FM All Exercises

Australian School of Business School of Actuarial Studies

Financial Mathematics Exercises

S1 2012 17 March 2012

Contents 1 Time Value of Money and Cash Flow Valuation 1.1

2

Time Value of Money . . . . . . . . . . . . . . . . . . . . . . . . . . .

2

Exercise 1.1 [int1] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.2 [int2] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.3 [int3] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.4 [int4] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.5 [int5] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.6 [int7] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.7 [int8] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.8 [int9] . . . . . . . . . . . . . . . . . . . . . . . . .

2

Exercise 1.9 [int10]

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6

Exercise 1.24 [new10] . . . . . . . . . . . . . . . . . . . . . . .

6


i


1.2

1.3

Level Annuities

Actuarial Studies UNSW

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6

Exercise 1.25 [ann1] . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.26 [ann2] . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.27 [ann3] . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.28 [annB1] . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.29 [annB2] . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.30 [annB3] . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.31 [annB4] . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.32 [annB5] . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.33 [annB6] . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.34 [annB7] . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.35 [ann4] . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.36 [ann5] . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.37 [ann6] . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.38 [ann7] . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.39 [annB8] . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.40 [annB9] . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.41 [ann13] . . . . . . . . . . . . . . . . . . . . . . .

8

Non-Level and Continuous Annuities

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Exercise 1.42 [ann8] . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.43 [ann9] . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.44 [ann12] . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.45 [new8]

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Exercise 1.46 [ann14] . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.47 [ann15] . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 1.50 [annB10] Exercise 1.51 [loaB2]

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Exercise 1.52 [annB11] Exercise 1.53 [loaB1]

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Exercise 1.54 [annB12]

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ii




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Exercise 1.59 [ann11] . . . . . . . . . . . . . . . . . . . . . . .

10


2 Life Contingencies

11

Exercise 2.1 [lif1]

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Exercise 2.2 [lif2]

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Exercise 2.3 [lif3]

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Exercise 2.4 [lif4]

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Exercise 2.5 [lif5]

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3 Loans and Investments 3.1

3.2

13

Loan Repayments . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

13

Exercise 3.1 [loa1] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 3.2 [loa2] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 3.3 [loa3] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 3.4 [loaB3] . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 3.5 [loaB4] . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 3.6 [loaB5] . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 3.7 [loa5] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 3.8 [loa6] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 3.9 [loa7] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 3.10 [loa8] . . . . . . . . . . . . . . . . . . . . . . . .

15

Investments

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15

Exercise 3.11 [loa4] . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 3.12 [loa9] . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 3.13 [new13] . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 3.14 [loa10]

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Exercise 3.15 [loaB6]

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Exercise 3.16 [loaB7]

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Exercise 3.18 [ann16] . . . . . . . . . . . . . . . . . . . . . . .

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iii



4 Interest Rate Risk 4.1

4.2

Term Structure of Interest Rates

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18

Exercise 4.1 [irr1] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 4.2 [irr2] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 4.3 [irr3] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 4.4 [irr4] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 4.5 [irr5] . . . . . . . . . . . . . . . . . . . . . . . . .

19

Price Sensitivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

19

Exercise 4.6 [irr6] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 4.7 [irr7] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 4.9 [lif6] 4.3

18

Immunisation

Exercise 4.10 [irr8]

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Exercise 4.11 [irr9]

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Exercise 4.12 [irr10] . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 4.13 [new12] . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 4.14 [irr11] . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 4.15 [irr12] . . . . . . . . . . . . . . . . . . . . . . . .

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5 Derivatives 5.1

24

Forwards, Futures and Swaps

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24

Exercise 5.1 [der1]

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Exercise 5.2 [der2]

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Exercise 5.3 [der3]

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Exercise 5.4 [der4]

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Exercise 5.5 [der5]

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Exercise 5.6 [der6]

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Exercise 5.7 [der7]

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Exercise 5.8 [der8]

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Exercise 5.9 [der9]

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Exercise 5.10 [der10] 5.2

Options

Exercise 5.11 [der11]

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iv




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6 Stochastic Interest Rates 6.1

6.2

6.3

27

IID Returns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

27

Exercise 6.1 [sto1] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 6.2 [sto2] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 6.3 [sto3] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 6.4 [sto4] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 6.5 [sto5] . . . . . . . . . . . . . . . . . . . . . . . . .

28

Lognormal Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 6.6 [sto6] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 6.7 [sto7] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 6.8 [sto8] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 6.9 [sto9] . . . . . . . . . . . . . . . . . . . . . . . . .

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Exercise 6.10 [sto10]

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Dependence and Further Concepts

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7 Solutions to Exercises

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7.1

Module 1

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7.2

Module 2

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7.3

Module 3

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7.4

Module 4

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7.5

Module 5

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7.6

Module 6

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82

1

Module 1 Time Value of Money and Cash Flow Valuation 1.1

Time Value of Money

Exercise 1.1:

[int1] Determine the interest earned during the

invested today under compound interest with with

i = 0.05,

th

5

year by

$100

and under simple interest

i = 0.05.

Exercise 1.2:

grow to

$275

in

to accumulate to

$275

at

[int2] At what rate of compound interest will

$200

5 years?

Exercise 1.3:

[int3] How many years does it take

an eective annual rate of

Exercise 1.4: now of

$275

$200

5%?

[int4] With compound interest at

i = 0.05,

what is the present value

in 5 years?

Exercise 1.5:

[int5] If

$150

grows to

$240

in

n

years, what will

$1000

grow to over

the same period?

Exercise 1.6: at least

5%

[int7] If funds invested today will earn

8%

for the next 10 years and

for the following 10 years, what is the most one must invest today to

accumulate $1 million in 20 years?

Exercise 1.7:

[int8] What level rate of interest is equivalent to

years followed by

Exercise 1.8:

5%

8%

for the next 10

for the following 10 years?

[int9] Assuming an eective rate of

cash ows below at times

t=0

and

t = 3.

2

i = 0.10,

nd the value of the


Exercise 1.9: (assuming


[int10] The following cash ows have a value of

7.7217

at time

t=0

i = 0.05):

Find the value of the following cash ows at time 0:

Exercise 1.10:

[int11] If

$100

is deposited at time

t=0

into an account earning

10% interest and $20 is withdrawn at t = 1 and 2, then how much can be withdrawn at t = 3?

Exercise 1.11:

[int12] At what rate of interest will

$100

accumulate to

$200

in 6

years?

Exercise 1.12:

[int13] (SOA Course 2 May 2000, Question 1) Joe deposits $10

11% per deposited n years

today and another $30 in ve years into a fund paying simple interest of year. Tina will make the same two deposits, but the $10 will be from today and the $30 will be deposited

2n

years from today. Tina's deposits earn

an annual eective rate of 9.15%. At the end of 10 years, the accumulated amount of Tina's deposits equals the accumulated amount of Joe's deposits. What is the value of

n?

Exercise 1.13:

[int15] (SOA Course 2 Nov 2000, Question 2) The following table

shows the annual eective interest rates being credited by an investment account,

3



by calendar year of investment. The investment year method is applicable for the rst 3 years, after which a portfolio rate is used. Calendar Year of

Investment

Investment

Year Rates

i1 10% 12% 8% 9% 7%

1990 1991 1992 1993 1994

i2 10% 5% (x − 2)% 11% 7%

Calendar Year of

Portfolio

Portfolio Rates

Rate

i3 x% 10% 12% 6% 10%

1993 1994 1995 1996 1997

8% (x − 1)% 6% 9% 10%

An investment of $100 is made at the beginning of years 1990, 1991, and 1992. The total amount of interest credited by the fund during the year 1993 is equal to 28.40. What is the value of

Exercise 1.14:

x?

[int19] You wish to buy a new home theatre system and have two

potential payment options.

Option A.

You pay $610 down (at

following year (at

Option B.

You pay $560 down (at

following year (at

t = 0),

$475 next year (at

t = 1)

and $340 the

t = 0),

$580 next year (at

t = 1)

and $274 the

t = 2). t = 2).

Assuming a compound interest accumulation function, determine the values of the rate of interest

r

Exercise 1.15:

for which Option A is preferred to Option B.

[int18] In return for a single payment of $1000, an investment bank

oers the following alternatives:

A.

A lump sum of $1330 after three years

B.

A lump sum of $1550 after ve years

C.

Four annual payments, each of amount $425, the rst payment being made after ve years

You wish to decide what alternative is the best. (a) Write down an equation of value for each alternative and nd the yield for each. (b) Assume that an investor selects alternative A and that after three years she invests the proceeds for a further two years at a xed rate of interest. How large must this rate of interest be in order for her to receive at least $1550 at the end of 5 years? Why do we ask this question?

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(c) Assume that an investor selects alternative B and that after ve years he wants to compare his proceeds with the value of alternative C. Determine the value of alternative C in 5 years time. What interest rate should you use?

Exercise 1.16:

[int20] Use an equation of value to determine the level annual pay-

ment (in arrears) equal in value to $1,000,000 (at time

t = 0)

at an interest rate of

13% p.a. eective, allowing for 5 payments.

Exercise 1.17:

[int6] If

Exercise 1.18:

[int14] (SOA Course 2 May 2001, Question 12) Bruce and Robbie

v = 0.94,

what are

d

each open up new bank accounts at time 0.

and

i?

Bruce deposits $100 into his bank

account and Robbie deposits $50 into his. Each account earns an annual eective discount rate of

d.

The amount of interest earned in Bruce's account during the

11th year is equal to

X.

The amount of interest earned in Robbie's account during

the 17th year is also equal to

Exercise 1.19:

X.

What is the value of

X?

[int24] Tina issues a 2-year promissory note for a face value of

$6000 and receives $4843.30 in return (ie. repay $6000 after 2 years).

she borrows $4843.30 and promises to

At the end of 6 months, 1 year, and 18 months, she

deposits $1000, $1000, and $2000 into her bank account and earns the same interest rate as the implied rate on the promissory note. Assuming interest is compounded semi-annually, determine how much extra money (in addition to the amount in her bank account) she will need to redeem (repay) the note for its face value in 2 years time.

Exercise 1.20:

[int25] A trust account quotes a nominal annual interest rate of 6%.

Interest is credited quarterly, on the last day of each March, June, September and December. Simple interest is paid for amounts on deposit for less than a quarter of a year. In 2001, Maria made 4 deposits of $1000 into her trust account every 1st day of March, June, September, December. By 31 December 2005, how much interest will Maria have earned from these deposits?

Exercise 1.21:

[int21] You are given the following interest options:

A.

an eective rate of discount of 5% per annum

B.

a nominal rate of interest of 5% per annum convertible semi-annually

C.

a nominal rate of interest of 5% per annum convertible monthly

D.

a nominal rate of discount of 5% per annum converted semi-annually

E.

a nominal rate of discount of 5% per annum converted monthly

F.

a force of interest of 5%

5



(a) Discuss dierences between the above interest rates by expressing each option as eective rates of interest. (b) How much will an investment of $10,000 accumulate to in

4 12

years based on

each of the above interest options?

Exercise 1.22: with

t

δ(t) = 0.04(1 + t)−1 , a(t):

[int22] Assume that the force of interest is

measured in years. Using an accumulation function

(a) Calculate the equivalent eective rate of interest for the period

t = 1 to t = 2.

(b) Calculate the equivalent eective rate of interest for the period

t = 2 to t = 3.

(c) Find the value at time at time

k.

[int23] A fund credits simple interest with

After time

k,

(a) Find the value of

i = 10%

that maximises

k

8%.

a(4).

from (a), nd the force of interest as function of time

t

0 ≤ t ≤ 4).

Exercise 1.24:

[new10] Suppose the force of interest is δ = i(m) against m and (ii) d(m) against

(on the same graph) (i)

1.2

from time zero

the fund accumulates at a constant force of interest of

k

(b) Using the value of (for

of an investment that accumulates to $200,000

t = 4.

Exercise 1.23: to time

t=2

0.05. Using Excel, plot m, for 0.5 < m < 50

Level Annuities

Exercise 1.25:

[ann1] You are given a combined annuity-immediate payable monthly

such that payments are $1000 p.a. for the rst 6 years and $400 p.a. for the next 4 years together with a lump sum of $2000 at the end of the 10 years. An interest rate of 12% p.a convertible monthly is assumed. (a) Find the present value of this annuity. (b) Calculate the amount of the level annuity-immediate payable for 10 years having the same present value as the payments in (a).

Exercise 1.26:

[ann2] Bill leaves an inheritance to four charities, Faith Foundation

(F), Hope Institution (H), Love Trust (L) and Peace, Inc. (P). The total inheritance is a series of level payments at the end of each year forever. During the rst 20 years, F, H, and L share each payment equally. All payments after 20 years revert to P. The present values of the shares of the four charities are known to be all equal. What is the implied eective rate of interest?

6


Exercise 1.27:


[ann3] Cathy must pay o a loan with ve annual payments of

$15,000 each. The rst loan payment is due 10 years from now. In order to accumulate the funds, she plans on making ten annual deposits of

C

into an account paying

eective annual interest of 6%. Having computed the least possible amount

C

(and

assuming she succeeded in her nancial mathematics course and thus it took her a negligible amount of time), she immediately makes the rst deposit. Calculate

C.

Exercise 1.28:

[annB1] Exercise 2.1.11 from Broverman 5th Ed (2.1.11 in 4th Ed).

Exercise 1.29:

[annB2] Exercise 2.1.19S from Broverman 5th Ed (2.1.19S in 4th

Ed).

Exercise 1.30:


Exercise 1.31:


Exercise 1.32:


Exercise 1.33:


Exercise 1.34:


Exercise 1.35:

[ann4] To settle a $100,000 death benet, Tim, the primary ben-

eciary, opted to take an annuity-immediate payable monthly for 25 years.

The

monthly payment was calculated using an eective annual interest rate of 3%. After making payments for 10 years, the insurance company decides to increase the monthly payments for the remaining 15 years by changing the eective annual interest rate to 5%. By how much will the monthly payment increase?

Exercise 1.36:

[ann5] Find the present value of a set of cash ows which pay $100

at the end of year 1, $200 at the end of year 2, $100 at end of year 3, $200 at the end of year 4, and so on ($100 at odd years, $200 at even), with the nal payment being at the end of the 20th year. The interest rate is 5% p.a. semi-annual compounding.

Exercise 1.37:

[ann6] Bob has inherited an annuity-due on which there remain 12

payments of $18,000 per year at an eective discount rate of 10%; the rst payment is due immediately.

He wishes to convert this to a 20-year annuity-immediate at

the same eective rate of discount. What will be the size of the payments under the new annuity?

Exercise 1.38:

[ann7] Broverman 5th Ed: 2.2.13 (2.2.13 in 4th Ed). Also solve for

the case if Smith repays the loan over 5 years (monthly payments).

Exercise 1.39:


7



Exercise 1.40:


Exercise 1.41:

[ann13] Given that

1.3

δ(t) =

1 , 20−t

t ≥ 0,

nd

s10 .

Non-Level and Continuous Annuities

Exercise 1.42:

[ann8] A loan of $4000 is being repaid by a 30-year increasing

annuity-immediate where payments increase each year and payments are in arrears. The initial payment is

P,

each subsequent payment is

P

larger than the preceding

payment. The annual eective interest rate is 4%. Calculate the value of the future payments (ie. the loan principal outstanding) after the ninth payment. Compare your result with the initial loan amount and explain it. Is such a payment pattern likely to exist in reality? Why?

Exercise 1.43:

[ann9] Nicole, a UNSW Business part-time student, expects an

increasing amount of income as she advances through her program but will need to borrow to cover her university costs. Accordingly, she plans to borrow a decreasing annual amount from a student credit loan during her 5 years at university, and to repay the loan with increasing amounts for 15 years after graduation. She borrows amounts

5X, 4X, 3X, 2X

and

X

at the beginning of each of 5 years, where the last

payment is paid at the beginning of her nal year. At the end of the rst year after graduation she pays $500, then increases the amount by $200 each year until a nal payment of $3300. If the eective annual interest rate assumed is 5%, determine

Exercise 1.44:

X.

[ann12] Paulo is saving madly to buy his rst home ten years from

now. He deposits to a fund each January 1 and July 1 for the years 2004 through 2014. The deposit he makes on each July 1 will be 10.25% greater than the one on the immediately preceding January 1. The amount he deposits on each January 1 (except for January 1, 2004) will be the same amount as the deposit made on the immediately preceding July 1. The fund will be credited with interest at a nominal annual rate of 10%, compounded quarterly. On December 31, 2014, the fund will have a balance of $110,000, an amount Paulo considers is enough for a home deposit and other miscellaneous expenses. Determine Paulo's initial deposit to the fund.

Exercise 1.45:

[new8] Using Excel, nd the present value of a 30-year annuity t3 + ln(10t + 12) at the end of year t, assuming an eective

immediate which pays interest rate of 5%.

Exercise 1.46:

[ann14] Value the following set of cashows at a rate of 10% p.a.: 2 1 2 1 1 and time 1 , $30 at times 1,1 and 2 , $40 at times $10 at time , $20 at times 3 3 3 3 3 2, 2 32 and 3 13 , $50 at times 3 and 3 23 , and $60 at time 4.

Exercise 1.47:

[ann15] Mary purchases an increasing annuity-immediate for $50,000

that makes twenty annual payments as follows:

8


(i) (ii)

P, 2P, . . . , 10P


in years 1 through 10; and

10(1.05)P, 10(1.05)2 P, . . . , 10(1.05)10 P

in years 11 through 20.

The annual eective interest rate is 7% for the rst 10 years and 5% thereafter. Calculate

P.

Exercise 1.48:

[new4] Suppose the interest rate is a constant 5% p.a. eective, and

the ination rate is a constant 3% p.a. Determine the (initial) annual payment from a 20 year annuity-immediate which is purchased at the fair price with $10,000 in the case of: (a) a xed annuity (level payments) (b) an ination-indexed annuity (c) Show that (a) and (b) are equally fair. Explain your calculations.

