1

F71AB – Financial Mathematics

Torsten Kleinow

2009/10

Office: Colin Maclaurin building, CM F11e-mail: t.kleinow@hw.ac.ukLectures: Tuesday 11:15 and 12:15 (PG201), Wednesday 15:15 (LT1)Tutorials: Wednesday 13.15 (NS136), Wednesday 17:15 (PG202)

2

Contents

• Interest and Present Value

• Cashflows

• Yields and Equation of Value

• Annuities, Loan schedules, Fixed income securities

• Inflation and Index Linked Gilts

• Force of Interest

• Discounted Cash Flows

• Macaulay Duration and Immunisation

• Forward Contracts

• Stochastic interest rate models

• Term structure of interest rates

3

1 Introduction

We will think about cashflows - amounts of money being paid or received atdifferent times - in a pictorial fashion.

In this course we will often use a timeline, similar to the one below.

-?

£2,000

? ? ?£300 £600 £600

1/1/02 1/10/02 1/1/03 1/7/03time

Above the line is a payment (to an account), on 1 January 2002, of £2,000.

Below the line are payments from the account (£300 on 1 October 2002, and of£600 on 1 January 2003 and 1 July 2003).

4

2 Interest and Present Value

2.1 Capital and Interest

At a given point of time, cash has a monetary value, but also has a time value.

For example, it is clear that you would rather have £110 now than £100 now, butis it so clear that you would be better off with £110 in 2 years’ time than £100now ?

The point is that £1 now is worth more than £1 at some time in the future.

Why is this ?

There are three reasons:-

• Deferred Use: You do not have the use of the £1 until you receive it, someoneelse does.

• Risk: The £1 you have been promised might never be paid (the companymight go bankrupt, the borrower might die, etc)

• Inflation: prices might rise, see Section 9

5

Therefore, if you deposit or invest your money now, you expect not only yourmoney back, but also a reward for letting someone else have the use of themoney in the deferred period. We call this reward interest.

The amount deposited or invested is called capital (or principal) and the rewardreceived for investing or depositing it is called interest.

Note: p.a. or per annum is used in this course to denote per year.

Example 2.1 You deposit £1,000 in a bank account for one year. The bank paysinterest of 3% p.a..

6 6

£1,030 = £1,000 + £30

capital interest

6

2.2 Effective Rate of Interest

Definition 2.2 Money grows at an effective rate of interest of i per time unit if Xgrows to X(1 + i)t in time t.

-?

1

?1 + i

start end

� -1 time unit

Example 2.3 • Invest £1 for 1 year at 5% p.a. effective. Then i = 0.05 andafter one year the investment is worth £1.05.

• Invest £100 for 6 months at 3.75% per half-year effective. Then i = 0.0375and the investment is worth £103.75 after half a year (= 6 months).

7

2.3 More than One Time Period

Suppose the effective rate of interest is i pa, how much will we receive if weinvest £1 for two years, three years, n years ?

-?

£1

? ?£1 + i £(1 + i)(1 + i)

0 1 2time

At time 0: £1At time 1: £(1 + i) (reinvest (1+i) at time 1)At time 2: £(1 + i)(1 + i) = (1 + i)2

...At time n: £(1 + i)n.

8

Now, let us consider n years in more detail.

Assuming £1 deposit gives us £(1 + i)n−1 at time n− 1, we withdraw it andre-invest it, then at time n we have

Principal (1 + i)n−1

Interest i(1 + i)n−1

Total (1 + i)n

By induction, we have shown that £1 will indeed accumulate (or grow) to£(1 + i)n if it is invested at effective rate i pa for n years.

9

Example 2.4 • Invest £50 for 7 years at 4% pa effective. 50× (1.04)7 = £65.80

• Invest £100 for 9 months at 2% per quarter year effective.100× (1.02)3 = £106.12

• Invest £80 for 1 year at 6.25% per half year effective. 80× (1.0625)2 = £90.31

Remarks 2.5 • This is a model of compound interest, where interest is earnedon interest previously earned, not just on the original capital (or principal).

• If interest is earned only on the original principal, it is called simple interestand after n time periods we have £(1 + in) rather than £(1 + i)n. This is notrealistic and we almost always work with compound interest. Unless otherwisestated, you should always assume that i is compound.

10

2.4 Compounding Over Any Time Period – Fractions of a TimeUnit

Suppose you invest £1 today at an effective rate of interest of i = 5% p.a.. Howmuch will you have after 1 month, 7 months, 12 months, 123 months?

Change the time unit to 1/12 year (1 month).

What is the effective rate of interest j per month which gives £1.05 at the end ofthe year (after 12 months).

Solve (1 + j)12 = 1 + i = 1.05 =⇒ 1 + j = (1 + i)1/12 = 1.051/12

Value of £1 at:time t=0: £1time t=1/12 (one month): £(1 + j) = (1 + i)1/12

time t=7/12 (seven months): £(1 + j)7 = (1 + i)7×1/12 = (1 + i)7/12

time t=1 = 12/12 (twelve months): £(1 + j)12 = (1 + i)12×1/12 = (1 + i)time t=123/12 (123 months): £(1 + j)123 = (1 + i)123×1/12 = (1 + i)123/12

11

A similar argument shows: if we invest £1 for t time units (0 ≤ t) at an effectiverate i per time unit, we get

£(1 + i)t (2.1)

at time t.

Example 2.6 • Invest £2,000 for 4 months at 10.5% p.a. effective.2, 000× (1.105)4/12 = £2, 067.68

• Accumulate £1,000 for 11 months at 7% p.a. effective.1, 000× (1.07)11/12 = £1063.98

• Invest £2,500 for 1 month at 2% per quarter year effective.2, 500× (1.02)1/3 = £2, 516.56

• Invest £1 for 2.7 years at 8.6% p.a. effective. 1× (1.086)2.7 = £1.25

• Invest £100 for 3.6 years at 4.1% per half-year effective.100× (1.041)(3.6×2) = £133.55

12

2.5 Changing the Time Period

Let i be the effective rate of interest per ti years (ti years = one time unit)Let j be the effective rate of interest per tj years (tj years = another time unit)

Then from (2.1) and the definition of the effective rate of interest we obtain thevalue after one year

(1 + i)1/ti = (1 + j)1/tj

⇒ (1 + i) = (1 + j)ti/tj or (1 + j) = (1 + i)tj/ti

13

Example 2.7 • What effective rate of interest per half year is equivalent to 7%p.a. effective?1.071/2 = 1.03440804→ 0.034408 per half-year to 6 decimal places

• What effective rate of interest per quarter year is equivalent to 15% per twoyears effective?1.151/4×1/2 = 1.01762374→ 0.017624 per quarter year to 6 decimal places

• What effective rate of interest per 1.7 years is equivalent to 1.3% per montheffective ?1.0131.7×12 = 1.301465576→ 0.301466 per 1.7 years to 6 decimal places

NOTE: It is recommended that rates of interest and related quantities arerounded to 6 decimal places, but that final answers representing currencyamounts are restricted to 2 decimal places.

14

2.6 Accumulation and Discounting

We say that money accumulates with interest. The amount to which £Xaccumulates with the addition of interest i is called its accumulation oraccumulated value.

Definition 2.8 If i is the effective rate of interest per time unit, we call(1 + i) the accumulation factor per time unit.

£X invested for t time units accumulates to £X(1 + i)t.(1 + i)t is the accumulation factor over t time units.

Accumulation is retrospective in nature; it tells us how much we have at the endof some time period. Often, what we want to know is the converse, that is, howmuch we need at the start of the period in order to meet some target at the end ofthe period. For example, we might need to repay a loan in 20 years’ time.

How much should we invest now, so that, with the addition of interest, we haveexactly what we need ?

15

Example 2.9 We need £1,000 in one year’s time. The bank offers interest at aneffective rate of 6% p.a.. How much should we invest now to meet this liability?

Clearly, we need £(1, 000/1.06) since £(1, 000/1.06)× 1.06 = £1, 000

Terminology

(i) We say that £(1, 000/1.06) is the discounted value of £1,000 over 1 year at6% p.a. effective.

(ii) We also say that £(1, 000/1.06) is the present value of £1,000 payable in oneyear’s time at 6% p.a. effective. We often write PV for Present Value.

(iii) When we invest to meet a future payment like this, we say we arediscounting it.

(iv) Discounting and Accumulation are opposite operations.

Definition 2.10 The discount factor v per time unit, at rate i p.a. effective is

v =1

1 + i.

16

Example 2.11 How much should we invest now to meet a liability of £2,000 in 2years’ time if the effective rate of interest is 6% p.a. ?

X × 1.062 = £2, 000⇒ X = 2, 000/1.062 = £1, 779.99

We say that £1,779.99 is the PV of £2,000 in 2 years or the discounted value of£2,000 in 2 years.

v = 1/1.06 = 0.943396 to 6 decimal places

Example 2.12 What is the PV of £10,000 in 5 years at an effective rate ofinterest of 3% per half-year ?

£10, 000/1.0310 = £7, 440.94 (= the PV of £10,000 in 5 years.

v = 1/1.03 = 0.970874 (to 6 d.p) per half year.

17

In general: Let i be the effective rate of interest per time unit. What is the PV of£1 in t time units ?

PV = £1

(1 + i)t= £vt

Example 2.13 Let v = 0.94 be the discount factor per half-year.

(i) What is the effective rate of interest per half-year ?

(ii) What is the equivalent discount factor per year ?

(i) v = (1 + i)−1 = 0.94⇒ i = 1/v − 1 = 0.063830 (6 d.p)

(ii) The equivalent effective rate of interest per annum, say j, is given by(1 + j) = (1 + i)2 = 1.0638302 ⇒ j = 0.131734 (6 d.p)⇒ discount factor = 1/(1 + j) = 0.883600 (6 d.p) = v2

18

2.7 Non-Constant Interest Rates

The effective rate of interest need not be the same during every time period, butwe will assume that the value in every future time period is known in advance,compare Section 15.

Example 2.14 The effective rate of interest per annum will be 5% during 2007,5.5% during 2008 and 6% during 2009. What will be the accumulation of £100invested on(i) 1/1/07 for 3 years ?(ii) 1/7/07 for 2 years ?(iii) 1/10/08 for 1 years ?

19

Solution(i) £100 invested on 1/1/07 grows to £105.00 by the end of the year. In thefollowing year, £105.00 grows by 5.5% to 105× 1.055, and subsequently grows to105× 1.055× 1.06 = £117.42(ii) This time, the £100 is invested in the middle of the first year, so only earnsinterest at 5% for 6 months (or half a year), followed by 5.5% in 2008 and a finalhalf-year at 6% during 2009. Hence the total accumulation is:100× 1.051/2 × 1.055× 1.061/2 = £111.30(iii) Using the same method as part (ii), £100 is invested at 5.5% for 3 months(=1/4 year) followed by 6% for the remaining 9 months (=3/4 year).ie 100× 1.0551/4 × 1.063/4 = £105.87

20

General Formula

Suppose that [0, T ] is partitioned into n time intervals by

0 = t0 < t1 < t2 . . . < tn = T

and the effective interest rate per time unit in the interval [tr−1, tr](r = 1, 2 . . . n)is ir.

Suppose a sum of X is invested at time 0. Then the accumulation from 0 to T ofX is

X(1 + i1)t1−t0(1 + i2)t2−t1 . . . (1 + in)tn−tn−1 = X

n∏r=1

(1 + ir)tr−tr−1

Similarly, the present value (PV) at time 0 of a sum Y at time T is

Yn∏r=1

1(1 + ir)tr−tr−1

= Y

n∏r=1

(1 + ir)tr−1−tr

21

Example 2.15 Starting now, the effective rate of interest will be 3% p.a. for 6months, 4% p.a. for the following one year and 5% p.a. for the following twoyears. What is the accumulation factor over the period [0,3.5] (years) ?

The time unit is 1 year

--

0 0.5 1.5 3.5

- -1.031/21.041 1.052

t0 t1 t2 t3

The accumulation factor = 1.031/2 × 1.04× 1.052 = 1.163672 to 6 d.p

22

Example 2.16 Starting now, the effective rate of interest will be 2% per half-yearfor 1 year, 1.5% per quarter year for 6 months and 8% p.a. for 18 months. Whatis the PV now of £1,000 due in 3 years’ time?

We need to choose one time unit and convert all effective rates to that time unit.In this case, we will choose the quarter year (3 months) and convert:2% per half year ≡ i1 = 0.009950 per quarter year1.5% per quarter year ≡ i2 = 0.015 per quarter year8% p.a. ≡ i3 = 0.019427 per quarter year(all figures to 6 d.p)PV = £1, 000× 1.009950−4 × 1.015−2 × 1.019427−6 = £831.25 rounded to 2 d.p

23

2.8 General Accumulation and Discount Factors

So far we have considered mostly accumulation or PV between now (representedby t=0) and some fixed future time. We now want to generalise that.

Definition 2.17 A(S, τ) is the accumulation at time τ of 1 invested at time S(τ > S).V (S, τ) is the discounted value at time S of 1 receivable at time τ (τ > S).

Example 2.18 The effective rate of interest will be 10% p.a. during 2007, 9%p.a. during 2008 and 8% p.a. during 2009. Find

(i) A(1/1/07, 1/1/10)

(ii) A(1/7/07, 1/7/09)

(iii) V (1/7/07, 1/7/09)

(iv) V (1/4/08, 1/10/09)

24

In time units of 1 year, we calculate:

(i) A(0, 3) = 1.294920

(ii) A(0.5, 2.5) = 1.188050

(iii) V (0.5, 2.5) = 0.841715

(iv) V (1.25, 2.75) = 0.884835

Remarks 2.19 (i) V (S, τ) = A(S, τ)−1

(ii) For S < R < τ holds:A(S, τ) = A(S,R)A(R, τ) (Accumulation from S to R then from R to τ)V (S, τ) = V (S,R)V (R, τ)

(iii) A(τ, τ) = V (τ, τ) = 1

25

3 Cashflows and Annuities

Definition 3.1 A series of certain cashflows is defined by:

(i) the times of payments (cashflows) t1, t2, . . . and

(ii) the amounts of payments Ci or C(ti) which will be paid at times ti. Theamounts can be positive or negative.

-? ?

C1 = +100

? ?C2 = −20

C3 = +50

C4 = −10

t1 t2 t3 t4time

26

What is the value of a series of cashflows at any time t assuming an effective rateof interest i p.a.?

Definition 3.2 The Present Value at any time t of a set of cashflows is

PVi(t) =∞∑n=1

Cn(1 + i)t−tn =∞∑n=1

Cnvtn−t

where i is the effective rate of interest and v = 1/(1 + i).

(i) PVi(t) depends on i and t as well as the amounts and times of the cash flows.

(ii) If t ≥ tn for all n, then PVi(t) represents the accumulation to time t ofinvestments C1, C2, . . . at times t1, t2, . . . assuming all investments earninterest at an effective rate i p.a.

(iii) If t ≤ tn for all n, then PVi(t) is the amount which should be invested at timet to pay C1, C2, . . . at times t1, t2, . . .

27

(iv) For all s and t:

PVi(s) = PVi(t)vt−s = PVi(t)(1 + i)s−t

(PVi(s) is the present value at time s of PVi(t))

(v) In terms of present values, and at a fixed effective rate of interest, the originalseries of cashflows is equivalent to a single payment of amount PVi(t) at timet, see the Principle of Equivalence (definition 4.1)

(vi) If two different series of cashflows have the same PV at one time, at a giveneffective rate of interest, they will have the same PV at any time (at thateffective rate of interest). This follows directly from (iv).

28

Example 3.3 Suppose i = 0.06 effective per time unit. Cashflows are as follows:

-? ?

C1 = +100

? ?C2 = −20

C3 = +50

C4 = −10

t1 = 1 t2 = 2 t3 = 3 t4 = 5time

What is:

(i) The accumulation at T = 6?

(ii) The present value at t0 = 0?

(iii) The present value at t = 3?

29

(i)4∑i=1

CiA(ti, 6) = 100A(1, 6)− 20A(2, 6) + 50A(3, 6)− 10A(5, 6)

= 100(1.06)5 − 20(1.06)4 + 50(1.06)3 − 10(1.06) = £157.52

(ii)4∑i=1

CiV (0, ti) = 100V (0, 1)− 20V (0, 2) + 50V (0, 3)− 10V (0, 5)

= 100(1.06)−1 − 20(1.06)−2 + 50(1.06)−3 − 10(1.06)−5 = £111.05

(iii) ∑ti≤3

CiA(ti, 3)+∑ti>3

CiV (t, ti) = 100A(1, 3)−20A(2, 3)+50A(3, 3)−10V (3, 5)

= 100(1.06)2 − 20(1.06) + 50− 10(1.06)−2 = £132.26

Note111.05× 1.066 = 157.52

132.26 = 157.52× 1.06−3 = 111.05× 1.063

30

3.1 Level Annuities Certain

An annuity is a regular series of cashflows, usually purchased from, or payableto, a financial institution. A pension is an example. When the cashflows arecertain, we call it an annuity-certain.

Simple annuities such as “1 per annum for n years” appear very often in financialproblems. The simplest has cashflows of 1 at the end of the next n time units.This is an annuity payable once per time unit in arrears for n time units (n aninteger, n > 0).

-?? ? ?

1 1 1 1

t = 0 1 2 ....... n− 1 ntime

31

Actuarial Notation

Let i be the rate of interest per time unit effective.

sin is the accumulation at time n of the annuity.ain is the present value at time 0 of the annuity.

When the rate of interest is constant and clear from the context we often omit thei and just write an and sn .

32

Revision – Geometric Series

Lemma 3.4 The sum of a geometric series:

a+ ar + ar2 + ....+ arn−1 = arn − 1r − 1

for r > 0, a ∈ R, n ∈ N

Proof. Let x = a+ ar + ar2 + .....+ arn−1

(r − 1)x = rx− x

= (ar + ar2 + ...+ arn−1 + arn)− (a+ ar + ...+ arn−1)

= arn − a

⇒ x =arn − ar − 1

= arn − 1r − 1

�

33

Theorem 3.5 The accumulated value and the present value of a level annuitycertain are given by

(i) accumulated value: sin =(1 + i)n − 1

i

(ii) present value at the start of the annuity: ain =1− vn

i

Proof.

(i)

sin =n∑r=1

CrA(tr, n) =n∑r=1

(1 + i)n−r

= (1 + i)n−1 + (1 + i)n−2 + . . .+ (1 + i) + 1

=(1 + i)n − 1(1 + i)− 1

(Geometric series with a = 1, r = 1 + i)

=(1 + i)n − 1

i

34

(ii) With v = 1/(1 + i):

ain =n∑r=1

CrV (0, tr) =n∑r=1

vr

= v + v2 + . . .+ vn−1 + vn

= v[1 + v + . . .+ vn−2 + vn−1

]= v

vn − 1v − 1

(Geometric series with a = r = v)

=v

1− v(1− vn) =

1− vn

i

Alternatively, we could use the result

ain = V (0, n)sin = vn(1 + i)n − 1

i=

1− vn

i

�

35

Example 3.6 Using a rate of interest of 12% p.a. effective, find:

(i) the accumulation at 20 years of £1,000 payable yearly in arrear for 20 years.

(ii) the present value now of £7,000 payable yearly in arrear for the next 15years.

(iii) the present value now of £500 payable at the end of each of the next 14half-years.

Solution

(i)

1, 000s20 = 1, 0001.1220 − 1

0.12= £72, 052.44

(ii)

7, 000a15 = 7, 0001− v15

i= 7, 000

1− 1.12−15

0.12= £47, 676.05

36

(iii) First we calculate the half-yearly effective rate, j say

j = 1.121/2 − 1 = 0.058301

to 6 decimal places.We want

500a.05830114

=1− 1.058301−14

0.58301= £4, 696.78

37

3.2 Level Annuities Due

An annuity-due is an annuity where the payments are received at the start ofeach time unit instead of at the end. We describe the payments as being inadvance rather than in arrear.

-? ? ? ?

1 1 11

t = 0 1 2 ....... n− 1 n

No paymentat time n

time

Actuarial Notation

Let i be the rate of interest per time unit effective.

sin is the accumulation at time n of the annuity.ain is the present value at time 0 of the annuity.We say “s-due” (or “s double-dot”) and “a-due” (or “a double-dot”)

38

3.3 The Rate of Discount

Define d = 1− v to be the effective rate of discount per time unit

Example 3.7 Given i = 0.04, find v and d.v = 1/1.04 = 0.961538d = 1− v = 0.038462

The idea can also be expressed as follows:Suppose the bank added interest to your account at the start of each time unitinstead of at the end, then how much interest would it add to be equivalent to arate of i per time unit effective ?

