theory of interest and life contingencies · pdf file1.4 present value and discount 9 ......

THEORY OF INTERESTAND LIFE CONTINGENCIES

WITH PENSION APPLICATIONS

A Problem-Solving Approach

Third Edition

Michael M. ParmenterASA, Ph.D.

ACTEX PublicationsWinsted, Connecticut

To my Mother and the memory of my Father

Copyright © 1988, 1999by ACTEX Publications

No portion of this book may be reproduced in any form or by anymeans without the prior written permission of the copyright owner.

Requests for permission should be addressed toACTEX LearningPO Box 715New Hartford CT 06057

Cover design by MUF

ISBN: 978-1-62542-839-4

________________________________

TABLE OF CONTENTS________________________________

CHAPTER ONEINTEREST: THE BASIC THEORY 11.1 Accumulation Function 1 1.2 Simple Interest 41.3 Compound Interest 61.4 Present Value and Discount 91.5 Nominal Rate of Interest 131.6 Force of Interest 16 Exercises 22

CHAPTER TWOINTEREST: BASIC APPLICATIONS 292.1 Equation of Value 292.2 Unknown Rate of Interest 332.3 Time-Weighted Rate of Return 34 Exercises 36

CHAPTER THREEANNUITIES 413.1 Arithmetic and Geometric Sequences 41 3.2 Basic Results 443.3 Perpetuities 523.4 Unknown Time and Unknown Rate of Interest 533.5 Continuous Annuities 573.6 Varying Annuities 58 Exercises 66

iv Table of Contents

CHAPTER FOURAMORTIZATION AND SINKING FUNDS 754.1 Amortization 754.2 Amortization Schedules 774.3 Sinking Funds 814.4 Yield Rates 83 Exercises 85

CHAPTER FIVEBONDS 935.1 Price of a Bond 93 5.2 Book Value 965.3 Bond Amortization Schedules 985.4 Other Topics 101 Exercises 105

CHAPTER SIXPREPARATION FOR LIFE CONTINGENCIES 1136.1 Introduction 1136.2 Probability and Expectation 1146.3 Contingent Payments 119 Exercises 123

CHAPTER SEVENLIFE TABLES AND POPULATION PROBLEMS 1277.1 Introduction 1277.2 Life Tables 1287.3 Analytic Formulae for 133jB7.4 The Stationary Population 1377.5 Expectation of Life 1417.6 Multiple Decrements 143 Exercises 148

CHAPTER EIGHTLIFE ANNUITIES 1558.1 Basic Concepts 1558.2 Commutation Functions 1618.3 Annuities Payable 166m>26C

8.4 Varying Life Annuities 1738.5 Annual Premiums and Reserves 178 Exercises 181

Table of Contents v

CHAPTER NINELIFE INSURANCE 1899.1 Basic Concepts 1899.2 Commutation Functions and Basic Identities 1929.3 Insurance Payable at the Moment of Death 1969.4 Varying Insurance 1999.5 Annual Premiums and Reserves 202 Exercises 210

CHAPTER TENSTATISTICAL CONSIDERATIONS 22310.1 Mean Variance 22310.2 Normal Distribution 23010.3 Central Limit Theorem 23310.4 Loss-at-Issue 235 Exercises 239

CHAPTER ELEVENMULTI-LIFE THEORY 24311.1 Joint-Life Actuarial Functions 24311.2 Last-Survivor Problems 25111.3 Reversionary Annuities 254 Exercises 256

CHAPTER TWELVEPENSION APPLICATIONS 263 Exercises 269

ANSWERS TO THE EXERCISES 275

INDEX 297

________________________________

PREFACE________________________________

It is impossible to escape the practical implications of compound interestin our modern society. The consumer is faced with a bewildering choiceof bank accounts offering various rates of interest, and wishes to choosethe one which will give the best return on her savings. A home-buyer isoffered various mortgage plans by different companies, and wishes tochoose the one most advantageous to him. An investor seeks to purchasea bond which pays coupons on a regular basis and is redeemable at somefuture date; again, there are a wide variety of choices available. Comparing possibilities becomes even more difficult when thepayments involved are dependent on the individual’s survival. Forexample, an employee is offered a variety of different pension plans andmust decide which one to choose. Also, most people purchase lifeinsurance at some point in their lives, and a bewildering number ofdifferent plans are offered. The informed consumer must be able to make an intelligent choicein situations like those described above. In addition, it is important that,whenever possible, she be able to make the appropriate calculationsherself in such cases. For example, she should understand why a givenseries of mortgage payments will, in fact, pay off a certain loan over acertain period of time. She should also be able to decide which portionof a given payment is paying off the balance of the loan, and whichportion is simply paying interest on the outstanding loan balance. The first goal of this text is to give the reader enough informationso that he can make an intelligent choice between options in a financialsituation, and can verify that bank balances, loan payments, bondcoupons, etc. are correct. Too few people in today's society understandhow these calculations are carried out. In addition, however, we are concerned that the student, besidesbeing able to carry out these calculations, understands why they work. Itis not enough to memorize a formula and learn how to apply it; youshould understand why the formula is correct. We also wish to presentthe material in a proper mathematical setting, so the student will see howthe theory of interest is interrelated with other branches of mathematics.

viii Preface

Let me explain why the phrase “Problem-Solving Approach”appears in the title of this text. We will prove a very small number offormulae and then concentrate our attention on showing how theseformulae can be applied to a wide variety of problems. Skill will beneeded to take the data presented in a particular problem and see how torearrange it so the formulae can be used. This approach differs frommany texts, where a large number of formulae are presented, and thestudent tries to memorize which problems can be solved by directapplication of a particular formula. We wish to emphasizeunderstanding, not rote memorization. A working knowledge of elementary calculus is essential for athorough understanding of all the material. However, a large portion ofthis book can be read by those without such a background by omittingthe sections dependent on calculus. Other required background materialsuch as geometric sequences, probability and expectation, is reviewedwhen it is required. Each chapter in this text includes a large number of examples andexercises. It should be obvious that the most efficient way for a studentto learn the material is for her to work all the exercises. Finally, let us stress that it is assumed that every student has acalculator (with a button) and knows how to use it. It is because ofyB

our ability to use a calculator that many formulae mentioned in oldertexts on the subject are now unnecessary. This book is naturally divided into two parts. Chapters 1-5 areconcerned solely with the Theory of Interest, and Life Contingencies isintroduced in Chapters 6-11. In Chapter 1 we present the basic theory concerning the study ofinterest. Our goal here is to give a mathematical background for thisarea, and to develop the basic formulae which will be needed in the restof the book. Students with a weak calculus background may wish toomit Section 1.6 on the force of interest, as it is of more theoretical thanpractical importance. In Chapter 2 we show how the theory in Chapter 1can be applied to practical problems. The important concept of equationof value is introduced, and many worked examples of numericalproblems are presented. Chapter 3 discusses the extremely importantconcept of annuities. After developing a few basic formulae, our mainemphasis in this chapter is on practical problems, seeing how data forsuch problems can be substituted in the basic formulae. It is in thissection especially that we have left out many of the formulae presentedin other texts, preferring to concentrate on problem-solving techniquesrather than rote memorization. Chapters 4 and 5 deal with further