Exercise 1.49:

[new1] A perpetuity has annual payments (in arrears) of 1, 3, 6,

10, 15, etc. For a constant force of interest of

δ = 0.05:

(a) Find the present value of the perpetuity analytically. (b) Verify your answer by nding the approximate present value using Excel by considering the 500 payments only. against time

t.

Plot the present value of each payment

Does this shape remind you of something?

Exercise 1.50:


Exercise 1.51:

[loaB2] Exercise 2.3.8 from Broverman 5th Ed (2.3.8 in 4th Ed).

Exercise 1.52:

[annB11] Exercise 2.3.15 from Broverman 5th Ed (2.3.15 in 4th

Ed).

Exercise 1.53:


Exercise 1.54:


Ed).

Exercise 1.55:


Ed).

Exercise 1.56:


Ed).

Exercise 1.57:


Ed).

9


Exercise 1.58:


[new2] Find the present value of a 10-year increasing annuity that

pays at an annual rate of 100, 200, . . . , 1000, given that the annual eective interest rate is 5% and: (a) payments are made annually in arrears (b) payments are made monthly in arrears (c) payments are made continuously

Exercise 1.59:

[ann11] A one-year deferred continuous varying annuity is payable 2 for 13 years. The rate of payment at time t is t − 1 per annum, and the force of 1 . Find the present value of the annuity. interest at time t is 1+t

10

Module 2 Life Contingencies Exercise 2.1:

[lif1]

Z

is the present value random variable for a special continuous

whole life insurance issued to

(x),

bt

paying

at death at

x+t

where:

bt = e0.05t For all

t,

it is given that

µx+t = 0.01

and

δt = 0.06.

Determine the expected value

and variance of

Z.

Exercise 2.2:

[lif2] Show that the following two denitions of the life annuity

are equivalent:

∞ X

a ¨k+1 k px qx+k =

∞ X

v k k px

k=0

k=0

Interpret both sides of the equation and explain why it has to be true.

Exercise 2.3:

[lif3] Prove the following identity:

d¨ ax + A x = 1

Exercise 2.4:

[lif4] You are given the following probabilities of death:

x

qx

0

0.10

1

0.05

2

0.10

3

0.20

4

0.40

5

0.70

6

1.00

Given a technical rate of interest of 5%, calculate by hand and using Excel: Pr[K(0) = k], e0 , Pr[K(2) = k], e2 , A2 , 2 A2 , A12:3 , A2:31 , A2:3 , a ¨ 2 , a2 , a ¨2:3

11

a ¨x


Exercise 2.5:


[lif5] (Gerber (1997), Exercise 17, p. 1356) Consider two independent

lives which are identical except that one is a smoker and the other is a non-smoker. It is known that: 1.

µx

2.

cµx

is the force of mortality for smokers for

and

c>1

is the force of mortality for non-smokers for

0 ≤ x < ω,

0 ≤ x < ω,

and

where

c

is a constant

Calculate the probability that the remaining lifetime of the smoker exceeds that of the non-smoker. Check for the reasonableness of your answer.

12

Module 3 Loans and Investments 3.1

Loan Repayments

Exercise 3.1:

[loa1] A bank decides to lend a company $15000 at a rate of interest

of 5% p.a. to be repaid by annual instalments over 5 years (in arrears). (a) Calculate the annual payment. (b) Calculate the loan outstanding at the end of the second payment using the retrospective method. (c) Calculate the loan outstanding at the end of the second payment using the prospective method. (d) At the end of the second year the bank tells you that from that time onwards the rate of interest charged is going to be increased to 7.5% p.a. If you still want to payo the loan by the end of the fth year what must your annual payment change to? (e) Having reviewed your company's free cash ows you decide that the amount calculated in (d) is not aordable. You renegotiate with the bank and they oer to extend your loan so that you can pay o the loan in an additional 4 years (instead of 3 years), at 7.6% p.a. What is the new annual repayment? (f ) Using Excel, setup a loan schedule for part (e) above.

Exercise 3.2:

[loa2] A loan of $20000 is to be repaid by 6 annual payments begin-

ning one year after the loan is made. The lender wants annual payments of interest only at a rate of 7% and repayments of the principal in a single lump sum at the end of 6 years. The borrower can accumulate the principal in a sinking fund earning an annual interest rate of 5%, and decides to do this by means of 6 level deposits starting one year after the loan is made. (a) What should the annual payment be?

13



(b) What if the sinking fund interest rate was 7%? (c) Suppose you can decide whether you can setup a sinking fund arrangement or to have a standard loan arrangement (ie. repay both capital and interest with each payment) to repay the loan. If the sinking fund rate was 5% which method would you prefer? (Hint: you should not need to do any extra calculations to decide this although you can use it to check your answer if you wish) (d) Model the cash ows of this sinking fund arrangement in a spreadsheet (for the 5% case).

Exercise 3.3:

[loa3] An individual borrows $5000 to buy a plasma TV. The sum

borrowed is repayable by 24 monthly instalments in arrears, which are calculated on the basis of a at rate of interest of 10% p.a. (a) Calculate the monthly repayment and the true (eective) annual rate of interest being charged. Do this by hand using Newton-Raphson with 5 iterations (starting at 10%), then using Excel with 10 iterations. (b) Just after making the 12th repayment, the outstanding loan is to be repaid. What is the outstanding balance which must be repaid at this time?

Exercise 3.4:


Exercise 3.5:


Exercise 3.6:


Exercise 3.7:

[loa5] A loan of $20000 is being repaid by monthly instalments of 1 years. Provide a schedule in 3 Excel showing the principal and interest contained in each of the last four monthly

principal and interest (18% p.a. nominal) over 8 instalments.

Exercise 3.8:

[loa6] A householder is paying o four debts by monthly payments

all at an eective rate of 1% per month (12% p.a. nominal). The monthly payments and respective terms to run are: Monthly Payment ($)

Terms to Run (Months)

4.36

11

17.20

15

35.00

12

20.24

18

The householder arranges a consolidation of these debts, with the total (sum) payments under the consolidated loan being equal to the total remaining payments under the existing loans.

14



Calculate the monthly instalment and the term to run of the consolidated loan so that the eective rate of interest involved will be unchanged.

Note that a nal

repayment may be required to ensure the loan is fully repaid. For this exercise, this nal repayment is assumed to be made one month after the last monthly instalment. You may nd Excel helpful in speeding up algebraic computation.

Exercise 3.9:

[loa7] Paul takes out a loan of $47,500 to purchase a new car. The

interest applicable is 12% p.a. (monthly compounding). Instead of paying o the loan using level instalments, he decides to pay it o using monthly payments over 3 years. The payments within each year is the same, but the payments in the 2nd year are 10% higher than the payments in the 1st year, and the payments in the 3rd year are 10% higher than the payments in the 2nd year. Set out the loan schedule for this loan in Excel.

Exercise 3.10:

[loa8] A recently married couple have decided to buy a new house

in Sydney. After an investigation of their nancial situation they nd that they will need to borrow $600,000 from the bank. The rate of interest charged is 6.75% p.a eective. (a) If they want to pay o the loan in 10 years using annual payments, how much would they have to pay in total over the 10 years? (b) If they want to pay o the loan in 10 years using monthly payments, how much would they have to pay in total over the 10 years? (c) Suppose they choose to follow (a). At the end of year 5 (just after the payment at time 5), interest rates increase to 7.25% p.a. eective. How much do they need to pay to settle the loan at that time?

3.2

Investments

Exercise 3.11:

[loa4] (McCutcheon & Scott, 1986, p. 158) A loan of $75,000 is to

be issued bearing interest at the rate of 8% per annum payable quarterly in arrears. The loan will be repaid at par (ie.

100 per 100 face value) in 15 equal annual

instalments, with the rst instalment being repaid ve years after the issue date. Find the price to be paid on the issue date by a purchaser of the whole loan who wishes to realise a yield of (a) 10% per annum eective, and (b) 10% per annum convertible half-yearly.

Exercise 3.12:

[loa9] (McCutcheon & Scott, 1986, p.

197) A loan of nominal

amount $500,000 was issued bearing interest of 8% per annum payable quarterly in arrears. The loan principal will be repaid at $105% by 20 annual instalments, each of nominal amount $25,000, the rst repayment being ten years after the issue date. An investor, liable to both income tax and capital gains tax, purchased the entire loan on the issue date at a price to obtain a net eective annual yield of 6%. Assume

15



that capital losses cannot oset capital gains for tax purposes. Find the price paid, given that his rates of taxation for income and capital gains are: (a) 40% and 30% respectively (b) 20% and 30% respectively Do this question in both Excel and

Exercise 3.13:

R.

[new13] Consider a 10-year bond of face value $100 and annual

coupons at a rate of 6%. (a) Write down an equation of value given the price of the bond is $80. (b) Using the Newton's Ralphson method in Excel, nd the yield to maturity of the bond using a precision of 0.01%. Use any appropriate initial estimate.

Exercise 3.14:

[loa10] (McCutcheon & Scott, 1986, p. 206) Two bonds (100 face

value) each have an outstanding term of four years. Redemption will be at par for both bonds. Interest is payable annually in arrears at the annual rate of 15% for the rst bond and 8% for the second bond. Interest payments have just been made and the prices of the bonds are $105.80 and $85.34 respectively. (a) Verify that an investor, liable for income tax at the rate of 35% and capital gains tax at the rate of 50% who purchases either of these bonds (but not both) will obtain a net yield on his transaction of 8% per annum. (b) Assume now that the investor is allowed to oset capital gains by capital losses. Show that, if the proportion of his available funds invested in the 8% bond is such that the overall capital gain is zero, he will achieve a net yield of combined transaction of 8.46% per annum.

Exercise 3.15:


Exercise 3.16:


Exercise 3.17:

[loa11] An investor purchased an Australian Government bond on

11 June 2006 paying a coupon 5.75% p.a with a maturity of 15 June 2011. The bond is ex-interest within 7 days prior to the coupon payment. Explain what is meant by ex-interest for an Australian government bond and describe the payments that the buyer will receive on an ex-interest Australian government bond. Determine the price paid for the Australian Government bond at a yield of 4.75% p.a on 11 June 2006.

Exercise 3.18:

[ann16] Outline the payments made on ination indexed bonds and

give an example of an investor who would invest in these nancial instruments.

16


Exercise 3.19:


[loa12] A loan of nominal amount $500,000 was issued bearing in-

terest of 8% per annum payable quarterly in arrears.

The loan will be repaid at

$110% by 10 annual instalments, each of nominal amount $50,000, the rst repayment being ten years after the issue date. An investor, liable to both income tax and capital gains tax, purchased the entire loan on the issue date at a price to obtain a net eective annual yield of 7%. Find the price paid, given that his rates of taxation for income and capital gains are both 15%. Do this question in Excel.

Exercise 3.20:

[loa13] A loan, on which interest is payable half-yearly, was issued

on 1 January 1974. The loan was to be redeemed with deferred annual payments (always on 1 January) in accordance with the following schedule: Amount redeemed

Redemption

in each year

rate

1 Jan 1984 to 1 Jan 1992 (inclusive)

$150 000

105%

1 Jan 1993 to 1 Jan 2003 (inclusive)

$250 000

110%

1 Jan 2004

$300 000

112%

Interest is payable at the rate of 7% p.a. until the payment on 1 July 2000 has been made and thereafter at 8% p.a. What was the issue price if a purchaser of the whole loan secured a yield of 6.5% p.a. eective on his or her investment? Do this question using Excel.

Exercise 3.21:

[loa14] A loan of nominal amount $1,200 was issued bearing interest

of 10% per annum payable annually in arrears. The loan will be repaid by 3 annual nominal payments of equal value, the rst repayment being two years after the issue date.

The actual repayment will be at $100% for the rst two instalments, and

$120% for the nal instalment. An investor, liable to both income tax and capital gains tax at 20%, purchased the entire loan on the issue date at a price to obtain a net eective annual yield of 8%. Find the price paid, given that it is greater than $1,200.

17

Module 4 Interest Rate Risk 4.1

Term Structure of Interest Rates

Exercise 4.1:

[irr1]

Consider the following spot interest rates that are quoted on a nominal p.a. basis (2) assuming interest compounds semi-annually (ie. they are i interest rates). Term (Years)

% p.a.

0.5

4.875180

1.0

5.031182

1.5

5.234408

2.0

5.448436

(a) Use these spot rates to calculate the value of a 6.75% bond paying semi-annual coupons maturing in two years time with a face value of $100. (b) Calculate the yield to maturity on this bond for the price calculated above. (c) Determine the par yield, as a semi-annual compounding yield, for one year and two year maturity bonds corresponding to the above rates. Interpret your result. (d) Determine the 6 month forward rates corresponding to these spot rates.

Exercise 4.2:

[irr2] Consider two 5 year bonds. One has a 9% coupon and sells for

101.00; the other has a 7% coupon and sells for 93.20. Find the price of a 5 year zero coupon bond.

Exercise 4.3:

[irr3] Let

s(t), 0 ≤ t ≤ ∞

denote a spot rate curve, that is, the t is e−s(t)t . Show explicitly that if the

present value of a dollar to be received at time spot rate curve is at and that

s(t) = r,

then all forward rates must be the same.

18


Exercise 4.4:


[irr4] The half year forward rates are as follows (semi-annual com-

pounding): Time Period

% p.a

0 0.5

5.00

0.5 1

5.50

1 1.5

6.00

1.5 2

6.10

2 2.5

6.25

2.5 3

7.00

Calculate the 1 year forward rates for time periods 0 1, 0.5 1.5, 1 2, 1.5 2.5, 2 3.

Exercise 4.5:

[irr5]

Consider the following spot rates (semi-annual compounding): Term (Years)

% p.a.

0.5

4.5000

1.0

5.2500

Time Period

% p.a.

1-1.5

7.5082

1.5-2

2.0290

and forward rates:

Calculate the value of a bond paying semi-annual coupons of 8% p.a., maturing in 2 years time.

4.2

Price Sensitivity

Exercise 4.6:

[irr6] Let

tinuous compounding)

δ

D(δ)

tinuously at the annual rate nominal in

n

be the duration, at a constant force of interest (con-

p.a., of a xed-interest security with interest payable con-

years time. Let

D per unit nominal and g = D/R. Show that: D(δ) =

Exercise 4.7:

redeemable at

R

per unit

¯a)n + nv n g(I¯ g¯ an + v n

[irr7] Consider a xed-interest security bearing interest of 5% p.a.

payable continuously and redeemable at par in

n

years time, where

n

is not neces-

sarily an integer. Assuming a constant continuously compounding force of interest of 7% p.a:

19



(a) Determine the duration of the security for

n = 20

and

n = 60.

(b) Note that the duration of the security, on the basis of a specied constant force of interest per annum

δ = 0.07,

δ , may be considered as a function of n.

Assuming

show that the duration is maximised when the following equation is

satised:

¯n ) = 0.07¯ an + 0.05(n − a

0.07 0.07 − 0.05

Hence (or otherwise) nd the maximum duration and the corresponding value of

n

at which the duration is maximised.

Exercise 4.8:

[new11] Consider a level

10-year annuity-immediate paying $1 at the

end of each year. (a) Write down the expressions which relate the modied duration and convexity with derivatives of the price (present value). (b) For

i = 5%,

nd the modied duration and the convexity.

(c) Using Excel and a Taylor's approximation, nd an approximate modied duration and convexity. Compare this with the answer in (b).

Exercise 4.9:

[lif6] Show that the Macaulay Duration of

Dx =

∞ X

a ¨x

is equal to:

wk k

k=0 where

4.3

wk

is given by:

vk k p wk = P∞ xl l=0 v l px

Immunisation

Exercise 4.10:

[irr8] (Boyle, 1992) Suppose the term structure of spot rates is level

for all maturities and equal to 8% p.a. Suppose that in the next instant, the term structure of interest rates will be either 9% p.a. for all maturities or 7% p.a. for all maturities. Consider the following strategy. An investor goes short a zero coupon bond with a 10-year maturity and a face value of 1000. Simultaneously, she uses

M5

and a

the value of liabilities

VL (i).

the proceeds to purchase a 5-year zero coupon bond with maturity value 15-year maturity bond with maturity value

M15 .

(a) Give expressions for the value of assets

VA (i) and

(b) What is meant by an arbitrage opportunity? (c) By suitable choice of

M5

and

M15

such that

VA (.08) − VL (.08)

is zero, demon-

strate that arbitrage prots are possible with parallel shifts to a at yield curve.

20


Exercise 4.11:


[irr9] Consider 3 coupon paying bonds

x, y

and

z.

You have

calculated their Price to be (Px , Py , Pz ), and their duration and convexity to be (Dx , Dy , Dz ) and (Cx , Cy , Cz ) respectively.

Consider a portfolio by buying 1 unit

of each bond. Derive a formula for the duration and convexity for the portfolio in terms of the price, duration, and convexity of the individual bonds.

Exercise 4.12:

[irr10] Consider a portfolio of insurance liabilities. Your best esti-

mate of the future outgoes (claims) are as follows Time

Outgo

1

3m

2

4m

3

3m

4

2m

Assume that the spot rate term structure is at and equal to 4.5%. Assume that the insurer can only invest in 2 ZCBs.

One matures in 0.5 years while the other

matures in 5 years. Find an immunisation strategy using the two bonds.

Using Excel, analyse: (a) What happens to the surplus if the yield shifts in a parallel fashion to 6.5%? (b) What happens to the surplus if the yield shifts in a parallel fashion to 2.5%? (c) What happens to the surplus if the yield curve twists and you are faced with the following spot rate curve? Time

Rate

0.5

3%

1

3.5%

1.5

4%

2

4.5%

2.5

5%

3

5.5%

3.5

6%

4

6.5%

4.5

7%

5

7.5%

(d) Determine the surplus if we are faced with a term structure where the

t-year

spot rate is given by:

st = α + β where

1 − e−λt λt

α = 0.06, β = 0.01, γ = −0.08, 21

+γ

and

1 − e−λt − e−λt λt

λ = 0.6

are known parameters.


Exercise 4.13:


[new12] The Nelson-Siegel class of term structure models are com-

monly used in practice to model the yield curve through time. One example is the model used in Exercise 4.12(e). Another simple example is the following yield curve, which describes the zero coupon yield of maturity

s τ = α + βt

1 − e−λτ λτ

(ie.