39

Bank account 1 - pays interest in advance. Let us suppose that for an investmentof 1 it paid interest of x immediately, so a total of 1 + x is available for investmentat the start of the period.

-?

?

1

x?1

If we assume that this x in interest can be withdrawn and invested elsewhere inBank account 2 - which pays interest in arrear, then the x would grow over theyear to x(1 + i) where i is the rate of interest and the 1 initially invested is still inbank account 1.

-?

x

?(1 + i)x

40

So, at the end of the year, we would have our initial investment of 1 plus theinterest in advance with added interest = x(1 + i). For this to be equivalent tointerest paid in arrears, we must solve

1 + x(1 + i) = 1 + i

⇒ x(1 + i) = i

⇒ x =i

1 + i=

1 + i

1 + i− 1

1 + i= 1− v = d

Interpretation: d is the effective rate of interest payable in advancecorresponding to the effective rate of interest i payable in arrear.

The following relationships are useful and should be remembered:

v =1

1 + i

d = 1− v = iv =i

1 + i

41

Theorem 3.8 The accumulated value and the present value of a annuity due(payment in advance) are given by

(i) accumulated value: sin =(1 + i)n − 1

d

(ii) present value at the start of the annuity: ain =1− vn

d

Proof.

(i)

sin =n∑r=1

(1 + i)r = (1 + i)n−1∑r=0

(1 + i)r = (1 + i)sin

= (1 + i)(1 + i)n − 1

i=

(1 + i)n − 1d

since d = i/1 + i

(ii)

ain = V (0, n)sin = vn(1 + i)n − 1

d=

1− vn

dsince vn = (1 + i)−n.

�

42

Example 3.9 (Compare example 3.6) Let the effective rate of interest per annumbe 12%. Find:

(i) si20

(ii) ai15

(iii) the present value at time 0 of £500 paid at the start of each of the next 14half-years.

Solution

(i)

si20

=1.1220 − 1

d= 1.12

1.1220 − 10.12

= 80.698735

(ii)

ai15

=1− 1.12−15

d= 1.12

1− 1.12−15

0.12= 7.628168

(iii) The natural time unit is the half-year, with effective rate of interest 0.058301per time unit (see example 3.6), then

500a0.05830114

= 500× 1.058301(1− 1.05830114

0.058301) = £4, 970.60

43

Theorem 3.10 Further Important Relationships

• sin = (1 + i)sin

• ain = (1 + i)ain

• sin = 1 + sin−1

• ain = 1 + ain−1

The proofs are easy and left to you.

44

3.4 Perpetuities

A perpetuity is an annuity that never ceases.

Example 3.11 (i) An annuity of £100 per annum, payable in arrear forperpetuity.

(ii) An annuity-due of £100 per month, payable (in advance) for perpetuity.

Definition 3.12 ai∞ is the present value of a perpetuity of 1 per time unit payablein arrear. We can define ai∞ exactly as before.

45

Theorem 3.13

ai∞ =1i

and ai∞ =1i

+ 1 =1d

Proof.

ai∞ = v + v2 + v3 + . . .+ vn + vn+1 + . . .

= limn→∞

(v + v2 + . . .+ vn) = limn→∞

ain

= limn→∞

1− vn

i

=1− limn→∞ vn

i= 1/i since v < 1

�

Example 3.14 a0.05∞ = 1

0.05 = 20 and a0.05∞ = 1

0.05 = 21

46

3.5 Deferred Annuities

Sometimes an annuity does not start immediately, but is deferred for a period oftime. For example, an employer might promise to pay an annuity to an employeewhen he/she retires.

Actuarial NotationWe use the notation “m|” before an annuity symbol to denote that the annuity isdeferred m time units. (Eg 10|ai5 ).

Example 3.15 Calculate, at an effective interest rate of 10% per annum, thepresent value at t = 0 of an annuity of 1 p.a., payable in arrear for 5 years,deferred for 10 years.

47

-? ?? ??

1 1111

t = 0 5. . . . . . 10 11 15time

Note The first payment is at t = 11, not t = 10.

10|ai5 =15∑t=11

V (0, t) = V (0, 10)15∑t=11

V (10, t) = v1015∑t=11

vt−10 = v10ai5

= 1.46

It should be clear that the general rule for this type of calculation is to calculatethe value at the end of the deferred period and then discount back to the start ofthe period.

48

3.6 Increasing Annuities

Simple Increasing Annuities

Definition 3.16 (Is)in is the accumulation at t = n of an annuity payable yearly inarrear, the amount being paid at time t(1 ≤ t ≤ n) being t.

-?? ? ?

1 2 n-1 n

t = 0 1 2 ....... n− 1 n

Payments

time

49

Definition 3.17 We also define in a similar way

(i) (Ia)in - the present value of the same series of payments

(ii) (Ia)in - as above but in advance

(iii) (Is)in - an accumulation of these payments, in advance

Theorem 3.18 Some Useful Results:

(i) (Is)in =sin − n

i

(ii) (Is)in =sin − nd

(iii) (Ia)in =ain − nvn

i

(iv) (Ia)in =ain − nvn

d

50

Proof.

(Is)in = (1 + i)n−1 + 2(1 + i)n−2 + . . .+ (n− 1)(1 + i) + n

=⇒ (1 + i)(Is)in = (1 + i)n + 2(1 + i)n−1 + . . .+ (n− 1)(1 + i)2 + n(1 + i)

=⇒ i(Is)in = (1 + i)(Is)in − (Is)in= (1 + i)n + (1 + i)n−1 + . . .+ (1 + i)2 + (1 + i)︸ ︷︷ ︸

sin

−n

⇒ (Is)in =sin − n

i

�

Example 3.19 Find (Ia)0.0510

v10 = 0.613913 and a10 = 8.108

⇒ (Ia)0.0510

= 1.05× 8.108− 10× 0.6139130.05

= 41.346

51

Compound Increasing Annuities

Another common form of increasing annuity is one that increases by a fixedpercentage each time unit.

Example 3.20 A person aged 65 receives a pension initially of £10 pa, payable inarrear for 35 years, and increasing by 5% each year. What is the present value ofthis annuity at an effective rate of interest of 7% pa ?

52

Solution

-?? ? ?

10 10 10 10×1.05 ×1.0533 ×1.0534

65 66 67 ....... 99 100age

At age 65, the present value of the payments, writing v = 11+i and i = 7% is

10v + 10(1.05)v2 + . . .+ 10(1.0533)v34 + 10(1.0534)v35

Now, define j such that 1 + j = 1+i1.05 = 1

(1.05v) , then the present value is:-

10v[1 + (1 + j)−1 + (1 + j)−2 + . . .+ (1 + j)−34

]But this expression is just 10v × aj

35= 241.68

53

Decreasing Annuities

A decreasing annuity is an annuity that starts at n in the 1st year and decreasesby 1 each time unit, so that an n-year decreasing annuity pays 1 in the nth year.

-? ? ? ?

n n-1 2 1

t = 0 1 2 ....... n− 1 ntime

(Da)in = nv + (n− 1)v2 + . . .+ 2vn−1 + vn

=n

1 + i+

n− 1(1 + i)2

+ . . .+2

(1 + i)n−1+

1(1 + i)n

=⇒ (1 + i)(Da)in = n+ (n− 1)v + . . .+ 2vn−2 + vn−1

=⇒ i(Da)in = n− v − v2 − . . .− vn−1 − vn = n− ain

⇒ (Da)n =n− an

i

54

Equally, we could consider the decreasing annuity to be a regular annuity of n+ 1per time unit less an increasing series of payments starting at 1 and increasing ton, so we could express:-

(Da)n = (n+ 1)an − (Ia)n

The usual notation applied for payments in advance or in arrears.

55

3.7 Annuities Payable More than Once per Time Unit

(see also Section 9)

Whilst interest rates are often quoted as“per annum”, it is quite common forpayments to be made at intervals of less than a year, for example quarterly (= 4times per year) or monthly (=12 times per year). Consequently, we need anotation to deal with this.

Definition 3.21 s(m)in is the accumulation after n time units of an annuity of 1 per

time unit payable m times per time unit, in arrear, at effective rate of interest i pertime unit.

Note that the payment is 1 per time unit so that each payment is of amount 1/m.

If we denote by j the effective rate of interest per 1/m time unit then

s(m)in =

1msjnm

56

Example 3.22 Find s(2)0.0710

s(2)0.0710 =

12s0.03440820

=12

(1.03440820 − 1)0.034408

= 14.054

We quote the value of an annuity to 3 decimal places.

Example 3.23 What is the accumulation at 1/1/2010 of an annuity of £1,000per month, payable in arrear from 1/1/2000. at an effective rate of interest of8.5% pa ?

Note The first payment would have been on 1/2/2000.

12, 000s(12)0.08510

= 12, 000112s0.006821120

= 1, 000(1.006821120 − 1)

0.006821= £184, 848.34

The same notation extends to present values and to annuities due, so that

• a(m)in = 1

majmn

• s(m)in = 1

m sjmn

• a(m)in = 1

m ajmn

57

4 Principal of Equivalence, Yields and Equation ofValue

4.1 The Principle of Equivalence

Any financial transaction involving a person or institution can be summarised in adiagram similar to the one below:

-? ?

Payments by

Payments tothe person

the personC1

? ?C2

C3

C4

t1 t2 t3 t4time

58

Actuaries use the Principle of Equivalence to compare two or more cashflowsand to assess whether one is worth more than the other.

Definition 4.1 (Principle of Equivalence) Consider a given effective rate of interesti, a sequence of cashflows: C1, C2, . . . with payments at times t1, t2, . . . and anothersequence of cashflows D1, D2, . . . with payments at times s1, s2, . . .. The twosequences of cashflows are said to be equivalent (or equal in value) if their values atany time t are the same, that is

∃ t ∈ R : PV Ci (t) = PV Di (t)

where PV Ci (t) and PV Di (t) denote the present values of the two sequences ofcashflows.

59

Remark 4.2 (i) It is assumed that the interest rate(s) are given, and that all valuesare calculated using the same rate(s). In definition 4.1 we have used the sameeffective rate of interest i for both present values.

(ii) If two series of cashflows have the same value at any time, t, then they musthave the same values at all times, since:Value at time t = Value at time s ×A(s, t) (t ≥ s)Value at time t = Value at time s ×V (t, s) (t ≤ s)

(iii) Since the value of a single payment of £X now is £X, we often find the price of aseries of cashflows to be the present value of these cashflows now.

(iv) We will be indifferent between two series of cashflows if their present values areequal.

60

Example 4.3 Show that the following three cashflows are all equivalent given aninterest rate of 5% effective per annum.

We will calculate our present values at time t = 0.

Series of cashflows 1 - One payment

-

Payments

6

10, 000(1.05)4

0 1 2 . . . 3 4years

The present value is

1.05−4 × 10, 000(1.05)4 = 10, 000

61

Series of cashflows 2 - Level annuity plus a lump sum payment

-

Payments

66 6 6

500 500 500 500 + 10, 000

0 1 2 . . . 3 4years

The present value is

500a5%4

+ 10, 000× 1.05−4 = 10, 000

62

Series of cashflows 3 - Level annuity

-

Payments

66 6 6

2,820.12 2,820.12 2,820.12 2,820.12

0 1 2 . . . 3 4years

The present value is

2, 820.12a5%4

= 10, 000.01

63

Example 4.4 You have been offered an investment that will pay you £200 at theend of each of the next 24 months. What is the maximum price should you payfor this if the interest rate is 0.5% per month effective ?

Answer - you would be willing to pay at most a price equivalent to the value ofthe cashflows.

The discount factor v = 1/1.005. Using the PV at time t = 0

PV = £200a0.5%24

= £2001− 1.005−24

0.005= £4, 512.57

ie you would be willing to pay up to £4,512.57.

64

4.2 Yields and the Equation of Value

Definition 4.5 Consider a series of cash flows C1, C2, . . . (positive and negative)payable at times t1, t2, . . .. The equation

PVi(0) =∞∑n=1

Cnvtn =

∞∑n=1

Cn

(1

1 + i

)tn= 0

is called Equation of Value.The solution i of this equation is called the Yield of the cashflows.

Whilst most equations of values will have multiple solutions, in most practicalsituations, it will have a unique positive real solution.

Example 4.6 An investor (A) lends to borrower (B) the sum of £1,000. B willrepay A by paying £200 after 1 year, £200 after 2 years and £300 after each ofyears 3, 4 and 5. Calculate the yield on the transaction for A.

65

-?

A pays

A receives ?? ? ? ?

1,000

200 200 300 300 300

0 1 2 3 4 5years

The yield is i p.a. effective where v = 1/(1 + i) and

1, 000 = 200v + 200v2 + 300(v3 + v4 + v5)

taking PVs at time 0.

The yield ≡ the rate of interest earned by A on this transaction. The transactionfor B is a mirror image of that for A and so the yield for B ≡ the yield for A, butwe refer to this as the rate of interest B pays.

66

4.3 Calculating the Yield

• The easiest way to solve EV is to use a computer.

• Linear Interpolation: find a value i1 with PVi1(0) > 0 and a value i2 withPVi2(0) < 0.

67

Result about similar triangles that:

i− i1i2 − i1

≈ f(i)− f(i1)f(i2)− f(i1)

-

6

i2i1 i

f(i1)

f(i)

f(i2) � -?

6

f(i2)− f(i1)

i2 − i1

� -?

6

f(i)− f(i1)

i− i1

and so i ≈ i1 − (i2 − i1)f(i1)

f(i2)− f(i1)

68

Example 4.7 In the situation of Example 4.6 above we have: i1 = 0.08, i2 = 0.09and i ≈ 0.08 + (0.01× 19.492

9.021+19.492 ) = 8.68% effective.

4.4 Prices and Yields

We are often faced with an investment opportunity (to buy shares, to invest in anew project etc.) which will produce some future cashflows C1, C2, . . . Cn attimes t1, t2, . . . tn.

Taking t0 as the time ”now”, we assume that all future cashflows are certain. Wethen need to decide whether to make the investment.

There are two obvious questions:-

(i) If we want to earn a given yield on the investment, what price should we pay?

(ii) If we pay a given price for the investment, what yield do we earn ?

69

We often use yields to compare alternative investment opportunities.

Example 4.8 We are considering two investments:

-?

A

?? ? ?

Price = 100

10 10 10 110

0 1 2 . . . 9 10years

70

-?

B

?? ? ?

Price = 100

8 8 8 130

0 1 2 . . . 9 10years

On the basis of yield, do we prefer investment A or investment B ?

The yield on A is 10% (obvious since the original investment is returned at theend of 10 years, and payments of interest are 10 pa for a 100 investment).

71

For investment B we need to solve the equation

f(i) = 100(1 + i)10 − 8[(1 + i)9 + (1 + i)8 + . . .+ (1 + i) + 1

]− 122 = 0

By trial and error:-f(0.09) = −6.807f(0.10) = +9.875By linear interpolation

i ≈ 0.09 +0.01× 6.8079.875 + 6.807

≈ 0.094(= 9.4%)

Hence we prefer A, which has the higher yield.

More carefully discussed in Chapters 11 and 12.

72

5 Loan Schedules and Mortgages

5.1 Three ways to Repay a Loan

You borrow £L from a bank, to be repaid by the end of n years. We say the termof the loan is n years and the amount of the loan is £L.

Given an effective rate of interest of i per annum, how could you repay the loan ?

73

Method 1 - As late as possible

-

6

Loan

Repayments

6

L

L(1 + i)n

0 1 2 . . . n− 1 nyears

After n years, you repay the entire loan and all the interest that accrued over theperiod, so you must repay a total of

L(1 + i)n

74

Method 2 - Repay interest only during the term, and repay the capital at theend of the term

-

6

Loan

Repayments

66 6 6

L

Li Li Li Li+ L

0 1 2 . . . n− 1 nyears

In the corporate sector a loan of this sort, where the borrower is a government ora company, is known as a bond or a fixed interest secuirty. We will discuss thesetypes of loan later in the course.

75

Method 3 - repay the loan and the interest in level instalments throughtoutthe term of the loan

-

6

Loan

Repayments

66 6 6

L

X X X X

0 1 2 . . . n− 1 nyears

In this section of the course, we will be discussing this (third) method.

76

Example 5.1 You borrow £10,000 for a term of 4 years. At an interest rate of 5%per annum effective, show that £2,820.12 paid yearly in arrear will exactly payoff the loan plus interest.

We set out the calculation in the form of a Loan Schedule.

Interest Capital Capital

Time(yrs) Repayment Content Content Outstanding

0 - - - 10,000.00

1 2,820.12 500.00 2,320.12 7,679.88

2 2,820.12 383.99 2,436.13 5,243.75

3 2,820.12 262.19 2,557.93 2,685.82

4 2,820.12 134.29 2,685.83 -0.01

77

In the first year, we must pay interest on the whole sum borrowed. This is5%×£10, 000 = £500. The total repayment is £2,820.12, so the balance, afterpaying the interest due, is £2,320.12 and this can be used to reduce the amountof the loan. This leaves an outstanding loan at the end of the first year of£7,679.88 (= £10, 000−£2, 320.12).

In the second year, interest amounts to 5%×£7, 679.88 = £383.99, since we onlypay interest on the reduced loan. The capital repaid is£2, 820.12−£383.99 = £2, 436.13. Hence the capital outstanding at the end ofthe second year is £7, 679.88−£2, 436.13 = £5, 243.75.

We continue like this in years 3 and 4, to discover that the loan is totally repaid atthe end of the 4th year.

(In practice, we would only repay £2,820.11 in the final year in order that theloan is repaid exactly.)

78Note:

To determine the amount of the repayments to repay the loan, we calculate:

£10, 000a5%4

= £2, 820.1183

However, as we cannot make repayments including fractions of a penny, we willrepay £2,820.12 and adjust with the final payment.

79

5.2 The Mathematics of the Loan Schedule

Time Interest Capital Capital

(yrs) Repayment Content Content Outstanding

0 - - - L = Xan

1 X iL = Xian Xvn L−Xvn = X(an − vn)

= X(1− vn) = Xan−1

2 X Xian−1 Xvn−1 X(an−1 − vn−1)

= X(1− vn−1) = Xan−2

...

r X Xian+1−r Xvn+1−r X(an+1−r − vn+1−r)

= X(1− vn+1−r) = Xan−r

...

n X iXv X(1− iv) 0

= Xv

80

Notice that the capital outstanding after the rth payment is Xan−r . This is thePV of future repayments.

General Principle - key to all loan schedule problems

The capital outstanding at any time is the present value at that time of all futurerepayments.

This is true even where the repayments and interest rates are not constant.

Example 5.2 In example 4.3, we can use this method to calculate directly thecapital outstanding after the first repayment.There were 4 payments in total, so 3 payments of £2,820.12 are outstanding afterthe first repayment.

81

The formula for the present value of these outstanding repayments is

£2, 820.12a5%3

= £7, 679.88

To calculate the capital content of the 3rd repayment, we calculate

£2, 820.12v4+1−3 = £2, 557.93

where v = 1.05−1

To calculate the interest content of the 3rd repayment, we can calculate thissimply from the previous result as £2, 820.12−£2, 557.93 = £262.19 since eachrepayment consists only of interest and capital (so the total repayment lessinterest component gives us the capital component) or we can calculate from theformula

iXan+1−r = 0.05×£2, 820.12× a5%4+1−3

= £262.19

82

5.3 Changing the term of a loan

It is common for a borrower to ask for a change in the term of a loan, forexample:-

• Extend or shorten the loan

• Miss one or more payments

• Repay part of the loan early

The lender will then need to calculate a new repayment amount.

Example 5.3 A borrower takes out a loan of £70,000 to be repayable by levelannual instalments over 20 years where the repayments are calculated at aneffective rate of interest of 7% pa.

83

(i) How much is each annual instalment ?

Using the Principle of Equivalence, we know that the loan of £70,000 mustequal the present value of the instalments, so, using X to denote theinstalment amount, we calculate

£70, 000 = Xa7%20⇒ X = £6, 607.50

(ii) What is the new repayment amount, if the borrower asks for the term to beextended by 5 years, immediately after he has made the 13th repayment ?

