11-29.
On
husband
January
1,
is
the present value of the benefits payable
after her death in
each of the two cases?
1998, Martha, aged 40, purchases a reversionary
annuity to benefit her child, aged 15.
before
December
payments
annuity,
will be 5000.
given
=
a\s
Any payment made on
or
10,000, and subsequent
2008, will be
31,
Find the net single premium for this
31.5,
(^4015
=
19.6,
a^^.jYi
—
^^-^
^"^
^40:15:TTj= ^^^'
1
1-30.
Marcella, aged 35, and Maria, aged 40, purchase a
life
insurance
policy which will pay annual benefits to the survivor upon the
women, first payment at the end of the year
of death. If Marcella dies first, Maria receives 2000 per year for
life.
If Maria dies first, Marcella receives
per year for life.
The policy is purchased by net annual premiums of 400, payable
at the beginning of each year as long as both women are alive.
death of one of the
X
Given
11-31.
Given
a^s
—
4=
15, ^40
120
=
— x,
13.6 and
(235:40
=
12.1,
findX
find the probability that a reversionary
annuity bought by Alphonse, aged 60, for his girlfriend Brigitte,
aged
19, will
be
in
payment
status after 10 years.
261
Multi-Life Theory
1
1-32.
Given
2(7;,^
-\-
la^y + 2^x1;^. + a^i^ = 24,
+ ay\x = 4,
+ 2ay\x = 22, find the present value of a 10,000
and
a^iy
a^\y
last-
survivor annuity to two individuals aged x and y.
1
1-33.
Tim
is
aged 40 and Yvette
is
aged 30. Tim purchases a deferred
month commencing at his age 65, with
payments of 50 per month continuing to Yvette for her life,
annuity, paying 100 per
at Tim's death after age 65. If Tim dies before age
no benefits are payable. Find the net single premium for this
beginning
65,
given
annuity
45^^
11-34.
=16.1, 45^^
Assume
in
will receive
=
Dt^q
=
191,
=157, D^s = 52, D^s =
=10-^ ^"^ = ^^
D/^q
12, 45^65
Question 33 that
50 per month for
39,
''
if
Tim
life
you answer the question now?
dies before age 65, Yvette
beginning
If not,
at
Can
her age 55.
what further data
is
required?
11-35.
Barney is aged 55 and Hilda is aged 65. An annuity of A' per
month at the beginning of each month is payable to Barney for
life.
If Barney dies before Hilda, a reversionary annuity of
Jf — 100 per month will continue to Hilda for life.
The net
single premium for this annuity is 50,000.
Find X given
4f=12,C» =
11-36.
8.5andC>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<^ =
2400 (4^'+ 4'/' -.<^)
=
+
36^<^d^^
200(115+105)+ 100(130)
=
D
57,000.
EXERCISES
12-1.
=
Given Djo
month.
2800,
the present value of all
payment occurs
12-2.
A^60
=
future benefits.
Assume
the first
at retirement.
Repeat Question
the first
12-3.
on a pension of 4000 per
7500 and A^6i = 7000, find
Pauline, aged 20, will retire at age 60
payment
the pension payments are 48,000 per year,
if
1
at the
Repeat Question 2
time of retirement.
the first
if
payment occurs one year
after the
time of retirement.
12-4.
Repeat Question
month
12-5.
12-6.
Is
it
1
after the date
if
the first monthly
possible to answer Question
only continue for a
maximum
additional information
is
Is
it
12-7.
1
if
the pension payments will
of 10 years?
If not,
what
required?
possible to answer Question
guaranteed into perpetuity?
is
payment occurs one
of retirement.
1
If not,
if the pension payments are
what additional information
required?
1000 per month, payable for 15
1999.
Half of this
benefit is financed by his company, and half is financed by a
defined contribution plan consisting of quarterly payments for
10 years beginning March 31, 1989. Assuming Miguel survives
to retirement and / = .06, find the quarterly payment to the
Miguel's retirement benefit
is
years certain, beginning on January 31,
defined contribution plan.
Chapter 12
270
12-8.
Repeat Question 7
retirement at
to
if
we no
longer assume that Miguel survives
Which commutation
age 65.
functions are
needed to answer the question?
12-9.
A
pension fund has a value of 1,000,000 on January
June 30, 1999, a withdrawal of 10,000
is
30, 1999, a further withdrawal of 20,000
made.
is
made.
1,
1999.
On
On September
On December
31, 1999, the fund has a value of 1,1 15,000.
Find
(a)
dollar-weighted rate of investment
(if possible) the
return for 1999.
Assume simple
interest for periods less
than a year.
Find
(b)
(if possible) the
time-weighted rate of investment
return for 1999.
12-10.
Repeat Question 9 given that the fund has values of 1,070,000
just before the June
30 withdrawal, and 1,100,000 just after the
September 30 withdrawal.
12-11.
A
pension fund has a value of 1,000,000 on January
1,
1999.
Withdrawals of 10,000 are made from the fund on March 31,
1999, October 31, 1999, and April 30, 2000. Immediately after
these withdrawals the fund has balances of 1,035,000, 1,085,000
On December
and 1,120,000.