τ -year

spot rate) as:

λ are constant parameters and βt is a time-varying parameter. It is assumed that α = 0.06 and λ = 0.6, while the values of βt for each time t are 2 independent and identically distributed with βt ∼ N (−0.01, 0.002 ). where

α

τ

and

(a) Using Excel, plot the spot curve (for maturities parameters and when

βt

0 < τ < 20)

for the given

is equal to its mean of -0.01.

(b) Why is the use of Fisher-Weil duration more realistic? (c) Consider the portfolio of insurance liabilities in Exercise 4.12, and again assume that the insurer can only invest in 2 ZCBs of maturity 0.5 and 5 years respectively. Find an immunisation strategy if the current value of

βt

is -0.01.

(d) Simulate 1000 outcomes for the yield curve in the next moment and determine the surplus in each case. Also plot a histogram for the surplus. Is your portfolio fully immunised? Why?

Exercise 4.14:

[irr11] Consider a portfolio of insurance liabilities. Your best esti-

mate of the future outgoes (claims) are as follows Time

Outgo

1

3m

2

4m

3

3m

4

2m

Assume that the spot rate is at and is equal to 4%. available for investment two coupon bonds.

Suppose the insurer has

One is a 4% coupon bond with 0.75

years till maturity. The second bond is a 8% coupon bond with 8 years till maturity. Find an immunisation strategy using the two bonds. (Derive your solutions without using Excel.)

Exercise 4.15:

[irr12] Suppose you have a liability of $2 million due at time 3. As-

sume that the spot rate is at and is equal to 6%. You have available for investment ve ZCBs, with maturities at 1, 2, 3, 4 and 5 respectively. (a) Suppose you wish to use an immunisation strategy using 2 bonds. Derive the portfolio. (b) Explain the term cash ow matching. Derive the portfolio (using 1, 2, 3, 4 or all 5 bonds) that corresponds to a cash ow matching strategy.

22



(c) Suppose the spot rate moves to 7% at. What happens to your surplus for the strategies in (a) and (b)? (d) Suppose the spot rate curve such that the spot rate for maturity

(3 + T )%.

T

is equal to

What happens to your surplus for the strategies in (a) and (b)?

23

Module 5 Derivatives 5.1

Forwards, Futures and Swaps

Exercise 5.1:

[der1] On 12 May 1987, the closing value of the S&P 500 Index was

293.3 and the December 1987 S&P 500 futures closing index, with delivery in 210 days, was 299.0. Calculate the theoretical futures index assuming transactions and storage costs are negligible, a constant annual continuously compounding interest rate of 7%, and that the S&P 500 portfolio pays dividends continuously at an annual rate of 3.5% of its market value on 12 May 1987. You may also assume that interest rates are deterministic so that the futures price is equal to the theoretical forward price.

Exercise 5.2:

[der2] Consider a forward contract to buy 6.5% coupon 6 year Trea-

sury bonds in 2 years time (immediately after the coupon then due has been paid). These bonds are assumed to be currently available as 6.5% 8 year Treasury bonds at a yield of 6.96% p.a. (semi-annual). Funding costs for the rst year are 6.5% p.a. (monthly compounding) and 7% p.a. (monthly compounding) for the second year. Determine the forward price and forward yield in two years time.

Exercise 5.3:

[der3] An investor holds a short position in a forward contract on

gold for delivery in 90 days at $450 an ounce.

The current spot price of gold is $420

an ounce and insurance and storage cost for gold are 2.5% p.a of the spot price, paid on delivery. Ninety day (simple) interest rates are 9.75% p.a. What is the value of this forward contract?

Exercise 5.4:

[der4] Consider a forward contract on 10000 shares, deliverable in 6

months time. The share is currently trading at $10.00. Assume that there will be a dividend payment of 0.40 per share in 3 months time. Funding costs for 6 months are 6% p.a. (monthly compounding). Transaction costs are 2% of the value of the shares purchased. Determine the forward price for sale of the shares in 6 months at which all net funding and other costs will be covered.

24


Exercise 5.5:


[der5] Explain the cost of carry formula:

Ft,T = St er(T −t) − der(T −t1 ) There are no storage costs but there is a payment to the holder of the spot asset (eg. dividend) of

Exercise 5.6:

d

at time

t1 .

[der6] Suppose the current spot and forward rates are as given in

Table 6.4 of Sherris (1996, p. 109). Calculate the implied repo rate (the risk-free rate implied by current spot and forward prices) associated with each forward contract (corresponding to the two 90-day forward rates). Also, for each forward contract, outline an arbitrage strategy you could use to realise your arbitrage prot

now.

Give

the amount of the prot in both instances. Use simple interest as the contracts are for terms less than one year.

Exercise 5.7:

[der7] Review Example 7.1 of Sherris (p. 133)

Exercise 5.8:


Exercise 5.9:


Exercise 5.10:

[der10] Consider an agreement where party A receives the spot

N units of a commodity each period while paying a xed amount X per N units. If the agreement is made for M periods (ie. at times t1 , t2 , .., tM ), derive a formula that can be used to determine the swap price X on inception of the

price for unit for

swap contract. Assume that no storage costs or dividends occur during the period, and the risk free interest rate is

5.2

r

(continuous compounding).

Options

Exercise 5.11:

[der11] In a one-period binomial model, it is assumed that the

current share price of

260

will either increase to

of one year. The annual risk-free interest rate is

285 or decrease to 250 at the 5% compounded continuously

end and

assume that this share pays no dividends. (a) Calculate the price of a one-year European call option with a strike price of

275

by replicating the payo with a portfolio of shares and bonds.

(b) Calculate the price of a one-year European put option with a strike price of

275

by replicating the payo with a portfolio of shares and bonds.

(c) Verify numerically that the put-call parity relationship holds in this case.

Exercise 5.12:

[der12] Assume that the stock price is currently $50, and will in-

crease or decrease by 10% at the end of the month.

The interest rate is 5% p.a.

(simple). Find the price of a call option with a strike price of $50.

25


Exercise 5.13:


[der13] In the pricing of forwards, futures, swaps and options, the

expected value of the underlying asset for these contracts has not appeared in the valuation. Explain why this is the case. (Do you nd it surprising?)

Exercise 5.14:

[der14] In a one-period binomial model, it is assumed that the

current share price of

260

will either increase to

of one year. The annual risk-free interest rate is

285 or decrease to 250 at the 4% compounded continuously

end and

assume that this share pays no dividends. Calculate the price of an option that pays the cash dierence between the square of the share price at the end of the year and

70225,

provided that the dierence is positive (ie. otherwise it pays nothing).

Exercise 5.15:

[der15] The current stock price is $20, and the risk free rate (simple)

is 5% p.a. One year call and put options with strike price $22 are priced at $1.2245 and $2.5000 respectively. Verify that there is an arbitrage opportunity, and identify the transactions required.

Exercise 5.16:

[der16] Suppose we want to price a call option on a share using

a binomial model of the share price. Consider a portfolio of the share and bond. Suppose however that the stock pays a xed dividend of $D on the maturity date of the option (ie at time

T)

and that the owner of the option will not receive the

share dividend. Derive a formula for the number of stocks and bond that need to be held to replicate this option payo.

Exercise 5.17:

[der17] Consider call and put (European) options (with the same

strike price) on gold. The spot price of gold is the option is at time

T.

G(0)

at time 0, and the maturity of

Storage costs of $c per unit of gold are payable at time

T.

By equating the cost at time 0 of two portfolios that have the same payo at time

T,

nd an updated version of the put-call parity that takes into account the storage

costs.

26

Module 6 Stochastic Interest Rates 6.1

IID Returns

Exercise 6.1:

[sto1] (Institute of Actuaries Examination April 2003) $1000 is in-

vested for 10 years. In any year, the yield on the investment will be 4% with probability 0.4, 6% with probability 0.2, 8% with probability 0.4, and is independent of the yield in any other year. (a) Calculate the mean accumulation at the end of 10 years. (b) Calculate the standard deviation of the accumulation at the end of 10 years. (c) Without carrying out any further calculations, explain how your answers to (a) and (b) would change (if at all) if: (i) the yields had been 5%, 6% and 7% instead of 4%, 6%, and 8% p.a. respectively. (ii) the investment had been made for 12 years instead of 10 years.

Exercise 6.2:

[sto2] (Institute of Actuaries Examination September 2003) In any

year t, the yield on a fund of investments has mean

jt

and standard deviation

st .

In

any year, the yield is independent of the yield in any other year. The accumulated value, after

n

years, of a unit sum of money invested at time 0 is

(a) Derive formulae for the mean and variance of years

Sn

if

jt = j

Sn . and

st = s

for all

t.

(b) Calculate the expected value of

S8

if

(c) Calculate the standard deviation of

Exercise 6.3:

j = 0.06.

S8

if

j = 0.06

and

s = 0.08.

[sto3] (Institute of Actuaries Examination September 2002) $10,000

is invested in a bank account which pays interest at the end of each year. The rate of interest is xed randomly at the beginning of each year and remains unchanged

27



until the beginning of the next year. The rate of interest applicable in any one year is independent of the rate applicable in any other year. During the rst year, the rate of interest per annum eective will be one of 3%, 4% or 6% with equal probability. During the second year, the rate of interest per annum eective will be either 5% with probability 0.7, or 4% with probability 0.3. (a) Assuming that interest is always reinvested in the account, calculate the expected accumulated amount in the bank account at the end of two years. (b) Calculate the variance of the accumulated account in the bank account at the end of two years.

Exercise 6.4:

[sto4] (Institute of Actuaries Examination April 2005) In any year,

the interest rate per annum eective on monies invested with a given bank is equally likely to be i1 or i2

(i1 > i2 ),

and is independent of the interest rates in all previous

years. (a) Express the mean and variance of the eective rate in a particular year in terms of

i1

and

i2 .

(b) The accumulated value at time bank at time $0.5 million.

Exercise 6.5:

t = 25

t = 0 has expected Find i1 and i2 .

years of $1 million invested with the

value $5.5 million and standard deviation

[sto5] (Institute of Actuaries Examination September 2000) An in-

surance company calculates the single premium for a contract paying $10,000 in ten years time as the present value of the benet payable at the expected rate of interest it will earn on its funds. The annual eective rate of interest over the whole of the next ten years will be 7%, 8% or 10% with probabilities 0.3, 0.5 and 0.2 respectively. (a) Calculate the single premium. (b) Calculate the expected prot at the end of the term of the contract.

6.2

Lognormal Model

Exercise 6.6:


surance company has just written contracts that require it to make payments to policyholders of $1,000,000 in ve years time. The total premiums paid by policyholders amounted to $850,000. The insurance company is to invest half the premium income in xed interest securities that provide a return of 3% per annum eective. The other half of the premium income is to be invested in assets that have an uncertain return. The return from these assets in year t, it , has a mean value of 3.5% per annum eective and a standard deviation of 3% per annum eective. The random variables

(1 + it )

(for

t = 1, 2, . . .)

are independent and lognormally distributed.

28



(a) Deriving all necessary formulae, calculate the mean and standard deviation of the accumulation of the premiums over the ve-year period. (b) A director of the company suggests that investing all the premiums in the assets with an uncertain return would be preferable because the expected accumulation of the premiums would be greater than the payments due to the policyholders. Explain why this still may be a more risky investment policy.

Exercise 6.7:

[sto7] (Institute of Actuaries Examination April 2002) A company

is adopting a particular investment strategy such that the expected annual eective rate of return from investments is 7% and the standard deviation of annual returns is 9%. Annual returns are independent and

it

is the return in the

tth

(1 + it ) is lognormally distributed where

year. The company has received a premium of $1,000 and

will pay the policyholder $1,400 after 10 years. (a) Calculate the expected value and standard deviation of an investment of $1,000 over 10 years, deriving all formulae that you use. (b) Calculate the probability that the accumulation of the investment will be less than 50% of its expected value in ten years time. (c) The company has invested $1,200 to meet its liability in 10 years time. Calculate the probability that it will have insucient funds to meet its liability.

Exercise 6.8:

[sto8] (Institute of Actuaries Examination September 2007) The ex-

pected eective annual rate of return from a bank's investment portfolio is 6% and the standard deviation of annual eective returns is 8%. The annual eective returns are independent and in year

(1 + it )

is lognormally distributed, where it is the return

t.

Deriving any necessary formulae: (a) Calculate the expected value of an investment of $2 million after ten years. (b) Calculate the probability that the accumulation of the investment will be less than 80% of the expected value.

Exercise 6.9:


vestment bank models the expected performance of its assets over a ve-day period. Over that period, the return on the bank's portfolio, and standard deviation of 0.2%. Calculate the value of

j

(1 + i)

i,

has a mean value of 0.1%

is lognormally distributed.

such that the probability that

i

is less than or equal to

j

is

0.05.

Exercise 6.10:

[sto10] (Institute of Actuaries Examination September 2004) The

expected annual eective rate of return from an insurance company's investments is

29



6% and the standard deviation of annual returns is 8%. The annual eective returns are independent and

tth

(1 + it ) is lognormally distributed,

where it is the return in the

year.

(a) Calculate the expected value of an investment of $1 million after ten years. (b) Calculate the probability that the accumulated of the investment will be less than 90% of the expected value.

6.3

Dependence and Further Concepts

Exercise 6.11:

[new3] Company A and Company B currently have a swap contract

where A pays (annually) a oating interest rate on a principal of $1 million to B in exchange for a xed rate of 5% from B to A. The oating rate assumed to be independent for each year, and with

it

(1 + it )

it

in year

t

is

is lognormally distributed

having mean 4% and standard deviation 2%.

(a) Find the expected value of the accumulated value after 10 years of the net receipts (from the perspective of A) from the swap. (b) Using Excel, simulate (1000 times) the interest rate for the next 10 years and verify your answer in (a).

In addition, nd the variance and plot the

distribution of the accumulated value. (c) In what situation may a company wish to enter into a xed-for-oating swap?

Exercise 6.12:

[new5] Suppose that the interest rate

yt

(for year t) follows a mean

reverting process dened by:

yt = µ + β(yt−1 − µ) + σεt where

εt ∼ N (0, 1)

iid for each time

µ = 0.05, β = 0.4, years time as S1 . Find:

It is also known that value of $1 in one (a)

t = 0, 1, . . .. and

σ = 0.01.

Denote the accumulated

E(S1 )

(b) Var(S1 ) (c)

Pr(S1 < 1.04)

if the interest rate last year was (i) 4% and (ii) 6%.

Exercise 6.13:

[new6] Suppose that the interest rate

yt

(for year t) follows a mean

reverting process dened by:

yt = µ + β(yt−1 − µ) + σεt It is also known that

µ = 0.05, β = 0.4, and σ = 0.01. Denote the accumulated S10 . Using 1000 simulations in Excel, estimate:

value of $1 in ten years time as

30


(a)


E(S10 )

(b) Var(S10 ) (c)

Pr(S10 < 1.55)

(d) Plot the histogram of

S10 .

if the interest rate last year was (i) 4% and (ii) 6%.

Exercise 6.14:

[new9] Suppose that the interest rates in each year are independent (1 + it ) ∼ LN (µ, σ 2 ) and it having mean 4% and

and identically distributed, with

standard deviation 2%. Denote as s ¨30 the expected value and variance of the 30 year accumulated value of an annual payment of $1 in advance. (a) Derive recursive formulae which can be used to nd (b) Using Excel and the formulae in (a), nd (c) Estimate

E(¨ s30 )

E(¨ s30 )

E(¨ s30 )

and Var(¨ s30 ).

and Var(¨ s30 ).

and Var(¨ s30 ) using simulation in Excel, and compare your

answers with those in (b).

31

Module 7 Solutions to Exercises 7.1

Module 1

Exercise 1.1

[int1]

Under compound interest: Accumulated amount at Accumulated amount at

t = 4: 100(1.05)4 = 121.5506 t = 5: 100(1.05)5 = 127.6282

Interest earned = 6.08 Under simple interest: Accumulated amount at Accumulated amount at

t = 4: 100(1 + 4 · 0.05) = 120 t = 5: 100(1 + 5 · 0.05) = 125

Interest earned = 5.00

Exercise 1.2

[int2]

200(1 + i)5 = 275 ⇒ i =

Exercise 1.3

5

ln(275/200) ln 1.05

= 6.527

years

[int4]

275v = 215.47,

Exercise 1.5

− 1 = 0.06576

[int3]

200(1.05)t = 275 ⇒ t =

Exercise 1.4

275 1/5 200

where

v=

1 1+i

[int5]

n

150(1 + i) = 240 ∴ 1000(1 + i)n = 1000 150(1 + i)n = 150

Exercise 1.6

1000 150

· 240 = 1600

[int7]

The most one requires will be the amount

X

which is sucient even when the

interest rate is always at its minimum of 5% in the latter 10 years (as this minimises

32



the interest earned). Therefore: 10 10

X(1.08) (1.05) ∴ X = 284360

Exercise 1.7

= 1000000

[int8]

The accumulation of $1 under the eective rate and the 8% and 5% rates should be equivalent: 20

(1 + i) = (1.08)10 (1.05)10 ∴ i = 0.06489

Exercise 1.8

[int9]

t = 0: 5 + 3v + v = 8.2307 Value at time 2 3

t = 3: 8.2307(1 + i) = 10.9550 Value at time 3

Alternatively:

5(1 + i)3 + 3(1 + i) + 2 = 10.9550

Exercise 1.9

[int10]

The 2nd set of cash ows contains two of the 1st set, one starting at

t = 1. Thus, 7.7217 + v7.7217 = 15.0757

another starting at

Exercise 1.10

t = 0

the value is given by:

[int11]

100 (1.10)3 − 20 (1.10)2 − 20 (1.10) = 86.90

Exercise 1.11

[int12]

6

100(1 + i) = 200 ∴ i = 0.12246

Exercise 1.12

[int13]

(SOA Course 2 May 2000, Question 1) Accumulated value of 1st account after 10 years:

10(1 + 0.11 · 10) + 30(1 + 0.11 · 5) = 67.50 Therefore, accumulated value of 2nd account after 10 years must also be 67.50:

10(1.0915)10−n + 30(1.0915)10−2n = 67.50 ∴ 30(1.0915)10 v 2n + 10(1.0915)10 v n − 67.50 = 0

33

and


n

∴v =

−10(1.0915)10 ±

= 0.8158 Since

v n > 0,

we have

Exercise 1.13


p

100(1.0915)20 + 120(1.0915)10 · 67.50 60(1.0915)10

− 1.1491

or

v n = 0.8158 ⇒ n =

ln 0.8158 ln v

= 2.325

[int15]

(SOA Course 2 Nov 2000, Question 2) For the rst investment (the $100 at beginning of 1990), the interest credited in 1993 is 8% of the accumulated value at the beginning of 1993 (ie. after 3 years):

100(1.10)(1.10)(1 + 0.01x)(0.08) = 9.68 + 0.0968x Similarly, for the second investment, the interest credited is:

100(1.12)(1.05)(0.10) = 11.76 and for the third investment:

100(1.08)(0.01(x − 2)) = 1.08x − 2.16 Therefore, the total interest credited in 1993 is:

19.28 + 1.1768x = 28.40 ∴ x = 7.7498

Exercise 1.14 Using

v=

[int19]

1 , we have the following present values for the two payment options: 1+r

P VA = 610 + 475v + 340v 2 P VB = 560 + 580v + 274v 2 P VB − P VA = −50 + 105v − 66v 2 > 0. Note that the quadratic −50 + 105v − 66v = 0 has no real roots. The quadratic is negative at v = 1, since −50 + 105 − 66 = −11 < 0. This implies that P VB < P VA for all values of r , so Option B is always preferred to A.