Just after the 13th repayment, there are 7 instalments left to pay, so theoutstanding loan is

Xa7%7

= £35, 609.73

Now, instead of paying this off over 7 years, the investor wants to pay this offover 12 years (= 7 + 5). To calculate the new instalment, the outstandingloan must be equivalent to the present value of the 12 outstandinginstalments, each of X ′ say. We calculate

£35, 609.73 = X ′a7%12⇒ X ′ = £4, 483.34

84

(iii) If instead of extending the term, the investor had asked to miss the 14th and15th repayments, what would each of the remaining 5 (= 7− 2) instalmentsbe? Over the 2 years when he did not make any repayments, the capitaloutstanding would have increased by 7%pa from £35,609.73 to35, 609.73(1.07)2 = 40, 769.58. We equate this amount to the 5 outstandingrepayments, where X ′′ is the new instalment amount

40, 769.58 = X ′′a7%5⇒ X ′′ = 9, 943.32

(iv) What is the new repayment amount, X ′′′, if the borrower had decided torepay £5,000 at time 13 together with his 13th repayment ?

The outstanding loan after the 13th repayment was £35,609.73. We reducethis by £5,000 and re-calculate the instalments over the remaining 7 years.

£35, 609.73−£5, 000 = X ′′′a7%7⇒ X ′′′ = £5, 679.73

85

5.4 Changing the Interest Rate

Often more than one interest rate is involved during the term of the loan. Thismay be because

(i) Interest rates were planned to change during the term, for example, amarketing device allowing borrowers to repay less during the initial period of aloan; or

(ii) The lender changes the interest rate for all borrowers to reflect changedmarket conditions. (This is normal practice in the UK)

86

Example 5.4 You borrow £50,000 from a bank, to be repaid over 25 years bylevel annual repayments based on an effective rate of interest of 3% pa.

Just after the 5th repayment has been paid, the bank raises its interest rate to 5%pa effective. What is the new repayment ?

First, we calculate the initial repayment amount,X.

£50, 000 = Xa3%25⇒ X = £2, 871.39

Next we calculate the outstanding loan just after the 5th repayment (ie wecalculate the present value of the outstanding 20 (= 25− 5) repayments).

= £2, 871.39a3%20

= £42, 719.03

Finally, we equate the outstanding loan to the present value of the newrepayment amount, X ′, on the new terms (ie at 5% pa interest).

£42, 719.03 = X ′a5%20⇒ X ′ = £3, 427.89

87

Example 5.5 You borrow £100,000 over 30 years, to be repaid by level annualrepayments. The rate of interest will be 6% pa effective for the first 15 years and7% pa effective thereafter. Find the repayment amount, X.

At t = 0, £100,000 must be equivalent to the present value of all futurerepayments, so we can write

100, 000 = Xa6%15

+Xa7%15v15

where v is calculated at 6%.

It is important to realise that the rate of interest of 6% applies for the first 15years, so to calculate the present value of the final 15 payments, we mustcalculate the annuity using the new interest rate, but during the deferred period,the original interest rate applies.

100, 000 = X(9.712249 + 0.417265× 9.107914)⇒ X = £7, 400.47

88

6 Fixed Interest Securities

6.1 Introduction and Terminology

A government or corporation can raise money in the capital markets by issuingfixed interest securities (FIS), also called bonds.

Investors will lend money to the issuer and in return will receive fixed interestpayments at fixed dates plus repayment of the loan at the end of the term.

-?

Loan

Repayments?? ? ?

L

Li Li Li Li+ L

0 1 2 . . . n-1 nyears

89

The loan is usually split into smaller units that can be traded on an exchange.

Example 6.1 A company raises £100,000,000 by issuing 1,000,000 bonds, eachone a loan of face value £100. These can be bought and sold on a stock exchange.

Definition 6.2 The nominal amount (also called face value) of a bond is theamount of the loan it represents (£100 in the example above).

The interest payments are called coupons, usually expressed as a percentage peryear of the nominal amount. They are always in arrear.

90

Example 6.3 (Example 6.1 continued)

Each bond of £100 nominal value carries coupons of 6% pa payable half-yearly.

-

Nominal

Coupons? ? ? ? ? ?

100

3 3 3 3 33

0 1 2 3 . . .years

Important Note - Coupons are usually expressed as the amount of interestpayable in a year, but are paid half-yearly (=twice per year) or quarterly (=4times per year).6% pa payable half yearly = £3 each half year10% pa payable quarterly = £2.50 each quarter year

91

Definition 6.4 The loan is repaid or redeemed at the end of the term. The date atwhich it is redeemed is known as the redemption date. Often the amount that isrepaid is different from the nominal amount. The redemption amount per 100nominal is the redemption rate, often expressed as a percentage.

Example 6.5 A company issues a 10-year bond, to be redeemed at 105%, withcoupon of 10% pa payable half-yearly. The nominal amount of each bond is £100.What repayments are made?

-

Nominal

Coupons? ? ? ? ?? ??

100

5 5 5 5 55 5 5+105

0 1 2 3 . . . 10years

92

Bonds may have many different names, for example

• Gilts (= gilt-edged securities) issued by the UK government

• T-bonds(= Treasury Bonds) issued by the US government

• Debentures issued by corporations

• Loan stocks issued by public sector corporations

6.2 Bond Prices

At any time, including the issue date, investors will decide what price to pay for abond. This is the basis of buying and selling bonds in the exchange. An importantcriterion is the yield compared to other investors.

93

Example 6.6 (Example 6.5 continued)An investor wishes to buy the bond in example 6.5 on its issue date, to give ayield of 11% pa effective (perhaps because he can obtain this yield from analternative investment). What price should he pay ?

Applying the equivalence principle, the investor will equate the present value offuture income at 11% pa with the unknown price, P, where

P = 5a0.05356520

+ 105(1.11)−10

Note: This is really a bond for 20 time units (= half years) to yield 5.3565% pertime unit (1.110.5 − 1 = 0.053565)

⇒ P = 60.4698 + 36.9794 = £97.45

This is less than the nominal amount of £100.

94

Remark 6.7 (i) The nominal amount of the loan is just a theoretical figure onwhich the coupon and redemption rates are based (usually £100).

(ii) The amount of capital the borrower can actually raise on the issue date is theprice that investors are willing to pay for the future income stream.

(iii) A loan is issued- at a premium if P > nominal amount- at par if P =nominal amount- at a discount if P < nominal amount

(iv) As well as the yield on alternative investments, investors will consider the creditrisk of the borrower, that is, the risk that they might default on interest orcapital payments. The greater the risk, the higher the yield they will require.

(v) There is an inverse relationship between yields and bond prices.

Once a bond is issued, it can be traded on an exchange at any time until it isredeemed. The prices will depend on:-- The remaining term to redemption- market conditions such as the yields obtainable from time to time.

95

Example 6.8 Returning to examples 6.5 and 6.6, just after the 10th coupon hasbeen paid, the investor sells the bond. At that time the market yield oncomparable 5-year bonds is 7% pa effective. What price will he obtain ?

P = 5a0.03440810

+ 105(1.07)−5 = 41.7074 + 74.8635 = £116.57

If the original investor paid £97.45, had received 10 coupon payments of £5 eachand sold the bond after 5 years for £116.57, what was the annual yield,i , on thetransaction ?

f(i) = 97.45(1 + i)10 − 5si10− 116.57 = 0

By trial and error, f(0.06) = −7.956, f(0.07) = +6.047⇒ i ≈ 0.06568 per halfyear⇒ yield ≈ 0.13568 ≈ 13.6%pa.

96

6.3 Bond Yields

There are two important yields on bond transcations

• The yield obtained by buying a bond and holding it to redemption is theredemption yield. This is what is quoted in financial newspapers. If all taxesare ignored, the yield is called the gross redemption yield or GRY.

In example 6.8, the first investor bought the bond on the issue date to obtaina GRY of 11% pa while the second investor bought the bond after 5 years toobtain a GRY of 7% pa.

• If a bond is sold before redemption, the investor will obtain (or realise) ayield that depends on both the buying price and the selling price. The firstinvestor above sold the bond after 5 years to realise a yield of approximately13.6% pa. (Clearly the realised and redemption yields are equal if the bond isheld to redemption).

97

6.4 Income Tax

Income tax may be payable by some investors on coupon payments (not onredemption payments). Tax will affect both yields and prices.

Example 6.9 An investor who pays income tax at 20% pa buys a 25-year bond tobe redeemed at par bearing semi-annual (= twice per year) coupons of 8% pa.What price will he pay to obtain a yield of 9% pa effective ?

SolutionFirst we must calculate the net coupon (ie after income tax) per £100 nominal ofthe bond. This is (1− 0.2)× (8/2) = 3.2 where 0.2 is the rate of income tax, 8 isthe annual coupon and 2 is the number of coupon payments per year.

98

-?

Price

Coupons? ? ? ? ?? ??

P

3.2 3.2 3.2 3.2 3.23.2 3.2 3.2+100

0 1 2 3 . . . 25years

P = 3.2a0.04403150

+ 100(1.09)−25 = 64.2481 + 11.5968 = £75.84

99

6.5 Capital Gains Tax (CGT)

Capital Gains Tax may be payable on any gain made on the sale or redemption ofa bond, where the gain relates to the difference between the price received onsale or redemption and the price originally paid. There is no relief from CGT onlosses.

Example 6.10 On 1 January 2002 an investor buys two bondsBond A - Price = £101Bond B - Price = £99

What tax does the investor pay on each bond if he sells them both one year laterfor £100 each ? The CGT rate is 30%.

Bond A Tax = 0.3×max{100− 101, 0} = 0Bond B Tax = 0.3×max{100− 99, 0} = 0.3

As there is no relief we cannot add the two together and say that we bought thetwo bonds for £200 and sold for £200 so there is no tax to pay. The calculationsmust be done independently for each bond.

100

It is easy to calculate a yield when the purchase and sale/redemption prices areknown, but it is not so easy to find the price for a given redemption yield becausethe price itself determines whether or not CGT will be payable.

Example 6.11 An investor liable to CGT at a rate of 30% buys a 15-year bondwith an annual coupon of 9% pa, to be redeemed at par. The price paid is £95 per£100 nominal. What redemption yield does the investor obtain?

Solution 100 > 95 so CGT is payable. The equation of value is:-

f(i) = 95(1 + i)15 − 9si15− (100− 0.3(100− 95)) = 0

By trial and error, f(0.09) = −16.712, f(0.10) = +12.386⇒ i ≈ 9.57%

101

What price should the investor pay to obtain a yield of 8% pa effective?

Solution We know that £95 gives a yield of about 9.6%, so P > 95. However, ifP ≥ 100 no CGT is payable, which changes the equation we use to calculate thepresent value:-

CGT payable P = 9a15 + (100− 0.3(100− P ))v15

No CGT payable P = 9a15 + 100v15

In general, a simple procedure is

• guess whether or not CGT is payable

• calculate P using the appropriate form of the present value

• check whether or not the initial guess was correct

102

Example 6.12 In the situation of example 6.11, let us guess that P > 100 then

P = 9a8%15

+ 100× 1.08−15 = 77.0353 + 31.5242 = £108.56

Clearly 108.56 > 100 so our initial guess was correct.

What price should the investor pay to obtain a yield of 10% pa effective?

Solution Here 10% > 9.6% so we guess that CGT will be payable, and we use theappropriate equation.

P = 9a10%15

+(100−0.3(100−P ))×1.10−15 ⇒ P (1−0.3(1.1)−15) = 9a10%15

+70×1.1−15

⇒ P = £91.81

This is less than £100 so our initial guess was correct.

In this example it was quite easy to guess because

• we knew the price to yield 9.6%

• coupons were annual instead of semi-annual

• the investor did not pay income tax

103

In general we would like an algorithm to tell us whether or not CGT will bepayable to avoid us guessing incorrectly and wasting time carrying out incorrectcalculations.

Theorem 6.13 Define:-

• n=outstanding term in the chosen time units

• d=coupon rate per £1 nominal in the same time units

• R=redemption amount per £1 nominal

• t1 = income tax rate

• t2 = rate of CGT

Let i∗(P ) be the redemption yield net of taxes (after taxes) if the price per £1nominal is P, then

(i) i∗(P ) is a decreasing function of P

(ii) P < R and CGT is payable if, and only if, i∗(P ) > d(1−t1)R

104

Proof. The equation of value is

P = d(1− t1)an + (R− t2 max(0, R− P ))vn

calculated at i∗(P )Let P1 and P2 be the prices of two other bonds.Consider P1 > P2 ≥ R, then no CGT is payable because you will have to pay moreto buy the bond than you will get back when you redeem it.

As the future income (i.e. from coupons and the redemption proceeds) isindependent of the price paid, it is clear that the yield must be lower if the pricepaid is higher, i.e. i∗(P1) < i∗(P2).

Now consider P2 < P1 ≤ R, then CGT is payable

P2 = d(1− t1)an + (R− t2(R− P2))vn at i∗(P2)

We add t2(P1 − P2)vn to both sides

⇒ P2 + t2(P1 − P2)vn = d(1− t1)an + (R− t2(R− P2 − P1 + P2))vn at i∗(P2)

105

But t2 < 1 and vn < 1 so the LHS < P1 ⇒ RHS < P1

and P1 = d(1− t1)an + (R− t2(R− P1))vn at i∗(P1)

and we have just shown that for P1 > P2

P1 > d(1− t1)an + (R− t2(R− P1))vn at i∗(P2)

So we can see that

g(i) = d(1− t1)an + (R− t2(R− P1))vn

is a decreasing function of i, and we have shown that

g(i∗(P2)) < g(i∗(P1))⇒ i∗(P2) > i∗(P1)

Suppose P ≥ R (so no CGT is payable), then

P = d(1− t1)an +Rvn at i∗(P )

⇒ P

R=d

R(1− t1)

1− vn

i+R

Rvn ≥ 1 at i∗(P )

106

⇒ d

R(1− t1)(1− vn) ≥ i(1− vn)

⇒ i ≤ d(1− t1)R

ie i∗(P ) ≤ d(1−t1)R with equality if P = R

Now if P < R, so CGT is payable, the result above with P = R and the first resultfor g(i) show that

i∗(P ) ≥ d(1− t1)R

�

107

Example 6.14 Consider a bond with term 10 years, coupon 10% pa payablehalf-yearly, to be redeemed at 120%. Find the price to yield 7% pa effectivepayable by an investor subject to CGT at 25% and (i) income tax at 30% (ii) noincome tax.

Solution(i) In time units of half-year, the desired yield is 3.4408% and the coupon rate is5% and

d(1− t1)R

=0.05(1− 0.3)

1.2= 0.029167 < 0.034408

Hence CGT is payable and (per £100 nominal)

P = 3.5a0.03440820

+[120− 0.25(120− P )

]1.07−10

P(1− 0.25(1.07)−10

)= 3.5a0.034408

20+ 90(1.07)−10

0.872913P = 50.0109 + 45.7514

=⇒ P = £109.70 (Check < 120)

108

(ii)d(1− t1)

R=

0.051.2

= 0.0416667 > 0.038804

Hence, no CGT is payable and (per £100 nominal)

P = 5a0.03440820

+ 120(1.07)−10 = 71.4442 + 61.0019 = £132.45

109

7 Measuring Fund Performance

Insurance companies and pension funds hold large funds on behalf of theirmembers. Members pay premiums or contributions into the fund, which isinvested by the manager, and later their benefits or pensions are paid out of thefund. So money is constantly being paid into and out of the fund.

Clearly the members expect their fund manager to obtain as good a rate of returnas possible. They can determine this by comparing the return with other forms ofinvestment, and with managers of other funds.

In this section, we will look at how we can measure the performance of a fundmanager, and consequently, how we can compare this with other investments.

110

7.1 The yield on a fund

Example 7.1 Suppose the payments into and out of the fund are as follows:

-? ?

Payments in

Payments out? ?

100 50

30 70

0 1 2 3years

Let us also suppose that the value of the fund at time t = 3 is 80. The yield is thesolution i∗ of:

f(i) = 100(1 + i)3 + 50(1 + i)2 − 30(1 + i)1.5 − 70(1 + i)0.5 − 80 = 0

We solve this to find i∗ = 8.56%

111

Remark 7.2 (i) With daily or even more frequent cashflows solving an equation ofvalue would be very complex.

(ii) Since cashflows can be positive or negative at any time, there might be nosolution to the equation, and a yield might not exist.

(iii) The yield calculated in this way is known (by fund managers) as the MoneyWeighted Rate of Return, but it is exactly the same as what we know as theyield or the Internal Rate of Return.

(iv) We abbreviate Money Weighted Rate of Return by MWRR.

112

Example 7.3 Consider the following fund:

-? ?

Payments in

Payments out? ?

100 1,000,000

1,000,000 200

0 1 2 3years

• Member A contributes £100 at time t = 0 and takes out £200 at time t = 3, sothe yield for member A is 200

100

1/3 − 1 ≈ 26%

• Member B contributes £1 million at time t = 1 and takes out £1 million attime t = 2. His yields is 0%.

113

• The MWRR is the solution i∗ to

f(i) = 100(1 + i)3 + 1, 000, 000(1 + i)2 − 1, 000, 000(1 + i)− 200 = 0

⇒ i∗ ≈ 0.01%

• Clearly the MWRR is dominated by the very large cashflows, which mightdistort the picture. In this example, the fund manager achieved very goodresults in years 1 and/or 3 but very poor results in year 2, but the MWRR onlyshows us the result in year 2 because this was the time when most moneywas invested in the fund. So MWRR is affected by the size of the cashflows.

• If investors A and B had invested different amounts of money, the MWRRwould have been different, but the fund manager would have done his jobjust as well.

So, what we need is a way of measuring the fund manager’s performance that isindependent of the amount of money he is given to invest.

114

7.2 The Linked Internal Rate of Return - LIRR

In Example 7.1 above, we had to calculate the yield over all three years becausewe had no information about the fund size except at the start (t = 0) and end(t = 3).

Suppose that we also know the fund size at other times.

-? ?

Payments in

Payments out

Fund Size

? ?

100 50

30 70

0 105 135 80

0 1 2 3years

0 1 2 3- years

115

Let i1 = yield in the period from t = 0→ t = 1Let i2 = yield in the period from t = 1→ t = 2Let i3 = yield in the period from t = 2→ t = 3

We can now calculate these:-

i1 = 0.05

f(i2) = 155(1 + i2)− 30(1 + i2)12 − 135 = 0⇒ i2 = 0.0713

f(i3) = 135(1 + i3)− 70(1 + i3)12 − 80 = 0⇒ i3 = 0.1482

We define the Linked Internal Rate of Return (=LIRR) to be the “average”yield, i, which satisfies

(1 + i)3 = (1 + i1)(1 + i2)(1 + i3)

This is the yield if we had invested 1 throughout at these three interest rates

⇒ LIRR = 8.90%

116

Remark 7.4 (i) The subdivision of the period affects the answer, and we can useany subdivision as long as we know the fund sizes. For example, suppose weknow these fund sizes at the times of the cashflows.

-? ?

Payments in

Payments out

Fund Size

? ?

100 50

30 70

0 105 160 140 80

0 1 2 3years

-

Let i1 = yield in the period from t = 0→ t = 1Let i2 = yield in the period from t = 1→ t = 1.5Let i3 = yield in the period from t = 1.5→ t = 2.5Let i4 = yield in the period from t = 2.5→ t = 3

117

We can now calculate these:-i1 = 0.05

f(i2) = 155(1 + i2)12 − 160 = 0⇒ i2 = 0.0656

f(i3) = 130(1 + i3)− 140 = 0⇒ i3 = 0.0769

f(i4) = 70(1 + i4)12 − 80 = 0⇒ i4 = 0.3061

So the LIRR, i, is calculated from

(1 + i)3 = (1 + i1)(1 + i2)12 (1 + i3)(1 + i4)

12 ⇒ LIRR = 0.1008

118

(ii) The LIRR is less influenced by very large cashflows. Suppose in Example 7.3 weknew these fund sizes:-

0 1 2 3years-

? ?

Payments in

Payments out

Fund sizes

? ?

100 1,000,000

1,000,000

1,000,150 200

200

100 150

Then i1 = 0.5, i2 = 0, i3 = 0.33333⇒ LIRR ≈ 26%

119

7.3 Time-Weighted Rate of Return = TWRR

Suppose we know the fund size at the time of every cashflow in or out of the fund.

For example, as before:-

-? ?

Payments in

Payments out

Fund Size

? ?

100 50

30 70

0 105 160 140 80

0 1 2 3years

-

Then we can calculate a growth factor (which is the same as an accumulationfactor), for each period, because there are no intervening cashflows.