31, 2000, the value of the fund
is
1,200,000.
(a)
Find
(if possible) the dollar- weighted rate
of investment
return for the two-year period from January
December
1,
1999 to
31, 2000.
time-weighted rate of return for the
(b)
Find
same
period.
(c)
Find
(if possible) the effective
(if possible) the
annual rate corresponding
to the rate calculated in part (a).
(d)
Repeat
(if possible) parts (a)
and
(b) for the year
1999
only.
12-12.
Redo Question
know
11
if,
in addition to the
that the fund has a value
1999. There
is
given information,
we
of 1,100,000 on December 31,
no deposit or withdrawal on
that date.
I
271
Pension Applications
12-13.
Roger, aged 25, earns 20,000 per year.
benefit,
beginning
This benefit
at
age 65,
It is
assumed
this firm until retirement.
year and
/
12-15.
=
cr^^
final year's salary.
first payment on the date
Roger survives and stays with
Roger's salary increases
at
4%
per
of future retirement
8.
Repeat Question 13
average
If
that
.07, find the present value
benefits given
12-14.
is
His annual retirement
of his
paid monthly, with the
is
of retirement.
=
60%
if
the retirement benefit
is
60%
of the
of the final 5 years' salaries.
Repeat Example 12.6 assuming that the monthly payments reach
a
maximum
life.
and
of 700 per month, and then remain
Use some of the values
=
5*75
=
A^75
1080, Nje
=
at that level for
=
980, N-jj
900
6180.
must choose between two actuarially equivalent
retirement options. Option I offers annual payments commencing at age 65; the initial payment will be 10,000 and each
subsequent payment will be 4% greater than the previous one.
Alternatively, she can choose early retirement by picking Option
II, which offers annual payments commencing immediately; the
initial payment will be K and each subsequent payment will be
12-16. Karen, aged 55,
4%
values Ds5
ees
12-17.
At 4%, we are given the
greater than the previous one.
=
13.
=
=
150, Des
80, N55
=
2000, Nes
offers a preretirement death benefit of
each completed year of service, payable
Tom
retire at
of
iV63
this
=
=
20,
2000
for
900, ^55
Find K.
Tom's company
death.
=
currently age 63,
is
age 65
if
benefit
was
at the
end of the year of
hired at age 30, and will
he survives to that age. Find the present value
given
30,000 and Ne4
=
M63
=-
28,000.
1000,
M64
=
960,
A^s
=
915,
Chapter 12
272
12-18.
Edward's normal retirement benefit
age 65
at
is
1
100 per month,
with 600 continuing as a reversionary annuity to his spouse.
Edward's
spouse
Edward
12-19.
is
Edward's
8.62 and
benefit
exactly 3 years older
on the date of his retirement, given a^2
his spouse
=
then
Find the value of the benefits to Edward and
than his spouse.
45^^
him,
before
dies
increases to 1500 per month.
If
4^65
=
9-^^'
^-^-
Brian's normal retirement benefit
ally equivalent joint
—
and
50%
is
An
300 per month.
to surviving
actuari-
spouse benefit pays
250 per month. Find the actuarially equivalent joint and 100%
to
surviving spouse benefit.
12-20. Greta, aged 55,
beginning
in
entitled to a pension paying
is
She
10 years.
is
2000 per month
offered the option of using part of
fund to pay her recently divorced husband, aged 60, a
this
annuity of 500 per month
the fund
would be used
commencing immediately. The
=
9.5, 45^^
=
8.5,
10 years.
D^s
=
monthly payments ofX, beginning
200 and Des
at
new payment
Find the
12-21. Leslie selects an optional pension benefit
which
=
85.
entitles her to
age 60, for 10 years certain
and thereafter as long as Leslie and her spouse are both
Leslie
is
5 years older
than her spouse.
benefit at the time of retirement
per month.
=
Deo
Find
280, D55.6O
X
=
is
a single-life annuity of
under the optional form, given
900, N^l^^
=
alive.
The normal pension
2500 and
7V^5^4
=
/
1
,000
=
.09,
2800.
X
is entitled to a monthly pension of
payment occurring immediately. In lieu of
a pension paying 250 per month for life, with
12-22. Allan retires at age 65 and
for
life,
this,
he
with the
may
first
elect
200 per month continuing
Alternatively, he
may
to his spouse for life after his death.
elect a pension
paying 260 per month,
with 180 per month continuing to his spouse for
death.
12-23.
of
pay her a reduced monthly pension
to
benefit, again beginning in
given 41f^
life
rest
life after his
FindX.
Can you solve Question 22 ifXis
not, state
a yearly pension for life?
what additional information
is
required.
If
273
Pension Applications
12-24.
The normal form of a pension
up
retiring in years
50%
with
to
benefit pays a
life
annuity to those
and including 1996, and a
life
annuity
continuing to a surviving spouse for those retiring
We
after 1996.
are given that
we
ees are married, and
will
=
/
.09 and that
assume
75%
of employ-
that the spouse
same age as the participant. Retirement occurs
The present value of all future benefits on January
the
always
is
at
1,
age 65.
1997,
is
10,000,000 for active participants and 2,000,000 for retired
participants.