Option A is preferred when

2

Exercise 1.15 [int18] (a) The equation of value

for each of the alternative is given as follows with the

corresponding yield (rate of interest)

r:

Alternative A:

1000(1 + r)3 = 1330 r = 9.9724% 34



Alternative B:

1000(1 + r)5 = 1550 r = 9.1607% Alternative C:

1 + 1000(1 + r)5 = 425 1 + 1+r

1 1+r

2

+

1 1+r

3 !

r = 8.5761% Note: for C, the yield must be found numerically (eg. using Solver in Excel - See Spreadsheet int18.xlsx)

(b) We need the accumulated value of $1330 at a rate r

to be at least $1550:

1330(1 + r)2 ≥ 1550 ∴ 1330r2 + 2660r − 220 ≥ 0 For equality, we have:

−2660 ±

r=

√

26602 + 4 · 1330 · 220 = 0.079543 2660

or

− 2.0795

∴ r ≥ 7.9543% We ask this question to compare A and B. If we chose A, then after 3 years, we would need to invest the $1330 at some interest rate (eg. by putting the money in the bank) for 2 years, after which it can be compared with the $1550 from B (at 5 years). Thus, we must be able to earn at least

7.95%

during this 2 year period for

A to be a better choice than B.

(c) The value of alternative C is: 1 425 1 + + 1+r

1 1+r

2

+

1 1+r

3 ! = 1508.97

which is less than the value of B ($1550). Note that we have used the interest rate from alternative C.

Exercise 1.16

[int20]

Equation of value:

P (v + v 2 + v 3 + v 4 + v 5 ) = 1000000 where

v=

1 . Therefore, we obtain 1.13

Exercise 1.17

P = 284314.54.

[int6]

d = 1 − v = 0.06 1 v = 1+i ⇒ i = v1 − 1 = 0.06383 Also:

i = d/v = 0.06383 35


Exercise 1.18


[int14]

(SOA Course 2 May 2001, Question 12) Note the relationship between accumulating using eective interest and discount rates:

1 + i = (1 − d)−1 Thus:

X = 100[(1 − d)−11 − (1 − d)−10 ] = 50[(1 − d)−17 − (1 − d)−16 ] For clarity, denote

R = (1 − d)−1 . 100R10 (R − 1) = 50R16 (R − 1) 2R10 = R16 R6 = 2 R = 21/6 ∴ X = 100(R11 − R10 ) = 38.8793

Exercise 1.19

[int24]

An equation of value to solve for the eective semi-annual interest rate

i

is

4843.30(1 + i)4 = 6000 whose only positive solution is

i = 0.055.

The total value of the three payments at

the time the note matures is

1000(1.055)3 + 1000(1.055)2 + 2000(1.055) = 4397.27 so she will need a top up of $1602.73 to redeem the note for $6000.

Exercise 1.20

[int25]

Each deposit accumulates under simple interest for 1 month, then under compound interest for the remaining quarters. Therefore, by the end of December 2005: March deposit accumulates to:

1 1000 1 + 0.06 12

19 0.06 1+ = 1333.59 4

June deposit accumulates to:

1 1000 1 + 0.06 12

18 0.06 1+ = 1313.88 4

September deposit accumulates to

1 1000 1 + 0.06 12

17 0.06 1+ = 1294.46 4 36



December deposit accumulates to

1 1000 1 + 0.06 12

16 0.06 1+ = 1275.33 4

Therefore, the total interest would be:

1333.59 + 1313.88 + 1294.46 + 1275.33 − 4000 = 1217.25

Exercise 1.21

[int21]

Using the relations for equivalent rates

m −m d(m) i(m) −1 = (1 − d) = 1 − = eδ 1+i= 1+ m m we have: Given rate

Equivalent eective rate

d = 0.05 i(2) = 0.05 i(12) = 0.05 d(2) = 0.05 d(12) = 0.05 δ = 0.05

Exercise 1.22

A(4.5) = 10000(1 + i)4.5 12596.32 12488.63 12517.38 12559.10 12529.11 12523.23

i

0.0526316 0.0506250 0.0511619 0.0519395 0.0513809 0.0512711

[int22]

Note that the accumulation function is given by:

Z a(t) = exp

t

δ(s)ds

Z = exp

0

(a)

(b)

0

t

0.04 ds 1+s

= (1 + t)0.04

a(2) − a(1) (3)0.04 − (2)0.04 = = 0.01635084 a(1) (2)0.04 a(3) − a(2) (4)0.04 − (3)0.04 = 0.01157375 = a(2) (3)0.04

(c) Since A(t) = A(0)(1 + t)0.04 , then A(0) = A(2) =

200000 (5)0.04

= 187530.19.

200000 0.04 (3) = 195954.86 (5)0.04

37

Therefore:


Exercise 1.23

[int23]

We have

Hence

(a) ie.

To maximize

k < 4;


( 1 + 0.1t a(t) = (1 + 0.1k) exp [0.08(t − k)]

t≤k t>k

( (1 + 0.1k) exp [0.08(4 − k)] a(4) = 1.4

k<4 k≥4

a(4),

dierentiate w.r.t.

k

(assuming

a(4) = (1 + 0.1k)e0.08(4−k) ,

will need to check this)

d a(4) = 0.1 exp [0.08(4 − k)] − 0.08 (1 + 0.1k) exp [0.08(4 − k)] dk and set to be zero. Solving for

k

(and noting that the exponential is positive) we

have:

0.1 = 0.08 (1 + 0.1k) ∴ k = 2.5 k = 2.5, a(4) > 1.4, so assumption of k < 4 is OK. Note that if the assumption of k < 4 was violated, then a(4) = 1.4 as opposed to our earlier expression 0.08(4−k) of (1 + 0.1k)e , and a(4) will have a maximum of 1.4 (at any k ≥ 4).

Check: at

(b) ( 1 + 0.1t t ≤ 2.5 a(t) = 1.25e0.08(t−2.5) t > 2.5 ( ln(1 + 0.1t) t ≤ 2.5 ln a(t) = ln 1.25 + 0.08(t − 2.5) t > 2.5 ( 0.1 t ≤ 2.5 d ln a(t) = 1+0.1t ∴ δ(t) = dt 0.08 t > 2.5

Exercise 1.24

[new10]

We use the result:

m −m i(m) d(m) −1 1+i= 1+ = (1 − d) = 1 − = eδ m m Therefore:

i(m) = m eδ/m − 1 d(m) = −m e−δ/m − 1 See spreadsheet new10.xlsx for plotting

i(m)

the one below.

38

and

d(m) .

The plot should be similar to



Nominal Interest/Discount Rates

0.046

0.048

0.050

Nominal Rate

0.052

0.054

i(m) d(m)

0

10

20

30

40

50

Compounding Frequency

Note that

δ = 0.05

and all the points on this plot yield the same eective rate of

interest!

Exercise 1.25

[ann1]

(a) Each monthly payment is rate

i

1 of the nominal amount. The eective monthly 12

is given by:

i=

0.12 = 0.01 12

Therefore:

72 120 1000 1 400 1 PV = a + a + 2000 12 72 0.01 1.01 12 48 0.01 1.01 72 120 1 72 1 48 ) ) 1 400 1 − ( 1.01 1 1000 1 − ( 1.01 + + = 2000 12 0.01 1.01 12 0.01 1.01 = 4262.50 + 618.34 + 605.99 = 5486.80 Alternatively, we can obtain the annual eective rate

j = 1.0112 − 1 = 0.1268

39

j:



Then, using annuities payable monthly:

6 10 1 1 (12) + 400a4 + 2000 PV = 1+j 1+j 6 10 1 − vj6 1 − vj4 1 1 = 1000 + + 400 2000 0.12 1+j 0.12 1+j = 5486.80 (12) 1000a6

(b) Assume level annuity payments payable monthly, then

X

1 120 ) 1 − ( 1.01 = 5486.80 0.01 ∴ X = 78.720

The quoted annual payment is then

Exercise 1.26

78.720 × 12 = 944.64.

[ann2]

The equation of value is based on the present value of the amount received by each charity (which are known to be equal):

P a = v 20 P a∞ 3 20 P 1 − v 20 P ∴ = v 20 3 i i 20 20 1 − v = 3v v 20 = 0.25 ∴ i = (0.25)−1/20 − 1 = 0.07177

Exercise 1.27

[ann3]

The accumulated value of the deposits 10 years after the rst deposit is:

AV = C s¨10 0.06 = C(1.06)10

1−

10 1 1.06

!

0.06

= 13.1808C The present value of the loan payments as at the end of 10 years is:

P V = 15000¨ a5 0.06 1 1 − 1.06 0.06

= 15000

5 !

= 63185.4568 where we have used a rate of 6% because that is the amount paid by the account (so the remaining funds in the bank will be earning 6% between years 10 and 15). Equating these values and solving for

C,

we obtain

40

C = 4793.75.



Exercises 1.281.34 [annB1annB7] See the Mathematics of Investment and Credit solutions manual.

Exercise 1.35

[ann4]

The original and nal monthly eective interest rates

i

and

j

are given by:

(1 + i)12 = 1.03 ⇒ i = 0.00246627 (1 + j)12 = 1.05 ⇒ j = 0.00407412 The original monthly payment

P

is given by:

P a300 i = 100000 ⇒ P =

100000 = 472.1087 a300 i

After 10 years, the remaining value of the annuity is:

P a180 i This will also be the present value of the new annuity (with annual payments of which is valued at the new interest rate

P 0 ),

j:

P a180 i = P 0 a180 j Therefore, the new payment is:

P0 = P

a180 i = 538.1869 a180 j

Thus, the payment increase is:

P 0 − P = 66.08

Exercise 1.36

[ann5]

The 1-year eective rate

i

and 2-year eective rate

j

are given by:

i = 1.0252 − 1 = 0.050625 j = 1.0254 − 1 = 0.103813 By drawing a cash ow diagram, it can be seen that the cash ow stream is a 20-year annuity with annual payments of $100, plus an additional $100 every 2nd year (ie. an additional 20-year annuity with biannual payments of $100). Thus:

P V = 100a20 i + 100a10 j = 1844.16

Exercise 1.37

[ann6]

A discount rate of 10% p.a. is equivalent to an interest rate of Therefore, the current cash ows are worth:

18000¨ a12 i = 18000 (1 + a11 i ) = 129162.10 We want annual payments of

X

where:

129162.10 = Xa20 i ∴ X = 16337.69 41

i = 11.11%

p.a.


Exercise 1.38


[ann7]

See solutions manual. For 5 years, monthly payment required is 296.94.

Exercises 1.391.40 [annB8annB9] See the solutions in the Broverman text and solutions manuals.

Exercise 1.41 s10

[ann13]

is the accumulated value (at time 10) of $1 paid at each time

t = 1, . . . , 10.

The

n is given by: 1 dt = exp [− ln(20 − 10) + ln(20 − n)] 20 − t 20 − n = exp ln 10 20 − n = 10

accumulated value (at time 10) of a single $1 paid at time

10

Z exp n

Therefore:

s10 =

10 X 20 − n n=1

Exercise 1.42

10

=

19 + 18 + . . . + 10 1 10(10 + 1) = 20 − = 14.5 10 10 2

[ann8]

The loan repayments are an increasing annuity with payments of

t = 1, 2, . . . , 30. t = 0):

times time

To determine the value of

P,

a ¨30 0.04 − 30v 30 4000 = P (Ia)30 0.04 = 0.04 ∴ P = 18.32

P, 2P, . . . , 30P

at

we use an equation of value (at

Immediately after the ninth payment, the outstanding loan is found by decomposing the remaining payments into a level annuity and an increasing annuity:

10P v + 11P v 2 + 12P v 3 + ... + 30P v 21 = 9P a21 0.04 + P (Ia)21 0.04 = 4774.80 This is greater than the original loan amount, and is because the earlier payments are (very) small relative to the latter payments. Therefore, the earlier payments are insucient to pay o the interest, let alone pay o part of the principal. Hence, this is also unlikely to exist in reality.

Exercise 1.43

[ann9]

The borrowings are a decreasing annuity, whereas the repayments can be decomposed into a level annuity of $300 and an increasing annuity starting at $200. An equation of value (at the end of year when nal loan amount is received) is:

X(Ds)5 0.05 (1 + i) = 300a15 0.05 + 200(Ia)15 0.05 42



Noting that:

a15 0.05 = 10.379658 − 15v 15 a ¨ (Ia)15 0.05 = 15 0.05 = 73.667689 0.05 (Ds)5 0.05 = (1 + i)5 (Da)5 0.05 = (1 + i)5 [6a5 0.05 − (Ia)5 0.05 ] a ¨5 0.05 − 5v 5 5 = (1 + i) 6a5 0.05 − 0.05 = 17.115531 We obtain:

X=

Exercise 1.44

300a15 0.05 + 200(Ia)15 0.05 = 993.11 (Ds)5 0.05 (1 + i)

[ann12]

The annual eective rate

i

is given by:

1 + i = 1.0254 ⇒ i = 0.103813 For convenience, we will determine the initial deposit by discounting all cash ows to time

t=0

(1/1/2004).

The entire series of deposits can be decomposed into two series: A. Deposits of

X, 1.1025X, . . . , (1.1025)10 X

B. Deposits of

1.1025X, (1.1025)2 X, . . . , (1.1025)11 X

where time

t

at times

t = 0, 1, . . . , 10 at times

t = 21 , 1 12 , . . . , 10 21

is measured in years.

These two series are (geometrically) increasing annuities, and can be present valued as geometric progressions:

P VA = X + 1.1025Xv + . . . + (1.1025)10 Xv 10 1 − (1.1025v)11 =X 1 − 1.1025v = 10.934815X P VB = v 1/2 (1.1025) X + 1.1025Xv + . . . + (1.1025)10 Xv 10 = v 1/2 (1.1025) P VA = 11.474726X Therefore, the initial deposit

X

is found as follows:

P VA + P VB = 110000v 11 ∴ X = 1656.19 43


Exercise 1.45


[new8]

The present value is $67989.65. Refer to Spreadsheet new8.xlsx.

Exercise 1.46

[ann14]

There are a number of ways to decompose the payments into annuity streams, which allow the present value to be determined more easily. A simple method is to consider 3 cash ow streams, each containing payments 1 year apart (which also means we will use the annual eective rate of 0.10): A. 10, 20, 30, 40 at times

t = 31 , 1 13 , 2 13 , 3 13

B. 20, 30, 40, 50 at times

t = 23 , 1 23 , 2 23 , 3 23

C. 30, 40, 50, 60 at times

t = 1, 2, 3, 4

Each of these streams is the combination of an increasing annuity and a level annuity. Therefore:

P VA = 10v 1/3 (I¨ a) 4 P VB = 10v 2/3 (I¨ a)4 + 10v 2/3 a ¨4 P VC = 10v(I¨ a)4 + 20v¨ a4 The annuity factors are given by:

(I¨ a)4 = 1 + 2v + 3v 2 + 4v 3 = 8.302780 1 − v4 a ¨4 = (1.1) 0.1 = 3.486852 Therefore, we obtain the present value:

P VA + P VB + P VC = 329.95 A more elegant method involves decomposing the original payments into: X. 10,20,30,. . . ,10,20,30 at all times (t

= 13 , 23 , 1, . . . , 4)

Y. 0,0,0,10,10,10,. . . ,30,30,30 at all times (t

= 13 , 23 , 1, . . . , 4)

The payments of X can be grouped by year, resulting in four payments of (10,20,30). Thus, X is a 4-year level annuity-due, with each payment being an increasing annuity:

P VX = 10(Ia)3 j a ¨4 i = 194.3179 44


where

i = 0.1


is the annual eective rate, and

j = (1.1)1/3 − 1

is the

1 -year eective 3

rate. Conversely, the payments of Y can are a 3-year increasing annuity, with each payment being a level annuity (omitting the rst three payments of 0):

P VY = 10a3 j (Ia)3 i = 135.6289 Summing these up, we obtain the same present value:

P VX + P VY = 329.95

Exercise 1.47

[ann15]

The present value (at time

t = 0)

of the rst ten payments is:

P (Ia)10 0.07 The present value (at time

t = 10)

of the last ten payments is:

10(10P ) = 100P since each payment is discounted at the same rate as the payment growth rate (each n n n payment is of amount 10(1.05) P which is worth 10(1.05) P v at time t = 10, and v n = (1.05)−n cancels with the factor (1.05)n ). Therefore, the present value of all payments at time

t=0

is:

P (Ia)10 0.07 + 100P (1.07)−10 = 50000 Since

(Ia)10 0.07 = 34.739133,

we get:

P = 584.29

Exercise 1.48

[new4]

(a) Let the annual payment be

X.

