bpp questions spring 2011

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PROfESSIONAL EDUCATIN

ExamM

Financial Economics

Question Answer Bank

Spring 2011 exams

Over 300 exam-style questions with full solutions to

help you prepare for the Financial Economics

segment of Society of Actuaries Exam M and

Exam 3F of the Casualty Actuarial Society



Exam MFE Financial Economics Question Answer Bank

Introduction

How to use this Question Answer Bank

Ths Question Answer Bank contains over 300 exam-style questions to help you prepare

thoroughly and efficiently for the Financial Economics segment of the Spring 2011 Exam M.

We recommend that you use the Question Answer Bank as you work through the course

and as part of your review. You should aim to answer as many questions as possibly under

realistic exam conditions without reference to your notes. You'll learn more by tring a

question under exam conditions and gettng it wrong than by referring to your notes.

Keep track of which questions you have attempted and whether you got those questions

right or wrong. If you have enough time before the exam, you should aim to have another

attempt at any questions that you got wrong.

Additional help from BPP

Practice exams

There are two BPP Exam MFE Practice Exams available for the Spring 2011 exam. You'll

also get full solutions, with advice on exam technique and references to the course of

reading.

Practice exams are a great way to test your preparation in the final weeks before the exam

using previously unseen questions.

Flashcards

BPP's user-friendly flashcards supplement the study program. They wil help you

remember the most important formulas, lists and other pertinent information. Our

flashcards are spiral bound to make them easy to use.

The most up-to-date information on upcoming BPP seminars with dates, locations and

seminar registration deadlines can be found on the BPP website at www bpptraíníng com

(9 BPP Professional Education: 2011 exams

Pagei



ExamMFE

Question & Answer Bank, Chapter 0

Question 3

You are given:

(i) The 6-month forward price of Stock X is 673.15.

(ii) The 1-year forward price of Stock X is 697.13.

(ii) Stock X does not pay dividends.

(iii) Stock Y has the same current price as Stock X and pays dividends at an annual

continuously-compounded rate of 3 .

Determine the 2-year forward price of Stock Y.

A 600.03

B 676.53

C 704.14

D 747.68

E 773.27

2 (9 BPP Professional Education




ExamMFE

ExamMFE

Question & Answer Bank

Chapter

1

Question 1

The interest rate for borrowing is 7.0%. The interest rate for lending is 6.0%. Both interest rates

are continuously compounded.

The ask price for Stock X is 80.00. The bid price for Stock X is 79.90. Stock X does not pay

dividends.

You are given the following information about a European put option on Stock X:

· The quoted ask price for the put option is 4.00.

· The quoted bid price for the put option is 3.95.

. The time until expiration is 1 year.

· The strike price is 72.

A 1-year European call option on Stock X has a strike price of 72. Determine which of the

following ask prices for the European call option permits arbitrage.

A 16.00

B 16.10

C 16.50

D 16.80

E 16.90

Question 2

You are given:

(i) At-year 100 strike European call option on stock X is currently sellng at 8.

(ii) At-year 100 strike European put option on stock X is currently sellng at 4.

(iii) The pre-paid forward price for the delivery of a share of stock X in t-years is 98.

Determine the price of a t-year zero-coupon bond that is certain to mature for 1,000.

A At most 925

B At least 925, but less than 935

C At least 935, but less than 945

D At least 945

E Cannot be determined from the given information.

(9 BPP Professional Education

1



ExamMFE


Question 3

The current exchange rate is 1.25 per Swiss franc. The interest rate on dollars is 8%, and the

interest rate on francs is 7%. Both interest rates are continuously compounded.

All of the options described below are European, and they expire in 1 year.

Determine which of the following has the most value.

A A dollar-denominated call option on one franc, with a strike price of 1.10

B A dollar-denominated put option on one franc, with a strike price of 1.10

C A franc-denominated put option on one dollar, with a strike price of 0.90909 francs

D 0.09 francs

E 0.11 dollars

Question

4

Stock X does not pay dividends.

An at-the-money European call on Stock X is 2.20 more expensive than an at-the-money

European put on Stock X. The call and the put both expire in 6 months.

The continuously-compounded interest rate is 9.0%.

Determine the current price of Stock X.

A 45.70

B 47.80

C 49.00

D 50.00

E 52.30

Question 5

The current price of a stock is 190. The stock does not pay dividends.

A European call option on the stock expires in 6 months and has a strike price of 200. The

premium for the call option is 15.94.

A European put option on the stock expires in 6 months and has a strike price of 200. The

premium for the put option is 16.18.

Calculate the cost of a conversion for a 1,000 T-bil that matures in 6 months.

A 904.84

B 923.00

C 948.80

D 950.00

E 951.20

2 9 BPP Professional Education




ExamMFE

Question 6

A common stock is currently priced at 52. You are given the following:

(i) The stock wil pay dividends equal to 1 at the end of each 3-month period and

beginnng 3 months from today

(ii) A 7-month European put option on the stock with a strike price of 50 currently sells for

2.95.

(iii) The continuously-compounded risk-free interest rate is 6% per year.

Determine the current price of a 7-month European call option on the stock with a strike price of

50.

A 1.76

B 2.77

C 4.47

D 4.71

E 4.92

Question 7

For a share of stock, you are given:

(i) The current stock price is 50.

(ii) ö = 0.08

(iii) The continuously-compounded risk-free interest rate is r = 0.04 .

(iv) The prices for 1-year European puts (P) under various strike prices (K) are shown in the

table below:

K

P

40

1.39

50

6.79

60

12.20

70

21.10

You own 4 special call options, one each with the strike prices in (iv). Each option can be

exercised immediately or one year from now.

Determine the highest strike price for which it is optimal to immediately exercise any of the

options.

A 40

B 50

C 60

D 70

E It is not optimal to exercise any of these call options


3



ExamMFE


Question 8

A certain stock pays continuous dividends at the rate 5=0.04. The continuously-compounded

risk-free rate of interest is r = 0.0798 .

A one-year at-the-money European put option on this stock costs 3 less than an at-the-money

European call option.

Determine the common strike price underlying these options.

A 77

B 78

C 79

D 80

E 81

Question 9

The exchange rate is 0.050 Japanese yen for 1 Venezuelan Bolivar.

The yen-denominated interest rate is 7 .

The Bolivar-denominated interest rate is 13%.

A 9-month European yen-denominated call on the Bolivar has a premium of 0.004603 yen.

A 9-month European yen-denomiated put on the Bolivar has a premium of 0.004793 yen.

Both the call and the put have the same strike price. Calculate the strike price, measured in yen.

A 0.045

B 0.048

C 0.051

D 0.053

E 0.055





ExamMFE

Question 10

You are given:

(i) The spot exchange rate is 920 for €l,OOO (euros).

(ii) The continuously-compounded risk-free interest rate for dollars is 6%.

(iii) The contiuously-compounded risk-free interest rate for euros is 3.2%.

(iv) The current price of a 1-year dollar-denominated European call option on one euro with

strike 0.90 is 0.0606. .

Determine the price of a 1-year dollar-denominated European put option on €l,OOO if the strike is

900.

A Less than 17.20

B At least 17.20, but less than 17.30

C At least 17.30, but less than 17.40

D At least 17.40, but less than 17.50

E At least 17.50

Question 11

Thee months ago, a 10-year bond was issued with a par value of 1,000. Its annual coupon rate

is 10%, and its coupons are paid semi-annually.

A 6-month European call option on the bond has a strike of 1,100, and its premium is 55.05.

The contiuously-compounded interest rate is 8%.

Find the value of a 6-month European put option on the bond that has a strike of 1,100.

A 12.16

B 14.23

C 34.78

D 83.79

E 95.87


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ExamMFE


Question 12

A European call option gives its owner the right to obtain a share of Stock A by giving up a share

of Stock B. The value of this call option is 3.

Another European call option gives its owner the right to obtain a share of Stock B by giving up a

share of Stock A. The value of this call option is 5.

Both call options have the same expiration date, which is at least 8 months from now.


Stock B pays a dividend of 2 in 6 months, and the current price of Stock B is 12.

Stock A does not pay dividends. Find the current price of Stock A.

A 7.93

B 8.07

C 8.14

D 11.93

E 12.07

Question 13

You are given:

(i) An option whose current price is 4 gives the option holder the right to exchange one

share of stock WX for one share of stock YZ one year from today.

(ii) The contiuously-compounded risk-free rate of interest is r = 0.065 .

(iii) The only dividend to be paid in the next 2-year period by stock YZ is 3 at a time

6 months from today.

(iv) The current price of stock YZ is 15.

(v) Stock WX pays no dividends and has a current price of 20.

Determine the price of an option giving the option holder the right to exchange one share of stock

YZ for one share of stock WX one year from today.

A 9.70

B 10.80

C 11.90

D 13.00

E Cannot be determined from the given information.




Question & Answer Bank, Chapter 1 ExamMFE

Question 14

A European call option on Stock X has a strike price of 58 and expires in 9 months.


The price of Stock X is 68. Stock X pays a 5 dividend in 6 months and again in one year.

Using the information above, determie the miimum possible premium for the call option.

A 0.00

B 3.72

C 8.36

D 9.45

E 10.00

Question 15

You are given:

(i) A European call option on stock A has a 58 strike and expires in 8 months.

(ii) The continuously-compounded risk-free rate of interest is r = 0.075 .

(iii) The current price of stock A is 72.

(iv) Dividends payable to the holder of stock A in the next year are 2 at the end of every

3-month period beginnng today.

Determine a lower bound for the value of the call in (i) based on the observation that this call is

deep in the money.

A 11.54

B 11.89

C 12.24

D 12.58

E 12.93


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ExamMFE


Question 16

A 1-year European call option has a strike price of 40 and a premium of 2.50.

The underlying stock has a price of 41 and does not pay dividends.


At time 0, an arbitrageur implements an arbitrage strategy involving the purchase or sale of

exactly one call option. The arbitrageur reinvests all arbitrage profits at the interest rate.

At the end of 1 year, the stock price is 39. Calculate the value of the arbitrageur's arbitrage

profits at the end of the year.

A 0.45

B 0.47

C 1.00

D 1.40

E 1.47

Question 17

Each of the following put option prices, P(K), is written as a function of its strike price.

P(100) =5

P(110) =13

P(120) = 20

Each of the puts has the same underlying asset and expires in 1 year.

The continuously-compounded interest rate is 11 %.

At time 0, an arbitrageur implements an arbitrage strategy involving the purchase or sale of two

of the 110-strike put options (possibly involving the other two options as well). The arbitrageur

reinvests all arbitrage profits at the interest rate.

At the end of 1 year, the price of the underlying asset is 112. Calculate the value of the

arbitrageur's arbitrage profits at the end of the year.

A 1.00

B 1.12

C 8.00

D 9.12

E 13.00




Question Answer Bank, Chapter 1

ExamMFE

Question 18

Consider the following assertions that have been made about vanila call and put options on

stocks:

1) The price of an American call option must always be greater than the price of an

otherwise identical European option.

(2) The price of an American option must always be greater than or equal to its intrinsic

value.

(3) The price of a European option must always be greater than or equal to its intrinsic value.

The markets are arbitrage-free and there are no bid-ask spreads.

Which of these assertions are necessarily correct?

A (2) only

B (3) only

C 1) and 2) only

D (2) and (3) only

E 1), 2) and 3)

Question 19

Four vanila call options with the same strike price are available in the same market on the same

currency:

. Option P is a 2-year American option.

· Option Q is a 1-year American option.

. Option R is a 2-year European option.

. Option S is a 1-year European option.

The market is arbitrage-free and there are no bid-ask spreads. The underlying currency earns a

non-zero interest rate.

Whch of the following is the strongest statement that can be made with certainty about the

option prices (which are indicated by the corresponding letters)?

A P,Q,R,S? 0

B P?R, Q?S

C P?Q, P?R, Q?S

D P?Q, P?R, Q?S, R?S

E P?R?Q?S


9



ExamMFE


Question

20

P( K) denotes the price of a 1-year American put option with strike price K .

For two such options, both on the same underlying index, the current market prices are:

P(5500) = 50

P(6000) = 170

According to the convexity inequality for American option prices which of the following

inequalities can be deduced for P(5800), the price of an option on the index with strike price

5800?

A P(5800) :: 94

B P(5800) ? 94

C 94:: P(5800) :: 122

D P(5800):: 122

E P 5800) ? 122

Question 21

P = P ( 5, K, T, 5, r) is the price of aT-year European put option with strike price K on stock X

that pays continuous dividends at the rate 5. 5 is the current stock price and r is the

continuously-compounded risk-free rate of interest.

C = C ( 5, K, T, 5, r) is the price of aT-year European call option with strike price K on stock

X.

Whch of the following relationships is incorrect if there are no arbitrage opportunities.

A

òC

-:;-1

òK

B

òP ~ 1

òK

C

òP _ òC ~ 0

òK òK

D

ò2p

-:;0

òK2

E

ò2C

-:;0

òK2

10





ExamMFE

Question

22

C(K) is the current price of a 1-year European call option on a certain stock where K is the

strike price. Let Ki = 40, K2 = 48, and K3 =60.

Suppose that 5 1) denotes the random price of a share of this stock in one year.

Which of the following positions in these three call options on the same stock wil have a payoff

one year from today that is positive if 40 ~ 5 (1) ~ 60 and is zero otherwise.

A Buy 10 calls with strike Ki, buy 10 calls with strike K3, and sell 20 calls with strike K2.

B Buy 11 calls with strike Ki, buy 9 calls with strike K3, and sell 20 calls with strike K2.

C Buy 11 calls with strike Ki, sell 9 calls with strike K3, and buy 20 calls with strike K2.

D Buy 12 calls with strike Ki, buy 8 calls with strike K3, and sell 20 calls with strike K2.

E Buy 12 calls with strike Ki, sell 8 calls with strike K3, and buy 20 calls with strike K2 .

Question

23

For the option position in Question 22, which of the following option price scenarios would result

in an arbitrage?

C(40)

C(48)

C(60)

A

10.00 7.50 5.00

B

10.00 7.50 5.50

C

10.00

7.00

5.50

D 9.00

7.50 5.50

E

9.00

7.25 4.00

Question

24

An options trader has the following holdings in his portfolio. All the options expire on

December 31 this year.

. 4,000 shares of Stock XYZ

. a long position in 3,000 call options on Stock XYZ with strike price 10

. a short position in 2,000 call options on Stock XYZ with strike price 15

. a long position in 6,000 put options on Stock XYZ with strike price 15

What is the minimum value this portfolio could have on the expiration date?

A 0

B 30,000

C 70,000

D 75,000

E 90,000


11



ExamMFE


Question 25

An investor currently has a holding of Stock XYZ, consisting of 20,000 shares, worth 100,000 in

total.

She wishes to ensure that the value of her portfolio in 6 months' time wil not be less than the

value it would have had if she had switched her entire portfolio to cash immediately.

The continuously compounded annualized interest rate earned on cash for the next 6 months is

4%.

Assuming that there are no transaction costs, which of the following strategies would achieve her

objective?

A Sell suffcient shares from the portfolio to purchase 24,000 6-month put options with a

strike price of 5.00 at a price of 0.25 each

B Sell suffcient shares from the portfolio to purchase 25,000 6-month put options with a


C Sell sufficient shares from the portfolio to purchase 18,750 6-month put options with a


D Sell suffcient shares from the portfolio to purchase 15,625 6-month put options with a


E None of these

Question

26

SOA Exam MFE May 2009

You are given:

(i) The current exchange rate is 0.011 /¥.

(ii) A four-year dollar-denominated European put option on yen with a strike price of 0.008

sells for 0.0005.

(iii) The continuously-compounded risk-free interest rate on dollars is 3%.

(iv) The continuously-compounded risk-free interest rate on yen is 1.5%.

Calculate the price of a four-year yen-denominated European put option on dollars with a strike

price of ¥125.

A ¥35

B ¥37

C ¥39

D ¥41

E ¥43





ExamMFE

Question

27


You are given:

i) C K, T) denotes the current price of a K -strike T -year European call option on a

nondividend-paying stock.

ii) P K, T) denotes the current price of a K -strike T -year European put option on the same

stock.

iii) 5 denotes the current price of the stock.

(iv) The continuously-compounded risk-free interest rate is r.

Which of the following is (are) correct?

I) 0:: C 50, T)-C 55, T):: 5e-rT

II) 50e-rT :: P 45, T) -C 50, T)+ 5:: 55e-rT

II) 45e-rT ::P 45,T)-C 50,T)+5::50e-rT

A (I) only

B II) only

C II) only

D (1) and (II) only

E I) and II) only

Question 28

SOA sample question

Consider a European call option and a European put option on a nondividend-paying stock. You

are given:

i) The current price of the stock is 60.

ii) The call option currently sells for 0.15 more than the put option.

iii) Both the call option and put option wil expire in 4 years.

iv) Both the call option and put option have a strike price of 70.

Calculate the continuously-compounded risk-free interest rate.

A 0.039

B 0.049

C 0.059

D 0.069

E 0.079


13



ExamMFE


Question

29

SOA sample question

Near market closing time on a given day, you lose access to stock prices, but some European call

and put prices for a stock are available as follows:

Strike Price

Call Price

Put Price

40

11 3

50 6

8

55

3

11

All six options have the same expiration date.

After reviewing the information above, John tells Mary and Peter that no arbitrage opportunities

can arise from these prices.

Mary disagrees with John. She argues that one could use the following portfolio to obtain

arbitrage profit: Long one call option with strike price 40; short three call options with strike price

50; lend 1; and long some calls with strike price 55.

Peter also disagrees with John. He claims that the following portfolio, which is different from

Mary s, can produce arbitrage profit: Long 2 calls and short 2 puts with strike price 55; long 1 call

and short 1 put with strike price 40; lend 2; and short some calls and long the same number of

puts with strike price 50.

Which of the following statements is true?

A Only John is correct.

B Only Mary is correct.

C Only Peter is correct.

D Both Mar and Peter are correct.

E N one of them is correct.





ExamMFE

Question 30

SOA sample question

Consider European and American options on a nondividend-paying stock.

You are given:

i) All options have the same strike price of 100.

ii) All options expire in six months.

(iii) The contiuously compounded risk-free interest rate is 10%.

You are interested in the graph for the price of an option as a function of the current stock price

In each of the following four charts I-IV, the horizontal axis,S, represents the current stock

price, and the vertical axis, íC, represents the price of an option.

t

II

Wö

95.1.2

51

1.0ø

95.12

IlL

iv.

..

¿t;

n

¡DO

95J2

s

'i

iun

95.t1

Match the option with the shaded region in which its graph lies. If there are two or more

possibilties, choose the chart with the smallest shaded region.

European Call

American Call

European Put

American Put

A

I

I

II

II

B

II

I

IV

II

C

II

I

II

II

D

II

II IV

II

E

II

II

IV

IV


15




ExamMFE

ExamMFE


Chapter 2

Question 1

Stock XYZ has a current price of 60. In 1 year, the stock price wil be either 75 or 55. The stock

does not pay dividends.

A 1-year European put option on Stock XYZ has a delta of -0.5. The put option has a payoff of

zero if the stock price reaches 75.


Determine the current value of the put option.

A 4.07

B 4.36

C 4.68

D 4.96

E 5 32

Question

2

Stock XYZ has a current price of 75. In 1 year, the stock price wil be either 100 or 60. The

stock does not pay dividends. The true probability that the stock wil be 100 in one year is 50%.

A 1-year European call option on Stock XYZ has a strike price of 80.


Determine the current value of the call option.

A 8.96

B 9.42

C 9.51

D 10.11

E 10 07


1



ExamMFE


Question 3

A one-step binomial tree model is to be used to estimate the prices of two European vanila

options with the same expiration date on a non-dividend-paying risky stock.

· Option C is a call option with a strike price of 100.

· Option P is a put option with a strike price of 80.

The current stock price is 100.

Consider the following assertions that have been made about the application of this model:

(1) It is possible that the calculated price of Option C is zero.

(2) It is possible that the calculated price of Option P is zero.

(3) It is always possible to replicate the payoff from Option P using a portfolio consisting of

positions in Option C and cash only.

Whch of these assertions are correct?

A (2) only

B (3) only

C 1) and 2) only

D (2) and (3) only

E 1), 2) and 3)

Question 4

For a one-period binomial model for a non-dividend paying stock, you are given:

(i) The period is 6 months.

(ii) The continuously-compounded risk-free rate of interest is 6.5%.

(iii) The stock currently sells for 56.

(iv) The payoff for a certain option position on this stock can be duplicated by holding 3

shares of stock and borrowing the present value of 105 due to be repaid at the end of

the 6-month period.

Determine the cost of establishing this option position based on the binomial modeL.

A 66.36

B 67.17

C 67.99

D 68.80

E 69.61





ExamMFE

Question 5

For a 1-period binomial model for stock prices, you are given:


(ii) The current price for a non-dividend paying stock is 56.

(iii) u =1.181, d=0.89

(iv) C is the current price of a 60-strike 6-month European call option.

Determine p, the true probability of an up move if the stock has a continuously-compounded

annual yield a = 0 12

A 0.43

B 0.47

C 0.51

D 0.55

E 0.59

Question 6

For a one-period binomial model for a stock price, you are given:

(i) Each period is 6 months.

(ii) The stock pays no dividends.

(iii) u = 1.27396

(iv) d = 0.83349

(v) The continuously-compounded annual return on the stock is a = 0.09.

Determine the difference p * -p where p * is the risk neutral probabilty of an up move and p

is the true probability of an up move.

A -0.035

B -0.015

C 0

D 0.015

E 0.035


3



ExamMFE


Question 7

A one-step binomial tree model is to be used to estimate the prices of two at-the-money European

vanila options on a non-dividend paying stock. The options have the same expiration date.

Consider the following assertions that have been made about risk-neutral and realistic

probabilties in the context of this model:

(1) The risk-neutral probability corresponding to an up movement wil always be less than

the risk-neutral probability corresponding to a down movement.

(2) Using realistic probabilities, the expected rate of return on the call option over the period

wil exceed the risk-free interest rate.

(3) Using realistic probabilties, the expected rate of return on the put option over the period

wil exceed the risk-free interest rate.

Whch of these assertions are correct?

A (1) only

B (2) only

C (1) and (2) only

D (1) and (3) only

E (1), (2) and (3)

Question 8

Stock XYZ has a current price of $65. The stock's volatilty is 20%. The continuously-

compounded expected return on the stock is a = 15 . The stock does not pay dividends.

The continuously-compounded risk-free rate of return is 8 .

A European call option on Stock XYZ expires in one year and has a strike price of $70.

Using a one-period binomial model, find the continuously-compounded expected rate of return

on the call option.

A 15.0%

B 31.7%

C 32.4%

D 41.6%

E 51.6%





ExamMFE

Question 9

Stock XYZ has a current price of 41. The stock's volatility is 30%. The continuously-

compounded expected return on the stock is a = 15 %. The stock does not pay dividends.

The continuously-compounded risk-free rate of return is 8%.

A European put option on Stock XYZ expires in one year and has a strike price of 40.

Using a one-period binomial model, find the continuously-compounded rate of return on the put

option.

A -14.1%

B -15.2

C 8.3

D 15.0

E 32.7


5



ExamMFE


Use the following information for Question 10 and Question 11.



(ii) The current price for a non-dividend paying stock is 56.

(iii) u =1.181, d=0.89

(iv) C is the current price of a 60-strike 6-month European call option.

(v) The contiuously-compounded risk-free rate of return is 5%.

Question 10

Determine the continuously-compounded annual yield rate r on the call option in Question 5.

A 0.50

B 0.53

C 0.56

D 0.59

E 0.62

Question 11

Determine C using the true up/down probabilties in the binomial model along with the

discountig based on the option yield rate r obtained in Question 10.

A Less than 2.70




E At least $2.85





ExamMFE

Question 12

One year from now, the price of Stock XYZ wil either be 106.34 or 58.36. The continuously-

compounded expected return on the stock is a = 17%. The true probability that the stock price

wil be 106.34 in one year is 56.23%. The stock does not pay dividends.

The continuously~compounded risk-free rate of return is 9 .

A European call option on Stock XYZ expires in one year and has a strike price of 75.

Determine the value of the call option.

A 8.73

B 11.57

C 12.19

D 14.87

E 17.62

Question 13

Stock XYZ has a current price of 11. In one year's time, the stock price wil be either 15 or 9.

The stock does not pay dividends.

A private investor is considering purchasing a 1-year European call option on Stock XYZ with a

strike price of 12.50.

The investor is subject to 40 capital gains tax on any gains made on stocks or options. The tax is

payable at the time of sale or expiration of an option. Losses cannot be offset against other gains.

The continuously-compounded interest rate for this investor is 4.0 .

Determine the fair option price for this investor.

A 0.58

B 0.70

C 0.82

D 1.02

E 1.63


7



ExamMFE


Question 14

The investor in Question 13 believes the expected real-world rate of return on the stock

(after paying capital gains tax) if he were to purchase the stock now and sell it after one year

would be 10 .

He decides to purchase the option now for 1.00.

Calculate his expected real-world rate of return (after tax) on the option.

A Between 0 and 5

B Between 5 and 10

C Between 10 and 15

D Between 15 and 25

E Between 25 and 50

Question 15

An alternative form of the one-step binomial model is constructed so that the risk-neutral

probabilties of II up and II down movements are both equal to 1/2.

The initial share price is denoted by 5 and the two possibilities for the final share price are

denoted by 5u and 5d (with u:; d).

The continuously-compounded risk-free interest rate is r.

A formula has been derived for calculating u that reflects correctly the volatility of the

underlying asset.

Whch of the following is the correct formula for d for this model?

A 1

B

1fu

C

u-l

u-er

D

(2 -u)er

E

2er -u

8





ExamMFE

Question 16

For a one-period binomial model for a non-dividend paying stock, you are given:


(ii) The stock currently sells for 56.

(iii) In 6 months the stock price wil be either 66.136 or 49.840.

(iv) C is the current price for a 6-month European call option where the strike price K

satisfies 51 ~ K ~ 65 .

Determine the increase in C if the current stock price were 57 rather than 56.

A 0.48

B 0.51

C 0.54D 0.57

E 0.60

Question 17


On April 30, 2007, a common stock is priced at 52.00. You are given the following:

(i) Dividends of equal amounts wil be paid on June 30, 2007 and September 30, 2007.

(ii) A European call option on the stock with strike price of 50.00 expiring in six months

sells for 4.50.

(iii) A European put option on the stock with strike price of 50.00 expiring in six months

sells for 2.45.

(iv) The continuously compounded risk-free interest rate is 6%.

Calculate the amount of each dividend.

A 0.51

B 0.73

C 1.01

D 1.23

E 1.45


9



ExamMFE


Question 18


For a one-period binomial model for the price of a stock, you are given:

(i) The period is one year.


(iii) u = 1.433 , where u is one plus the rate of capital gain on the stock if the price goes up.

(iv) d = 0.756, where d is one plus the rate of capital loss on the stock if the price goes down.

