fm all exercises

Australian School of BusinessSchool of Actuarial Studies

Financial Mathematics

Exercises

S1 2012

17 March 2012

Contents

1 Time Value of Money and Cash Flow Valuation 2

1.1 Time Value of Money . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

Exercise 1.1 [int1] . . . . . . . . . . . . . . . . . . . . . . . . . 2

Exercise 1.2 [int2] . . . . . . . . . . . . . . . . . . . . . . . . . 2

Exercise 1.3 [int3] . . . . . . . . . . . . . . . . . . . . . . . . . 2

Exercise 1.4 [int4] . . . . . . . . . . . . . . . . . . . . . . . . . 2

Exercise 1.5 [int5] . . . . . . . . . . . . . . . . . . . . . . . . . 2

Exercise 1.6 [int7] . . . . . . . . . . . . . . . . . . . . . . . . . 2

Exercise 1.7 [int8] . . . . . . . . . . . . . . . . . . . . . . . . . 2

Exercise 1.8 [int9] . . . . . . . . . . . . . . . . . . . . . . . . . 2

Exercise 1.9 [int10] . . . . . . . . . . . . . . . . . . . . . . . . 3

Exercise 1.10 [int11] . . . . . . . . . . . . . . . . . . . . . . . 3

Exercise 1.11 [int12] . . . . . . . . . . . . . . . . . . . . . . . 3

Exercise 1.12 [int13] . . . . . . . . . . . . . . . . . . . . . . . 3

Exercise 1.13 [int15] . . . . . . . . . . . . . . . . . . . . . . . 3

Exercise 1.14 [int19] . . . . . . . . . . . . . . . . . . . . . . . 4

Exercise 1.15 [int18] . . . . . . . . . . . . . . . . . . . . . . . 4

Exercise 1.16 [int20] . . . . . . . . . . . . . . . . . . . . . . . 5

Exercise 1.17 [int6] . . . . . . . . . . . . . . . . . . . . . . . . 5

Exercise 1.18 [int14] . . . . . . . . . . . . . . . . . . . . . . . 5

Exercise 1.19 [int24] . . . . . . . . . . . . . . . . . . . . . . . 5

Exercise 1.20 [int25] . . . . . . . . . . . . . . . . . . . . . . . 5

Exercise 1.21 [int21] . . . . . . . . . . . . . . . . . . . . . . . 5

Exercise 1.22 [int22] . . . . . . . . . . . . . . . . . . . . . . . 6

Exercise 1.23 [int23] . . . . . . . . . . . . . . . . . . . . . . . 6

Exercise 1.24 [new10] . . . . . . . . . . . . . . . . . . . . . . . 6

i

Financial Mathematics � Exercises Actuarial Studies � UNSW

1.2 Level Annuities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

Exercise 1.25 [ann1] . . . . . . . . . . . . . . . . . . . . . . . . 6

Exercise 1.26 [ann2] . . . . . . . . . . . . . . . . . . . . . . . . 6

Exercise 1.27 [ann3] . . . . . . . . . . . . . . . . . . . . . . . . 7

Exercise 1.28 [annB1] . . . . . . . . . . . . . . . . . . . . . . . 7

Exercise 1.29 [annB2] . . . . . . . . . . . . . . . . . . . . . . . 7

Exercise 1.30 [annB3] . . . . . . . . . . . . . . . . . . . . . . . 7

Exercise 1.31 [annB4] . . . . . . . . . . . . . . . . . . . . . . . 7

Exercise 1.32 [annB5] . . . . . . . . . . . . . . . . . . . . . . . 7

Exercise 1.33 [annB6] . . . . . . . . . . . . . . . . . . . . . . . 7

Exercise 1.34 [annB7] . . . . . . . . . . . . . . . . . . . . . . . 7

Exercise 1.35 [ann4] . . . . . . . . . . . . . . . . . . . . . . . . 7

Exercise 1.36 [ann5] . . . . . . . . . . . . . . . . . . . . . . . . 7

Exercise 1.37 [ann6] . . . . . . . . . . . . . . . . . . . . . . . . 7

Exercise 1.38 [ann7] . . . . . . . . . . . . . . . . . . . . . . . . 7

Exercise 1.39 [annB8] . . . . . . . . . . . . . . . . . . . . . . . 7

Exercise 1.40 [annB9] . . . . . . . . . . . . . . . . . . . . . . . 7

Exercise 1.41 [ann13] . . . . . . . . . . . . . . . . . . . . . . . 8

1.3 Non-Level and Continuous Annuities . . . . . . . . . . . . . . . . . . 8

Exercise 1.42 [ann8] . . . . . . . . . . . . . . . . . . . . . . . . 8

Exercise 1.43 [ann9] . . . . . . . . . . . . . . . . . . . . . . . . 8

Exercise 1.44 [ann12] . . . . . . . . . . . . . . . . . . . . . . . 8

Exercise 1.45 [new8] . . . . . . . . . . . . . . . . . . . . . . . 8

Exercise 1.46 [ann14] . . . . . . . . . . . . . . . . . . . . . . . 8

Exercise 1.47 [ann15] . . . . . . . . . . . . . . . . . . . . . . . 8

Exercise 1.48 [new4] . . . . . . . . . . . . . . . . . . . . . . . 9

Exercise 1.49 [new1] . . . . . . . . . . . . . . . . . . . . . . . 9

Exercise 1.50 [annB10] . . . . . . . . . . . . . . . . . . . . . . 9

Exercise 1.51 [loaB2] . . . . . . . . . . . . . . . . . . . . . . . 9

Exercise 1.52 [annB11] . . . . . . . . . . . . . . . . . . . . . . 9

Exercise 1.53 [loaB1] . . . . . . . . . . . . . . . . . . . . . . . 9

Exercise 1.54 [annB12] . . . . . . . . . . . . . . . . . . . . . . 9

Exercise 1.55 [annB13] . . . . . . . . . . . . . . . . . . . . . . 9

Exercise 1.56 [annB14] . . . . . . . . . . . . . . . . . . . . . . 9

ii


Exercise 1.57 [annB15] . . . . . . . . . . . . . . . . . . . . . . 9

Exercise 1.58 [new2] . . . . . . . . . . . . . . . . . . . . . . . 10

Exercise 1.59 [ann11] . . . . . . . . . . . . . . . . . . . . . . . 10

2 Life Contingencies 11

Exercise 2.1 [lif1] . . . . . . . . . . . . . . . . . . . . . . . . . 11

Exercise 2.2 [lif2] . . . . . . . . . . . . . . . . . . . . . . . . . 11

Exercise 2.3 [lif3] . . . . . . . . . . . . . . . . . . . . . . . . . 11

Exercise 2.4 [lif4] . . . . . . . . . . . . . . . . . . . . . . . . . 11

Exercise 2.5 [lif5] . . . . . . . . . . . . . . . . . . . . . . . . . 12

3 Loans and Investments 13

3.1 Loan Repayments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

Exercise 3.1 [loa1] . . . . . . . . . . . . . . . . . . . . . . . . . 13

Exercise 3.2 [loa2] . . . . . . . . . . . . . . . . . . . . . . . . . 13

Exercise 3.3 [loa3] . . . . . . . . . . . . . . . . . . . . . . . . . 14

Exercise 3.4 [loaB3] . . . . . . . . . . . . . . . . . . . . . . . . 14

Exercise 3.5 [loaB4] . . . . . . . . . . . . . . . . . . . . . . . . 14

Exercise 3.6 [loaB5] . . . . . . . . . . . . . . . . . . . . . . . . 14

Exercise 3.7 [loa5] . . . . . . . . . . . . . . . . . . . . . . . . . 14

Exercise 3.8 [loa6] . . . . . . . . . . . . . . . . . . . . . . . . . 14

Exercise 3.9 [loa7] . . . . . . . . . . . . . . . . . . . . . . . . . 15

Exercise 3.10 [loa8] . . . . . . . . . . . . . . . . . . . . . . . . 15

3.2 Investments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15

Exercise 3.11 [loa4] . . . . . . . . . . . . . . . . . . . . . . . . 15

Exercise 3.12 [loa9] . . . . . . . . . . . . . . . . . . . . . . . . 15

Exercise 3.13 [new13] . . . . . . . . . . . . . . . . . . . . . . . 16

Exercise 3.14 [loa10] . . . . . . . . . . . . . . . . . . . . . . . 16

Exercise 3.15 [loaB6] . . . . . . . . . . . . . . . . . . . . . . . 16

Exercise 3.16 [loaB7] . . . . . . . . . . . . . . . . . . . . . . . 16

Exercise 3.17 [loa11] . . . . . . . . . . . . . . . . . . . . . . . 16

Exercise 3.18 [ann16] . . . . . . . . . . . . . . . . . . . . . . . 16

Exercise 3.19 [loa12] . . . . . . . . . . . . . . . . . . . . . . . 17

Exercise 3.20 [loa13] . . . . . . . . . . . . . . . . . . . . . . . 17

Exercise 3.21 [loa14] . . . . . . . . . . . . . . . . . . . . . . . 17

iii


4 Interest Rate Risk 18

4.1 Term Structure of Interest Rates . . . . . . . . . . . . . . . . . . . . 18

Exercise 4.1 [irr1] . . . . . . . . . . . . . . . . . . . . . . . . . 18

Exercise 4.2 [irr2] . . . . . . . . . . . . . . . . . . . . . . . . . 18

Exercise 4.3 [irr3] . . . . . . . . . . . . . . . . . . . . . . . . . 18

Exercise 4.4 [irr4] . . . . . . . . . . . . . . . . . . . . . . . . . 19

Exercise 4.5 [irr5] . . . . . . . . . . . . . . . . . . . . . . . . . 19

4.2 Price Sensitivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

Exercise 4.6 [irr6] . . . . . . . . . . . . . . . . . . . . . . . . . 19

Exercise 4.7 [irr7] . . . . . . . . . . . . . . . . . . . . . . . . . 19

Exercise 4.8 [new11] . . . . . . . . . . . . . . . . . . . . . . . 20

Exercise 4.9 [lif6] . . . . . . . . . . . . . . . . . . . . . . . . . 20

4.3 Immunisation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

Exercise 4.10 [irr8] . . . . . . . . . . . . . . . . . . . . . . . . 20

Exercise 4.11 [irr9] . . . . . . . . . . . . . . . . . . . . . . . . 21

Exercise 4.12 [irr10] . . . . . . . . . . . . . . . . . . . . . . . . 21

Exercise 4.13 [new12] . . . . . . . . . . . . . . . . . . . . . . . 22

Exercise 4.14 [irr11] . . . . . . . . . . . . . . . . . . . . . . . . 22

Exercise 4.15 [irr12] . . . . . . . . . . . . . . . . . . . . . . . . 22

5 Derivatives 24

5.1 Forwards, Futures and Swaps . . . . . . . . . . . . . . . . . . . . . . 24

Exercise 5.1 [der1] . . . . . . . . . . . . . . . . . . . . . . . . 24

Exercise 5.2 [der2] . . . . . . . . . . . . . . . . . . . . . . . . 24

Exercise 5.3 [der3] . . . . . . . . . . . . . . . . . . . . . . . . 24

Exercise 5.4 [der4] . . . . . . . . . . . . . . . . . . . . . . . . 24

Exercise 5.5 [der5] . . . . . . . . . . . . . . . . . . . . . . . . 24

Exercise 5.6 [der6] . . . . . . . . . . . . . . . . . . . . . . . . 25

Exercise 5.7 [der7] . . . . . . . . . . . . . . . . . . . . . . . . 25

Exercise 5.8 [der8] . . . . . . . . . . . . . . . . . . . . . . . . 25

Exercise 5.9 [der9] . . . . . . . . . . . . . . . . . . . . . . . . 25

Exercise 5.10 [der10] . . . . . . . . . . . . . . . . . . . . . . . 25

5.2 Options . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

Exercise 5.11 [der11] . . . . . . . . . . . . . . . . . . . . . . . 25

Exercise 5.12 [der12] . . . . . . . . . . . . . . . . . . . . . . . 25

iv


Exercise 5.13 [der13] . . . . . . . . . . . . . . . . . . . . . . . 26

Exercise 5.14 [der14] . . . . . . . . . . . . . . . . . . . . . . . 26

Exercise 5.15 [der15] . . . . . . . . . . . . . . . . . . . . . . . 26

Exercise 5.16 [der16] . . . . . . . . . . . . . . . . . . . . . . . 26

Exercise 5.17 [der17] . . . . . . . . . . . . . . . . . . . . . . . 26

6 Stochastic Interest Rates 27

6.1 IID Returns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

Exercise 6.1 [sto1] . . . . . . . . . . . . . . . . . . . . . . . . . 27

Exercise 6.2 [sto2] . . . . . . . . . . . . . . . . . . . . . . . . . 27

Exercise 6.3 [sto3] . . . . . . . . . . . . . . . . . . . . . . . . . 27

Exercise 6.4 [sto4] . . . . . . . . . . . . . . . . . . . . . . . . . 28

Exercise 6.5 [sto5] . . . . . . . . . . . . . . . . . . . . . . . . . 28

6.2 Lognormal Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

Exercise 6.6 [sto6] . . . . . . . . . . . . . . . . . . . . . . . . . 28

Exercise 6.7 [sto7] . . . . . . . . . . . . . . . . . . . . . . . . . 29

Exercise 6.8 [sto8] . . . . . . . . . . . . . . . . . . . . . . . . . 29

Exercise 6.9 [sto9] . . . . . . . . . . . . . . . . . . . . . . . . . 29

Exercise 6.10 [sto10] . . . . . . . . . . . . . . . . . . . . . . . 29

6.3 Dependence and Further Concepts . . . . . . . . . . . . . . . . . . . 30

Exercise 6.11 [new3] . . . . . . . . . . . . . . . . . . . . . . . 30

Exercise 6.12 [new5] . . . . . . . . . . . . . . . . . . . . . . . 30

Exercise 6.13 [new6] . . . . . . . . . . . . . . . . . . . . . . . 30

Exercise 6.14 [new9] . . . . . . . . . . . . . . . . . . . . . . . 31

7 Solutions to Exercises 32

7.1 Module 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

7.2 Module 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

7.3 Module 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

7.4 Module 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

7.5 Module 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

7.6 Module 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 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 5th year by $100invested today under compound interest with i = 0.05, and under simple interestwith i = 0.05.