Given
12-25.
change
—
8
and
value as of January
in the present
to the
a^^
in the
d(^^\^
1
at
age 65.
an equivalent benefit paying
with the
4o6 ^
and
7^75
first 5
'^•^'^^'
=
1580.
997, of future benefits due
=
is
a
life
annuity of 1000 per
Alternatively, he decides to select
X per
years guaranteed.
^65
1
6.5, find the increase
normal form.
Roland's normal retirement benefit
month, beginning
,
=
500, A^65
=
month beginning
Find
X given
4950, D^o
=
at
that
age 70,
/
365, D^s
=
=
.06,
200
ANSWERS
TO THE EXERCISES
CHAPTER 1
1.
a)
20,720.00
b) 22,216.24
c)
14,202.52
2.
a) 23,590.81
b) 22,110.67
c)
24,273.33
5.
b)
No
a)
Amount of interest earned
6.
c)
The
total interest earned,
in the n'^
A(n)
16,497.82
d)
year of the investment.
— A{0),
is
the
sum of amounts
earned each year.
d)
No. The
7.
a)
3|
8.
700.00
9.
14,710.39
10.
.12988
12.
(a)
13.
a)
681.47
b) .136364
14.
a)
7092.84
b) 6501.66
15.
b)
The present value
/V
are rates, not amounts.
b) .08333
c)
t
2.96899
c)
d) .06991
681.47
years in the past
is
/.
_!
.
y
1
—
it
will turn
negative in j years.
18.
Expression (c)
19.
2700.00
20.
C
is /V/^;
all
the others are i^d.
most advantageous to the investor (effective annual rate of
.12603); A is most advantageous to Acme Trust (effective annual
is
rate of. 12551).
21.
3947.80
22.
a)
.12360
b) .11486
c)
.11711
d) .00976
Answers
276
b) .14
.07
23.
a)
24.
24
25.
7
29.
a)
33.
(1+0''
34.
lna
36.
1-6+2! - 3T +
- + - +
b) .13926
.12222
+
3f-
+ d'
Inb
/^
/3
c) .12793
d) .13171
-••
/-6=2T+3T+4T +
is
.
'?
4!
42.
fi
6
d) .13232
Indlnc
/
-
c) .14490
(-*?)""']
[>-
37.
/
•
"*
,
and
8-d=^
c)
2113.17
c)
.08049
larger.
CHAPTER 2
1.
6053.01
2.
5852.81
3.
1377.92
4.
3.86582 years
5.
/
6.
a) 4.26
7.
342.84
8.
19.05945 years
9.
20,279.91
10.
a)
11.
«<'<^
12.
.12928
13.
.097
14.
a)
316.02
b) .17923
15.
a]
.07
b) .08313
=
to Exercises
^ and
/
=
11,168.05
b) 2.68938
b) 4112.94
2!
3!
^
4!
so
Answers
277
to Exercises
16.
.20
17.
a)
.27105
b) .27027
18.
a)
.21053
b) .20513
c)
No, we need the fund balance on July
c) .26316
1,
1997.
CHAPTER 3
1.
a) 82;
297
b) 23;
616
c)
_3_ 2,516,583
^^ 4I6' 1,048,576
3.
1098.41
4.
86.49
5.
22,240.31
6.
a)
5746.64
e) This
year
at
/
103; 286
r.
.
^^
11,487.56
b)
d) 215,233,605; 442,865
59,049
4,017,157
8,388,608' 1,492,992
8443.70
c)
d) 32.7312 years
not possible; the present value of an annuity of 1000 per
is
=
.08 cannot exceed 12,500.00,
which
is
less than 3
our current value.
<
n
<
they are equal
^
12.
a-^
13.
865.57
18.
3896.13
19.
a)
619.14
b) 648.58
c) 850.82
20.
a)
791.44
b) 811.94
c)
21.
19,856.34
25.
s-^;
a)
14,245,831.44]
b)
13,889,685.65
> at
c)
14,599,612.40
1
4%
if
/
0.
986.14
a)
13,479,564.11
b)
13,395,316.83
c)
13,248,156.69
> at
6%
times
Answers
278
26.
^2-
<
.05155
(c)
is
greatest for
(a)
is
greatest for .05155
(b)
is
greatest for
^+
2x
33
x^
/
/
>
<
/
<
.06667
.06667
5=
—y
-\-2x-y
34.
1810.03
35.
32.8125
36.
.10248
37.
49,999.50
39.
a)
40.
.20
41.
10,760.60
42.
546.84
43.
.04881
44.
2,147,717.73
45.
.07177
46.
a)
47.
17;
48.
159; 56.01
49.
24; 573.46
50.
12;
51.
.005637
54.
.08795
55.
.09952
58.
5973.56
59.
(d)
60.
719.85
b) 766.67
666.67
24
c) 6
569.88
369.94
<
(c)
<
(a)
=
(b)
<
(e)
c)
381.17
to Exercises
Answers
279
to Exercises
63.
12,652.20
64.
18,377.37
66.
887.16
67.
b)
68.
^)
-^<^
b)
^>
p-q
^^
p-q
CHAPTER 4
1795.93; 2741.89
1.
a)
27,629.66
b)
2.
a)
2665.34
b) 373.15; 26.85
3.