(v) The continuously compounded annual expected return on the stock is 10%.

Calculate the true probabilty of the stock price going up.

A 0.52

B 0.57

C 0.62

D 0.67

E 0.72

Question 19


For a one-year straddle on a nondividend-paying stock, you are given:

(i) The straddle can only be exercised at the end of one year.

(ii) The payoff of the straddle is the absolute value of the difference between the strike price

and the stock price at expiration date.

(iii) The stock currently sells for 60.00.


(v) In one year, the stock wil either sell for 70.00 or 45.00.

(vi) The option has a strike price of 50.00.

Calculate the current price of the straddle.

A 0.90

B 4.80C 9.30

D 14.80

E 15.70





ExamMFE

Question

20


You are given the following regarding stock of,Widget World Wide WWW):

(i) The stock is currently sellng for 50.

(ii) One year from now the stock wil sell for either 40 or 55.

(iii) The stock pays dividends continuously at a rate proportional to its price. The dividend

yield is 10%.

The contiuously-compounded risk-free interest rate is 5%.

Whle reading the Financial Post, Michael notices that a one-year at-the-money European call

written on stock WWW is sellng for 1.90. Michael wonders whether this call is fairly priced.

He uses the binomial option pricing model to determine if an arbitrage opportunity exists.

What transactions should Michael enter into to exploit the arbitrage opportunity (if one exists)?

A No arbitrage opportunity exists.

B Short shares of WW, lend at the risk-free rate, and buy the call priced at 1.90.

C Buy shares of WWW, borrow at the risk-free rate, and buy the call priced at 1.90.

D Buy shares of WWW, borrow at the risk-free rate, and short the call priced at 1.90.

E Short shares of WW, borrow at the risk-free rate, and short the call priced at 1.90.


11



ExamMFE


Question 21


The following one-period binomial stock price model was used to calculate the price of a one-

year 10-strike call option on the stock.

~

Su =12

So =10

~

Sd =8

You are given:

(i) The period is one year.

(ii) The true probability of an up-move is 0.75.

(iii) The stock pays no dividends.

(iv) The price of the one-year 10-strike call is 1.13.

Upon review, the analyst realizes that there was an error in the model construction and that Sd

the value of the stock on a down-move, should have been 6 rather than 8. The true probability of

an up-move does not change in the new model, and all other assumptions were correct.

Recalculate the price of the call option.

A 1.13

B 1.20

C 1.33

D 1.40

E 1.53





ExamMFE

Question

22

SOA sample question

You are given the following inormation about a securities market:

(i) There are two nondividend-paying stocks, X and Y.

(ii) The current prices for X and Yare both 100.

(iii) The continuously compounded risk-free interest rate is 10%.

(iv) There are three possible outcomes for the prices of X and Y one year from now:

Outcome

X

Y

1

200

0

2

50

0

3

0

300

Let Cx be the price of a European call option on X, and Py be the price of a European put option

on Y. Both options expire in one year and have a strike price of 95.

Calculate Py - C X .

A 4.30

B 4.45

C 4.59

D 4.75

E 4.94

Question 23

SOA sample question

A discrete-time model is used to model both the price of a nondividend-paying stock and the

short-term interest rate. Each period is one year.

At time 0, the stock price is So = 100 and the effective annual interest rate is ro = 5 .

At time 1, there are only two states of the world, denoted by u and d. The stock prices are

Su = 110 and Sd = 95. The effective annual interest rates are ru = 6% and rd = 4%.

Let C(K) be the price of a 2-year K -strike European call option on the stock.

Let P(K) be the price of a 2-year K -strike European put option on the stock.

Determie P(108)-C(108).

A -2.85

B -2.34

C -2.11

D -1.95

E -1.08


13



ExamMFE


Question

24

SOA sample question

You use the usual method in McDonald and the following inormation to construct a one-period

binomial tree for modeling the price movements of a nondividend-paying stock. (The tree is

sometimes called a forward tree).


(ii) The intial stock price is 100.

(iii) The stock's volatility is 30%.

(iv) The continuously compounded risk-free interest rate is 4 .

At the beginnng of the period, an investor owns an American put option on the stock. The

option expires at the end of the period.

Determine the smallest integer-valued strike price for which an investor wil exercise the put

option at the beging of the period

A 114

B 115

C 116

D 117

E 118





ExamMFE

ExamMFE


Chapter

3

Use the following information for Question 1 to Question 6.



(ii) The current price for a non-dividend paying stock is $60.

(iii) The up price after one period is Su =$70.860 and the down price after one period is

Sd = 53.400.

Question 1

Determine (J, the annual volatilty of the stock.

A 0.10

B 0.15

C 0.20

D 0.25

E 0.30

Question

2

Determine r, the continuously-compounded risk-free rate of interest.

A 0.030

B 0.035

C 0.040

D 0.045

E 0.050


1



ExamMFE







Sd =$53.400.

Question 3

Determine p * , the risk-neutral probabilty of an up move.

A 0.450

B 0.455

C 0.460

D 0.465

E 0.470

Question

4

Determine the price of a one-year at-the-money European put option on one share of this stock.

A 3.35

B 3.40

C 3.45

D 3.50

E 3.55

Question 5

Determine the price of a one-year at-the-money American put option on one share of this stock.

A 3.35

B 3.40

C 3.45

D 3.50

E 3.55





ExamMFE






Sa = 53.400.

Question 6

Determine the cost to an investor who establishes a European call bull spread consisting of a

62 strike call option and a 77 strike call option.

A 3.29

B 3.39

C 3.49

D 3.59

E $3.69

Question 7

For a two-period binomial model for a non-dividend paying stock, you are given:


(ii) The price for a one-year at-the-money European put option is 1.70.

(iii) The current stock price is $30.

(iv) Su = 35.430 , Sa = 26.70

Determine the price for a one-year at-the-money European call option.

A 2.92

B 2.98

C 3.04

D 3.10

E $3.16


3



ExamMFE


Question 8

Stock XYZ has a current price of 50. Its annual volatility is 30%. The stock does not pay

dividends.

The stock price follows the binomial model, with 3 periods, each of length 4 months.

A 1-year American call option on Stock XYZ has a strike price of 70.


Determine the current value of the call option.

A 1.98

B 2.18

C 9.37

D 10.36

E 12.22

Question 9

Stock A has a current price of 50. Its anual volatilty is 20%. The stock does not pay dividends.

The stock price follows the binomial model, with each period being 1 year in length.

A 2-year American put option on Stock A has a strike price of 55.


Determine the current value of the put option.

A 4.04

B 5.41

C 5.86

D 6.36

E 7.17





ExamMFE

Question 10

A stock index has a current value of 70. Its volatility is 30%. The index pays dividends

continuously at an annual rate of 4 .

The continuously-compounded interest rate is 7%.

A 9-month American call option on the index has a strike price of 60.

Use a 3-period binomial model to find the price of the American call option.

A 2.11

B 12.22

C 13.11

D 14.18

E 17 05

Question 11

The current exchange rate is 3.50 per Kuwaiti dinar. The dinar has a volatility of 15% relative to

the dollar. The continuously-compounded Kuwaiti interest rate is 5.5%.

The continuously-compounded US interest rate is 8.0%.

A 6-month American call option on the Kuwaiti dinar has a strike price of 3.50.

Use a 2-period binomial model to calculate the price of the American call option.

A 0.12

B 0.16

C 0.23

D 0.34

E 0 41

Question 12

The current price of Stock A is 12. Its volatilty is 15%. It pays dividends continuously at an

anual rate of 3.0 .


A 2-month European put option on Stock A has a strike price of 13. Calculate the price of the

put option using a 2-period binomial modeL.

A 0.80

B 0.97

C 1.00

D 1.06

E 1 25


5



ExamMFE


Question 13

A stock index is currently 100. The volatility of the index is 35%. The continuously-compounded

dividend rate on the index is 4 .

An American put option on the index has a strike price of 98. The put option expires in 2 years.

The contiuously-compounded interest rate is 7%.

Use a binomial model with 2 time periods to find the current value of the option.

A 12.94

B 13.54

C 13.88

D 20.66

E 25.39

Question 14

A one-year option on a stock that pays a continuous dividend yield is to be valued using a

recombining binomial tree with 12 time steps of equal length.

The up factor to be applied to the asset price at each step is u = 1.056928 and the corresponding

down factor is d = 0.950406 .

Determine the annualized volatility (j that has been assumed in the calculations.

A 5.3

B 18.4

C 21.2

D 36.8

E 63.7





ExamMFE

Question 15

The current price of gold is 600 per ounce. Storing gold requires an insurance premium to be

paid continuously at the rate of 0.5% per annum (so that ö = -0.005).

A 1-year European vanlla put option has a strike price of 600 per ounce.

The continuously-compounded risk-free interest rate over the next year is 5% per annum.

Thee calculations of the value of this option have been carried out using a two-step, a three-step

and a four-step binomial tree, each assuming an annualized volatilty of 15%. The results of the

calculations are denoted by P2' P3 and P4, respectively.

Which of the following inequalities gives the correct relationship between the values of P2, P3

and P4?

A P2 ~ P3 ~ P4

B P2 ~ P4 ~ P3

C P3 ~ P2 ~ P4

D P3 ~ P4 ~ P2

E P4 ~ P3 ~ P2

Question 16

Stock XYZ has a current price of 50. The stock's volatility is 30% per annum. The continuously-

compounded expected return on the stock is a = 17% per annum. The stock does not pay

dividends.

The contiuously-compounded risk-free rate of return is 6%.

An American call option on Stock XYZ expires in one year and has a strike price of 48. The call

option is valued using a two-period binomial modeL.

Calculate the continuously-compounded annual expected return on the call option at the initial

node, assuming that the option wil be sold after 6 months.

A 42.1

B 44.2

C 45.8

D 46.1

E 48.3


7



ExamMFE


Question 17

Stock XYZ has a current price of 50. The stock's volatility is 30% per annum. The continuously-

compounded expected return on the stock is a = 17% per annum. The stock does not pay

dividends.


A European call option on Stock XYZ expires in one year and has a strike price of 48. The call

option is valued using a two-period binomial modeL.

Calculate the continuously-compounded expected return on the call option at the intial node,

assuming that the option wil be held until expiration.

A 44.2

B 45.8

C 46.1

D 55.3

E 57.8

Question 18

Determine which of the sets of parameters below gives rise to arbitrage in a Cox-Ross-Rubinstein

binomial tree.

A

B

C

D

E

r=5%

r=6%

r=7%

r=8%

r=9%

h=l

h=2

h=O.l

h =0.4

h = 0.5

Cí=30%

Cí= 20%

Cí=5%

Cí=50%

Cí=5%

Question 19

The current price of a stock is 100. Its volatility is 30%. The stock does not pay dividends. The

continuously-compounded expected rate of return on the stock is a = 10% .


An American call option on the stock expires in one year and has a strike price of 95.

Use a Cox-Ross-Rubinstein binomial tree with 2 periods to find the value of the call option.

A 18.01

B 18.58

C 19.11

D 19.52

E 21.06





ExamMFE


A multi-step binomial tree with time steps of one week has been used to calculate the value of a

one-year stock option. The initial part of the tree, showing the first two weeks, is shown below.

The value of the underlying stock, which does not pay dividends, is shown inside each box and

the corresponding value of the option is shown above.

9.96

54.44

8.45

52.17

7.12

7.02

50.00

50.10

5.85

48.01

4.74

46.10

Question

20

Estimate ~(S, 0) , the value of delta at the initial node.

A 0.570

B 0.598

C 0.625

D 0.677

E 0 750

Question 21

Estimate r( Sh ,h) , the value of gamma at the end of the first week.

A 0.0150

B 0.0192

C 0.0215

D 0.0242

E 0 0258


9



ExamMFE


Question

22

You are given the following formula:

B(S,O)

C(udS ,2h) - eL1(S, 0) -le2Qs, 0) -C(s, 0)

where e = udS - 5

2h

Estimate B(S, 0), the value of theta at the intial node.

A -4.2

B -3.5

C -2.1

D -0.7

E 0

Question 23

In a simple random walk model of stock prices, the price of a stock is assumed to change by 0.01

during each trading day. The probability that the price moves up on a given day is 51 % and the

probability that it moves down is 49%. Movements on different days are assumed to be

statistically independent.

Stock V AR has a current price of 5.00. A trader is concerned about potential losses on the stock

over the next year which consists of 250 trading days. He wishes to determie the smallest value

of L such that:

Pr(S -5.00 ~ -L):: 0.01

where 5 denotes the stock price at the end of the year.

Use a normal distribution to approximate the value of L.

A Less than 0.30

B Between 0.30 and 0.35

C Between 0.35 and 0.40

D Between 0.40 and 0.45

E More than 0.45





ExamMFE

Question

24

A trader holds an American put option on a stock.

Consider the following events that could occur:

(1) The company announces unexpectedly that it wil pay a special dividend to all

stockholders.

(2) The central bank announces unexpectedly that it wil increase the interest rate it applies

byl%.

(3) The prices of all stocks in the market unexpectedly become more volatile.

Whch of these events, when considered in isolation, would make the trader more likely to

exercise the option early?

A (2) only

B (3) only

C 1) and 2) only

D (2) and (3) only

E 1), 2) and 3)

Question

25


For a two-period binomial model for stock prices, you are given:


(ii) The current price for a nondividend-paying stock is 70.00.

(iii) u = 1.181, where u is one plus the rate of capital gain on the stock per period if the price

goes up.

(iv) d = 0.890 , where d is one plus the rate of capital loss on the stock per period if the price

goes down.

(v) The contiuously compounded risk-free interest rate is 5%.

Calculate the current price of a one-year American put option on the stock with a strike price of

80.00.

A 9.75

B 10.15

C 10.35

D 10.75

E 11.05


11



ExamMFE


Question 26


You use the usual method in McDonald and the following information to construct a binomial

tree for modeling the price movements of a stock. (Ths tree is sometimes called a forward tree.)

(i) The length of each period is one year.

(ii) The current stock price is 100.

(iii) The stock's volatilty is 30%.

(iv) The stock pays dividends continuously at a rate proportional to its price. The dividend

yield is 5 .

(v) The continuously-compounded risk-free interest rate is 5 .

Calculate the price of a two-year 100-strike American call option on the stock.

A 11.40

B 12.09

C 12.78

D 13.47

E 14.16

Question

27

SOA sample question

For a two-period binomial modeL, you are given:

(i) Each period is one year.

(ii) The current price for a nondividend-paying stock is 20.

(iii) u = 1.2840, where u is one plus the rate of capital gain on the stock per period if the

stock price goes up.

(iv) d = 0.8607, where d is one plus the rate of capital loss on the stock per period if the stock

price goes down.

(v) The continuously compounded risk-free interest rate is 5 .

Calculate the price of an American call option on the stock with a strike price of 22.

A 0B 1

C 2

D 3

E 4





ExamMFE

Question

28

SOA sample question

Consider a 9-month dollar-denominated American put option on British pounds.

You are given that

i) The current exchange rate is 1.43 US dollars per pound.

ii) The strike price of the put is 1.56 US dollars per pound.

iii) The volatilty of the exchange rate is a = 0.3.

(iv) The US dollar continuously compounded risk-free interest rate is 8 .

(v) The British pound continuously compounded risk-free interest rate is 9 .

Using a three-period binomial modeL, calculate the price of the put.

A 0.17

B 0.19

C 0.21

D 0.23

E 0.25


13



ExamMFE


For Question 29 and Question 30, consider the following three-period binomial tree model for a

stock that pays dividends continuously at a rate proportional to its price. The length of each

period is 1 year, the continuously compounded risk-free interest rate is 10 , and the continuous

dividend yield on the stock is 6.5 .

~

585.9375

~

468.75

~

375

--

328.125

300

--

262.5

~

--

210

~

~

183.75

--

147

--

102.9

Question

29

SOA sample question

Calculate the price of a 3-year at-the-money American put option on the stock.

A 15.86

B 27.40

C 32.60

D 39.73

E 57.49

Question 30

SOA sample question

Approximate the value the value of gamma at time 0 for the 3-year at-the-money American put

on the stock

A 0.0038

B 0.0041

C 0.0044

D 0.0047

E 0.0050





ExamMFE

Question 31

SOA sample question

You are to price options on a futures contract. The movements of the futures price are modeled

by a binomial tree. You are given:

i) Each period is 6 months.

ii) ufd = 4/3, where u is one plus the rate of gain on the futures price if it goes up, and d is

one plus the rate of loss if it goes down.

iii) The risk-neutral probability of an up move is 1/3.

(iv) The initial futures price is 80.


Let C1 be the price of a 1-year 85-strike European call option on the futures contract, and Cii be

the price of an otherwise identical American call option.

Determine C II C i .

A 0

B 0.022

C 0.044

D 0.066

E 0.088


15




ExamMFE

ExamMFE


Chapter

4

Note: When carrying out calculations based on the Black-Scholes modeL, Exam MFE candidates

are instructed to roùnd to 2 decimal places before using the tables for the normal

probabilty distribution. For example, to find the value of N(0.1247), you would round

.the 0.1247 to 2 decimal places and look up N(0.12), which equals 0.5478. Ths approach

saves time, but can lead to some inaccuracy in the final answers, but this won't matter in

the exam. However, for interest, we show some accurate answers as well after relevant

solutions.

Question 1

Stock XYZ has a current price of 50. The stock's volatilty is 25%. The stock pays dividends at a

continuously-compounded rate of 4%.


A European call option on Stock XYZ expires in six months and has a strike price of 52.

Use the Black-Scholes model to find the value of the call option.

A 2.99

B 3.42

C 4.13

D 4.58

E 5 00

Question

2

Stock XYZ has a current price of 50. The stock's volatilty is 25%. The stock wil pay a dividend

of 6 in 9 months.


A European put option on Stock XYZ expires in six months and has a strike price of 52.

Use the Black-Scholes model to find the value of the put option.

A 3.22

B 3.50

C 4.13

D 5.90

E 7 08


1



ExamMFE


Question 3

You are given:

(i) The current price of a share of stock YZ is 100.

(ii) The contiuously-compounded risk-free interest rate is r = 0.045 .

(iii) The continuous dividend rate for stock YZ is ö = 0.02 .

(iv) Annual volatility for stock YZ is (5 = 0.30 .

(v) P is the price for a 98 strike, 3-month European put option on a share of stock YZ.

Assuming that the Black-Scholes framework is valid, determine P.

A 0.85

B 2.98

C 4.67

D 5.96

E 7.26

Question

4

For an at-the-money, 6-month European call option on a share of stock, you are given:

(i) 8.00 is the current price of this call option.

(ii) The contiuously-compounded risk-free interest rate is r = 0.06 .


(iv) The annual volatilty of this stock is (5 = 0.50

Assuming the Black-Scholes formula holds, determie the current stock price.

A Less than 49



D At least 51, but less than 52

E At least 52





ExamMFE

Question 5

Within the Black-Scholes framework, you are given:

(i) At time t, 5 (t) denotes the price of a share of stock paying no dividends.

(ii) 5(0)= 90

(iii) P is the price of a 2-year European put option with strike price 90 e2r where r is the

risk-free annual rate of interest.

(iv) var(ln(S(t))) = 0.25t

Determine P.

A Less than 22




E At least 34

Question 6

For a six-month at-the-money European call option on a stock, you are given:

(i) The strike price is 50.00.

(ii) The only dividend during the six-month period is 2.00 to be paid in 3 months.

(iii) (Y = 0.30

(iv) The continuously-compounded risk-free interest rate is r = 0.05.

Using the Black-Scholes approach, determine the price of this call option.

A 2.96

B 3.33

C 3.69

D 4.07

E 4.44


3



ExamMFE


Question 7

Stock XYZ has a current price of 50. The stock's volatility is 25%. The stock pays dividends at a



A European put option on Stock XYZ expires in six months and has a strike price of 52.

Using the Black-Scholes model, find the value of delta for the put option.

A -0.521

B -0.510

C -0.503

D -0.498

E -0.471

Question 8

Stock XYZ has a current price of 75. The stock's volatilty is 30%. The stock pays dividends at a



Use the Black-Scholes model to find the price of a 1-year European call option with a strike price

of 80, where the underlying asset is a futures contract maturing at the same time as the call

option.

A 4.11

B 8.01

C 8.22

D 9.20

E 10.36





ExamMFE

Question 9

A European put option maturing in 9 months is priced using the Black-Scholes modeL. The strike

price is 80. The values of di and d2 are:

di = 1 13

d2 =0.87


The current price of the underlying stock is 100. The stock pays no dividends.

An investor has decided to replicate the option using the stock and the risk-free asset. Determine

the amount that must be invested in the risk-free asset to replicate the option.

A 9.53

B 12.62

C 14.18

D 14.48

E 18 79

Question 10

A European put option maturing in 3 months is priced using the Black-Scholes modeL. The strike

price is 95. The values of di and d2 are:

di = 0 513

d2 =0.390

The price of the put option is 2.54.


The underlying stock pays dividends at a constant rate and has a current price of 100.

Determine the value of delta, A, for the put option.

A -0.697

B -0.582

C -0.333

D -0.304

E -0.299


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(i) The current price of a share of stock XY is 40.

(ii) The contiuously-compounded risk-free rate of interest is r = 0.08 .


(iv) Annual volatilty of the stock is (F = 0.30

(v) For a three-month 40-strike European call option on stock XY:

price = 2.7804 , delta = 0.5824 , gamma = 0.0652

(vi) Dave has a position consisting of one written call option (as in (v)) and one written three-

month 40-strike European put option.

Question 11

Determine delta for Dave s position.

A -1.00

B -0.16

C 0.00

D 0.16

E 1.00

Question 12

Using the delta value of Dave's position estimate the change in value of his position if the stock

price dropped to 39.625 immediately after the position was established.

A -0.38

B -0.06

C 0.00

D 0.06

E 0.38





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(ii) The continuously-compounded risk-free rate of interest is r = 0.08.


(iv) Annual volatility of the stock is (Y = 0.30





Question 13

Determine gamma for Dave's position

A -0.26

B -0.13

C 0.00

D 0.13

E 0.26

Question 14

Determine the ratio of the volatilty of Dave s position to the volatilty of Stock XY.

A Less than 1.2




E At least 1.5


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(iv) Annual volatilty of the stock is (J = 0.30





Question 15

Determine the annual volatilty of Dave's position.

A 0.30

B 0.33

C 0.36

D 0.39

E 0.42

Question 16

Determine the expected annual continuously-compounded rate of return on Dave's position if

the expected annual continuously-compounded rate of return on Stock XY is a = 0.15.

A Less than 8

B At least 8 , but less than 12

C At least 12 , but less than 16

D At least 16 , but less than 20

E At least 20





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(iv) Annual volatility of the stock is (J = 0.30




month 40-strike European put option on the same stock.

Question 17

Determine the Sharp ratio for Dave's position if the expected annual continuously-compounded

rate of return on Stock XY is a = 0.15 .

A Less than 0.24




E At least 0.33

Question 18

Stock XYZ has a current price of 41. The stock's volatilty is 30%. The stock wil pay a dividend

of 4 in 6 months.


A European put option on Stock XYZ expires in nie months and has a strike price of 45.

Use the Black-Scholes model to find the value of the put option.

A 2.01

B 4.16

C 4.93

D 5.85

E 7.24


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Question 19

The current exchange rate is 1.05 per Euro. The continuously-compounded interest rate in the

US is 6%. The continuously-compounded interest rate in Europe is 3.2%.

The volatility is (J = 10 .

Find the value of a 1-year European call on the Euro with a strike price of 1.00.

A 0.029

B 0.050

C 0.067

D 0.088

E 0.142

Question

20

A portfolio consists of three options on Stock X: Option A, Option B, and Option C.

Some of the characteristics of the three options are described in the table below. As indicated by

the question marks, some of the inormation about the options is not provided.

A B

C

Type

Call

? Put

Price

5.8868

3.8117 2.3441

Delta

?

0.4830 -0.2710

Gamma

?

0.0332

?

Vega

0.1501

?

0.1325

Elasticity

4.3272

5.0687 ?

The current value of the portfolio is 12.0426.

Find the option elasticity of the portfolio.

A 1.59

B 2.82

C 4.62

D 7.56

E 8.46





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Question 21

For a 1-year 80-strike European put option withi the Black-Scholes framework, you are given:


(ii) The contiuously-compounded risk-free interest rate is 6% per year.

(iii) Annual volatility for the stock is (j = 0.50 .

(iv) The stock pays no dividends.

Calculate the annual volatility of this put option.

A 1.16

B 1.22

C 1.28

D 1.34

E 1.43

Question

22

Consider a call option on Stock XYZ that does not pay dividends. State which of the following

Greeks wil usually have a negative value.

A Delta

B Gamma

C Theta

D Vega

E Rho

Question

23

Stock XYZ has a current price of 87. The stock's volatility is 20%. The stock wil not pay

dividends for 5 months.

The continuously-compounded risk-free rate of return is 8% per annum.

An investor purchases a 3-month European call option and a 3-month European put option, both

with the same strike price of 75.

Use the Black-Scholes model to find the delta of the investor's portfolio.

A 0.0417

B 0.9164

C 0.9489

D 0.9586

E 1


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Question

24

Stock XYZ has a current price of 110. The stock pays dividends at a contiuously-compounded

rate of 12% per annum. The continuously-compounded risk-free rate of retu is 6% per annum.

A European call option on Stock XYZ that expires in six months and has a strike price of $120 is

priced at 4.16.

Using the Black-Scholes modeL, find the stock's implied volatility.

A 30%

B 32%

C 35%

D 37%

E 40%

Question 25

For options on stock indices, volatility skew is a term that could be used to describe which of

the following observations about implied volatilties?

A Implied volatilties always increase when buying and sellng two options that differ only

in their times until maturity.

B Implied volatilities increase as put option prices increase.

C Implied volatilities only change because the strike price does not.

D Implied volatilties decline as the strike price increases.

E Implied volatilties decrease as call option prices increase.

Question

26

Whch of the following is not one of the assumptions of the Black-Scholes model when pricing

options on stocks?

A The stock prices are normally distributed.

B Stock prices do not jump in the modeL.

C The effective risk-free interest rate has a known constant value.

D All taxes are zero.

E It does not cost any extra to borrow when already deeply in debt.




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Question

29


For a European call option on a stock within the Black-Scholes framework, you are given:

(i) The stock price is 85.

(ii) The strike price is 80.

(iii) The call option wil expire in one year.

(iv) The contiuously compound risk-free interest rate is 5.5%.

(v) (J = 0.50

(vi) The stock pays no dividends.

Calculate the volatilty of this call option.

A 50%

B 69%

C 123%

D 139%

E 278%

Question 30


Let S(t) denote the price at time t of a stock that pays no dividends. The Black-Scholes

framework holds. Consider a European call option with exercise date T, T:; 0 , and exercise

price S O)erT, where r is the continuously compounded risk-free interest rate.