Preface ix

applications of the material in Chapters 1 through 3, namelyamortization, sinking funds and bonds. Chapter 6 begins with a review of the important concepts ofprobability and expectation, and then illustrates how probability can becombined with the theory of interest. In Chapter 7 we introduce lifetables, discussing how they are constructed and how they can be applied.Chapter 8 is concerned with life annuities, that is annuities whosepayment are conditional on survival, and Chapter 9 discusses lifeinsurance. These ideas are generalized to multi-life situations in Chapter10. Finally, Chapter 11 demonstrates how many of these concepts areapplied in the extremely important area of pension plans. Chapters 1 through 6 have been used for several years as the textmaterial for a one semester undergraduate course in the Theory ofInterest, and I would like to thank those students who pointed out errorsin earlier drafts. In addition, I am deeply indebted to Brenda Crewe andWanda Heath for an excellent job of typing the manuscript, and to mycolleague, Dr. P. P. Narayanaswami, for his invaluable technicalassistance. Chuck Vinsonhaler, University of Connecticut, was stronglysupportive of this project, and introduced me to the people at ACTEXPublications, for which I owe him a great deal. Dick London did thetechnical content editing, Marilyn Baleshiski provided the electronictypesetting, and Marlene Lundbeck designed the text's cover. I wouldlike to thank them for taking such care in turning a very roughmanuscript into what I hope is a reasonably comprehensive yet friendlyand readable text book for actuarial students.

St. John’s, Newfoundland Michael M. ParmenterDecember, 1988

__________________________________PREFACE TO THE

REVISED EDITION__________________________________

In the fifteen months since the original edition of this text was published,a number of comments have been received from teachers and studentsregarding that edition. We are pleased to note that most of the comments have been quitecomplimentary to the text, and we are making no substantial modifica-tions at this time. A significant, and thoroughly justified, criticism of the originaledition is that time diagrams were not used to illustrate the examplesgiven in the second half of the text, and that deficiency has been rectifiedby the inclusion of thirty-five additional figures in the Revised Edition. Thus it is fair to say that there are no contained in thenew topicsRevised Edition, but rather that the pedagogy has been strengthened. Forthis reason we prefer to call the new printing a Revised Edition, ratherthan a Second Edition. In addition we have corrected the errata in the original edition. Wewould like to thank all those who took the time to bring the variouserrata to our attention.

February, 1990 M.M.P.

__________________________________PREFACE TO THE

THIRD EDITION__________________________________

It is now more than ten years since the original publication of thistextbook. In that time, several very significant developments haveoccurred to suggest that a new edition of the text is now needed, andthose developments are reflected in the modifications and additions madein this Third Edition. First, improvements in calculator technology give us betterapproaches to reach numerical results. In particular, many calculatorsnow include iteration algorithms to permit direct calculation of unknownannuity interest rates and bond yield rates. Accordingly, the olderapproximate methods using interpolation have been deleted from thetext. Second, with the discontinued publication of the classic textbookLife Contingencies by C.W. Jordan, our text has become the only onepublished in North America which provides the traditional presentationof contingency theory. To serve the needs of those who still prefer thistraditional approach, including the use of commutation functions and adeterministic life table model, we have chosen to include various topicscontained in Jordan’s text but not contained in our earlier editions.These include insurances payable at the moment of death (Section 9.3),life contingent accumulation functions (Section 8.2), the table of uniformseniority concept for use with Makeham and Gompertz annuity values(Section 11.1), simple contingent insurance functions (Section 11.1), andan expansion of the material regarding multiple-decrement theory(Section 7.6). Third, actuaries today are interested in various concepts of financebeyond those included in traditional interest theory. To that end we haveintroduced the ideas of real rates of return, investment duration, modifiedduration, and so on, in this Third Edition.

12 Preface

Fourth, the new edition provides a gentle introduction to the moremodern stochastic view of contingency theory, in the completely newChapter 10, to supplement the traditional presentation. In connection with the expansion of topics, the new editioncontains over forty additional exercises and examples. As well, thenumerical answers to the exercises have been made more precise and theerrata in the previous edition have been corrected. We would like tothank everyone who brought such errata to our attention. With the considerable modifications made in the new edition, webelieve this text is now appropriate for two major audiences: pensionactuaries, who wish to understand the use of commutation functions anddeterministic contingency theory in pension mathematics, and universitystudents, who seek to understand basic contingency theory at an intro-ductory level before undertaking a study of the more mathematicallysophisticated stochastic contingency theory. As with the original edition of this text, the staff at ACTEXPublications has been invaluable in the development of this new edition.Specifically I would like to thank Denise Rosengrant for her textcomposition and typesetting work, and Dick London, FSA, for histechnical content editing.

February, 1999 M.M.P

__________________________________CHAPTER ONE

INTEREST: THE BASIC THEORY__________________________________

1.1 ACCUMULATION FUNCTION

The simplest of all financial transactions is one in which an amount ofmoney is invested for a period of time. The amount of money initiallyinvested is called the and the amount it has grown to after theprincipaltime period is called the at that time.accumulated value This is a situation which can easily be described by functionalnotation. If is the length of time for which the principal has beentinvested, then the amount of money at that time will be denoted by ( ).A tThis is called the For the moment we will onlyamount function.consider values 0, and we will assume that is measured in years. t t We remark that the initial value (0) is just the principal itself.A In order to compare various possible amount functions, it isconvenient mathematically to define the from theaccumulation function

amount function as ( ) . We note that (0) 1 and that ( ) is( )(0)a t a A t A t

Aœ œ

just a constant multiple of ( ) , namely ( ) ( ) where (0) isa t A t k a t k Aœ † œthe principal. What functions are possible accumulation functions? In theory,any function ( ) with (0) 1 could represent the way in which moneya t a œaccumulates with the passage of time. Certainly, however, we wouldhope that ( ) is increasing. Should ( ) be continuous? That depends on a t a t the situation; if ( ) represents the amount owing on a loan years aftera t tit has been taken out, then ( ) may be continuous if interest continues toa taccumulate for non-integer values of . However, if ( ) represents thet a tamount of money in your bank account years after the initial depositt(assuming no deposits or withdrawals in the meantime), then ( ) willa t stay constant for periods of time, but will take a jump whenever interestis paid into the account. The graph of such an ( ) will be a stepa tfunction. We will normally assume in this text that ( ) is continuous; ita t is easy to make allowances for other situations when they turn up.