7595.17
4.
797.34
5.
a)
6.
750.43
7.
a)
True
b) False
54.82; 842.53
d) 179,205.00
9.
True
b) 84,077.57
e)
1101.04
2jih
10.
538.58
11.
.04535
12.
a) v,3(l
13.
Bi\
14.
3544.22
+
/
•
i-c
Sy^)
^i/)
+ (A- B)(\
1.14)
15.
4
16.
c)
36.83
1+y
^JY - X-Sy^.
^^
c)
110.26; 787.09
Answers
280
17.
to Exercises
Payment
Interest
Principal
1
1285.79
700.00
585.79
4414.21
2
1285.79
617.99
667.80
3746.41
Year
Outstanding Balance
5000.00
18.
3
1285.79
524.50
761.29
2985.12
4
1285.79
417.92
867.87
2117.25
5
1285.79
296.42
989.37
1127.88
6
1285.79
157.90
1127.88
Payment
Interest
Principal
Year
Outstanding Balance
50,000.00
1
4537.82
3250.00
1287.82
48,712.18
2
4537.82
3166.29
1371.53
47,340.65
3
4537.82
3077.14
1460.68
45,879.97
4
4537.82
2982.20
1555.62
44,324.35
5
4537.82
2881.08
1656.74
42,667.61
6
4537.82
2773.39
1764.43
40,903.18
7
4537.82
2658.71
1879.11
39,024.07
8
4537.82
2536.56
2001.26
37,022.81
9
4537.82
2406.48
2131.34
34,891.47
10
4537.82
2267.95
2269.87
32,621.60
11
4537.82
2120.40
2417.42
30,204.18
12
4537.82
1963.27
2574.55
27,629.63
13
4537.82
1795.93
2741.89
24,887.74
14
4537.82
1617.70
2920.12
21,967.62
15
4537.82
1427.90
3109.92
18,857.70
16
4537.82
1225.75
3312.07
15,545.63
17
4537.82
1010.47
3527.35
12,018.28
18
4537.82
781.19
3756.63
8,261.65
19
4537.82
537.01
4000.81
4,260.85
20
4537.82
276.97
4260.85
Answers
19.
Year
281
to Exercises
Payment
Interest
Principal
Outstanding Balance
2462.10
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
20.
358.08
-
58.08
2615.82
366.21
-66.21
2682.03
375.48
-
75.48
2757.51
386.05
13.95
2743.56
384.10
15.90
2727.66
381.87
18.13
2709.53
379.33
20.67
2688.86
376.44
23.56
2665.30
400
400
400
600
600
373.14
26.86
2638.44
369.38
30.62
2607.82
365.09
34.91
2572.91
360.21
239.79
326.64
273.36
2333.12
2059.76
600
600
600
600
600
288.37
311.63
1748.13
244.74
355.26
1392.87
195.00
405.00
987.87
138.30
461.70
526.17
73.66
526.34
300
300
300
300
300
344.69
400
400
400
400
400
350.95
44.69
2506.79
50.95
2557.74
Payment
Interest
Principal
1
200.00
160.00
40.00
2
200.00
153.60
46.40
Year
Outstanding Balance
1000.00
960.00
913.60
3
200.00
146.18
53.82
859.78
4
200.00
137.56
62.44
797.34
5
200.00
127.57
72.43
724.91
6
200.00
115.99
84.01
640.90
7
200.00
102.54
97.46
543.44
8
200.00
86.95
113.05
430.39
9
200.00
68.86
131.14
299.25
10
200.00
47.88
152.12
147.13
70.67
23.54
147.13
Answers
282
21.
Year
Payment
to Exercises
Principal
Interest
^
-v"
1
1
v"
'n-ii
2
1-v"-'
v"^l
3
l-v"-2
v"-2
4
l-v"-3
V«-3
5
l_v"-4
V"-*
20
19
-v"
1
v^-
'n-4|
a
19
'/7-20I
1-V2
n-2
n1
n
26.
27.
1
a)
1086.91
c)
.07483
b)
^
^
v3
v-^
1
-V
V
1886.91
Sinking Fund Deposit
is
1912.22;
New Total
Payment
is
6349.48
Old Total Payment was 6278.78
28.
a)
.042775
b) .125
29.
a)
115.00
b) 34.06
c)
375.63
d)
30.
a)
380.08
b) .09590
c)
565.12
d) 87.30
31.
a)
556.50
b) .13164
c)
554.34
32.
a)
749.85
b) .17488
c)
.19
33.
a)
374.67
b) .18805
c)
.17687
34.
9839.98
35.
2760.22
36.
423.63
37.
a)
b) 4911.00
c)
7140.71
4545.92
15'^
d) .13148;. 12083
Answers
283
to Exercises
CHAPTER 5
1.
1067.95
2.
978.94
4.
a)
5.
1120.42
b) 27 { years
.035
7.
^-'2P
8.
.03293
9.
.07267
2 years, assuming an exact
number of years
10.
1
11.
76.82
13.
a)
1049.93
14.
a)
1037.68; 982.68
b) 990.83; ;>90.83
c)
1015.19; 996.85
d)
16.