You are given:

(i) 5(0) = 100

(ii) T = 10

(iii) var (In S(t) L = O.4t, t:; 0

Determine the price of the call option.

A 7.96

B 24.82

C 68.26

D 95.44

E There is not enough inormation to solve the problem.

14 (Ç BPP Professional Education




ExamMFE

Question

31


For a six-month European put option on a stock, you are given:

(i) The strike price is 50.00.

(ii) The current stock price is 50.00.

(iii) The only dividend during this time period is 1.50 to be paid in four months.

(iv) (J = 0.30

(v) The contiuously compounded risk-free interest rate is 5%.

Under the Black-Scholes framework, calculate the price of the put option.

A 3.50

B 3.95

C 4.19

D 4.73

E 4.93

Question 32


Assume the Black-Scholes framework.

Eight months ago, an investor borrowed money at the risk-free interest rate to purchase a one-

year 75-strike European call option on a nondividend-paying stock. At that time, the price of the

call option was 8.

Today, the stock price is 85. The investor decides to close out all positions.

You are given:

(i) The contiuously-compounded risk-free rate interest rate is 5%.

(ii) The stock's volatilty is 26 %.

Calculate the eight-month holding profit.

A 4.06

B 4.20

C 4.27

D 4.33

E 4.47


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Question 33


Assume the Black-Scholes framework. Consider a one-year at-the-money European put option on

a nondividend-paying stock.

You are given

(i) The ratio of the put option price to the stock price is less than 5 .

(ii) Delta of the put option is -0.4364.

(iii) The continuously-compounded risk-free interest rate is 1.2 .

Determine the stock's volatilty.

A 12

B 14

C 16

D 18

E 20

Question 34


Consider a one-year 45-strike European put option on a stock S. You are given:

(i) The current stock price, S(O), is 50.00.

(ii) The only dividend is 5.00 to be paid in nie months.

(iii) var

LIn

Fti (S)J = O.01xt, 0:: t:: 1.

(iv) The continuously-compounded risk-free interest rate is 12 .

Under the Black-Scholes framework, calculate the price of 100 units of the put option.

A 1.87

B 18.39

C 18.69

D 19.41

E 23.76





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Question 35

SO sample question

You are considering the purchase of 100 units of a 3-month 25-strike European call option on a

stock.

You are given:

(i) The Black-Scholes framework holds.

(ii) The stock is currently sellng for 20.

(iii) The stock's volatility is 24 .

(iv) The stock pays dividends continuously at a rate proportional to its price. The dividend

yield is 3 .


Calculate the price of the block of 100 options.

A 0.04

B 1.93

C 3.50

D 4.20

E 5.09


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Question 36

SOA sample question

You are considering the purchase of a 3 month 41.5 strike American call option on a

nondividend-paying stock.

You are given:

(i) The Black-Scholes framework holds.

(ii) The stock is currently selling for 40.

(iii) The stock's volatilty is 30 .

(iv) The current call option delta is 0.5.

Determine the current price of the option.

A 20-20.453 C5 e-x2 /2dx

B 20-16.138 f~5 e-x2 /2dx

C 20 40.453 f~5 e x2 /2dx

D 16.138 f~5 e-x2 /2dx-20.453

E 40.453 f~5 e x2 /2dx 20.453





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Question 37

SOA sample question

You are to estimate a nondividend-paying stock's annualized volatilty using its

prices in the past nine months.

Month

Stock Price ( / shar~

1

80

2

64

3

80

4

64

5

80

6

100

7

80

8

64

9

80

Calculate the historical volatility for this stock over the period.

A 83%

B 77%

C 24%

D 22%

E 20%

Question 38

SOA sample question

Consider a forward start option which, 1 year from today, wil give its owner a 1-year European

call option with a strike price equal to the stock price at that time.

You are given:

(i) The European call option is on a stock that pays no dividends.

(ii) The stock's volatility is 30%.

(iii) The forward price for delivery of 1 share of the stock 1 year from today is 100.


Under the Black-Scholes framework, determine the price today of the forward start option.

A 11.90

B 13.10

C 14.50

D 15.70

E 16.80


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Question 39

SOA sample question

Assume the Black-Scholes framework. Consider a stock, and a European call option and a

European put option on the stock. The current stock price, call price, and put price are 45.00,

4.45, and 1.90, respectively.

Investor A purchases two calls and one put. Investor B purchases two calls and writes three puts.

The current elasticity of Investor A s portfolio is 5.0. The current delta of Investor B s portfolio is

3.4.

Calculate the current put-option elasticity.

A -0.55

B -1.15

C -8.64

D -13.03

E -27.24

Question 40

SOA sample question

Assume the Black-Scholes framework. You are given:

(i) S(t) is the price of a nondividend-paying stock at time t.

(ii) 5(0) = 10



At time t = 0 , you write a one-year European option that pays 100 if (5(1)) 2 is greater than 100

and pays nothing otherwise.

You delta-hedge your commitment.

Calculate the number of shares of the stock for your hedgig program at time t = 0 .

A 20

B 30

C 40

D 50

E 60





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Question

41

SO sample question

You compute the delta for a 50-60 bull spread with the following inormation:

(i) The contiuously compounded risk-free rate is 5 .

(ii) The underlying stock pays no dividends.

(iii) The current stock price is 50 per share.

(iv) The stock's volatilty is 20%.

(v) The time to expiration is 3 months.

How much does the delta change after 1 month, if the stock price does not change?

A increases by 0.04

B increases by 0.02

C does not change, within rounding to 0.01

D decreases by 0.02

E decreases by 0.04

Question

42

SO sample question

You own one share of a nondividend-paying stock. Because you worry that its price may drop

over the next year, you decide to employ a rollng insurance strategy, which entails obtainng one

3-month European put option on the stock every three months, with the first one being bought

immediately.

You are given:

(i) The continuously compounded risk-free interest rate is 8 .

(ii) The stock's volatility is 30%.

(iii) The current stock price is 45.

(iv) The strike price for each option is 90% of the then-current stock price.

Your broker wil sell you the four options but wil charge you for their total cost now.

Under the Black-Scholes framework, how much do you now pay your broker?

A 1.59

B 2.24

C 2.85

D 3.48

E 3.61


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Question

43

SOA sample question

Assume the Black-Scholes framework. Consider a derivative security of a stock.

You are given:

(i) The continuously compounded risk-free interest rate is 0.04.

ii) The volatilty of the stock is J .

iii) The stock does not pay dividends.

iv) The derivative security also does not pay dividends.

v) S t) denotes the time- t price of the stock.

iv) The time- t price of the derivative security is S t) rk/02 ,where k is a positive constant.

Find k

A 0.04

B 0.05

C 0.06

D 0.07

E 0.08





Question

44

ExamMFE

SOA sample question

Assume the Black-Scholes framework. Consider a 1-year European contingent claim on a stock.

You are given:

(i) The time-O stock price is 45.

(ii) The stock's volatility is 25 .

(iii) The stock pays dividends continuously at a rate proportional to its price. The dividend

yield is 3 .


(v) The time-l payoff of the contingent claim is as follows:

payoff

42

42

Calculate the time-O contingent-claim elasticity.

A 0.24

B 0.29

C 0.34

D 0.39

E 0.44

8 1 )


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ExamMFE


ChapterS


For a share of stock currently sellng for 40, you are given:

(i) The continuously-compounded risk-free annual rate of interest is r = 0.08.

ii) The stock pays no dividends.

iii) The following data for three-month European call options on a share of this stock:

40-Strke Call

45-Strike Call

Price

2.7847

1.3584

Delta

0.5825 0.3285

Gamma

0.0651

0.0524

Theta (per day)

-0.0173 -0.0129

Question 1

Suppose that you are a market maker who just sold 100 of the 40-strike call options to an investor,

and suppose that you wil delta-hedge the position with shares of stock. Any capital needed for

the hedged position can be borrowed at the risk-free rate of interest.

Determine your action.

A Sell 58.25 shares and lend the amount 3,721.53.

B Sell 58.25 shares and borrow the amount 3,721.53.

C Buy 58.25 shares and lend the amount 2,051.53.

D Buy 58.25 shares and borrow the amount 2,051.53.

E Buy 6.51 shares and lend the amount 27.07.


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ExamMFE


Use the following information for Question 1 through to Question 5.


(i) The continuously-compounded risk-free annual rate of interest is r = 0.08 .


(iii) The following data for three-month European call options on a share of this stock:

40-Strike Call

45-Strike Call

Price

2.7847

1.3584

Delta

0.5825

0.3285

Gamma

0.0651

0.0524

Theta (per day)

-0.0173

-0.0129

Question 2

Using delta, gamma, and theta, approximate the 40-strike option value after one day if the stock

price moves to 41.25.

A 3.46

B 3.49

C 3.52

D 3.55

E 3.58

Question 3

Using delta, gamma, and theta to approximate the change in option value, determine the

overnight profit/loss (i.e. one day) for the position in Question 5.1 if the stock price moves to

41.25.

A The overnight loss is 3.81

B The overnight loss is 2.28.

C The overnight profit is 0.00.

D The overnight profit is 2.28.

E The overnight profit is 3.81.





ExamMFE

Use the following information for Question 1 through to Question 5.


(i) The continuously-compounded risk-free annual rate of interest is r = 0.08 .


(iii) The following data for three-month European call options on a share of this stock:

40-Strike Call

45-Strke Call

Price

2.7847

1.3584

Delta

0.5825

0.3285

Gamma

0.0651

0.0524

Theta per day)

-0.0173

-0.0129

Question 4

Suppose that you are a market maker who just sold 100 of the 40-strike call options to an investor.

You wil take a delta and gamma neutral position including shares of stock along with a number

of 45-strike calls.


A Buy 58.25 shares and sell 100 call options with a 45-strike.

B Buy 58.25 shares and buy 100 call options with a 45-strike.

C Buy 17.44 shares and sell 124.24 call options with a 45-strike.

D Buy 17.44 shares and buy 124.24 call options with a 45-strike.

E Sell 17.44 shares and buy 124.24 call options with a 45-strike.

Question 5

Suppose that you are a market maker who just sold 50 of the three-month 40-strike European put

options to an investor, and suppose that you wil delta-hedge the position with shares of stock.

Any capital needed for the hedged position can be borrowed at the risk-free rate of interest.


A Sell 20.875 shares and lend the amount 934.63.

B Sell 20.875 shares and borrow the amount 934.63.

C Buy 20.875 shares and lend the amount 934.63.

D Buy 20.875 shares and borrow the amount 934.63.

E Buy 20.875 shares and borrow the amount 735.37.


3



ExamMFE


Question 6


(i) The current price of a three-month 90-strike European call option is 2.7168.

(ii) The annual volatility of the stock is (J = 0.30.

(iii) The stock currently sells for 80 and it pays no dividends.

(iv) The continuously-compounded risk-free rate of interest is r = 0.08 .

(v) The option in (i) has a delta value of 0.3285 and a gamma value of 0.0262.

(vi) C = C (5, T - t) is the option price as a function of the stock price and the remainng time

to maturity T - t .

Estimate the change in the option price if one day elapses while the stock price remains at 80

and other parameters are unchanged.

A The price drops by at most 0.01.

B The price drops by at least 0.01, but less than 0.02.

C The price drops by at least 0.02, but less than 0.03.

D The price drops by at least 0.03, but less than 0.04.

E The price drops by at least 0.05.





ExamMFE

Question 7

A market maker sells a call option and then delta-hedges the position with shares of the

underlying stock. With hourly re-hedgig the standard deviation of 1-hour returns is 0.60. With

a strategy of re-hedgig every two days the standard deviation of 2-day returns is D.

Determine D using Boyle and Emmanuel's modeL.

A 3.10

B 4.16

C 10.86

D 17.28

E 28.80

Question 8

A market-maker sells an option and then delta-hedges her position.

If she re-hedges hourly, then the hourly standard deviation of returns is 0.087.

If she re-hedges daily, then the daily standard deviation of returns is 2.088.

Using Boyle and Emanuel s formulas, calculate the daily standard deviation of returns if she re-

hedges hourly.

A 0.06

B 0.36

C 0.43

D 1.45

E 1.88


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ExamMFE


Question 9

A market maker writes 50 European call options with strike 70 and purchases 60 European call

options with strike 75 on the same stock.

For these two options you are given:


(ii)

70 strike call

75-strike call

Price

6.060

3.990

Delta

0.569

0.432

Rho

0.084

0.065

Vega

0.137

0.137

Determine the investment required of the market maker to delta-hedge this position in options.

A 88.70

B 116.03

C 177.10

D 249.40

E 376.00

Question 10

You are given the following inormation about a European put option (where B is expressed as

an annualized rate):

i1 =-0.337

B=-1.017

r=0.073

The stock underlying the put option has a current price of 40 and a volatility of 25% per annum.

The stock does not pay dividends.


Calculate the price of the European put option.

A 0.21

B 1.15

C 12.45

D 28.11

E 39.41





ExamMFE

Question 11

You are given the following information about a European put option:

Price = 0.158

L1 =-0.048

8=-1.346 (per

annum)

r=O.013

The stock underlying the put option has a current price of 50. The stock does not pay dividends.

The continuously-compoundèd risk-free rate of return is 6% per annum.

A market-maker writes 100 of the put options and then delta-hedges. The next day the stock

price falls to 49.

Estimate the market-maker's profit.

A -1.06

B -0.98

C -0.32

D -0.24

E 1.06

Question 12

You are given the following information about a European option:

Price = 3.35

L1 = 0.569

r=0.052

8 = -7.409 per annum)

The stock underlying the option has a current price of 50. The stock does not pay dividends.


Seven days later, the price of the underlying stock is 53. Estimate the new price of the option.

A 1.73

B 4.68

C 4.99

D 5.15

E 5.43


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Question 13

An investor buys a 70-75 call bull spread.

The current price of the underlying stock is 70, and the underlying stock does not pay

dividends.

The 70 strike call and the 75 strike call are described below.

70 strike call

75 strike call

Price

6.060

3.990

Delta

0.569

0.432

Rho

0.084

0.065

Vega

0.137

0.137

What investment is required for the investor to delta-hedge the position?

A -11.67

B -9.59

C -7.52

D -0.60

E 0.14





ExamMFE

Question 14

A market-maker has written a 91-day call option with a strike price of 95.


dividends.

You are given the following inormation about this and another call option on the underlying

stock.

95-strke call

105-strike call

Matuity

91 days

180 days

Price

9.39 7.33

Delta

0.6917

0.4963

Gamma

0.0235

0.0189

Vega

0.1757

0.2801

Theta per day)

-0.0372

-0.0291

Rho

0.1490

0.2086

Calculate the investment required for the market-maker to de1ta- and gamma-hedge the position.

A 2.30

B 7.18

C 24.34

D 52.59

E 59 78


9



ExamMFE


Question 15

A market-maker has written a 91-day call option with a strike price of 95.


dividends.

You are given the following inormation about options on the underlying stock.

95-strike call

105-strike call

100-strike put

Maturity

91 days

180 days

30 days

Price

9.39

7.33

3.22

Delta

0.6917

0.4963

-0.4638

Gamma

0.0235

0.0189

0.0462

Vega

0.1757

0.2801

0.1139

Theta per day)

-0.0372

-0.0291

-0.0638

Rho

0.1490

0.2086

-0.0408

Calculate the investment required for the market-maker to delta- and gamma- and rho-hedge the

position.

A 3.22

B 6.92

C 9.30

D 37.35

E 37.85





ExamMFE

Question 16

An investor buys a 70-75 put bear spread.


dividends.


The 70-strike put and the 75-strike put are described below.

70-strike put

75-strike put

Price

5.02 7.87

Delta

-0.431 -0.568

Rho

-0.088

-0.119

Vega

0.137

0.137

The investor delta-hedges the put-bear spread. Calculate the cost of the stock that the investor

purchases in order to delta-hedge the put-bear spread.

A 0.14

B 2.85

C 9.59

D 30.17

E 39 76


11



ExamMFE


Question 17

An investor buys a 70-75 put bear spread on 1,000 shares of stock.


dividends.


The 70-strike put and the 75-strike put are described below.

70-strike put

75-strike put

Price

5.02

7.87

Delta

-0.431

-0.568

The investor delta-hedges the put-bear spread.

After 1 day, the stock price has risen to 72, and the puts have the following new prices and

deltas:

70-strike put

75-strike put

Price

4.19

6.77

Delta

-0.376

-0.513

Calculate the profit earned by the investor.

A 1.95

B 2.05

C 2.74

D 4.00

E 6 05

12 ~ BPP Professional Education




ExamMFE

Question 18

A market-maker writes a 180-day call option with a strike price of 95.


dividends.

You are given the following inormation about call options on the underlying stock.

95-strike call 100-strike call

Maturity

180 days

180 days

Price

12.26

9.56

Delta

0.6793

0.5880

Gamma

0.0170

0.0187

Vega

0.2510

0.2761

Theta per day)

-0.0286

-0.0295

Rho

0.2746

0.2428

Calculate the investment required for the market-maker to delta- and gamma- and vega-hedge

the position.

A 10.91

B 12.70

C 14.48

D 21.48

E 23 17

Question 19

A market-maker has the following portfolio relating to Stock XYZ:

long 100 call options

short 500 shares of stock

Stock XYZ does not pay dividends.

The call options have a delta of ~ = 0.23 and a gamma of r = 0.0567. Put options with the same

strike price and expiration date are also available.

Whch of the following portfolio adjustments would allow the market-maker to delta- and

gamma-hedge this portfolio?

A sellng 100 put options and buying 400 shares of stock

B buying 100 put options and sellng 100 shares of stock

C sellng 77 put options and buying 100 shares of stock

D sellng 619.48 put options and buying 477 shares of stock

E sellng 619.48 put options and sellng 400 shares of stock


13



ExamMFE


Question 20

A market-maker's portfolio has a delta of 100.

The current price of non-dividend-paying Stock XYZ is 50 and European call options on

Stock XYZ are available (strike price 50), priced at 5. The delta of the call option is 0.5.

The market-maker delta-hedges his portfolio, which generates a cash amount of 1,800. The

transaction involved is:

A sell 10 units of stock and sell 260 call options

B sell 10 units of stock and sell 180 call options

C sell 20 units of stock and buy 40 call options

D sell 20 units of stock and sell 160 call options

E sell 45 units of stock and buy 90 call options

Question 21

For a delta-hedged position, which of the following would not be a good strategy to reduce the

risk of extreme price movements causing losses?

A Reduce the gamma of the position

B Static replication using other options

C Buy out-of-the-money options as insurance

D Create a financial product designed to insure against large movements

E Purchase a gamma-neutral portfolio

Question

22

A trader uses a particular trading strategy. Based on past experience, he has found that not

rebalancing his portfolio always leads to a financial loss. The average loss incurred is a direct

function of the time t in weeks since the portfolio was last rebalanced Ths function is

L(t) = t(3t+5).

The cost of rebalancing the portfolio each time is 20 and the portfolio is rebalanced at regular

intervals.

Calculate the optimum number of times x to rebalance the portfolio to minimize the loss over an

n-week period.

A

n.J

x=-

m

x = 3-J +5

5+m

x=

n2.J

m

=-

n.J

3+m

x

n

B

C

D

E

14





ExamMFE

Question 23


For two European call options, Call-I and Call-II, on a stock, you are given:

Greek

Call-I

Call-II

Delta 0.5825

0.7773

Gamma

0.0651

0.0746

Vega 0.0781

0.0596

Suppose you just sold 1000 units of Call-I.

Determine the numbers of units of Call-II and stock you should buy or sell in order to both delta-

hedge and gamma-hedge your position in Call-I.

A buy 95.8 units of stock and sell 872.7 units of Call-II

B sell 95.8 units of stock and buy 872.7 units of Call-II

C buy 793.1 units of stock and sell 692.2 units of Call-II

D sell 793.1 units of stock and buy 692.2 units of Call-II

E sell 11.2 units of stock and buy 763.9 units of Call-II

Question

24


Assume that the Black-Scholes framework holds. The price of a nondividend-paying stock is

30.00. The price of a put option on this stock is 4.00.

You are given:

(i) L' = -0.28

(ii) í = 0.10

Using the delta-gamma approximation, determine the price of the put option if the stock price

changes to 31.50.

A 3.40

B 3.50

C 3.60

D 3.70

E 3.80


15



ExamMFE


Question

25


Assume the Black-Scholes framework. Consider a stock and a derivative security on the stock.

You are given:

(i) The continuously-compounded risk-free interest rate r is 5.5 .

(ii) The time- t price of the stock is S(t).

(iii) The time- t price of the derivative security is ert In (S(t)) .

(iv) The stock's volatilty is 30 .

(v) The stock pays dividends continuously at a rate proportional to its price.

(vi) The derivative security does not pay dividends.

Calculate d, the dividend yield on the stock.

A 0.00

B 0.01

C 0.02

D 0.03

E 0.04

Question 26


Assume that the Black-Scholes framework holds. Consider an option on a stock.

You are given the following inormation at time 0:

(i) The stock price is 5(0), which is greater than 80.

(ii) The option price is 2.34.

(iii) The option delta is -0.181.

(iv) The option gamma is 0.035.

The stock price changes to 86.00. Using the delta-gamma approximation, you find that the option

price changes to 2.21.

Determine 5 0).

A 84.80

B 85.00

C 85.20

D 85.40

E 85.80

16 (Ç BPP Professional Education




ExamMFE

Question

27

SOA sample question

Assume that the Black-Scholes framework holds. Let S(t) be the price of a nondividend-paying

stock at time t, t? o. The stock's volatility is 20 , and the continuously compounded risk-free

interest rate is 4 .

You are interested in contingent claims with payoff being the stock price raised to some power.

For 0:: t ~ T, consider the equation

FtT L S(Tf J = S( t)X

where the left-hand side is the prepaid forward price at time t of a contingent claim that pays

S tf at time T. A solution for the equation is x = 1.

Determine another x that solves the equation.

A -4

B -2

C -1

D 2

E 4


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ExamMFE


Question

28

SOA sample question

Several months ago, an investor sold 100 units of a one-year European call option on a

nondividend-paying stock. She immediately delta-hedged the commitment with shares of the

stock, but has not ever re-balanced her portfolio. She now decides to close out all positions.

You are given the following inormation:

(i) The risk-free interest rate is constant.

(ii)

Several months ago

Now

Stock price

40.00

50.00

Call option price

8.88

14.42

Put option price

1.63

0.26

Call option delta

0.794

The put option in the table above is a European option on the same stock and with the same

strike price and expiration date as the call option.

Calculate her profit.

A 11

B 24

C 126

D 217

E 240

18 f BPP Professional Education




ExamMFE

Question

29

SOA sample question

Consider the Black-Scholes framework. A market-maker, who delta-hedges, sells a three-month

at-the-money European call option on a nondividend-paying stock.

You are given:

(i) The contiuously compounded risk-free interest rate is 10 .

(ii) The current stock price is 50.

(iii) The current call option delta is 0.6179.

(iv) There are 365 days in the year.

If, after one day, the market-maker has zero profit or loss, determine the stock price move over

the day

A 0.41

B 0.52

C 0.63

D 0.75

E 1.11

Question 30

SOA sample question

A market-maker sells 1,000 1-year European gap call options, and delta-hedges the position with

shares.

You are given:

(i) Each gap call option is written on 1 share of a nondividend-paying stock.

(ii) The current price of the stock is 100.


(iv) Each gap call option has a strike price of 130.

(v) Each gap call option has a payment trigger of 100.

(vi) The risk-free interest rate is 0 .

Under the Black-Scholes framework, determine the initial number of shares in the delta-hedge.

A 586

B 594

C 684

D 692

E 797


19



ExamMFE


Question 31

SOA sample question

Consider a European call option on a nondividend-paying stock with exercise date T, T:; 0 .

Let S(t) be the price of one share of the stock at time t, t? O. For 0:: t:: T , let C(s, t) be the

price of one unit of the call option at time t, if the stock price is s at that time.

You are given:

(i)

dS t) =O.ldt+CídZ t),

where Cí is

a

positive constant

and Z t)) is

a Brownian

motion.

S(t)

(ii)

dC (S(t), t)

C S t),t) r S t),t)dt+Cíc S t),t)dZ t), O::t::T

(iü)

C(S(0),O)=6

(iv)

At time t = 0, the cost of shares required to delta-hedge one unit of the call option is 9.

(v)

The contiuously compounded risk-free interest rate is 4 .

Determine r S O),O).

A 0.10

B 0.12

C 0.13

D 0.15

E 0.16

20 Ç BPP Professional Education




ExamMFE

ExamMFE


Chapter

6


The following data concerns a six-month binomial model for stock prices where each period is

three months in duration:

(i) The current price of a share of stock ABD is 41.


(iii) The continuously-compounded risk-free annual rate of interest is r = 0.08 .

(iv) Annual volatility of the stock is (J = 0.30 .

(v) Su = 48.5975 , Sd = 36.0019 , Suu = 57.6029 , Sud= 42.6732

Question 1

Determine the risk-neutral probabilty that the stock price is 42.6732 six months from today.

A 0.125

B 0.249

C 0.331

D 0.414

E 0.497

Question

2

Suppose the stock price moves down in the first three months, and then moves up in the next

period of three months.

Determine the payoff at expiration for a 40-strike six-month Asian arithmetic price put option on

a share of stock ABD where the average price is calculated as the average of the 3-month price

and the 6-month price.

A 0.00

B 0.33

C 0.66

D 1.33

E 1.66


1



ExamMFE



The following data concerns a six-month binomial model for stock prices where each period is

three months in duration:

(i) The current price of a share of stock ABD is 41.


(iii) The continuously-compounded risk-free annual rate of interest is r = 0.08 .

(iv) Annual volatility of the stock is (Y = 0.30 .

(v) Su = 48.5975 , Sd = 36.0019 , Suu = 57.6029 , Sud= 42.6732

Question 3

Using the 2-period binomial modeL, determine the price of a 40-strike six-month Asian arithmetic

price call option on a share of stock ABD where the average price is calculated as the average of

the 3-month price and the 6-month price.

A Less than 4.00




E At least 4.75

Question

4

Using the 2-period binomial modeL, determine the price of a 40-strike six-month Asian geometric

price call option on a share of stock ABD where the average price is calculated as the average of

the 3-month price and the 6-month price.

A Less than 4.00




E At least 4.75

2 (g BPP Professional Education




ExamMFE

Question 5

A market maker has just made available the following types of Asian options:

Average price options (strike price = 100)

Series AP12:

Series GP12:

Series AP250:

Arithmetic average, sampled monthly

Geometric average, sampled monthly

Arithmetic average, sampled daily

Average strike options

Series GS12:

Series GS250:

Geometric average, sampled monthly

Geometric average, sampled daily

The options are all European put options on the same underlying asset. They all expire at the

end of 1 year

The underlying asset has no associated cashfows.

Identif the correct relationships between the prices of these four options.