Exercise 1.2: [int2] At what rate of compound interest will $200 grow to $275 in5 years?

Exercise 1.3: [int3] How many years does it take $200 to accumulate to $275 atan e�ective annual rate of 5%?

Exercise 1.4: [int4] With compound interest at i = 0.05, what is the present valuenow of $275 in 5 years?

Exercise 1.5: [int5] If $150 grows to $240 in n years, what will $1000 grow to overthe same period?

Exercise 1.6: [int7] If funds invested today will earn 8% for the next 10 years andat least 5% for the following 10 years, what is the most one must invest today toaccumulate $1 million in 20 years?

Exercise 1.7: [int8] What level rate of interest is equivalent to 8% for the next 10years followed by 5% for the following 10 years?

Exercise 1.8: [int9] Assuming an e�ective rate of i = 0.10, �nd the value of thecash �ows below at times t = 0 and t = 3.

2


Exercise 1.9: [int10] The following cash �ows have a value of 7.7217 at time t = 0(assuming 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 earning10% interest and $20 is withdrawn at t = 1 and 2, then how much can be withdrawnat t = 3?

Exercise 1.11: [int12] At what rate of interest will $100 accumulate to $200 in 6years?

Exercise 1.12: [int13] (SOA Course 2 May 2000, Question 1) Joe deposits $10today and another $30 in �ve years into a fund paying simple interest of 11% peryear. Tina will make the same two deposits, but the $10 will be deposited n yearsfrom today and the $30 will be deposited 2n years from today. Tina's deposits earnan annual e�ective rate of 9.15%. At the end of 10 years, the accumulated amountof Tina's deposits equals the accumulated amount of Joe's deposits. What is thevalue of n?

Exercise 1.13: [int15] (SOA Course 2 Nov 2000, Question 2) The following tableshows the annual e�ective 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 ofInvestment

InvestmentYear Rates

Calendar Year ofPortfolio Rates

PortfolioRate

i1 i2 i31990 10% 10% x% 1993 8%1991 12% 5% 10% 1994 (x− 1)%1992 8% (x− 2)% 12% 1995 6%1993 9% 11% 6% 1996 9%1994 7% 7% 10% 1997 10%

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

Exercise 1.14: [int19] You wish to buy a new home theatre system and have twopotential payment options.

Option A. You pay $610 down (at t = 0), $475 next year (at t = 1) and $340 thefollowing year (at t = 2).

Option B. You pay $560 down (at t = 0), $580 next year (at t = 1) and $274 thefollowing year (at t = 2).

Assuming a compound interest accumulation function, determine the values of therate of interest r for which Option A is preferred to Option B.

Exercise 1.15: [int18] In return for a single payment of $1000, an investment banko�ers 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 madeafter �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 foreach.

(b) Assume that an investor selects alternative A and that after three years sheinvests the proceeds for a further two years at a �xed rate of interest. Howlarge must this rate of interest be in order for her to receive at least $1550 atthe end of 5 years?

Why do we ask this question?

4


(c) Assume that an investor selects alternative B and that after �ve years he wantsto compare his proceeds with the value of alternative C. Determine the valueof 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 of13% p.a. e�ective, allowing for 5 payments.

Exercise 1.17: [int6] If v = 0.94, what are d and i?

Exercise 1.18: [int14] (SOA Course 2 May 2001, Question 12) Bruce and Robbieeach open up new bank accounts at time 0. Bruce deposits $100 into his bankaccount and Robbie deposits $50 into his. Each account earns an annual e�ectivediscount rate of d. The amount of interest earned in Bruce's account during the11th year is equal to X. The amount of interest earned in Robbie's account duringthe 17th year is also equal to X. What is the value of X?

Exercise 1.19: [int24] Tina issues a 2-year promissory note for a face value of$6000 and receives $4843.30 in return (ie. she borrows $4843.30 and promises torepay $6000 after 2 years). At the end of 6 months, 1 year, and 18 months, shedeposits $1000, $1000, and $2000 into her bank account and earns the same interestrate as the implied rate on the promissory note. Assuming interest is compoundedsemi-annually, determine how much extra money (in addition to the amount in herbank account) she will need to redeem (repay) the note for its face value in 2 yearstime.

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 andDecember. Simple interest is paid for amounts on deposit for less than a quarter ofa year. In 2001, Maria made 4 deposits of $1000 into her trust account every 1st dayof March, June, September, December. By 31 December 2005, how much interestwill Maria have earned from these deposits?

Exercise 1.21: [int21] You are given the following interest options:

A. an e�ective 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 di�erences between the above interest rates by expressing each optionas e�ective rates of interest.

(b) How much will an investment of $10,000 accumulate to in 412years based on

each of the above interest options?

Exercise 1.22: [int22] Assume that the force of interest is δ(t) = 0.04(1 + t)−1,with t measured in years. Using an accumulation function a(t):

(a) Calculate the equivalent e�ective rate of interest for the period t = 1 to t = 2.

(b) Calculate the equivalent e�ective rate of interest for the period t = 2 to t = 3.

(c) Find the value at time t = 2 of an investment that accumulates to $200,000at time t = 4.

Exercise 1.23: [int23] A fund credits simple interest with i = 10% from time zeroto time k. After time k, the fund accumulates at a constant force of interest of 8%.

(a) Find the value of k that maximises a(4).

(b) Using the value of k from (a), �nd the force of interest as function of time t(for 0 ≤ t ≤ 4).

Exercise 1.24: [new10] Suppose the force of interest is δ = 0.05. Using Excel, plot(on the same graph) (i) i(m) against m and (ii) d(m) against m, for 0.5 < m < 50

1.2 Level Annuities

Exercise 1.25: [ann1] You are given a combined annuity-immediate payable monthlysuch that payments are $1000 p.a. for the �rst 6 years and $400 p.a. for the next4 years together with a lump sum of $2000 at the end of the 10 years. An interestrate 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 yearshaving 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 inheritanceis 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 e�ective 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 accumu-late the funds, she plans on making ten annual deposits of C into an account payinge�ective annual interest of 6%. Having computed the least possible amount C (andassuming she succeeded in her �nancial mathematics course and thus it took her anegligible 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 4thEd).






Exercise 1.35: [ann4] To settle a $100,000 death bene�t, Tim, the primary ben-e�ciary, opted to take an annuity-immediate payable monthly for 25 years. Themonthly payment was calculated using an e�ective annual interest rate of 3%. Af-ter making payments for 10 years, the insurance company decides to increase themonthly payments for the remaining 15 years by changing the e�ective annual in-terest 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 $100at the end of year 1, $200 at the end of year 2, $100 at end of year 3, $200 at the endof year 4, and so on ($100 at odd years, $200 at even), with the �nal payment beingat 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 12payments of $18,000 per year at an e�ective discount rate of 10%; the �rst paymentis due immediately. He wishes to convert this to a 20-year annuity-immediate atthe same e�ective rate of discount. What will be the size of the payments under thenew annuity?

Exercise 1.38: [ann7] Broverman 5th Ed: 2.2.13 (2.2.13 in 4th Ed). Also solve forthe case if Smith repays the loan over 5 years (monthly payments).


7



Exercise 1.41: [ann13] Given that δ(t) = 120−t , t ≥ 0, �nd s10 .

1.3 Non-Level and Continuous Annuities

Exercise 1.42: [ann8] A loan of $4000 is being repaid by a 30-year increasingannuity-immediate where payments increase each year and payments are in arrears.The initial payment is P , each subsequent payment is P larger than the precedingpayment. The annual e�ective interest rate is 4%. Calculate the value of the futurepayments (ie. the loan principal outstanding) after the ninth payment. Compareyour result with the initial loan amount and explain it. Is such a payment patternlikely to exist in reality? Why?

Exercise 1.43: [ann9] Nicole, a UNSW Business part-time student, expects anincreasing amount of income as she advances through her program but will need toborrow to cover her university costs. Accordingly, she plans to borrow a decreasingannual amount from a student credit loan during her 5 years at university, and torepay the loan with increasing amounts for 15 years after graduation. She borrowsamounts 5X, 4X, 3X, 2X and X at the beginning of each of 5 years, where the lastpayment is paid at the beginning of her �nal year. At the end of the �rst year aftergraduation she pays $500, then increases the amount by $200 each year until a �nalpayment of $3300. If the e�ective annual interest rate assumed is 5%, determine X.

Exercise 1.44: [ann12] Paulo is saving madly to buy his �rst home ten years fromnow. He deposits to a fund each January 1 and July 1 for the years 2004 through2014. The deposit he makes on each July 1 will be 10.25% greater than the one onthe 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 theimmediately preceding July 1. The fund will be credited with interest at a nominalannual rate of 10%, compounded quarterly. On December 31, 2014, the fund willhave a balance of $110,000, an amount Paulo considers is enough for a home depositand 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 annuityimmediate which pays t3 + ln(10t + 12) at the end of year t, assuming an e�ectiveinterest rate of 5%.

Exercise 1.46: [ann14] Value the following set of cash�ows at a rate of 10% p.a.:$10 at time 1

3, $20 at times 2

3and time 11

3, $30 at times 1,12

3and 21

3, $40 at times

2, 223and 31

3, $50 at times 3 and 32

3, and $60 at time 4.

Exercise 1.47: [ann15] Mary purchases an increasing annuity-immediate for $50,000that makes twenty annual payments as follows:

8


(i) P, 2P, . . . , 10P in years 1 through 10; and

(ii) 10(1.05)P, 10(1.05)2P, . . . , 10(1.05)10P in years 11 through 20.

The annual e�ective 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. e�ective, andthe in�ation rate is a constant 3% p.a. Determine the (initial) annual payment froma 20 year annuity-immediate which is purchased at the fair price with $10,000 inthe case of:

(a) a �xed annuity (level payments)

(b) an in�ation-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 byconsidering the 500 payments only. Plot the present value of each paymentagainst time t. Does this shape remind you of something?


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 4thEd).






9


Exercise 1.58: [new2] Find the present value of a 10-year increasing annuity thatpays at an annual rate of 100, 200, . . . , 1000, given that the annual e�ective interestrate 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 payablefor 13 years. The rate of payment at time t is t2 − 1 per annum, and the force ofinterest at time t is 1

1+t. Find the present value of the annuity.

10

Module 2

Life Contingencies

Exercise 2.1: [lif1] Z is the present value random variable for a special continuouswhole life insurance issued to (x), paying bt 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 valueand variance of Z.

Exercise 2.2: [lif2] Show that the following two de�nitions of the life annuity axare equivalent:

∞∑k=0

ak+1 kpxqx+k =∞∑k=0

vkkpx

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

Exercise 2.3: [lif3] Prove the following identity:

dax + Ax = 1

Exercise 2.4: [lif4] You are given the following probabilities of death:

x qx0 0.101 0.052 0.103 0.204 0.405 0.706 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,

2A2, A12:3

, A 12:3

, A2:3 , a2, a2, a2:3

11


Exercise 2.5: [lif5] (Gerber (1997), Exercise 17, p. 1356) Consider two independentlives which are identical except that one is a smoker and the other is a non-smoker.It is known that:

1. µx is the force of mortality for non-smokers for 0 ≤ x < ω, and

2. cµx is the force of mortality for smokers for 0 ≤ x < ω, where c is a constantand c > 1

Calculate the probability that the remaining lifetime of the smoker exceeds that ofthe 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 interestof 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 theretrospective method.

(c) Calculate the loan outstanding at the end of the second payment using theprospective method.

(d) At the end of the second year the bank tells you that from that time onwardsthe rate of interest charged is going to be increased to 7.5% p.a. If you stillwant to payo� the loan by the end of the �fth year what must your annualpayment change to?

(e) Having reviewed your company's free cash �ows you decide that the amountcalculated in (d) is not a�ordable. You renegotiate with the bank and theyo�er to extend your loan so that you can pay o� the loan in an additional 4years (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 interestonly at a rate of 7% and repayments of the principal in a single lump sum at theend of 6 years. The borrower can accumulate the principal in a sinking fund earningan annual interest rate of 5%, and decides to do this by means of 6 level depositsstarting 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 arrangementor to have a standard loan arrangement (ie. repay both capital and interestwith each payment) to repay the loan. If the sinking fund rate was 5% whichmethod would you prefer?(Hint: you should not need to do any extra calculations to decide this � al-though you can use it to check your answer if you wish)

(d) Model the cash �ows of this sinking fund arrangement in a spreadsheet (forthe 5% case).

Exercise 3.3: [loa3] An individual borrows $5000 to buy a plasma TV. The sumborrowed is repayable by 24 monthly instalments in arrears, which are calculatedon the basis of a �at rate of interest of 10% p.a.

(a) Calculate the monthly repayment and the true (e�ective) annual rate of inter-est 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 3.7: [loa5] A loan of $20000 is being repaid by monthly instalments ofprincipal and interest (18% p.a. nominal) over 81

3years. Provide a schedule in

Excel showing the principal and interest contained in each of the last four monthlyinstalments.

Exercise 3.8: [loa6] A householder is paying o� four debts by monthly paymentsall at an e�ective rate of 1% per month (12% p.a. nominal). The monthly paymentsand respective terms to run are:

Monthly Payment ($) Terms to Run (Months)4.36 1117.20 1535.00 1220.24 18

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

14


Calculate the monthly instalment and the term to run of the consolidated loan sothat the e�ective rate of interest involved will be unchanged. Note that a �nalrepayment 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. Theinterest applicable is 12% p.a. (monthly compounding). Instead of paying o� theloan using level instalments, he decides to pay it o� using monthly payments over3 years. The payments within each year is the same, but the payments in the 2ndyear are 10% higher than the payments in the 1st year, and the payments in the 3rdyear are 10% higher than the payments in the 2nd year. Set out the loan schedulefor this loan in Excel.

Exercise 3.10: [loa8] A recently married couple have decided to buy a new housein Sydney. After an investigation of their �nancial situation they �nd that they willneed to borrow $600,000 from the bank. The rate of interest charged is 6.75% p.ae�ective.