Time
b)
Coupon
1077.93
1091.93
c)
<
1052.36; 999.79
Interest
,
Principal
Book Value
1055.08
17.
1
35.00
26.38
8.62
2
35.00
26.16
8.84
1037.62
3
35.00
25.94
9.06
1028.56
4
35.00
25.71
9.29
1019.27
5
35.00
25.48
9.52
1009.75
6
35.00
25.25
9.75
1000.00
Coupon
Interest
Principal
Book Value
Time
1046.46
973.79
1
35.00
38.95
2
35.00
39.11
3
35.00
39.27
4
35.00
39.44
5
35.00
39.62
6
35.00
39.81
-3.95
-4.11
-4.27
-4.44
-4.62
-4.81
977.74
981.85
986.12
990.56
995.18
1000.00
Answers
284
18.
Coupon
Time
Interest
to Exercises
Book Value
Principal
1067.95
17
45.00
42.00
3.00
1046.93
45.00
40.73
4.27
1013.88
20.
5048.19
21.
172.05
22.
865.04
23.
68.34
24.
a)
903.47
b) 869.48
25.
a)
12.7; 12.0
b)
26.
a)
929.76
b) 953.74
27.
a)
1046.94
b) .12963
28.
a)
875.91
b)
1128.21
29.
.06064
30.
a) 89.53
b)
111.41
31.
9272.10
32.
a) 50.00
b)
f
33.
a)
100.00
15.25
b) Infinite
c)
1000.00
c)
^^^.^i _
^
I
Answers
285
to Exercises
CHAPTER 6
0^
1.
^^
e)
2.
b)i
3.
d)i
«)|
4.
5.
1;
6.
20
26
7.
§
21
37
8.
43.75
9.
11
10.
.221
11.
6141.30
12.
2192.98; 1696
13.
.0574
14.
.6875
15.
10,258.31
16.
108,888,888.89
17.
2749.31
18.
2343.75
19.
116.18
20.
.1987
21.
.012
22.
a)
.
858.02
b)
1120.88
c)
.08237
F
2704
Answers
286
to Exercises
CHAPTER 7
1.
Age
dx
£x
qx
1000
700
.7
1
300
2
3
126
90
84
126
.3
210
1^
b)
2.
a)
3.
.13043
6.
X
^
c)
.4
1
^2i^
d)
1^
e)
_h_
_d^
p^
q^
1000
100
.90
.10
150
.83
.16
150
.80
.20
300
.50
.50
4
900
750
600
300
180
.40
.60
5
120
120
.00
1.00
1
2
3
1
-
f
1^)'
6
7.
a)
.95
b) .0975
c)
.01073
d) .0486
i^3
8.
^
11.
.16
12.
.32
13.
a)
14
1^.
^
Px
16.
/ix
21.
a)
90
b)
90
22.
a)
.5
b)
^
23.
a)
.99830
b) .99154
e)
.00169
f)
ho
1000
b)0
c)8
119-x
- 120-x' ^^~
= —
Ins
d) .77
e)
— Ixlnw —
c)
.00505
f)
-05263
g) .07465
\
\
120-;c' ^^
c)
.3
~ 120-x
c^ {In c) {In g)
.2857
d) .1385
80
c)
.00170
d) .00846
Answers
25.
to Exercises
287
.12857
1000^35
^^-
ux
^^
hs-T/so-.eTes
'^
-.645)
1000(^33 -.4^30
.
""^
T35
- ATso -
29.
.0194
32.
540,000
33.
ego—
35.
63.63
36.
a)
70
b) 80
37.
a)
60
b) 20
38.
a)
80
b)
39
40
+
40.
a)
^^^^\~
41
4.5;
^40
=
e^o
-
400^50
hs-ATso-.6Te5
.6Te5
5 (both exactly
and approximately)
54.375
c)
d) 42.2
10
^50
+
T-js
^''^
—
+
10^0
35^75
b) 45
[ip^^^<i'S
ii)
(4;>ro')(.i^.'+.i:')
1"^ ^30
-
.(r)
'
^30
^30
decrement
iv)
+
-
(5ProO(^is'+.is')
.(r)
^30
(c)
i
2
only acts on ages x
43.
1
42
44.
103
45.
a)
.737
,(r)
"30
b) .194
c)
.0163
<
50
Answers
288
to Exercises
CHAPTER 8
b) 267.01
618.16
c)
225.53
1.
a)
2.
287,223.64
3.
1110.50
4.
a)
36,923.08
b) 34,528.49
5.
a)
11,773.36
b)
6.
a)
36,693.42
b) 34,404.64
d) 520.54
e)
10,138.45
3C
7.
500)_^(1.13r2'2,/'45
/=1
8.
1296.93
9.
1567.84
10.
4099.37
13.
1356.61
14.
2441.99
15.
a)
16.
a) 55,647.06
.
1799.00
b) 53,882.35
20,000iV48
b)
17.
^
20,000(Z)38
19.
All
20.
(a), (b), (d)
21.
5027.59
22.
2.8167
23.
a) .3863
24.
a)
81,400
28.
29.
270.02
30.