2 Chapter 1

In Figure 1.1 we have drawn graphs of three different types ofaccumulation functions which occur in practice:

a(t)

(0,1)(0,1)

(a) (c)

(0,1)

a(t)a(t)

(b)

FIGURE 1.1

Graph (a) represents the case where the of interest earned isamountconstant over each year. On the other hand, in cases like (b), the amountof interest earned is increasing as the years go on. This makes moresense in most situations, since we would hope that as the principal getslarger, the interest earned also increases; in other words, we would liketo be in a situation where “interest earns interest”. There are manydifferent accumulation functions which look roughly like the graph in(b), but the exponential curve is the one which will be of greatest interestto us. We remarked earlier that a situation like (c) can arise wheneverinterest is paid out at fixed periods of time, but no interest is paid ifmoney is withdrawn between these time periods. If the amount ofinterest paid is constant per time period, then the “steps” will all be of thesame height. However, if the amount of interest increases as theaccumulated value increases, then we would expect the steps to get largerand larger as time goes on. We have used the term interest several times now, so perhaps it istime to define it!

Interest Accumulated Value Principalœ

This definition is not very helpful in practical situations, since weare generally interested in comparing different financial situations todetermine which is most profitable. What we require is a standardizedmeasure for interest, and we do this by defining the effective rate ofinterest i (per year) to be the interest earned on a principal of amount 1over a period of one year. That is,

i a . .œ (1) 1 (1 1)

Interest: The Basic Theory 3

We can easily calculate using the amount function ( ) instead ofi A t a t A t k a t . ( ), if we recall that ( ) ( ) Thusœ †

(1) 1 (1 2)(1) (0) (1) (0)(0) (0)i a . .a a A A

a Aœ œ œ

Verbally, the effective rate of interest per year is the amount ofinterest earned in one year divided by the principal at the beginning ofthe year. There is nothing sacred about the term “year” in this definition.We can calculate an effective rate of interest over any time period bysimply taking the numerator of the above fraction as being the interestearned over that period. More generally, we define the effective rate of interest in the nth

year by

(1 3)( ) ( 1) ( ) ( 1)( 1) ( 1)i . .A n A n a n a n

A n a nn œ œ

Note that , calculated by (1.3), is the same as defined by either (1.1) ori i1

(1.2).

Example 1.1

Consider the function ( ) 1a t t t .œ 2

(a) Verify that (0) 1 a .œ(b) Show that ( ) is increasing for all 0 a t t . (c) Is ( ) continuous? a t(d) Find the effective rate of interest for ( )i a t .(e) Find i .n

Solution(a) (0) (0) (0) 1 1 a .œ œ2

(b) Note that ( ) 2 1 0 for all 0, so ( ) is increasing.a t t t a t w œ (c) The easiest way to solve this is to observe that the graph of ( ) is a a t

parabola, and hence ( ) is continuous (or recall from calculus thata t all polynomial functions are continuous).

(d) (1) 1 3 1 2i a .œ œ œ

(e) ( ) ( 1) 1 [( 1) ( 1) 1]( 1) ( 1) ( 1) 1

i a n a n n n n n a n n n

œ œ

n

2 2

2

.n n n

œ

212

4 Chapter 1

1.2 SIMPLE INTEREST

There are two special cases of the accumulation function ( ) that we will a texamine closely. The first of these, , is used occasionally,simple interestprimarily between integer interest periods, but will be discussed mainlyfor historical purposes and because it is easy to describe. The second ofthese, , is by far the most important accumulationcompound interestfunction and will be discussed in the next section. Keep in mind that inboth of these cases ( ) is continuous, and also that there are some a t practical settings where modifications must be made. Simple interest is the case where the graph of ( ) is a straight line.a t Since (0) 1, the equation must therefore be of the general forma œa t bt b i( ) 1 for some . However, the effective rate of interest isœ given by (1) 1 , so the formula isi a bœ œ

( ) 1 , 0 (1 4)a t it t . .œ

(1,1 )i(0,1)

( ) 1a t it

t

FIGURE 1.2

Remarks

1 This is case (a) graphed in Figure 1.1. In this situation, the amount. of interest earned each year is constant. In other words, only theoriginal principal earns interest from year to year, and interestaccumulated in any given year does not earn interest in futureyears.

2. The formula ( ) 1 applies to the case where the principal isa t itœ A a(0) (0) 1. More generally, if the principal at time 0 is equalœ œto , the amount at time will be ( ) (1 ).k t A t k itœ


3. We noted above that the “ ” in ( ) 1 is also the effectivei a t itœ rate of interest for this function. Note however that

1 [1 ( 1)]1 ( 1)œ i in i n

i nn

(1 5)1 ( 1)œ ii n . .

Observe that is not constant. In fact, decreases as getsi i nn n

larger, a fact which should not surprise us. If the amount ofinterest stays constant as the accumulated value increases, thenclearly the effective rate of interest is going down.

4. Clearly ( ) 1 is a formula which works equally well for alla t itœ values of , integral or otherwise. However, problems can developtin practice, as illustrated by the following example.

Example 1.2

Assume Jack borrows 1000 from the bank on January 1, 1996 at a rate of15% simple interest per year. How much does he owe on January 17,1996?SolutionThe general formula for the amount owing at time in general is tA t . t t( ) 1000(1 15 ), but the problem is to decide what value of œ should be substituted into this formula. An obvious approach is to takethe number of days which have passed since the loan was taken out anddivide by the number of days in the year, but should we count thenumber of days as 16 or 17? Getting really picky, should we worryabout the time of day when the loan was taken out, or the time of daywhen we wish to find the value of the loan? Obviously, any value of istonly a convenient approximation; the important thing is to have aconsistent rule to be used in practice. Two techniques are common:(a) The first method is called , and with it we useexact simple interest

. (1.6) 365t number of daysœ

When counting the number of days it is usual to count the last day,

but not the first. In our case this would lead to so Jack16365t œ

owes 1000 1 (.15) 1006.58.16365’ “Š ‹ œ

6 Chapter 1

(b) The second method is called (or ordinary simple interestthe ), and with it we useBanker’s Rule

. (1.7)360t number of daysœ

The same procedure as above is used for calculating the number of

days. In our case, we would have so the debt is16360t œ

1000 1 (.15) 1006.67.16360’ “Š ‹ œ

The common practice in Canada is to use exact simple interest, whereasordinary simple interest is used in the United States and in internationalmarkets.