120

g1 =105100

g2 =160155

g3 =140130

g4 =8070

The total growth factor, as if we had invested 1 for the whole period, is:-

105100× 160

155× 140

130× 80

70= 1.334

The Time-Weighted Rate of Return (TWRR), i, is defined as the “average”annual return corresponding to the total growth factor where

(1 + i)3 = 1.334⇒ TWRR = 10.08%

Note: The TWRR ≡ LIRR when the subdivision is determined by the times of thecashflows.

121

Example 7.5 We consider two funds:

-? ?

Payments in

Fund Size

10 989

10.5 110 1250

0 0.5 1 1.5 2years

122

-?

6

Payments in

Fund Size

1000 -1090

1050 11000 12.5

0 0.5 1 1.5 2years

(Note that the payment out has been drawn as a negative payment in)

The yield for the two funds over each year (10% in the first year and 25% in thesecond year) is the same for the two funds.

123

MWRR TWRR LIRR

(= LIRR : 0→ 1, 1→ 2) (0→ 12 ,

12 → 2)

Fund A 24.82% 17.26% 21.07%

Fund B 10.13% 17.26% 10.08%

Remark 7.6 (i) The fund’s performances are the same but only the TWRR picksthis up.

(ii) The MWRR is influenced by the amount of money invested at any time. Fund Bhas more money invested in the 1st year when the return was lowest soMWRR(B) < MWRR(A).

(iii) LIRR is influenced by the subdivision.

(iv) TWRR is the best measure but requires more data, especially for an active fundthat is receiving regular contributions and making regular payments during theyear.

124

Theorem 7.7 (Calculation of TWRR)

-? ? ? ? ?

Net cashflows

Fund at t−

C0 C1 C2 C3 Cn−1

F (1) F (2) F (3)F (0) = 0 F (n− 1)F (n)

0 t1 t2 t3 . . . tn−1 tnyears

By t− we mean the time just before the cashflow made at time t.

The TWRR = i (per annum effective) where

(1 + i)tn =n∏k=1

F (k)F (k − 1) + C(k − 1)

125

Proof. Let ik = yield p.a effective in the time interval [tk−1, tk]. Then

(1 + i)tn =n∏k=1

(1 + ik)tk−tk−1

Consider the period tk−1 → tk

(Ck−1 + Fk−1)(1 + ik)tk−tk−1 = Fk

⇒ (1 + i)tn =n∏k=1

FkFk−1 + Ck−1

�

126

7.4 Unitised Funds

A unitised fund is an investment fund divided initially into a number of units thatcan be bought by investors. The price of each unit will move in line with themovement in the value of the fund as a whole. Investors can buy and sell unitsaccording to the rules set out by the manager of the unitised fund. For example, itmay only be possible to buy and sell units on certain dates during the year. Someunits will pay a dividend (income) during the year, but this can be directlyre-invested to purchase more units.

Example 7.8 The following table shows the unit prices on 1 April each year. Allincome is reinvested in the units.

Year 1997 1998 1999 2000 2001 2002 2003

Price(£) 1.56 1.81 2.07 1.70 1.81 1.95 2.10

So an investment of £1,560 would buy 1,000 units on 1 April 1997. On 1 April2000, these would be worth £1,700.

127

Assume that at any 1 April, units can be bought and sold at the prices shown(note that in practice, there is often a difference in the buying and selling pricewith the selling (bid) price being lower than the buying (offer) price).

An investor bought 200 units in 1997, 1998 and 1999, and bought 600 moreunits in 2000. He sold all 1,200 units in 2003.

(i) Calculate the MWRR(ii) Calculate the TWRR

Solution

-?? ? ?

6

200× 1.56200× 1.81

200× 2.07600× 1.70

1, 200× 2.10

t = 0= 1/4/97

1 2 3 4 5 6time

128

(i) MWRR = i pa effective where

200[1.56(1 + i)6 + 1.81(1 + i)5 + 2.07(1 + i)4 + 3(1.70)(1 + i)3] = 1, 200× 2.10

⇒ i = 4.55%pa

(ii) The TWRR for any investor over the 6 year period is j per annum effectivewhere

(1 + j)6 =2.101.56

⇒ j = 5.08%pa

We can check this calculation, using the long method:-

(1 + i)6 =200× 1.81200× 1.56

× 400× 2.07400× 1.81

× 600× 1.70600× 2.07

× 1200× 2.101200× 1.70

=1.811.56

× 2.071.81

× . . .× 2.101.70

=2.101.56

129

8 Nominal Rates of Interest

Throughout this section the time unit will be one year.

8.1 Nominal Interest

Consider a rate of interest i per annum.

• Effective interest means that i is paid at the year end on 1 invested at thestart of the year.

• Nominal interest of i(m) per annum compounding m times a year means thati(m)

m is paid at the end of each fraction 1/m of a year.

Example 8.1 At a rate of 8% per annum effective

130

-?

?

1

1 + 0.08

t = 0 t = 1time

i = 0.08

At a rate of 8% per compounding quarterly

-?

? ? ? ?

1

1 + 0.020.02 0.02 0.02

t = 0 t = 10.25 0.50 0.75time

i(4) = 0.08

131

i(m) per annum compounding mthly or m times per year is identical to i(m)

m

per 1/m year effective.

It is usually easier to change to the appropriate time unit and use the effectiverate of interest, but you might still come across this (pre-calculator) notation insome of the text-books. The relationship between the effective rate i pa and thenominal rate i(m) pa is:

1 + i =(

1 +i(m)

m

)mExample 8.2 The following are equivalent:-

Effective Rate Nominal Rate

i = 0.06 i(2) = 0.059126

i = 0.12 i(12) = 0.113866

i = 0.082432 i(4) = 0.08

i = 0.092025 i(2) = 0.09

132

8.2 Nominal Rates of Discount

Recall that d = 1− v is the effective rate of discount per annum. It could beinterpreted as the amount of interest payable at the start of the year equivalent toi at the end of the year.

The nominal discount rate d(m) represents interest of d(m)

m paid at the start ofeach 1/m of a year.

Example 8.3 d = 0.08 pa effective

-?

?

1

1 + 0.08(1 + i)0.08 -

t = 0 t = 1time

133

d(4) = 0.08

-?

? ? ? ? ?

? ? ? ??

- 1 + 0.02si′

4

1

10.02 0.02 0.02 0.02

t = 0 t = 10.25 0.50 0.75time

where i′ is the effective rate of interest per quarter year.

d(m) is exactly the same as a discount factor of 1− d(m)

m per 1m of a year.

Example 8.4 d(4) = 0.08⇒ v1 = 0.98 per quarter year⇒ v = 0.984 per annum⇒ i = 1

0.984 − 1 = 0.084166 effective per annum

134

9 Inflation

Literature: John McCutcheon and William F. Scott: An introduction to themathematics of finance, 2003 : pp. 179–186

Inflation is the fall of purchasing power of money. It is measured with referenceto an index (Retail Price Index (UK), Consumer Price Index (US)).

9.1 Retail Price Index

• The RPI measures the monthly change in the prices of a specified set of goodsand services.

• It is published one month in arrear (usually second or third Tuesday).

• In the UK the RPI is widely used to measure inflation.

135

time

rela

tive

valu

e of

inde

x (J

an 8

7 =

100

)

1987 1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006

100

110

120

130

140

150

160

170

180

190

RPI (UK)

CPI (US)

Figure 1: The RPI (UK inflation index) and the CPI (US inflation index). The CPIis standardized to have a value of 100 in January 1987. (source: ONS, www.statistics.gov.uk

and U.S. Department of Labor, BLS, www.bls.gov)

136

The RPI (as well as any other inflation index) is time dependent.

Notation:

• Inflation index at time t: Q(t) (time is measured in years)

• Rate of inflation per year: q(t) =Q(t)

Q(t− 1)− 1

• Average inflation rate per year between time s and time t: q with:

(1 + q)t−s =Q(t)Q(s)

=⇒ q =(Q(t)Q(s)

) 1t−s

− 1

We will also use the notation: Q(Jan 05), q(May 96)=Q(May 96)Q(May 95)

− 1, and

RPI(t), RPI(Jan 2003) if we use a particular inflation index.

137

time

annu

al c

hang

e in

%

1988 1989 1990 1991 1992 1993 1994 1995 1996 1997 1998 1999 2000 2001 2002 2003 2004 2005 2006

12

34

56

78

910

RPI (UK)

CPI (US)

Figure 2: The figure shows the graphs of q(t) for the RPI and CPI values in Figure1.

138

J F M A M J J A S O N D

1987 100.0 100.4 100.6 101.8 101.9 101.9 101.8 102.1 102.4 102.9 103.4 103.3

1988 103.3 103.7 104.1 105.8 106.2 106.6 106.7 107.9 108.4 109.5 110.0 110.3

1989 111.0 111.8 112.3 114.3 115.0 115.4 115.5 115.8 116.6 117.5 118.5 118.8

1990 119.5 120.2 121.4 125.1 126.2 126.7 126.8 128.1 129.3 130.3 130.0 129.9

1991 130.2 130.9 131.4 133.1 133.5 134.1 133.8 134.1 134.6 135.1 135.6 135.7

1992 135.6 136.3 136.7 138.8 139.3 139.3 138.8 138.9 139.4 139.9 139.7 139.2

1993 137.9 138.8 139.3 140.6 141.1 141.0 140.7 141.3 141.9 141.8 141.6 141.9

1994 141.3 142.1 142.5 144.2 144.7 144.7 144.0 144.7 145.0 145.2 145.3 146.0

1995 146.0 146.9 147.5 149.0 149.6 149.8 149.1 149.9 150.6 149.8 149.8 150.7

1996 150.2 150.9 151.5 152.6 152.9 153.0 152.4 153.1 153.8 153.8 153.9 154.4

1997 154.4 155.0 155.4 156.3 156.9 157.5 157.5 158.5 159.3 159.5 159.6 160.0

1998 159.5 160.3 160.8 162.6 163.5 163.4 163.0 163.7 164.4 164.5 164.4 164.4

1999 163.4 163.7 164.1 165.2 165.6 165.6 165.1 165.5 166.2 166.5 166.7 167.3

2000 166.6 167.5 168.4 170.1 170.7 171.1 170.5 170.5 171.7 171.6 172.1 172.2

2001 171.1 172.0 172.2 173.1 174.2 174.4 173.3 174.0 174.6 174.3 173.6 173.4

2002 173.3 173.8 174.5 175.7 176.2 176.2 175.9 176.4 177.6 177.9 178.2 178.5

2003 178.4 179.3 179.9 181.2 181.5 181.3 181.3 181.6 182.5 182.6 182.7 183.5

2004 183.1 183.8 184.6 185.7 186.5 186.8 186.8 187.4 188.1 188.6 189.0 189.9

2005 188.9 189.6 190.5 191.6 192.0 192.2 192.2 192.6 193.1 193.3 193.6

Table 1: The monthly values of the RPI since January 1987

139

First interpretation

The RPI measures price inflation, e.g. a set of goods costing £1,595 in January1998 would have cost £1,831 in January 2004.The increase from Jan 98 to Jan 2004 (6 years) was:(

18311595

− 1)

= 0.148 = 14.8%

The average inflation rate, q, p.a. was:

(1 + q)6 = 1.148 =⇒ q = 6√

1.148− 1 = 0.023 = 2.3%

140

Second interpretation

The RPI measures the change in the value (purchasing power) of money: £1 inJan 1998 is the same as £1.148 in Jan 2004.In general: £X at time s has the purchasing power of

£XRPI(t)RPI(s)

at time t.

Example 9.1 Consider a life-insurance contract paying £1 million in 40 years.

Inflation rate q Purchasing power today

2 % 1, 000, 000Q(0)Q(40)

= 1, 000, 0001

1.0240= 452, 890.42

3 % 1, 000, 0001

1.0340= 306, 556.84

4 % 1, 000, 0001

1.0440= 208, 289.04

141

Example 9.2 (Real yields vs. monetary yields)(Compare FM 1, Section 4 (Yields) and Section 5 (Bonds))

Bond issued in January 1994

Term 5 years

Coupon 9 % p.a.

Payment frequency annually

Issue price 100 %

Redemption rate 105 %

Question: What is the yield to an investor not subject to tax?a) in monetary termsb) in real terms with reference to the RPI

142

t = 0 t = 1 t = 2 t = 3 t = 4 t = 5

−100 9 9 9 9 114 = 9+105

Answer to a): Interest rate i p.a. effective where 100 = 9a5 + 105v5

=⇒ i = 9.82 % p.a. effective

a5 denotes the present value at time 0 of an annuity of 1 paid once per year inarrear,

a5 =1− v5

i

and v = 1/(1 + i) denotes the discount factor per year.

143

Answer to b):

time t 0 1 2 3 4 5

year 94 95 96 97 98 99

cash flows -100 9 9 9 9 114

RPI 141.3 146 150.2 154.4 159.5 163.4

real value of -100141.3146

9141.3150.2

9141.3154.4

9141.3159.5

9141.3163.4

114

the cash flows = -100 = 8.71 = 8.47 = 8.24 = 7.97 = 98.58

Real yield i′ p.a. effective:

100 = 8.71v + 8.47v2 + 8.24v3 + 7.97v4 + 98.58v5

=⇒ i′ = 6.65%

144

The average inflation rate, q over the period from January 1994 to January 1999is given by

(1 + q)5 =Q(Jan 99)Q(Jan 94)

=163.4141.3

=⇒ q = 2.95%

Note:

• i′ = 6.65 6= i− q = 9.82− 2.95 = 6.87

• The “base date” at which purchasing power is calculated (January 1994) isirrelevant for the calculation of i′. i′ doesn’t change if we express the adjustedpayoff in terms of purchasing power in January 1998, or at any other time.Why?

• What is the real yield of a similar bond issued in January 1987?

145

In general:

Given a series of cash flows C(tk) at time tk, k = 1, . . . , n and the relevant valuesof an appropriate inflation index Q(tk) the real yield i′ is such that

n∑k=1

C(tk)Q(0)Q(tk)

(1 + i′)−tk = 0 . (9.2)

Note:

• Equation (9.2) is satisfied if

n∑k=1

C(tk)1

Q(tk)(1 + i′)−tk = 0 .

The real yield is therefore independent of the date at which the purchasingpower is calculated.

• The inflation index will not be known for future cash flows.

146

• If we assume a constant rate of inflation q p.a., then cash flows C(tk) at timetk have purchasing power at time 0:

C(tk)Q(0)Q(tk)

= C(tk)Q(0)

Q(0)(1 + q)tk= C(tk)(1 + q)−tk

If follows that for net cash flows C(tk) at time tk, k = 1, . . . , n (t1 = 0) thereal yield i′ is such that

n∑k=1

C(tk)(1 + q)−tk(1 + i′)−tk = 0

orn∑k=1

C(tk)(1 + i)−tk = 0

with (1 + i) = (1 + q)(1 + i′)

147

Conclusion:If we assume a constant rate of inflation q p.a., the monetary yield i and the realyield i′ are related by the equation

(1 + i) = (1 + q)(1 + i′) .

General procedure for calculating real yieldsSuppose net cash flows C(tk) at time tk, k = 1, . . . , n, (t1 = 0).

(i) Convert all payments C(tk) into the same purchasing power unit (e.g. at timet = 0) by multiplication by Q(0)/Q(tk) to get CR(tk) = C(tk)× (Q(0)/Q(tk))

(ii) Solve the equation of value

n∑k=1

CR(tk)(1 + i′)−tk = 0 .

148Note:

• In some cases we will know the true inflation rate for all transactions. This isonly possible if the real yield is calculated retrospectively

• If we have some past payments and some future payments, we use the knownRPI values for the past payments and make assumptions about the RPI forfuture payments

149

9.2 Index-Linked Gilts

More information: UK Debt Management Office (DMO) Website:(http://www.dmo.gov.uk/gilts/indexlink/f2lin.htm)

Interest payments and the redemption proceeds are adjusted in line with inflation(measured by the RPI). The reasons for this type of bond being issued are:

• to protect investors against inflation risk and

• to help pension funds to provide index-linked benefits since index-linkedliabilities can be matched with an index-linked asset.

The first index-linked stock was issued by the UK Government in March 1981.

150

The monetary amount of an interest payment D(tk) at time tk of a stock withannual coupon rate C payable m times a year is

D(tk) = 100C

m

Q(tk − s)Q(t0)

per £100 nominal (9.3)

where t0 is the base date and adjustments are made according to an inflationindex at time tk − s. Q(t0) is called the base inflation figure (base RPI figure).

In the UK s is 8 months for index-linked gilts issued before April 2005 and s isequal to 3 months for all index-linked gilts issued since April 2005.

151

Example 9.3 Consider the 4 18% Index-linked Treasury Stock 2030

Interest dates 22 January, 22 July

Redemption date 22 July 2030

All adjustments are made according to the RPI eight months prior to the payment(May for the payment next Jan. and Nov. for the payment next July).The base RPI figure is 135.1 (Oct. 1991)

An eight-month time lag for indexation means that an investor who purchasessuch a treasury stock knows, at the time of purchase, the monetary amount of thenext coupon payment.

152

Bank of England press release (13 Dec 2005):

“4 1/8% Index-Linked Treasury Stock 2030In accordance with the terms of the prospectus dated 12 June 1992, the Bank ofEngland announces that the rate of interest payable on the above stock for theinterest payment due on 22 July 2006 will be £2.9555 per £100 nominal of stock.” (source: http://www.bankofengland.co.uk/publications/news/2005/170.htm)

payment (22 July 06) =4.125

2× RPI(Nov 05)

RPI(Oct 91)

=4.125

2× 193.6

135.1= 2.9555885

≈ 2.9555 (rounded down to four decimal places)

153

The interest rate payment on 22 Jan. 2007:

4.1252× RPI(May 06)

135.1

Redemption on 22 July 2030

100× RPI(Nov 2029)135.1

per £100 nominal.

The monetary yield an investor will obtain depends on future inflation.

154Assume:

• Inflation rate from Nov. 2005 is 2.5 % p.a., that is:

Q(Dec 2005) = Q(Nov 2005)(1.0251/12) = 193.6(1.0251/12) ≈ 194

• quoted price on 22 July 2005 was 215.57(source: UK Debt Management Office, www.dmo.gov.uk/gilts/f2gilts.htm)

Question: Calculate the (monetary) yield an investor who purchased the stock on22 July 2005 will obtain using the figures above.

155

Date Cash flow

22 July 05 -215.57

22 Jan 06 D =4.125

2× Q(May 05)

135.1=

4.1252× 192

135.1

22 July 06 D =4.125

2× Q(Nov 05)

135.1=

4.1252× 193.6

135.1

22 Jan 07 D =4.125

2× Q(May 06)

135.1=

4.1252× 193.6× 1.0250.5

135.1

22 July 07 D =4.125

2× Q(Nov 06)

135.1=

4.1252× 193.6× 1.0251

135.1...

22 July 2030 D =(

100 +4.125

2

)× Q(Nov 29)

135.1

156

Monetary yield i p.a. effective is the solution of

0 = −215.57 +4.125

2× 192

135.1v0.5 +

4.1252× 193.6

135.1v1

+4.125

2× 193.6

135.1× 1.0250.5 × v1.5

+4.125

2× 193.6

135.1× 1.0251 × v2

...

+(

100 +4.125

2

)× 193.6

135.1× 1.02524 × v25

157

0 = −215.57 +4.125

2× 192

135.1v0.5 +

4.1252× 193.6

135.1v1

+4.125

2× 193.6

135.1× v ×

{1.0250.5v0.5 + 1.0251v1 + . . .+ 1.02524v24

}︸ ︷︷ ︸

= 2a(2)j

24with (1 + j)−1 = 1.025(1 + i)−1

+100× 193.6135.1

× 1.02524 × v25

=⇒ i = 4.06%

158

Remark:

• We can use different inflation assumptions.

– q = 3.0 % =⇒ i = 4.54%

– Or we can use different inflation rates for different periods, such as

∗ 2.5 % up to 22 January 2007 and∗ 3.0 % from January 2007 up to maturity.

• Because of the RPI time lag s, index-linked bonds do not provide totalinflation protection, in particular, the risk of inflation during the last 8months (3 months) is not reflected

159

Instead of calculating the yield of an index-linked gilt for a given inflation rate,we could also calculate the constant rate of inflation q that gives a specifiedmonetary yield. This rate is called the “break even rate of inflation”.

Example 9.4 The break-even rate of inflation for a yield of 4.2% is between2.5% (i = 4.06%) and 3% (i = 4.54%).