267.10
13,256.14
b)
•
c)
20,000(7^48
v'^dj^
+
57,647.06 d) .06
- A'62 + N^)
Ns3)
are equal to each other, but (c)
is
different
b) .568
b) 81,000
c)
81,800
d)No
628.04
Answers
289
to Exercises
31.
263,780.49
32.
279,822.15
33.
78,214.11
34.
82,970.66
35.
65,699.85; 69,695.35
36.
461.82
38.
a^-^
39.
a)
40.
18,804.27
41.
105,600
42.
27,466.67
43.
62,933.33
44.
3733.33
52.
2165.03
53.
197.46
54.
216.84
55.
334.86
56.
1.4286
=
10,000.00
_
2
2
b) 9502.13
CHAPTER 9
10,255.34
1.
a) 30,769.23
b)
2.
a) 25,527.41
b) 7912.74
3.
a)
30,087.74
b)
9994.20
4.
a)
32,302.58
b)
11,528.11
5.
a)
31,842.58
b)
11,146.28
.
c)
^\a^^
9295.00
^
Answers
290
6.
a) 23,977.21
9.
77,482.30
10.
84,582.90
11.
87,343.87
15.
(l-io/?.)v^o
17.
7466.67
18.
a)
19.
41,346.15
2Q
b)
10,511.82
+ (ioA)v^%+io
48,076.92
b) 40,828.40
- MgQ + Z)80 + Mo -
50,000(^30
.0476
Mjp)
b)
19,800.00
17,164.35
b)
15,965.94
a)
44,444.44
b) 41,457.53
c) 48,178.08
28.
a)
27,708.65
b) 22,196.28
c)
29.
9648.56
30.
33,778.45
31.
118,181.82
32.
55,554.16
33.
a)
^^
^^'
ln(2fjL
37.
40,726.69
38.
a) 28,571.43
40.
29,600.00
41.
11,800.00
42.
26,200.00
43.
31,920.00
22.
a)
23.
6148.89
24.
22,461.17
25.
79,348.05
26.
a)
27.
c)
19,456.91
34,946.58
I
Decrease
-\-
to Exercises
6)
—
b) Increase
Infi
JTTS
b) 57,243.34
Answers
..
291
to Exercises
1000(/?o-/?io)H-40,OOOM5i
46.
651.36
47.
9270.83
49.
a)
50.
a) 74.07
52.
1318.18
53.
1966.67
54.
206.55
55.
215.84
56.
1179.25
57.
1182.76
58.
20,176.47
59.
3431.88
681.82
b) 824.18
c) 59.55
b) 61.39
61.
62.
a)
^3
.
b) 9274.78
1118.64
.
5000i?20
.
5000/?20
c)
15,040.63
d) .63586
30
25,000M25
+
150,OOOM5o
5000/?25
+
^)
- (y^T^) fcs +
—
5000/?50
w.
2N,o)
" 3(^^^)^50
65.
a)
285.32
b)
1995.23
c)
2103.56
d)
66.
a)
437.22
b) 3057.45
c)
3223.45
d)
67.
.64
69.
.143865
72.
a)
.03788
b) .03861
c) .04
73.
a)
.1739
b) .1875
d) .04
Answers
292
to Exercises
CHAPTER 10
.07505
1.
.25;
2.
250; 75,051.58
3.
.24244; .07846
4.
.27267; .09398
5.
.00756; .000256
6.
.26115; .08198
7.
.21429; .07410
8.
.21350; .07441
9.
.12987; .04807
10
^^-
i3'
11.
8.47826; 10.276
12.
8.96897; 10.303
13.
9.07; 3.14
14
122- 5
^
45
15.
^
M
+ ^'
16.
a) 30;
17.
a)
18.
.72193
19.
a)
(/i
300
30
'
+ 2^)(/x +
6)2
b) 32.83;
1077.9
^
b) 22.76
(-.2979, .7979)
The upper end of
b) (-.5719, 1.0719)
the interval in (a)
is
of some use, but a direct
calculation gives a higher probability; (b)
20.
4.47
21.
a)
23.
Question
b) .701
.601
19:
a)
not useful.
d) .787
c) .451
(-297.9,797.9)
Question 21: a) 601
is
b) 701
b) (-571.9, 1071.9)
c)
451
d) 787
Answers
to Exercises
24.
a)
.628
25.
250; 75.05
26.
250; 75,050
27.
250; 7505
28.
2955.80
29.
2644.14
30.
3044.04
31.
a)
-.125; .2461
b) 0;
32.
a)
-.1111; .2469
b) -.0926;
34.
-.01; .1133
35.
86;
293
b) .732
d) .822
c) .471
I
.0675
.3097
109
CHAPTER 11
1-
a) (40P\5)i40P25f
C)
+
UoPlsf
b)
1
-
2(40/?i5)(40/?25)(l
(40;?15)(40;?25)^
-40^25)
d) 2(35;7i5)(20/?25)(l -20^25)
A
^-
6.
145-2/
(75
a)
-
0(70
-
{\-2sP22f
+
6(l-28/'22)^(28/?22
+
b) 20(1
C)
[1
-
-
28^22)^ (28J^22)^
(28P22 -29/^22 )]^
7.