1.3 COMPOUND INTEREST

The most important special case of the accumulation function ( ) is thea tcase of compound interest. Intuitively speaking, this is the situationwhere money earns interest at a fixed effective rate; in this setting, theinterest earned in one year earns interest itself in future years. If is the effective rate of interest, we know that (1) 1 , so 1 i a i œ becomes 1 after the first year. What happens in the second year? iConsider the 1 as consisting of two parts, the initial principal 1 and ithe interest earned in the first year. The principal 1 will earn interest in i the second year and will accumulate to 1 The interest will also i. iearn interest in the second year and will grow to (1 ) Hence the totali i . amount after two years is 1 (1 ) (1 ) . By continuing this i i i i œ 2

reasoning, we see that the formula for ( ) isa t

( ) (1 ) , 0 (1 8)a t i t . .œ t

(1,1 )i(0,1)

( ) (1 ) ta t i

t

FIGURE 1.3


Remarks

1. This is an example of the type of function shown in part (b) of thegraph in Figure 1.1.

2. The formula ( ) (1 ) applies to the case where the principala t iœ t

is (0) (0) 1 More generally, if the principal at time 0 isA a . œ œequal to , the amount at time will be ( ) (1 ) .k t A t k iœ t

3. Observe that the “ ” in (1 ) is the effective rate of interest.i i t

More generally,

i i i. .i iin

n n

nœ œ(1 ) (1 )

(1 )1 1 (1 9)

œ

1

1

Hence, in this case is the same for all positive integers . Wei nn

shouldn’t be surprised, since this fits with our idea that, in com-pound interest, the effective rate of interest is constant.

4. Mathematically, any value of , whether integral or not, can betsubstituted into ( ) (1 ) . This is an easier task for us todaya t iœ t

than it was fifty years ago; we just have to press the appropriatebuttons on our calculators! Again, there are problems determiningwhat value of should be used, but we can deal with them as wetdid in the last section.

In practical situations, however, a very different solution issometimes used in the case of compound interest. To find theamount of a loan (for example) when is a fraction, first find thetamounts for the integral values of immediately before andt immediately after the fractional value in question. Then use linearinterpolation between the two computed amounts to calculate therequired answer.

This is equivalent to saying that compound interest is usedfor integral values of , and simple interest is used between integraltvalues. In Figure 1.4, the solid line represents ( ) (1 ) ,a t iœ t

whereas the dotted lines indicate the graph of ( ) if lineara tinterpolation is used.

(1,1 )i(0,1)

t

( )a t

FIGURE 1.4

8 Chapter 1

As we will see later, this common procedure benefits the lender ina financial transaction, and (consequently) is detrimental to the borrowerif she has to repay the loan at a duration between integral values.

Example 1.3

Jack borrows 1000 at 15% compound interest.(a) How much does he owe after 2 years?(b) How much does he owe after 57 days, assuming compound interest

between integral durations?(c) How much does he owe after 1 year and 57 days, under the same

assumption as in part (b)?(d) How much does he owe after 1 year and 57 days, assuming linear

interpolation between integral durations?(e) In how many years will his principal have accumulated to 2000?Solution(a) 1000(1.15) 1322.50.2 œ

(b) The most suitable value for is , and the accumulated value is57365t

1000(1.15) 1022.07.57

365 œ

(c) 1000(1.15) 1175.381 57365 œ

(d) We must interpolate between (1) (1000)(1 15) 1150 00 andA . .œ œA . . . (2) 1000(1 15) 1322 50 The difference between theseœ œ2

values is (2) (1) 172 50 The portion of this differenceA A . . œwhich will accumulate in 57 days, assuming simple interest, isŠ ‹57

365 (172.50) 26.94. Thus the accumulated value after 1 yearœ

and 57 days is 1150.00 26.94 1176.94. Observe that the œborrower owes more money in this case than he does in part (c).

(e) We seek such that 1000(1.15) 2000, or that (1.15) 2.t t tœ œUsing logs we obtain

4.9595 years.log 2log 1.15t œ œ

To close this section, we will compare simple interest and compoundinterest to see which gives the better return. In Figure 1.5, graphs forboth simple interest and compound interest are drawn on the same set ofaxes.


1 it

(1 )ti

(1,1 )i(0,1) t

a(t)

FIGURE 1.5

We know that the exponential function (1 ) is always concave i t

up (because the second derivative is (1 ) [ (1 )] , which is greater i ln i t 2

than zero), whereas 1 is a straight line. These facts tell us that the it only points of intersection of these graphs are the obvious ones, namely(0,1) and (1, 1 ). They also give us the two important relationships i

(1 ) 1 , for 0 1 (1 10) i it t .t

and

(1 ) 1 , for 1. (1.11) i it tt

Hence we conclude that compound interest yields a higher return thansimple interest if 1, whereas simple interest yields more if 0 1.t t The first of these statements does not surprise us, since for 1, we t have interest as well as principal earning interest in the (1 ) case. The i t

second statement reminds us that, for periods of less than a year, simpleinterest is more beneficial to the lender than compound interest, a factwhich was illustrated in Example 1.3.

1.4 PRESENT VALUE AND DISCOUNT

In Section 1.1 we defined accumulated value at time as the amount thattthe principal accumulates to over years. We now define the t presentvalue t years in the past as the amount of money that will accumulate tothe principal over years. In other words, this is the reverse procedure oftthat which we have been discussing up to now.

| | |PresentValue

Principal AccumulatedValue

qqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

FIGURE 1.6

10 Chapter 1

For example, 1 accumulates to 1 over a single year. How much imoney is needed, at the present time, to accumulate to 1 over one year?We will denote this amount by , and, recalling that accumulates tov vv i v i(1 ), we have (1 ) 1. Therefore œ

v . .iœ 1

1 (1 12)

These two accumulations are shown in Figure 1.7.

| | |i i

11 1

1 0 1

1

qqqqqqqqqqqqqqqqqqqqqqqqqqqqq

FIGURE 1.7

From now on, unless explicitly stated otherwise, we will assumethat we are in a compound interest situation, where ( ) (1 ) Ina t i . œ t

this case, the present value of 1, years in the past, will be .1( )

t vl i

ttœ

We summarize this on the time diagram shown in Figure 1.8.

| | | | |

1(1 ) (1 )

1 0 1

11 11

AMOUNTTIME

qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

t t

i ii i t t

FIGURE 1.8

Observe that, since (1 ) , the function ( ) (1 )v i a t it t tœ œ

expresses all these values, for both positive and negative values of .tHence (1 ) gives the value of one unit (at time 0) at any time , past or i tt

future. The graph is shown in Figure 1.9.