(a) If they want to pay o� the loan in 10 years using annual payments, how muchwould 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 muchwould 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 paymentat time 5), interest rates increase to 7.25% p.a. e�ective. How much do theyneed 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 tobe 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 annualinstalments, 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 whowishes to realise a yield of (a) 10% per annum e�ective, and (b) 10% per annumconvertible half-yearly.

Exercise 3.12: [loa9] (McCutcheon & Scott, 1986, p. 197) A loan of nominalamount $500,000 was issued bearing interest of 8% per annum payable quarterly inarrears. The loan principal will be repaid at $105% by 20 annual instalments, eachof 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 entireloan on the issue date at a price to obtain a net e�ective annual yield of 6%. Assume

15


that capital losses cannot o�set 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 R.

Exercise 3.13: [new13] Consider a 10-year bond of face value $100 and annualcoupons 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 ofthe bond using a precision of 0.01%. Use any appropriate initial estimate.

Exercise 3.14: [loa10] (McCutcheon & Scott, 1986, p. 206) Two bonds (100 facevalue) each have an outstanding term of four years. Redemption will be at par forboth bonds. Interest is payable annually in arrears at the annual rate of 15% forthe �rst bond and 8% for the second bond. Interest payments have just been madeand 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 capitalgains tax at the rate of 50% who purchases either of these bonds (but notboth) will obtain a net yield on his transaction of 8% per annum.

(b) Assume now that the investor is allowed to o�set capital gains by capitallosses. 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 ofcombined transaction of 8.46% per annum.



Exercise 3.17: [loa11] An investor purchased an Australian Government bond on11 June 2006 paying a coupon 5.75% p.a with a maturity of 15 June 2011. The bondis �ex-interest� within 7 days prior to the coupon payment. Explain what is meantby �ex-interest� for an Australian government bond and describe the payments thatthe buyer will receive on an �ex-interest� Australian government bond. Determinethe price paid for the Australian Government bond at a yield of 4.75% p.a on 11June 2006.

Exercise 3.18: [ann16] Outline the payments made on in�ation indexed bonds andgive 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 repay-ment being ten years after the issue date. An investor, liable to both income tax andcapital gains tax, purchased the entire loan on the issue date at a price to obtain anet e�ective annual yield of 7%. Find the price paid, given that his rates of taxationfor 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 issuedon 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 Redemptionin 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 beenmade and thereafter at 8% p.a. What was the issue price if a purchaser of the wholeloan secured a yield of 6.5% p.a. e�ective on his or her investment? Do this questionusing Excel.

Exercise 3.21: [loa14] A loan of nominal amount $1,200 was issued bearing interestof 10% per annum payable annually in arrears. The loan will be repaid by 3 annualnominal payments of equal value, the �rst repayment being two years after the issuedate. 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 capitalgains tax at 20%, purchased the entire loan on the issue date at a price to obtain anet e�ective 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. basisassuming interest compounds semi-annually (ie. they are i(2) interest rates).

Term (Years) % p.a.0.5 4.8751801.0 5.0311821.5 5.2344082.0 5.448436

(a) Use these spot rates to calculate the value of a 6.75% bond paying semi-annualcoupons 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 yearand two year maturity bonds corresponding to the above rates. Interpret yourresult.

(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 for101.00; the other has a 7% coupon and sells for 93.20. Find the price of a 5 yearzero coupon bond.

Exercise 4.3: [irr3] Let s(t), 0 ≤ t ≤ ∞ denote a spot rate curve, that is, thepresent value of a dollar to be received at time t is e−s(t)t. Show explicitly that if thespot 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.a0 � 0.5 5.000.5 � 1 5.501 � 1.5 6.001.5 � 2 6.102 � 2.5 6.252.5 � 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 4.50001.0 5.2500

and forward rates:

Time Period % p.a.1-1.5 7.50821.5-2 2.0290

Calculate the value of a bond paying semi-annual coupons of 8% p.a., maturing in2 years time.

4.2 Price Sensitivity

Exercise 4.6: [irr6] Let D(δ) be the duration, at a constant force of interest (con-tinuous compounding) δ p.a., of a �xed-interest security with interest payable con-tinuously at the annual rate D per unit nominal and redeemable at R per unitnominal in n years time. Let g = D/R. Show that:

D(δ) =g(I a)n + nvn

gan + vn

Exercise 4.7: [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 interestof 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 speci�ed constantforce of interest per annum δ, may be considered as a function of n. Assumingδ = 0.07, show that the duration is maximised when the following equation issatis�ed:

0.07an + 0.05(n− an ) =0.07

0.07− 0.05

Hence (or otherwise) �nd the maximum duration and the corresponding valueof n at which the duration is maximised.

Exercise 4.8: [new11] Consider a level 10-year annuity-immediate paying $1 at theend of each year.

(a) Write down the expressions which relate the modi�ed duration and convexitywith derivatives of the price (present value).

(b) For i = 5%, �nd the modi�ed duration and the convexity.

(c) Using Excel and a Taylor's approximation, �nd an approximate modi�ed du-ration and convexity. Compare this with the answer in (b).

Exercise 4.9: [lif6] Show that the Macaulay Duration of ax is equal to:

Dx =∞∑k=0

wkk

where wk is given by:

wk =vkkpx∑∞l=0 v

llpx

4.3 Immunisation

Exercise 4.10: [irr8] (Boyle, 1992) Suppose the term structure of spot rates is levelfor all maturities and equal to 8% p.a. Suppose that in the next instant, the termstructure of interest rates will be either 9% p.a. for all maturities or 7% p.a. for allmaturities. Consider the following strategy. An investor goes short a zero couponbond with a 10-year maturity and a face value of 1000. Simultaneously, she usesthe proceeds to purchase a 5-year zero coupon bond with maturity value M5 and a15-year maturity bond with maturity value M15.

(a) Give expressions for the value of assets VA(i) and the value of liabilities VL(i).

(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 pro�ts are possible with parallel shifts to a �at yieldcurve.

20


Exercise 4.11: [irr9] Consider 3 coupon paying bonds x, y and z. You havecalculated 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 unitof each bond. Derive a formula for the duration and convexity for the portfolio interms 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 Outgo1 3m2 4m3 3m4 2m

Assume that the spot rate term structure is �at and equal to 4.5%. Assume thatthe insurer can only invest in 2 ZCBs. One matures in 0.5 years while the othermatures 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 withthe following spot rate curve?

Time Rate0.5 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-yearspot rate is given by:

st = α + β

(1− e−λt

λt

)+ γ

(1− e−λt

λt− e−λt

)where α = 0.06, β = 0.01, γ = −0.08, and λ = 0.6 are known parameters.

21


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 themodel used in Exercise 4.12(e). Another simple example is the following yield curve,which describes the zero coupon yield of maturity τ (ie. τ -year spot rate) as:

sτ = α + βt

(1− e−λτ

λτ

)where α and λ are constant parameters and βt is a time-varying parameter. It isassumed that α = 0.06 and λ = 0.6, while the values of βt for each time t areindependent and identically distributed with βt ∼ N(−0.01, 0.0022).

(a) Using Excel, plot the spot curve (for maturities 0 < τ < 20) for the givenparameters and when βt 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 as-sume that the insurer can only invest in 2 ZCBs of maturity 0.5 and 5 yearsrespectively. 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 determinethe surplus in each case. Also plot a histogram for the surplus. Is your portfoliofully 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 Outgo1 3m2 4m3 3m4 2m

Assume that the spot rate is �at and is equal to 4%. Suppose the insurer hasavailable for investment two coupon bonds. One is a 4% coupon bond with 0.75years 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 withoutusing 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 theportfolio.

(b) Explain the term cash �ow matching. Derive the portfolio (using 1, 2, 3, 4 orall 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 forthe strategies in (a) and (b)?

(d) Suppose the spot rate curve such that the spot rate for maturity T is equal to(3 + T )%. 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 was293.3 and the December 1987 S&P 500 futures closing index, with delivery in 210days, was 299.0. Calculate the theoretical futures index assuming transactions andstorage costs are negligible, a constant annual continuously compounding interestrate of 7%, and that the S&P 500 portfolio pays dividends continuously at an annualrate of 3.5% of its market value on 12 May 1987. You may also assume that interestrates are deterministic so that the futures price is equal to the theoretical forwardprice.

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 bondsat 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 ongold for delivery in 90 days at $450 an ounce. The current spot price of gold is $420an ounce and insurance and storage cost for gold are 2.5% p.a of the spot price, paidon delivery. Ninety day (simple) interest rates are 9.75% p.a. What is the value ofthis forward contract?

Exercise 5.4: [der4] Consider a forward contract on 10000 shares, deliverable in 6months time. The share is currently trading at $10.00. Assume that there will be adividend payment of 0.40 per share in 3 months time. Funding costs for 6 monthsare 6% p.a. (monthly compounding). Transaction costs are 2% of the value of theshares purchased. Determine the forward price for sale of the shares in 6 months atwhich all net funding and other costs will be covered.

24


Exercise 5.5: [der5] Explain the cost of carry formula:

Ft,T = Ster(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 d at time t1.

Exercise 5.6: [der6] Suppose the current spot and forward rates are as given inTable 6.4 of Sherris (1996, p. 109). Calculate the implied repo rate (the risk-free rateimplied 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 pro�t now. Givethe amount of the pro�t in both instances. Use simple interest as the contracts arefor terms less than one year.

Exercise 5.7: [der7] Review Example 7.1 of Sherris (p. 133)



Exercise 5.10: [der10] Consider an agreement where party A receives the spotprice for N units of a commodity each period while paying a �xed amount X perunit for 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 theswap contract. Assume that no storage costs or dividends occur during the period,and the risk free interest rate is r (continuous compounding).

5.2 Options

Exercise 5.11: [der11] In a one-period binomial model, it is assumed that thecurrent share price of 260 will either increase to 285 or decrease to 250 at the endof one year. The annual risk-free interest rate is 5% compounded continuously andassume that this share pays no dividends.

(a) Calculate the price of a one-year European call option with a strike price of275 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 of275 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, theexpected value of the underlying asset for these contracts has not appeared in thevaluation. 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 thecurrent share price of 260 will either increase to 285 or decrease to 250 at the endof one year. The annual risk-free interest rate is 4% compounded continuously andassume that this share pays no dividends. Calculate the price of an option that paysthe cash di�erence between the square of the share price at the end of the year and70225, provided that the di�erence 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 $1.2245and $2.5000 respectively. Verify that there is an arbitrage opportunity, and identifythe transactions required.

Exercise 5.16: [der16] Suppose we want to price a call option on a share usinga 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 dateof the option (ie at time T ) and that the owner of the option will not receive theshare dividend. Derive a formula for the number of stocks and bond that need tobe held to replicate this option payo�.

Exercise 5.17: [der17] Consider call and put (European) options (with the samestrike price) on gold. The spot price of gold is G(0) at time 0, and the maturity ofthe option is at time T . 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 timeT , �nd an updated version of the put-call parity that takes into account the storagecosts.

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 prob-ability 0.4, 6% with probability 0.2, 8% with probability 0.4, and is independent ofthe 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 anyyear t, the yield on a fund of investments has mean jt and standard deviation st. Inany year, the yield is independent of the yield in any other year. The accumulatedvalue, after n years, of a unit sum of money invested at time 0 is Sn.

(a) Derive formulae for the mean and variance of Sn if jt = j and st = s for allyears t.

(b) Calculate the expected value of S8 if j = 0.06.

(c) Calculate the standard deviation of S8 if j = 0.06 and s = 0.08.

Exercise 6.3: [sto3] (Institute of Actuaries Examination September 2002) $10,000is invested in a bank account which pays interest at the end of each year. The rateof 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 yearis independent of the rate applicable in any other year.

During the �rst year, the rate of interest per annum e�ective will be one of 3%,4% or 6% with equal probability. During the second year, the rate of interest perannum e�ective 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 ex-pected 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 theend of two years.

Exercise 6.4: [sto4] (Institute of Actuaries Examination April 2005) In any year,the interest rate per annum e�ective on monies invested with a given bank is equallylikely to be i1 or i2 (i1 > i2), and is independent of the interest rates in all previousyears.

(a) Express the mean and variance of the e�ective rate in a particular year interms of i1 and i2.

(b) The accumulated value at time t = 25 years of $1 million invested with thebank at time t = 0 has expected value $5.5 million and standard deviation$0.5 million. Find i1 and i2.

Exercise 6.5: [sto5] (Institute of Actuaries Examination September 2000) An in-surance company calculates the single premium for a contract paying $10,000 in tenyears time as the present value of the bene�t payable at the expected rate of interestit will earn on its funds. The annual e�ective rate of interest over the whole of thenext 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 pro�t at the end of the term of the contract.

6.2 Lognormal Model

Exercise 6.6: [sto6] (Institute of Actuaries Examination September 2005) An in-surance company has just written contracts that require it to make payments topolicyholders of $1,000,000 in �ve years time. The total premiums paid by policy-holders amounted to $850,000. The insurance company is to invest half the premiumincome in �xed interest securities that provide a return of 3% per annum e�ective.The other half of the premium income is to be invested in assets that have an uncer-tain return. The return from these assets in year t, it, has a mean value of 3.5% perannum e�ective and a standard deviation of 3% per annum e�ective. The randomvariables (1 + it) (for t = 1, 2, . . .) are independent and lognormally distributed.

28


(a) Deriving all necessary formulae, calculate the mean and standard deviation ofthe accumulation of the premiums over the �ve-year period.

(b) A director of the company suggests that investing all the premiums in theassets with an uncertain return would be preferable because the expectedaccumulation of the premiums would be greater than the payments due to thepolicyholders.

Explain why this still may be a more risky investment policy.

Exercise 6.7: [sto7] (Institute of Actuaries Examination April 2002) A companyis adopting a particular investment strategy such that the expected annual e�ectiverate of return from investments is 7% and the standard deviation of annual returnsis 9%. Annual returns are independent and (1 + it) is lognormally distributed whereit is the return in the tth year. The company has received a premium of $1,000 andwill pay the policyholder $1,400 after 10 years.

(a) Calculate the expected value and standard deviation of an investment of $1,000over 10 years, deriving all formulae that you use.

(b) Calculate the probability that the accumulation of the investment will be lessthan 50% of its expected value in ten years time.

(c) The company has invested $1,200 to meet its liability in 10 years time. Cal-culate the probability that it will have insu�cient funds to meet its liability.