Therefore:

Xa20 0.05 = 10000 ∴ X = 802.4259 (b) Let the rst annual payment be

X.

Therefore:

(1.03)Xv + (1.03)2 Xv 2 + . . . + (1.03)20 Xv 20 = 10000 1 − (1.03v)20 = 10000 ∴ 1.03Xv 1 − 1.03v ∴ X = 608.1346

45



(c) To be fair, we need the present value of both payment streams to be equal (to 10000). In nominal terms:

20 X

k 608.13(1.03)k v0.05 =

k=1

20 X

k 802.43v0.05 = 10000

k=1

or in real terms:

20 X

608.13vrk

k=1 where

1.05 1.03

r=

Exercise 1.49

=

20 X

802.43(1.03)−k vrk = 10000

k=1

− 1.

[new1]

(a) The annual eective interest rate is We note that the payment at time

i = eδ − 1 = 0.05127.

n is 1+2+. . .+n.

Therefore, the perpetuity

can be decomposed into a level perpetuity of 1 (1st payment at time 1), plus a level perpetuity of 2 (1st payment at time 2), etc. This can be interpreted as an increasing perpetuity, with each regular payment being a level perpetuity itself:

¨∞ + 3v 3 a ¨∞ + . . . P V = v¨ a∞ + 2v 2 a 2 3 =a ¨∞ v + 2v + 3v + . . . =a ¨∞ (Ia)∞ Therefore:

1 a ¨∞ PV = d i 1 = 2 di = 8200 where

d = iv = 0.04877.

(b) See spreadsheet new1.xlsx. The plot should be similar to the one below.

46

60 0

20

40

PV of payment

80

100

120



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0

100

200

300

400

500

Time

Aside: The graph is actually related to the Gamma function. The present value of the

nth

payment is:

1 2 n 1 n 1 2 −δn 1 −δn n v + nv = n e + ne 2 2 2 2 Thus, we require:

∞ X

2 −δn

ne

∞ X

and

ne−δn

n=1

n=1

These can be approximated using the Gamma function by noting that:

Z Γ(k) = (k − 1)! =

∞

xk−1 e−x dx ≈

0

∞ X

xk−1 e−x

k=1

Therefore, the present value is:

1 PV = 2

∞ X

n2 e−δn +

Zn=1∞

∞ X

! ne−δn

n=1

Z ∞ 1 2 −δx −δx ≈ x e dx + xe dx 2 0 0 Z Z 1 1 ∞ 2 −u 1 ∞ −u = u e du + 2 ue du 2 δ3 0 δ 0 1 2! 1! = + 2 δ3 δ2

47

where

u = δx


The force of interest is

δ = 0.05,


which results in:

P V ≈ 8200

Exercises 1.501.57 [annB10annB15] See the solutions in the Broverman text and solutions manuals.

Exercise 1.58 We require

[new2]

(12)

100(Ia)10 , 100(Ia)10

100(I¯ a)10 .

, and

The monthly eective rate is:

j = (1 + i)1/12 − 1 = 0.004074 and therefore the nominal rate (payable monthly) is:

i(12) = 12j = 0.04889 The force of interest is:

δ = ln(1 + i) = 0.04879 Therefore:

a ¨10 − 10v 10 i = 3937.38 a ¨ − 10v 10 = 10 (12) i = 4026.81 a ¨ − 10v 10 = 10 δ = 4035.01

100(Ia)10 =

(12)

100(Ia)10

100(I¯ a)10

Exercise 1.59

[ann11]

The present value of the annuity is given by:

Z PV =

14

(t2 − 1)v t dt

1

v t varies (continuously) with time t: Z t 1 1 t v = exp − ds = exp (− ln(1 + t)) = 1+t 0 1+s

where the discount factor

Therefore, we have:

Z PV = 1

14 2

t −1 dt = t+1

Z

48

14

(t − 1)dt = 84.50 1


7.2


Module 2

Exercise 2.1

[lif1]

Note that if the force of mortality is constant, then the survival function is exponential:

Z t µx+s ds = e−µt t px = exp − 0

Therefore:

Z

∞

bt v t t px µx+t dt

E[Z] = Z0 ∞

e0.05t e−0.06t e−0.01t 0.01dt 0 Z ∞ = 0.01 e−0.02t dt =

0

1 = 2 2

Z

∞

e0.1t e−0.12t e−0.01t 0.01dt 0 Z ∞ e−0.03t dt = 0.01

E[Z ] =

0

1 = 3

∴ Var(Z) = E[Z 2 ] − (E[Z])2 =

Exercise 2.2

1 1 1 − = 3 4 12

[lif2]

Writing out the (contingent) annuity-certain in terms of the interest rate:

∞ X

a ¨k+1 k px qx+k =

k=0

k=0 This gives us a

∞ X 1 − v k+1

v

d

k px qx+k

term, which is similar to the RHS. We don't want the

write it in terms of

p:

49

q,

so we


=

∞ X 1 − v k+1 d

k=0

=


∞ X 1 − v k+1 d

k=0

k px

(1 − px+k )

(k px − k+1 px )

"∞ # ∞ X 1 X = (k px − k+1 px ) − v k+1 (k px − k+1 px ) d k=0 " ∞ !k=0 ∞ # ∞ ∞ X X X X 1 = −v v k k px + v k+1 k+1 px k px − k px d k=1 k=0 k=0 " k=0 ∞ # ∞ X X 1 1−v v k k px + v k k px = d k=0 k=1 " # ∞ ∞ X X 1 −v v k k px + = v k k px d k=0 k=0 since

v 0 0 px = 1.

Therefore:

∞ 1−v X k v k px = d k=0

=

∞ X

v k k px

k=0 The LHS is an annuity certain with a term equal to the future lifetime of the individual.

The RHS considers each annual payment separately, noting that the

individual will receive it if he/she is alive. Both describe the cash ows of a term annuity, and therefore they must be equal.

Exercise 2.3 Let

K(x)

[lif3]

be the curtate lifetime random variable of

(x).

i h a ¨x = E a ¨K(x)+1 = = =

∞ X

a ¨k+1 Pr [K(x) = k]

k=0 ∞ X

a ¨k+1 k| qx k=0 ∞ k+1 X k=0

1−v d

k| qx

"∞ # ∞ X 1 X v k+1 k| qx = k| qx − d k=0 k=0 =

1 − Ax d 50

We have:



∴ d¨ ax + Ax = 1

Exercise 2.4

[lif4]

The required values are:

Pr[K(0) = k] = k p0 qk = {0.1, 0.045, 0.0855, 0.1539, 0.2462, 0.2586, 0.1108, 0} e0 = E[K(0)] =

6 X

k Pr[K(0) = k]

k=0

= 3.6203 Pr[K(2) = k] = k p2 q2+k = {0.1, 0.18, 0.288, 0.3024, 0.1296, 0} e2 = E[K(2)] =

4 X

for

k Pr[K(2) = k]

k=0

= 2.1816 A2 =

4 X

v k+1 Pr[K(2) = k]

k=0

= 0.8576 2

A2 =

4 X

v 2(k+1) Pr[K(2) = k]

k=0

= 0.7379 A12:3 =

2 X

v k+1 Pr[K(2) = k]

k=0

A2:31 A2:3

= 0.5073 = v 3 3 p2 = 0.3732 = A12:3 + A2:31 = 0.8805

a ¨2 =

4 X

v k k p2

k=0

= 2.9900 a2 = a ¨2 − 1 = 1.9900 a ¨2:3 =

2 X

v k k p2

k=0

= 2.5102 See Excel spreadsheet lif4.xlsx for the calculations.

51

k = 0, . . . , 5

for

k = 0, . . . , 7


Exercise 2.5 Let

T = T (x)


[lif5] be the (random) lifetime of the non-smoker and

T S = T S (x)

be the

lifetime of the smoker. Note that we are comparing newborns. We want to nd:

S

Z

Pr T > T =

∞

Pr T S > t fT (t) dt

0 where we have conditioned on the future lifetime Note that:

T

to construct the integral.

Z t S Pr T > t = t p0 = exp − cµu du = (t p0 )c S

0 where t p0

= Pr(T > t).

Also, the density of

fT (t) =

T

is given by:

d d d FT (t) = (1 − S(t)) = − t p0 dt dt dt

S(t) is equivalent to t p0 . Z ∞ S Pr T > T = − (t p0 )c (t p0 )0 dt "0 #∞ (t p0 )c+1 =− c+1

since the survival function

Therefore:

0

1 = 1+c

(because t p0 is of the form

e−

·t

[parameter]

)

The answer is reasonable because:

•

If

c = 1,

then the two lives are identical so the probability should be

1/2

(which is the case here)

•

If

c>1

(as it is assumed), then the mortality of the smoker is higher, so the

probability that he outlives the non-smoker should be less than

1/2

(which is

also the case here)

7.3

Module 3

Exercise 3.1

[loa1]

(a) The repayments involve annual payments (in arrears) of

X,

ie. an annuity-

immediate:

Xa5 = 15000 X = 3464.62 (b) The retrospective method considers the accumulated value of past cash ows:

OB2 = 15000(1.05)2 − Xs2 = 9435.02 52



(c) The prospective method considers the present value of future cash ows:

OB2 = Xa3 = 9435.02 (d) Let the new annual payment be

Y.

The new future repayments must be able

to repay the outstanding balance calculated in (b) and (c). Therefore:

Y a3 0.075 = OB2 = 9435.02 Y = 3628.12 (e) Let the renegotiated payment be

Z.

We have:

Za4 0.076 = OB2 = 9435.02 Z = 2823.31 (f ) See the tutorial solution spreadsheet (loa1.xls).

Exercise 3.2

[loa2]

(a) The annual payment consists of two components: the interest repayment, and the payment into the sinking fund. The interest payment is simply 7% of the loan:

20000(0.07) = 1400 The sinking fund payment is an amount

X

such that the sinking fund will

accumulate to the $20000 principal after 6 years:

Xs6 0.05 = 20000 ∴ X = 2940.35 Therefore, the annual repayment is:

1400 + 2940.35 = 4340.35 (b) Similarly to (a), we have:

Xs6 0.07 = 20000 ∴ X = 2795.92 Therefore, the annual repayment is 4195.92. (c) The standard loan arrangement would require annual payments of $4195.92 as the two methods are equivalent when the sinking fund and loan interest rates are the same. Hence, we would prefer the standard loan arrangement, as it involves lower annual repayments. In general, the standard loan is preferable if the sinking fund rate is below the loan interest rate (which is usually true loan interest rate higher than savings interest rate). (d) See the tutorial solution spreadsheet (loa2.xls).

53


Exercise 3.3


[loa3]

(a) Total interest to be repaid:

I = Lf n = 5000 · 0.10 · 2 = 1000 Monthly loan repayments are:

R= The monthly eective rate

j

L+I 6000 = = 250 N 24

is the solution to the following equation of value:

Ra24 j = 5000 Using Newton-Raphson, we dene the function

f (j) = Ra24 j − 5000 = 250

f

as follows:

1 − (1 + j)−24 j

− 5000

The derivative is:

0

f (j) = 250 Starting with

24j(1 + j)−25 − (1 − (1 + j)−24 ) j2

j0 = 0.10: j1 = j0 − j2 = j1 − j3 = j2 − j4 = j3 − j5 = j4 −

f (j0 ) f 0 (j0 ) f (j1 ) f 0 (j1 ) f (j2 ) f 0 (j2 ) f (j3 ) f 0 (j3 ) f (j4 ) f 0 (j4 )

This results in an annual eective rate

i

= −0.062716 = −0.022528 = 0.004878 = 0.014291 = 0.015125 of:

i = (1 + j5 )12 − 1 = 19.74% Using spreadsheet loa3.xlsx for 10 iterations of Newton Ralphson. The solution is

j = 0.015131.

Therefore, the eective annual rate

i

is:

i = (1 + j)12 − 1 = 19.75% (b) The outstanding balance (using the prospective method) is:

250a12 j = 2724.66 54



Exercises 3.43.6 [loaB3loaB5] See the Mathematics of Investment and Credit solutions manual.

Exercise 3.7

[loa5]

The eective monthly rate is

0.18 12

= 0.015.

X

Therefore, the monthly repayment

is

given by:

Xa100 0.015 = 20000 ⇒ X = 387.41 As we are only required to provide the last four instalments (ie. 97th100th payments), we can begin the loan schedule at time balance at time

t = 96

t = 96

(months). The outstanding

is:

OB96 = Xa4 0.015 = 1493.23 The loan schedule for times

t = 97, . . . , 100

can then be found using the following

recursive equations:

It = OBt−1 · 0.015 P Rt = X − It OBt = OBt−1 − P Rt Therefore, we obtain: Time

Interest

Principal Repaid

Outstanding Balance

(t)

(It )

(P Rt )

(OBt )

96

?

?

1493.23

97

22.40

365.01

1128.22

98

16.92

370.49

757.73

99

11.37

376.05

381.69

100

5.73

381.69

0

The full loan schedule is shown in the tutorial spreadsheet provided (loa5.xls).

Exercise 3.8

[loa6]

The present value of the original four loans is:

P V = 4.36a11 0.01 + 17.2a15 0.01 + 35a12 0.01 + 20.24a18 0.01 = 1009.51 The total (sum) payments is:

11 (4.36) + 15 (17.2) + 35 (12) + 18 (20.24) = 1090.28 For the new consolidated loan, denote the term to run (in months) as

n

n.

Since

must be an integer, the consolidated loan cannot have level repayments with a

further restriction that these repayments must sum to the original total (1090.28).

55



Therefore, the loan will be assumed to involve monthly repayments of additional nal payment of

Y

X,

with an

to account for any remaining outstanding balance.

We have two equations to describe the two constraints (repayments must sum to original total; eective rate must remain the same):

nX + Y = 1090.28 Xan 0.01 + Y v n+1 = 1009.51 Here, we have three variables and two equations. Therefore, we must x one variable, after which we can determine the other two as a function of the rst variable. A logical choice would be to set

X and Y n, v n+1 , an 0.01 choices of n: get

n

to dierent values (as it has to be an integer) and

by solving the simultaneous equations (noting that the coecients are all known once

n

is known).

n

X

Here are the results for various

Y

11

69.25

328.56

12

72.45

220.86

13

74.02

128.05

14

74.51

47.13

15

74.29

-24.11

16

73.60

-87.40

17

72.61

-144.05

18

71.41

-195.13

Intuitively, the nal payment should not be negative and should be as small as possible as it represents an extra `top-up' payment to ensure the loan is fully repaid to the nearest cent. The purpose of the nal payment is not to repay a signicant portion of the loan. and so we choose

Hence, the solution where

n = 14

with

X = 74.51

and

n = 14 is the most appropriate, Y = 47.13. In other words, the

consolidated annuity is of $74.51 for 14 months, with a nal payment of $47.13 at the end of 15th month. Note: the necessity of the nal payment table) that there is no solution where

n

Y

can be veried by observing (from the

is an integer and

Y = 0.

The table can be found in spreadsheet loa6.xlsx.

Exercise 3.9

[loa7]

The initial payment 0.12 = 0.01: 12

X

can be found as follows, noting the monthly eective rate is

Xa12 0.01 + (1.1)Xv 12 a12 0.01 + (1.1)2 Xv 24 a12 0.01 = 47500 ∴ X = 1440.80 The loan schedule is shown in the tutorial spreadsheet provided (loa7.xls).

56


Exercise 3.10 (a) Let

X


[loa8]

be the annual payment. We have:

Xa10 0.0675 = 600000 ⇒ X = 84441.97 Hence total payment is (b) Let

Y

10X = 844419.69.

be the monthly payment. The monthly eective rate

i

is given by:

1 + i = (1.0675)1/12 ⇒ i = 0.005458 Therefore:

Y a120 i = 600000 ⇒ Y = 6828.08 Hence the total payment is

120Y = 819369.39, which is 25050.30 less than the

total payment in (a). (c) The amount to settle the loan is the outstanding balance at time

t = 5.

This

will not change when the interest rate changes, as the outstanding balance is determined from the past (ie.

using the recursive relationships of a loan

schedule). Therefore, the amount will be

OB5

under the old interest rate of

6.75%:

OB5 = Xa5 0.0675 = 348558.74 Note that this implies the prospective method (based on the future) becomes incorrect when interest rates change, while the retrospective method (based on the past) remains correct. If we used the prospective method, we would have calculated an outstanding balance of:

Xa5 0.0725 = 343923.44 This is the present value of future repayments (as payments now have to be discounted at the new rate), but it is not the outstanding balance.

Exercise 3.11

[loa4]

Note that we cannot nd a solution analytically due to the complexity of the loan (dierent period for interest and principal (re)payments; non-equal actual payments). Therefore, we will model the loan using a spreadsheet (loa4.xls).

t = 0, 41 , . . . , 19. The interest (It ), principal repaid (P Rt ) outstanding balance (OBt ), and nominal/actual payments (N P Rt , AP Rt )

The time step will be one quarter, ie.

are described using a set of recursive relationships:

It = OBt− 1 · 0.02 4

P Rt = AP Rt − It OBt = OBt− 1 − P Rt 4 75000 N P Rt = = 5000 (t = 5, 6, . . . , 19) 15 AP Rt = 5000 (t = 5, 6, . . . , 19) 57



0.08 = 0.02, and the 4 nominal and actual payments are equal since the loan is repaid at par. where we note that the quarterly eective interest rate is

The net cash ows received by the purchaser are the payments of interest principal

(AP Rt ).

i

The price for a required eective yield

these cash ows at the rate

i.

(It )

and

is found by discounting

Therefore:

X 1 t P = (It + AP Rt ) 1+i t where the summation is taken over For For

t = 0, 14 , . . . , 19.

i = 0.10, theprice is P = 66636.60. 2 i = 1 + 0.10 − 1 = 0.1025, the price 2

Exercise 3.12

is

P = 65565.63.