(0,1)

( )a t(1 )ti

t

FIGURE 1.9


Example 1.4

The Kelly family buys a new house for 93,500 on May 1, 1996. Howmuch was this house worth on May 1, 1992 if real estate prices haverisen at a compound rate of 8% per year during that period?SolutionWe seek the present value, at time 4, of 93,500 at time 0. This ist œ

93,500 68,725.29.11.08Š ‹4

œ

What happens to the calculation of present values if simple interestis assumed instead of compound interest? The accumulation function isnow ( ) 1 Hence, the present value of one unit years in the pasta t it. tœ is given by , where (1 ) 1 Thus the present value isx x it . œ

x . it .œ 1

1 (1 13)

The time diagram for this case is shown in Figure 1.10.

| | | | |

1 11 1 1 1

1 0 1

1 AMOUNT

TIME

qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

t t

it i i it

FIGURE 1.10

In Exercise 15, you are asked to sketch the graph of this situation.Unlike the compound interest case, this graph changes dramatically as itpasses through the point (0, 1). We now turn our attention to the concept of discount. For themoment we will not assume compound interest, since any accumulationfunction will be satisfactory. Imagine that 100 is invested, and that one year later it hasaccumulated to 112. We have been viewing the 100 as the “startingfigure,” and have imagined that interest of 12 is added to it at the end ofthe year. However, we could also view 112 as the basic figure, andimagine that 12 is deducted from that value at the start of the year. Fromthe latter point of view, the 12 is considered an amount of discount. Students sometimes get confused about the difference betweeninterest and discount, but the important thing to remember is that theonly difference is in the point of view, not in the underlying financialtransaction. In both situations we have 100 accumulating to 112, andnothing can change that.

12 Chapter 1

Since discount focuses on the total at the end of the year, it isnatural to define the effective rate of discount, , asd

d . .aaœ

(1) 1(1) (1 14)

In other words, standardization is achieved by dividing by (1) insteada of (0), as was done in (1.2) to define the effective rate of interest .a i More generally, the effective rate of discount in the year isnth

given by

d . .a n a na nn œ ( ) ( 1)

( ) (1 15)

(Compare this with the definition of , given by (1.3).)in Now we will derive some basic identities relating to . Oned iidentity follows immediately from the definition of , namely,d

d a ia i i

iœ œ

(1) 1 (1 ) 1

(1) 1 1 . (1.16) œ

Since 1 1, this tells us that . i d i Immediately from the above we obtain

1 1 (1 17)1 11 œ œ œd v. .i

i i

Actually, this identity is exactly what we would expect from thedefinition of . The fact that 1 accumulates to 1 over one year is thed dexact analogy of 1 accumulating to 1 over the same period. i Solving either of the above identities for , we obtain i

i . .ddœ 1 (1 18)

The reader will be asked to derive other identities in the exercisesand to give verbal arguments in support of them. We note that allidentities derived so far hold for any accumulation function. For the restof this section, it will be assumed that ( ) (1 ) .a t iœ t


In Section 1.3 we learned that to find the accumulated value yearst in the future we multiply by (1 ) , whereas to find the present value i tt

years in the past we multiply by . However, identity (1.17) tells1 (1 ) i t

us that 1 . Hence, if is involved, the rules for present and11 d diœ

accumulated value are reversed: present value is obtained by multiplying

by (1 ) , and accumulated value by multiplying by 1(1 )

d . d

tt

Example 1.5

1000 is to be accumulated by January 1, 1995, at a compound rate ofdiscount of 9% per year.(a) Find the present value on January 1, 1992.(b) Find the value of corresponding to .i dSolution(a) 1000(1 .09) 753.57. œ3

(b) 098910991 i . .d .

d .œ œ œ

Example 1.6

Jane deposits 1000 in a bank account on August 1, 1996. If the rate ofcompound interest is 7% per year, find the value of this deposit onAugust 1, 1994.SolutionSome students think that the answer to this question should be 0, becausethe money hasn’t been deposited yet! However, in a mathematical sense,we know that money has value at all times, past or future, so the correct

answer is 1000 873.44. 1 1.07Š ‹2

œ

1.5 NOMINAL RATE OF INTEREST

We will assume ( ) (1 ) throughout this section and, unless stateda t iœ t

otherwise, in all remaining sections of the book.

Example 1.7

A man borrows 1000 at an effective rate of interest of 2% per month.How much does he owe after 3 years?

14 Chapter 1

SolutionWhat we want is the amount of the debt after three years. Since theeffective interest rate is given per month, three years is 36 interestperiods. Thus the answer is 1000(1.02) 2039.89.36 œ

The point of the above example is to illustrate that effective ratesof interest need not be given per year, but can be defined with respect toany period of time. To apply the formulae developed to this point, wemust be sure that is the number of in anyt effective interest periods particular problem. In many real-life situations, the effective interest period is not ayear, but rather some shorter period. Perhaps the lender tries to keep thisfact hidden, as it might be to his benefit to do so! For example, supposeyou want to take out a mortgage on a house and you discover a rate of12% per year. When you dig a little, however, what you find out is thatthis rate is “convertible semiannually”, which means that it is really 6%effective per half-year. Is that the same thing? Not at all. Consider whathappens to an investment of 1. After half a year it has accumulated to1.06. After one year (two interest periods) it has become(1.06) 1.1236. So, over a one-year period, the amount of interest2 œgained is .1236, which means the effective rate of interest per year isactually 12.36%. Although it may not be clear from the advertising,many mortgage loans are convertible semiannually, so the effective rateof interest is higher than the rate quoted. As another example, consider a well-known credit card whichcharges 18% per year convertible monthly. This means that the actual

rate of interest is .015 effective per month. Over the course of a.1812 œ

year, 1 will accumulate to (1.015) 1.1956, so the effective rate of12 œinterest per year is actually 19.56%. The 18% in the last example is called a rate of interest,nominalwhich means that it is convertible over a period other than one year. Ingeneral, we use the notation to denote a nominal rate of interesti ( )m

convertible times per year, which implies an effective rate of interestm

of per of a year. If is the effective rate of interest per year, itim m i( )m

th

follows that

1 1 (1 19) œi . .im” •( )m m


Example 1.8

Find the accumulated value of 1000 after three years at a rate of interestof 24% per year convertible monthly.SolutionThis is really 2% effective per month, so the answer is the same asExample 1.7, namely 1000(1.02) 2039.89.36 œ