Exercise 6.8: [sto8] (Institute of Actuaries Examination September 2007) The ex-pected e�ective annual rate of return from a bank's investment portfolio is 6% andthe standard deviation of annual e�ective returns is 8%. The annual e�ective re-turns are independent and (1 + it) is lognormally distributed, where it is the returnin year 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 lessthan 80% of the expected value.

Exercise 6.9: [sto9] (Institute of Actuaries Examination September 2000) An in-vestment bank models the expected performance of its assets over a �ve-day period.Over that period, the return on the bank's portfolio, i, has a mean value of 0.1%and standard deviation of 0.2%. (1 + i) is lognormally distributed.

Calculate the value of j such that the probability that i is less than or equal to j is0.05.

Exercise 6.10: [sto10] (Institute of Actuaries Examination September 2004) Theexpected annual e�ective rate of return from an insurance company's investments is

29


6% and the standard deviation of annual returns is 8%. The annual e�ective returnsare independent and (1 + it) is lognormally distributed, where it is the return in thetth 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 lessthan 90% of the expected value.

6.3 Dependence and Further Concepts

Exercise 6.11: [new3] Company A and Company B currently have a swap contractwhere A pays (annually) a �oating interest rate on a principal of $1 million to Bin exchange for a �xed rate of 5% from B to A. The �oating rate it in year t isassumed to be independent for each year, and (1 + it) is lognormally distributedwith it having mean 4% and standard deviation 2%.

(a) Find the expected value of the accumulated value after 10 years of the netreceipts (from the perspective of A) from the swap.

(b) Using Excel, simulate (1000 times) the interest rate for the next 10 yearsand verify your answer in (a). In addition, �nd the variance and plot thedistribution 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 meanreverting process de�ned by:

yt = µ+ β(yt−1 − µ) + σεt

where εt ∼ N(0, 1) iid for each time t = 0, 1, . . ..

It is also known that µ = 0.05, β = 0.4, and σ = 0.01. Denote the accumulatedvalue of $1 in one years time as S1. Find:

(a) 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 meanreverting process de�ned 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 accumulatedvalue of $1 in ten years time as S10. Using 1000 simulations in Excel, estimate:

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 independentand identically distributed, with (1 + it) ∼ LN(µ, σ2) and it having mean 4% andstandard deviation 2%. Denote as s30 the expected value and variance of the 30year accumulated value of an annual payment of $1 in advance.

(a) Derive recursive formulae which can be used to �nd E(s30 ) and Var(s30 ).

(b) Using Excel and the formulae in (a), �nd E(s30 ) and Var(s30 ).

(c) Estimate E(s30 ) and Var(s30 ) using simulation in Excel, and compare youranswers with those in (b).

31

Module 7

Solutions to Exercises

7.1 Module 1

Exercise 1.1 [int1]

Under compound interest:Accumulated amount at t = 4: 100(1.05)4 = 121.5506Accumulated amount at t = 5: 100(1.05)5 = 127.6282Interest earned = 6.08

Under simple interest:Accumulated amount at t = 4: 100(1 + 4 · 0.05) = 120Accumulated amount at t = 5: 100(1 + 5 · 0.05) = 125Interest earned = 5.00

Exercise 1.2 [int2]

200(1 + i)5 = 275⇒ i =(275200

)1/5 − 1 = 0.06576

Exercise 1.3 [int3]

200(1.05)t = 275⇒ t = ln(275/200)ln 1.05

= 6.527 years

Exercise 1.4 [int4]

275v5 = 215.47, where v = 11+i

Exercise 1.5 [int5]

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

150150(1 + i)n = 1000

150· 240 = 1600

Exercise 1.6 [int7]

The most one requires will be the amount X which is su�cient even when theinterest rate is always at its minimum of 5% in the latter 10 years (as this minimises

32


the interest earned). Therefore:X(1.08)10(1.05)10 = 1000000∴ X = 284360

Exercise 1.7 [int8]

The accumulation of $1 under the e�ective rate and the 8% and 5% rates should beequivalent:(1 + i)20 = (1.08)10(1.05)10

∴ i = 0.06489

Exercise 1.8 [int9]

Value at time t = 0:5 + 3v2 + v3 = 8.2307

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

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 = 0 andanother starting at t = 1. Thus, the value is given by:7.7217 + v7.7217 = 15.0757


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


100(1 + i)6 = 200∴ i = 0.12246


(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)10v2n + 10(1.0915)10vn − 67.50 = 0

33


∴ vn =−10(1.0915)10 ±

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

60(1.0915)10

= 0.8158 or − 1.1491

Since vn > 0, we have vn = 0.8158⇒ n = ln 0.8158ln v

= 2.325


(SOA Course 2 Nov 2000, Question 2)

For the �rst investment (the $100 at beginning of 1990), the interest credited in1993 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


Using v = 11+r

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

PVA = 610 + 475v + 340v2

PVB = 560 + 580v + 274v2

Option A is preferred when PVB − PVA = −50 + 105v − 66v2 > 0. Note that thequadratic −50 + 105v − 66v2 = 0 has no real roots. The quadratic is negative atv = 1, since −50 + 105− 66 = −11 < 0. This implies that PVB < PVA for all valuesof r, so Option B is always preferred to A.


(a) The equation of value for each of the alternative is given as follows with thecorresponding 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:

1000(1 + r)5 = 425

(1 +

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 - SeeSpreadsheet 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:

r =−2660±

√26602 + 4 · 1330 · 220

2660= 0.079543 or − 2.0795

∴ r ≥ 7.9543%

We ask this question to compare A and B. If we chose A, then after 3 years, wewould need to invest the $1330 at some interest rate (eg. by putting the money inthe bank) for 2 years, after which it can be compared with the $1550 from B (at 5years). Thus, we must be able to earn at least 7.95% during this 2 year period forA to be a better choice than B.

(c) The value of alternative C is:

425

(1 +

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 ratefrom alternative C.


Equation of value:P (v + v2 + v3 + v4 + v5) = 1000000

where v = 11.13

. Therefore, we obtain P = 284314.54.


d = 1− v = 0.06v = 1

1+i⇒ i = 1

v− 1 = 0.06383

Also: i = d/v = 0.06383

35



(SOA Course 2 May 2001, Question 12)

Note the relationship between accumulating using e�ective interest and discountrates:

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


An equation of value to solve for the e�ective 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 atthe 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.


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

March deposit accumulates to:

1000

(1 + 0.06

1

12

)(1 +

0.06

4

)19

= 1333.59

June deposit accumulates to:

1000

(1 + 0.06

1

12

)(1 +

0.06

4

)18

= 1313.88

September deposit accumulates to

1000

(1 + 0.06

1

12

)(1 +

0.06

4

)17

= 1294.46

36


December deposit accumulates to

1000

(1 + 0.06

1

12

)(1 +

0.06

4

)16

= 1275.33

Therefore, the total interest would be:

1333.59 + 1313.88 + 1294.46 + 1275.33− 4000 = 1217.25


Using the relations for equivalent rates

1 + i =

(1 +

i(m)

m

)m= (1− d)−1 =

(1− d(m)

m

)−m= eδ

we have:

Given rate Equivalent e�ective rate i A(4.5) = 10000(1 + i)4.5

d = 0.05 0.0526316 12596.32

i(2) = 0.05 0.0506250 12488.63

i(12) = 0.05 0.0511619 12517.38

d(2) = 0.05 0.0519395 12559.10

d(12) = 0.05 0.0513809 12529.11δ = 0.05 0.0512711 12523.23


Note that the accumulation function is given by:

a(t) = exp

(∫ t

0

δ(s)ds

)= exp

(∫ t

0

0.04

1 + sds

)= (1 + t)0.04

(a)a(2)− a(1)

a(1)=

(3)0.04 − (2)0.04

(2)0.04= 0.01635084

(b)a(3)− a(2)

a(2)=

(4)0.04 − (3)0.04

(3)0.04= 0.01157375

(c) Since A(t) = A(0)(1 + t)0.04, then A(0) = 200000(5)0.04

= 187530.19. Therefore:

A(2) =200000

(5)0.04(3)0.04 = 195954.86

37



We have

a(t) =

{1 + 0.1t t ≤ k

(1 + 0.1k) exp [0.08(t− k)] t > k

Hence

a(4) =

{(1 + 0.1k) exp [0.08(4− k)] k < 4

1.4 k ≥ 4

(a) To maximize a(4), di�erentiate w.r.t. k (assuming a(4) = (1 + 0.1k)e0.08(4−k),ie. k < 4; will need to check this)

d

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

and set to be zero. Solving for k (and noting that the exponential is positive) wehave:

0.1 = 0.08 (1 + 0.1k)

∴ k = 2.5

Check: at k = 2.5, a(4) > 1.4, so assumption of k < 4 is OK. Note that if the as-sumption of k < 4 was violated, then a(4) = 1.4 as opposed to our earlier expressionof (1 + 0.1k)e0.08(4−k), and a(4) will have a maximum of 1.4 (at any k ≥ 4).

(b)

a(t) =

{1 + 0.1t t ≤ 2.5

1.25e0.08(t−2.5) t > 2.5

ln a(t) =

{ln(1 + 0.1t) t ≤ 2.5

ln 1.25 + 0.08(t− 2.5) t > 2.5

∴ δ(t) =d

dtln a(t) =

{0.1

1+0.1tt ≤ 2.5

0.08 t > 2.5

Exercise 1.24 [new10]

We use the result:

1 + i =

(1 +

i(m)

m

)m= (1− d)−1 =

(1− d(m)

m

)−m= eδ

Therefore:

i(m) = m(eδ/m − 1

)d(m) = −m

(e−δ/m − 1

)See spreadsheet new10.xlsx for plotting i(m) and d(m). The plot should be similar tothe one below.

38


0 10 20 30 40 50

0.04

60.

048

0.05

00.

052

0.05

4

Nominal Interest/Discount Rates

Compounding Frequency

Nom

inal

Rat

e

i(m)d(m)

Note that δ = 0.05 and all the points on this plot yield the same e�ective rate ofinterest!

Exercise 1.25 [ann1]

(a) Each monthly payment is 112

of the nominal amount. The e�ective monthlyrate i is given by:

i =0.12

12= 0.01

Therefore:

PV =1000

12a72 0.01 +

(1

1.01

)72400

12a48 0.01 +

(1

1.01

)120

2000

=1000

12

(1− ( 1

1.01)72

0.01

)+

(1

1.01

)72400

12

(1− ( 1

1.01)48

0.01

)+

(1

1.01

)120

2000

= 4262.50 + 618.34 + 605.99

= 5486.80

Alternatively, we can obtain the annual e�ective rate j:

j = 1.0112 − 1 = 0.1268

39


Then, using annuities payable monthly:

PV = 1000a(12)

6+

(1

1 + j

)6

400a(12)

4+

(1

1 + j

)10

2000

= 1000

(1− v6j0.12

)+

(1

1 + j

)6

400

(1− v4j0.12

)+

(1

1 + j

)10

2000

= 5486.80

(b) Assume level annuity payments payable monthly, then

X

(1− ( 1

1.01)120

0.01

)= 5486.80

∴ X = 78.720

The quoted annual payment is then 78.720× 12 = 944.64.


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

P

3a20 = v20Pa∞

∴P

3

(1− v20

i

)= v20

P

i

1− v20 = 3v20

v20 = 0.25

∴ i = (0.25)−1/20 − 1 = 0.07177


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

AV = Cs10 0.06

= C(1.06)10

(1−

(1

1.06

)100.06

)= 13.1808C

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

PV = 15000a5 0.06

= 15000

(1−

(1

1.06

)50.06

)= 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 C = 4793.75.

40


Exercises 1.28�1.34

[annB1�annB7] See the Mathematics of Investment and Credit solutions manual.


The original and �nal monthly e�ective 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:

Pa300 i = 100000⇒ P =100000

a300 i= 472.1087

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

Pa180 i

This will also be the present value of the new annuity (with annual payments of P ′),which is valued at the new interest rate j:

Pa180 i = P ′a180 j

Therefore, the new payment is:

P ′ = Pa180 ia180 j

= 538.1869

Thus, the payment increase is:

P ′ − P = 66.08


The 1-year e�ective rate i and 2-year e�ective 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-yearannuity with annual payments of $100, plus an additional $100 every 2nd year (ie.an additional 20-year annuity with biannual payments of $100). Thus:

PV = 100a20 i + 100a10 j = 1844.16


A discount rate of 10% p.a. is equivalent to an interest rate of i = 11.11% p.a.Therefore, the current cash �ows are worth:

18000a12 i = 18000 (1 + a11 i) = 129162.10

We want annual payments of X where:

129162.10 = Xa20 i

∴ X = 16337.69

41



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


[annB8�annB9] See the solutions in the Broverman text and solutions manuals.


s10 is the accumulated value (at time 10) of $1 paid at each time t = 1, . . . , 10. Theaccumulated value (at time 10) of a single $1 paid at time n is given by:

exp

(∫ 10

n

1

20− tdt

)= exp [− ln(20− 10) + ln(20− n)]

= exp

[ln

(20− n

10

)]=

20− n10

Therefore:

s10 =10∑n=1

20− n10

=19 + 18 + . . .+ 10

10= 20− 1

10

10(10 + 1)

2= 14.5


The loan repayments are an increasing annuity with payments of P, 2P, . . . , 30P attimes t = 1, 2, . . . , 30. To determine the value of P , we use an equation of value (attime t = 0):

4000 = P (Ia)30 0.04 =

(a30 0.04 − 30v30

0.04

)∴ P = 18.32

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

10Pv + 11Pv2 + 12Pv3 + ...+ 30Pv21 = 9Pa21 0.04 + P (Ia)21 0.04

= 4774.80

This is greater than the original loan amount, and is because the earlier paymentsare (very) small relative to the latter payments. Therefore, the earlier payments areinsu�cient to pay o� the interest, let alone pay o� part of the principal. Hence, thisis also unlikely to exist in reality.