[loa9]

This question requires a spreadsheet model due to the complexity of the capital gains tax. A spreadsheet model can be setup using the standard recursive formulae, along with the following formulae for tax purposes:

ITt = τI · It

+

CGTt = (τCG · CGt ) = τCG

N P Rt AP Rt − P 500000

+

The capital gain is determined as the actual repayment less the purchase price. The N P Rt portion of the loan repaid by AP Rt is of the entire loan, and this portion of 500000 N P Rt P . the loan was purchased at a price of 500000 The price is the present value of the net cash ows at the eective yield of 6%:

P =

X t

v t CFt =

X

v t (AP Rt + It − ITt − CGTt )

t

The spreadsheet model can be found in loa9.xls. The solution is (a) $439,608.30 and (b) $538,334.94. Note that each component of the price can be expressed as follows (where

X

v t AP Rt = 25000(1.05)v 10 a ¨10

t

X

t

v (It − ITt ) = (1 − τI ) (0.08)

t

X

v t OBt− 1

4

t

X t

25000 v CGTt = 0.3 25000(1.05) − P 500000 t

+

v 10 a ¨10

where:

X

(4)

(4)

(4)

v t OBt− 1 = 500000a10 + 475000v 10 a1 + . . . + 25000v 28 a1 4

t (4)

(4)

(4)

= 500000a10 + 25000v 10 a1 (Da)19 58

i = 0.06):


Exercise 3.13


[new13]

(a) The equation of value is:

80 = 6a10 + 100v 10 (b) Newton-Raphson nds the root of

f (x) = 0

xn = xn−1 − where

xn

is the root obtained from the

recursively as follows:

f (xn−1 ) f 0 (xn−1 )

nth

iteration.

The equation we need to solve is:

1 − v 10 + 100v 10 − 80 0=6 i 6 6(1 + i)−10 = − + 100(1 + i)−10 − 80 i i

Thus, dene:

f (x) =

6 6(1 + x)−10 − + 100(1 + x)−10 − 80 x x

See Spreadsheet new13.xlsx for the iterations in Excel.

The solution is

i =

9.1349%.

Exercise 3.14

[loa10]

You can (and should) do this question both by hand and by spreadsheet.

(a) The capital gain is already known as the price is given and the bond only has one capital repayment at the end. For the 15% bond on its own, the capital gain is:

100 − 105.80 < 0 Therefore there will be no CGT. The equation of value is then:

105.80 = 15 (1 − 0.35) a4 j + 100 (1 + j)−4 where

j

is the annual eective yield. The solution

j = 8.00%

can easily be

veried by substitution. For the 8% bond, the capital gain is:

100 − 85.34 = 14.66 > 0 Therefore there will be CGT. The equation of value is then:

85.34 = 8(1 − 0.35)a4 j + 100 − 14.66(0.5) (1 + j)−4 59


The solution

j = 8.00%


can easily be veried by substitution.

Therefore both bonds (on their own) give an eective yield of 8% p.a. The solution by spreadsheet is provided in loa10.xls. The equations to be used are:

N P Rt AP Rt It OBt ITt

= 100 (t = 4) = N P Rt = i · OBt−1 = OBt−1 − N P Rt = (0.35)It N P Rt P CGTt = (0.50) AP Rt − 100 X P = v t (AP Rt + It − ITt − CGTt ) t

i is the interest v = (1 + j)−1 .

where and

rate (15% or 8%) and

We require the yield such that in a solution of

P

j

is the annual eective yield,

is (a) 105.80 or (b) 85.34. These both result

j = 8.00%.

(b) Instead of putting $100 in one single bond, we now put and

$ (100 − x)

into the 15% bond

into the 8% bond. This means that we buy:

x 105.8 and

$x

of the 15% bond with face value 100

100 − x 85.34

of the 8% bond with face value 100

We want to nd the value

x

such that the net capital gain is 0.

Note that

this would minimise our net CGT liability, as we have minimised the capital gain and we do not get a refund from capital losses (therefore no advantage in having a negative CG).

CG =

100 − x x (100 − 105.8) + (100 − 85.34) = 0 105.8 85.34 ∴x=

100(14.66) 85.34 5.8 + 14.66 105.8 85.34

= 75.81

The equation of value is then:

100 =

100 − x x 15(1 − 0.35)a4 j + 100vj−4 + 8(1 − 0.35)a4 j + 100vj−4 105.8 85.34

The solution

j = 8.46%

can be veried by substitution.

For the spreadsheet model, we note that the combined portfolio can be considered as a single bond (since interest payments are still level, and face value

60



is still $100). The price is $100, since we have zero capital gain (ie. purchasing a par bond). The interest rate becomes:

i = 0.15α + 0.08(1 − α) x is the proportion of the rst bond purchased, and 105.8 is the proportion of the second bond purchased. where

α=

1−α =

100−x 85.34

The rest of the spreadsheet remains the same. The yield can be solved numerically to obtain

j = 8.46%.

Exercises 3.153.16 [loaB6loaB7] See the Mathematics of Investment and Credit solutions manual.

Exercise 3.17

[loa11]

When an Australian government bond is ex-interest, the owner of the bond at the date of the bond going ex-interest receives the next coupon payment.

The buyer

does not receive the next coupon payment. The buyer receives the maturity face value and all coupons except the next payment. Price is:

f

v d (C + Gan + 100v n ) where the interest rate used is the semi-annual eective rate

i=

0.0475 2

= 0.02375,

and:

f = number of days to next coupon date (11 Jun 2006 to 15 Jun 2006) d = number of days between coupon dates (15 Dec 2005 to 15 Dec 2006) C = next coupon payment G = regular semi-annual coupon payment n = is the number of coupons to maturity Hence

f = 4, d = 182, C = 0, G =

5.75 2

= 2.875,

and

n = 10.

Price on 11 June 2006

is (next coupon is not received):

4 P = v 182 2.875a10 + 100v 10 = 0.999484 (2.875 × 8.808866 + 100 × 0.790789) = 0.999484 × 104.404433 = 104.351

Exercise 3.18

[ann16]

See Sherris (p. 4043). Users that may be interested include superannuation funds and anyone with long term liabilities that rise in line with ination (such as inationlinked life annuities, or some general insurance liabilities). See also: http://en.wikipedia.org/wiki/Ination-indexed_bond

61


Exercise 3.19


[loa12]

Using a spreadsheet model similar to Exercise 3.12, we obtain a price of $516,099.04. The spreadsheet is provided in loa12.xls.

Exercise 3.20

[loa13]

A spreadsheet model can be used with the following equations (where time 0, 21 , . . . , 30 is in years):

t =

( 0.035 t ≤ 26.5 it = 0.04 t > 26.5   150000 t = 10, 11, . . . , 18 N P Rt = 250000 t = 19, 20, . . . , 29   300000 t = 30   1.05 t = 10, 11, . . . , 18 Rt = 1.10 t = 19, 20, . . . , 29   1.12 t = 30 AP Rt = N P Rt × Rt It = it · OBt− 1 2

OBt = OB − N P Rt X vjt (AP Rt + It ) P = t− 12

t

j = 0.065 is the required P eective yield. Also note that the original outstandbalance (face value) is t N P Rt = 4, 400, 000.

where ing

The spreadsheet is provided in loa13.xls. The price is $4,797,748.66.

Exercise 3.21 The price

P

[loa14]

is the present value of all cash ows at the net eective yield of

This question is not too complex, and can be solved by hand.

j = 0.08.

The complexity

arises from capital gains tax. However, note that the capital gain for the rst two repayments (at times

t = 2, 3)

is:

400 −

400 P <0 1200

P > 1200. Therefore, the only possibility of a capital gain is with repayment (at time t = 4). For the 3rd repayment, the capital gain is:

since we are given the 3rd

400 1 P = 480 − P 1200 3 P < 1440. Therefore, the capital

400(1.2) − which will only be positive if repayment is:

( 0.20 480 − 13 P CGT4 = 0 62

P < 1440 P ≥ 1440

gains tax for the 3rd



Thus, we can solve this question by considering two cases:

(i) For

P < 1440,

P < 1440 and P ≥ 1440.

there is CGT, and the price is given by:

P = (1 − 0.20) 0.10 1200v + 1200v 2 + 800v 3 + 400v 4 1 2 3 4 + 400(v + v + 1.2v ) − 0.20 480 − P v 4 3 which can be solved to obtain assumption of (ii) For

P ≥ 1440,

P < 1440,

P = 1249.47.

This is consistent with the

so it is a valid solution.

there is no CGT, and the price is given by:

P = (1 − 0.20) 0.10 1200v + 1200v 2 + 800v 3 + 400v 4 + 400(v 2 + v 3 + 1.2v 4 ) = 1258.80 This is not consistent with the assumption of

P ≥ 1440,

so it is not a valid

solution.

Therefore, the price of the loan is $1249.47.

7.4

Module 4

Exercise 4.1

[irr1]

(a) From the spot rates we can work out the discount factors:

v(0.5) = v(1) = v(1.5) = v(2) =

0.04875180 1+ 2

−1

0.05031182 1+ 2

−2

0.05234408 1+ 2

−3

0.05448436 1+ 2

−4

= 0.976204 = 0.951525 = 0.925421 = 0.898067

and hence the value of the bond is:

P =

6.75 [v(0.5) + v(1) + v(1.5) + v(2)] + 100v(2) = 102.467 2

(b) To determine the yield (eective yearly), we need to solve:

(2)

6.75a2 i + 100vi2 = 102.467 This can be done numerically to obtain

i = 5.50537%

(using the function

written earlier in Exercise 3.13). The equivalent semi-annual eective yield is j = (1 + i)1/2 − 1 = 2.7158% (ie. i(2) = 5.43161%).

63



(c) The par yield of the 1 year coupon bond can be determined by solving:

100i [v(0.5) + v(1)] + 100v(1) = 100 or equivalently:

i[v(0.5) + v(1)] + v(1) = 1 ∴i=

1 − v(1) = 0.025146 v(0.5) + v(1)

The equivalent nominal rate (payable semi-annually) is

i(2) = 5.0292%.

Similarly, the par yield of the 2 year coupon bond is:

1 − v(2) = 0.027173 v(0.5) + v(1) + v(1.5) + v(2) which is equivalent to a nominal yield (payable semi-annually) of

i(2) = 5.4346%.

This means that if these bonds have coupon rates equal to these par yields, then their price will be 100 using the spot rates of this question. (d) Recall the following relationship between spot rates and forward rates, which holds due to no arbitrage (ie. accumulation of $1 with certainty must be the same under spot and forward rates):

(1 + st )t = (1 + st−1 )t−1 (1 + ft−1,t ) This can be rewritten as:

ft−1,t

(1 + st )t −1 = (1 + st−1 )t−1

Using the above equation in semi-annual time steps, we obtain:

f0, 1 = s0.5 = 0.024376 2

f 1 ,1 2

f1,1.5 f1.5,2

(1 + s1 )2 − 1 = 0.025937 = (1 + s0.5 ) (1 + s1.5 )3 = − 1 = 0.028207 (1 + s1 )2 (1 + s2 )4 = − 1 = 0.030459 (1 + s1.5 )3

where the spot and forward rates are semi-annual eective rates. These forward rates are equivalent to nominal p.a. rates (payable semi-annually) of 4.8752%, 5.1873%, 5.6415%, and 6.0919%.

64


Exercise 4.2


[irr2]

A zero coupon bond can be constructed by purchasing a combination of the two bonds (ie.

x

y

9% bonds and

7% bonds).

To have a net coupon each period of zero, we require:

9x + 7y = 0 To have a face value of $100 on maturity, we require:

x+y =1 Solving these simultaneously, we obtain

x = −3.5

and

y = 4.5.

The price of the

ZCB is therefore:

P = −3.5 × 101.00 + 4.5 × 93.20 = 65.9

Exercise 4.3 For all

t,

[irr3]

by equating the accumulation of $1 with certainty, we obtain:

Z

t−1

Z

s(x)dx

s(x)dx exp (ft−1,t ) = exp

exp

t

0

0 t−1

Z

Z s(x)dx + ft−1,t =

∴ 0

t

s(x)dx 0

∴ ft−1,t = rt − r(t − 1) = r which is a at forward rate curve.

Exercise 4.4

[irr4]

ft−1,t as the nominal forward rate (semi-annual compounding) for time period (t − 1, t). The forward rates can be found by equating the accumulation of $1 over Denote

the corresponding time periods:

2 f0,1 1+ 2 2 f0.5,1.5 1+ 2 2 f1,2 1+ 2 2 f1.5,2.5 1+ 2 2 f2,3 1+ 2

= 1+ = 1+ = 1+ = 1+ = 1+

1+ f0.5,1 1+ 2 f1,1.5 1+ 2 f1.5,2 1+ 2 f2,2.5 1+ 2 f0,0.5 2

f0.5,1 2

f1,1.5 2

f1.5,2 2

f2,2.5 2

f2.5,3 2

65

= (1.025)(1.0275) = 1.053188 = (1.0275)(1.03) = 1.058325 = (1.03)(1.0305) = 1.061415 = (1.0305)(1.03125) = 1.062703 = (1.03125)(1.035) = 1.067344



These can be solved to obtain:

f0,1 = 5.2498% f0.5,1.5 = 5.7498% f1,2 = 6.0500% f1.5,2.5 = 6.1750% f2,3 = 6.6247%

Exercise 4.5

[irr5]

The value of the bond is given by:

P = 4 [v(0.5) + v(1) + v(1.5) + v(2)] + 100v(2) where

v(t) is the discount factor associated with the t-year (semi-annual compound-

ing) spot rate:

1 (1 + s2t )2t

v(t) =

v(0.5) = 0.977995 and v(1) = 0.949497 are calculated by substituting the given spot rates, whereas v(1.5) and v(2) are determined using spot and forward rates: 1

v(1.5) = 1+

s1.5 3 2

=

1 2 1 + s21 1+

f1,1.5 2

1

= 1+

0.0525 2 2

1+

0.075082 2

= 0.915142 v(2) =

1 1+

s2 2

4 =

1 1+

s1 2

2

1+

f1,1.5 2

1+

1+

0.0525 2 2

1+

0.075082 2

= 0.905951

Exercise 4.6

1+

0.020290 2

1

=

The resulting price is

f1.5,2 2

P = 105.591.

[irr6]

We have:

P =D·a ¯n + Re−δn ∂P ∂ =D a ¯n − nRe−δn ∂δ ∂δ

66



where:

Z n ∂ ∂ a ¯n = e−δt dt ∂δ ∂δ 0 Z n ∂ −δt = e dt 0 ∂δ Z n te−δt dt =− 0

¯a)n = −(I¯ Therefore, the duration is given by:

1 ∂P · P ∂δ ¯a)n − nRe−δn −D(I¯ =− D·a ¯n + Re−δn ¯a)n + nv n g(I¯ = g¯ an + v n

D(δ) = −

where

g = D/R.

Exercise 4.7

[irr7]

(a) Consider a nominal amount of $1 of bond. The coupon income is payable continually to redemption at time 4.6):

D(δ) = at

δ = 0.07,

n.

g = 0.05 p.a.,

The duration is (from Exercise

¯a)n + nv n g(I¯ g¯ an + v n

which results in 11.592 for

n = 20

and 14.345 for

(b) To nd the maximum duration, we need to dierentiate

n = 60.

D(δ) w.r.t. n (and set

this to zero). To do so, it is useful to note the following intermediate results:

∂ n v ∂n ∂ nv n ∂n ∂ a ¯n ∂n ∂ ¯ I¯ a n ∂n

= v n ln v = −δv n = nv n ln v + v n = (1 − δn) v n 1 ∂ 1 (1 − v n ) = (δv n ) = v n δ ∂n δ 1 ∂ 1 = (¯ an − nv n ) = (v n + (1 − δn)v n ) = nv n δ ∂n δ =

Therefore:

¯a)n + nv n (gv n − δv n ) (g¯ an + v n ) (gnv n + (1 − δn)v n ) − g(I¯ ∂ D(δ) = ∂n (g¯ an + v n )2

67



Setting this to zero:

¯a)n + nv n (gv n − δv n ) 0 = (g¯ an + v n ) gnv n + (1 − δn)v n − g(I¯ ¯a)n + nv n (g − δ) v n = (g¯ an + v n ) 1 + (g − δ)n v n − g(I¯ ¯a)n − nv n = (g¯ an + v n ) + (g − δ) gn¯ an + nv n − g(I¯ ¯a)n = (g¯ an + v n ) + (g − δ) gn¯ an − g(I¯ ¯a)n ): v n and (I¯ h i g an ) + (g − δ) gn¯ an − (¯ an − nv n ) 0 = (g¯ an + 1 − δ¯ δ h i g an − n + nδ¯ = 1 + (g − δ) a ¯n + gn¯ an − (¯ an ) δ i h g an − n) = 1 + (g − δ) a ¯n − (¯ δ h i g ∴ (δ − g) a ¯n − (¯ an − n) = 1 δ g 1 ∴a ¯n + (n − a ¯n ) = δ δ−g δ ¯n ) = ∴ δ¯ an + g(n − a δ−g as required (g = 0.05, δ = 0.07). This can then be solved numerically to obtain n = 64.349 and a corresponding duration of 14.349. Noting that we require an equation in terms of

Exercise 4.8

a ¯n

(ie. without

[new11]

(a) The modied duration and convexity are:

1 ∂P P ∂i 1 ∂ 2P C= P ∂i2

MD = −

(b) For a 10 year annuity, we note that:

1 − (1 + i)−10 i ∂P 10i(1 + i)−11 + (1 + i)−10 − 1 = ∂i i2 2 2 ∂ P 2(−55i (1 + i)−12 − 10i(1 + i)−11 − (1 + i)−10 + 1) = ∂i2 i3 i = 0.05 we see that: P (i) =

For

P (0.05) = 7.7217 1 ∂P MD = − P ∂i = 4.856 1 ∂ 2P C= P ∂i2 = 35.602 68



(c) The rst and second derivatives of the price are approximated by:

∂P P (i + h) − P (i − h) = ∂i 2h ∂ 2P P (i + h) − 2P (i) + P (i − h) = ∂i2 h2 h is a small increment in the interest rate. For this question, we shall take h = 0.005. See Excel spreadsheet new11.xlsx. For h = 0.005, modied duration is 4.8576 and convexity is 35.6093. where

Exercise 4.9

[lif6]

Recall that:

a ¨x =

∞ X

∞ X v k px = (1 + i)−k k px k

k=0

k=0

To nd the duration, we dierentiate w.r.t.

i:

∞ ∞ X X d k(1 + i)−k−1 k px = − kv k+1 k px a ¨x = − di k=0 k=0 Therefore the modied duration of

a ¨x

is:

P∞ k k=0 kv k px P MDx = v ∞ l l=0 v l px and the Macaulay Duration is:

P∞ ∞ k X MDx k=0 kv k px P Dx = = wk k, = ∞ l v l=0 v l px k=0

where

v k k px P wk = ∞ l . l=0 v l px

Again, this is the weighted average of the payment maturities, where the weights take into account both the time value of money and the probabilities of survival (since the payments are contingent to the survival of

Exercise 4.10

(x)).