A AP12 ~ AP250, AP12 ~ GP12, GS12 ~ GS250

B AP12 ~ AP250 AP12:; GP12 GS12:; GS250

C AP12 :; AP250 AP12 ~ GP12 GS12 ~ GS250

D AP12 :; AP250, AP12 ~ GP12, GS12:; GS250

E AP12 :; AP250 AP12:; GP12 GS12 ~ GS250


3



ExamMFE


Question 6

Stock XYZ has a current price of 50. Its annual volatility is 30%. The stock does not pay

dividends.

The stock price follows the binomial model, with 3 periods, each of length 1 month.

A 3-month Asian average strike call option on Stock XYZ uses an arithmetic average based on the

stock price at the end of each of the 3 months. Exercise is only permitted on the expiry date.

The continuously-compounded risk-free interest rate is 6% per annum.

Determine the current value of the Asian option.

A 1.09

B 1.25

C 1.31

D 1.56

E 1.78

Question 7

A derivatives trader has a portfolio that involves the following European barrier options, which

have now reached their expiry date.

· Option A is a down-and-out call option with a barrier of 5 and a strike price of 10.

· Option B is an up-and-in put option with a barrier of 10 and a strike price of 15.

· Option C is a deferred down-rebate option with a barrier of 5 and a payment amount of

20.

The trader's portfolio currently consists of 10000 units of Option A, a short holding of 5000 units

of Option Band 5000 units of Option C.

The payoffs for all the options are based on the same underlying asset over the same period. The

underlying asset price started at 7, finshed at 8, and attained a maximum of 11 and a

minimum of 4 over this period.

Calculate the total payoff from the barrier options in the trader's portfolio.

A -35,000

B 40,000

C 65,000

D 100,000

E 135,000





ExamMFE

Question 8

A bank creates a market in the following 1-year European options on January 1:

. Option X is a vanilla call option with a strike price of 10.

. Option Y is an up-and-out call option with a barrier of 15 and a strike price of 10.

. Option Z is an up-and-in call option with a barrier of 15 and a strike price of 10.

The payoffs for all the options are based on the same underlying asset, whose price is 5 on

January 1

Trader A purchases 1000 units of Option X on January 1

Trader B purchases 1000 units of Option Y on January 1 and 1000 units of Option Z on

February 1.

Let VA and VB denote the value of each trader's portfolio (ignoring the costs of acquiring the

options) when the options expire on December 31.

Assuming that the traders make no adjustments to their positions during the year, the most

accurate statement that can be made about the relationship between V A and VB is:

A VA =VB

B VA?' VB

C VA:; VB

D VA = VB -10000

E The precise relationship between V A and VB cannot be established without the

additional assumption that the market is arbitrage-free.


5



ExamMFE


Question 9

You are given:

(i) The current price of a share of stock is 80.

(ii) The continuously-compounded risk-free annual rate of interest is r = 0.06 .

(iii) Bob establishes a position in options based on the stock in (i). The position consists of a

purchased up and in six-month 85-strike European call option on 10 shares and a

written up and out, 6-month, 85-strike European call option on 10 shares.

(iv) The barrier associated with the options in (iii) is 90.

(v) Bob invests 45 to establish this position.

Determine Bob's profit/loss at maturity on this position if the stock's highest value over the six-

month period is 88.25, its lowest value is 78, and its value at maturity is 87.75.

A - 75.28

B - 73.87

C - 71.37

D - 20.28

E - 18.87





ExamMFE

Question 10

For a share of stock BBP, you are given:

(i) On April

1, 2007 the price is 80 per share.


(iii) The risk-free annual rate of interest is r = 0.06 .

(iv) The annual volatility of the stock is (j = 0.30 .

(v) On April 1, 2007, an investor can purchase for 2.89 an option giving the right to

purchase for 4.50 on July 1, 2007 an 85-strike, three-month European call option on a

share of BBP.

Determie the price on April 1, 2007 for an option giving the right to sell for 4.50 on July 1, 2007

an 85-strike, three-month European call option on a share of BBP.

A Less than 1.45




E At least 1.90

Question 11

Six months ago a trader paid 15,000 to purchase a compound option that gave him the option of

purchasing after 3 months for a price of 75,000 a vanilla option on 1000 ounces of gold.

The vanlla option is a 3-month European put option with a strike price of 600 per ounce.

After 3 months the price of gold was 600 and the price of a 3-month European put option on

1 ounce of gold was 100.

The trader has not sold any of the options under consideration. Interest and the cost of storing

gold may be ignored.

Calculate the trader's overall profit if the gold price is now 500 per ounce.

A - 90,000

B - 65,000

C - 15,000

D + 10,000

E + 25,000


7



ExamMFE


Question 12

Stock XYZ has a current price of 50. The stock's volatility is 25% per annum. The stock pays

dividends at a continuously-compounded rate of 4 % per annum.


Option A is a European call option on Stock XYZ that expires in 12 months and has a strike price

of

50.

Option B is a European compound call option on Option A that expires in 6 months and has a

strike price of 10. Its current price is 0.90.

Option C is a European compound put option on Option A that expires in 6 months and has a

strike price of 10.

Use the Black-Scholes model to find the current price of Option C.

A 3.58

B 4.90

C 5.04

D 5.19

E 5.50

Question 13

Stock XYZ has a current price of 60. In 1 year, the stock price wil be either 75 or 55. The stock

does not pay dividends.

A 1-year European gap put option on Stock XYZ has a trigger price of 80 and a strike price of

65.

The contiuously-compounded risk-free rate is 7.0% per annum.

Determine the current value of this option.

A 0

B 0.61

C 0.65

D 4.96

E 5.32

8 (§ BPP Professional Education




ExamMFE

Question 14

Stock XYZ has a current price of 50. The stock's volatility is 25% per annum. The stock pays

dividends at a continuously-compounded rate of 4 per annum.

The continuously-compounded risk-free rate of return is 7 per annum.

A European gap call option on Stock XYZ expires in six months and has a trigger price of 50.

The strike price has been set so that, according to the Black-Scholes model, the initial price of the

option wil be zero.

Determine the payoff from this option, rounded to the nearest dollar, if the stock price in

6 months is 55.

A 3

B 1

C -1

D -3

E -4


You are given:

(i) A stock paying no dividends currently sells for 80.

(ii) The value of delta for aT-year, 100-strike European put option on a share of the stock in

(i) is -0.8.

(iii) The value today of 1 paid in T years if S(T) :;100 is 0.12.

Question 15

Determine the delta value for a T -year, 100-strike European call option on a share of this stock.

A 0.2

B 0.4

C 0.6

D 0.8

E Cannot be determined


9



ExamMFE



You are given:

(i) A stock paying no dividends currently sells for 80.

(ii) The value of delta for aT-year, 100-strike European put option on a share of the stock in

(i) is -0.8.

(iii) The value today of 1 paid in T -years if 5 (T) :; 100 is 0.12.

Question 16

Determine the price of a T-year European call option that pays S(T) - 95 if S(T) :;100 and

nothing otherwise.

A 4.00

B 4.20

C 4.40

D 4.60

E 4.80

Question 17

Stock A and Stock B are both currently priced at 50.

Stock A has a volatility of 20 per annum and pays dividends at a continuously-compounded

rate of 5 per annum.

Stock B has a volatility of 30 per annum and pays dividends at a continuously-compounded

rate of 3 per annum.

The correlation between the log-share prices is 0.5.


An exchange option permits the holder to receive 1 share of Stock B in exchange for 1 share of

Stock A in 6 months.

Calculate the current value of the exchange option using the Black-Scholes modeL.

A 0.50

B 2.34

C 3.39

D 3.41

E 3.96





ExamMFE


For a model with two stocks, you are given:

(i) Sj (t) denotes the price of stock j at time t.

(ii) Si (0) = 120, 52 (0) = 135

(iii) var(ln(Si(t)))=

0.16t ,var(ln(S2(t)))=0.09t

(iv) Stock 1 pays dividends continuously at the rate ói = 0.05 while stock 2 pays no

dividends.

(v) The risk-free annual rate of interest is r = 0.075.

(vi) The

correlation between In(Si(t)) and In(S2(t)) is

0.7.

Question 18

Determine the price for an option that gives an investor the right to exchange a share of stock 2

for a share of stock 1 in six months.

A 3.31

B 3.50

C 3.89

D 4.18

E 4.47

Question 19

Determine the price for a claim equal to mint Si (0.5),52 (0.5)) payable in six months.

A At most 114

B At least 114, but at most 116

C At least 116, but at most 118

D At least 118, but at most 120

E At least 120


11



ExamMFE


Question

20

Options 1 and 2 are European call options with strike price K on the same underlying asset.

Both options are up-and-out barrier options with the same period of cover. The barrier level for

Option 1 is Hi and the barrier level for Option 2 is H2, where H2 :; Hi.

Let max St and min St denote the highest and lowest asset prices attained over the period, and

let ST denote the asset price at the expiration date.

A portfolio consists of a long position in Option 1 and a short position in Option 2.

Identiy the correct payoff function for this portfolio of options from the expressions below, in

which I() denote an indicator function taking the value 1 for a true event and 0 for a false event.

A max(O,ST-KJxI(Hi ~minSt ~H2)

B max O,ST -KJXI Hi ~ maxSt ~ H2)

C max O,K -ST JXI Hi ~ maxSt ~ H2)

D min(O,K-ST J xI(Hi ~ minSt ~ H2)

E min

(0,

K -ST J xI(Hi ~ maxSt ~ H2)

Question 21

A special one-year derivative has a payoff equal to the excess, if any, of the arithmetic average

over the geometric average. This payoff is to be based on observations of the price of Stock XYZ

taken at the end of each month.

Calculate the payoff of the option if the 12 observations of the price of Stock XYZ are:

20

13

19 26

26

29 28

29

32 34

37

41

A

0.00

B

1.00

C

1.20

D

1.14

E

1.16

12





ExamMFE

Question

22

Asian option A is a one-year geometric average price put option with a strike price of 50.

Asian option B is a one-year arithmetic average strike call option.

Both options have payoffs based on 12 observations of the price of Stock XYZ, taken at the end of

each month.

An investor buys option A and sells option B.

Calculate the investor's combined payoff from these options in one year, rounded to the nearest

dollar, if the 12 observations of the price of Stock XYZ are:

42 50

43

32 27

35 32

38

49 56 57

48

A

- 14

B

- 2

C

+ 1

D

+ 2

E

+ 3

Question 23

Stock XYZ has a current price of 40 and a volatilty of 25% per annum. The stock price follows

geometric Brownian motion.

The stock does not pay dividends over the next 6 months and the contiuously-compounded

risk-free rate is 6% per annum.

Consider a four-month up-and-out put option with a strike price of 45 and barrier level H.

Find the limiting value of this option as H becomes very large:

A 0.00

B 0.90

C 4.71

D 4.95

E 44.11


13



ExamMFE


Question

24

Define a compound-compound option to be an option where the underlying asset is a compound

option.

In terms of call and put types, how many possible combinations are there for a compound-

compound option?

A 4

B 8

C 9

D 12

E 16

Question

25

Which of the following items of information is not used to price a compound option?

A the price of the underlying asset of the underlying option

B the term of the underlying asset of the underlying option

C the nature call or put) of the underlying option

D the strike price of the compound option

E the nature call or put) of the compound option





ExamMFE

Question

26


For a stock, you are given:

(i) The current stock price is 50.00.

(ii) ö = 0.08

(iii) The continuously compounded risk-free interest rate is r = 0.04 .

(iv) The prices for one-year European calls (C) under various strike prices (K) are shown

below:

K C

40

9.12

50

4.91

60

0.71

70

0.00

You own four special put options each with one of the strike prices listed in (iv). Each of these

put options can only be exercised immediately or one year from now.

Determine the lowest strike price for which it is optimal to exercise these special put option(s)

immediately.

A 40

B 50

C 60

D 70

E It is not optimal to exercise any of these put options


15



ExamMFE


Question

27


Let S(t) denote the price at time t of a stock that pays dividends continuously at a rate

proportional to its price. Consider a European gap option with expiration date T, T:; 0 .

If the stock price at time T is greater than 100, the payoff is S(T) - 90 ; otherwise, the payoff is

zero.

You are given:

(i) 5(0) = 80

(ii) The price of a European call option with expiration date T and strike price 100 is 4.

(iii) The delta of the call option in (ii) is 0.2.

Calculate the price of the gap option.

A 3.60B 5.20

C 6.40

D 10.80

E There is not enough information to solve the problem





ExamMFE

Question

28


You have observed the following monthly closing prices for stock XYZ:

Date

Stock Price

January 31, 2008

105

February 29, 2008

120

March 31, 2008

115

April 30, 2008

110

May 31,2008

115

June 30 2008

110

July 31, 2008

100

August 31, 2008

90

September 30,2008

105

October 31, 2008

125

November 30,2008

110

December 31 2008

115

The following are one-year European options on stock XYZ. The options were issued on

December 31 2007.

i) An arithmetic average Asian call option the average is calculated based on monthly

closing stock prices) with a strike of 100.

(ii) An up-and-out call option with a barrier of 125 and a strike of 120.

iii) An up-and-in call option with a barrier of 120 and a strike of 110.

Calculate the difference in payoffs between the option with the largest payoff and the option with

the smallest payoff.

A 5

B 10

C 15

D 20

E 25


17



ExamMFE


Question

29


Your company has just written one milion units of a one-year European asset-or-nothig put

option on an equity index fund.

The equity index fund is currently trading at 1,000. It pays dividends continuously at a rate

proportional to its price; the dividend yield is 2 . It has a volatility of 20 .

The option's payoff wil be made only if the equity index fund is down by more than 40 at the

end of one year.

The continuously-compounded risk-free interest rate is 2.5 .

Using the Black-Scholes model determine the price of the asset-or-nothing put options.

A 0.2 millon

B 0.9 million

C 2.7 millon

D 3.6 millon

E 4.2 millon

Question 30


You are given the following inormation about a nondividend-paying stock:


(ii) The stock-price process is a geometric Brownian motion.

(iii) The continuously-compounded expected return on the stock is 10 .

(iv) The stock's volatilty is 30 .

Consider a nie-month 125-strike European call option on the stock.

Calculate the probability that the call wil be exercised.

A 24.2

B 25.1

C 28.4

D 30.6

E 33.0





ExamMFE

Question 31

SOA sample question

Consider a chooser option (also known as an as-you-like-it option) on a nondividend-paying

stock. At time 1, its holder wil choose whether it becomes a European call option or a European

put option, each of which wil expire at time 3 with a strike price of 100.

The chooser option price is 20 at time t = 0 .

The stock price is 95 at time t = o. Let C(T) denote the price of a European call option at time

t = 0 on the stock expiring at time T, T:; 0 , with a strike price of 100.

You are given:

(i) The risk-free interest rate is O.

(ii) C(1) = 4 .

Determine C(3).

A 9

B 11

C 13

D 15

E 17


19



ExamMFE


Question 32

SOA sample question

Prices for 6 month 60 strike European up and out call options on a stock S are available. Below is

a table of option prices with respect to various H, the level of the barrier. Here, S O) = 50.

H

Price of up and out call

60

0

70

0.1294

80

0.7583

90

1.6616

00

4.0861

Consider a special 6-month 60-strike European knock-in, partial knock-out call option that

knocks in at Hl = 70, and partially knocks out at H2 = 80. The strike price of the option is 60.

The following table summarizes the payoff at the exercise date:

Hl Not

Hit

Hl Hit

H2 Not

hit

H2Hit

0

2xmax (5(0.5)- 60,0)

max (5(0.5)-60,0)

Calculate the price of the option.

A 0.6289

B 1.3872

C 2.1455

D 4.5856

E It cannot be determined from the information given above.

20 § BPP Professional Education




ExamMFE

Question 33

SOA sample question

Assume the Black-Scholes framework. For a European put option and a European gap call

option on a stock, you are given:

(i) The expir date for both options is T .

(ii) The put option has a strike price of 40.

(iii) The gap call option has strike price 45 and payment trigger 40.

(iv) The time-O gamma ofthe put option is 0.07.

(v) The time-O gamma of the gap call option is 0.08.

Consider a European cash-or-nothing call option that pays 1000 at time T if the stock price at

that time is higher than 40.

Find the time-O gamma of the cash-or-nothing call option.

A -5

B -2

C 2

D 5

E 8

Question 34

SOA sample question

Assume the Black-Scholes framework. For a stock that pays dividends continuously at a rate

proportional to its price, you are given:


(ii) The stock s volatilty is 0.2.

(iii) The continuously compounded expected rate of stock-price appreciation is 5 .

Consider a 2-year arithmetic average strike option. The strike price is

A 2) =t S 1)+S 2))

Calculate var A 2)) .

A 1.51

B 5.57

C 10.29

D 22.29

E 30.57


21



ExamMFE


Question 35

SOA sample question

Assume the Black-Scholes framework. Consider two nondividend-paying stocks whose time- t

prices are denoted by Si t) and 52 t), respectively.

You are given:

(i) Si (0) = 10 and 52 (0) = 20 .

(ii) Stock 1 s volatilty is 0.18.

(iii) Stock 2 s volatility is 0.25.

(iv) The correlation between the continuously compounded returns of the two stocks is -0.40.


(vi) A one-year European option with payoff max1min(2Si(1),S2(1)J-17,O) has a current

(time-O) price of 1.632.

Consider a European option that gives its holder the right to sell either two shares of Stock 1 or

one share of Stock 2 at a price of 17 one year from now.

Calculate the current time-O) price of this option.

A 0.66

B 1.12

C 1.49

D 5.18

E 7.86





ExamMFE

ExamMFE


Chapter 7

Question 1

Z(t) is Brownian motion.

Calculate p( Z(10) :;11 Z(5) = -1) .

A 0

B 0.187

C 0.326

D 0.345

E 0.814

Question 2

Let si, s2 ,s :; 0 .

State which one of the following statements is not a property of Brownian motion, Z(t).

A Z(O) =0

B The increments Z(t+s)-Z(t) have a normal distribution with mean 0 and variance s.

C The increments Z(t+si) - Z(t) and Z(t) -Z(t-S2) are statistically independent.

D The increments Z( t + s) - Z( t) form a martingale.

E Z(t) has continuous sample paths.

Question 3


State which of the following processes is a martingale with respect to Z(t) .

A

exp(-Z(t))

B eXPL -Z t)+ltJ

C eXPL -Z t)-ltJ

D

eXPL -Z t)+lt2J

E

eXPL -z t)-lt2J


1



ExamMFE


Question 4

You are given:

S t) = 55eo.045t + O.5Z t) , Z t) =

standard

Brownan motion

Use Ito s Lemma to determine dS(t).

A dS(t) = 0.045dt + O.5dZ(t)

B dS(t) = 0.045S(t)dt + 0.5S(t)dZ(t)

C dS(t) = 0.17 dt + 0.5dZ(t)

D dS(t) = 0.17S(t)dt + O.5S(t)dZ(t)

E dS(t) = 0.125S(t)dt + 0.25S(t)dZ(t)

Question 5

You are given:

i) dS t) = as t)dt + as t)dZ t)

ii) X S,t)=3é t)+txS t)

Determine the drift of the process X (S, t) .

A S(t)(( 3a+ta2S(t) )é(t) +at + 1 J

B S(t)(3(a+a2S(t))eS(t)+at+1J

C S(t)L3aeS(t)+at+1J

D X(t)( (3a+to.2S(t) )é(t) +at J

E X(t)((3a+t(j2X(t))+at+1J





ExamMFE

Question 6

You are given:

dX(t)=0.09dt+

0.4dZ t) ,X O)=O

Z (t) = standard Brownian motion

Determine var X 2)) .

A 0.16

B 0.32

C 0.40

D 0.64

E 0.80

Question 7

You are given:

dX t)=0.09dt+0.4dZ t), X O)=O

z ( t) = standard Brownian motion

Determine Pr(X(2):; 0.15).

A 0.52

B 0.54

C 0.56

D 0.58

E 0.60


3



ExamMFE


Question 8

r(t), the continuously-compounded interest rate at time t is modeled by the diffusion process:

dr(t) = (0.06-r(t))dt+0.OldZ(t)

where Z t) is Brownian motion.

The theoretical price, V(t), of a derivative whose value is linked to r(t) is:

V t) = exp L -tr t) J

Determine the diffusion model satisfied by V(t).

A dV t) =

(-0.06t+(t-l)r(t))dt-0.OltdZ(t)

V(t)

B dV(t) = 1-0.06t+0.0001t2 +tr(t)Jdt-O.OltdZ(t)

V t) L

C dV t) = 1-0.06t+

0.0001t2 + t-l)r t)Jdt-O.OltdZ t)

V t) L

D dV(t) =1-0.06t+0.00005t2 +tr(t)Jdt-O.01tdZ(t)

V t) L

E dV t) = 1-0.06t+

0.00005t2 + t-l)r t)Jdt-O.OltdZ t)

V t) L

Question 9

You are given:

i) dX t) = 0.09X t)dt + O.4X t)dZ t) , Z t) = standard Brownian motion

ii) X O) = 55

Determine X t) .

A X t)

=

55eo.01t+O.4Z(t)

B X (t) = 55 eO'Olt + o.i6Z(t)

C X t)

=

55eo.05t+O.4Z(t)

D X t)

=

55eo.05t+O.i6Z(t)

E X t)

=

55eo.09t+O.4Z(t)





ExamMFE

Question 10

You are given:

(i) dX(t) = 0.09X(t)dt + O.4X(t)dZ(t) , Z(t) = standard Brownian motion

(ii) X(0)=55

Determine Pr(X(2):; 65).

A 0.28

B 0.31

C 0.34

D 0.37

E 0.40

Question 11

G(t) , the spot price of gold at time t is modeled by the diffusion process:

dG(t) = 0.06dt+0.ldZ(t)

G(t)


Calculate the probability that the gold price at time 5 exceeds 1000, given that it is 500 at time O.

A 3.1

B 3.9

C 5.0

D 13.9

E 20.1

Question 12


Identify the correct statement about the drift of the process Z(t)4 -6Z(t)2 .

A The drift is always zero.

B The drift is always positive.

C The drift is never positive.

D The drift is positive when IZ(t)1 :;1, but not otherwise.

E The drift is positive when Z t):;.J, but not otherwise.


5



ExamMFE


Question 13

S(t) is geometric Brownian motion.

The processes U t)

, V t) and W t) are defined as follows:

U(t) = S(t)+ 100

V(t) = 10S(t)

W(t) = S(t)2

Identify which of these processes are also geometric Brownian motion.

A U t)

, V t) and W t)

B U(t) and V(t) only

C V t) and W t) only

D W t) only

E none of them


You are given:

(i)

x (t) is the value of one Canadian dollar in U.5. dollars at time t.

(ii)

X(O) = 0.85

(iii)

dX t) .

-) = O.Oldt + 0.20dZ t) ,Z t) = standard Brownian motion

(t

(iv)

G t) = X t)eO.02 T-t) is the forward price in U.S. dollars per Canadian dollar and T is

the time until maturity of the forward contract.

Question 14

Determine Pr(X(l) :; 0.80).

A 0.58

B 0.60

C 0.62

D 0.64

E 0.66

6 g BPP Professional Education




ExamMFE


You are given:

(i)

X(t) is the value of one Canadian dollar in U.S. dollars at time t.

(ii)

X(O) = 0.85

(iii)

dX(t)

X (t) = 0.01 dt + 0.20 d Z (t) ,Z ( t) = standard Brownian motion

(iv)

G t) = X t)eO.02 T-t) is the forward price in U.S. dollars per Canadian dollar and T is

the time until maturity of the forward contract.

Question 15

Using Ito's Lemma, determine the stochastic differential equation for G(t).

A

dG(t)

0.02 dt + 0.20 dZ(t)

=

(t)

B

dG(t)

0.01 dt + 0.20 dZ(t)

G(t)

C

dG(t)

G t) = -0.01 dt + 0.20 dZ t)

D

dG(t)

G t) = 0.20 dZ t)

E

dG(t)

G t) = 0.04 dZ t)

Question 16

At time t = 0.5 years, determine the probability that the forward price exceeds 0.88 on a T = 1

year forward contract for the u.s. equivalent of one Canadian dollar.

A Less than 0.43




E At least 0.49


7



ExamMFE


Question 17

For two stocks, you are given:

i) Stock 1 pays dividends continuously at the rate Öi =0.02 . Stock 2 pays no dividends.

ii) S¡ t) is the price of stock i at time t.

iii) Z t) is standard Brownian motion.

iv) Both stocks follow geometric Brownian motion with:

dSi ;) = 0.09dt + 0.25dZ t) , dS2t) = 0.13dt + b dZ t)

Si t 52 t)

(v) The continuously-compounded risk-free rate of interest is r = 0.045 .

Determie b .

A Less than 0.26




E At least 0.32

Question 18

For a diffusion process describing the continuously-compounded risk-free rate of interest X t) ,

you are given:

dX t) = a 0.06 - X t)) dt + 0.20dZ t) , a:; 0 , X O) = 0.05

Z t) = standard Brownian motion

Determine lim X t) .

t~oo

A 0.050

B 0.052

C 0.058

D 0.060

E The limit does not exist.





ExamMFE

Question 19

For a diffusion process describing the continuously-compounded risk-free rate of interest X (t) ,

you are given:

dX(t)=a(0.06-X(t))dt +O.2OdZ(t), a:;

0 ,X(O)

=

0.05

Z ( t) = standard Brownian motion

Determine lim ECX(t)J.

t--CO

A 0.050

B 0.052

C 0.058

D 0.060

E The limit does not exist.

Question

20


X(t) is a diffusion process with stochastic differential equation:

dX(t) =Â(a-X(t))dt+O dZ(t)

where the Greek letters represent parameters with constant values.

Two other diffusion processes, S(t) and N(t)

, are defined by the equations:

N(t) = eÂt (X(t) -a)

S(t) = N(tf

The drift of the process N(t) is denoted by flN(t) and the drift of the process S(t) is denoted by

fls(t). Find flN(t)+ fls(t).

A 0

B eÂt

C O eÂt

D O e2Ât

E 0 2e2Ât


9



ExamMFE


Question 21

Wi t), W2 t) and W3 t) are

independent

Brownian

motions.

Let:

Zi (t) = Wi (t)

Z2(t) = aWi (t)+bW2(t)

Z3 t) =

cWi(t)+

dW2(t)+

eW3(t)

It is required to find

values

for the

constants a, b, c, d and e so

that Zi t), Z2 t) and Z3 t)

are Brownian motions with Pi2 = 0.8, Pi3 = 0.4 and P23 = 0.5 ,where Pij denotes the correlation

between Zi(t) and Zj(t).

Determine the correct value for the constant e.

A 0

B 0.5

C .J0.75

D 1

E ß

Question

22

Which of the following statements is not always correct?

A Any linear combination of martingales is a martingale.

B If you subtract one martingale from another then you always get zero drift.

C If X (t) has a constant drift of ll then X ( t) - llt is a martingale.

D The product of two martingales is a martingale.