The borrowings are a decreasing annuity, whereas the repayments can be decom-posed into a level annuity of $300 and an increasing annuity starting at $200. Anequation 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

(Ia)15 0.05 =a15 0.05 − 15v15

0.05= 73.667689

(Ds)5 0.05 = (1 + i)5(Da)5 0.05

= (1 + i)5 [6a5 0.05 − (Ia)5 0.05]

= (1 + i)5(

6a5 0.05 −a5 0.05 − 5v5

0.05

)= 17.115531

We obtain:

X =300a15 0.05 + 200(Ia)15 0.05

(Ds)5 0.05(1 + i)= 993.11


The annual e�ective rate i is given by:

1 + i = 1.0254 ⇒ i = 0.103813

For convenience, we will determine the initial deposit by discounting all cash �owsto 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)10X at times t = 0, 1, . . . , 10

B. Deposits of 1.1025X, (1.1025)2X, . . . , (1.1025)11X at times t = 12, 11

2, . . . , 101

2

where time t is measured in years.

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

PVA = X + 1.1025Xv + . . .+ (1.1025)10Xv10

= X

(1− (1.1025v)11

1− 1.1025v

)= 10.934815X

PVB = v1/2(1.1025)[X + 1.1025Xv + . . .+ (1.1025)10Xv10

]= v1/2(1.1025)PVA

= 11.474726X

Therefore, the initial deposit X is found as follows:

PVA + PVB = 110000v11

∴ X = 1656.19

43



The present value is $67989.65.

Refer to Spreadsheet new8.xlsx.


There are a number of ways to decompose the payments into annuity streams, whichallow the present value to be determined more easily.

A simple method is to consider 3 cash �ow streams, each containing payments 1year apart (which also means we will use the annual e�ective rate of 0.10):

A. 10, 20, 30, 40 at times t = 13, 11

3, 21

3, 31

3

B. 20, 30, 40, 50 at times t = 23, 12

3, 22

3, 32

3

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:

PVA = 10v1/3(Ia)4

PVB = 10v2/3(Ia)4 + 10v2/3a4PVC = 10v(Ia)4 + 20va4

The annuity factors are given by:

(Ia)4 = 1 + 2v + 3v2 + 4v3

= 8.302780

a4 = (1.1)

(1− v4

0.1

)= 3.486852

Therefore, we obtain the present value:

PVA + PVB + PVC = 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 annu-ity:

PVX = 10(Ia)3 j a4 i = 194.3179

44


where i = 0.1 is the annual e�ective rate, and j = (1.1)1/3− 1 is the 13-year e�ective

rate.

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

PVY = 10a3 j(Ia)3 i = 135.6289

Summing these up, we obtain the same present value:

PVX + PVY = 329.95


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 (eachpayment is of amount 10(1.05)nP which is worth 10(1.05)nPvn at time t = 10, andvn = (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


(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)2Xv2 + . . .+ (1.03)20Xv20 = 10000

∴ 1.03Xv

(1− (1.03v)20

1− 1.03v

)= 10000

∴ X = 608.1346

45


(c) To be fair, we need the present value of both payment streams to be equal (to10000).

In nominal terms:

20∑k=1

608.13(1.03)kvk0.05 =20∑k=1

802.43vk0.05 = 10000

or in real terms:

20∑k=1

608.13vkr =20∑k=1

802.43(1.03)−kvkr = 10000

where r = 1.051.03− 1.


(a) The annual e�ective interest rate is i = eδ − 1 = 0.05127.

We note that the payment at time n is 1+2+. . .+n. Therefore, the perpetuitycan be decomposed into a level perpetuity of 1 (1st payment at time 1), plus alevel perpetuity of 2 (1st payment at time 2), etc. This can be interpreted asan increasing perpetuity, with each regular payment being a level perpetuityitself:

PV = va∞ + 2v2a∞ + 3v3a∞ + . . .

= a∞(v + 2v2 + 3v3 + . . .

)= a∞ (Ia)∞

Therefore:

PV =

(1

d

)(a∞i

)=

1

d2i= 8200

where d = iv = 0.04877.

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

46


●●●●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●●

0 100 200 300 400 500

020

4060

8010

012

0

Time

PV

of p

aym

ent

Aside: The graph is actually related to the Gamma function. The present value ofthe nth payment is:

1

2n2vn +

1

2nvn =

1

2n2e−δn +

1

2ne−δn

Thus, we require:∞∑n=1

n2e−δn and∞∑n=1

ne−δn

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

Γ(k) = (k − 1)! =

∫ ∞0

xk−1e−xdx ≈∞∑k=1

xk−1e−x

Therefore, the present value is:

PV =1

2

(∞∑n=1

n2e−δn +∞∑n=1

ne−δn

)

≈ 1

2

(∫ ∞0

x2e−δxdx+

∫ ∞0

xe−δxdx

)=

1

2

(1

δ3

∫ ∞0

u2e−udu+1

δ2

∫ ∞0

ue−udu

)where u = δx

=1

2

(2!

δ3+

1!

δ2

)

47


The force of interest is δ = 0.05, which results in:

PV ≈ 8200


[annB10�annB15] See the solutions in the Broverman text and solutions manuals.


We require 100(Ia)10 , 100(Ia)(12)

10, and 100(Ia)10 . The monthly e�ective 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:

100(Ia)10 =a10 − 10v10

i= 3937.38

100(Ia)(12)

10=a10 − 10v10

i(12)

= 4026.81

100(Ia)10 =a10 − 10v10

δ= 4035.01


The present value of the annuity is given by:

PV =

∫ 14

1

(t2 − 1)vtdt

where the discount factor vt varies (continuously) with time t:

vt = exp

(−∫ t

0

1

1 + sds

)= exp (− ln(1 + t)) =

1

1 + t

Therefore, we have:

PV =

∫ 14

1

t2 − 1

t+ 1dt =

∫ 14

1

(t− 1)dt = 84.50

48


7.2 Module 2

Exercise 2.1 [lif1]

Note that if the force of mortality is constant, then the survival function is expo-nential:

tpx = exp

(−∫ t

0

µx+sds

)= e−µt

Therefore:

E[Z] =

∫ ∞0

btvttpxµx+tdt

=

∫ ∞0

e0.05te−0.06te−0.01t0.01dt

= 0.01

∫ ∞0

e−0.02tdt

=1

2

E[Z2] =

∫ ∞0

e0.1te−0.12te−0.01t0.01dt

= 0.01

∫ ∞0

e−0.03tdt

=1

3

∴ Var(Z) = E[Z2]− (E[Z])2 =1

3− 1

4=

1

12

Exercise 2.2 [lif2]

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

∞∑k=0

ak+1 kpxqx+k =∞∑k=0

(1− vk+1

d

)kpxqx+k

This gives us a v term, which is similar to the RHS. We don't want the q, so we

write it in terms of p:

49


=∞∑k=0

(1− vk+1

d

)kpx (1− px+k)

=∞∑k=0

(1− vk+1

d

)(kpx − k+1px)

=1

d

[∞∑k=0

(kpx − k+1px)−∞∑k=0

vk+1 (kpx − k+1px)

]

=1

d

[(∞∑k=0

kpx −∞∑k=1

kpx

)− v

∞∑k=0

vkkpx +∞∑k=0

vk+1k+1px

]

=1

d

[1− v

∞∑k=0

vkkpx +∞∑k=1

vkkpx

]

=1

d

[−v

∞∑k=0

vkkpx +∞∑k=0

vkkpx

]since v00px = 1. Therefore:

=1− vd

∞∑k=0

vkkpx

=∞∑k=0

vkkpx

The LHS is an annuity certain with a term equal to the future lifetime of theindividual. The RHS considers each annual payment separately, noting that theindividual will receive it if he/she is alive. Both describe the cash �ows of a termannuity, and therefore they must be equal.

Exercise 2.3 [lif3]

Let K(x) be the curtate lifetime random variable of (x). We have:

ax = E[aK(x)+1

]=∞∑k=0

ak+1 Pr [K(x) = k]

=∞∑k=0

ak+1 k|qx

=∞∑k=0

[1− vk+1

d

]k|qx

=1

d

[∞∑k=0

k|qx −∞∑k=0

vk+1k|qx

]

=1− Axd

50


∴ dax + Ax = 1

Exercise 2.4 [lif4]

The required values are:

Pr[K(0) = k] = kp0qk

= {0.1, 0.045, 0.0855, 0.1539, 0.2462, 0.2586, 0.1108, 0} for k = 0, . . . , 7

e0 = E[K(0)] =6∑

k=0

k Pr[K(0) = k]

= 3.6203

Pr[K(2) = k] = kp2q2+k

= {0.1, 0.18, 0.288, 0.3024, 0.1296, 0} for k = 0, . . . , 5

e2 = E[K(2)] =4∑

k=0

k Pr[K(2) = k]

= 2.1816

A2 =4∑

k=0

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

= 0.8576

2A2 =4∑

k=0

v2(k+1) Pr[K(2) = k]

= 0.7379

A12:3 =

2∑k=0

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

= 0.5073

A 12:3 = v33p2 = 0.3732

A2:3 = A12:3 + A 1

2:3 = 0.8805

a2 =4∑

k=0

vkkp2

= 2.9900

a2 = a2 − 1 = 1.9900

a2:3 =2∑

k=0

vkkp2

= 2.5102

See Excel spreadsheet lif4.xlsx for the calculations.

51


Exercise 2.5 [lif5]

Let T = T (x) be the (random) lifetime of the non-smoker and T S = T S(x) be thelifetime of the smoker. Note that we are comparing newborns. We want to �nd:

Pr(T S > T

)=

∫ ∞0

Pr(T S > t

)fT (t) dt

where we have conditioned on the future lifetime T to construct the integral.

Note that:

Pr(T S > t

)= tp

S0 = exp

(−∫ t

0

cµudu

)= (tp0)

c

where tp0 = Pr(T > t). Also, the density of T is given by:

fT (t) =d

dtFT (t) =

d

dt(1− S(t)) = − d

dttp0

since the survival function S(t) is equivalent to tp0. Therefore:

Pr(T S > T

)= −

∫ ∞0

(tp0)c (tp0)

′ dt

= −

[(tp0)

c+1

c+ 1

]∞0

=1

1 + c(because tp0 is of the form e−[parameter]·t)

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 theprobability that he outlives the non-smoker should be less than 1/2 (which isalso 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 ableto 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, andthe payment into the sinking fund. The interest payment is simply 7% of theloan:

20000(0.07) = 1400

The sinking fund payment is an amount X such that the sinking fund willaccumulate 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 asthe two methods are equivalent when the sinking fund and loan interest ratesare the same. Hence, we would prefer the standard loan arrangement, as itinvolves lower annual repayments. In general, the standard loan is preferableif 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 = Lfn = 5000 · 0.10 · 2 = 1000

Monthly loan repayments are:

R =L+ I

N=

6000

24= 250

The monthly e�ective rate j is the solution to the following equation of value:

Ra24 j = 5000

Using Newton-Raphson, we de�ne the function f as follows:

f(j) = Ra24 j − 5000 = 250

(1− (1 + j)−24

j

)− 5000

The derivative is:

f ′(j) = 250

(24j(1 + j)−25 − (1− (1 + j)−24)

j2

)Starting with j0 = 0.10:

j1 = j0 −f(j0)

f ′(j0)= −0.062716

j2 = j1 −f(j1)

f ′(j1)= −0.022528

j3 = j2 −f(j2)

f ′(j2)= 0.004878

j4 = j3 −f(j3)

f ′(j3)= 0.014291

j5 = j4 −f(j4)

f ′(j4)= 0.015125

This results in an annual e�ective rate i of:

i = (1 + j5)12 − 1 = 19.74%

Using spreadsheet loa3.xlsx for 10 iterations of Newton Ralphson. The solutionis j = 0.015131. Therefore, the e�ective 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 3.4�3.6

[loaB3�loaB5] See the Mathematics of Investment and Credit solutions manual.

Exercise 3.7 [loa5]

The e�ective monthly rate is 0.1812

= 0.015. Therefore, the monthly repayment X isgiven by:

Xa100 0.015 = 20000⇒ X = 387.41

As we are only required to provide the last four instalments (ie. 97th�100th pay-ments), we can begin the loan schedule at time t = 96 (months). The outstandingbalance at time t = 96 is:

OB96 = Xa4 0.015 = 1493.23

The loan schedule for times t = 97, . . . , 100 can then be found using the followingrecursive equations:

It = OBt−1 · 0.015

PRt = X − ItOBt = OBt−1 − PRt

Therefore, we obtain:

Time Interest Principal Repaid Outstanding Balance(t) (It) (PRt) (OBt)96 ? ? 1493.2397 22.40 365.01 1128.2298 16.92 370.49 757.7399 11.37 376.05 381.69100 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:

PV = 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. Sincen must be an integer, the consolidated loan cannot have level repayments with afurther restriction that these repayments must sum to the original total (1090.28).

55


Therefore, the loan will be assumed to involve monthly repayments of X, with anadditional �nal payment of Y to account for any remaining outstanding balance.

We have two equations to describe the two constraints (repayments must sum tooriginal total; e�ective rate must remain the same):

nX + Y = 1090.28

Xan 0.01 + Y vn+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. Alogical choice would be to set n to di�erent values (as it has to be an integer) andget X and Y by solving the simultaneous equations (noting that the coe�cientsn, vn+1, an 0.01 are all known once n is known). Here are the results for variouschoices of n:

n X Y11 69.25 328.5612 72.45 220.8613 74.02 128.0514 74.51 47.1315 74.29 -24.1116 73.60 -87.4017 72.61 -144.0518 71.41 -195.13

Intuitively, the �nal payment should not be negative and should be as small aspossible as it represents an extra `top-up' payment to ensure the loan is fully repaidto the nearest cent. The purpose of the �nal payment is not to repay a signi�cantportion of the loan. Hence, the solution where n = 14 is the most appropriate,and so we choose n = 14 with X = 74.51 and Y = 47.13. In other words, theconsolidated annuity is of $74.51 for 14 months, with a �nal payment of $47.13 atthe end of 15th month.

Note: the necessity of the �nal payment Y can be veri�ed by observing (from thetable) that there is no solution where n is an integer and Y = 0.

The table can be found in spreadsheet loa6.xlsx.

Exercise 3.9 [loa7]

The initial payment X can be found as follows, noting the monthly e�ective rate is0.1212

= 0.01:

Xa12 0.01 + (1.1)Xv12a12 0.01 + (1.1)2Xv24a12 0.01 = 47500

∴ X = 1440.80

The loan schedule is shown in the tutorial spreadsheet provided (loa7.xls).