Assume a normal government bond has a yield of 4.2%.

Assume: future rate of inflation = 2.5%⇒ buy the government bondAssume: future rate of inflation = 3.0%⇒ buy the index-linked gilt

160

Question: Calculate the break-even rate of inflation for the “4 18% Index-Linked

Treasury Stock 2030” if the monetary yield for a normal government bond is4.2%.

Solution:

(1 + i) = 1.042 =⇒ v =1

1.042

Equation of Value:

0 = −215.57 +4.125

2× 192.0

135.1v0.5 +

4.1252× 193.6

135.1v1

+4.125

2× 193.6

135.1× v ×

{(1 + q)0.5v0.5 + (1 + q)1v1 + . . .+ (1 + q)24v24

}︸ ︷︷ ︸

= 2a(2)j

24with (1 + j)−1 = (1 + q)(1 + i)−1

+100× 193.6135.1

× (1 + q)24 × v25

161

Solve this equation for q to get the break-even rate of inflation:

=⇒ q = 2.65%

Assume: future inflation rate < 2.65%⇒ buy the government bondAssume: future inflation rate ≥ 2.65%⇒ buy the index-linked gilt

162

KE Y D E V E LO P M E N T S I N T H E G I LT S M A R K E T AP R I L 2004 T O JA N UA RY 2005ABreakeven in f lat ion rates

A.5 Breakeven inflation rates (BEIRs) fell for most of the financial year to end-January2005, indicating a modest underperformance of index-linked gilts relative to conventionals.However, these declines came after BEIRs had reached record highs in late-May/early-June of3.11 per cent (10-year) and 3.10 per cent (30-year). 30-year breakevens moved sharply higheragain from end-November, reportedly reflecting continuing pension fund demand for long-dated index-linked gilts. Over the period to end-January 10-year BEIRs fell by 13bps to 2.80per cent whilst 30-year rates fell by only 1bp to 2.92 per cent. Chart A4 below sets out 10-yearand 30-year breakeven inflation rates for the financial year to end-January 2005.

Chart A4: 10-year and 30-year breakeven inflation ratesChart A4: 10-year and 30-year breakeven inflation rates

per

cent

2.75

2.80

2.85

2.90

2.95

3.00

3.05

3.10

3.15

25-Jan-0526-Nov-0427-Sep-0429-Jul-0430-May-0431-Mar-04

Source: Debt Management Office

30-year10-year

46 Debt and Reserves Management Report 2005-06

Figure 3: The Break-Even Rate of Inflation calculated by the Debt ManagementOffice. (source: DMO, Debt and reserves management report 2005-06)

163

Example 9.5 What is the real yield of the “4 18% Index-Linked Treasury Stock

2030” if we assume that the rate of inflation will be 2.5% p.a. from Nov 05?

Date real cash flow (Pur. Power at 22 July 05)

22 July 05 -215.57

22 Jan 064.125

2× Q(May 05)

135.1× Q(July 05)Q(Jan 06)

=4.125

2× 192

135.1× 192.2

193.6× 1.0251/6

22 July 064.125

2× 193.6

135.1× 192.2

193.6× 1.0251/6+0.5

22 Jan 07 =4.125

2× 193.6× 1.0250.5

135.1× 192.2

193.6× 1.0251/6+1

=4.125

2× 192.2

135.1× 1

1.0251/6+0.5

......

22 July 2030(

100 +4.125

2

)× 192.2

135.1× 1

1.0251/6+0.5

164

Equation of Value for the real yield i′ and v = 1/(1 + i′)

0 = −215.57

+4.125

2× 192

135.1× 192.2

193.6× 1.0251/6× v0.5

+4.125

2× 192.2

135.1× 1

1.0251/6+0.5× v1

+4.125

2× 192.2

135.1× 1

1.0251/6+0.5× v1.5

...

+4.125

2× 192.2

135.1× 1

1.0251/6+0.5× v25

+100× 192.2135.1

× 11.0251/6+0.5

× v25

165

0 = −215.57

+4.125

2× 192

135.1× 192.2

193.6× 1.0251/6× v0.5

+4.125

2× 192.2

135.1× 1

1.0251/6+0.5× v0.5 ×

{v0.5 + v1 + . . .+ v24.5

}︸ ︷︷ ︸

2a(2)i′

24.5

+100× 192.2135.1

× 11.0251/6+0.5

× v25

=⇒ i′ = 1.52%

We could replace Q(May 05) = 192 by Q(Nov 05)(1.025−0.5) = 191.22 (this is anapproximation) and get

166

0 = −215.57

+4.125

2× 193.6× 1.025−0.5

135.1× 192.2

193.6× 1.0251/6× v0.5

+4.125

2× 192.2

135.1× 1

1.0251/6+0.5× v0.5 ×

{v0.5 + v1 + . . .+ v24.5

}+100× 192.2

135.1× 1

1.0251/6+0.5× v25

0 = −215.57 +4.125

2× 192.2

135.1× 1

1.0251/6+0.5×{v0.5 + v1 + . . .+ v25

}︸ ︷︷ ︸

2a(2)i′

25

+100× 192.2135.1

× 11.0251/6+0.5

× v25

=⇒ i′ = 1.52%

167

Remark: The real yield can also be calculated from

(1 + i) = (1 + q)(1 + i′)

1 + i′ =1 + i

1 + q=

1.04061.025

= 1.01522 =⇒ i′ = 1.52%

168

10 Force of Interest and Continuous Cash Flows

Literature: John McCutcheon and William F. Scott: An introduction to themathematics of finance, 2003 : 2.4, 2.7, 3.1, 3.5, 3.6

10.1 The force of Interest

A sum of money is invested at some time, it grows continuously. At time t itsamount is A(t).

How can we measure how fast the amount of money is growing at time t?

169

Time t (in years)

Acc

umul

atio

n A

(t)

0 2 4 6 8 10

100

105

110

115

120

125

170

For time t and s > 0 assume that the nominal yield with simple compounding ofthis investment during the period [t, t+ s] is i(t, s) p.a..

A(t+ s) = A(t)(1 + i(t, s)s)

i(t, s) =(A(t+ s)A(t)

− 1)

1s

=A(t+ s)−A(t)

A(t)1s

=1

A(t)A(t+ s)−A(t)

s

−→s→01

A(t)dA(t)dt

171

Definition 10.1 The force of interest (per annum) at time t is

δ(t) =1

A(t)dA(t)dt

Remark:

δ(t) =d

dt(logA)(t) ⇒

∫ t2

t1

δ(t)dt = logA(t2)− logA(t1)

Important general formula:

A(t2) = A(t1) exp(∫ t2

t1

δ(t)dt)

(10.4)

If δ(t) and the initial capital a are given, A(t) is the solution of the differentialequation

dA(t)dt

= δ(t)A(t) A(0) = a

172

Remark:

(i) Only assumption: A is strictly positive and smooth (e.g. continuouslydifferentiable)

(ii) If A(0) (initial investment at time 0) and the force of interest δ(t) for t ≥ 0are known, we can calculate the future amount of the investment at anyfuture time.

If the force of interest is constant, δ(t) = δ for all t ≥ 0 the accumulation of £1invested at

time 1 : A(1) = 1 exp(∫ 1

0

δds

)= exp(δ)

time 2 : A(2) = 1 exp(∫ 2

0

δds

)= exp(2δ)

time t : A(t) = 1 exp(∫ t

0

δds

)= exp(δt)

173

Therefore:

A(t) = exp(δt) = exp(δ)t

Define: i = exp(δ)− 1 =⇒ A(t) = (1 + i)t

Important result:

If the force of interest p.a. is constant, δ(t) = δ then the effective rate of interestp.a. is given by

(1 + i) = exp(δ)

174

More generally:

• a constant force of interest δ p.a.

• an effective rate of interest i p.a.

• a nominal rate of interest i(p) p.a. convertible p times per year

• a nominal rate of discount d(q) p.a. convertible q times per year

• an effective rate of discount d p.a.

are all equivalent if the following holds:

exp(δ) = (1 + i) =(

1 +i(p)

p

)p=(

1− d(q)

q

)−q= (1− d)−1

175

Second Approach

Assume that a nominal interest rate i(m) p.a. compounding m times a year isconstant for all m, i(m) = δ. this is not a nominal rate of interest compoundingm times, it is a nominal interest p.a. for an investment

over a period of 1/m years, because we also receive the redemption after the end of each period

Number of payments per year

1 2 3 . . . m −→ ∞

(1 + δ) (1 + δ/2)2 (1 + δ/3)3 . . . (1 + δ/m)m −→ eδ

Accumulation of 1 at the end of the year (t = 1)

For m→∞ (continuous compounding) we call δ the force of interest.

176

With the same argument as above we find that the accumulation of £1 after half ayear is

e12 δ = exp

{12δ

}More general the accumulation at time t of £1 invested in time s < t is

eδ(t−s) = exp{δ(t− s)}

The force of interest usually depends on time.

Divide one year into m equally sized time intervals [(k − 1)/m, k/m],k = 1, . . . ,m. Assume that the force of interest δ(k) is constant during the timeinterval [(k − 1)/m, k/m] but depends on k for all k = 1, . . . ,m,

177

We invest 1 at time 0. At time t the accumulated value is

A(t) = exp(δ(1)/m)× exp(δ(2)/m)× . . .× exp(δ(k)/m)

with k/m ≤ t < (k + 1)/m.

A(t) = exp(δ(1)m

+ . . .+δ(k)m

)= exp

(1m

k∑l=1

δ(l)

)

= exp

1m

∑l : l/m≤t

δ(l)

−→ exp(∫ t

0

δ(s)ds)

(m→∞)

Interpretation:

The force of interest is a ”nominal rate of interest compounding infinitely often”per year.

178

Assume a (possible time dependent) force of interest δ(t) p.a. .

Let t1 < t2, it follows from the general formula (10.4) that an investment of

C exp(−∫ t2

t1

δ(s)ds)

at time t1 will accumulate to C at time t2.

Therefore,

exp(−∫ t2

t1

δ(s)ds)

is the present value at time t1 of £1 due at time t2 and

exp(∫ t2

t1

δ(s)ds)

is the accumulation at time t2 of £1 due at time t1.

179

Example 10.2 Find the accumulated amount at time 20 of a cash flow of £1,000now and a further £1,000 after 10 years from now, assuming a force of interestδ(t) p.a. where

δ(t) = 0.06 + 0.0005t− 0.00005t2

Solution:

Time 0 10

Cash flow 1,000 1,000

AV at time 20 1, 000 exp(∫ 20

0

δ(s)ds)

1, 000 exp(∫ 20

10

δ(s)ds)

= 1, 000× exp(

3.53

)= 1, 000× exp

(1.675

3

)=⇒ Accumulated value at time 20 is

1000× exp(

3.53

)+ 1000× exp

(1.675

3

)= £4, 959.03

180

10.2 Continuous Cash Flows

Consider £10,000 p.a. paid in

• 1 annual instalment at the end of each year of £10,000 or

• 2 half yearly instalments at the end of each half year of £5,000 or

• 12 monthly instalments of £833 13 at the end of each month or

• 365 daily instalments of £27.40 at the end of each day or

• every dt of a year of £10,000 dt

Note: We ignore interest for the moment.

181

More generally:

Let M(t) denote the total payment made between time 0 and time t. Then, therate of payment is

ρ(s) = M ′(s) for all s

For 0 ≤ t1 ≤ t2, the total payment between time t1 and time t2 is

M(t2)−M(t1) =∫ t2

t1

M ′(s)ds =∫ t2

t1

ρ(s)ds

Remark: If the rate of payment ρ(s) = ρ is constant over one year then the totalpayment over this year is equal to the rate of payment, since

M(t2)−M(t1) =∫ t2

t1

ρds = ρ (t2 − t1)︸ ︷︷ ︸= 1 year

= ρ

182

Present Values and Accumulated Values

For times s and t with s < t:

Let V (s, t) denote the present value at time s of 1 unit paid at time t.

Let A(s, t) denote the accumulation at time t of 1 unit paid at time s.

Remark: For a constant force of interest δ(u) = δ per annum:

A(s, t) = exp(∫ t

s

δ(u)du)

= exp (δ(t− s))

V (s, t) = exp(−∫ t

s

δ(u)du)

= exp (−δ(t− s))

183

Consider a cash flow paid continuously at rate

ρ(t) p.a.

for T years starting at time 0.

Total payment between time t and time t+ dt is

M(t+ dt)−M(t) = M ′(t)dt = ρ(t)dt

if dt is very small.

We treat the amount paid between t and t+ dt as paid in t.

184

Amount ρ(t)dt

PV at time 0 ρ(t)V (0, t)dt

AV at time T ρ(t)A(t, T )dt

Present value at time 0 of the entire cash flow between time 0 and time T :∫ T

0

ρ(t)V (0, t)dt =∫ T

0

ρ(t) exp(−∫ t

0

δ(u)du)dt

(”sum up” all present values for entire cash flow)

Accumulated value at time T of the entire cash flow between time 0 and time T :∫ T

0

ρ(t)A(t, T )dt =∫ T

0

ρ(t) exp

(∫ T

t

δ(u)du

)dt

185

Example 10.3 Find the present value of a cash flow of £1,000 p.a. payablecontinuously for 10 years, assuming a constant force of interest δ p.a.

Solution: ρ(t) = 1, 000, V (0, t) = exp(−δt)

PV =∫ 10

0

ρ(t)V (0, t)dt

=∫ 10

0

1, 000 exp(−δt)dt = 1, 000[

exp(−δt)−δ

]100

= 1, 0001− exp(−10δ)

δ= 1, 000

1− v10

δ

where v =1

1 + iand i = exp(δ)− 1

186

Second approach (different perspective)

Assume a force of interest δ(t) p.a.

(i) Let A(t) denote the accumulated value at time t of C invested at time 0.A fulfills the following differential equation

dA(t)dt

= A(t)δ(t)︸ ︷︷ ︸interest earned

, A(0) = C

Note:

• δ(t) =1

A(t)dA(t)dt

(t)

• For a small time interval [t, t+ ∆t] we have approximately

A(t+ ∆t)−A(t) = A(t)δ(t)∆t

187

(ii) Let A(t) denote now the accumulated value at time t of a continuous cashflow paid at rate ρ(s) p.a. between time 0 and time t.Note: For a small time interval [t, t+ ∆t] the total payments made areapproximately ρ(t)∆t, which we treat to be paid at time t.If ∆t is small, we have approximately

A(t+ ∆t)−A(t) = A(t)δ(t)∆t︸ ︷︷ ︸interest earned on A(t)

+ ρ(t)∆t︸ ︷︷ ︸new investment

× (1 + δ(t)∆t)︸ ︷︷ ︸accumulation

=⇒ A(t+ ∆t)−A(t)∆t

≈ A(t)δ(t) + ρ(t)(1 + δ(t)∆t)

∆t→ 0 yields the following differential equation

dA(t)dt

= A(t)δ(t)︸ ︷︷ ︸interest earned

+ ρ(t)︸︷︷︸new investment

, A(0) = 0 (10.5)

188

The solution of (10.5) is (see tutorial)

A(t) =∫ t

0

ρ(u) exp(∫ t

u

δ(s)ds)du

Remark: We will always assume in this module that ρ(t) and δ(t) satisfyappropriate conditions such that a solution A(t) exists.

189

Comparison

(i) One paymentdA(t)dt

= A(t)δ(t)︸ ︷︷ ︸interest earned

, A(0) = C

A(t) = C exp(∫ t

0

δ(s)ds)

(ii) Continuous payment

=⇒ dA(t)dt

= A(t)δ(t)︸ ︷︷ ︸interest earned

+ ρ(t)︸︷︷︸new investment

, A(0) = 0

A(t) =∫ t

0

ρ(u) exp(∫ t

u

δ(s)ds)du

190

Different types of cash flows

(i) One payment C at time 0:

A(t) = C exp(∫ t

0

δ(s)ds)

(ii) n payments C(tk) at times tk < t, k = 1, . . . , n:

A(t) =n∑k=1

C(tk) exp(∫ t

tk

δ(s)ds)

(iii) Continuous payment at rate ρ(s):

A(t) =∫ t

0

ρ(u) exp(∫ t

u

δ(s)ds)du

even more general, introduce here a combination of discrete and continuous payments, see the definitionA(t, i) in the def. of DPP (section 3.3)

191

Some Standard Cash Flows

Assume a constant force of interest δ p.a. .Let i = exp(δ)− 1 denote the equivalent effective rate of interest p.a.

(i) Continuous payment at rate 1 p.a. for n years:an denotes the present value at time 0 (at an effective rate of interest i p.a. /force of interest δ p.a.)sn denotes the accumulation at time n (at an effective rate of interest i p.a. /force of interest δ p.a.)Important formula:

an =∫ n

0

1(

11 + i

)tdt =

∫ n

0

1vtdt =∫ n

0

exp(−δt)dt

=[−1δ

exp(−δt)]t=nt=0

=1− exp(−δn)

δ=

1− vn

δ

=i

δan

192

sn =∫ n

0

1(1 + i)n−tdt =∫ n

0

(1 + i)tdt

=∫ n

0

exp(δt)dt =[

1δ

exp(δt)]t=nt=0

=exp(δn)− 1

δ=

(1 + i)n − 1δ

=i

δsn = (1 + i)nan

Note:

an =i

δan

In particular

a1 =i

δa1

Therefore, the cash flows

• continuous payment at rate δ p.a. for one year

• one payment of i/δ at the end of one year

have the same present value.

193Picture on black board

(ii) (Ia)n = PV(0) of the following cash flow:

t = 1 t = 2t = 0

1 p.a. 2 p.a. n p.a.

t = n−1 t = n

(Ia)n =n∑k=1

∫ k

k−1

k exp(−δt)dt =n∑k=1

k

[−1δ

exp(−δt)]t=kt=k−1

=n∑k=1

k

δ

{exp

(− δ(k − 1)

)− exp(−δk)

}=

1δ

n∑k=1

kvk {exp(δ)− 1}︸ ︷︷ ︸=i

=i

δ(Ia)n =

an − nvn

δ

194

(iii) (I a)n is the present value of a continuous cash flow where the rate ofpayment is

ρ(t) = t

rate ofpayment

0 t n time (in years)

see tutorial sheet

195

Example 10.4 Find the accumulated value at time 15 of a cash flow payingcontinuously at a rate ρ(t) p.a. where

ρ(t) =

10 for 0 ≤ t < 5

15 for 5 ≤ t < 10

20 for 10 ≤ t < 15

196

Solution:

PV(15) =∫ 5

0

10 exp[(15− t)δ]dt

+∫ 10

5

15 exp[(15− t)δ]dt

+∫ 15

10

20 exp[(15− t)δ]dt

= exp(15δ){

10a5 + 15v5a5 + 20v10a5

}= a5

{10(1 + i)15 + 15(1 + i)10 + 20(1 + i)5

}

Accumulation at time n

(Is)n = (1 + i)n(Ia)n

(I s)n = (1 + i)n(I a)n

197

11 Discounted Cash Flows

11.1 Yields

Consider the following cash flows (representing, for example, profits frominvestments in a company or venture)

time 0 1 2 · · · 9 10

cash flows -100 10 10 · · · 10 110

Yield: 10%, since 0.1 is the solution of the equation of value:

f(i) = −100(1 + i)10 + 10sin + 100 = 0 (11.6)

Rearranging:

100(1 + i)10 = 10sin + 100

or

100 = 10ain + 100(1 + i)−10

198

Interpretation of the yield in terms of the principle of equivalence:If money can be invested to earn 10% then we would be willing to exchange 100now for the future cash flow.

It is necessary to be able to invest money at 10%.

Compare the two cash flows:1. Investment of 100 into the bank account at time 02. Investment into the company and investment of all payments into the bankaccount

time 0 1 2 · · · 9 10

Cash Flow 1 100(1 + i)10 bank account

Cash Flow 2 10 10 · · · 10 110 investment

10sin + 100 bank account

Therefore, saying that the yield of the original cash flows is 10% is correct as amathematical statement (see eq. (11.6)), but it only leads to an interpretation interms of the principle of equivalence if market interest rates are 10%.