E If Z t) is Brownian motion and n is a positive odd number, then E L Z t t J is always

zero.

Question 23

Z(t) is Brownian motion. Determine which of the following processes is not a martingale.

A 12t-3L Z(t)+4Z(t)2J

B e -8t-4Z(t)

C 6t Z t)- 2Z t)3

D elrZ(íCt)

E Z t)3 -3tZ t)2





ExamMFE

Question 4

Consider the stochastic differential equation:

dX(t) = Y(t)dt+~Y(t) dZ(t)


Whch of the following statements about the process V (t) = e-2X(t) is correct?

A V (t) is a martigale because it has no drift at alL.

B V (t) is not a martingale because X (t) has a drift of Y (t) .

C V t) is a martigale because V t) has a positive drift of e -2Y t) .

D V (t) has a volatility of -2X (t )~Y (t) and so cannot be a martingale.

E V(t) is not a martingale because V(t) has a drift of e-2Y(t).

Question

25

Consider the stochastic process X (t) = Z (t) e -rZ(t) where Z(t) is Brownian motion.

Find the stochastic differential equation for X (t) .

A dX (t) = re -rZ(t) C 0.5Z( t) -1 J dt+C 1- rZ( t) J e-rZ(t)dZ (t)

B dX (t) = re-rZ(t) C O.5rZ( t) -1 J dt+C 1- rZ (t) J e -rZ(t)dZ(t)

C dX t) = O.5re-rZ t) L l-r2Z t)Jdt-CrZ t)-lJe-rZ t)dZ t)

D dX (t) = r2e -rZ(t) C l-Z( t) Jdt+ rZ(t)e -rZ(t)dZ (t)

E dX t) = C 1- rZ t) J e-rZ t)dZ t) - re-rZ t) C 1-0.5rZ t) J dZ t)

Question

26

Whch of the following statements is incorrect?

A The Sharpe ratio wil increase if the risk-free rate decreases.

B The Sharpe ratio wil decrease if the expected rate of return decreases.

C Assets whose prices are not perfectly correlated cannot have the same Sharpe ratio.

D The Sharpe ratio equals the risk premium of an asset divided by the volatilty of that

asset.

E The Sharpe ratio wil decrease if the volatilty of the asset increases.


11



ExamMFE


Question

27


Consider a model with two stocks. Each stock pays dividends continuously at a rate proportional

to its price.

5 j (t) denotes the price of one share of stock j at time t.

Consider a claim maturing at time 3. The payoff of the claim is Maximum Si 3), 52 3)) .

You are given:

(i) Si (0) = 100

(ii) 52 (0) = 200

(iii) Stock 1 pays dividends of amount 0.05Si (t)dt between time t and time t+ dt .

(iv) Stock 2 pays dividends of amount 0.152 (t)dt between time t and time t+ dt .

(v) The price of a European option to exchange Stock 2 for Stock 1 at time 3 is 10.

Calculate the price of the claim.

A 96

B 145

C 158

D 200

E 234





ExamMFE

Question

28



(i)

S t) is the value of one British pound in US. dollars at time t.

(ii)

dS(t)

-= O.ldt+O.4dZ t)

S(t)

(iii)

The continuously compounded risk-free interest rate in the US. is r = 0.08.

(iv)

The continuously compounded risk-free interest rate in Great Britain is r* = 0.10 .

(v)

G t) = S t)e r-r*) T -t) is the forward price in US. dollars per British pound, and T is the

maturity time of the currency forward contract.

Based on Itô's Lemma, which of the following stochastic differential equations is satisfied by

G t) ?

A dG t) =

G(T)(0.12dt+0.4dZ(t))

B dG(t) = G(T) (0.10dt+0.4dZ(t))

C dG t) = G T) 0.08dt+ O.4dZ t))

D dG(t) = G(T) (0.12dt+ 0.16dZ(t))

E dG t) =

G(T)(0.10dt+0.16dZ(t))


13



ExamMFE


Question

29


Consider two nondividend-paying assets X and Y, whose prices are driven by the same Brownian

motion Z. You are given that the assets X and Y satisfy the stochastic differential equations:

dX(t) = 0.07dt+

0.12dZ(t)

X(t)

dY(t)

-=Gdt+HdZ(t)

Y(t)

where G and H are constants.

You are also given:

(i) d(lnY(t))=0.06dt+O dZ(t)

(ii) The contiuously compounded risk-free interest rate is 4 .

(iii) 0 ~ 0.25

Determine G.

A 0.065

B 0.070

C 0.075

D 0.100

E 0.120





ExamMFE

Question 30


X(t) is an Ornstein-Uhlenbeck process defined by:

dX t) = 2 4-X t)) dt+8dZ t)

where Z(t) is a standard Brownian motion.

1

Let Y(t)=-.

X(t)

You are given that:

dY t) =a Y t))dt+ ß Y t))dZ t)

for some functions a y) and ß y).

Determine a 1i).

A -9

B -1

C 4

D 7

E 63

Question 31


Two nondividend-paying assets have the following price processes:

dSi (t) = 0.08dt + 0.2dZ(t)

Si (t)

dS2 (t) = 0.0925dt -0.25dZ(t)

52 (t)

where Z(t) is a standard Brownian motion. An investor is to synthesize the risk-free asset by

allocating 1000 between the two assets.

Determine the amount to be invested in the first asset Si .

A 333.33

B 444.44

C 555.56

D 666.67

E 750.00


15



ExamMFE


Question 32


Assume the Black-Scholes framework. For t? 0 , let S(t) be the time- t price of a nondividend-

paying stock. You are given:

(i) 5(0) = 0.5

(ii) The stock price process is:

dS(t) = 0.05dt+0.2dZ(t)

S(t)

where Z(t) is a standard Brownian motion.

(iii) EL S(lt J = 1.4, where a is a negative constant.

(iv) The continuously-compounded risk-free interest rate is 3 .

Consider a contingent claim that pays S(lt at time 1.

Calculate the time-O price of the contingent claim.

A 1.345

B 1.359

C 1.372

D 1.401

E 1.414

Question 33


Consider an arbitrage-free securities market model, in which the risk-free interest rate is constant.

There are two nondividend-paying stocks whose price processes are:

Si t) = Si 0)e°lt+O.2Z t)

52 (t) = 52 (0)eO.i25t+0.3Z(t)

where Z(t) is a standard Brownian motion and t? 0 .

Determine the continuously-compounded risk-free interest rate.

A 2

B 5

C 8

D 10

E 20





ExamMFE

Question 34

SOA sample question

Consider the Black-Scholes framework. Let S(t) be the stock price at time t, t? 0 .

Define X(t) = In S(t) .

You are given the following three statements concerning X(t).

i) X t), t? 0) is an arithmetic Brownian motion.

ii) var X t+h)-X t))

=

o.2h, t?O, h:;O.

iii) lim fi X jT)_X U-l)T)J2 = J2T

n--00 j=il n n

A Only (i) is true

B Only (ii) is true

C Only i) and ii) are true

D Only (i) and (iii) are true

E i), ii) and iii) are true

Question 35

SOA sample question

Consider the Black-Scholes framework. You are given the following three statements on

variances, conditional on knowing S(t), the stock price at time t.

(i)

var(X(t+h)-X(t)) = (J2h, h:; 0

(ii)

varr dS t) IS t)J = J2dt

L S t)

(iii)

var(S(t+dt) I S(t)) = S(t)2(J2dt

A

B

C

D

E

Ony i) is true

Only ii) is tre

Only (i) and (ii) are true

Only (ii) and (iii) are true

i), ii) and iii) are true


17



ExamMFE


Question 36

SOA sample question

Consider two nondividend-paying assets X and Y. There is a single source of uncertainty which

is captured by a standard Brownian motion (Z(t)). The prices of the assets satisfy the stochastic

differential equations:

dX t) =0.07dt+0.12dZ t) and dY t) =

Adt+BdZ(t)

X t) Y t)

where A and B are constants.

You are also given:

i) d 1n Y t)) = ,udt+

0.085dZ(t)

(ii) The continuously compounded risk-free interest rate is 4 .

Determine A .

A 0.0604

B 0.0613

C 0.0650

D 0.0700

E 0.0954

Question 37

SOA sample question

Let iZ(t)J be a standard Brownian motion. You are given:

i) U t) =2Z t)-2

ii) V t)= Z t))2_t

iii) W t) = t2 Z t) - 21 s Z s) ds

Which of the processes defined above has/have zero drift?

A iV(t)J only

B iW(t)J only

C iU t)J and iV t)J only

D iV(t)J and iW(t)J only

E All three processes have zero drift.





ExamMFE

Question 38

SOA sample question

Consider the stochastic differential equation:

dX(t)=Â(a-X(t))dt+tTdZ(t), t?O

where Â, a and tT are positive constants, and t Z t) J is a standard Brownian motion.

The value of X O) is known.

Find a solution.

A X(t) = X(O)e-..t +a(l-e-..t)

B X t) = X O)+ 1 ads

+ 1 tTdZ s)

C X t) = X O)+ 1 aX s)ds+ 1 tTX s)dZ s)

D X(t)=X(O)+a(e..t -1)+ 1tTeÂsdZ(s)

E X t)=X O)e-..t +a l-e-..t)+ 1tTe-.. t-S)dZ s)


19



ExamMFE


Question 39

SOA sample question

At time t = 0 , Jane invests the amount of W(O) in a mutual fund. The mutual fund employs a

proportional investment strategy: There is a fixed real number ø, such that, at every point of

time, 100Ø of the fund's assets are invested in a nondividend paying stock and 100(1- Ø) in a

risk-free asset.

You are given:

(i) The continuously compounded rate of return on the risk-free asset is r.

ii) The price of the stock, S t), follóws a geometric Brownian motion

dS t) = adt + adZ t), t? 0

S(t)

where i Z t) J is a standard Brownian motion.

Let W(t) denote Jane's fund value at time t, t? 0 .

Which of the following equations is true?

A dW t) =

(Øa+(l-Ø)r)dt+adZ(t)

W(t)

B W(t) = W(O) expHØa+(l-Ø)r)t+øaZ(t)J

C W t) = W O) expt øa+ l-Ø)r -lJøa2 )t+øaZ t))

D W t)=

W(O)CS(t)jS(O)Jø e(i-Ø)rt

E W t) = W O)CS t)jS O)Jø eXPL 1-Ø) r+lJøa2)tJ

20 g BPP Professional Education




ExamMFE

Question 40

SOA sample question

The cubic variation of the standard Brownian motion for the time interval (0, T) is defined

analogously to the quadratic variation as

lim f iZUhJ-ZC(j-l)h JJ3

n~oo j=i

where h = T/nh = Tin.

What is the distribution of the cubic variation?

A

N(O,O)

B

N(O, T' )

C N(O, T)

D

N(O, T3/2)

E N(-~T/2, T)

Question 41

SOA sample question

The stochastic process i R(t)J is given by

R(t) = R(O)e-t +0.05(1-e-t )+0.11 es-t ~R(s) dZ(s)

where fZ t)J is a standard Brownian motion.

Define X(t)=(R(t))2.

Find dX t).

A L O.l~X(t) -2X(t)Jdt+0.2(X(t))3/4 dZ(t)

B L O.l1~X(t) -2X(t)Jdt+0.2(X(t))3/4 dZ(t)

C L 0.12~X t) -2X t) J dt+0.2 X t)f/4 dZ t)

D 10.01+(0.1-2R(0))e-ti~X(t)dt+0.2(X(t)f/4 dZ(t)

E 0.1-2R 0))e-t ~X t)dt+0.2 X t)f/4 dZ t)


21



ExamMFE


Question 42

SOA sample question

The price of a stock is governed by the stochastic differential equation:

dS t) = 0.03dt+ O.2dZ t)

S(t)

where 1 Z(t)) is a standard Brownian motion. Consider the geometric average

G = (5(1) x 5(2) x 5(3) )1/3

Find the variance of In G .

A 0.03

B 0.04

C 0.05

D 0.06

E 0.07





ExamMFE

Question

43

SOA sample question

Let x(t) be the dollar/euro exchange rate at time t. That is, at time t, €1 = $x(t).

Let the constant r be the dollar-denominated continuously compounded risk-free interest rate.

Let the constant r€ be the euro-denominated continuously-compounded risk-free interest rate.

You are given

dx(t)

-= (r-r€)dt+O dZ(t)

x(t)

where iZ(t)J is a standard Brownian motion and 0 is a constant.

Let y t) be the euro/ dollar exchange rate at time t. Thus, y t) = 1jx t) .

Which of the following equation is true?

A

dy(t) = (r€ -r)dt-O dZ(t)

y(t)

dy(t) =(r€ -r)dt+O dZ(t)

y(t)

dy(t) 2

-=(r€ -r-i¡zO )dt-O dZ(t)

y(t)

dy(t) =(r€ -r+1j0 2)dt+O dZ(t)

y(t)

dy(t) = (r€ _ r + 0 2 )dt - O dZ(t)

y(t)

B

c

D

E


23



ExamMFE


Question

44

SOA sample question

The prices of two nondividend-paying stocks are governed by the following stochastic

differential equations:

dSi(t) =0.06dt+0.02dZ(t)

Si (t)

dS2(t) =

O.03dt+

kdZ(t)

52 (t)

where 1 Z t) ì is a standard Brownian motion and k is a constant.

The current stock prices are Si 0) = 100 and 52 0) = 50 .

The continuously-compounded risk-free interest rate is 4 .

You now want to construct a zero-investment, risk-free portfolio with the two stocks and risk-

free bonds.

If there is exactly one share of Stock 1 in the portfolio, determine the number of shares of Stock 2

that you are now to buy. (A negative number means shorting Stock 2.)

A -4

B -2

C -1

D 1

E 4

Question

45

SOA sample question


You are given the following information for a stock that pays dividends continuously at a rate

proportional to its price.

(i) The current stock price is 0.25.

(ii) The stock s volatility is 0.35.

(iii) The continuously compounded expected rate of stock-price appreciation is 15 .

Calculate the upper limit of the 90 lognormal confidence interval for the price of the stock in

6

months.

A 0.393

B 0.425

C 0.451

D 0.486

E 0.529





ExamMFE

Question 46

SO sample question


You are given:

(i) S(t) is the price of a stock at time t.

(ii) The stock pays dividends continuously at a rate proportional to its price. The dividend

yield is 1 .

(iii) The stock-price process is given by

dS(t) = 0.05dt+

0.25dZ(t)

S(t)

where tZ(t)) is a standard Brownian motion under the true probability measure.

(iv) Under the risk-neutral probabilty measure, the mean of Z(O.5) is -0.03.

Calculate the continuously compounded risk-free interest rate.

A 0.030

B 0.035

C 0.040

D 0.045

E 0.050


25



ExamMFE


Question

47

SOA sample question


Let S(t) be the time- t price of a stock that pays dividends continuously at a rate proportional to

its price.

You are given:

(i)

dS t) -

- = fldt + 0.4dZ t)

S(t)

where 1 Z t) J is a standard Brownian motion under the risk-neutral probability measure

ii) for 0:: t :: T , the time- t forward price for a forward contract that delivers the square of

the stock price at time Tis:

Ft,T (52) = S2(t)exp(0.18(T -t))

Calculate fl .

A 0.01

B 0.04

C 0.07

D 0.10

E 0.40





ExamMFE

Question 48

SOA sample question

Define:

i) W t) = t2

ii) X t)= t), where t) is the greatest integer part of tt; for example, 3.14)=3, 9.99)=9,

and 4)= 4.

iii) Y t) = 2t+0.9Z t), where iZ t),t ?01 is a standard Brownian motion.

Let vl (U) denote the quadratic variation of a process U over the time interval (0, T) .

Rank the quadratic variations of W, X and Y over the time interval (0,2.4).

A

2 2 2

V2.4 W) ~ V2.4 Y) ~ V2.4 X)

B

Vf.4 W) ~ Vf.4 X) ~ Vf.4 Y)

C Vf.4 X) ~ Vf.4 W) ~ Vf.4 Y)

D

Vù (X) ~ Vf.4 (Y) ~ Vf.4 (W)

E

None of the above

Question

49

SOA sample question

Let S t) denote the time- t price of a stock. Let Y t) = S t))2. You are given:

dY t) = 1.2dt - O.5dZ t), Y O) = 64 ,

Y(t)

where i Z t), t ? 01 is a standard Brownian motion.

Let (L, U) be the 90 lognormal confidence interval for 5(2).

Find U.

A 27.97B 33.38

C 41.93

D 46.87

E 53.35


27




ExamMFE

ExamMFE


Chapter

8

Question 1

It has been proposed to model P( r, t, T), the price at time t of a zero-coupon bond maturing at

time T, by the diffusion process:

dP(r, t, T) = Lr(t) + ß.Jr(t)(T - t) J P(r, t, T)dt+ (j.JT - tP(r, t, T)dZ(t)

where Z(t) is Brownian motion and ß and (j are positive constants.

The following objections have been leveled at this proposed process:

1. The Sharpe ratio is not a constant.

2. The Sharpe ratio is not independent of T .

3. Bond price volatility doesn t decrease as maturity approaches.

Whch of these statements are factually correct for this model and are valid objections for a

realistic model of zero-coupon bond prices?

A None of them

B 2 only

C 1 and 2 only

D 1 and 3 only

E All of them

Question

2

Identify which of the following statements are correct for the Vasicek modeL.

1. Negative interest rates are possible.

2. The short rate exhibits mean-reversion.

3. The Sharpe ratio is assumed to be constant.

A None of them

B 2 only

C 1 and 2 only

D 2 and 3 only

E All of

them


1



ExamMFE


Question 3

Identify which of the following statements are correct for the Cox-Ingersoll-Ross modeL.

1. Negative interest rates are possible.

2. The short rate exhibits mean-reversion.

3. The process for the short rate is an Ornstein-Uhlenbeck process.

A None of them

B 2 only

C 1 and 2 only

D 2

and 3 only

E All of

them

Question

4

A researcher intends to use the Vasicek interest rate model to calculate the price at time 0 of a

European option expiring at time T on a zero-coupon bond.

· Let rt denote the instantaneous short rate at time t.

· Let (Yt denote the volatility of the short rate at time t.

· Let l1 denote the long-term value of the short rate in the real world and let l1 * denote

the long-term value of the short rate in a risk-neutral world.

· Let a denote the mean reversion rate, which is assumed to have a value of 0.2.

Which of the following gives the most accurate description of the inputs relating to the Vasicek

model whose values need to be specified before this calculation can be carried out?

A ro, (Yo, l1

B ro, (Yo, l1 *

C rt (0:: t:: T), (Yo, l1 *

D rt (0:: t :: T), (Yt (0:: t :: T), l1

E rT (YT l1

Question 5

Whch of the following gives the most accurate description of the behavior assumed for the

future instantaneous volatilty of the short rate of interest according to the Cox-Ingersoll-Ross

model?

A The volatilty has a constant value throughout that is known at the outset.

B The volatilty has a constant value throughout that is not known at the outset.

C The volatilty varies stochastically independently of the value of the short rate.

D The volatilty varies stochastically and is linked to the initial value of the short rate.

E The volatilty varies stochastically and is linked to the future value of the short rate.





ExamMFE


Let P(r(t),t,T) denote the price at time t of 1 to be paid with certainty at time T:; t when the

short rate at time t is r (t) .

For a CIR interest rate model you are given:

P(0.04,O,3) = 0.91792

P(0.05,3,6) = 0.89990

Question 6

Determine B(t,T) where T=t+3.

A Less than 1.85




E At least 2.15

Question 7

Determine 100P(0.06, t ,T) where T=t+3.

A 86.54

B 87.10

C 87.66D 88.22

E 88.78


3



ExamMFE


Question 8

The short rate of interest r(t) is modeled using the Itô process defined by the SDE:

dr t) = 0.2tO.05-r t))dt+0.OldZ t), t? 0


Let m(t) = E(r(t)) denote the expected value of the short rate at time t.

A formula for m t) can be found by solving a differential equation satisfied by m t)

, which can

be deduced from the SDE for r(t).

Find the formula for m t).

A 0.05

B 0.05(1-e-'.2t)

C 0.05+ r 0)-0.05)eO.2t

D 0.05+(0.05-r(0))e-O.2t

E r O)e-O.2t +0.05 1-e-0.2t)





ExamMFE

Question 9

Thee continuous-time models are being considered for modeling the short rate of interest:

Modell:

Model

2:

dr = a(b - r)dt + O dz

dr = a(b - r)dt + O JTdz

dr = a(b(t) - r)dt + O dz

odel

3:

The following observations have been made about these models:

Observation (1)

With Models 1 and 2 it is not possible to select values for the parameters that wil ensure

that the models exactly reproduce the current observed bond prices in all cases.

Observation (2)

With Models 1 and 3, for certain combinations of parameter values, it is possible for the

calculated value of a zero coupon bond that pays 100 on maturity to exceed 100.

Observation (3)

Because all three models are stochastic in nature, they can only be used to find the

distribution of bond prices, not a precise value.

Whch of these observations are correct?

A (2) only

B (3) only

C (1) and (2) only

D (2) and (3) only

E (1), (2) and (3)

Question 10

A lender is considering entering into an FRA at time 0, whereby the interest receivable on 10m

would be fixed from time 2 to time 3.

The n -year annual effective spot rates are equalto 0.06 + O.01n for n = 1,2,3 and the annualized

volatility of the two-year forward price of a one-year bond is 10%.

Rather than commit to the FRA, the lender decides to purchase a 2-year at-the-money call option

on it. Calculate the price of the call option.

A $Om

B $0.237m

C $0.422m

D $0.517m

E $0.603m


5



ExamMFE


Question 11

An interest rate caplet wil pay l,OOO,OOOmax(R2 -0.05,0) at the end of year 2, where R2

denotes the effective annual interest rate applicable to the period from time 1 year to time 2 years.

The price at time 0 of a l-year zero coupon bond is 0.95.

The log of the price at time 1 of a zero-coupon bond (with a nominal value of 1) maturing at

time 2 is normally distributed with variance ¿i = 0.0001. The price at time 0 of this bond is 0.90.

Calculate the value at time 0 of this caplet, to 2 signiicant figures.

A 2,500

B 4,600

C 5,000

D 5,800

E 6,700

Question 12

A 3-year interest rate cap wil pay 1,000,

000 max Rt -0.05,0) at the end of year t t = 1,2,3),

where Rt denotes the effective annual interest rate applicable to the period from time t -1 years

to time t years.

The price at time 0 of a l-year zero coupon bond is 0.95.

The log of the price at time 1 of a zero-coupon bond (with a nominal value of 1) maturing at time

2 is normally distributed with variance O'f = 0.0001. The price at time 0 of this bond is 0.90.

The log of the price at time 2 of a zero-coupon bond maturing at time 3 is normally distributed

with variance 0'1 = 0.0003. The price at time 0 of this bond is 0.85.

Calculate the value at time 0 of this interest rate cap, to the nearest 100.

A 13,000

B 15,000

C 18,600

D 20,700

E 21,300





ExamMFE


You are given the following current bond prices:

(i)

Bond maturity (years)

1 2

3

4

Zero-coupon bond price

0.9434

0.8817

0.8163

0.7488

(ii) The bond forward price is lognormally distributed with volatility 0 =0.05.

Question 13

Determie the one-year forward price for a zero-coupon bond maturing for 1,000 at the end of

the third year.

A $798

B $865

C $871

D $878

E $882

Question 14

Using the Black formula, determie the price of a one-year at-the-money European put option on

a zero-coupon bond maturing for 1,000 at the end of the thid year.

A Less than 2.30




E At least 2.60

Question 15

Consider a binomial interest rate tree where:

. the continuously-compounded interest rate at time 0 is 8% per annum

. at each step, the interest rate moves up or down by 2%

. the risk-neutral probability of an up move is 0.5.

Calculate the implied continuously-compounded forward rate applicable between times 2 and 3.

A 7.31%

B 7.59%

C 7.92%

D 8.00%

E 8.04%


7



ExamMFE



You are given:

i) The interest rates for the first two periods of a three-period binomial tree are:

ro = 6.000%

ru = 7.704%

rd = 4.673%

ii) All interest rates are effective annual rates of interest.

iii) The risk-neutral probabilty that the rate goes up from one period to the next is 0.50.

Question 16

Determine ruu + rud + rdd .

A 0.189

B 0.191

C 0.193

D 0.195

E 0.197

Question 17

You use the binomial interest rate model to evaluate a 7.0% interest rate cap on a 5,000 three-

year loan where the loan interest payments are made at the end of each year and the principal is

repaid at maturity.

Determine the value of this interest rate cap.

A At

most

45

B Greater than 45, but less than 65

C Greater than 65, but less than 85

D Greater than 85, but less than 105

E At least 105





ExamMFE


You are given the following current market data for zero-coupon risk-free bonds:

Maturity

Bond Price

Yield Volatility in 1 year

1 year

0.9524

n/a

2 years

0.8985

8

A 2 period Black Derman Toy model is constructed from this data so that the risk neutral

probability of rates going up is 0.50:

ro

ru

rd

Question 18

Determine ro.

A 0.049

B 0.050

C 0.051

D 0.052

E 0 053

Question 19

Determine rd.

A 0.055

B 0.057

C 0.060

D 0.063

E 0 065

Question

20

Determine ru.

A 0.055

B 0.057

C 0.060

D 0.063

E 0 065


9



ExamMFE



You are given the following current market data for zero-coupon risk-free bonds:

Maturity

Bond Price

Yield Volatilty in 1 year

1 year

0.9524

nja

2 years

0.8985

8%

A 2-period Black-Derman-Toy model is constructed from this data so that the risk neutral

probabilty of rates going up is 0.50:

ro

ru

rd

Question 21

Using the Black-Derman-Toy tree given above determine the price for an option to purchase in

one year for 940 a one-year, zero-coupon bond maturing for 1,000.

A 3.65

B 4.16

C 5.66

D 7.30

E 8.32





ExamMFE


The following current market data is available for zero-coupon bonds:

Matuity

Bond Price

Yield Volatility

(years)

( )

in 1 year

1

0.9346

nja

2

0.8495

10%

3

0.7722

12%

The final column shows the standard deviation of the natural log of the yield for the bond in one

year's time.

Question

22

A Black-Derman-Toy tree is to be calibrated to the above market data, using up and down

probabilties of Vi at each node.

Identif the higher interest rate in the tree at time 1, ie the rate applying from time 1 to time 2 at

the upper node.

A 6.998

B 9.027

C 9.976

D 10.002

E 11.026

Question 23

A Black-Derman- Toy tree is to be calibrated to the above market data, using up and down

probabilities of Vi at each node.

Identify both the highest interest rate in the tree at time 2 (ie applying from time 2 to time 3) and

the volatilty parameter at time 2.