56


Exercise 3.10 [loa8]

(a) Let X be the annual payment. We have:

Xa10 0.0675 = 600000⇒ X = 84441.97

Hence total payment is 10X = 844419.69.

(b) Let Y be the monthly payment. The monthly e�ective 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 thetotal payment in (a).

(c) The amount to settle the loan is the outstanding balance at time t = 5. Thiswill not change when the interest rate changes, as the outstanding balanceis determined from the past (ie. using the recursive relationships of a loanschedule). Therefore, the amount will be OB5 under the old interest rate of6.75%:

OB5 = Xa5 0.0675 = 348558.74

Note that this implies the prospective method (based on the future) becomesincorrect when interest rates change, while the retrospective method (basedon the past) remains correct. If we used the prospective method, we wouldhave calculated an outstanding balance of:

Xa5 0.0725 = 343923.44

This is the present value of future repayments (as payments now have to bediscounted at the new rate), but it is not the outstanding balance.


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

The time step will be one quarter, ie. t = 0, 14, . . . , 19. The interest (It), principal re-

paid (PRt) outstanding balance (OBt), and nominal/actual payments (NPRt, APRt)are described using a set of recursive relationships:

It = OBt− 14· 0.02

PRt = APRt − ItOBt = OBt− 1

4− PRt

NPRt =75000

15= 5000 (t = 5, 6, . . . , 19)

APRt = 5000 (t = 5, 6, . . . , 19)

57


where we note that the quarterly e�ective interest rate is 0.084

= 0.02, and thenominal and actual payments are equal since the loan is repaid at par.

The net cash �ows received by the purchaser are the payments of interest (It) andprincipal (APRt). The price for a required e�ective yield i is found by discountingthese cash �ows at the rate i. Therefore:

P =∑t

(1

1 + i

)t(It + APRt)

where the summation is taken over t = 0, 14, . . . , 19.

For i = 0.10, the price is P = 66636.60.

For i =(1 + 0.10

2

)2 − 1 = 0.1025, the price is P = 65565.63.


This question requires a spreadsheet model due to the complexity of the capitalgains 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

(APRt −

NPRt

500000P

)+

The capital gain is determined as the actual repayment less the purchase price. Theportion of the loan repaid by APRt is

NPRt

500000of the entire loan, and this portion of

the loan was purchased at a price of NPRt

500000P .

The price is the present value of the net cash �ows at the e�ective yield of 6%:

P =∑t

vtCFt =∑t

vt (APRt + It − ITt − CGTt)

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 i = 0.06):∑t

vtAPRt = 25000(1.05)v10a10∑t

vt (It − ITt) = (1− τI) (0.08)∑t

vtOBt− 14∑

t

vtCGTt = 0.3

(25000(1.05)− 25000

500000P

)+

v10a10

where: ∑t

vtOBt− 14

= 500000a(4)

10+ 475000v10a

(4)

1+ . . .+ 25000v28a

(4)

1

= 500000a(4)

10+ 25000v10a

(4)

1(Da)

(4)

19

58



(a) The equation of value is:

80 = 6a10 + 100v10

(b) Newton-Raphson �nds the root of f(x) = 0 recursively as follows:

xn = xn−1 −f(xn−1)

f ′(xn−1)

where xn is the root obtained from the nth iteration.

The equation we need to solve is:

0 = 6

(1− v10

i

)+ 100v10 − 80

=6

i− 6(1 + i)−10

i+ 100(1 + i)−10 − 80

Thus, de�ne:

f(x) =6

x− 6(1 + x)−10

x+ 100(1 + x)−10 − 80

See Spreadsheet new13.xlsx for the iterations in Excel. The solution is i =9.1349%.


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 hasone 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 e�ective yield. The solution j = 8.00% can easily beveri�ed 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 veri�ed by substitution.

Therefore both bonds (on their own) give an e�ective yield of 8% p.a.

The solution by spreadsheet is provided in loa10.xls. The equations to be usedare:

NPRt = 100 (t = 4)

APRt = NPRt

It = i ·OBt−1

OBt = OBt−1 −NPRt

ITt = (0.35)It

CGTt = (0.50)

(APRt −

NPRt

100P

)P =

∑t

vt (APRt + It − ITt − CGTt)

where i is the interest rate (15% or 8%) and j is the annual e�ective yield,and v = (1 + j)−1.

We require the yield such that P is (a) 105.80 or (b) 85.34. These both resultin a solution of j = 8.00%.

(b) Instead of putting $100 in one single bond, we now put $x into the 15% bondand $ (100− x) into the 8% bond. This means that we buy:

x

105.8of the 15% bond with face value 100

and100− x85.34

of the 8% bond with face value 100

We want to �nd the value x such that the net capital gain is 0. Note thatthis would minimise our net CGT liability, as we have minimised the capitalgain and we do not get a refund from capital losses (therefore no advantage inhaving a negative CG).

CG =x

105.8(100− 105.8) +

100− x85.34

(100− 85.34) = 0

∴ x =100(14.66)

85.345.8

105.8+ 14.66

85.34

= 75.81

The equation of value is then:

100 =x

105.8

(15(1− 0.35)a4 j + 100v−4j

)+

100− x85.34

(8(1− 0.35)a4 j + 100v−4j

)The solution j = 8.46% can be veri�ed by substitution.

For the spreadsheet model, we note that the combined portfolio can be con-sidered 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. purchasinga par bond). The interest rate becomes:

i = 0.15α + 0.08(1− α)

where α = x105.8

is the proportion of the �rst bond purchased, and 1−α = 100−x85.34

is the proportion of the second bond purchased.

The rest of the spreadsheet remains the same. The yield can be solved numer-ically to obtain j = 8.46%.


[loaB6�loaB7] See the Mathematics of Investment and Credit solutions manual.


When an Australian government bond is �ex-interest�, the owner of the bond at thedate of the bond going ex-interest receives the next coupon payment. The buyerdoes not receive the next coupon payment. The buyer receives the maturity facevalue and all coupons except the next payment.

Price is:v

fd (C +Gan + 100vn)

where the interest rate used is the semi-annual e�ective rate i = 0.04752

= 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 = 5.752

= 2.875, and n = 10. Price on 11 June 2006is (next coupon is not received):

P = v4

182

(2.875a10 + 100v10

)= 0.999484 (2.875× 8.808866 + 100× 0.790789)

= 0.999484× 104.404433

= 104.351


See Sherris (p. 40�43). Users that may be interested include superannuation fundsand anyone with long term liabilities that rise in line with in�ation (such as in�ation-linked life annuities, or some general insurance liabilities). See also:http://en.wikipedia.org/wiki/In�ation-indexed_bond

61



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


A spreadsheet model can be used with the following equations (where time t =0, 1

2, . . . , 30 is in years):

it =

{0.035 t ≤ 26.5

0.04 t > 26.5

NPRt =

150000 t = 10, 11, . . . , 18

250000 t = 19, 20, . . . , 29

300000 t = 30

Rt =

1.05 t = 10, 11, . . . , 18

1.10 t = 19, 20, . . . , 29

1.12 t = 30

APRt = NPRt ×Rt

It = it ·OBt− 12

OBt = OBt− 12−NPRt

P =∑t

vtj (APRt + It)

where j = 0.065 is the required e�ective yield. Also note that the original outstand-ing balance (face value) is

∑tNPRt = 4, 400, 000.

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


The price P is the present value of all cash �ows at the net e�ective yield of j = 0.08.

This question is not too complex, and can be solved by hand. The complexityarises from capital gains tax. However, note that the capital gain for the �rst tworepayments (at times t = 2, 3) is:

400− 400

1200P < 0

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

400(1.2)− 400

1200P = 480− 1

3P

which will only be positive if P < 1440. Therefore, the capital gains tax for the 3rdrepayment is:

CGT4 =

{0.20

(480− 1

3P)

P < 1440

0 P ≥ 1440

62


Thus, we can solve this question by considering two cases: P < 1440 and P ≥ 1440.

(i) For P < 1440, there is CGT, and the price is given by:

P = (1− 0.20)[0.10

(1200v + 1200v2 + 800v3 + 400v4

)]+ 400(v2 + v3 + 1.2v4)− 0.20

(480− 1

3P

)v4

which can be solved to obtain P = 1249.47. This is consistent with theassumption of P < 1440, so it is a valid solution.

(ii) For P ≥ 1440, there is no CGT, and the price is given by:

P = (1− 0.20)[0.10

(1200v + 1200v2 + 800v3 + 400v4

)]+ 400(v2 + v3 + 1.2v4)

= 1258.80

This is not consistent with the assumption of P ≥ 1440, so it is not a validsolution.

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) =

(1 +

0.04875180

2

)−1= 0.976204

v(1) =

(1 +

0.05031182

2

)−2= 0.951525

v(1.5) =

(1 +

0.05234408

2

)−3= 0.925421

v(2) =

(1 +

0.05448436

2

)−4= 0.898067

and hence the value of the bond is:

P =6.75

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

(b) To determine the yield (e�ective yearly), we need to solve:

6.75a(2)

2 i+ 100v2i = 102.467

This can be done numerically to obtain i = 5.50537% (using the functionwritten earlier in Exercise 3.13). The equivalent semi-annual e�ective yield isj = (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)

v(0.5) + v(1)= 0.025146

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)

v(0.5) + v(1) + v(1.5) + v(2)= 0.027173

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, whichholds due to no arbitrage (ie. accumulation of $1 with certainty must be thesame 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 + st−1)t−1 − 1

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

f0, 12

= s0.5 = 0.024376

f 12,1 =

(1 + s1)2

(1 + s0.5)− 1 = 0.025937

f1,1.5 =(1 + s1.5)

3

(1 + s1)2 − 1 = 0.028207

f1.5,2 =(1 + s2)

4

(1 + s1.5)3 − 1 = 0.030459

where the spot and forward rates are semi-annual e�ective rates. These for-ward rates are equivalent to nominal p.a. rates (payable semi-annually) of4.8752%, 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 twobonds (ie. x 9% bonds and y 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 theZCB is therefore:

P = −3.5× 101.00 + 4.5× 93.20 = 65.9

Exercise 4.3 [irr3]

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

exp

(∫ t−1

0

s(x)dx

)exp (ft−1,t) = exp

(∫ t

0

s(x)dx

)

∴∫ t−1

0

s(x)dx+ ft−1,t =

∫ t

0

s(x)dx

∴ ft−1,t = rt− r(t− 1) = r

which is a �at forward rate curve.

Exercise 4.4 [irr4]

Denote 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 overthe corresponding time periods:(

1 +f0,12

)2

=

(1 +

f0,0.52

)(1 +

f0.5,12

)= (1.025)(1.0275) = 1.053188(

1 +f0.5,1.5

2

)2

=

(1 +

f0.5,12

)(1 +

f1,1.52

)= (1.0275)(1.03) = 1.058325(

1 +f1,22

)2

=

(1 +

f1,1.52

)(1 +

f1.5,22

)= (1.03)(1.0305) = 1.061415(

1 +f1.5,2.5

2

)2

=

(1 +

f1.5,22

)(1 +

f2,2.52

)= (1.0305)(1.03125) = 1.062703(

1 +f2,32

)2

=

(1 +

f2,2.52

)(1 +

f2.5,32

)= (1.03125)(1.035) = 1.067344

65


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:

v(t) =1

(1 + st2

)2t

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

v(1.5) =1(

1 + s1.52

)3 =1(

1 + s12

)2 (1 + f1,1.5

2

)=

1(1 + 0.0525

2

)2 (1 + 0.075082

2

)= 0.915142

v(2) =1(

1 + s22

)4 =1(

1 + s12

)2 (1 + f1,1.5

2

)(1 + f1.5,2

2

)=

1(1 + 0.0525

2

)2 (1 + 0.075082

2

) (1 + 0.020290

2

)= 0.905951

The resulting price is P = 105.591.

Exercise 4.6 [irr6]

We have:

P = D · an +Re−δn

∂P

∂δ= D

∂

∂δan − nRe−δn

66


where:

∂

∂δan =

∂

∂δ

∫ n

0

e−δtdt

=

∫ n

0

∂

∂δe−δtdt

= −∫ n

0

te−δtdt

= −(I a)n

Therefore, the duration is given by:

D(δ) = − 1

P· ∂P∂δ

= −−D(I a)n − nRe−δn

D · an +Re−δn

=g(I a)n + nvn

gan + vn

where g = D/R.

Exercise 4.7 [irr7]

(a) Consider a nominal amount of $1 of bond. The coupon income is g = 0.05 p.a.,payable continually to redemption at time n. The duration is (from Exercise4.6):

D(δ) =g(I a)n + nvn

gan + vn

at δ = 0.07, which results in 11.592 for n = 20 and 14.345 for n = 60.

(b) To �nd the maximum duration, we need to di�erentiate D(δ) w.r.t. n (and setthis to zero). To do so, it is useful to note the following intermediate results:

∂

∂nvn = vn ln v = −δvn

∂

∂nnvn = nvn ln v + vn = (1− δn) vn

∂

∂nan =

1

δ

∂

∂n(1− vn) =

1

δ(δvn) = vn

∂

∂n

(I a)n

=1

δ

∂

∂n(an − nvn) =

1

δ(vn + (1− δn)vn) = nvn

Therefore:

∂

∂nD(δ) =

(gan + vn) (gnvn + (1− δn)vn)−(g(I a)n + nvn

)(gvn − δvn)

(gan + vn)2

67


Setting this to zero:

0 = (gan + vn)(gnvn + (1− δn)vn

)−(g(I a)n + nvn

)(gvn − δvn)

= (gan + vn)(1 + (g − δ)n

)vn −

(g(I a)n + nvn

)(g − δ) vn

= (gan + vn) + (g − δ)[gnan + nvn − g(I a)n − nvn

]= (gan + vn) + (g − δ)

[gnan − g(I a)n

]Noting that we require an equation in terms of an (ie. without vn and (I a)n ):

0 = (gan + 1− δan ) + (g − δ)[gnan −

g

δ(an − nvn)

]= 1 + (g − δ)

[an + gnan −

g

δ(an − n+ nδan )

]= 1 + (g − δ)

[an −

g

δ(an − n)

]∴ (δ − g)

[an −

g

δ(an − n)

]= 1

∴ an +g

δ(n− an ) =

1

δ − g

∴ δan + g(n− an ) =δ

δ − gas required (g = 0.05, δ = 0.07). This can then be solved numerically toobtain n = 64.349 and a corresponding duration of 14.349.