199

Example 11.1 Consider the cash flows of two investments

time 0 1 2 3 4 5 6 7 8 9 10

Investment A -100 10 10 10 10 10 10 10 10 10 110

Investment B -100 21 21 21 21 121

The yield of both cash flows is 10% Check, 10 × 1.1 + 10 = 21

If market interest rates are 10% then:

• we would pay 100 for either investment

• we are indifferent between investment A and B

200

If market interest rates are 15%:

time 0 1 2 3 4 5 6 7 8 9 10

Investment A -100 10 10 10 10 10 10 10 10 10 110

Investment B -100 21 21 21 21 121

Investment into bank account 100× 1.1510 = 404.56

Investment A 100 + 10s0.1510

= 303.04

Investment B 100 + 21s0.32255

= 298.32

0.3225 = 1.152 − 1, we get an interest rate of 0.3225 per two yearsConclusion:

• we would not pay 100 for either investment

• we rank investment A above investment B

201

If market interest rates are 5%:

time 0 1 2 3 4 5 6 7 8 9 10

Investment A -100 10 10 10 10 10 10 10 10 10 110

Investment B -100 21 21 21 21 121

Investment into bank account 100× 1.0510 = 162.89

Investment A 100 + 10s0.0510

= 225.78

Investment B 100 + 21s0.10255

= 228.85

Conclusion:

• we would pay 100 for either investments

• we rank investment B above investment A

What exactly are ”market rates of interest”?

These are the yields of fixed income securities, usually those issued by thegovernment. It is more difficult in practice, but we ignore this

202

11.2 Net Present Value

Given a series of cash flows C1, . . . , Cn at times t1, . . . , tn, there net present valuefunction is

NPV(i) =n∑r=1

Cr(1 + i)−tr

i.e. their discounted value as a function of the interest rate.

Given a continuous stream of cash flows with rate of payment ρ(t) during thetime period [0, T ], its net present value function is

NPV(δ) =∫ T

0

ρ(t) exp {−δt} dt

i.e. its discounted value as a function of the force of interest.

203

Remarks:

• By comparing NPV’s at the market interest rate we can choose betweeninvestments. (Principle of Equivalence)

• By comparing NPV functions we can see how sensitive outcomes are tochanges in (market) interest rates. important, if interest rates are not exactly known, but uncertain

Example 11.2 Consider again the investments A and B from the previous sectionand another investment C with the following cash flows:

time 0 1 2 3 4 5 6 7 8 9 10

Investment A -100 10 10 10 10 10 10 10 10 10 110

Investment B -100 21 21 21 21 121

Investment C -100 259.37

204

From the previous section we have

NPVA(i) = −100 + 10ai10

+ 100(1 + i)−10

NPVB(i) = −100 + 21{

(1 + i)−2 + (1 + i)−4 + (1 + i)−6

+(1 + i)−8 + (1 + i)−10}

+ 100(1 + i)−10

The yield of C calculated as the solution of the Equation of Value is 10% and theNPV function is

NPVC(i) = −100 + 259.37(1 + i)−10

205

NP

V

Market Interest Rate

A

B

C

0.0 0.05 0.10 0.15 0.20

-40

-20

020

4060

8010

0

206

Remarks:

(i) The yield exists⇐⇒ NPV(i)=0 (NPV(δ)=0) has a solution (equation ofvalue), but the NPV-function always exists even if the yield does not. picture on black

board

(ii) NPV-functions of two investments may cross more than once picture on black board

(iii) NPVs are additive: NPVA+B = NPVA + NPVB

(iv) Consider the following NPVs (of two investments X and Y):

NPVX = −100 + 21v2 + 110v3

NPVY = −100 + 212.72 ∗ v10

describe the cash flowsIf the market interest rate is close to j = 0.07 we might beindifferent between X and Y. But Y is clearly much more sensitive to a changein the interest rate. For some reasons we would try to avoid this kind of risk=⇒ ranking of NPVs (or the principle of equivalence) are not the only factorsto be considered in practice.

207

NP

V

Market Interest Rate

Y

X

0.0 0.05 0.10 0.15 0.20

-50

050

100

j

208

(v) If the yield exists (i.e. unique i∗ > −1 and that NPV(i∗) = 0) it is called theinternal rate of return or IRR. This is an alternative name most often used incorporate finance.

Example 11.3 Consider again the investments A and B from the previous section

time 0 1 2 3 4 5 6 7 8 9 10

Investment A -100 10 10 10 10 10 10 10 10 10 110

Investment B -100 21 21 21 21 121

Which of the investments A or B would you prefer if the market interest rate was

(i) 5%

(ii) 20%

209

Answer to (i):

NPVA(0.05) = 10a0.0510

+ 100(1.05)−10 − 100 = 38.61

NPVB(0.05) = 21a0.10255

+ 100(1.05)−10 − 100 = 40.49

=⇒ prefer BAnswer to (ii):

NPVA(0.2) = 10a0.210

+ 100(1.2)−10 − 100 = −41.92

NPVB(0.2) = 21a0.445

+ 100(1.2)−10 − 100 = −43.83

=⇒ prefer A

210

11.3 The Discounted Payback Period

Most investments involve an initial outlay followed by income or profits.

If an investment has a positive NPV (at rate i) there must be a time when theinitial outlay is recoverd, allowing for interest.

This time is the discounted payback period (DPP), sometimes called thebreak-even point.

211

Definition 11.4 Let A(t, i), t ∈ [0, T ], be the accumulated value (at rate of interesti) of a set of Cash Flows {Ctk , tk ∈ [0, T ]} and a continuous payment stream at rateρ(s), i.e.

A(t, i) =∑tk≤t

Ctk(1 + i)t−tk +∫ t

0

ρ(s)(1 + i)t−sds .

The discounted payback period (DPP) is the first time t with A(t, i) ≥ 0, i.e.

DPP(i) = min{t ∈ [0, T ]

∣∣∣ A(t, i) ≥ 0}

If ρ(s) = 0, the DPP is

DPP(i) = min{t ∈ {t1, . . . , tn}

∣∣∣ A(t, i) ≥ 0}

Remark:

A(t, i) ≥ 0 ⇐⇒ A(t, i)(1 + i)−t ≥ 0

A(t, i)(1 + i)−t is the NPV of all cash flows received up to time t (at rate ofinterest i).

212

Example 11.5 You invest 100 at time t = 0, and will receive cash flows of 30 attimes t = 1, 2, 3, 4, 5. The market interest rate is 5% p.a. effective. Find NPV(0.05)and the discounted payback period (DPP).

NPV(0.05) = 30a0.055− 100 = 29.88

The DPP is the first time t such that

30a0.05t− 100 ≥ 0 =⇒ 30

1− vt

0.05− 100 ≥ 0 =⇒ 1− vt

0.05≥ 10

3

=⇒ vt ≤ 0.83333 . . . =⇒ 1.05t ≥ 1.2

=⇒ t ≥ ln(1.2)ln(1.05)

= 3.73 =⇒ DPP(0.05) = 4 years

213

We can think of an investment like this as a loan schedule with a ”loan” of 100and ”repayments” of 30 that do more than payoff the loan. The DPP is the firsttime when the outstanding capital is ≤ 0.

Time ”Repayment” ”Interest” ”Capital” ”Capital outstanding”

0 100

1 30 5 25 75

2 30 3.75 26.25 48.75

3 30 2.44 27.56 21.19

4 30 1.06 28.94 -7.75

5 30 -0.39 30.39 -38.14

↑ ↑Interest earned 38.14× 1.05−5 = 29.88

rather than owed = NPV(0.05)

214

Example 11.6 An investment provides a continuous payment stream at a rate ofpayment of ρ(t) = 20 exp(−0.04t) for 10 years. The initial cost for this investmentare £100. The market force of interest is δ = 0.05.

NPV(0.05) = −100 +∫ 10

0

20 exp(−0.04s) exp(−0.05s)ds

= −100 + 20∫ 10

0

exp(−0.09s)ds

= −100 + 201

0.09[1− exp(−0.9)] = 32

NPV-function:

NPV(δ) = −100 + 201

0.04 + δ[1− exp{−10(0.04 + δ)}]

215

DPP(0.05) is the solution of

0 = −100 + 20∫ t

0

exp(−0.09s)ds

= −100 + 201

0.09[1− exp(−0.09t)]

100× 0.0920

= 1− exp(−0.09t)

1− 100× 0.0920

= exp(−0.09t)

t = − 10.09

ln{

1− 100× 0.0920

}= 6.64 (years)

DPP-function

DPP(δ) = − 10.04 + δ

ln{

1− 100(0.04 + δ)20

}

216

NP

V

Market Force of Interest

0.0 0.05 0.10 0.15 0.20 0.25

-20

020

4060

DP

P

Market Force of Interest

0.0 0.05 0.10 0.15 0.20 0.25

010

2030

4050

End week 4

217

11.4 Discounted Cash Flows

Complex investment decisions can be evaluated by working out the net cash flowsand discounting them to the present. These can include money that is borrowedto finance projects.

Example 11.7 Suppose you can both borrow and invest money at 6% p.a.effective. You borrow £100,000,000, the loan to be repaid by level annualrepayments over 10 years. The money finances a project that will make a profit of£18,000,000 at the end of the 2nd to 10d inclusive.

(i) Find the net cash flows.

(ii) What is the accumulated profit at an interest rate of 0.06 at time t = 10?

(iii) What is NPV(0.06)?

Solution for (i): The loan repayment is

100ma0.0610

= 13.587m

218

Hence net cash flows are:

time 0 1 2 3 4 5 6 7 8 9 10

bank +100 -13.587 -13.587 . . . -13.587

project -100 18 . . . 18

net 0 -13.587 4.413 . . . 4.413

Solution for (ii): The accumulation (in million £) is:

−13.587× (1.06)9 + 4.413s0.069

= 27.756

Solution for (iii): The NPV (in million £) is:

NPV(0.06) = 27.756× (1.06)−10 = 15.499

Methods like these are called ”Discounted cash flow methods” (DCF) (eventhough we often use accumulation too).

219

Suppose, however, that the interest rates at which we can lend or borrow aredifferent (more realistic).

Example 11.8 We wish to borrow 100 to invest in a project that will returnprofits of 18 at the end of each of the next 10 years. You can borrow money at 8%p.a. effective and lend money (or deposit) at 6% p.a. effective. Suppose yourepay the loan by level annual repayments. Find

(i) the net cash flow

(ii) the accumulated profit at t = 10

(i): loan repayment:100a0.0810

= 14.90

The net cash flow each year is: 18 - 14.90 = 3.10

(ii): The accumulated profit is 3.10s0.0610

= 40.86

220

We do not calculate a NPV here. When borrowing and lending rates are different,NPV and IRR (yields) are not always meaningful, because they are functions ofone rate of interest.

The investor can do better in this example. It does not make sense to invest netcash flow to earn interest of 0.06% while capital outstanding on the loan iscosting 0.08%. It would be better to devote all the profits to repaying the loan assoon as possible.

What is the accumulated profit at time t = 10 if the loan is repaid as soon aspossible?

221

We must find the DPP at 8%, i.e. the first time t such that

18s0.08t≥ 100(1.08)t

=⇒s0.08t

1.08t= a0.08

t≥ 100

18=⇒ 1− vt

0.08≥ 100

18

=⇒ 1− vt ≥ 818

=⇒ vt ≤ 1018

=⇒ 1.08t ≥ 1810

= 1.8

=⇒ t ≥ ln(1.8)ln(1.08)

= 7.63

Therefore, the DPP is t = 8.Now, 18s0.08

8= 191.66 and 100(1.08)8 = 185.09

Hence at time t = 8 the loan is repaid with (191.66 - 185.09) = 6.37 to spare.This and the two remaining profits are available to invest at 6%.

Profit at time t = 10: 6.37× (1.06)2 + 18× 1.06 + 18 = 44.24 > 40.86 (profit fromlevel loan repayment)

222

Note that NPV would have no meaning here but we can always find theaccumulated profit from a project in this way.

When there is only one rate of interest for borrowing and lending the pattern ofloan repayments do not matter.

223

Example 11.9 An investor wish to build a factory costing £100.000.000 toproduce computer games consols. It is expected to produce profits of£50.000.000 at the end of the second, third, fourth and fifth year after which itwill be written off because the next generation of consols will have arrived.

The project can be financed in two ways:

(A) The money can be borrowed at an interest rate of 7% p.a. effective. Theprofits will be used to pay off the loan as soon as possible, and once that isdone further profits will be invested to earn 5% p.a. effective.

(B) The investor can issue a bond with annual coupon of 6%, payable yearly inarrear, with term 5 years. It will be issued at par but must be redeemed at110%. If the investor needs to borrow any money it will be at 7% p.a.effective, while any surplus profits can be invested to earn 5% p.a. effective.

Which method should the investor choose?

224

We find the total profit at time t = 5 in each case.

(A) At time t = 1 the loan has accumulated to £107m. It will be repaid by timet+ 1, where t is the earliest time at which

50s0.07t≥ 107(1.07)t

=⇒ a0.07t≥ 107

50=⇒ 1− vt ≥ 0.1498 =⇒ vt ≤ 0.8502

=⇒ 1.07t ≥ 1.1762 =⇒ t ≥ ln 1.1762ln 1.07

≈ 2.4 =⇒ t = 3

50s0.073− 107(1.07)3 = 29.665

Hence the DDP is 4 years and the net cash flow at time t = 4 is 29.665.The final (accumulated) profit is

29.665× 1.05 + 50 = 81.15 (million)

225

(B) The investor must borrow £6m at time t = 1 to pay the first coupon. Hence attime t = 2 the net profit is

50− 6− 6× 1.07 = 37.58

and at times t = 3, 4, 5 the net profit is

50− 6 = 44

Hence the final profit after redeeming the loan is

37.58(1.05)3 + 44s0.053− 110 = 72.21

Hence choose A.

226

Example 11.10 An investment into a cinema provides a continuous paymentstream at a rate of payment

ρ(t) =

50000t for the first three years, t ∈ [0, 3]

150000 for t ∈ (3, 6]

The investor believes that the cinema has a value of 1,000,000 at the end of yearsix.The investor can lend money at a force of interest of δ1 = 0.05 and can borrowmoney at a force of interest of δ2 = 0.08.The cinema costs 1,200,000 of which the investor provides 1,000,000. He willborrow the remaining 200,000 from a bank.

227

What is the discounted payback period of the loan if the investor uses all cashflows associated with the cinema to repay the loan as soon as possible?

DPP is the solution of:

200, 000 =∫ t

0

ρ(t) exp(−0.08t)dt

For t ∈ [0, 3] (first three years):

200, 000 = 50000(I a)t

= 500001

0.08[at − tv

t]

= 500001

0.08

[1− vt

0.08− tvt

]For t = 3 we get:

50000(I a)3 = 192, 042.60 < 200, 000 =⇒ DPP(0.08) > 3

228

For t ∈ [3, 6] we have to find the solution of

200, 000 =∫ t

0

ρ(s) exp(−0.08s)ds

=∫ 3

0

ρ(s) exp(−0.08s)ds+∫ t

3

ρ(s) exp(−0.08s)ds

= 192, 042.60 +∫ t

3

150, 000 exp(−0.08s)ds

= 192, 042.60 + 150, 0001

0.08[exp(−0.08× 3)− exp(−0.08t)]

DPP(0.08) =1

−0.08log{

exp(−0.08× 3)− 200, 000− 192, 042.60150, 000

0.08}

= 3.068 years

229

All cash flows from the cinema after 3.068 years are profits.

(The investor can lend money at a force of interest of δ1 = 0.05,.... The investorbelieves that the cinema has a value of 1,000,000 at the end of year six.)

Accumulated profits after six years:

A(6) = −1, 000, 000 exp(0.05× 6)

+∫ 6

3.068

150, 000 exp(0.05(6− t))dt+ 1, 000, 000

= −1, 000, 000 exp(0.05× 6)

+150, 0001

−0.05

{exp(0.05× (6− 6))− exp(0.05× (6− 3.068))

}+1, 000, 000

= 123, 813.33

230

12 Duration, Convexity and Immunization

Literature: John McCutcheon and William F. Scott: An introduction to themathematics of finance, 2003 : 10.1 -10.3, 10.5 - 10.7

12.1 Duration

Consider a set of discrete future cash flows Ctk to be paid at time tk,0 < t1 < . . . < tn < T and a continuous cash flow at rate of payment ρ(t) fort ∈ [0, T ].

The net present value of these cash flows at constant force of interest δ is

NPV(δ) =n∑k=1

Ctk exp(−δtk) +∫ T

0

ρ(t) exp(−δt)dt

= NPV(i)

with effective rate of interest i p.a. and exp(δ) = 1 + i.

231

Definition 12.1 The (Macaulay) duration τ(δ) of these cash flows at force ofinterest δ is

τ(δ) = − 1NPV(δ)

× d

dδNPV(δ)

=∑nk=1 tkCtk exp(−δtk) +

∫ T0tρ(t) exp(−δt)dt∑n

k=1 Ctk exp(−δtk) +∫ T0ρ(t) exp(−δt)dt

Interpretation:τ(δ) measures the proportionate change in the NPV(δ) due to a small change inthe force of interest δ.If the force of interest changes from δ0 to δ1 then:

NPV(δ1)− NPV(δ0)NPV(δ0)

≈ −(δ1 − δ0)τ(δ0)

This is a good approximation for immediate small changes in δ.

232

Remarks:

(i) τ measures the sensitivity of the NPV with respect to δ.

(ii) τ is also called discounted mean term and volatility (other definitions ofvolatility exist - McCutcheon & Scott (2003) use this one)

(iii) τ is the weighted average maturity of all cash flows, weight of cash flow attime tk is

wtk =Ctk exp(−δtk)

NPV(δ)for a set of cash flows

respectively

w(t) =ρ(t) exp(−δt)

NPV(δ)for continuous cash flows

233

Example 12.2 Calculate the duration τ at force of interest δ of an n-yearZero-Coupon-bond.

NPV(δ) = exp(−δn)d

dδNPV(δ) = −n exp(−δn)

=⇒ τ(δ) = − 1NPV(δ)

× d

dδNPV(δ)

= nexp(−δn)exp(−δn)

= n

234

Example 12.3 Two investments providing the following two sets of cash flows

Time (years) Cash Flow 1 Cash Flow 2

1 50 6.7

2 25 13.4

3 12 26.8

4 6 53.6

Calculate the net present value and the duration at δ = 0.05.

δ = 0.05 =⇒ i = exp(δ)− 1 = 0.051

NPV1(i) = 85.4 = NPV2(i)

τ1(i) =

{1× 50× v + 2× 25× v2 + 3× 12× v3 + 4× 6× v4

}NPV1(i)

= 1.68

τ2(i) =

{1× 6.7× v + 2× 13.4× v2 + 3× 26.8× v3 + 4× 53.6× v4

}NPV2(i)

= 3.22

235

Dur

atio

n

Force of Interest

0.0 0.05 0.10 0.15 0.20 0.25 0.30

1.50

1.55

1.60

1.65

1.70

Figure 4: Duration of Cash Flow 1 in Example 12.3.

236

Dur

atio

n

Force of Interest

0.0 0.05 0.10 0.15 0.20 0.25 0.30

3.00

3.05

3.10

3.15

3.20

3.25

Figure 5: Duration of Cash Flow 2 in Example 12.3.

237

Example 12.4 Two fixed-income securities (FIS)

A: term 10 years, coupon 12% p.a. paid annually in arrear, redemption at par

B: term 25 years, coupon 3% p.a. paid annually in arrear, redemption 120%

τ should depend on δ, since the derivative is always taken with respect to δ and not with respect to i

τA(δ) ={

1× 12× v + 2× 12× v2 + . . .+ 10× 12× v10

+10× 100× v10}× 1

12a10 + 100v10

={

12(Ia)10 + 1, 000v10}× 1

12a10 + 100v10

τB(δ) ={

3(Ia)25 + 120× 25× v25}× 1

3a25 + 120v25

238

δ Price A Price B τA τB

NPVA NPVB

4% 163.97 90.61 7.1451 17.7975

3% 176.21 108.66

5% 152.75 76.13

Change in Price change in % −(δ1 − δ0)τ(0.04)

force of interest A B A B

4%→ 3% +7.465% +19.92% +7.1451 +17.7975

4%→ 5% -6.843% -15.98% -7.1451 -17.7975

239

Theorem 12.5 Assume two cash flows with present value V1 and V2 and durationτ1, τ2 at a given force of interest. If V1 + V2 6= 0, the duration of both cash flowscombined is

τ1&2 =V1τ1 + V2τ2V1 + V2

Proof. V1&2 = V1 + V2 anddV

dδ= −τV

Therefore,

τ1&2 = − 1V1&2

dV1&2

dδ= − 1

V1 + V2

d(V1 + V2)dδ

= − 1V1 + V2

(dV1

dδ+V2

dδ

)= − 1

V1 + V2(−τ1V1 − τ2V2)

�

240

12.2 Convexity and Immunization

Set-up: Financial Institution (e.g. Life Office)Future net Liabilities: Ltk at time tkAssets: income Asn at time sn

Assume

NPVA(δ0) =∑n

Asn exp(−δ0sn) = NPVL(δ0) =∑k

Ltk exp(−δ0tk) (12.7)

where δ0 denotes the current force of interest.