16,647.06
8.
17,647.06
9.
17,817.94
10.
.1506
11.
.1919
+
15(1
)
+
1
20(1
-
5(1 -28^22 )'^(28j^22)^
-
28/?22)^ (28/^22)^
28^22)' (2SP22)'
+ 6(1 - 28/?22)(28P22)^ + (2SP22f
+ 6[1 - (28^^22 -29/?22)]^(28/?22 " 29j^22)
Answers
294
12.
.3084
13.
117,647.06
14.
1085.38
16.
io/?5o(l
18.
T2%:n\
19.
%x
20.
.97286
21.
202,865.76
22.
a) 52,941.18
23.
83.43
24.
2824.32
25.
No;
27.
260,400.00
28.
74,400.00; 37,200.00
29.
105,500.00
30.
772.41
31.
.15017
32.
125,000.00
33.
4319.08
34.
Yes; 4736.73
35.
323.06
36.
11,640.00
37.
50,280.00
39.
3.67
40.
129,360.00
41.
2952.00
=
-Peo)
+
5^..;^
io/?6o(l
-.Pio)
+ Wn\
<a, <a^
a^y
iV66
+
b)
18,853.70
and Dss are needed
-
(\oP5o){\oP6o)(^
to Exercises
-peoPio)
Answers
295
to Exercises
CHAPTER 12
1.
124,642.86
2.
128,571.43
3.
120,000.00
4.
123,928.57
5.
No;
6.
No;
7.
1110.66
8.
i)55,A'56,^65andA^65
9.
a)
.14646
b) Impossible
10.
a)
.14646
b) .14598
11.
a)
.23418
b) .23368
c)
.11094
d)
12.
a)
.23418
b) .23368
c)
.11094
d) .12111;. 12086
13.
29,595.14
14.
27,404.47
15.
23,365.57
16.
3512.20
17.
2850.00
18.
131,760.00
19.
214.29 per month
20.
685.12
21.
907.69
22.
350.00
23.
No, we need
24.
703,125.00
25.
1535.05
A^7o
andDjo
/
to
know
a^s or al\
Not possible
INDEX
Banker's Rule,
Absolute annual rate of
Accumulated value
amortization schedule,
1
of annuity-certain,
of life annuity, 163
1
102-104
callable,
coupon,
of discount on a bond,
function,
98-100
45,49
Accumulation
1
1
2
,
Amortization
93
93-94
salesman's method,
loan,
93-94
yield rate,
Book
75
107
110
serial,
method of repaying a
93
definition of,
price,
106
Alternate formula,
6
Bond
decrement, 147
value of bond,
96-98
of premium on a bond,
98-100
Amortized value, 98
function,
1
Annuity
Central Limit Theorem,
233-234
Central rate
140
of death,
continuous,
57-58
decreasing,
60-61
44
definition of,
102-104
Callable bond,
for bonds, 98-100
Amount
102
Call date,
77-81
schedule,
145-146
of decrement,
Chebyshev's
Common
231
rule,
stock,
1 1
Commutation functions
due, 49-51
immediate, 44-48
annuities,
increasing, 60
continuous,
172
varying, 58-65
insurances,
192-194, 199
49-51,159-160
Annuity-due,
payable
m times a year,
170
varying,
Average ages
1
77
in stationary
population,
Average value,
142-143
1 1
joint-life,
161-163, 174
248
170,249
Comparison date, 29
m'^^>\
Compound
discount
definition of,
Compound
interest
definition of, 6
12
Index
298
245
Contingent insurances,
119,120
Contingent payments,
life
Effective rate
annuity,
of discount,
171-172
204
Continuous premium,
Convexity,
12
of interest,
2-3,5,7
Endowment
1 1
Coupon
192
insurance,
93
bond,
rate,
15-16
rate of,
104
57-58
Continuous annuity,
Continuous
nominal
Duration,
93
modified,
Current value,
net annual premium,
21
net single premium,
192
155-156
107
pure,
51
Equated time,
32, 109
29-32
Equation of value,
Equivalent rates of interest
Death-rate
annual,
and discount,
128
^;r,
140
central, W;f,
instantaneous,
Decreasing
life
136-137
/i;r,
Decreasing annuity,
60-61
annuity,
176-177
Exact simple
12
5
interest,
Expectation of
life,
141-142,229,245
Expected value, 1 1
Expenses, 180,206-207
Decreasing term insurance,
Face value
201
of bond
Decrement
145-146
central rate of,
145-146
force of,
Deferred insurance,
Deferred
life
190
Flat price of a bond,
191
158-159,
annuity,
DeMoivre,
of decrement,
of
127,134
interest,
accumulation
of,
of,
101
1
94
compound, 12
on a bond,
definition of,
12
effective rate of,
force of,
function,
22
9- 1
1
145-146
22
16-22
of mortality,
Discount
98
Force
of discount,
169
amount
93
of insurance
136-137
247
Formulas of mortality
deMoivre's, 134
Gompertz', 137
Makeham's, 137
joint-life,
Makeham's second,
Fractional premium,
Future lifetime,
141
150
205-206
Index
299
Gompertz' formula,
Gross premiums,
137
Last-survivor functions,
251-254
180,
206-207
Life Annuity
171-172
continuous,
Immediate life annuity,
156-159
156-159
immediate,
174-175
174-175
increasing,
199-201
Increasing insurance,
Insurance
245, 247-248
joint-life,
last-survivor,
201
decreasing term,
deferred,
92
1
96- 1 99