Remark

An alternative method of solving Example 1.8 is to find , the effectiveirate of interest per year, and then proceed as in Section 1.3. We would

have 1 1 (1 02) 1 26824, and the answeri . . imœ Š ( )

12m m‹ œ œ

would be 1000 (1.26824) 2039.88.3 œ

Notice the difference of .01 in the two answers. This is becausenot enough decimal places were kept in the value of , and some error icrept in. Of course, if you use the memory in your calculator it isunlikely that this type of error will occur. Nevertheless, the first solutionis still preferable; time spent on unnecessary calculations can besignificant in examination situations. It will be extremely important in later sections of the text to be ableto convert from one nominal rate of interest to another whose convertiblefrequency is different. Here is an example of this.

Example 1.9

If 15, find the equivalent nominal rate of interest convertiblei . (6) œsemiannually.Solution

We have 1 1 , so 2[(1.025) 1] .15378.2 615Š œ i . i

(2) 2 6(2) 3‹ Š ‹ œ œ

In the same way that we defined a nominal rate of interest, wecould also define a nominal rate of discount, , as meaning and( )m

effective rate of discount of per of a year. Analogous to identitydm m( )m

th

(1.19), it is easy to see that

1 1 (1 20) œ d . .dm” •( )m m

16 Chapter 1

Since 1 , we conclude that11 œ d i

1 1 (1 ) 1 , (1 21)” • œ i dm n i d .( ) ( )

1m nm n

œ œ ” •

for all positive integers and .m n

Example 1.10

Find the nominal rate of discount convertible semiannually which isequivalent to a nominal rate of interest of 12% per year convertiblemonthly.Solution

” •1 1 , so 1 (1 01) 942045, from2 12 2 d i d . . (2) (12) (2)2 12

6

œ œ œ” •which we find 2(1 942045) 11591d . . .(2) œ œ

1.6 FORCE OF INTEREST

We note before starting this section that it is somewhat theoretical, and isindependent of the rest of the text. Anyone wishing to proceed directlyto more practical problems can safely omit this material. In particular,more background knowledge is required for a full understanding herethan is required for any other section; students with only a sketchyknowledge of calculus might omit this on first reading. Assume that the effective annual rate of interest is .12, and thati œwe want to find nominal rates equivalent to . The formulai i( )m

i m i( ) 1/m mœ ‘(1 ) 1 , which comes from identity (1.19), is used tocalculate these values which are shown in Table 1.1.

TABLE 1.1

mi

1 2 5 10 50.12 .1166 .1146 .1140 .1135( )m

We observe that decreases as gets larger, a fact which wei m( )m

will be able to prove later in this section. We also observe that the valuesof are decreasing very slowly as we go further and further along; ini( )m


the language of calculus, seems to be approaching a limit. This is, ini( )m

fact, what is happening, and we can use L’Hopital’s rule to see what thelimit is. There is no need to assume .12 in our derivation, so wei œproceed with arbitrary . i

lim i lim m i lim im m m

m mm

mÄ_ Ä_ Ä_

( ) 1/1/

1œ ‘(1 ) 1 (1.22)(1 ) 1

œ

Since (1.22) is of the form , we take derivatives top and bottom, cancel,00

and obtain

lim i lim i ln i ln im m

m m

Ä_ Ä_

( ) 1/œ † œ (1 ) (1 ) (1 ), (1.23) ‘

since (1 ) 1lim i .m

m

Ä_ œ1/

This limit is called the and is denoted by , so weforce of interest $have

(1 ). (1.24)$ œ ln i

In our example, (1.12) .11333. The reader should compare this$ œ ln œwith the entries in Table 1.1. Intuitively, represents a nominal rate of interest which is$convertible , a notion of more theoretical than practicalcontinuously importance. However, can be a very good approximation for when$ i( )m

m is large (for example, a nominal rate convertible daily), and has theadvantage of being very easy to calculate. We note that identity (1.24) can be rewritten as

e i$ œ 1 . (1.25)

The usefulness of this form is shown in the next example. Again westress the importance of being able to convert a rate of interest with agiven conversion frequency to an equivalent rate with a differentconversion frequency.

18 Chapter 1

Example 1.11

A loan of 3000 is taken out on June 23, 1997. If the force of interest is14%, find each of the following:(a) The value of the loan on June 23, 2002.(b) The value of i.(c) The value of i .(12)

Solution(a) The value 5 years later is 3000(1 ) . Using 1 , we œ i e i5 $

obtain 3000( ) 3000 6041 26e e . .. .14 5 7œ œ

(b) 1 15027i e . .œ .14 œ

(c) 1 1 , so we have the result12Š œ œi i e (12) 12

.14‹ 12( 1) .14082.i e(12) .14/12œ œ

Remark

Note that if we tried to solve part (a) by first obtaining .15027 (as ini œpart (b)), and then calculating 3000(1.15027) , we would get 6041.16, an5

answer differing from our first answer by .10. There is nothing wrongwith this second method, except that not enough decimal places werecarried in the value of to guarantee an accurate answer. Let us repeat an iearlier admonition: it is always wise to do as few calculations asnecessary. Observe that [(1 ) ] (1 ) (1 ), where stands forD i i ln i D † t tœthe derivative with respect to Hence we see thatt.

(1 ) (1 26)[(1 ) ] [ ( )](1 ) ( )$ œ œ

Hln i . .i D a ti a t œ

t

t

Let us see why this fact happens to be true. Recall from the definition of

the derivative that [ ( )] , so( ) ( )D a t lim a t h a thœ

hÄ0

D a t a t h a t a t h a t hlim lim . .[ ( )] ( ) ( )

( ) ( ) (1 27)œ œ

†h h

a t h a ta t

Ä Ä0 0

( ) ( )( )


The term in (1.27) is just the effective rate of interest( ) ( )( )

a t h a ta t

over a very small time period , so is the nominal annualh h

a t h a t a t

( ) ( )( )

rate corresponding to that effective rate, which agrees with our earlierdefinition of .$ The above analysis does more than that, however. It also indicateshow the force of interest should be defined for arbitrary accumulationfunctions. First, let us observe that (1 ) is independent of $ œ ln i t.However, this is a special property of compound interest, correspondingto a constant . For arbitrary accumulation functions, we define theinforce of interest at time , , byt $t

, (1 28)[ ( )]( )$t œ

D a ta t .

since we would normally expect to depend on $t t. For certain functions, it is more convenient to use the equivalentdefinition

[ ( ( ))] (1 29)$t œ D ln a t . .