Problem: How to choose the assets so that an instantaneous change in the forceof interest from δ0 to δ1 will not result in NPVA(δ1) < NPVL(δ1)?

”Perfect Matching”: Choose s1 = t1, .... and Atk = Ltk for all k (not a practicalsolution)

241

Redington’s theory of immunization

Theorem 12.6 Assume that the NPV of the assets is equal to the NPV of theliabilities,

NPVA(δ0) = NPVL(δ0) ”NPV matching” (12.8)

Choose assets such that:

τA(δ0) = τL(δ0) ”Duration matching” (12.9)

and (”Convexity condition”)∑k

s2nAsn exp(−δ0sn) >∑k

t2kLtk exp(−δ0tk) (12.10)

respectively ∫ T

0

t2ρA(t) exp(−δ0t)dt >∫ T

0

t2ρL(t) exp(−δ0t)dt (12.11)

Then there exists an ε > 0 such that for all δ1 with |δ1 − δ0| < ε the following holds

NPVA(δ1) > NPVL(δ1)

242

Interpretation: If the conditions (12.8), (12.9) and (12.10) (discrete CF) or(12.11) (continuous CF) are fulfilled, then an immediate small change in theforce of interest δ0 → δ1 will give: NPVA(δ1) > NPVL(δ1).

Proof. Define f(δ) = NPVA(δ)− NPVL(δ)

Equation (12.8) =⇒ f(δ0) = 0Equation (12.9) =⇒ f ′(δ0) = 0Equations (12.10) and (12.11) =⇒ f ′′(δ0) > 0

It follows that f has a local minimum at δ0 and f(δ0) = 0.

f(δ) > f(δ0) = 0

for all δ close to δ0.

243

delta

f(de

lta)

delta_0 = ln(1.04)

0.02 0.03 0.04 0.05 0.06

010

0020

0030

0040

00

�

244

Definition 12.7 For any set of cash flows {Ctk} and a continuous cash flow at rateof payment ρ(t)

CONV(δ) =1

NPV(δ)d

dδ2NPV(δ)

=∑nk=1 t

2kCtk exp(−δtk) +

∫ T0t2ρ(t) exp(−δt)dt∑n

k=1 Ctk exp(−δtk) +∫ T0ρ(t) exp(−δt)dt

is the convexity of the cash flows.

Remarks:

(i) If conditions (12.8), (12.9), (12.10) and (12.11) hold, the financialinstitution is said be immunized (in the sense of Redington) against smallchanges in the force of interest (or rate of interest)

(ii) The financial institution is immunized against immediate small changes inthe force/rate of interest assuming a flat yield curve. (Generalizes toimmediate parallel shifts of a yield curve)

245

(iii) Suppose (12.8) and (12.9) hold, then (12.10) is equivalent to∑k

(tk − τA(δ0))2Atk exp(−δ0tk) >∑k

(tk − τL(δ0))2Ltk exp(−δ0tk) (12.12)

=⇒ The spread of the assets around the common τ is greater than the spreadof the liabilities. A similar result holds for continuous cash flows.

(iv) Suppose (12.8) holds, then (12.10) and (12.11) are equivalent to

CONVA(δ0) > CONVL(δ0)

where CONVA and CONVL denote the convexities of the cash flowsassociated with the assets and the liabilities, respectively.

246

Example 12.8 A Life Office has to pay £1m in 10 years. Current force of interestδ0 = ln(1.04) (i.e. rate of interest i=4%). The available securities are a 5 yearsand a 15 years Zero-Coupon-Bond. The office has assets of £675,560 =£1, 000, 000(1.04)−10 = NPVL.How much should the Life Office invest into each Zero-Coupon-Bond?

change 675,560 into 675,564

Solution:

(i) Note that NPVA = NPVL

(ii) Price of each Zero-Coupon-Bond per £100 nominal isP5 = 82.193 = 100(1 + i)−5, P15 = 55.526

247

(iii) Suppose £X is invested into 5-year Zero-Coupon-Bond and £(675, 560−X) isinvested into 15-year Zero-Coupon-Bond.Choose X such that

τA(δ0) = τL(δ0) at δ0 = ln(1.04)

τL(δ0) = 10

τA(δ0) =1

675, 560

{5X

P5100(1 + i)−5 + 15

675, 560−XP15

100(1 + i)−15

}=

{5X + 15(675, 560−X)

} 1675, 560

=! 10 = τL(δ0)

=⇒ X = £337, 780

=⇒ 12 assets in 5-year Zero-Coupon-Bond

=⇒ 12 assets in 15-year Zero-Coupon-Bond

(iv) Check that convexity condition (12.10) holds. Then the financial institutionis immunized.

248

Remark: (Check yourself)

(i) With the solution above we have:

• If interest rates move to 3% p.a. effective:NPV(3%) of (assets-liabilities) ≈ +£864.

• If interest rates move to 5% p.a. effective:NPV(5%) of (assets-liabilities) ≈ +£700.

(ii) If all money would have been invested in 5-years Zero-Coupon-Bond:

• If interest rates move to 3% p.a. effective:NPV(3%) of (assets-liabilities) ≈ -£35,094.

• If interest rates move to 5% p.a. effective:NPV(5%) of (assets-liabilities) ≈ +£30,086.

=⇒ Immunization works against profits as well as losses.

249

delta

f(de

lta)

delta_0 = ln(1.04)

0.02 0.03 0.04 0.05 0.06

010

0020

0030

0040

00

Figure 6: NPVA − NPVL for i between 2% and 6% (δ ∈ [ln 1.02, ln 1.06]).

250

delta

f(de

lta)

0.05 0.10 0.15 0.20 0.25 0.30 0.35

010

000

3000

050

000

Figure 7: NPVA − NPVL for i between 0% and 40%.

251

13 Arbitrage and Forward Contracts

Literature: John C. Hull: Options, Futures and Other Derivatives, 2000

We will now consider securities that have random (uncertain) future prices.Trading in these securities yields random future cash flows.

252

13.1 Arbitrage and the No-Arbitrage Assumption

Loosely: Arbitrage is a risk-free trading profit.

Formally: An arbitrage opportunity exists in a financial market if either:

(i) an investor can make a deal which would give an immediate profit with norisk of future loss or

(ii) an investor can make a deal at no initial cost, has no risk of future loss and anon-zero probability of future profit.

include something about arbitrage with martingales, no free lunch, bounded from below, ...

253

Example 13.1 Bank A charges interest at 5% p.a. effective on loansBank B provides interest at 6% p.a. effective on deposits=⇒ Arbitrage opportunity:Borrow as much as possible from A, deposit it with B, repay A eventually andmake profit.No initial costs, but non-zero probability (=1) of profit.

Pricing of financial instruments often involves short-selling. Short-selling meansselling a security that is not owned with the intention of buying it later.Short-selling yields profits when the price of the security goes down and a loss ifthe price goes up.

254

Example 13.2 Your broker sells on your behalf shares of a security which areowned by another client. All income of that security during the period of lendinghas to be given to the owner. Shares must be bought back and given back to theowner. Margins are required for security reasons.

Example 13.3 ”Arbitrage opportunity” involving short-sellingTwo assets traded over 1 time period on a stock exchange.Suppose that during this period market prices either go up or down.

Price at time 1

Asset Price now Market falls Market rises

1 7 4 8

2 9 6 12

255

Trading strategy:

At time 0:

• Sell 3 units of asset 1

• Buy 2 units of asset 2

=⇒ income (at time 0):

3× 7︸ ︷︷ ︸”(short-)selling asset 1”

− 2× 9︸ ︷︷ ︸”buying asset 2”

= 21− 18 = 3

256

At time 1:

• Buy 3 units of asset 1

• Sell 2 units of asset 2

=⇒ income (at time 1) if market falls:

2× 6︸ ︷︷ ︸”selling asset 2”

− 3× 4︸ ︷︷ ︸”buying asset 1”

= 0

=⇒ income (at time 1) if market rises:

2× 12︸ ︷︷ ︸”selling asset 2”

− 3× 8︸ ︷︷ ︸”buying asset 1”

= 0

=⇒ Arbitrage

No arbitrage, if at time 0:”price of asset 1” = 2

3 × ”price of asset 2”

257

No Arbitrage Assumption

no arbitrage opportunity can exist

Remarks:

(i) Reasonable assumption in liquid markets. Trades of arbitrageurs willeliminate arbitrage opportunities (by force of supply and demand)

(ii) No arbitrage implies: If two sets of assets have exactly the same payout(value at some future time), they must have the same value now.Example above: 3 × asset 1 = 2 × asset 2 at time 1.Hence at time 0: price of asset 1 = 2

3 price of asset 2.

(iii) Replicating portfolio for a given set of assets= portfolio with the same value at some future time.No arbitrage =⇒ replicating portfolio must have the same value as the givenset of assets at any intermediate time.

(iv) No arbitrage and replicating portfolio are key ideas in modern finance.

258

13.2 Forward Price and Hedging

Set-up:Asset: current price is known, future price is uncertain. (could be an (equity)share, fixed amount of foreign currency, ...)

Example 13.4 A British company knows that it will receive US-$ 1 million in 3months.

Problem: Unknown future exchange rate!

Company contacts bank and agrees to sell US-$ 1 million in 3 month for the nowagreed exchange rate 0.6854.

259

Definition 13.5 A forward contract is an agreement to buy -or sell respectively-an asset at a certain future time (= delivery time) for a certain price.The price agreed at time t = 0 to be paid at delivery time t = T is called thedelivery price. It is chosen such that the value of the forward contract is zero attime 0.

time 0 time T

Agreement made, no payments asset supplied, payments

Definition 13.6 The forward price for a particular forward contract at aparticular time (t ≥ 0) is the delivery price that would make the forward contract tohave zero value at that time t ≥ 0.

Example 13.7 (Example 13.4 continued) 1 month later: delivery price =0.6854, forward price will be different (in general)!

260

General Terminology

The following terminology is used in the industry.Long position in forwardcontract = agree to buy the asset

Short position in forward contract = agree to sell the asset

risk-free rate of interest = rate of interest that can be earned without assumingany risks (e.g. treasury bills)

Question: How to calculate the forward price of an asset? We will apply theNo-Arbitrage assumption to calculate the forward price of an asset.

261

Assumptions:

• No arbitrage

• no transaction costs

• known risk-free force of interest δ, at which money can be invested andborrowed

Notation:

T is the delivery time (in years)

St price of the asset at time t ∈ [0, T ]

K forward price

262

13.3 Forward Contracts on Securities with no Income

Example 13.8 stock: S0 = £30risk-free force of interest δ = 0.05 p.a.No dividends within the next two years.

What is the two-years forward price for this asset?

263

(i) Assume, the two-year forward price is K=£35 (delivery time T = 2)

An arbitrageur can adopt the following strategy:

today (time t=0) in 2 years (time T = 2)

1. Borrow £30 at δ =0.05 for 2 years

+30 −30e2×0.05

2. Buy the stock -30 0

3. Enter a f.c. to sell 1stock for £35 (=K)in 2 years

0 +35

0 35− 33.16 = 1.84

=⇒ Arbitrage

Conclusion: Any forward price > £33.16 would allow this arbitrage strategy.

264

(ii) Assume, the two-year forward price is K=£31 (delivery time T = 2)

An arbitrageur can adopt the following strategy:

today (time t=0) in 2 years (time T = 2)

2. Short the stock +30 0

1. Lend £30 at δ =0.05 for 2 years

−30 +30e2×0.05

3. Enter a f.c. to buy 1stock for £31 (=K)in 2 years

0 -31

0 33.16− 31 = 2.16

=⇒ Arbitrage

Conclusion: Any forward price < £33.16 would allow this arbitrage strategy.

265Summary:

• Any forward price > £33.16 provides an arbitrage opportunity.

• Any forward price < £33.16 provides an arbitrage opportunity.

Therefore, the fair forward price is 33.16. This is the only price that does notprovide an arbitrage opportunity.

General formula for the forward price:

K = S0 exp(δT )

Example 13.9 price of stock: £134.75risk-free force of interest: δ = 3.99% p.a. for 3-months investments

=⇒ 3-months forward price:

K = £134.75 exp(

0.0399× 312

)= £136.10

266

Hedging

Consider the investor who has agreed to supply the asset at time T for theforward price K = S0 exp(δT ).

Strategy 1: Do nothing before time T . Buy the asset at time T and deliver if forthe price K.

Receive: at time t = 0: 0at time t = T : K − ST ( could be negative)

Strategy 2: Borrow K exp(−δT ) (= S0) at the risk-free force of interest δ, to berepaid at time T , and buy the asset now. Receive K at time T and repay theloan.

Receive: at time t = 0: 0at time t = T : 0

267

Remarks:

• Strategy 2 =⇒ No chance of profit and no risk of loss.

• Strategy 2 is called a static hedge for the forward contract.

• In general a hedge portfolio is a portfolio which eliminates or reduces therisk associated with an investment.

• There is a relationship between pricing and hedging.

268

13.4 Forward Contracts on Securities that Provide a KnownCash Income

Assume, the asset will pay a known cash dividend C at a known time t ∈ [0, T ].Show:

K = (S0 − C exp(−δt)) exp(δT )

(see tutorial)

Example 13.10 (example 13.9 continued)S0 = £134.75, δ = 3.99% p.a. for 3-months investmentsa dividend of £10.00 is paid in one month time.

=⇒ 3-months forward price:

K = £(

134.75− 10.00 exp(−0.0399

12

))exp

(0.0399× 3

12

)= £126.03

269

13.5 Forward Contracts on Securities that Provide a KnownDividend Yield

Example 13.11 Forward contract on currencies

Notation: S0 = current price of one unit of foreign currencyδf = risk-free force of interest in foreign currency.

(i) Assume

K > S0 exp {(δ − δf )T}

270Strategy:

CF today (time t=0) CF in 2 years at T = 2

Borrow S0 exp(−δfT ) indomestic currency at rateδ p.a. for T years

S0 exp(−δfT ) −S0 exp(−δfT ) exp(δT )

Buy exp(−δfT ) of the for-eign currency and investat rate δf p.a. for T years

−S0 exp(−δfT )

Enter a f.c. to sell 1 unit offoreign currency at T

0 K

0 K − S0 exp((δ − δf )T )

K 6= S0 exp((δ − δf )T ) =⇒ Arbitrage

Therefore, K = S0 exp((δ − δf )T ) is the only forward price that does not allowany arbitrage opportunity.

271

In General:

Assume the asset provides a known dividend yield, paid continuously at rateρ(t) = gSt at time t (g > 0), i.e. starting with 1 unit at time t = 0 and reinvestingthe income immediately will accumulate to exp(gT ) units of the asset.

Then:

K = S0 exp {(δ − g)T}

Example 13.12 on 9 Feb 2004

US-$ 1 = £0.5385UK-risk-free force of interest δ = 3.99% p.a.US-risk-free force of interest δf = g = 0.962% p.a. for 3-months investments

3-months forward price:

K = 0.5385 exp{

(0.0399− 0.00926)× 312

}= 0.5426

272

13.6 The Value of a Forward Contract

Notation: ft = value of a forward contract at time t, t ∈ [0, T ]

Remark:

• f0 = 0

• fT =

ST −K (long position)

K − ST (short position)

273

Long Forward Contract Consider the following two portfolios at time t.

Portfolio A: Buy existing long forward contract for price ft with deliveryprice Kl. Invest Kl exp(−δ(T − t)) at the rate δ p.a. for (T − t) years.Value of portfolio A: ft +Kl exp(−δ(T − t))

Portfolio B: Enter a new forward contract to buy the asset for delivery price= forward price = K at time T . Invest K exp(−δ(T − t)) at rate δ p.a. for(T − t) years.Value of portfolio B: K exp(−δ(T − t))

at Time T :Value of portfolio A: ST −Kl +Kl = ST

Value of portfolio B: ST −K +K = ST

274

No arbitrage =⇒

Value of portfolio A at time t = Value of portfolio B at time t

ft +Kl exp(−δ(T − t)) = K exp(−δ(T − t))

ft = (K −Kl) exp(−δ(T − t))

Short Forward Contract Similar arguments:

ft = (Kl −K) exp(−δ(T − t))

275

Example 13.13 3-month forward contract on US-$.Current exchange rate: 1 US-$ = 0.535 £3-month UK risk-free interest rate: 4.72% =⇒ δ = log(1.0472)3-month US risk-free interest rate: 2.4% =⇒ δf = log(1.024)

Current forward price (delivery price):

Kl = S0 exp {(δ − δf )T}

= 0.535 exp{(

log(1.0472)− log(1.024)) 3

12

}= 0.5380049

276

One month later:

Exchange rate: 1 US-$ = 0.52 £2-month UK risk-free interest rate: 4.82% =⇒ δ = log(1.0482)2-month US risk-free interest rate: 2.56% =⇒ δf = log(1.0256)Forward Price:

K = S0 exp {(δ − δf )T}

= 0.52 exp{(

log(1.0482)− log(1.0256)) 2

12

}= 0.5218925

Price of Short Forward Contract:

ft = (Kl −K) exp(−δ(T − t))

= (0.5380049− 0.5218925) exp(− log(1.0482)× 2

12

)= 0.01598648

277

14 The Term Structure of Interest Rates

-time 0 1 2 3 4

int. rate (random) i1 i2 i3 i4

force (random) δ(t)6

ZCB (known) P2

?

100ZCB (known) P4

?

100Bond (known) P C C C C +R

The market rate of interest may (and usually does) depend on the term of theinvestment - this is why we are interested in the term structure of interest rates.

278

14.1 Zero-Coupon Bonds and Spot Rates

Definition 14.1 An n-year zero-coupon bond is a security which pays 1 after nyears and nothing else, i.e. a fixed-interest security, redeemed at par, with term n

years and no coupon payments.

In the UK market, a zero-coupon bond is sometimes called a “gilt strip”, a giltwhich has been stripped of its interest payments, or indeed just one of the couponpayments.

Definition 14.2 The n-year spot-rate of interest yn is the redemption yield p.a.of an n-year zero-coupon bond. Let Pn be the price now of an n-year zero-couponbond, then, for n > 0:

EoV: Pn = 1× (1 + yn)−n ⇒ yn = P− 1n

n − 1

Example 14.3 What is the spot-rate of interest, y5, for a 5-year zero couponbond currently priced at 0.78 ? y5 = 0.78

15 − 1 = 5.095%

279

Consider a Fixed Interest Security which pays redemption proceeds of R at timen, and dividends (coupons), d, at times 1, 2, . . . , n− 1, n. With no-arbitrage, theprice of this security, P , must be equal to the price of a series of zero couponbonds.

-?

?? ? ?

P

d d d d+R

0 1 2 . . . n− 1 ntime

P = d(P1 + P2 + . . .+ Pn) +RPn

= d((1 + y1)−1 + (1 + y2)−2 + . . .+ (1 + yn)−n) +R(1 + yn)−n

If this were not true then we could make money by buying a series ofzero-coupon bonds and selling the fixed-interest security (or vice versa) to makea profit (⇒ arbitrage).

280

14.2 Yield Curves

Yield curves are graphical representation of the redemption yields of bonds withrespect to the remaining term of the bonds. They can be calculated for ZCBs aswell as for bonds with attached coupons.

• Yield curves show graphically the relationship between term to redemptionand yield.

• Yields are a reflection of prices. They are determined in the market bydemand and supply.

• The yield curves for bonds with different coupons might be different becauseof factors affecting the investors who buy and sell them, for example, the taxtreatment of coupons.

• Yield curves vary from day to day, and, depending on market expectationscould be increasing, or decreasing in shape, or even humped (see below).

281

-

6

0 5 10 15 20 25 303.63.73.83.94.04.14.24.34.44.54.64.7

Term

GRY(%)

Source: The Times 29/10/02

Redemption Yields on Gilts - Oct 2002

282

Explanations for the shape of the yield curve

There are three key theories that explain the shape of the yields curve. These are:-

(i) Expectations Theory

This says that long-term interest rates reflect investors’ expectations ofshort-term interest rates in the future. This means that the long-term rate iscomposed of all the short term rates up to that long term.