[irr8]

(a) The present value of assets (5-year and 15-year bonds) and liabilities (10-year ZCB) are:

VA (i) = M5 v 5 + M15 v 15 VL (i) = 1000v 10 where

v = (1 + i)−1 .

(b) A zero cost portfolio that provides a positive (non negative) prot with zero chance of loss.

69


(c) To ensure that


VA (i) = VL (i),

we require:

M5 + M15 v 10 = 1000v 5 The duration of the assets and liabilities are:

P tCt v t 5M5 v 5 + 15M15 v 15 DA (i) = Pt = t M5 v 5 + M15 v 15 t Ct v 10v t DL (i) = t = 10 v Therefore, to ensure that

DA (i) = DL (i),

we require:

5M5 v 5 + 15M15 v 15 = 10 M5 v 5 + M15 v 15 which simplies to:

M5 = M15 v 10 M15 = 734.66. We can check that this is an arbitrage opportunity by looking at VA (i) − VL (i)

Solving these simultaneous equations, we get

M5 = 340.29

and

for each parallel shift in the at yield curve:

i

VA (i) − VL (i)

7%

0.55

8%

0

9%

0.45

This shows that the surplus will be positive regardless of the direction in which interest rates shift (as long as we have a parallel shift in a at yield curve).

Exercise 4.11

[irr9]

See Sherris p. 87.

70


Exercise 4.12 At

i = 4.5%,


[irr10]

we have (for the liabilities):

VL =

X

Ct v t

t

= 3v + 4v 2 + 3v 3 + 2v 4 = 10.84 P tCt v t DL = t VL 3v + 8v 2 + 9v 3 + 8v 4 = 10.84 = 2.29 P t(t + 1)Ct v t+2 CL = t VL 3 6v + 24v 4 + 36v 5 + 40v 6 = 10.84 = 7.84 For the assets:

VA = M0.5 v 0.5 + M5 v 5 0.5M0.5 v 0.5 + 5M5 v 5 DA = VA Ensuring that

VA = VL

and

DA = DL ,

we get:

M0.5 v 0.5 + M5 v 5 = 10.84 0.5M0.5 v 0.5 + 5M5 v 5 = 24.79 These can be solved to obtain

M0.5 = 6.6803

and

M5 = 5.3647.

The eect of a change in interest rates on the surplus is determined in Excel Spreadsheet irr10.xlsx. The results are shown in the following table: Rate

VA

VL

S

6.5%

10.39

10.38

0.01

4.5%

10.84

10.84

0.00

2.5%

11.34

11.33

0.01

Twist (d)

10.32

10.67

0.35

Twist (e)

10.86

10.89

0.03

The negative surplus for the twist scenarios illustrate the failure of immunisation to protect against non-parallel shifts in the yield curve.

Exercise 4.13

[new12]

(a) Refer to Excel Spreadsheet new12.xlsx (Worksheet 1) to nd a plot similar to the one below.

71



0.054 0.050

0.052

Spot Rate

0.056

0.058

Spot Curve

0

5

10

15

20

Term

(b) Fisher-Weil duration relaxes the assumption that the yield curve is at.

In

practice, yield curves are rarely at. Therefore, dierent interest rates must be used when present valuing cash ows of dierent maturities. By splitting the cash ows into ZCBs of dierent maturities, we can value the cash ows by valuing the ZCBs (which is done using spot rates). The time-weighted value (duration) is also calculated using these principles. (c) The immunisation strategy is found using the Fisher-Weil duration (since the yield curve is no longer at):

VL =

X

Ct (1 + st )−t

t

= 10.609 P tCt (1 + st )−t DL = t VL = 2.275 P −(t+2) t t(t + 1)Ct (1 + st ) CL = VL = 7.572 For the assets:

VA = M0.5 (1 + s0.5 )−0.5 + M5 (1 + s5 )−5 0.5M0.5 (1 + s0.5 )−0.5 + 5M5 (1 + s5 )−5 DA = VA 72


Ensuring that

VA = VL

and


DA = DL ,

we get:

M0.5 (1 + s0.5 )−0.5 + M5 (1 + s5 )−5 = 10.609 0.5M0.5 (1 + s0.5 )−0.5 + 5M5 (1 + s5 )−5 = 24.134 These can be solved (with Solver in Excel - see spreadsheet new12.xlsx) to obtain

M0.5 = 6.588

and

M5 = 5.516.

(d) See Excel Spreadsheet new12.xlsx for the simulations and histogram similar to that of below.

0

50

Frequency

100

150

Histogram of Surplus

−0.02

−0.01

0.00

0.01

0.02

Surplus

As shown from the results and the graph, the portfolio is not fully immunised as there is a chance we can make a loss.

Exercise 4.14 At

i = 4%,

[irr11]

we have:

VL = 10.96 DL = 2.29 CL = 7.94

73


Letting

P1


P2

be the price of the 0.75 year bond, and

be the price of the 8 year

bond, we have (assuming annual coupons and $100 face value):

P1 D1 C1 P2 D2 C2 Let

x1

= 100.99 = 0.75 = 1.21 = 126.93 = 6.43 = 49.29

be the number of units of the 0.75 yr bond, and

x2

be the units of the 8 yr

bond. Therefore, we have:

V A = x1 P 1 + x2 P 2 x1 P1 D1 + x2 P2 D2 DA = VA Ensuring that:

VA = VL DA = DL we obtain two simultaneous equations which can be solved to give

x2 = 2.345.

These correspond to investing

7.983

and

2.977

x1 = 7.905

and

in the two bonds respec-

tively. A check of convexity shows that

Exercise 4.15 (a) At

CA − CL = 7.49 > 0,

so the portfolio is immunised.

[irr12]

i = 6%,

we have:

VL = 1.68 DL = 3 CL = 10.68 To form an immunised portfolio of bonds, it is clear that we need to use 2 or more bonds with maturities on either side of 3 (as the portfolio duration would be an average). Furthermore, to maximise the asset convexity (as immunisation suggests) we would wish to use the 1 year and 5 year bonds (Barbell strategy). Using the 1 year and 5 year bonds, and solving using methods similar to previous exercises, we obtain an investment strategy of $0.8396m into both bonds (ie. face values of 0.8900 and 1.1236). The convexity of the assets would be

14.240 > 10.680.

(b) Cashow matching involves nding assets that match the liability CF perfectly without considering the costs. Since we have ZCBs available, we would simply invest in a 3 year ZCB with a face value of $2m.

74



(c) If interest rates increase to 7%, then for the immunised portfolio:

VA − VL = 0.8900v + 1.1236v 5 − 2v 3 = 0.0003 For the CF matching, we have:

VA − VL = 2v 3 − 2v 3 = 0 Therefore, both methods have surpluses of approximately zero (exact for CF matching). (d) If interest rates twist, then the surplus of the immunised portfolio is:

0.8900

1 1.04

+ 1.1236

1 1.08

5

−2

1 1.06

3 = −0.0588

The CF matching surplus is still exactly zero.

7.5

Module 5

Exercise 5.1

[der1]

The theoretical futures index is determined as follows (accumulated value of spot index and storage costs):

ft0 = St er(T −t) − d¯ sT −t = 293.3e0.07(210/365) − (0.035)(293.3)¯ s210/365 0.07 = 305.353447 − (0.035)(293.3)(0.587085) = 299.326724 This is close to the observed index of 299.0 (only 0.11% dierence). Aside: The dierence between the actual and theoretical futures price is usually quite small (less than 1%), except for major events (eg. 1987 nancial crisis). The 0 dierence (ft − St ) is called the futures to cash spread or basis, whereas (ft − St ) is the theoretical basis.

Exercise 5.2

[der2]

The forward price will be the accumulated cost (purchase and funding) less coupons received. The price of the bond today is:

3.25a16 i + 100vi16 = 97.214171 where

i=

0.0696 2

= 0.0348

is the semi-annual eective rate.

This price (purchasing cost) is accumulated 2 years at the funding costs:

" 97.214171

0.065 1+ 12

12 12 # 0.07 1+ = 111.223057 12 75



The accumulated value of coupons received at the funding costs is:

3.25s2 i1 (1 + i2 )2 + 3.25s2 i2 = 13.700113 where

i1 = 1 +

0.065 6 12

−1

and

i2 = 1 +

0.07 6 12

−1

are the semi-annual eective

funding costs. The forward price is therefore

111.223057 − 13.700113 = 97.522944

The forward yield is the yield on the (6-year) bond which is locked in by the purchase of the forward contract:

97.522944 = 3.25a12 j + 100vj12 which results in

Exercise 5.3

j = 3.50%

per half-year, equivalent to an eective

7.013%

p.a.

[der3]

The current 90-day forward price implied by the market is (accumulated spot price and costs):

f90

90 = 420 1 + (0.0975) 365

+ (0.025) (420)

90 = 432.69 365

However, our short contract is for a forward price of $450, ie. we have an agreement to sell gold for $450 in 90 days time. Thus, based on today's market conditions, we should be able to make a prot of prot is worth:

Exercise 5.4

450 − 432.69 = 17.31 in 90 days time.

Today, this

17.31 90 = $16.90 1 + (0.0975) 365

[der4]

(Also see Example 6.9 in Sherris, p. costs less receipts (dividends).

112) The forward price is the accumulated

The cost of purchasing the shares now (including

transaction costs) is:

1.02 (10000 · 10) = 102000 This is accumulated to account for funding costs over the next 6 months:

6 0.06 102000 1 + = 105098.506 12 The accumulated dividends are:

0.06 0.4 (10000) 1 + 12 Therefore the forward price is

3 = 4060.3005

105098.506 − 4060.3005 = $101038.21

76


Exercise 5.5


[der5]

The cost of carry formula is based on no-arbitrage arguments. If it is violated, for r(T −t) example if Ft,T > St e − der(T −t1 ) , then one could short a forward, purchase the asset with funds borrowed at the risk-free rate, invest the dividend at the risk-free rate (at time

t1 ),

and exercise the forward contract at time

positive prot (at time

T)

T.

This would lock in a

with certainty at zero cost.

See lecture notes for further details. It may be helpful to draw a diagram.

Exercise 5.6

[der6]

For the 1st contract, the forward rate for days 30120 is 7.3%, whereas the 120-day spot rate is 7.4%. Therefore, for no-arbitrage, we require a repo rate

r

given by:

90 120 30 r 1+ · 0.073 = 1 + · 0.074 1+ 365 365 365 ∴ r = 0.075639 However, the 30-day spot rate is 7.5% (< 7.5639%). Therefore, we could: Action Borrow $1m at 30-day spot (7.5%) Short forward for $1,006,164 (7.3%) Invest $999,948 in 120-day spot (7.4%) Total

(t = 0)

(t = 30)

(t = 120)

1,000,000

1,006,164

0

0

1,006,164

1,024,275

999,948

0

1,024,275

52

0

0

This results in a guaranteed prot of $52 at zero cost. For the 2nd contract, the forward rate for days 90180 is 7.0%, whereas the 180-day spot rate is 7.4%. Similarly, we have:

90 90 180 1+ r 1+ · 0.070 = 1 + · 0.074 365 365 365 ∴ r = 0.076677 The 90-day spot rate is 7.4% (< 7.6677%). Therefore: Action Borrow $1m at 90-day spot (7.4%) Short forward for $1,018,247 (7.0%) Invest $999,352 in 180-day spot (7.4%) Total

(t = 0)

(t = 30)

(t = 120)

1,000,000

1,018,247

0

0

1,018,247

1,035,822

999,352

0

1,035,822

648

0

0

This results in a guaranteed prot of $648 at zero cost.

Exercise 5.7

[der7]

See details in Sherris Example 7.1 (p. 133)

77


Exercise 5.8


[der8]


Exercise 5.9

[der9]


Exercise 5.10

[der10]

Let the random future commodity price be time

Sti .

The swap provides for party A at

ti : Sti − X

The swap allows to get the dierence between the spot price at time without actually exchanging the commodity. The swap price

X

ti

and

X

is set such that the

initial value of the swap is 0. To value the swap, consider what happens if party A enters into a series of short forward positions (which cost nothing to enter in) to replicate the swap. In order to settle the forward contract (buy the commodity and sell it to the long party at the pre-specied price at the same time), the payo is

F0,ti − Sti where

ti .

F0,ti

represents the forward price set at time 0 (now) for settlement at time

The initial value of this forward is 0. Hence for times

t1 , t2 , . . . , tM

the net eect

of the swap and forward must be equal to 0:

0=

=

=

M X i=1 M X i=1 M X

e−rti [(Sti − X) + (F0,ti − Sti )] F0,ti e−rti − Xe−rti S0 − Xe−rti

i=1

= M S0 − X

M X

e−rti

i=1

M S0 ∴ X = PM −rti i=1 e

Exercise 5.11 (a) Let

[der11]

(hC , BC )

be the units of stocks and bonds we hold in our portfolio to

replicate the payo of the call.

275, ( max (285 − 275, 0) = 10 C1 = max (250 − 275, 0) = 0

For the call option with exercise price

78

the payo function is : if stock price goes up if stock price goes down



Hence we want to nd a portfolio such that the value at time 1 is:

C10

( hC 285 + BC e0.05 = 10 = hC 250 + BC e0.05 = 0

if stock price goes up if stock price goes down

Solving these equations:

10 285 − 250 BC = e−0.05 (250) hC =

−10 285 − 250

Since this portfolio pays the same amount as the option at time 1 regardless of the stock price, to have an arbitrage free market this portfolio must be worth the same as the option at time 0. Hence:

C0 = hC 260 + BC = 6.34 275, ( max (275 − 285, 0) = 0 P1 = max (275 − 250, 0) = 25

(b) For the put option with exercise price

the payo function is if stock price goes up if stock price goes down

Hence we want to nd a portfolio such that the value at time 1 is:

( hP 285 + BP e0.05 = 0 P10 = hP 250 + BP e0.05 = 25



−25 285 − 250 BC = e−0.05 (285) hC =

25 285 − 250

Since this portfolio pays the same amount as the option at time 1 regardless of the stock price, to have an arbitrage free market this portfolio must be worth the same as the option at time 0. Hence:

P0 = hP 260 + BP = 7.93 (c) The put-call parity relationship states that (in words): Call Value

+ Discounted

Strike Price

= Put

value

+ Share

ie.

C0 + Xe−rt = P0 + S0 This is satised, since (for

t = 1)

:

C0 + Xe−rt = 6.34 + 275e−0.05 = 267.93 P0 + S0 = 7.93 + 260 = 267.93 79

Price


Exercise 5.12


[der12]

For the call option with exercise price

50,

the payo function is

( max (55 − 50) = 5 C1 = max (45 − 50, 0) = 0 Hence we want to nd a portfolio holding


(h, B)

units of the stock and bond, such

that the value at time 1 is:

( 1 =5 h55 + B 1 + 0.05 12 0 C1 = 1 h45 + B 1 + 0.05 12 = 0



5 55 − 45 (45) −5 B= 1 55 − 45 1 + 0.05 12 h=

Therefore:

C0 = h50 + B = 2.59

Exercise 5.13

[der13]

The pricing is done by ensuring that there are no arbitrage properties, rather than through any particular probability assessments (except to decide where the stock can go to, eg. up to 55 or down to 45). In terms of pricing, there is no dierence between a model where the stock price can go up to 55 with probability 99% and a model where the stock price can go up to 55 with probability 2%. An important observation is that these contracts are priced as a function of the current underlying asset price, that is, they are relative prices. The probability of movements in the asset price do impact the underlying asset price but once this is known, these other contracts are determined by no-arbitrage based on the current underlying asset price.

Exercise 5.14

[der14]

This option (which is no longer a simple call or put option) pays:

( max (2852 − 2652 , 0) = 11000 X1 = max (2502 − 2652 , 0) = 0 Hence we want to nd a portfolio holding


(h, B)

units of the stock and bond such

that the value at time 1 is:

( h285 + Be0.04 = 11000 X10 = h250 + Be0.04 = 0

80





11000 285 − 250 −11000 −0.04 B=e (250) 285 − 250 h=

Since this portfolio pays the same amount as the option at time 1, to have an arbitrage free market this portfolio must be worth the same as the option at time 0. Hence:

X0 = h260 + B = 6223.81

Exercise 5.15

[der15]

Put call parity does not hold since:

22 K = 1.2245 + = 22.1769 1+i 1.05 P0 + S0 = 20 + 2.5 = 22.5

C0 +

Therefore we want to buy the LHS (call and bond of face value 22 and maturity 1) and short the RHS (put and share). The cost of this transaction today is:

22.1769 − 22.5 < 0 This is a prot, since it is a negative cost. At time 1, the cash ows of the put, stock, call and bond will all cancel out.

Exercise 5.16

[der16]

The stock is assumed to either go up to su or down to ert at time t, where r is the risk free rate.

sd .

The bond starts o as

1

and rises to

Suppose we want to price this derivative that pays

xu

or

xd .