A Highest interest rate = 14.237%, volatility parameter = 18.991 %

B Highest interest rate = 13.058%, volatility parameter = 14.029%

C Highest interest rate = 12.782%, volatilty parameter = 18.991 %

D Highest interest rate = 12.591 %, volatilty parameter = 12.000%

E Highest interest rate = 7.450 , volatilty parameter = 14.029


11



ExamMFE


The next three questions use some of the interest rates that are derived in the solutions to Question 22 and

Question 23 so you might like to check the answers to these against the solutions before proceeding.



Maturity

Bond Price

Yield Volatility

(years)

( )

in 1 year

1

0.9346

nja

2

0.8495

10%

3

0.7722

12%


year's time.

Question

24

Use a Black-Derman-Toy tree to calculate the value of an option to buy 100 nominal of a 2-year

zero-coupon bond at time 1 for a price of 82.

A 1.13

B 1.46

C 1.72

D 2.01

E 2.24

Question

25

Use a Black-Derman-Toy tree to calculate the value of an option to sell 100 nominal of a l-year

zero-coupon bond at time 2 for a price of 92.

A 0.70

B 0.92

C 1.04

D 1.16

E 1.38





ExamMFE



Matuity

Bond Price

Yield Volatilty

(years)

( )

in 1 year

1

0.9346

n/a

2

0.8495

10%

3

0.7722

12%


year's time.

Question

26

The price of an option to buy 100 nominal of a l-year zero-coupon bond at time 1 for a price of

91 is 0.34. Calculate the price of an equivalent put option.

A 0.24

B 0.39

C 0.44

D 0.47

E 0.51

Question

27



Bond maturity (years)

1 2

Zero-coupon bond price

0.9434

0.8817

A European call option that expires in 1 year gives you the right to purchase a l-year bond for

0.9259.

The bond forward price is lognormally distributed with volatilty (Y = 0.05 .

Using the Black formula, calculate the price of the call option.

A 0.011

B 0.014

C 0.017

D 0.020

E 0.022


13



ExamMFE


Question

28


You use a binomial interest rate model to evaluate a 7.5% interest rate cap on a 100 three-year

loan.

You are given:

(i) The interest rates for the binomial tree are as follows:

ro =6.000

ru =7.704

rd =4.673

ruu =9.892

rud = rdu = 6.000%

rdd =3.639

(ii) All interest rates are annual effective rates.

(iii) The risk-neutral probability that the annual effective interest rate moves up or down is 1/2.

(iv) The loan interest payments are made anually.

Using the binomial interest rate model, calculate the value of this interest rate cap.

A 0.57

B 0.96

C 1.45

D 1.98

E 2.18





ExamMFE

Question

29


Let P(r, t, T) denote the price at time t of 1 to be paid with certainty at time T, t:: T , if the

short rate at time t is equal to r.

For a Vasicek model you are given:

P(0.04, 0,2) = 0.9445

P(0.05, 1,3) = 0.9321

P(r*, 2,4) = 0.8960

Calculate r * .

A 0.04

B 0.05

C 0.06

D 0.07

E 0 08

Question 30


You are given the following three-period interest rate tree. Each period is one year. The risk-

neutral probability of each up-move is 70%. The interest rates are continuously-compounded

rates.

ruu = 18%

ru = 15%

~

~

o =12

~

~

~

rud = 12%

rdu = 12%

rd =9%

~

rdd = 6%

Consider a European put option that expires in 2 years, giving you the right to sell a one-year

zero-coupon bond for 0.90. This zero-coupon bond pays 1 at maturity.

Determine the price of the put option.

A 0.012

B 0.018

C 0.021

D 0.024

E 0 029


15



ExamMFE


Question 31


You are to use a Black-Derman- Toy model to determine Fo,2 (P(2,3) J, the forward price for time-

2 delivery of a zero-coupon bond that pays 1 at time 3.

In the Black-Derman-Toy model, each period is one year. The following effective anual interest

rates are given:

rd =30

ru =60

rdd = 20

ruu = 80

Determine 1000xFo,2 (P(2,3)J.

A 667

B 678

C 690

D 709

E 712





ExamMFE

Question 32


You are given:

(i) The true stochastic process of the short-rate is given by:

dr(t) = (0.008 - O.lr(t)) dt + 0.05dZ(t),

where Z(t) is a standard Brownian motion under the true probabilty measure.

(ii) The risk-neutral process of the short-rate is given by:

dr(t) = (0.013 - O.lr(t)) dt + 0.05dZ(t),

where Z(t) is a standard Brownian motion under the risk-neutral probabilty measure.

(iii) For t:: T , let P( r, t, T) be the price at time t of a zero-coupon bond that pays 1 at time

T, if the short-rate at time t is r. The price of each zero-coupon bond follows an Itô

process:

dP(r(t),t,T)

a(r(t), t, T) dt -q (r(t), t, T) dZ(t)

, t:: T

P r t), t, T)

Calculate a 0.04, 2, 5) .

A 0.041

B 0.045

C 0.053

D 0.055

E 0.060


17



ExamMFE


Question 33

SOA sample question

You are using the Vasicek one-factor interest-rate model with the short-rate process calibrated as

dr(t) = 0.6 (b - r(t) J dt + O dZ(t)

For t:: T ,let P(r, t, T) be the price at time t of a zero-coupon bond that pays $1 at time T, if the

short-rate at time t is r. The price of each zero-coupon bond in the Vasicek model follows an Itô

process

dP(r(t),t,TJ

J =a r t),t,TJdt-q r t),t,TJdZ t), t::T

r t),t,T

You are given that a 0.04, 0,2) = 0.04139761 .

Find a(0.05, 1,4) .

A 0.042

B 0.045

C 0.048

D 0.050

E 0.052

Question 34

SOA sample question

The Cox-Ingersoll-Ross CIR) interest-rate model has the short-rate process:

dr(t) = a (b - r(t) J dt + O ~r(t)dZ(t)

where tZ(t)J is a standard Brownian motion.

For t:: T ,let P(r, t, T) be the price at time t of a zero-coupon bond that pays $1 at time T, if the

short-rate at time t is r. The price of each zero-coupon bond in the CIR model follows an Itô

process:

dP(r(t),t,TJ

P r t),t,TJ a r t),t,TJdt-q r t),t,TJdZ t), t::T

You are given a 0.05,7,9) = 0.06.

Find a(0.04, 11, 13) .

A 0.042

B 0.045

C 0.048

D 0.050

E 0.052





ExamMFE

Question 35

SO sample question

You are given the following incomplete Black-Derman-Toy interest rate tree model for the

effective annual interest rates:

~

16.8%

~

17.2%

~

12.6%

--

9%

--

13.5%

~

-

9.3%

~

~

11%

--

--

Calculate the price of a year-4 caplet for the notional amount of 100. The cap rate is 10.5%.

A 0.97

B 1.09

C 1.21

D 1.33

E 1 45


19



ExamMFE


Question 36

SOA sample question

You are given:

i) The true stochastic process of the short-rate is given by:

dr(t) = (0.09 - O.5r(t)) dt + 0.3dZ(t)

where t Z( t)J is a standard Brownian motion under the true probability measure.

ii) The risk-neutral process of the short-rate is given by:

dr(t) = (0.15-0.5r(t)) dt+ O (r(t)) dZ(t)

where t Z( t) J is a standard Brownian motion under the risk-neutral probabilty measure.

iii) g r, t) denotes the price of an interest-rate derivative at time t, if the short rate at that

time is r. The interest-rate derivative does not pay any dividend or interest.

iv) g R t), t) satisfies:

dg (r(t), t) = lJ (r(t), g (r(t), t)) dt+ O.4g( r(t), t) dZ(t) .

Determine lJ(r,g).

A (r-0.09)g

B (r-0.08)g

C (r-O.03)g

D (r+0.08)g

E (r+0.09)g





ExamMFE

Question 37

SOA sample question

The following is a Black Derman Toy binomial tree for effective anual interest rates.

6

5

..

~

~

o

~

~

3

~

rud

2

Compute the volatility in year 1 of the 3-year zero-coupon bond generated by the tree.

A 14

B 18

C 22

D 26

E 30


21



ExamMFE


Question

38

SOA sample question

You are given the following market data for zero-coupon bonds with a maturity payoff of $100.

Maturity (years)

Bond price ( )

Volatilty in year 1

1

94.34

NjA

2

88.50

10%

A 2-period Black-Derman-Toy interest tree is calibrated using the data from above:

ro

~

ru

rd

Calculate rd' the effective annual rate in year 1 in the down state.

A 5.94

B 6.60

C 7.00

D 7.27

E 7.33

Question 39

SOA sample question

For t:: T ,let P(t, T, r) be the price at time t of a zero-coupon bond that pays $1 at time T, if the

short-rate at time t is r.

You are given:

(i) P(t,T,r)=

A(t,T)

exp

L-B(t,T)rJ for

some functions A(t,T) and B(t,T).

(ii) B(O,

3) = 2.

Based on P(O,

3, 0.05), you use the delta-gamma approximation to estimate P(O,

3, 0.03), and

denote the value as PEst (0,3,0.03) .

Find PEst (0,3, 0.03)

P(O, 3,0.05)

A 1.0240

B 1.0408

C 1.0416

D 1.0480

E 1.0560





ExamMFE

ExamMFE


Chapter

9

Some of the questions below have been taken from past SOA exam questions in other subjects.

In some cases these have been modified to be consistent with the current MFE exam syllabus.

Question 1

SOA Exam C May 2007

You are given:

i) A computer program simulates n = 1000 pseudo- U O,

1) variates.

ii) The variates are grouped into k = 20 ranges of equal length.

20

iii) The sum of squares of the observed numbers is ¿ Of = 51,850 .

j=l

iv) The chi-square goodness-of-fit test for U O,l) is performed.

Determine the result of the test.

0 E)2

Note that the chi-squared test statistic is calculated as L ~ ' where 0 is the observed frequency

and E is the expected frequency, and that Xf9 (1 ) = 36.19.

A Do not reject Ho at the 0.10 signicance leveL.

B Reject Ho at the 0.10 signicance level, but not at the 0.05 significance leveL.

C Reject Ho at the 0.05 signicance level, but not at the 0.025 significance leveL.

D Reject Ho at the 0.025 significance level, but not at the 0.01 significance leveL.

E Reject Ho at the 0.01 significance leveL.


1



ExamMFE


Question

2

Suppose that X follows a two-parameter Pareto distribution with distribution function:

F X)=l- -. ìa

B+x)

where a = 2.5 and B = 50 . The inversion method is to be used to simulate values of X .

Which of the following is the correct formula expressing the simulated value of X in terms of the

random number u taken from a U O,l) distribution?

A x=50(1-(1-u)2.5)

B X=50((1-ur-o.40 -1)

C X=50 1- 1- u )0.40)

D X=50((1+ufO.40 +1)

E X=50 1+u)2.S -1)

Question 3

You are given the following random sample of values from a Uniform O,l) distribution:

0.64

0.38

0.83

0.84

0.08

0.39

0.93

0.92

0.28

0.07

Use these values to obtain 10 simulated values from a binomial distribution with parameters

n = 2 and p = 0.75. Determie the mean of your simulated random sample.

A 0.5

B 1.2

C 1.4

D 1.5

E 1.8





ExamMFE

Question

4

You are given the following random sample of values from a Uniform O,l) distribution:

0.64

0.38

0.83

0.84

0.08

0.39

0.93

0.92

0.28

0.07

Use these values to obtain a random sample of size 10 from a Pareto distribution with parameters

e = 400 and a = 6. Determine the mean of your simulated random sample rounded to the

nearest 10.

A 70

B 80

C 90

D 100

E 110

Question 5

You are given the following random sample of values from a Uniform

(0,1 ) distribution:

0.64

0.38

0.83

0.84

0.08

0.39

0.93

0.92

0.28

0.07

Use these values to produce 10 simulated values from an inverse exponential distribution with

distribution function F x) = e-B / x, where e = 100. Determie an estimate of the median of this

distribution.

A Less than 50

B Between 50 and 100

C Between 100 and 250

D Between 250 and 500

E Greater than 500

Question 6

SOA Exam C May 2005

A company insures 100 people age 65. The annual probabilty of death for each person is 0.03.

The deaths are independent.

Use the inversion method to simulate the number of deaths in a year. Do this three times using

the following random numbers from the uniform distribution on 0,1):

ui = 0.20 u2 = 0.03 u3 = 0.09

Calculate the average of the simulated values.

A 1/3

B 1

C 5/3

D 7/3

E 3


3



ExamMFE


Question 7

Insurance for a city s snow removal costs covers four winter months. The insurer assumes that

the city's monthly costs are independent and normally distributed with mean 15,000 and

standard deviation 2,000.

To simulate four months of claim costs, the insurer uses the inversion method (where small

random numbers correspond to low costs) to simulate the city s monthly cost.

The four numbers drawn from the uniform distribution on (0,1) are:

0.5398

0.1151

0.0013

0.7881

Calculate the insurer's total simulated claim cost.

A 53,000

B 53,200

C 53,400

D 53,600

E 53,800

Question 8

You want to use simulation to estimate the mean of a distribution. You have generated the

following values from the distribution:

102

113

145

120

137 150

144

156

Estimate the additional number of simulations that you would need to perform to ensure that the

standard error of the estimator X is at most 3.

A 19

B 33

C 34

D 41

E 42





ExamMFE

Question 9

The sample mean xn = xi +... + xn ) / n for a sample of size n is used to estimate E X). For this

n

sample the value of s2 is computed as L (Xi - xn)2 / (n -1) .

i=i

It is desired that the estimate is within 2 of the true mean with 95 confidence.

The sample of size n is simulated by the inversion method. For which of the following situations

have a sufficient number of simulations been performed?

1.

2.

3.

2 .

n = 490, xn =50, s =125

n = 500 , xn = 50 , s2 = 135

n =500, xn =48, s2 =125

A

B

C

D

E

1

only

2

only

3 only

1 and 3 only

2 and 3 only

Question 10

SOA Exam C November 2006

You are plannng a simulation to estimate the mean of a non-negative random variable. It is

known that the population standard deviation is 20 larger than the population mean.

Use the central limit theorem to estimate the smallest number of trials needed so that you wil be

at least 95 confident that the simulated mean is within 5 of the population mean.

A 944

B 1299

C 1559

D 1844

E 2213

Question 11

SOA Exam C May 2007

Simulation is used to estimate the value of the cumulative distribution function at 300 of the

exponential distribution with mean 100.

Determine the minimum number of simulations so that there is at least a 99 probabilty that the

estimate is within i: of the correct value.

A 35

B 100

C 1418

D 2013

E 3478


5



ExamMFE


Question 12

The 70th percentile of the lognormal distribution with parameters ¡. = 6 and 52 = 0.5 is:

A 585

B 590

C 595

D 600

E 605

Question 13

A lognormal distribution has 10th percentile equal to 100 and 90th percentile equal to 2,500. Find

the 50th percentile.

A 500

B 700

C 900

D 1,100

E 1,300

Question 14

You are given that St follows the lognormal stock price model, and:

(i) The current price is So = 55 .

(ii) The expected price in 6 months is E( 50.5)=58.

(iii) The stock pays no dividends in the next 6 months.

Determine the continuously-compounded expected annual rate of return earned by this stock.

A 5.3

B 7.7

C 9.4

D 10.6

E 11.2





ExamMFE

Question 15


(i) Pr(SO.5 ::74) =0.8413

(ii) Pr (50.5 :: 56) = 0.1587

Determine the volatility of the stock.

A 0.18

B 0.20

C 0.22

D 0.24

E 0.26

Question 16


(i) E(S2) = 110

(ii) var ( 52 ) = 2,420

Determine the volatility of the stock.

A 0.18

B 0.22

C 0.26

D 0.30

E 0.34

Question 17

The price of a stock follows the risk-neutral lognormal price modeL. You are given:

(i) The continuously-compounded risk free rate is r=0.05 .

(ii) There are no dividends paid by this stock.

(iii) The annual volatilty of this stock is 0 =0.30.

(iv) The risk-neutral stock price is simulated by the inversion method.

(v) The current stock price is 70.

Determine the average simulated price in 6 months using the following three random numbers

from the uniorm distribution on (0,1):

ui = 0.6293, u2 = 0.1611, u3 = 0.5000

A 67.44

B 68.12

C 69.01

D 70.84

E 71.55


7



ExamMFE


Question 18

Consider the following statements about the calculation of option prices using the risk-neutral

lognormal stock price modeL.

I A put option that has a given strike price and expiration time wil always be more

valuable than the corresponding call option

II The option prices for calls and puts wil always be an increasing function of time to

expiration.

II The Black-Scholes formula can

in some cases give a negative theoretical option price.

Whch of these statements are true?

A II only

B II only

C I and II only

D II and II only

E None of them

Question 19

SOA sample question

The price of a stock in seven consecutive months is:

Month

Price

1

54

2

56

3

48

4

55

5

60

6

58

7

62

Calculate the annualized expected return of the stock

A Less than 0.28




E At least 0 31





ExamMFE

Question

20

SOA sample question

The price of a non-dividend-paying stock is to be estimated using simulation. It is known that:

(i) The time- t stock price, St, follows the lognormal distribution:

In ~ ~ N( (a- 0-2 )t, 0-2t)

So

(ii) So = 50, a = 0.15, and 0- = 0.30.

The following are three Uniform(O,l) random numbers

0.9830 0.0384 0.7794

Use each of these three numbers to simulate a time-2 stock price.

Calculate the mean of the three simulated prices.

A Less than 75




E At least 115

Question 21

You are given the following inormation for a stock with current price 0.25:

(i) The price of the stock is lognormally distributed with continuously-compounded

expected annual rate of return a = 0.15.

(ii) The dividend yield of the stock is zero.

(iii) The annual volatilty of the stock is 0- = 0.35

Determine the upper bound of the two-tailed 90 confidence interval for the price of the stock in

6

months.

A 0.393

B 0.425

C 0.451

D 0.486

E 0.529


9



ExamMFE


Question 22

For a European call option on a share of stock following the lognormal stock price model, you are

given:

i) The current price is 60.

ii) The strike price is 62.

iii) There are no dividends.

iv) The time until expiration of the option is 0.5 years.

(v) 62 = 0.16 where 6 is the annual volatility of the stock.

(vi) The continuously-compounded expected annual rate of return earned by this stock is

a=0.09.

Determine the probability that the option expires in the money.

A 0.46

B 0.48

C 0.50

D 0.52

E 0.54

Question 23

For a European call option on a share of stock following the lognormal stock price model, you are

given:

i) The current price is 60.

ii) The strike price is 62.

iii) There are no dividends.

iv) The time until expiration of the option is 0.5 years.

(v) 62 = 0.16 where 6 is the annual volatility of the stock.

vi) Under the risk-neutral probabilty measure, the continuously-compounded expected

annual rate of return earned by this stock is r = 0 09

Determine the price of the call option

A 1.8

B 2.6

C 4.7

D 6.1

E 7.1





ExamMFE

Question

24

SOA sample question

Michael uses the following method to simulate 8 standard normal random variates:

Step 1: Simulate 8 Uniform O,l) random numbers Ui, U2,..., Us.

Step 2: Apply the stratified sampling method to the random numbers so that Ui and Ui+4 are

transformed to random numbers Vi and Vi+4 that are uniformly distributed over the

interval i -1)/4, i/4)

, i = 1,2,3,4. In each of the four quartiles, a smaller value of U

results in a smaller value of V .

Step 3: Compute 8 standard normal random variates by Zi = N-i (Vi) ,where N-i is the inverse

of the cumulative standard normal distribution function.

Michael draws the following 8 Uniform(O,l) random numbers:

i

1

2 3

4 5

6 7

8

U.

0.4880 0.7894 0.8628 0.4482 0.3172 0.8944 0.5013 0.3015

Find the difference between the largest and the smallest simulated normal random variates.

A 0.35

B 0.78

C 1.30

D 1.77

E 2.50


11



Question & Answer Bank - Solutions, Chapter 0

ExamMFE

Exam MFE Question & Answer Bank

Chapter 0 Solutions

Solution 1

Answer: C

Difficulty level: ..

The total payoff from this option portfolio is given by:

max S -70, 0) - 2x max S - 85, 0) - 2x max 75 - 5, 0)

Considering different ranges for the stock price in one year gives:

5::70: -2x 75-S) = 25 -150

maximum payoff = -10

70~S ::75:

S-70-2x 75-S) = 35 -220

maximum payoff = 5

75~S::85:

5-70

maximum payoff = 15

5:;85:

S-70-2x S -85) = 100-5

maximum payoff = 15

The maximum payoff is therefore 15, which occurs at a stock price of 85.

Payoff function

20

15

10

5

o

-5

-10

-15

-20

-25

-30

-35

Solution 2

Answer: E

Difficulty level: .

The pre-paid forward rate is found by subtracting the present value of the dividend payments, discounted

at the risk-free interest rate, from the current share price.

The pre-paid forward price is:

Pre-paid forward price = 5000 -100e -D.05 2 / 12)-0.06 i / i2) _ 100e -D.05 2 / 12)-0.06 4/ i2)

= 5000 - 98.6755 - 97.2064

= 4,804.12


1



ExamMFE

Question Answer Bank - Solutions, Chapter 0

Solution 3

Answer: C

Difficulty level:

The 6-month and l-year forward prices for Stock X are related to its current spot price Xo by:

FO~O 5 = XoerxO 5 = 673 15

and Fo~i = Xoerxi = 697.13

where r is the continuously-compounded annual risk-free rate of interest.

Taking logs of both equations gives:

In Xo + 0.5r = In( 673.15) = 6.5120

In Xo + r = In 697.13) = 6.5470

Solving these simultaneous equations in Xo and r gives:

r = 0.0700 and Xo = 650.00

The 2-year forward price for Stock Y is then given by:

Frl2 = Yoe(r-O)x2 = 650.00xe(O.07-0.03)x2 = 704.14

2





ExamMFE

Exam MFE Question Answer Bank

Chapter 1 Solutions

Solution 1

Answer:

A


There is a shortcut for answering this question. An ask price tells us the price at which we can buy the call

option. So any arbitrage must involve buying the option. The arbitrageur therefore wants the ask price to

be as low as possible. If the ask price is low enough, arbitrage is available. Therefore, based on the

structure of this question, it is clear that the lowest of the choices will be the correct answer. That is, if

arbitrage were available when buying at a price of 16.10, then arbitrage would also be available when

buying at a price of16.00.

The ask price for the call option is the price that an investor must pay to purchase the call option.

If arbitrage is possible, then the investor can purchase the call option for the ask price as part of a

position that requires no net investment from the investor.

Put-call parity tells us that purchasing a European call option produces the same cash flows as

buying a share of stock, buying a European put option, and borrowing cash.

CEur(K,T) = So + PEur(K, T)_Ke-rT

The cash flows produced by the call option can be acquired by purchasing the call option, and the

same set of cash flows can be sold by selling a share of stock, sellng a put option, and lending

cash. If the price of acquiring the cash flows is less than the price received from sellng the cash

flows, then arbitrage is available:

cask (K T) ~ Sbid + pbid (K T) _ Ke-riT

Eur 0 Eur

~ arbitrage

Ths can be rewritten as:

CÊ~r(72,l) ~ 79.90+3.95-72e-0.06

CÊ~r 72,1) ~ 16.043

Solution 2

Answer:C

Difficulty level: .

According to put-call parity, we have:

C - P = Fo~ (stock X) - K e-rt ~ e-rt =0. 94 ~ l,OOOe-rt = 940

'- '- '- '-

4 98 iOOe-rt


1



ExamMFE


Solution 3

Answer: A

Difficulty level: ....

Let's begi by defining some of the variables:

xo = 1.25 (current underlying asset price for dollar-denominated options)

K = 1.10 (current strike price for dollar-denominated options)

1

-=0.80

xo

. = 0.90909

K

ri =0.07

r =0.08

(current underlying asset price for franc-denominated options)

current strike price for franc-denominated options)

interest rate earned on francs)

(inerest rate earned on dollars)

From put-call parity, we can observe that the dollar-denominated call has more value than the

dollar-denominated put:

C (K T) n (K T) - -r¡T K -rT

Eur , - rEur , - xoe - e

CEur (1.10,1)- PEur (1.10,1) = 1.25e-o.07 _1.10e-O.08

CEur(1.10,l) - PEur (1.10,1) = 0.15006

CEur 1.10,1) = 0.15006+ PEur 1.10,1)

=? CEur(1.10,l):; PEur(1.10,l) and CEur(1.10,l):; 0.15006

So the value of the call described in Choice A is 0.15006 dollars greater than the put described in

Choice B. This rules out Choices Band E.

Furthermore, the value of 0.09 francs is:

0.09x1.25 = 0.1125 dollars

Since 0.1125 dollars is less than 0.15006 dollars, we can also rule out Choice D.

The value of the franc-denominated put is Pi 0.8,0.90909,1) francs, which is equivalent to:

1.25x Pi (0.8,0.90909,1) dollars

The following formula tells us that the dollar-denominated call option is equal to 1.1 of the franc-

denominated put options, so the dollar denominated call option is more valuable than the franc-

denominated put option:

C$ (xo, K, T) = xoKPi (~,. , T)

Xo K

C$ (1.25,1.1,1) = 1.25xl.lx Pi (0.8,0.90909,1)

C$ (1.25,1.1,1) = 1.1XL 1.25x Pi (0.8,O.90909,l)J

C (1.25,1.1,1) :;1.25x Pi (0.8,0.90909,1)

Therefore we can also rule out Choice C, and Choice A must be the correct answer.

2

§ BPP Professional Education




ExamMFE

Here's another way to rule out Choice C. The payoff of the put described in Choice Cis:

Payoff from put described in Choice C = Max

( 0.90909- x~ ,0 )

francs

= Max(0.90909xi -1,0) dollars

Max(xi -1.10,0) d 11

= 0 ars

.10

Payoff from call described in Choice A

1.10

Therefore the value of the call described in Choice A must be greater than the value of the put described in

Choice C.

Solution 4

Answer: D

Difficulty level: .

The difference in the cost of the options can be substituted into the formula for put-call parity:

CEur (K, T) - PEur(K, T) = So - Ke-rT

2.20 = So - Ke -(0.09)(0.5)

For an at-the-money option, K = So ' so:

2.20 = So - SOe -(0.09)(0.5)

2.20 = So ( 1- e -(0.09)(0.5) )

So =50.00

Solution 5

Answer: E

Difficulty level: .

The creation of a synthetic T-bil maturing for K by buying the stock, buying a put, and selling a

call is called a conversion. Ths strategy can be ilustrated using put-call parity:

Ke-rT =SO+PEur(K,T)-CEur(K,T)

This equation allows us to find the present value of 200:

200e-rx0.5 = 190+ 16.18-15.94

200e-rx0.5 = 190.24

The present value of 1,000 is 5 times the present value of 200:

200e-rx0.5 = 190.24

l,OOOe-rxO.5 = 5x190.24

1, OOOe -rxO.5 = 951.20


3



ExamMFE


Solution 6

Answer: D

Difficulty level: .

According to put-call parity we have:

C - e = So - PV (Dividends) -

2.95 52 -0015 -0030

e + e .