207-210
reserves,
158
173-178
1
4 1 - 1 43
Life insurance (see Insurance)
190,193,195
life,
155-180
Life expectancy,
199-201
varying,
single-life,
temporary,
varying,
191, 193
term,
165-166
select,
253
moment of death,
254-256
reversionary,
245,248
last survivor,
179-180
reserves,
increasing, 199-201
joint-life,
166-172
a year,
1
253
m times
payable
191
endowment,
whole
159-160
due,
annuity,
life
158-159
deferred,
60
Increasing annuity,
Increasing
decreasing, 176-177
Life table,
127-133
180,206-207
Loss-at-issue, 235-238
Loading,
Interest
amount of,
compound,
1
6-9
conversion period,
definition of,
14
effective rate of, 2-3, 5, 7
1 6-22
force of,
measurement of a fund,
33-35
rate of,
simple,
4-6
Iteration,
for
4,
107
137,150
Market price of a bond, 98
Mathematical expectation,
Maturity date,
nominal
Interpolation,
Makeham's formulas
bond,
1
13-16
Mean,
223-230
230
Method of equated time,
Median,
33,55,132
55-56
109
Modified duration,
Joint-life functions,
1 1
93, 102
243-251
105
Mortality
central rate of,
1
40
32,
Index
300
136-137,247
force of,
134,137,150
128,130
formulas
rate of,
of,
Mortality table (see Life Table)
143-147
Multiple decrements,
98
flat,
98
market,
Principal,
2
1,
Probability
114-116
general,
Net amount at risk, 220
Net annual premium, 178-1 79,
202-207
Net single premium, 156,157
Nominal rate
15-16
of discount,
of interest,
Normal
of death,
of survival,
interest,
6
variables
263-269
137-140
of,
230
of,
second moment
of,
225
standard deviation of,
230
223-230
variance of,
Rate
Preferred stock,
absolute,
1 1
Premium
147
annual death
amortization of,
98-100
204
fractional, 205-206
gross, 180,206-207
continuous,
178-179,
net annual,
rate,
128
central death rate,
140
central,
net single,
of decrement,
145-146
Rate of return
dollar-weighted,
202-207
33-34,
265
156, 157
on an bond,
95
m times a year,
definition of,
time-weighted,
265
Redemption
date,
Redemption value,
205-206
34-35,
102
93
Reserve
Present value
9-10
annuities,
179-180
insurance, 207-210
Price
bond,
223-238
235-238
223-230
loss-at-issue,
median
Population, stationary,
payable
75
Pure endowment, 155-156
52-53,61-62
Perpetuity,
145
130
232
93
Pension plans,
128, 130
of decrement,
mean
Par value,
252-253
last survivor,
Random
Ordinary simple
244-245
joint-life,
Prospective method,
13-16
distribution,
229
density function,
93-94
prospective formulas,
207
Index
301
Uniform
210
successive years,
207
terminal,
distribution of deaths,
138-139, 141,215
Retrospective method,
75
Variance, 223-230
Reversionary annuities,
254-256
Varying annuities,
Varying insurance,
Second moment,
Select-and-ultimate tables,
165
Whole
Sequences
arithmetic,
41-42
geometric,
42-44
5
Sinking fund
81
method of repaying a
loan,
81-83
230
Standard deviation,
Stationary population,
137-140
Successive approximation,
55-56
128
Survival function,
Table of uniform seniority
251,259
Makeham, 250,262
Gompertz,
annuity,
158, 169, 175
Term
insurance,
decreasing,
increasing,
Terminal age,
191,193
201
199-201
134
190
93-94
using sinking fund, 85
ordinary, 6
definition of,
insurance,
of an investment, 83-85
6
4-6
definition of,
life
life
bond,
Banker's Rule,
Temporary
105
Yield rate
interest
exact,
Volafility,
165
Select mortality,
Simple
199-201
Varying life annuities,
173-178
225
110
Serial bond,
58-65
ABOUT THE
An
treatment of comand basic aspects of
introductory
pound
iife
TEXT:
interest
The theory
contingencies.
trated
by
is illus-
and
examples,
numerous exerA knowledge of algebra is
reinforced
cises.
worked
through
required; a familiarity with introductory
calculus and probability
useful but
is
not essentia!.
ABOUT THE AUTHOR:
Michael M. Parmenter holds degrees
:Qm__yniversity of Toronto and UniHe has taught
of Alberta.
matics and actuarial science for
-six years at Memorial Unrv^er'
,
^
'f
II!
i
t
ill
25
I
I
I
Newfoundland,
and
is
the
hor of a textbook on discrete
matics and graph theory.
;;
tj
755 40L
^5109
Theory cf literett ^
1-56698-::33-9
-
•_TC>0
-
^^..
SD16
9999e
liiiim
y\j-j^fj
II III
HI
III
Hill