We also remark that, since ( ) ( ) , it follows thatA t k a t œ †

[ ( ( ))] (1 30)[ ( )]( )$t œ œ

D A tA t D ln A t . .

Example 1.12

Find in the case of simple interest.$t

Solution

$t œ œ

D itit it

i . (1 )1 1

We now have a method for finding the force of interest, , given$t

any accumulation function ( ) What if we are given instead, anda t . $t

wish to derive ( ) from it? a t To start with, let us write our definition of from (1.29) using a$t

different variable, namely [ ( ( ))], where now means the$r œ D ln a r Dderivative with respect to . Integrating both sides of this equation fromr0 to , we obtaint

20 Chapter 1

r dr ln a r dr D ln a œœ( t t

r

t

0 0 0 [ ( ( ))] ( ( ))$ ( ¹

ln a t ln aœ ( ( )) ( (0))

( ( )), (1 31)œ ln a t .

since (0) 1 and 1 0. Then taking the antilog we have a lnœ œ

( ) (1 32) a t e . .œ ' tr0 $ dr

Example 1.13

Prove that if is a constant (i.e., independent of ), then ( ) (1 )$ r a t iœ t

for some . iSolutionIf , the right hand side of (1.32) is ( ) Hence the c e e e . $r

c dr ct c tœ œ œ' t0

result is proved with 1i e .œ c

Example 1.14

Prove that ( ) ( ) (0) for any amount function ( )( n

t0

A t dt A n A A t .$ œ

SolutionThe left hand side is

dt A t dt D A t dtA tD A t

A tœœ( n n n

t0 0 0

( ) ( ) [ ( )][ ( )]

( )$ ( (” •

A t A n A œ œ ( ) ( ) (0) as required.¹n

0

The identity in the above example has an interesting verbalinterpretation. The term represents the effective rate of interest at$t dt time for the infinitesimal “period of time” . Hence ( ) t dt A t dt$t

represents the amount of interest earned in this period, and ( )' nt0 A t dt$

represents the total amount of interest earned over the entire period, anumber which is clearly equal to ( ) (0)A n A . We now return to the compound interest case where we havea t i( ) (1 ) . It is interesting to write some of the formulae alreadyœ t

developed as power series expansions. For example (1 )$ œ ln ibecomes

. (1.33)2 3 4$ œ âi i i i2 3 4


Convergence is a concern here, but as long as | 1, which is usuallyi | the case, the above series does converge. Another important formula was 1, which becomesi e œ $

. (1.34) 2! 3!i œ â$ $ $

2 3

Since all terms on the right hand side are positive, this allows us toconclude immediately that . We note in passing that this seriesi $converges for all .$

Next let us expand the expression (1 ) , which1i d dddœ œ 1

becomes

i d d d d d d d . .œ (1 ) (1 35) â œ â2 3 2 3

Again this shows us very clearly that . We also note that we musti dhave | 1 for this series to converge. In fact, trying to put 2d | d œ

yields an amusing result: the left hand side is 2, whereas21 2i œ œ

the right hand side becomes 2 2 2 , all of which are positive 2 3 âterms. Thus we have “proven” that 2 is a positive number! Next let us expand as a function of . From (1.19) we havei i( )m

i m i( ) 1/m mœ [(1 ) 1], so

m i i i i mœ â ( )

1 1 1 1 12 3m m m m m m” •1 11 ( 1) ( )( 1)( 2)

2! 3!

i i i . .

œ â ” •1 1 1

2 3m m m1 ( 1)( 2)2! 3! (1 36)

Again, this converges for 1 | i | . Why are we interested in power series expansions? Well, we havealready seen that they sometimes allow us to easily conclude facts likei $ (although they certainly aren’t needed for that). They also give usa quick means of calculating some of these functions, since often onlythe first few terms of the series are necessary for a high degree ofaccuracy. If you ask your calculator to do this work for you instead, itwill oblige, but the program used for the calculation will often be avariation of one of those described above.

22 Chapter 1

As a final example, let us expand in terms of . We haved( )m $

(1 ) , (1 37)1” • dm i e .( )

1m m

œ œ $

so

d m eœ ( ) /m m[1 ]$

m mœ â

– — Š ‹1 1( ) ( )

2! 3!$

$ $m m

2 3

m m m mœ â” •$ $ $2 3

2 32! 3!

. .m mœ â$ $ $2 3

22! 3!(1 38)

From this we easily see that In other words, there is nolim d . m

m

Ä_

( ) œ $

need to define a force of discount, because it will turn out to be the sameas the force of interest already defined.

EXERCISES

1.1 Accumulation Function; 1.2 Simple Interest;1.3 Compound Interest

1-1. Alphonse has 14,000 in an account on January 1, 1995. (a) Assuming simple interest at 8% per year, find the

accumulated value on January 1, 2001. (b) Assuming compound interest at 8% per year, find the

accumulated value on January 1, 2001. (c) Assuming exact simple interest at 8% per year, find the

accumulated value on March 8, 1995. (d) Assuming compound interest at 8% per year, but linear

interpolation between integral durations, find the accumula-ted value on February 17, 1997.


1-2. Mary has 14,000 in an account on January 1, 1995. (a) Assuming compound interest at 11% per year, find the

accumulated value on January 1, 2000. (b) Assuming ordinary simple interest at 11% per year, find the

accumulated value on April 7, 2000. (c) Assuming compound interest at 11% per year, but linear

interpolation between integral durations, find the accumula-ted value on April 7, 2000.