K e-rt

'-

50 e -0.06x7 /12

~ C = 4.71

Solution 7

Answer:

A


From put-call parity the prices for corresponding 1-year European call options are:

C = P +50 e-0.08 - K e-o.04

K

C

maxtO , S-KJ

40

9.11

10.00

50

4.91

0

60

0.71

0

70

0.00

0

It is optimal to immediately exercise only when K = 40 .

Solution 8

Answer: D

Difficulty level: .

Since the options are at-the-money we have S=K . So from put-call parity, we have:

C - P = Se-ôt - Ke-rt = K( e-ôt _ e-rt) ~

3=K(e-0.04 _e-o.0798) ~ K=80

Solution 9

Answer: B

Difficulty level: ...

The call and the put are denominated in yen, so we wil use yen as the base currency. The

underlying asset is one Bolivar. Put-call parity allows us to solve for K:

( )_n ( )_ -rfT_K-rT

Eur K,T rEur K, T - xoe e

0.004603 - 0.004793 = 0.05e -0.13(0.75) - Ke -0.07(0.75)

K=0.048

4





ExamMFE

Solution 10

Answer: A


Let C denote the call price for 1 euro with strike 0.90 and let P be denote the corresponding put

price.

According to put-call parity we have:

-rit -rt

C - P = Xo e - K e ~

0.0606 - P = 0.92e-O.032 - 0.90e-O.06 ~

1000P = 17.16

Solution 11

Answer: B


We must use the interest rate provided to calculate the value of the bond. The first coupon

payment is made in 3 months, and the subsequent payments are made every 6 months thereafter:

Bo = eO.08(0.25) ( 50a2õ1408i + 1, OOOe -O.08(iO) )

= 1.0202(50(13.4933)+ 449.3290)

= 1,146.6989

One coupon payment is made during the next 6 months, and its present value is:

PVO,O.5 Coupons) = 50e-O.08 0.25) = 49.0099

Substituting into put-call parity:

CEur K, T) - PEur K, T) = Bo - PVO,T Coupons) - Ke-rT

55.05- PEur (1100,0.5) = 1,146.6989 - 49.0099-1,100e-O.08(0.5)

PEur (K, T) = 14.23

Solution 12

Answer: B


The first trick is to notice that the second call is the same as a put on Stock A.

The second trick is to realize that we do not need to know the length of time until the options expire.

We do, however, need to notice that Stock B pays a dividend before the options mature.

The second call is a call on Stock B, but it is also a put on Stock A:

CEur(BO,Ao,T)=PEur(AO,BO,T) ~ PEur(AO,BO,T)=5

The prepaid forward price of stock B is:

F6,T(B) = So - PVo,T(Div) = 12 - 2e-O.07(0.5) = 10.069


5



ExamMFE


The general form of put-call parity can now be used to solve for the prepaid forward price of

Stock

A:

CEur (Ao,Bo, T) - PEur(AO,BO' T) = F6,T(A) - F6,T(B)

3-5 = F6,T A)-10.069

F6,T(A) = 8.069

Since Stock A does not pay dividends, its price is equal to its prepaid forward price:

F6,T(A) = Ao

~ Ao =8.069

Solution 13

Answer: C

Difficulty level: .

The option in (i) is a European call with strike asset stock WX. We are asked about the

corresponding put option. From put call parity we have:

C - P = F6i (stock YZ) - F6i (stock WX) ~ P = 11.90

- - -

i5 _ 3e-o.065x0.5 20

Solution 14

Answer: C

Difficulty level: .

The minimum price of a European call option is described by:

Max O,PVO,T FO,T)-PVO,T K)) :: CEur S,K,T)

The present value of the forward price is the same thing as the prepaid forward price. Since the

option expires in 9 months, we can ignore the second dividend:

PVO,0.75 FO,O.75) = F6,0.75

(X) = 68_5e-o.075(0.5) = 63.184

This value can be substituted into the expression for the minimum value of a European call

option:

Max O,63.184-58e-D.075 .75)) :: CEur 68,58,O.75)

8.356:: CEur 68,58,O.75)

Solution 15

Answer: E


Since the call is deep in the money, the corresponding put is way out of the money and therefore

nearly worthless (i.e. P is a small positive number). From put-call parity we have:

C = P + 50-

~ -

: ° 72

tV ( dividends ~

2( e-0.075 / 4 + e -0.075/2)

K e-rt

'-

8 e -0.075(8/12)

:; 12.93

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ExamMFE

Solution 16

Answer: E


The trick is to recognize that the call option is priced too low. Since it is priced too low, the arbitrageur

buys the call option.

To avoid arbitrage, the price of the call option must satisfy:

Max O, PVo,i Fo,i)- PVo,i K)) :: CEur 41,40,l)

Max(O,41-40e-O.05) :: CEur(41,40,l)

2.95:: CEur(41,40,l)

Since the call option has a premium of $2.50, its price is too low and arbitrage is possible.

The arbitrageur buys the call option, sells the stock and invests the present value of the strike

price. This produces a time 0 cash flow of:

-2.50+41-40e-O.05 =0.451

How does the arbitrageur know what to do? By looking at the inequality. It could be restated as the

position that consists of buying the stock and borrowing the present value of $40 should cost less than the

strategy of buying a call option. Since it costs more, do the opposite - sell the stock and lend the present

value of 40. And since the price of the call option is relatively low, buy it.

At the end of 1 year, the stock price is $39, so the call option expires worthless and the payoff

from the whole portfolio is $1:

0-39+40 =1

Accumulating the time 0 cash flow forward at 5 % and adding it to the $1 cash flow at time 1, the

arbitrageur's accumulated profits are:

0.451eo.05 + 1.000 = 1.474

Solution 17

Answer: D


According to Theorem 1.3, arbitrage is possible if:

P K2)-P Ki) P K3)-P K2)

:;

K2 -Ki K3 -K2

Therefore arbitrage is possible since:

13-5

110-100

20-13

:;

120-110

=: 0.8:;0.7

The first step to setting up the arbitrage strategy is to determine Â:

Â = K3 -K2

K3-Ki

120 -110 = 0.5

120-100

The arbitrage calls for selling the 110-strike put. Since the arbitrageur sells 2 of the 110-strike

puts, the arbitrageur purchases 1 of each of the other puts:

2Â = 2(0.5) = 1

2 1- Â) = 2 1- 0.5) = 1


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ExamMFE


At time 0, the arbitrageur receives a net cash flow of 1:

-5+ 2(13)-20 = 1

At time 1, when the underlying asset has a price of 112, the 100-strike and the 110-strike puts

expire worthless. The 120-strike put has a payoff of 8:

120-112 =8

Accumulating the time 0 cash flow forward at 11 % and adding it to the $8 cash flow at time 1, the

arbitrageur's accumulated profits are:

leo.1l+8.000 = 9.116

Solution 18

Answer: A


Assertion (1) is not necessarily true. If the underlying stock is not expected to pay a dividend

during the life of the option, the price of an American call option wil be exactly equal to the price

of an otherwise identical European option.

Assertion (2) is true - the price of an American option must always be greater than or equal to its

intrinsic value. If the option is currently in the money, the holder can choose to exercise and wil

receive the intrinsic value. If it is out of the money, the holder can (and wil ) choose to do

nothing, and wil receive no payment at that time. Either way, the payment is at least the

intrinsic value.

Remember that the intrinsic value is equal to the payoff function evaluated based on the current value of

the underlying asset, ie it is max(St -K,O) for a call option and max(K-St,O) for a put

option.

Assertion (3) is not always true. It is possible for the price of a European option to be less than its

intrinsic value. In other words, a European option can have a negative time value. Ths can

happen, for example, with a European call option with a long life on a stock that has a high

dividend yield. It is possible that the dividends the option holder is "missing out" on before the

maturity date have a sufficiently high value that they exceed any potential increase in the share

price over the remaining life of the option and the interest saved on the strike payment. The

option holder would like to exercise straight away and receive all the future dividends, but is not

allowed to because the option is European.

Solution 19

Answer: C


P?Q?

A 2-year American option must be worth at least as much as an otherwise identical 1-year option.

If we held a 2-year option, we could decide to exercise it by the end of the first year anyway,

which would make it equivalent to having a 1-year American option. Restricting our choices in

this way cannot increase the value. So P? Q .

The general conclusion is that American options (calls or puts) with longer lives are worth at least as much

as their shorter counterparts.

8





ExamMFE

R? S?

The same is not necessarily true of European options. If the underlying currency pays a very

high interest rate, it is possible that the interest lost while waiting til the end of the second year

with the 2-year option can reduce the value of the option to below that of the 1-year option.

In normal circumstances, the inequality R? 5 would usually hold, but it is not guaranteed.

If however, the underlying asset does not generate any cash

flows (as for a non-dividend-paying stock), it

would be true that R? 5, because the European options would have the same values as their American

counterparts so this would be the same inequality as P? Q.

P?R, Q?S?

An American option must be worth at least as much as an otherwise identical European option,

since a holder of an American option could simply choose to ignore the early exercise feature. So

P ? Rand Q? 5 .

R?Q?

There is no simple connection between the value of a 1-year American option (which has an early

exercise feature) and a 2-year European option (which has a longer life). So we cannot conclude

that R? Q.

The set of inequaliies P Q R 5 ? 0 would also hold since the owner of an option could simply pretend

that they didn't own the option. They would then have no obligation to make any payments, which would

make the option's value equal to zero. However, the inequalities in C are stronger, which overrides

answer A.

Solution 20

Answer:D Difficulty level: .

For put options with strike prices satisfying Ki ~ K2 ~ K3, the convexity inequality states that:

P K2)-P Ki) ~ P K3)-P K2)

K2 -Ki K3-K2

Here we have:

Ki = 5500, K2 = 5800 and K3 = 6000

So the convexity inequality tells us that:

P(5800) - 50 ~ 170 - P(5800)

5800 - 5500 6000 - 5800

P(5800)-50 ~ 170-P(5800)

300 200

:

Ç: 200 (P(5800) - 50) :: 300 (170 - P(5800))

500P 5800):: 61,000

P 5800):: 61,000 = 122

500

Ç:

Ç:

It may be possible to establish other upper and lower bounds for the put with strike 5800, but these do not

follow from the convexity inequality.


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ExamMFE


Solution 21

Answer: C

For the following, assume that Ki ~ K2 ~ K3.


(i) This statement is true since 0 ~ C (5, Ki, T, 5, r) - C (5, K2 , T, 5, r) ~ K2 - Ki.

(ii) Ths statement is tre since 0 ~ P( 5, K2, T, 5, r) - P( 5, Ki, T, 5, r) ~ K2 - Ki.

(iii) This statement is false due to put-call parity:

C - P = 5 e -õt _ K e -rT ~ òC _ òP = _ e -rT ~ 0

òK òK

(iv) This statement is true due to the following:

C (5, Ki, T, 5, r) - C (5, K2 , T, 5, r)

?

K2 -Ki

C (5, K2, T, 5, r) - C (5, K3, T, 5, r)

K3 -K2

C (5, K2, T, 5, r) - C (5, Ki, T, 5, r) C (5, K3 T, 5, r) - C (5, K2 , T, 5, r)

~ ::

-~ ~-~

chord slope is increasing so the graph is. concave up (positive second derivative)

(v) This statement is true due to the following:

P( 5 ,K2 ,T, 5, r) - P( 5 ,Ki ,T, 5, r) p( 5 ,K3, T, 5, r) - P( 5 ,K2 ,T, 5, r)

~

~-~ - ~-~

chord slope is increasing so the graph is. concave up (positive second derivative)

Solution 22

Answer: D


Here is the payoff function formula for a position consisting of one 40-strike call, a 48-strike

calls, and b 60-strike calls:

payoff =

o

5(1)-40

5(1)-40 + a(S(1)-48)

5(1)-40 + a(S(1)-48) + b(S(1)-60)

o

5(1)-40

1 + a)S l) -40 -48a

(1 + a + b )5(1) -40 -48a - 60b

if 5(1):: 40

if 40 ~ 5(1) ~ 48

if 48 ~ 5 (1) ~ 60

if 60 ~ 5(1)

if 5(1):: 40

if 40 ~ 5 (1) ~ 48

if 48 ~ 5 ( 1) ~ 60

if 60 ~ 5(1)

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ExamMFE

Now choose a such that the payoff is zero for 5 1) = 60 :

~-40 + a ~-48J = 0 ~ a = -20/12

60 60

Now choose b=8/12 (i.e. l+a+b=O) so that the payoff does not depend on 5(1) for

5 1):; 60.

The graph of this piecewise linear payoff function versus 5 (1) looks like:

Payoff

S(l)

40 48 60

Since a =- 20 / 12 and b = 8/12 with a position consisting of one call with K = 40 , a proportional

payoff graph would result if we construct a position consistig of:

· Buy 12 calls with strike Ki = 40, buy 8 calls with strike K3 = 60,

· and sell 20 calls with strike K2 = 48 .

Solution 23

Answer: E Difficulty level: ..

The cost of establishing the position is:

12 C 40) - 20 C 48) + 8 C 60)

We thus are looking for an answer choice where this cost is less than or equal to zero. Working by

elimination we see that answer choice E is correct. Remember from your reading that portfolios

of this type can result in an arbitrage if the call prices do not satisfy the convexity relationship:

C Ki) - C K2) C K2) - C K3)

:; ~ C K) is decreasing and concave up

K2 -Ki K3-K2

-chord slope -chord slope

Solution 24

Answer: C

Difficulty level: .

If 5 denotes the stock price on the expiration date, the payoff function for this portfolio is:

f(S) = 40005 +3000max(S -10,O)-2000max(S -15,O)+6000max(15-S,O)

To find the mimum value, we need to evaluate this function at the extreme edges,S = 0 and

5 = 00 , and at the two strike prices, 5 = 10 and 5 = 15 .


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ExamMFE


This is because the graph of the payoff function for a portfolio of vanila options consists of straight-line

segments joining these points.

f(O) = 0+0-0+ 6000x15 = $90,000

f(10) = 4000xl0+0-0+6000x5 = $70,000

f(15) = 4000x15+3000x5-0+0 = $75,000

f(oo) = lim 140005+3000(5-10)-2000(5 -15)+0) = lim 50005 = 00

s~oo s~oo

We have to calculate the last one as a limit to avoid getting indeterminate expressions of the form 00 - 00 .

So the minimum value is $70,000, corresponding to a stock price on December 31 of $10.

Solution 25

Answer: C


The graph of the payoff function for a portfolio containing vanila options on the same underlying asset

and/or the underlying asset is always continuous and piecewise linear - that is, it consists of straight-

line sections joined together. In each of the cases (A) to (D), the investor's portfolio wil consist of a

reduced holding of stock and a long position in put options with a strike price K (say). So it is suffcient to

check whether the portfolio value exceeds the accumulated cash value at the three critical values of the final

share price: 0, K and infinity.

The current stock price is $100,000 = $5.00. So Portfolio A consists of 24,000 put options (strike

20,000

5.00), which cost 24,OOOx 0.25 = 6,000, and a reduced holding of 20,000- 6~~~0 = 18,800

shares.

We can carry out similar calculations for Portfolios B, C and D.

Let 5 denote the stock price in 6 months' time. As 5 -- 00, the put options in each of the

portfolios wil have no value and the long position in the stock wil make the portfolio value tend

to infinity as well. To ensure that the investor meets her objective, we need to check that, when

5 = 0 and 5 = K , the portfolios also exceed the accumulated cash value of

100,OOOeO.02 = 102,020. The results of these calculations are shown in the table below.

Number

of

Number of put

Payoff when 5 = 0

Payoff when 5 = K

shares held

options held

Portfolio A

18,800

24,000

$120,000./

$94,000 x

(K =5.00)

(price 0.25)

Portfolio B

19,000

25,000

$131,250./

$99,750 x

(K =5.25)

(price 0.20)

Portfolio C

19,400

18,750

$103,125./

$106,700./

(K =5.50)

(price 0.16)

Portfolio D

19,750

15,625

$89,844 x

$113,562./

(K =5.75)

(price 0.08)

Portfolio C is the only one that satisfies both constraints.

In each case the investor's portfolio is at least partly dependent on the share price, and so constitutes a risky

portfolio. It is only possible to exceed the risk-free rate of return with Portfolio C because the market is not

arbitrage-free. The put options in C (and possibly some of the others too) are underpriced .

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ExamMFE

Solution 26

Answer: E


The option we are pricing is an option to sell dollars (equivalent to buying yen). So we need to

apply the put-call parity relationship here.

From (ii) we know that an option to sell ¥1 for 0.008 (a put option on the yen) costs 0.0005.

The put-call parity relationship for currency options is:

-riT -rT

C-P=xOe -Ke

Here, C and P are the prices of call and put options (to buy and sell ¥1, respectively), xo is the current

price of¥l in dollars and rf is the foreign (Japanese) risk-fee interest rate.

So an option to buy ¥1 for 0.008 (a call option on the yen) costs:

-riT T

C = P+xOe -Ke-r

= 0.0005+ 0.011e-0.0i5(4) -0.008e-0.03(4)

= 0.003764

We can scale this up so that the 0.008 becomes 1 by dividing through by 0.008. This tells us

that an option to buy ¥125 (=~) for 1 costs 0.4705 ( 0.003764). This price expressed in

0.008 0.008

yen is:

$0.4705 = ¥ 0.4705 = ¥43 (rounded to the nearest yen)

0.011

Solution 27

Answer: E


The expression in the middle of the inequality in (I) corresponds to the value of a portfolio

consisting of a long position in a 50-strike call and a short position in a 55-strike call. The payoff

at time T from this portfolio is max (ST - 50,0)- max (ST - 55,0).

Consider the value of this payoff for different ranges of ST:

· ST~50: 0-0=0

· 50:: ST :: 55: (ST -50)-0 = ST -50, which satisfies 0:: ST -50:: 5 in this range

· ST :; 55: (ST -50)-(ST -55) = 5

So, whatever the value of ST we have:

0:: Payoff:: 5

In an arbitrage-free market the current value of the portfolio equals the discounted expected

payoff. It follows that 0:: max(ST -50,O)-max(ST -55,0):: 5e-rT . So inequality (I) is correct.


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ExamMFE


The expression in the middle of the inequalities in (II) and (II) corresponds to the value of a

portfolio consisting of a long position in a 45-strike put, a short position in a 50-strike call and a

long position in the stock. The payoff at time T from this portfolio is

max 45 - ST ,0) - max ST - 50,0) + ST. Consider the value of this payoff for different ranges of

ST:

.

ST ~45:

(45-ST )-O+ST = 45

.

45::ST::50:

O-O+ST =ST

.

ST :;50:

0- ST - 50) + ST = 50

So:

45 :: Payoff :: 50

It follows that 45e -rT :: max 45 - ST , 0) - max ST - 50,0) + ST :: 50e -rT. So inequality II) is

correct, but inequality II) is not.

Since we are not told how the stock price behaves, we can select a model (eg a binomial tree) in which ST is

always less than 45. This wil then contradict the left-hand inequality in (II.

Solution 28

Answer: A

Difficulty level: .

We can solve this using the information given and the put-call parity relationship:

r:Eur (K, T)- PEur (K, T~ = So - Ke-rT

0.15

~ 0.15 = 60-70e-4r

r = _lIn 60-0.15 = 0.039

4 70

So the continuously-compounded interest rate is 0.039.

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ExamMFE

Solution 29

Answer: D


One way to find an arbitrage strategy is to try to set up a portfolio that eliminates any net exposure to the

underlying asset. So the unspecifed number of 55-strike calls in Mary s strategy is probably going to be 2,

so that the long positions (one 40-strike call + two 55-strike calls) offset the short positions (three 50-strike

calls).

This approach wil ensure that the graph of the option payoff as a function of the stock price is flat at both

ends and the payoff will be constrained to fall within a certain range.

If Mary uses a long position in two of the 55-strike calls, the cost of setting up her option portfolio

is:

11-3x 6+2x 3+ 1 =0

. -

all options cash

So this is a zero-cost portfolio.

This confirms that we've probably chosen the right number of 55-strike calls to use.

The payoffs from Mary's option portfolio for different ranges of the stock price (ignoring the 1

cash she lends) are calculated in the following table:

MARY

Position

S~40 40~S~50

50 ~ 5 ~55

55~S

Call 40

+1

0

5-40 5-40

5-40

Call 50

-3

-0

-0

-3(5 -50) -3(5-50)

Call 55

+2

0

0

0

2(5-55)

Total

0

0

5-40 (:;0)

110-25 (:; 0)

0

We've shown strict inequalities (~) in the table, but you could equally have used weak inequalities (::).

Since the payoffs at the critical points (eg when 5 = 40) form a continuous function with no jumps, we

don't have to worry about the distinction. In fact, checking that the values either side of each boundary

match up is a good way to check that you have not made any mistakes.

We see that the payoffs from this option portfolio wil always be at least zero. If we include the

cash, the total payoff wil be at least 1 plus any interest on the cash.

So Mary s strategy is an arbitrage.

Using a similar logic, we can tr an exposure of short three calls and long three puts for the 50-

strike options in Peter's portfolio.

The cost of setting up Peter's portfolio is then:

2x 3-2x 11+ 11- 3-3x 6+3x 8+ 2 =0

\ i -

ptions cash

So this is also a zero-cost portfolio.


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ExamMFE


The payoffs from Peter's portfolio (ignoring the 2 cash he lends) are as follows:

PETER

Position

S~40

40~S~50

50~ 5 ~55

55~S

Call 55

+2

0

0

0

2(5-55)

Put 55

-2

-2(55-5)

-2(55-5)

-2(55-5) 0

Call 40

+1

0

5-40

5-40

5-40

Put 40

-1

-(40-5)

0

0

0

Call 50

-3

0

0

-3(5-50)

-3(5-50)

Put 50

+3

3(50-5)

3(50-5)

0

0

Total

0

0

0

0

0

We see that the payoffs from the option portfolio wil always be exactly zero. If we include the

cash, the total payoff wil be at least 2 plus any interest on the cash.

So Peter's strategy is also an arbitrage.

Solution 30

Answer: D


For a European call option, the strongest general inequalities we have are:

max O, PVo,T Fo,T )-PVo,T K)):: CEur S,K, T):: 5

ie max O, 5 -100e-O.05):: CEur 5,100, 0.5):: 5

~

5.i2

Ths corresponds to Diagram II.

For a European put option, the strongest general inequalities we have are:

max(O, PVo,T(K)-PVO,T(Fo,T)):: PEur(S,K, T):: K

ie max O,100e-D.05 -5):: PEur S,100,O.5):: 100

~

5.i2

This corresponds to Diagram iv.

For an American call option, the strongest general inequalities we have are:

max

5 -K, CEur S,K, T)J:: CArner S,K, T):: 5

ie max S-100, CEur S,100, T)J ::CArner S,100, T):: 5

Since 5 -100 ~ 5 - 95.12, the narrowest region we can pin down for the American call option is

the same as for the European call option, ie Diagram II.

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~ BPP Professional Education




ExamMFE

For an American put option, the strongest general inequalities we have are:

max K -5, PEur S,K, T)J:: PAmer 5, K, T):: K

ie max 100-S, PEur S,100, T)J:: PAmer

(5,100, T):: 100

Since 100-5:; 95.12-5, this narrows the possibilties down further compared to the European

put option and we get Diagram III.


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ExamMFE


Chapter 2 Solutions

Solution 1

Answer: D

Difficulty level: .

In the up state, the put option payoff is zero. The replicating strategy must reproduce the put

option payoff, so we can solve for B, the amount to lend at 7%:

75 -0.5) + BeO.07 = 0

B =34.96

The value of the put option is the cost of lending 34.96 minus the amount obtained from selling

0.5 shares of stock:

~S + B = -0.5 60)+34.96 = 4.96

Solution 2

Answer: A

Difficulty level: .

The true probability is a red herring. We need the risk-neutral probability to find the value of the call

option.

The risk-neutral probabilty of the stock moving up in price is:

p*

e(r-o)h -d

u-d

Se(r-o)h -Sd

=

Su-Sd

75eo.05 -60 = 0.4711

100-60

Since the call option wil be worth $20 if the stock price increases and $0 if the stock price

declines, the current value of the call option is:

C - 0.4711 20)+ 1-0.4711) 0) - 6

- 005 - 8.9

.

Alternatively, you can calculate the value of the option by constructing a replicating portfolio that involves

purchasing ô shares of stock and lending an amount B, where ô = 0.5 and B = -28.54. The value of the

call option is then ôS + B = 0.5 75) - 28.54 = 8.96.

Solution 3

Answer: D


Assertion (1) is not correct. The binomial model assumes that the market is arbitrage-free. Ths

means that the parameter u must be chosen so that we do not have Su ~ 100 . (100 is the current

stock price.) If this situation were possible, the stock would always fall in value over the period,

following either an up or a down step. We would then be able to create an arbitrage

opportunity by short-sellng the stock and holding the resulting cash until the end of the period.

So the payoff from Option C wil be positive at the up node (and possibly at the down node

too). The current value of the option, which is calculated as the discounted risk-neutral expected

value of the payoff, must therefore also be positive.


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ExamMFE


Assertion (2) is correct. If Sd:; 80 (and hence also Su:; 80), Option P wil be out-of-the-money

(with a payoff of zero) at both the final nodes in the tree. The current value of the option, which

is calculated as the discounted risk-neutral expected value of the payoff, would then equal zero

(according to this model).

This situation is most likely to occur if the underlying asset has low volatility and the put option is close to

its expiration date.

Assertion (3) is correct. Since Option C cannot have a zero payoff at both nodes, the payoff at the

upper and lower nodes wil have different values (with a higher value at the upper node). It wil

therefore always be possible to reproduce any two payoffs (and in particular the payoffs for

Option P) using a suitable combination of Option C and cash, by solving a set of two

simultaneous equations.

In this case, there wil always be a unique solution to this pair of simultaneous equations to replicate any

pair of payoffs. If Option P has zero value at both nodes, as in Assertion (2), the replicating portfolio

consists of zero units of Option C and zero units of cash

Solution 4

Answer: A

Difficulty level: .

According to the law of one price, the cost of the position is:

3S-105e-D.5r =3x56-105xe-D.0325 =66.36

Solution 5

Answer: E

Difficulty level: .

Seah

'-

56eo.06

Su x p + Sa x l-p) ~ P = 0.59050

- -

6.136 49.840

Solution 6

Answer: A


The equation for the true probabilty is:

Seah = Suxp + Sdx l-p) ~

eO.09x0.5 = 1.27396 P + 0.83349 1- p) ~

p = 0.4825

The risk neutral probability of an "up" move is determined from the given inormation as

follows:

u = erh + d.J , d = erh - d.J ~ erh = .J = 1.03045 ~

= erh - d = 1.03045 - 0.83349 = 0.4472

P u - d 1.27396 - 0.83349

So:

p * -p = 0.4472 - 0.4825 = - 0.0353

2




Question & Answer Bank - Solutions, Chapter 2 ExamMFE

Solution 7

Answer: C


We assume here that the stock price tree used in the binomial model is constructed from the

forward prices, which is the standard approach.