For a call option, we

have:

xu = (su − K)+ xd = (sd − K)+ Consider a portfolio that holds

φ

stocks and

ψ

bonds. The value of this portfolio

today is:

V (0) = φs0 + ψ1 The value of this portfolio at time

T

is:

φ (su + D) + ψerT φ (sd + D) + ψerT Therefore, we need

φ and ψ

if the stock goes up if the stock goes down

such that the payos are identical (so that this portfolio

will have the same value as our derivative):

φ (su + D) + ψerT = xu φ (sd + D) + ψerT = xd 81




xu − xd su − sd xu − xd −rT ψ=e xu − (su + D) su − sd φ=

The value of the portfolio with these

φ

and

ψ

will be the same as the value of our

derivative.

Exercise 5.17

[der17]

Consider two investment portfolios:

(K − c) e−rT

•

Portfolio A: one call plus cash of

•

Portfolio B: one put plus one unit of gold

The value of portfolio A at expiry is given by:

( GT − K + K − c = GT − c 0+K −c=K −c

GT > K GT ≤ K

if if

(i.e. option is exercised) (i.e. option is worthless)

The value of portfolio B at expiry is given by:

( 0 + GT − c = GT − c K − GT + GT − c = K − c

GT > K GT ≤ K

if if

(i.e. option is exercised) (i.e. option is worthless)

Therefore, both portfolios have a payo at expiry of:

max {K, GT } − c Because the two portfolios have equal payo at expiry and the options cannot be exercised before expiry, the portfolios must also have equal value for any time In particular, at issue (t

= 0),

we must have:

c0 + (K − c)e−rT = p0 + S0 or:

c0 + Ke−rT = p0 + S0 + ce−rT

7.6

Module 6

Exercise 6.1

[sto1]

(a) The mean accumulation is:

E(1000S10 ) = 1000E [(1 + i1 ) . . . (1 + i10 )] = 1000E(1 + i1 ) . . . E(1 + i10 ) 10

= 1000 [E(1 + i)] = 1000(1.06)10 = 1790.85

82

(by independence)

(identically distributed)

t < T.



(b) The variance of the accumulation is: Var(1000S10 )

2 = 10002 E S10 − [E (S10 )]2

where:

2 E S10 = E (1 + i1 )2 . . . (1 + i10 )2 10 = E (1 + i)2 10 = 0.4(1.04)2 + 0.2(1.06)2 + 0.4(1.08)2 = 1.1239210 Therefore: Var(1000S10 )

= 9145.60

and:

σ(1000S10 ) = 95.63 (c)

(i) The mean accumulation depends only on the mean interest rate, which is not changed. However, the variance of the accumulation will be lower, as the variance in interest rates is lower. (ii) The mean will be larger as we are accumulating over a longer period. The standard deviation will also be larger, as investing in a longer term will result in a greater spread of possible accumulated amounts.

Exercise 6.2

[sto2]


E(Sn ) = E [(1 + i1 ) · · · (1 + in )] = E(1 + i1 ) · · · E(1 + in ) (by independence) = [E(1 + i)]n (identically distributed) = (1 + j)n The variance of the accumulation is: Var(Sn )

= E Sn2 − [E (Sn )]2

where:

E Sn2 = E (1 + i1 )2 · · · (1 + in )2 n = E (1 + i)2 n = E 1 + 2i + i2 n = 1 + 2j + j 2 + s2 noting that

E(X 2 ) = Var(X) + [E(X)]2 Var(Sn )

(b) (c)

for a random variable

= 1 + 2j + j 2 + s

E(S8 ) = 1.59385 √ σ(S8 ) = 2.65844 − 2.54035 = 0.34364 83

2 n

− (1 + j)2n

X.

Therefore:


Exercise 6.3


[sto3]


E(10000S2 ) = 10000E [(1 + i1 )(1 + i2 )] = 10000E [1 + i] E [1 + i2 ] (iid) 1 1 1 = 10000 1.03 + 1.04 + 1.06 (0.7 · 1.05 + 0.3 · 1.04) 3 3 3 = 10923.70 (b) The variance of the accumulation is: Var(10000S2 )

= 100002 E S22 − [E (S2 )]2

where:

E S22 = E (1 + i1 )2 (1 + i2 )2 = E (1 + i1 )2 E (1 + i2 )2 1 1 1 2 2 2 = 1.03 + 1.04 + 1.06 0.7 · 1.052 + 0.3 · 1.042 3 3 3 = 1.193465 Therefore: Var(10000S2 )

Exercise 6.4

= 19278

[sto4]

(a) The mean and variance are:

i1 + i2 2 2 2 Var(i) = E(i ) − [E(i)] 1 1 = (i21 + i22 ) − (i1 + i2 )2 2 4 1 2 = (i1 − i2 ) 4 E(i) =

(b) The mean and variance of the accumulated value can be determined using the formulae in Exercise 6.2:

E (Sn ) = (1 + j)n Var(Sn )

= 1 + 2j + j 2 + s2

n

− (1 + j)2n

Therefore:

5.5 = (1 + j)25 ⇒ j = 0.0705686 25 0.52 = 1 + 2j + j 2 + s2 − (1 + j)50 ⇒ s2 = 0.000377389 84


Solving for

i1

and

i2 :


i1 + i2 = j = 0.0705686 2 ∴ i1 + i2 = 0.1411372 1 (i1 − i2 )2 = s2 = 0.000377389 4 ∴ i1 − i2 = 0.0388530

Therefore, we obtain

Exercise 6.5

i1 = 0.089995

and

i2 = 0.051142.

[sto5]

(a) The single premium

X

is given by:

X [1 + E(i)]10 = 10000 ∴ X [0.3(1.07) + 0.5(1.08) + 0.2(1.10)]10 = 10000 ∴ X = 4589.26 (b) Expected prot is:

E(π) = E X(1 + i)10 − 10000 = X 0.3(1.07)10 + 0.5(1.08)10 + 0.2(1.10)10 − 10000 = 42.94

Exercise 6.6

[sto6]

(a) The mean of accumulated premiums is:

E(P ) = 425000(1.03)5 + 425000E [(1 + i1 ) · · · (1 + i5 )] = 425000(1.03)5 + 425000 [E(1 + i)]5 = 425000(1.03)5 + 425000(1.035)5 = 997458

85

(since

1 + it

iid)



The standard deviation is found as follows:

Var(P )

= Var 425000(1.03)5 + 425000

5 Y

! (1 + ik )

k=1

= 4250002 Var

5 Y

! (1 + ik )

k=1 5 Y

= 4250002 E

(1 + ik )2

! − 4250002 (1.035)10

k=1

E

5 Y

! (1 + ik )2

5 = E (1 + i)2

k=1

5 = E 1 + 2i + i2 5 = 1 + 2E(i) + [E(i)]2 + Var(i) = 1 + 2(0.035) + (0.035)2 + (0.03)2 = 1.416534 p ∴ σ(P ) = Var(P ) = 32743.21

5

(b) Investing all premiums in the risky assets is likely to be more risky because although there may be a higher probability of the assets accumulating to more than $1 million, the standard deviation would be twice as high so the probability of a large loss would also be greater.

Exercise 6.7

[sto7]


E(1000S10 ) = 1000E [(1 + i1 ) · · · (1 + i10 )] = 1000E(1 + i1 ) · · · E(1 + i10 ) 10

= 1000 [E(1 + i)] = 1000(1.07)10 = 1967.15

(by independence)


The variance of the accumulation is: Var(1000S10 )

2 = 10002 E S10 − [E (S10 )]2

where:

2 E S10 = E (1 + i1 )2 · · · (1 + i10 )2 10 = E (1 + i)2 10 = 1 + 2E(i) + [E(i)]2 + Var(i) 10 = 1 + 2(0.07) + (0.07)2 + (0.09)2 = 1.115310 86



Therefore, we have:

p

σ(1000S10 ) =

Var(1000S10 )

= 531.65

PrQ(1000S10 < 0.5(1967.15)). The distribution of the accumuS10 = 10 k=1 (1 + ik ) is determined as follows:

(b) We want to nd lation term

(1 + ik ) ∼ LN (µ, σ 2 ) ∴ ln(1 + ik ) ∼ N (µ, σ 2 ) ∴

10 X

ln(1 + ik ) ∼ N (10µ, 10σ 2 )

k=1

∴ exp

10 X

! ln(1 + ik )

=

k=1

10 Y

(1 + ik ) ∼ LN (10µ, 10σ 2 )

k=1

where: 1

E(1 + i) = 1.07 = eµ+ 2 σ

2

2 2 + i) = 0.092 = e2µ+σ eσ − 1 2 2 2 σ ∴ 0.09 = (1.07) e − 1

Var(1

∴ σ 2 = 0.007050 1 ∴ µ = ln 1.07 − (0.007050) = 0.064134 2 Therefore:

Pr (1000S10 < 0.5(1967.15)) = Pr (S10 < 0.983575) ln 0.983575 − 10µ √ = Pr Z < 10σ where Z ∼ N (0, 1) = Pr (Z < −2.4778) = 0.00661 (c) The probability required is:

Pr (1200S10

7 < 1400) = Pr S10 < 6 ln 76 − 10µ = Pr Z < √ 10σ = Pr (Z < −1.8349) = 0.03326

87


Exercise 6.8


[sto8]


E(S10 ) = E [(1 + i1 ) · · · (1 + i10 )] = E(1 + i1 ) · · · E(1 + i10 ) 10

= [E(1 + i)] = (1.06)10

(by independence)


Therefore:

E (2, 000, 000 S10 ) = 2, 000, 000 (1.06)10 = 3, 581, 695 (b) The distribution of

S10

is lognormal:

(1 + ik ) ∼ LN (µ, σ 2 ) ∴ ln(1 + ik ) ∼ N (µ, σ 2 ) ∴

10 X

ln(1 + ik ) ∼ N (10µ, 10σ 2 )

k=1

∴ S10 =

10 Y

(1 + ik ) = exp

k=1

10 X

! ln(1 + ik )

∼ LN (10µ, 10σ 2 )

k=1

where: 1

E(1 + i) = 1.06 = eµ+ 2 σ

2

2 2 + i) = 0.082 = e2µ+σ eσ − 1 2 ∴ 0.082 = (1.06)2 eσ − 1

Var(1

∴ σ 2 = 0.0056798 1 ∴ µ = ln 1.06 − (0.0056798) = 0.055429 2 Therefore, the required probability is:

Pr (S10 < 0.8 E(S10 )) = Pr S10 < 0.8(1.06)10 = Pr (S10 < 1.4327) ln 1.4327 − 10µ √ = Pr Z < 10σ = Pr (Z < −0.8171) = 0.207

Exercise 6.9 We have

[sto9]

(1 + i) ∼ LN (µ, σ),

where: 1

2

E(1 + i) = eµ+ 2 σ = 1.001 2 2µ+σ 2 Var(1 + i) = e eσ − 1 = 0.0022 88



µ

These can be solved simultaneously to obtain

2

and

σ2:

Var(1 + i) = e e −1 2 ∴ 0.0022 = (1.001)2 eσ − 1 µ+ 21 σ 2

σ2

∴ σ 2 = 0.000003992 1 ∴ µ = ln 1.001 − (0.000003992) = 0.0009975 2 Therefore, we can nd

j

as follows:

0.05 = Pr(i < j) = Pr(1 + i < 1 + j) = Pr(ln(1 + i) < ln(1 + j)) ln(1 + j) − µ = Pr Z < σ ln(1 + j) − µ = −1.645 σ ∴ ln(1 + j) = −0.00228921 ∴

∴ j = e−0.00228921 − 1 = −0.2287%

Exercise 6.10

[sto10]

Note: This question is similar to Exercise 6.8. (a) The mean accumulation is:

E(S10 ) = E [(1 + i1 ) · · · (1 + i10 )] = E[1 + i1 ] · · · E[1 + i10 ] 10

= [E(1 + i)] = (1.06)10

(by independence)


Therefore:

E (1, 000, 000 S10 ) = 1, 000, 000 (1.06)10 = 1, 790, 848 (b) The distribution of

S10

is lognormal:

(1 + ik ) ∼ LN (µ, σ 2 ) ∴ ln(1 + ik ) ∼ N (µ, σ 2 ) ∴

10 X

ln(1 + ik ) ∼ N (10µ, 10σ 2 )

k=1

∴ S10 =

10 Y

(1 + ik ) = exp

k=1

10 X k=1

89

! ln(1 + ik )

∼ LN (10µ, 10σ 2 )



where: 1

E(1 + i) = 1.06 = eµ+ 2 σ

2

2 2 + i) = 0.082 = e2µ+σ eσ − 1 2 2 2 σ ∴ 0.08 = (1.06) e − 1

Var(1

∴ σ 2 = 0.0056798 1 ∴ µ = ln 1.06 − (0.0056798) = 0.055429 2 Therefore, the required probability is:

Pr (S10 < 0.9 E(S10 )) = Pr S10 < 0.9(1.06)10 = Pr (S10 < 1.61172) ln 1.61172 − 10µ √ = Pr Z < 10σ = Pr (Z < −0.32304) = 0.373

Exercise 6.11

[new3]

(a) The accumulated value based on a principal of $1 is:

S10 = (0.05 − i1 )(1 + i2 ) · · · (1 + i10 ) + · · · + (0.05 − i10 ) The mean of

S10

is:

E(S10 ) = E[(0.05 − i1 )(1 + i2 ) · · · (1 + i10 ) + · · · + (0.05 − i10 )] = E[0.05 − i1 ]E[1 + i2 ] · · · E[1 + i10 ] + · · · + E[0.05 − i10 ] (by independence)

= 0.01(1.04)9 + 0.01(1.04)8 + · · · + 0.01(1.04) + 0.01 = 0.01s10 0.04 = 0.120061 (b) The distribution of the interest rate in each year is:

1 + i ∼ LN (µ, σ 2 ) where: 1

E(1 + i) = 1.04 = eµ+ 2 σ 2

2

2µ+σ 2

=⇒ Var(1 + i) = 0.02 = e 2 =⇒ 0.022 = (1.04)2 eσ − 1 0.022 2 =⇒ σ = ln 1 + 1.042 1 =⇒ µ = ln 1.04 − σ 2 2 90

σ2

e

−1



The simulation can be found in Spreadsheet new3.xlsx.

3 0

1

2

Density

4

5

Simulated Density of S

−0.2

−0.1

0.0

0.1

0.2

0.3

0.4

S

(c) A company who has an exposure to a oating rate may want to hedge this interest rate risk by using a xed-for-oating swap. For example, if the company has a liability to repay a loan at a oating rate, they could enter into a swap to pay a xed rate and receive a oating rate. The payment and receipt of the oating rate will net to zero, allowing the company to pay a net rate which is xed.

Exercise 6.12

[new5]

(a) The mean is:

E(S1 ) = E(1 + y1 ) = E(1 + µ + β(y0 − µ)) = 1.05 + 0.4(y0 − 0.05) Therefore

E(S1 ) = 1.046

and

E(S1 ) = 1.054

for

y0 = 0.04

respectively. (b) The variance is: Var(S1 )

= Var(1 + y1 ) = Var(1 + µ + β(y0 − µ) + σε1 ) = σ2 = 0.012 91

and

y0 = 0.06



(c) The probability is:

Pr(S1 < 1.04) = Pr(1 + y1 < 1.04) = Pr(µ + β(y0 − µ) + σε1 < 0.04) −0.01 − 0.4(y0 − 0.05) = Pr ε1 < 0.01 where ε1 ∼ N (0, 1) Pr(S1 < 1.04) = 0.2743 y0 = 0.06 respectively.

Therefore and

Exercise 6.13

and

Pr(S1 < 1.04) = 0.0808

for

y0 = 0.04

[new6]

The simulation can be found in Spreadsheet new6.xlsx. The histograms should be similar to the ones below.

150 100 50 0

Frequency

200

250

Histogram of S (4%)

1.4

1.5

1.6

1.7 S

92

1.8

1.9



150 100 0

50

Frequency

200

250

Histogram of S (6%)

1.4

1.5

1.6

1.7

1.8

1.9

S

Exercise 6.14

[new9]

(a) Denoting the mean and standard deviation of

j = 0.04

and

s = 0.02),

it

as

j

and

s

respectively (ie.

the formulae can be derived as in the lecture notes:

s¨n = (1 + y1 ) · · · (1 + yn ) + (1 + y2 ) · · · (1 + yn ) + · · · + (1 + yn ) = (1 + yn ) [(1 + y1 ) · · · (1 + yn−1 ) + · · · + (1 + yn−1 ) + 1] = (1 + yn )(¨ sn−1 + 1) ∴ E(¨ sn−1 ) = E[(1 + yn )(¨ sn + 1)] = E(1 + yn )E(¨ sn−1 + 1) by independence = (1 + j)[E(¨ sn−1 ) + 1] sn−1 + 1)2 E s¨2n = E (1 + yn )2 (¨ by independence = E (1 + yn )2 E (¨ sn−1 + 1)2 2 2 = E 1 + 2yn + yn E s¨n−1 + 2sn−1 + 1 = 1 + 2j + j 2 + s2 E s¨2n−1 + 2E (sn−1 ) + 1 The variance can then be determined using: Var (¨ sn−1 )

= E s¨2n−1 − [E (¨ sn−1 )]2

(b) The recursion can be found in Excel spreadsheet new9.xlsx.

E (¨ s30 ) = 58.33

and

V ar (¨ s30 ) = 17.16.

(c) The distribution of the interest rate in each year is:

1 + i ∼ LN (µ, σ 2 ) 93

We see that



where: 1

E(1 + i) = 1.04 = eµ+ 2 σ

2

=⇒ Var(1 + i) = 0.022 = e2µ+σ 2 =⇒ 0.022 = (1.04)2 eσ − 1 0.022 2 =⇒ σ = ln 1 + 1.042 1 =⇒ µ = ln 1.04 − σ 2 2

2

2 eσ − 1

The simulation can be found in spreadsheet new9.xlsx.

0.04 0.02 0.00

Density

0.06

0.08

Simulated Density of Annuity

45

50

55

60 s30

94

65

70

75

FM All Exercises

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