1-3. For the ( ) function given in Example 1.1, prove that fora t i in n1 all positive integers .n

1-4. Consider the function ( ) 1 ( 2 ) , 0, 0a t i i t i t .œ È 2 2

(a) Show that (0) 1 and (1) 1 . a a iœ œ (b) Show that ( ) is increasing and continuous for 0 a t t . (c) Show that ( ) 1 for 0 1, but ( ) 1 fora t it t a t it

t . 1 (d) Show that ( ) (1 ) if is sufficiently large.a t i t t

1-5. Let ( ) be a function such that (0) 1 and is constant for all .a t a i nœ n

(a) Prove that ( ) (1 ) for all integers 0a t i t .œ t

(b) Can you conclude that ( ) (1 ) for all 0a t i t ?œ t

1 6 Let ( ) be an amount function. For every positive integer ,- . A t ndefine ( ) ( 1)I A n A n .n œ

(a) Explain verbally what representsI n

(b) Prove that ( ) (0)A n A I I I . œ â1 2 n

(c) Explain verbally the result in part (b). (d) Is it true that ( ) (0) ? Explain.a n a i i i œ 1 2 â n

1-7. (a) In how many years will 1000 accumulate to 1400 at 12%simple interest?

(b) At what rate of simple interest will 1000 accumulate to 1500in 6 years?

(c) Repeat parts (a) and (b) assuming compound interest insteadof simple interest.

24 Chapter 1

1-8. At a certain rate of simple interest, 1000 will accumulate to 1300after a certain period of time. Find the accumulated value of 500 at

a rate of simple interest as great over twice as long a period of23

time.

1-9. Find the accumulated value of 6000 invested for ten years, if thecompound interest rate is 7% per year for the first four years and11% per year for the last six.

1-10. Annual compound interest rates are 13% in 1994, 11% in 1995 and15% in 1996. Find the effective rate of compound interest per yearwhich 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 2in years, 2 grows to 3 in years, and 1 grows to 5 in years.x y zProve that 6 grows to 10 in years.z x y

1-12. If 1 grows to in periods at compound rate per period and 1K x igrows to in periods at compound rate 2 per period, which oneK y iof the following is always true? Prove your answer.

(a) 2x y (b) 2 x yœ (c) 2 x y (d) y xœ È (e) 2y x

1.4 Present Value and Discount

1-13. Henry has an investment of 1000 on January 1, 1998 at acompound annual rate of discount 12.d .œ

(a) Find the value of his investment on January 1, 1995. (b) Find the value of corresponding to .i d (c) Using your answer to part (b), rework part (a) using insteadi

of . Do you get the same answer?d

1-14. Mary has 14,000 in an account on January 1, 1995. (a) Assuming compound interest at 12% per year, find the

present value on January 1, 1989. (b) Assuming compound discount at 12% per year, find the

present value on January 1, 1989. (c) Explain the relative magnitude of your answers to parts (a)

and (b).


1-15. (a) Sketch a graph of ( ) with its extension to present value ina tthe case of simple interest.

(b) Explain, both mathematically and verbally, why 1 is not itthe correct present value years in the past, when simple tinterest is assumed.

1-16. Prove that is constant in the case of compound interest.dn

1-17. Prove each of the following identities mathematically. For parts(a), (b) and (c), give a verbal explanation of how you can see thatthey are correct.

(a) (d) 11 1d iv d iœ œ

(b) 1 (e) 1 12 2d v d ii dœ Š ‹ Š ‹ œ

(c) (f) 1 1 i d id i d d i œ œ È È1 18. Four of the following five expressions have the same value (for-

i 0). Which one is the exception?

(a) (b) (c) ( ) (d) (e)(1 )

( )1 i d d i i d i dd

di d

v3

2

23 3 2

1-19. The interest on for one year is 216. The equivalent discount on L Lfor one year is 200. What is ?L

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 convertiblequarterly.

Account B: compound interest at 11.97% per year convertible 5times per year.

Account C: compound discount at 11.8% per year convertible10 times per year.

Which account is most advantageous to the investor? Whichaccount is most advantageous to Acme Trust?

26 Chapter 1

1-21. Phyllis takes out a loan of 3000 at a rate of 16% per yearconvertible 4 times a year. How much does she owe after 21months?

1-22. The Bank of Newfoundland offers a 12% mortgage convertiblesemiannually. Find each of the following:

(a) (b) (c) i d i(4) (12)

(d) The equivalent effective rate of interest per month.

1-23. 100 grows to 107 in 6 months. Find each of the following: (a) The effective rate of interest per half-year. (b) (c) (d) i i d(2) (3)

1 24 Find such that 1 .1

1 - . n i

n œ

( )6

8

n i

i

(6)

(8)

1-25. Express as a function of .d i(7) (5)

1-26. Show that 1 1 1 1 .3 30 5v di i dŒ Œ Œ È œ (3) (30) (5)

1-27. Prove that .i d i d(4) (8) (8) (4)

1-28. (a) Prove that .i d i dm

( ) ( )( ) ( )

m mm m

œ

(b) Prove that .1 1 1d i m( ) ( )m m œ

1.6 Force of Interest

1-29. Find the equivalent value of in each of the following cases.$ (a) .13i œ (b) 13d .œ (c) .13i(4) œ (d) 13d .(5) œ


1-30. In Section 1.3, it was shown that for 0 1, (1 ) 1 . t i itt

Show that 1 (1 ) is maximized at [ ].1 œit i t ln i lnt$

$

1-31. Assume that the force of interest is doubled. (a) Show that the effective annual interest rate is more than

doubled. (b) Show that the effective annual discount rate is less than

doubled.

1-32. Show that .50.lim iiÄ0 2

œ$$

1-33. Find ( ) if 04(1 ) a t . t .$t œ 1

1 34 Obtain an expression for if ( )- . A t ka b c .$tt t dœ 1 3 t

1 35 Using mathematical induction, prove that for all positive integers- .

n v i n d ddv dv, ( ) (1 )( 1)!, where denotes derivative

nn

n1$ œ

with respect to .v

1-36. Express as a power series expansion in terms of .v $

1-37. Express as a power series expansion in terms ofd i.

1 38 Prove that if - . i i n m.( ) ( )n m

1 39 Prove that for all 1- . d d i i n . ( ) ( )n n$

1 40 Show that ( ) ( ) , where is the derivative with( )( )- . D D D A t

A t$ $t tœ 2

2

respect to .t

1-41. Show that .2 4 6 $ œ âd i d i d i

2 2 3 3

1-42. Which is larger, or Prove your answer.i d? $ $

28 Chapter 1

1-43. (a) Write a computer program which will take a given value of iand output values of for a succession of values of i m.( )m

(b) Extend the program in part (a) to also give you a value for ,$using .$ œ lim i

m

m

Ä_

( )

1-44. Write a computer program which will take a given value of and$output the equivalent value of ( )i. Use the power series expansion.

theory of interest and life contingencies · pdf file1.4 present value and discount 9 ......

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