(ii) Liquidity Preference Theory

It is generally accepted that the higher the risk an investor bears, the higherthe expected return needs to be to entice him to take the extra risk. Whenshort-term interest rates move, the price of a long-dated bond will be affectedfar more than a short-term bond (the Macaulay duration of a long-term bondis higher than the duration of a short term bond). This means that investorswill expect a higher rate of interest on a long-dated bond than a short-datedone, because of the extra risk.

283

For example, consider 2 bonds that will be redeemed at par, one in 3 years,the other in 20 years. Assuming only one coupon payment is made every yearand that the coupon rate is 9% per annum, we can calculate their presentvalue at a market interest rate of 5% pa.

PV(3-year bond) = 9a3 + 100v3 = 110.89PV(20-year bond) = 9a20 + 100v20 = 149.85

Now, if interest rate rise to 6%, the price of the 3-year bond falls to 108.02(= a fall in price of 2.6%) whereas the 20-year bond falls to 134.41 (a fall of10.3%).

So, the longer bond is affected more by the interest rate movement.

(iii) Market Segmentation Theory

Different investors are attracted to different types of bond. For example, apension fund is more interested in long-term and index-linked bonds as theymore closely “match” the pension fund’s liabilities. On the other hand, aninvestor not liable to income tax would be more interested in high couponbonds. The yield for each “segment” of the market is then determined bywhere demand equals supply.

284

14.3 Forward Rates

Definition 14.4 The forward rate ft,r is the effective annual rate of interest agreednow (=time 0) for an investment made at time t for a period of r years.

-0 t t+ r

agreementno payment

ft,r is paid

What this means is that the present value at time t of a unit amount at time t+ r

is vr where v = (1 + ft,r)−1.

Remember that this is a future rate, agreed now to apply from time t to t+ r.Next year, the same rate is likely to be different for the same time period, whichwould then be one year closer. We would still need to calculate vr but v will havechanged to reflect changes in perception from time 0 to time 1, and the newforward rate will be ft−1,r. Compare this argument to the change of forwardprices and the value of forward contracts.

285

The relationship between spot rates and forward rates

Clearly f0,r = yr

Consider £1 invested now for t+ r years.

Using spot rates, this will accumulate to (1 + yt+r)t+r = P−1t+r

Alternatively, if we invest the £1 for t years at the spot rate and then from time tto t+ r at the forward rate, the accumulations must be equal,

(1 + yt)t(1 + ft,r)r = P−1t (1 + ft,r)r

No arbitrage =⇒ £1 invested now for t+ r years must always give the samepayoff at time t+ r:

P−1t+r = P−1

t (1 + ft,r)r

⇒ (1 + ft,r)r =PtPt+r

=(1 + yt+r)t+r

(1 + yt)t

286

It is also clear that £1 invested at the spot rate for t years must accumulate to thesame value as £1 invested at the forward rate from times0→ 1, 1→ 2, . . . , t− 1→ t i.e.(1 + yt)t = (1 + f0,1)(1 + f1,1)(1 + f2,1) . . . (1 + ft−1,1)

Note Suppose at a particular time the (effective annual) rate of interest does notdepend on time,i.e. yt = fs,r = i for all t, s, r ≥ 0then the yield curve at that time is flat (=level).

Definition 14.5 The n-year Par Yield is the coupon rate payable annually per £1nominal which would give a bond with term n years a price of £1 (for a given termstructure) assuming the bond is redeemed at par.

That is, for a given zero-coupon yield curve, the n-year par yield, cn is given by

1 = cn[(1 + y1)−1 + (1 + y2)−2 + . . .+ (1 + yn)−n] + (1 + yn)−n

287

Note In terms of zero-coupon bond prices, we have

cn =1− Pn∑nk=1 Pk

Example 14.6 Consider the following spot interest rates

Term (in years) Spot Rate (in % pa)

1 5

2 5.5

3 6

The 3-year par yield = c3 where

1 = c3(1.05−1 + 1.055−2 + 1.06−3) + 1.06−3 ⇒ c3 = 5.96% pa

288

14.4 Continuous Time Rates

Definition 14.7 The t-year spot force of interest is

Yt = −1t

lnPt .

Note that Pt = exp−tYt

Yt is also called(i) the continuously compounded spot rate, or(ii) the continuous time spot rate.

Relationship between the t-year spot rate and the t-year continuouslycompounded spot rate:

1Pt

= (1 + yt)t = expYt.t

Hence yt = expYt −1 (c.f. i = expδ −1)

Example 14.8 yt = 5%⇒ Yt = 0.04879 = 4.879%

289

Definition 14.9 The continuous time forward rate, Ft,r is the force of interestequivalent to the annual rate of interest ft,r.

Hence, we obtain the following relationships

(1 + ft,r)r = exp(Ft,rr) =⇒ 1 + ft,r = exp(Ft,r)

We have

exp(Ft,rr) = (1 + ft,r)r =(1 + yt+r)t+r

(1 + yt)t= exp(Yt+r(t+ r)− Ytt)

Therefore

Ft,r =(t+ r)Yt+r − tYt

r

We also have that

exp(Ft,rr) = (1 + ft,r)r =PtPt+r

⇒ Ft,r =1r

ln(

PtPt+r

)(∗)

290

Definition 14.10 The instantaneous forward rate Ft is defined as

Ft = limr→0

Ft,r

F0 is also called the short rate of interest.

Note: f0,t = yt ⇒ F0,t = YtHence F0 = limt→0 Yti.e. F0 is the continuously compounded interest rate consistent with animmediately maturing zero-coupon bond.

This is relevant because some stochastic interest rate models, model the shortrate (these are known as short rate models).

From (*) we obtain:-

Ft = limr→0

Ft,r = limr→0

lnPt − lnPt+rr

= − d

dtlnPt = − 1

Pt

d

dtPt

Using −Ft = ddt lnPt we get

−∫ t

0

Fsds = lnPt − lnP0

291

As lnP0 = ln 1 = 0, we have

Pt = exp−∫ t0 Fsds

Compare this to the expression in a deterministic market with force of interest δt:

Pt = exp−∫ t0 δsds

Important We have only considered the term structure for a particular point intime. Ft today (at time 0) might be completely different from Ft next year (attime t = 1) and also completely different from Ft−1 at time 1, since bond pricesmight be stochastic. A deterministic force of interest, δt, and Ft as consideredabove are therefore different things, although the last two equations lookedalmost identical.

292

15 Stochastic Interest Rates

The financial contracts that are of interest to actuaries are often very long term,and the rates of interest over the long term play a significant part in theircalculations.

Up to now we have generally looked at models where interest rates haveremained constant, or we have assumed that we know, in advance, what theinterest rates will be (for example, we have assumed that the interest rate will be10% per annum for the first 2 years and 8% per annum thereafter).

But, in reality, we cannot know the rate of interest in the future. And, in fact, thefurther into the future we go, the less certain we can be.

One solution is to use a Stochastic Interest Rate Model.

In essence, we model the uncertainty about future interest rates by assuming thatfuture interest rates are random variables.

293

15.1 Model 1

We wish to invest for 10 years starting at time 1, but are uncertain about futureinterest rates.

We will assume that interest rates will be constant during the years from time 1 totime 11, but that the rate might be 2%pa, 4%pa or 6% pa, each being equallylikely.

Let it represent the interest earned over the year t− 1→ t

We can observe i1.

Prob(it = 2%) = 13 for t = 2, . . . , 11

Prob(it = 4%) = 13 for t = 2, . . . , 11

Prob(it = 6%) = 13 for t = 2, . . . , 11

294

In this model the interest rate is the same every year for t = 2, ...11 so theexpected value of the interest rate is

E[annual interest rate] = 13 (0.02 + 0.04 + 0.06) = 0.04

i.e. the expected interest rate is 4% pa.

We can also calculate the variance using the formula

V ar[it] = E[(it − E[it])2

]=

13

[(0.02− 0.04)2 + (0.04− 0.04)2 + (0.06− 0.04)2

]= 0.0002667

⇒ SD ≈ 0.0163

where SD = standard deviation.

295

Accumulation

Now, suppose we had invested £1 for 10 years at time 1, how much will we haveat the end of year 11?

We will have

S(1)11 =

1.0210 with probability 1/3

1.0410 with probability 1/3

1.0610 with probability 1/3

So the expected amount of money is:-

E[S(1)11 ] =

13

[1.0210 + 1.0410 + 1.0610] = 1.497

SD[S(1)11 ] = 0.234

Note: This is not the same as 1.0410 = 1.480 because the expected accumulationis not equal to the accumulation at the expected rate of interest.

296

Remark: Jensen’s inequality:

Let f : R 7→ R be a convex function (f ′′ ≥ 0) then:

E[f(X)] ≥ f(E[X]

)for all random variables X for which

E[|X|] <∞ and E[∣∣f(X)

∣∣] <∞Here: X = i and f(x) = (1 + x)10, f ′′(x) = 10× 9× (1 + x)8 > 0

E[f(X)] = E[(1 + i)10] = 1.497 > 1.480 = (1 + E[i])10

297

Present Value

How much should we invest at time 1 to have £1 in 11 years’ time ?

Answer: we should invest

A1 =

1.02−10 with probability 1/3

1.04−10 with probability 1/3

1.06−10 with probability 1/3

E[A1] = 0.685 6= 1.04−10 = 0.676SD[A1] = 0.107

298

15.2 Model 2

The interest rate each year will be

it =

2% p.a. with probability 1/3

4% p.a. with probability 1/3

6% p.a. with probability 1/3

The rate each year is independent of the rate in any other year. This is differentfrom Model 1 because rates in different years may be different.

More formally, we say that the interest rate in year t, t = 2, . . . , 11 is it where it isa sequence of i.i.d random variables with the above distribution.

[Note i.i.d = independent and identically distributed]

Whilst it is still not realistic to assume that an interest rate could randomly jumpfrom one year to the next to one of the three rates, it is perhaps more realisticthan assuming that the rate will be constant throughout the 10-year term.

In each year: E[it] = 0.04 and SD[it] = 0.0163 as for Model 1.

299

Accumulation

Consider £1 invested for 10 years at time 1.

How much is accumulated at the end of year 11 (after 10 years)?

It isS

(2)11 = (1 + i2)(1 + i3) . . . (1 + i11)

because at the end of the first year £1 is invested at rate i2 and grows to 1 + i2.This is invested the following year at rate i3, so the fund now grows to(1 + i2)(1 + i3) etc.

E[S

(2)11

]= E

[11∏t=2

(1 + it)

]=

11∏t=2

E[1 + it]

by independence

=11∏t=2

1.04 = 1.0410 ≈ 1.480

which is not the same as for Model 1.

300

Furthermore, we have

E[S

(2)2

11

]=

11∏t=2

E[(1 + it)2

]= E

[(1 + it)2

]10since the interest rates in each year are independent. SinceE[i2t]

= V ar[it] + E[it]2 (because V ar[it] = E[i2t ]− E[it]2) we obtain

E[(1 + it)2

]= E

[1 + 2it + i2t

]= 1 + 2E[it] + V ar[it] + E[it]2

= 1 + 2× 0.04 + 0.01632 + 0.042 = 1.081866

andE[S

(2)2

11

]= 1.08186610 = 2.19651 .

We therefore obtain for the variance of S(2)11

V ar[S

(2)2

11

]= E

[S

(2)2

11

]− E

[S

(2)11

]2= 2.19651− 1.0420 = 0.005388

⇒ SD[S

(2)11

]= 0.0734 (c.f SD

[S

(1)11

]= 0.234)

301

Present Values

How much should we invest at time 1 in order to have 1 at the end of year 11?

A(2)1 = (1 + i1)−1(1 + i2)−1 . . . (1 + i10)−1

(1 + it)−1 =

1.02−1 with probability 1/3

1.04−1 with probability 1/3

1.06−1 with probability 1/3

E[(1 + it)−1

]= 0.96178

E[(1 + it)−2

]= 0.92524

E[A

(2)1

]=

11∏t=2

E[(1 + it)−1

]by independence

= 0.9617810 = 0.67726 (c.f E[Y1] = 0.685)

302

E[A

(2)1

2]= E

[11∏t=2

(1 + it)−2

]=

11∏t=2

E[(1 + it)−2

]= 0.9252410 = 0.45977

V ar[A

(2)1

]= 0.00109

SD[A

(2)1

]= 0.033 (c.f SD[Y1] = 0.107)

303

15.3 Independent Annual Rates of Interest

Model

Let it be the rate of interest pa in year t. We assume that {it}∞t=1 is a sequence ofi.i.d random variables. Let Sn be the accumulation at time n of £1 invested attime 0 and let An be the accumulation at time n of:

£1 invested at time 0+ £1 invested at time 1

+...

+ £1 invested at time n− 1.

304

Therefore, An is the accumulation of a constant annuity of £1 payable in advanceand we clearly have

Sn = (1 + i1)(1 + i2) . . . (1 + in) =n∏t=1

(1 + it) .

And

An =

(1 + i1)(1 + i2) . . . (1 + in)

+ (1 + i2) . . . (1 + in)

+...

+ (1 + in)

=

n∑t=1

n∏k=t

(1 + ik)

Given a distribution for it, and knowledge of the moments of it:-(a) Can we calculate the distributions of An and/or Sn?

– In general, no.

(b) Can we calcualte the moments of An and/or Sn ?– Yes

305

The Moments of Sn

We can calculate the first moment.

E[Sn] = E

[n∏t=1

(1 + it)

]by independence

= {E[1 + it]}n as i.d

In general, the mth Moment is given by

E[Smn ] = E

[(n∏t=1

(1 + it)

)m]= E

[n∏t=1

(1 + it)m]

=n∏t=1

E[(1 + it)m] = {E[(1 + it)m]}n

⇒ All moments of Sn can be calculated if we know the distribution and momentsof it.

306

The Moments of An

The First Moment:

E[An] = E

[n∑t=1

n∏k=t

(1 + ik)

]

=n∑t=1

n∏k=t

E[1 + ik] (independence)

=n∑t=1

(E[1 + it])n−t+1 (identically distributed)

=n∑t=1

(E[1 + i1])t =n∑t=1

(1 + E[i1])t

= sn at i

wherei = E[it] = E[i1] = . . . = E[in]

307

The Second Moment - E[A2n]:

An = (1 + in)(1 +An−1)

An−1 and in are independent, and hence so are (1 +An−1) and (1 + in)

E[A2n] = E[(1 + in)2(1 +An−1)2] = E[(1 + in)2]E[(1 +An−1)2]

= E[1 + 2in + i2n]E[1 + 2An−1 +A2n−1]

= (1 + 2E[in] + E[i2n])(1 + 2E[An−1] + E[A2n−1])

The only unknown part of this is E[A2n−1] but we know E[A2

1] becauseA1 = (1 + i) so we can calculate the others recursively, using E[A2

1] we calculateE[A2

2] etc...

308

The mth Moment

We use a similar recursive method.

Amn = (1 + in)m(1 +An−1)m

E[Amn ] = E[(1 + in)m]E[(1 +An−1)m]

using the fact that the it are iid

= E[(1 + in)m]m∑k=0

(m!

k!(m− k)!

)E[(An−1)k]

using the Binomial expansion for (1 + x)m where, in this case, x = An−1.

We can calculate E[Amn ] recursively for n = 2, 3, . . . having first calculatedE[Akn−1] for k = 1, 2, . . . ,m.

Note that we have looked at accumulations, An, Sn in this section, but we couldjust as easily have looked at present values.

309

Example 15.1 The amount we need to invest now to pay £1 at times0, 1, 2, . . . , n− 1 is

1 + (1 + i1)−1 + (1 + i1)−1(1 + i2)−1 + . . .+ (1 + i1)−1 . . . (1 + in−1)−1

=n−1∑t=0

t∏k=0

(1 + ik)−1 (i0 = 0)

310

15.4 The Independent Log-Normal Model

In this sections we assume that (1 + it) ∼ log-normal with parameters (µ, σ2),that is, the force of interest δt = ln(1 + it) ∼ N(µ, σ2)

We can then use standard results (based on MGF of N(µ, σ2)):

E[1 + it] = E[exp δt] = exp(µ+

12σ2

)

E[(1 + it)m] = E[exp(mδt)] = exp(µm+

12m2σ2

)V ar[1 + it] = exp

(2µ+ σ2

)[exp

(σ2)− 1]

We are still assuming that {it}t≥1 are i.i.d. This is a special case of the model inthe previous section.

Note that (1 + it) is distributed over (0,+∞) so, although we can have negativeinterest rate it, we cannot have negative accumulations.

311

Reminder(1) X ∼ N(µ, σ2) then X−µ

σ ∼ N(0, 1)(2) X1, . . . , Xn i.i.d; Xi ∼ N(µ, σ2)⇒

∑ni=1Xi ∼ N(nµ, nσ2)

Mathematical Properties

Sn =n∏t=1

(1 + it) =⇒ ln(Sn) =n∑t=1

ln(1 + it) ∼ N(nµ, nσ2)

Hence Sn ∼ log-normal(nµ, nσ2)

Furthermore

E[Sn] = exp(nµ+

12nσ2

)(MGF again)

V ar[Sn] = exp(2nµ+ nσ2

) [exp

(nσ2

)− 1]

Let Xn be the present value of £1 payable at time n, then

Xn =n∏t=1

(1 + it)−1

312But

ln[(1 + it)−1] = − ln(1 + it) ∼ N(−µ, σ2)

HenceXn ∼ log-normal (−nµ, nσ2)

Conclusion - for the independent log-normal model the accumulation and thepresent value of a single payment have a known distribution.

313

Example 15.2 Assume the {it}∞t=1 are i.i.d and that (1 + it) ∼ log-normal (µ, σ2).Given that E[it] = 0.05 and V ar[it] = 0.0882,(i) Calculate µ and σ2

(ii) Calculate the probability S12 > 1.8(iii) Calculate E(A2) and V ar(A2)

Solution(i)

E[1 + it] = exp(µ+

12σ2

)= 1.05

V ar[it] = V ar[1 + it] = exp(2µ+ σ2

) [exp

(σ2)− 1]

= 0.0882

=⇒ 0.0882 = 1.052[

exp(σ2)− 1]⇒ σ2 ≈ 0.007000

µ+12σ2 = ln 1.05⇒ µ ≈ 0.04529

314

(ii)S12 ∼ log-normal (12µ, 12σ2)

prob[S12 > 1.8] = prob[ln(S12) > ln 1.8] = prob

ln(S12)− 12µ√12σ2︸ ︷︷ ︸

∼N(0,1)

>ln 1.8− 12µ√

12σ2︸ ︷︷ ︸≈0.153

= 1− Φ(0.153) ≈ 1− Φ(0.15) ≈ 1− 0.5596 = 0.4404

where Φ is the distribution function for the standard normally distributed randomvariable.

315

(iii)

E[A2] = s0.052

= 2.1525

A2 = (1 + i2)(1 +A1) = (1 + i2)(2 + i1)⇒ A22 = (1 + i2)2(2 + i1)2

=⇒ E[A22] = E[(1 + i2)2]E[(2 + i1)2]

=[V ar(1 + i2) + (E[1 + i2])2

]E[4 + 4i1 + i21

]=

[0.0882 + 1.052

][4 + 4(0.05) + V ar[i1] +

(E[i1]2

) ]≈ 4.6744

⇒ V ar[A2] = E[A22]− (E[A2])2 ≈ 0.04114

316

Example 15.3 {it}∞t=1 are i.i.d. Each it has the following distribution:

it =

0.02 with probability 1/3

0.05 with probability 1/3

0.08 with probability 1/3

X represents the accumulated amount at time 2 of £2 invested at time 0 and £1invested at time 1. Calculate E[X] and V ar[X].

317

Solution

X = 2(1 + i1)(1 + i2) + (1 + i2) = (1 + i2)(3 + 2i1)

E[it] =13

(0.02 + 0.05 + 0.08) = 0.05

E[i2t ] =13

(0.022 + 0.052 + 0.082) = 0.0031

E[X] = E[1 + i2]E[3 + 2i1] = 1.05(3 + 2(0.05)) = 3.255

E[X2] = E[(1 + i2)2]E[(3 + 2i1)2] = E[1 + 2i2 + i22]E[9 + 12i1 + 4i21]

= (1 + 2(0.05) + 0.0031)(9 + 12(0.05) + 4(0.0031)) = 10.6034

⇒ V ar[X] = 10.6034− 3.2552 = 0.0084