(a) The modi�ed duration and convexity are:

MD = − 1

P

∂P

∂i

C =1

P

∂2P

∂i2

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

P (i) =1− (1 + i)−10

i∂P

∂i=

10i(1 + i)−11 + (1 + i)−10 − 1

i2

∂2P

∂i2=

2(−55i2(1 + i)−12 − 10i(1 + i)−11 − (1 + i)−10 + 1)

i3

For i = 0.05 we see that:

P (0.05) = 7.7217

MD = − 1

P

∂P

∂i= 4.856

C =1

P

∂2P

∂i2

= 35.602

68


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

∂P

∂i=P (i+ h)− P (i− h)

2h∂2P

∂i2=P (i+ h)− 2P (i) + P (i− h)

h2

where h is a small increment in the interest rate. For this question, we shalltake h = 0.005. See Excel spreadsheet new11.xlsx. For h = 0.005, modi�edduration is 4.8576 and convexity is 35.6093.

Exercise 4.9 [lif6]

Recall that:

ax =∞∑k=0

vkkpx =∞∑k=0

(1 + i)−kkpx

To �nd the duration, we di�erentiate w.r.t. i:

d

diax = −

∞∑k=0

k(1 + i)−k−1kpx = −∞∑k=0

kvk+1kpx

Therefore the modi�ed duration of ax is:

MDx = v

∑∞k=0 kv

kkpx∑∞

l=0 vllpx

and the Macaulay Duration is:

Dx =MDx

v=

∑∞k=0 kv

kkpx∑∞

l=0 vllpx

=∞∑k=0

wkk, where wk =vkkpx∑∞l=0 v

llpx

.

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

Exercise 4.10 [irr8]

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

VA (i) = M5v5 +M15v

15

VL (i) = 1000v10

where v = (1 + i)−1.

(b) A zero cost portfolio that provides a positive (non negative) pro�t with zerochance of loss.

69


(c) To ensure that VA (i) = VL (i), we require:

M5 +M15v10 = 1000v5

The duration of the assets and liabilities are:

DA(i) =

∑t tCtv

t∑tCtv

t=

5M5v5 + 15M15v

15

M5v5 +M15v15

DL(i) =10vt

vt= 10

Therefore, to ensure that DA(i) = DL(i), we require:

5M5v5 + 15M15v

15

M5v5 +M15v15= 10

which simpli�es to:M5 = M15v

10

Solving these simultaneous equations, we get M5 = 340.29 and M15 = 734.66.We can check that this is an arbitrage opportunity by looking at VA (i)−VL (i)for each parallel shift in the �at yield curve:

i VA (i)− VL (i)7% 0.558% 09% 0.45

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


See Sherris p. 87.

70



At i = 4.5%, we have (for the liabilities):

VL =∑t

Ctvt

= 3v + 4v2 + 3v3 + 2v4

= 10.84

DL =

∑t tCtv

t

VL

=3v + 8v2 + 9v3 + 8v4

10.84= 2.29

CL =

∑t t(t+ 1)Ctv

t+2

VL

=6v3 + 24v4 + 36v5 + 40v6

10.84= 7.84

For the assets:

VA = M0.5v0.5 +M5v

5

DA =0.5M0.5v

0.5 + 5M5v5

VA

Ensuring that VA = VL and DA = DL, we get:

M0.5v0.5 +M5v

5 = 10.84

0.5M0.5v0.5 + 5M5v

5 = 24.79

These can be solved to obtain M0.5 = 6.6803 and M5 = 5.3647.The e�ect of a change in interest rates on the surplus is determined in Excel Spread-sheet irr10.xlsx. The results are shown in the following table:

Rate VA VL S6.5% 10.39 10.38 0.014.5% 10.84 10.84 0.002.5% 11.34 11.33 0.01Twist (d) 10.32 10.67 �0.35Twist (e) 10.86 10.89 �0.03

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


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

71


0 5 10 15 20

0.05

00.

052

0.05

40.

056

0.05

8

Spot Curve

Term

Spo

t Rat

e

(b) Fisher-Weil duration relaxes the assumption that the yield curve is �at. Inpractice, yield curves are rarely �at. Therefore, di�erent interest rates mustbe used when present valuing cash �ows of di�erent maturities. By splittingthe cash �ows into ZCBs of di�erent maturities, we can value the cash �ows byvaluing 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 theyield curve is no longer �at):

VL =∑t

Ct(1 + st)−t

= 10.609

DL =

∑t tCt(1 + st)

−t

VL= 2.275

CL =

∑t t(t+ 1)Ct(1 + st)

−(t+2)

VL= 7.572

For the assets:

VA = M0.5(1 + s0.5)−0.5 +M5(1 + s5)

−5

DA =0.5M0.5(1 + s0.5)

−0.5 + 5M5(1 + s5)−5

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) toobtain M0.5 = 6.588 and M5 = 5.516.

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

Histogram of Surplus

Surplus

Fre

quen

cy

−0.02 −0.01 0.00 0.01 0.02

050

100

150

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


At i = 4%, we have:

VL = 10.96

DL = 2.29

CL = 7.94

73


Letting P1 be the price of the 0.75 year bond, and P2 be the price of the 8 yearbond, we have (assuming annual coupons and $100 face value):

P1 = 100.99

D1 = 0.75

C1 = 1.21

P2 = 126.93

D2 = 6.43

C2 = 49.29

Let x1 be the number of units of the 0.75 yr bond, and x2 be the units of the 8 yrbond. Therefore, we have:

VA = x1P1 + x2P2

DA =x1P1D1 + x2P2D2

VA

Ensuring that:

VA = VL

DA = DL

we obtain two simultaneous equations which can be solved to give x1 = 7.905 andx2 = 2.345. These correspond to investing 7.983 and 2.977 in the two bonds respec-tively.

A check of convexity shows that CA−CL = 7.49 > 0, so the portfolio is immunised.


(a) At 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 ormore bonds with maturities on either side of 3 (as the portfolio duration wouldbe an average). Furthermore, to maximise the asset convexity (as immunisa-tion suggests) we would wish to use the 1 year and 5 year bonds (Barbellstrategy). Using the 1 year and 5 year bonds, and solving using methods sim-ilar to previous exercises, we obtain an investment strategy of $0.8396m intoboth bonds (ie. face values of 0.8900 and 1.1236). The convexity of the assetswould be 14.240 > 10.680.

(b) Cash�ow matching involves �nding assets that match the liability CF perfectlywithout considering the costs. Since we have ZCBs available, we would simplyinvest 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 + 1.1236v5 − 2v3 = 0.0003

For the CF matching, we have:

VA − VL = 2v3 − 2v3 = 0

Therefore, both methods have surpluses of approximately zero (exact for CFmatching).

(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 spotindex and storage costs):

f ′t = Ster(T−t) − dsT−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% di�erence).

Aside: The di�erence between the actual and theoretical futures price is usuallyquite small (less than 1%), except for major events (eg. 1987 �nancial crisis). Thedi�erence (ft − St) is called the �futures to cash spread� or basis, whereas (f ′t − St)is the �theoretical basis�.

Exercise 5.2 [der2]

The forward price will be the accumulated cost (purchase and funding) less couponsreceived.

The price of the bond today is:

3.25a16 i + 100v16i = 97.214171

where i = 0.06962

= 0.0348 is the semi-annual e�ective rate.

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

97.214171

[(1 +

0.065

12

)12(1 +

0.07

12

)12]

= 111.223057

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

12

)6 − 1 and i2 =(1 + 0.07

12

)6 − 1 are the semi-annual e�ectivefunding 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 purchaseof the forward contract:

97.522944 = 3.25a12 j + 100v12j

which results in j = 3.50% per half-year, equivalent to an e�ective 7.013% p.a.

Exercise 5.3 [der3]

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

f90 = 420

(1 + (0.0975)

90

365

)+ (0.025) (420)

90

365= 432.69

However, our short contract is for a forward price of $450, ie. we have an agreementto sell gold for $450 in 90 days time. Thus, based on today's market conditions, weshould be able to make a pro�t of 450− 432.69 = 17.31 in 90 days time. Today, thispro�t is worth:

17.31

1 + (0.0975) 90365

= $16.90

Exercise 5.4 [der4]

(Also see Example 6.9 in Sherris, p. 112) The forward price is the accumulatedcosts less receipts (dividends). The cost of purchasing the shares now (includingtransaction costs) is:

1.02 (10000 · 10) = 102000

This is accumulated to account for funding costs over the next 6 months:

102000

(1 +

0.06

12

)6

= 105098.506

The accumulated dividends are:

0.4 (10000)

(1 +

0.06

12

)3

= 4060.3005

Therefore the forward price is 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, forexample if Ft,T > Ste

r(T−t)− der(T−t1), then one could short a forward, purchase theasset with funds borrowed at the risk-free rate, invest the dividend at the risk-freerate (at time t1), and exercise the forward contract at time T . This would lock in apositive pro�t (at time T ) 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 30�120 is 7.3%, whereas the 120-dayspot rate is 7.4%. Therefore, for no-arbitrage, we require a repo rate r given by:(

1 +30

365r

)(1 +

90

365· 0.073

)=

(1 +

120

365· 0.074

)∴ r = 0.075639

However, the 30-day spot rate is 7.5% (< 7.5639%). Therefore, we could:

Action (t = 0) (t = 30) (t = 120)Borrow $1m at 30-day spot (7.5%) 1,000,000 �1,006,164 0Short forward for $1,006,164 (7.3%) 0 1,006,164 �1,024,275Invest $999,948 in 120-day spot (7.4%) �999,948 0 1,024,275Total 52 0 0

This results in a guaranteed pro�t of $52 at zero cost.

For the 2nd contract, the forward rate for days 90�180 is 7.0%, whereas the 180-dayspot rate is 7.4%. Similarly, we have:(

1 +90

365r

)(1 +

90

365· 0.070

)=

(1 +

180

365· 0.074

)∴ r = 0.076677

The 90-day spot rate is 7.4% (< 7.6677%). Therefore:

Action (t = 0) (t = 30) (t = 120)Borrow $1m at 90-day spot (7.4%) 1,000,000 �1,018,247 0Short forward for $1,018,247 (7.0%) 0 1,018,247 �1,035,822Invest $999,352 in 180-day spot (7.4%) �999,352 0 1,035,822Total 648 0 0

This results in a guaranteed pro�t 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 Sti . The swap provides for party A attime ti:

Sti −XThe swap allows to get the di�erence between the spot price at time ti and Xwithout actually exchanging the commodity. The swap price X is set such that theinitial value of the swap is 0.

To value the swap, consider what happens if party A enters into a series of shortforward positions (which cost nothing to enter in) to replicate the swap. In order tosettle the forward contract (buy the commodity and sell it to the long party at thepre-speci�ed price at the same time), the payo� is

F0,ti − Stiwhere F0,ti represents the forward price set at time 0 (now) for settlement at timeti. The initial value of this forward is 0. Hence for times t1, t2, . . . , tM the net e�ectof the swap and forward must be equal to 0:

0 =M∑i=1

e−rti [(Sti −X) + (F0,ti − Sti)]

=M∑i=1

(F0,tie

−rti −Xe−rti)

=M∑i=1

(S0 −Xe−rti

)= MS0 −X

M∑i=1

e−rti

∴ X =MS0∑Mi=1 e

−rti


(a) Let (hC , BC) be the units of stocks and bonds we hold in our portfolio toreplicate the payo� of the call.

For the call option with exercise price 275, the payo� function is :

C1 =

{max (285− 275, 0) = 10 if stock price goes up

max (250− 275, 0) = 0 if stock price goes down

78


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

C ′1 =

{hC285 +BCe

0.05 = 10 if stock price goes up

hC250 +BCe0.05 = 0 if stock price goes down

Solving these equations:

hC =10

285− 250

BC = e−0.05 (250)

(−10

285− 250

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

C0 = hC260 +BC = 6.34

(b) For the put option with exercise price 275, the payo� function is

P1 =

{max (275− 285, 0) = 0 if stock price goes up


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

P ′1 =

{hP285 +BP e

0.05 = 0 if stock price goes up

hP250 +BP e0.05 = 25 if stock price goes down


hC =−25

285− 250

BC = e−0.05 (285)

(25

285− 250

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

P0 = hP260 +BP = 7.93

(c) The put-call parity relationship states that (in words):

Call Value + Discounted Strike Price = Put value + Share Price

ie.C0 +Xe−rt = P0 + S0

This is satis�ed, since (for t = 1) :

C0 +Xe−rt = 6.34 + 275e−0.05 = 267.93

P0 + S0 = 7.93 + 260 = 267.93

79



For the call option with exercise price 50, the payo� function is

C1 =

{max (55− 50) = 5 if stock price goes up


Hence we want to �nd a portfolio holding (h,B) units of the stock and bond, suchthat the value at time 1 is:

C ′1 =

{h55 +B

(1 + 0.05 1

12

)= 5 if stock price goes up

h45 +B(1 + 0.05 1

12

)= 0 if stock price goes down


h =5

55− 45

B =(45)(

1 + 0.05 112

) ( −5

55− 45

)Therefore:

C0 = h50 +B = 2.59


The pricing is done by ensuring that there are no arbitrage properties, rather thanthrough any particular probability assessments (except to decide where the stockcan go to, eg. up to 55 or down to 45). In terms of pricing, there is no di�erencebetween a model where the stock price can go up to 55 with probability 99% anda model where the stock price can go up to 55 with probability 2%. An importantobservation is that these contracts are priced as a function of the current underlyingasset price, that is, they are relative prices. The probability of movements in theasset price do impact the underlying asset price but once this is known, these othercontracts are determined by no-arbitrage based on the current underlying asset price.


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

X1 =

{max (2852 − 2652, 0) = 11000 if stock price goes up

max (2502 − 2652, 0) = 0 if stock price goes down

Hence we want to �nd a portfolio holding (h,B) units of the stock and bond suchthat the value at time 1 is:

X ′1 =

{h285 +Be0.04 = 11000 if stock price goes up

h250 +Be0.04 = 0 if stock price goes down

80



h =11000

285− 250

B = e−0.04 (250)

(−11000

285− 250

)Since this portfolio pays the same amount as the option at time 1, to have anarbitrage free market this portfolio must be worth the same as the option at time0. Hence:

X0 = h260 +B = 6223.81


Put call parity does not hold since:

C0 +K

1 + i= 1.2245 +

22

1.05= 22.1769

P0 + S0 = 20 + 2.5 = 22.5

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 pro�t, since it is a negative cost.

At time 1, the cash �ows of the put, stock, call and bond will all cancel out.


The stock is assumed to either go up to su or down to sd. The bond starts o� as 1and rises to ert at time t, where r is the risk free rate.

Suppose we want to price this derivative that pays xu or xd. For a call option, wehave:

xu = (su −K)+

xd = (sd −K)+

Consider a portfolio that holds φ stocks and ψ bonds. The value of this portfoliotoday is:

V (0) = φs0 + ψ1

The value of this portfolio at time T is:

φ (su +D) + ψerT if the stock goes up

φ (sd +D) + ψerT if the stock goes down

Therefore, we need φ and ψ such that the payo�s are identical (so that this portfoliowill have the same value as our derivative):

φ (su +D) + ψerT = xu

φ (sd +D) + ψerT = xd

81



φ =xu − xdsu − sd

ψ = e−rT(xu −

(xu − xdsu − sd

)(su +D)

)The value of the portfolio with these φ and ψ will be the same as the value of ourderivative.


Consider two investment portfolios:

• Portfolio A: one call plus cash of (K − c) e−rT

• 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 if GT > K (i.e. option is exercised)

0 +K − c = K − c if GT ≤ K (i.e. option is worthless)

The value of portfolio B at expiry is given by:{0 +GT − c = GT − c if GT > K (i.e. option is exercised)

K −GT +GT − c = K − c if GT ≤ K (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 beexercised before expiry, the portfolios must also have equal value for any time t < T .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) (by independence)

= 1000 [E(1 + i)]10 (identically distributed)

= 1000(1.06)10

= 1790.85

82


(b) The variance of the accumulation is:

Var(1000S10) = 10002[E(S210

)− [E (S10)]

2]where:

E(S210

)= E

[(1 + i1)

2 . . . (1 + i10)2]

= E[(1 + i)2

]10=[0.4(1.04)2 + 0.2(1.06)2 + 0.4(1.08)2

]10= 1.1239210

Therefore:Var(1000S10) = 9145.60

and:σ(1000S10) = 95.63

(c) (i) The mean accumulation depends only on the mean interest rate, whichis 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. Thestandard deviation will also be larger, as investing in a longer term willresult 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(S2n

)− [E (Sn)]2

where:

E(S2n

)= E

[(1 + i1)

2 · · · (1 + in)2]

=(E[(1 + i)2

])n=[E(1 + 2i+ i2

)]n=(1 + 2j + j2 + s2

)nnoting that E(X2) = Var(X) + [E(X)]2 for a random variable X. Therefore:

Var(Sn) =(1 + 2j + j2 + s2

)n − (1 + j)2n

(b) E(S8) = 1.59385

(c) σ(S8) =√

2.65844− 2.54035 = 0.34364

83


Exercise 6.3 [sto3]


E(10000S2) = 10000E [(1 + i1)(1 + i2)]

= 10000E [1 + i]E [1 + i2] (iid)

= 10000

(1

31.03 +

1

31.04 +

1

31.06

)(0.7 · 1.05 + 0.3 · 1.04)

= 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

31.032 +

1

31.042 +

1

31.062

)(0.7 · 1.052 + 0.3 · 1.042

)= 1.193465

Therefore:Var(10000S2) = 19278

Exercise 6.4 [sto4]

(a) The mean and variance are:

E(i) =i1 + i2

2Var(i) = E(i2)− [E(i)]2

=1

2(i21 + i22)−

1

4(i1 + i2)

2

=1

4(i1 − i2)2

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

E (Sn) = (1 + j)n

Var(Sn) =(1 + 2j + j2 + s2

)n − (1 + j)2n

Therefore:

5.5 = (1 + j)25 ⇒ j = 0.0705686

0.52 =(1 + 2j + j2 + s2

)25 − (1 + j)50 ⇒ s2 = 0.000377389

84


Solving for i1 and i2:i1 + i2

2= j = 0.0705686

∴ i1 + i2 = 0.1411372

1

4(i1 − i2)2 = s2 = 0.000377389

∴ i1 − i2 = 0.0388530

Therefore, we obtain i1 = 0.089995 and i2 = 0.051142.

Exercise 6.5 [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 pro�t 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 (since 1 + it iid)

= 425000(1.03)5 + 425000(1.035)5

= 997458

85


The standard deviation is found as follows:

Var(P ) = Var

(425000(1.03)5 + 425000

5∏k=1

(1 + ik)

)

= 4250002Var

(5∏

k=1

(1 + ik)

)

= 4250002E

(5∏

k=1

(1 + ik)2

)− 4250002 (1.035)10

E

(5∏

k=1

(1 + ik)2

)=(E[(1 + i)2

])5=[E(1 + 2i+ i2

)]5=[1 + 2E(i) + [E(i)]2 + Var(i)

]5=(1 + 2(0.035) + (0.035)2 + (0.03)2

)5= 1.416534

∴ σ(P ) =√Var(P ) = 32743.21

(b) Investing all premiums in the risky assets is likely to be more risky becausealthough there may be a higher probability of the assets accumulating to morethan $1 million, the standard deviation would be twice as high so the proba-bility 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) (by independence)

= 1000 [E(1 + i)]10 (identically distributed)

= 1000(1.07)10

= 1967.15

The variance of the accumulation is:

Var(1000S10) = 10002[E(S210

)− [E (S10)]

2]where:

E(S210

)= E

[(1 + i1)

2 · · · (1 + i10)2]

= E[(1 + i)2

]10=[1 + 2E(i) + [E(i)]2 + Var(i)

]10=[1 + 2(0.07) + (0.07)2 + (0.09)2

]10= 1.115310

86


Therefore, we have:

σ(1000S10) =√Var(1000S10) = 531.65

(b) We want to �nd Pr (1000S10 < 0.5(1967.15)). The distribution of the accumu-lation term S10 =

∏10k=1(1 + ik) is determined as follows:

(1 + ik) ∼ LN(µ, σ2)

∴ ln(1 + ik) ∼ N(µ, σ2)

∴10∑k=1

ln(1 + ik) ∼ N(10µ, 10σ2)

∴ exp

(10∑k=1

ln(1 + ik)

)=

10∏k=1

(1 + ik) ∼ LN(10µ, 10σ2)

where:

E(1 + i) = 1.07 = eµ+12σ2

Var(1 + i) = 0.092 = e2µ+σ2(eσ

2 − 1)

∴ 0.092 = (1.07)2(eσ

2 − 1)

∴ σ2 = 0.007050

∴ µ = ln 1.07− 1

2(0.007050) = 0.064134

Therefore:

Pr (1000S10 < 0.5(1967.15)) = Pr (S10 < 0.983575)

= Pr

(Z <

ln 0.983575− 10µ√10σ

)where Z ∼ N(0, 1)

= Pr (Z < −2.4778)

= 0.00661

(c) The probability required is:

Pr (1200S10 < 1400) = Pr

(S10 <

7

6

)= Pr

(Z <

ln 76− 10µ√

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) (by independence)

= [E(1 + i)]10 (identically distributed)

= (1.06)10

Therefore:

E (2, 000, 000S10) = 2, 000, 000 (1.06)10 = 3, 581, 695

(b) The distribution of S10 is lognormal:

(1 + ik) ∼ LN(µ, σ2)

∴ ln(1 + ik) ∼ N(µ, σ2)

∴10∑k=1

ln(1 + ik) ∼ N(10µ, 10σ2)

∴ S10 =10∏k=1

(1 + ik) = exp

(10∑k=1

ln(1 + ik)

)∼ LN(10µ, 10σ2)

where:

E(1 + i) = 1.06 = eµ+12σ2

Var(1 + i) = 0.082 = e2µ+σ2(eσ

2 − 1)

∴ 0.082 = (1.06)2(eσ

2 − 1)

∴ σ2 = 0.0056798

∴ µ = ln 1.06− 1

2(0.0056798) = 0.055429

Therefore, the required probability is:

Pr (S10 < 0.8E(S10)) = Pr(S10 < 0.8(1.06)10

)= Pr (S10 < 1.4327)

= Pr

(Z <

ln 1.4327− 10µ√10σ

)= Pr (Z < −0.8171)

= 0.207

Exercise 6.9 [sto9]

We have (1 + i) ∼ LN(µ, σ), where:

E(1 + i) = eµ+12σ2

= 1.001

Var(1 + i) = e2µ+σ2(eσ

2 − 1)

= 0.0022

88


These can be solved simultaneously to obtain µ and σ2:

Var(1 + i) =(eµ+

12σ2)2 (

eσ2 − 1

)∴ 0.0022 = (1.001)2

(eσ

2 − 1)

∴ σ2 = 0.000003992

∴ µ = ln 1.001− 1

2(0.000003992) = 0.0009975

Therefore, we can �nd j as follows:

0.05 = Pr(i < j)

= Pr(1 + i < 1 + j)

= Pr(ln(1 + i) < ln(1 + j))

= Pr

(Z <

ln(1 + j)− µσ

)

∴ln(1 + j)− µ

σ= −1.645

∴ 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.


E(S10) = E [(1 + i1) · · · (1 + i10)]

= E[1 + i1] · · ·E[1 + i10] (by independence)

= [E(1 + i)]10 (identically distributed)

= (1.06)10

Therefore:

E (1, 000, 000S10) = 1, 000, 000 (1.06)10 = 1, 790, 848

(b) The distribution of S10 is lognormal:

(1 + ik) ∼ LN(µ, σ2)

∴ ln(1 + ik) ∼ N(µ, σ2)

∴10∑k=1

ln(1 + ik) ∼ N(10µ, 10σ2)

∴ S10 =10∏k=1

(1 + ik) = exp

(10∑k=1

ln(1 + ik)

)∼ LN(10µ, 10σ2)

89


where:

E(1 + i) = 1.06 = eµ+12σ2

Var(1 + i) = 0.082 = e2µ+σ2(eσ

2 − 1)

∴ 0.082 = (1.06)2(eσ

2 − 1)

∴ σ2 = 0.0056798

∴ µ = ln 1.06− 1

2(0.0056798) = 0.055429

Therefore, the required probability is:

Pr (S10 < 0.9E(S10)) = Pr(S10 < 0.9(1.06)10

)= Pr (S10 < 1.61172)

= Pr

(Z <

ln 1.61172− 10µ√10σ

)= Pr (Z < −0.32304)

= 0.373


(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, σ2)

where:

E(1 + i) = 1.04 = eµ+12σ2

=⇒ Var(1 + i) = 0.022 = e2µ+σ2(eσ

2 − 1)

=⇒ 0.022 = (1.04)2(eσ

2 − 1)

=⇒ σ2 = ln

(1 +

0.022

1.042

)=⇒ µ = ln 1.04− 1

2σ2

90


The simulation can be found in Spreadsheet new3.xlsx.

−0.2 −0.1 0.0 0.1 0.2 0.3 0.4

01

23

45

Simulated Density of S

S

Den

sity

(c) A company who has an exposure to a �oating rate may want to hedge thisinterest rate risk by using a �xed-for-�oating swap. For example, if the com-pany has a liability to repay a loan at a �oating rate, they could enter into aswap to pay a �xed rate and receive a �oating rate. The payment and receiptof the �oating rate will net to zero, allowing the company to pay a net ratewhich is �xed.


(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 and y0 = 0.06respectively.

(b) The variance is:

Var(S1) = Var(1 + y1)

= Var(1 + µ+ β(y0 − µ) + σε1)

= σ2

= 0.012

91


(c) The probability is:

Pr(S1 < 1.04) = Pr(1 + y1 < 1.04)

= Pr(µ+ β(y0 − µ) + σε1 < 0.04)

= Pr

(ε1 <

−0.01− 0.4(y0 − 0.05)

0.01

)where ε1 ∼ N(0, 1)

Therefore Pr(S1 < 1.04) = 0.2743 and Pr(S1 < 1.04) = 0.0808 for y0 = 0.04and y0 = 0.06 respectively.


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

Histogram of S (4%)

S

Fre

quen

cy

1.4 1.5 1.6 1.7 1.8 1.9

050

100

150

200

250

92


Histogram of S (6%)

S

Fre

quen

cy

1.4 1.5 1.6 1.7 1.8 1.9

050

100

150

200

250


(a) Denoting the mean and standard deviation of it as j and s respectively (ie.j = 0.04 and s = 0.02), the formulae can be derived as in the lecture notes:

sn = (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]

E(s2n)

= E[(1 + yn)2(sn−1 + 1)2

]= E

[(1 + yn)2

]E[(sn−1 + 1)2

]by independence

= E(1 + 2yn + y2n

)E(s2n−1 + 2sn−1 + 1

)=(1 + 2j + j2 + s2

) [E(s2n−1

)+ 2E (sn−1 ) + 1

]The variance can then be determined using:

Var (sn−1 ) = E(s2n−1

)− [E (sn−1 )]2

(b) The recursion can be found in Excel spreadsheet new9.xlsx. We see thatE (s30 ) = 58.33 and V ar (s30 ) = 17.16.

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

1 + i ∼ LN(µ, σ2)

93


where:

E(1 + i) = 1.04 = eµ+12σ2

=⇒ Var(1 + i) = 0.022 = e2µ+σ2(eσ

2 − 1)

=⇒ 0.022 = (1.04)2(eσ

2 − 1)

=⇒ σ2 = ln

(1 +

0.022

1.042

)=⇒ µ = ln 1.04− 1

2σ2

The simulation can be found in spreadsheet new9.xlsx.

45 50 55 60 65 70 75

0.00

0.02

0.04

0.06

0.08

Simulated Density of Annuity

s30

Den

sity

94

fm all exercises

Documents