Since the up and down probabilities must add up to 1, Assertion (1) is equivalent to saying that the

up probability is always less than liz.

We can show algebraically that Assertion (1) is correct.

The risk-neutral up probabilty is given by the formula:

e(r-Ô)h -d

p*=

u-d

where:

u = e(r-ô)h+a.J

d = e(r-Ô)h-a.J

So we

have:

e(r-Ô)h _ e(r-Ô)h-a.J

p* = e(r-Ô)h+a.J _ e(r-Ô)h-a.J

1_e-a.J

ea.J _e-a.J

If we multiply the numerator and denominator by ea.J , this becomes:

ea.J -1 ea.J -1

p*=

e2a.J -1 (ea.J + 1)( ea.J -1)

1

ea.J +1

We've used the diference of two squares identity x2 -1 = (x+ l)(x-l) here.

Since (j and h are both positive, it follows that ea.J :;1 and therefore p* ~ i¡i .

Assertion (2) is correct.

A call option gives an investor leveraged (or geared ) exposure to the underlying asset. A 10%

movement in the price of the underlying asset wil result in a percentage movement exceeding

10% in the price of the call option. So the call option is a more risky investment than the

underlying asset and wil therefore command a higher expected rate of return in the real world.

Assertion (3) is not correct. A put option, on the other hand, acts as a form of insurance. Its value

moves in the opposite direction to the underlying asset. If there is a 10% movement in the

underlying asset price, a portfolio consisting of the underlying asset and a put option wil

experience price movements smaller than 10%. The portfolio is therefore less risky and has a

lower expected return than the underlying asset itself.

You can come to the same conclusions for Assertions (2) and (3) if you consider the deltas or the Sharpe

ratios for the call and put option, which are studied in later chapters.


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ExamMFE


Solution 8

Answer: D


The values of u and dare:

u = e(r-å)h+o.J = e(0.os-0.00)i+O.2,J = 1.3231

d = e(r-å)h-o-.J = e(Oos-0.00)i-O.2,J = 0.8869

The risk-neutral probabilty of the stock going up is:

* e(r-å)h -d

p =

u-d

e o.os-O)i _ 0 8869

1.3231- 0.~869 = 0.4502

The stock wil either go up to 86.003 or down to 57.650:

Stock: Call option:

86.003

16.003

65.000

?

57.650

0.000

The value of the call option can be found using risk-neutral pricing:

C 11 . 1 - (0.4502)16.003+(1-0.4502)0.000 -6650

option va ue - 0 os - .

e .

The true probabilty of the stock going up is:

e(a-å)h -d

p=

u-d

e 0.i5-0)i -08869

. = 0.6302

1.3231- 0.8869

We can now solve for the continuously-compounded expected return on the call option:

r 0.6302(16.003) + (1- 0.6302)(0.000)

e

6.650

= In 0.6302 16.003) + 1- 0.6302) 0.000))

6.650

r=0.416

Solution 9

Answer: B



u = e(r-å)h+o.J = e(0.os-0.00)i+O.3,J = 1.4623

d = e(r-å)h-o-.J = e(0.os-0.00)i-O.3,J = 0.8025

The risk-neutral probabilty of the stock going up is:

p*

e r-å)h -d

u-d

e o.os-O)i _ 0.8025

0.4256

1.4623 - 0.8025

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ExamMFE

The stock wil either go up to 59.954 or down to 32.903:

Stock: Put option:

59.954 0.000

41.000

?

32.903

7.097

The value of the put option can be found using risk-neutral pricing:

P . 1 (0.4256)0.000+(1-0.4256)7.097 3.763t option va ue = 0 08

e .

The true probabilty of the stock going up is:

e(a-o)h _ d e(0.15-0)1 - 08025

P = . 0.5446

-d 1.4623-0.8025

We can now solve for the continuously-compounded expected return on the put option:

eY = 0.5446(0.000) + (1- 0.5446)(7.097)

3.763

= in( 0.5446(0.000)+ (1- 0.5446)(7.097))

3.763

r=-0.152

Solution 10

Answer: B


C -Cd 6.136 - 0

Ll = u ( ) = 0.37653 :: 56 Ll = 21.086

S(u-d) 56 1.181 - 0.89

B = e-rh x uCd - dCu = 0.97531 x 0 - 0.89x6.136 = -18.304

u-d 1.181-0.89

::C = SLl + B =2.781

- -

1.086 -18.304

The yield r on the call option, which has initial value 5 Ll + B , can be found by considering the

accumulated values of the two components 5 Ll and B at the end of the period:

eyh x(S Ll + B) eah x 5 Ll + erh x B

yh 5 Ll

: e = eah x

'- 5 Ll + B

1.06184 ~

7.58154

+ erh x B

'- 5 Ll + B

1.02532 '-

-6.58154

:: r = 0.52901

Note: The last line expresses the fact that the yield on the call option is a weighted average of the yields on

the stock and cash. The parameter values a = 0.12 and h = 0.5 come from the earlier question.


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ExamMFE


Solution 11

Answer: C


C = e-o.5y x Cu x P

- - -

-u.26450 6.136 0.59050

+ e-O.5yx Cd x (i-p) = 2.781

'-

o

Note: Suppose we had used a = 0.15 instead of a = 0.12 as above. We would obtain diferent values of p

and r. However, we would stil get the same option price C = 2.781 using the new option yield rate and

the new true probability of an up move. In other words, the price of the option does not depend on a.

In particular, if we let a = r ( risk neutral pricing ), we wil still obtain the correct option price.

Solution 12

Answer: C

Difficulty level: .

The current value of the stock is:

puSeÔh + (1- p)dSeÔh

5=

eah

(0.5623)106.34 + (1- 0.5623)58.36 = 72.00

eO.17

The risk-neutral probabilty of an upward movement is:

0.09 58.36

e --

72.00 _

106.34 58.36 - 0.4256

---

2.00 72.00

' e(r-Ô)_d

p' =

u-d

If the stock price increases to $106.34, then the call option wil pay:

106.34 -75.00 = 31.34

The value of the call option is:

C 11 . 1 _ (0.4256)31.34 + (1- 0.4256)0.00 option va ue - 0 09

e .

12.19

Solution 13

Answer: D


The future values of investments held by an investor who is subject to tax wil be reduced by the amount of

tax payable. As a result, the risk-neutral probabilities for an investor who is subject to tax wil usually

difer from those for a tax-exempt investor. However, if the investor pays the same rate of tax on all gains

(including interest payments, which is not the case here), it turns out that the tax elements cancel out and

the risk-neutral probabilities are unchanged.

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ExamMFE

Since this investor is subject to capital gains tax (CGT), the value of the stock at the upper node

must be reduced by the amount of tax payable on the gain. So the appropriate tree to use for

modeling the stock for this investor is:

15 - 0.4 15 - 11) = 13.4

II

y~

- p* 9

We can find the risk-neutral probabilties for this investor using the fact that the discounted

expected value of the stock must equal its initial value:

e-o.04 C13.4p*+9(1-p*)J = 11

So:

lleO.04 -9

p* = 13.4-9

0.556572

Now let the fair value of the option for this investor be x .

At the upper node, the value of the option wil equal max O,15 -12.50) = 2.50. So this investor

wil need to pay 0.4 2.50-x) in tax. At the lower node, the option wil have no value and no tax

wil be payable.

So the tree for the option looks like this:

2.50 - 0.4 2.50 - x) = 1.50 + O.4x

x

y~

- p* 0

As usual, the value of the option must equal its discounted expected future value. So we have:

e-o.04 C(1.50+ O.4x)p * +0(1- p*) J = x

x = 1.50p * = 1.02

eO.04 - O.4p *

So:


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ExamMFE


Solution 14

Answer: E


The appropriate tree to use for modeling the stock for this investor, based on real-world

probabilities is:

15 - 0.4(15 - 1 I) = 13.4

/

I

~

9

Since the real-world expected rate of return on the stock for this investor is 10 , the discounted

expected value of the stock, calculated using the real-world probabilities, must equal its initial

value:

e-0.10 (13.4p+9(1- p) J = 11

11 0.10 9

e - = 0.717473

P 13.4-9

o:

Now let the real-world expected rate of return on the option for this investor be r.

At the upper node, the value of the option wil equal max(O,15 -12.50) = 2.50 and this investor

wil need to pay 0.4(2.50 -1.00) = 0.60 in tax. So the tree for the value of the option for this

investor looks like this:

2.50 - 0.4(2.50 - 1.00) = 1.90

/

.00

~

o

Since the value of the option must equal its discounted expected future value, we have:

e-Y (1.90p+ 0(1- p) J = 1.00

~ r=31.0

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ExamMFE

Solution 15

Answer: E


The model assumed here uses the following tree for the share price:

Su

/

~

Sd

We can find the relationship between u and d required for this model to be arbitrage-free by

using the fact that the discounted expected value of the stock calculated using the risk-neutral

probabilties) must equal its initial value:

e-r LtxSu+tXSd J = 5

We can then cancel the 5' s and rearrange to find the formula for d:

e-r Ltu+td J = 1

:: d=2er-u

Solution 16

Answer: C


We

have:

1.181 = 66.136 = u = e r-J)h + d.J

56

:: 1.05109 = ud= e2(r-J)h = e2rx0.5 :: r = 0.0498

e r-J)h _ d 0.5xO.0498 -0 89

*= = e . =0.46470

u - d 1.181 - 0.89

0.89 = 49.840 = d = e r-J)h - d.J

56

An increase in the stock price of $1 wil cause the up stock price to increase by u=1.181 . So the

call payoff in the up state wil increase by 1.181 since it was already in the money in the up

state. The revised stock price in the down state is 57d=50.73 which means the call is stil out of

the

money.

So the increase in the call price due to the 1$ increase in the stock price is the actuarial value of

the increased payoff:

1.181xp*xe-o.sr = 0.535


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ExamMFE


Solution 17

Answer: B


Let 0 denote the amount of each dividend payment. We are carrying out the calculation on

Apri130. So the dividend payments are due in 2 months time and in 6 months time.

From put-call parity, we know that:

C + Ke-rT = P+ 5 -PV(Oividends)

Substituting the numerical values given:

4.50+50e -o.06xi62 = 2.45+52.00- o( e -O.06Xi~ + e -O.06xi52 )

~ 0= 1.428 = 0.73

1.965

Solution 18

Answer: A

Difficulty level: .

If the initial stock price is 5, the two possible values in the binomial tree are Su = 1.4335 and

Sd = 0.7565. The discounted expected value of the stock price must equal the intial value, so

that:

e-O.10 Cl.433Sp+0.756S(1-p)J= 5

Canceling the 5 s and rearranging gives:

eO.10 0.756

p= =0.52

1.433-0.756

Solution 19

Answer: E

Difficulty level: .

The payoff from the straddle in one year's time is 151 - 501.

The tree of stock prices (in the gray boxes) and option values (in italics) looks like this:

Now 1 year

20

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ExamMFE

The risk-neutral up probabilty is:

Serh -Sd

p*=

Su-Sd

60eO.08 -45

70-45

0.7999

The value of the straddle can be calculated as the discounted expected value of the payoff:

V = e-O.08 L 20p * +5 1- p*) J = 15.69

The value (rounded to the nearest 10(t) is $15.70.

Solution 20

Answer: B


We can calculate the theoretical value of the call option based on the binomial modeL. The up

and down ratios for the stock price are u = 55/50 = 1.1 and d = 40/50 = 0.8. So the risk-neutral

up probability is:

e(r-å)h -d

p*=

u-d

e O.05-0.10)x1 _ 0 8

. = 0.5041

1.1-0.8

The payoffs from the option are 5 at the up node and zero at the down node. The option

price can then be calculated as the discounted expected value of the payoffs:

e-D.05x1 L5xp * +Ox l- p*) J = 2.40

Since this is less than the actual market price (1.90), the option appears to be underpriced. So we

can create an arbitrage by purchasing the option and at the same time hedging the position by

sellng the stock (since the prices of the call option and the stock wil tend to move in the same

direction). This wil create a net cashfow of -1.90+50 =48.10:; 0, which the trader can lend at

the risk-free rate. So the correct action is B.

Solution 21

Answer: D


The option payoffs are 2 at the up node and 0 at the down node (whether calculated correctly or

not).

The original calculation of the option price was:

-r I * (*)J * r

L 2Pwrong + 0 1- Pwrong = 1.13 :: Pwrong = 0.565e

where P~rong is the risk-neutral probabilty calculated as:

* er -dwrong

Pwrong = d

u- wrong

er -0.8 * r 08

:: O.4Pwrong = e - .

1.2-0.8

An alternative way to derive this equation is to note that the discounted expected payoff from the stock

itself calculated using the risk-free interest rate and risk-neutral probabilities, must equal the current stock

price. This leads to the following equation, which can be rearranged to give the same equation:

e-r (12P~rong +8(1-P~rong)J = 10


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ExamMFE


Solving these two equations simultaneously, we get:

0.4 (0.565er) = er - 0.8 =; er = ~ = 1.0336

0.774

The correct calculation of the risk-neutral up probability should have been:

p* = er -d = 1.0336-0.6 0.7227

u-d 1.2-0.6

The correct calculation of the option price is then:

e-r r 2p*+0(1-p*)J =~X2xO.7227 = 1.3983

L 1.0336

Solution 22

Answer: A


Denote the risk-neutral probabilties associated with the 3 outcomes by Pi, P2 and P3. As these

are probabilties, we know that Pi + P2 + P3 = 1 .

We also know that the price of any asset in the market is equal to its discounted expected payoff.

Asset Y itself wil pay 0, 0 or 300 in one year's time. So its value now is:

e-rx Op1 +OP2 +300p3)

=100

=; P3

100eO.1 = 0.3684

300

Asset X itself wil pay 200, 50 or 0 in one year's time. So its value now is:

e -r X (200p1 + 50P2 + 0P3 ) = 100 =; 4P1 + P2 = 2eO.1

But P2 = 1-P1 -P3 = 1-P1 -0.3684 = 0.6316-P1

So:

4P1 + 0.6316 - Pi) = 2eO.1

2eO.1 - 0.6316

=; Pi = 3 0.5262 and P2 = 0.1054

We can now calculate the current prices of the two options. The call option pays 105, 0 or o. So

its price is:

Cx = e-r xl05P1 = e-O.1 xl05x0.5262 = 49.99

The put option pays 95, 95 or O. So its price is:

Py = e-r X

(95p1 +95p2) = e-D.1 x95x(0.5262+ 0.1054) = 54.29

So: Py -Cx =54.29-49.99=4.30





Alternatively, you can say that Py - C X is the value of a combined holding of 1 long put and 1 short call.

You can then find the value of this combined holding, which has payoffs of -10, 95 and O.

Solution 23

Answer: B


The tree of asset prices (with the interest rates shown in parentheses) is:

110 (6 )

100 (5 )

95 (4 )

The 5% here means that the effctive annual interest rate during the first year is assumed to be 5%.


110 95

u=-=l.l and d=-=0.95

100 100

The risk-neutral probabilty of the stock price going up in the first year is:

* e(r-o)h -d

p =

u-d

1.05-0.95

=

1.1-0.95

2

3

Since we don't know how the stock price moves after time 1, we can't work out the option prices

P(108) and C(108) directly. However, we can find the value of P(108) - C(108) by using the put-

call parity relationship. Ths tells us that:

P(108) - C(108) = 108x Discount factor -100

where Discount factor represents the 2-year discount factor calculated at the risk-free interest

rate.

Note that we know the interest rate will be 5% in the first year, but in the second year there are two

possible values. The rate we need to use for discounting in the second year wil be a weighted average of

6 and4 .

We can find the two-year risk-free discount factor by noting that it wil be the same as the value

at time 0 of a zero-coupon bond that always pays 1 at time 2.

So: Discountfactor=p* X~X~+(1-P*)X~X~=0.9042

, 1.05 1.06 1.05 1.04,

Value of a 2-year bond

So the put-call parity equation tells us that:

P(108)-C(108) = 108xO.9042-100 = -2.34


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ExamMFE


Solution 24

Answer:B



u = e(r-o)h+a.. = e(O.04-0)(O.25)+O.3.. = eO.16 = 1.1735

d = e(r-O)h-a.. = e(O.04-0)(O.25)-O.3.. = e-o.14 = 0.8694

The tree of asset prices (with the option payoffs shown in parentheses) is:

117.35 0)

100

86.94 K -86.94)

In a sensible one-step binomial model for a put option, the payoff will be 0 at the upper node and wil take a

positive value at the lower node.

The risk-neutral probability of the stock price going up is:

' e( r-o)h - d

p' =

u-d

eO.01 - 0.8694

1.1735-0.8694

0.4626

We can then calculate the intial value of the option, assuming it is a European option, as the

discounted risk-neutral expectation of the payoff, which is:

e-O.01 LP* XO+( 1-p* )X(K -86.94)J = 0.5320(K -86.94)

Since this is an American option, there is an opportunity to exercise it at the initial node. We wil

do this if the exercise value at that point exceeds the European value we have just calculated,

that is, if:

K -100:; 0.5320 (K -86.94)

Ç: K - 0.5320 K :;100 - 0.5320 (86.94)

Ç:

K:; 100-0.5320(86.94) 114.85

1-0.5320

This means that we would exercise it if K = 115, but not if K = 114.

14





ExamMFE


Chapter 3 Solutions

Solution 1

Answer:C Difficulty level: .

1.181 = 70.860 = u = e r-o)h + a-.J , 0.89 = 53.400 = d = e r-o)h - a-.J

60 60

1.181 _ u _ 2a-.J 1 (1.181) - 2 lCh - 2 ~5 - 0200

:= - - - - e := n - - Jvrt - JVU.O := J - .

0.89 d 0.89

Solution 2

Answer: E


1.181 = 70.860 = u = /r-o)h + a-.J 0.89 = 53.400 = d = e(r-o)h - a-.J

60 60

:= 1.05109 = ud = l r-o)h = e2rx0.5 := r = 0.0498

Solution 3

Answer: D

Difficulty level: .

e(r-o)h _ d

p* =

u - d

0.5xO.0498 -089

e . = 0.46470

1.181- 0.89


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ExamMFE


Solution 4

Answer:B


From the given up and down prices:

u = 70.860 = 1.181

60

d = 53.400 = 0.89

60

Now we can compute the 3 possible prices for the stock at the end of the second 6-month period:

~

83.686 = 70.860u

( 0.000)

60.00

70.860

53.400

~

63.065 = 70.860d = 53.400u

(0.000)

47.526 = 53.400d

(12.474)

The put option payoff max to, 60 - 5 (1) J is given in brackets at each node in the diagram.

We can now determine the put option values at the end of the first period:

60.00

70.860

(0.000)

53.400

(6.513)

6.513=Ox p* x e-O.5r +

12.474

x

(1-p*)xe-D.5r

'- '- '- '-

.4647 0.97539 0.5353 0.97539

Finally, we can determine the option value at time 0:

p= Ox p* x e-0.5r +

6.513x 1-p*)xe-D.5r =3.401

'- '- '- '-

.4647 0.97539 0.5353 0.97539

Note: For a European option one can calculate the price using only the option values at the final nodes

along with binomial probabilities. It is only necessary to work backward through the entire tree when the

option style is American. Here is an ilustration of this for this question:

p = 0 x(p*)2x e-r + 0 x2(p*)(1-p*)x e-r + 12.474 x(1-p*)2x e-r

= 12.474 x0.535302xe-0.0498 = 3.401

2

Ç BPP Professional Education




ExamMFE

Solution 5

Answer: C


In the binomial model it is assumed that an American put can only be exercised at the end of a

period. So we need only compare the exercise values at time 0.5 with the option values at time 0.5

in the diagram in the previous solution. Notice in the down state in this diagram that the value of

holding the option is 6.513, whereas the exercise value is 60 - 53.400 = 6.600 . So we replace 6.513

by the higher exercise value and then work backward as in the previous solution:

60.00

70.860

(0.000)

53.400

(6.600)

PA = Ox p* x e-D.5r +6.600x(1-p*)xe-D.5r =3.446

- - - ~

.4647 0.97539 0.5353 0.97539

Solution 6

Answer: D


For a call bull spread the low strike call (K = 62 ) is bought and the high strike call (K = 77 ) is

sold. We need to compute the position payoff at the final nodes and then work backwards

through the tree. The position payoff is computed as follows:

maxiO ,5 1) - 621- maxiO ,5 1) - 771

121.686-6.686 = 15 if 5(1) = 83.686

= 1.065 - 0 = 1.065 if 5(1) = 63.065

0-0=0 if

5(1)

=47.526

Here is the tree from Solution 4 (modified to include the payoffs just determined) along with

other crucial numbers calculated earlier:

~

83.686 = 70.860u

( 15.000)

60.00

70.860

53.400

~

63.065 = 70.860d = 53.400u

(1.065)

47.526 = 53.400d

(0.000)

p* = 0.4647 ,e-O.5r = 0.9753


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ExamMFE


Here are the position values at the end of the first period:

Up: 15xp*x e-O.5r + 1.065x(1-p*)xe-D.5r = 7.355

Down: 1.065x p*x e-D.5r + O.Ox(l-p*)xe-D.Sr = 0.483

So the value price) of the position at time 0 is:

7.355xp*x e-O.5r + 0.483x(1-p*)xe-0.5r = 3.586

Note: Once again the option style is European so the price could be determined using payoffs at the final

nodes along with binomial probabilities.

Solution 7

Answer: E


We can use put-call parity along with results gleaned from the given information:

C = P + 5 - Ke-r

'- '- -

1.70 30 30xO.9S139

= 3.16

The discount factor 0.95139 was calculated from the equations:

1.181 = 35.430 = u = erh + iJ..

30

:: 1.05109 = ud = e2rh = e2rxO.S = er

0.89 = 26.70 = d = erh - iJ..

30

:: e-r = 0.95139

Solution 8

Answer: A


This problem is simplified by the fact that the call option is in-the-money at only one of the possible end-of-

year prices.

An American call option on a stock that doesn't pay dividends has the same price as a European

call option, so we do not need to check to see if early exercise is optimal (since we know it won't

be).

The factors for the stock price are:

u = e(r-o)h+I.. = e(0.10-0.00)(1/3)+0.3.J1/3 = 1.22941

d = e(r-o)h-iJ.. = e(0.10-0.00)(1/3)-0.3.f = 0.86947

The recombining tree is partially filed out below. It is not completely filed out because many of

the final stock prices are clearly less than 70, so the call option wil have a payoff of zero at those

nodes.

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ExamMFE

(61.47

50

D

92.91

65.71

D

D

A European call option with a strike price of 70 has only one payoff, and it occurs only if the

stock moves upward in each of the three periods:

Call option payoff = 92.91-70 = 22.91

The risk-neutral probabilty of each upward movement is:

* e(r-o)h_d

p =

u-d

e O.10-0.00) 1/3) _ 0.86947

1.22941- 0.86947

0.4568

The tree of prices for the call option is therefore:

/4.47

.98~

0.00

22.91

0.00

0.00

0.00

0.00

Here's a shortcut that we can take when there are no cash flows prior to maturity.

Alternatively, since the only cash flow occurs at time T = 1, we can quickly find the value of the

call option as the discounted expected value of that one cash flow:

C = (0.4568)3 x

22.91

eO.10

1.98

Solution 9

Answer: B



u = e(r-O)h+c.J = e(o.OS-O.OO)(1)+O.2v = 1.32313

d = e(r-o)h-O .J = e(o.oS-O.OO)(1)-O.2v = 0.88692


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ExamMFE


The risk-neutral probability of an upward movement is:

* e r-o)h -d

p =

u-d

e O.08-0.00)1 _ 0.88692

1.32313 - 0.88692

0.4502

The tree of stock prices is:

87.53

66.16

50.00

58.68

44.35

39.33

The corresponding tree of put prices is:

0.00

0.00

5.41

0.00

10.65

15.67

Notice that it is optimal to exercise the American put early at the node shown in bold.

Solution 10

Answer: C


Each of the periods is 3 months long:

h = T = 0.75 = 0.25

n 3


u = e r-o)h+a.J = e O.07-0.04) 0.25)+O.3.J0.25 = 1.17058

d = e r-o)h-a.J = e O.07-0.04) O.25)-O.3.J = 0.86719


* e r-o)h_d

p. =

u-d

e O.07-0.04)O.25 -0.86719

1.17058-0.86719

0.4626

The index and call option values at each node are:

Index:

81.94

83.18

Call option:

36.00

22.38

13.11

11.39

5.57

0.74

52.28

12.28

95.92

23.18

70.00

71.06

60.70

61.62

1.62

52.64

45.65

0.00

In this example, there are no nodes where the exercise value exceeds the continuation value (the value

calculated by discounting the next two nodes). So it is never optimal to exercise the option early.

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ExamMFE

Solution 11

Answer: B Difficulty level: ..

With options on currencies, we treat the foreign interest rate as the dividend yield of the

underlying asset.


u = e(r-o)h+O .J = e(O08-0.055)(0.25)+0.15.J = 1.08464

d = e(r-o)h-O .J = e(0.08-0.055)(0.25)-0.15.J = 0.93356


* e(r-o)h_d

p =

u-d

e 0.08-0.055)0.25 _ 0 93356

. = 0.48126

1.08464 - 0.93356

The currency values and the values of the American call option at each node are:

Dinar:

Call option:

4.118 0.618

3.796

0.314

3.500 3.544

0.159

0.044

3.267 0.021

3.050

0.000

Solution 12

Answer: B Difficulty level: ..

Each of the periods is 1 month long:

h= T = 2/12 =~

n 2 12


u = e(r-o)h+O .J = e(0.06-0.03) I 12+0.15-/1 112 = 1.04687

d = e(r-o)h-O .J = e(0.06-0.03)

/12-0.15-/1

/12

= 0.96002


* e(r-o)h -d

p =

u-d

e(0.06-0.03)

/12 -096002

. = 0.4892

1.04687 -0.96002

The prices of the stock and the European put option at each node are:

Stock:

Put option:

13.151

0.000

12.562

0.478

12.000 12.060

0.966 0.940

11.520

1.444

11.060 1.940


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ExamMFE


Solution 13

Answer: C


Each of the periods is 1 year long:

h=T=~=l

n 2


u = e(r-o)h+CY.J = e(O.07-0.04)(1)+O.3S.J = 1.46228

d = e(r-o)h-CY.J = e(O07-0.04)(1)-O.3S.J = 0.72615


* e r-o)h -d

p =

u-d

e O07-0.04)1_0 72615

. = 0.4134

1.46228-0.72615

The index and put option values at each node are:

Index:

Put

option:

213.83

0.00

146.23

0.00

100.00

106.18

13.88

0.00

72.61

25.39

52.73

45.27

Solution 14

Anwer: B

Difficulty level: .

The formulas for the up and down factors in this model are:

u = e r-o)h+CY.J

d = e r-o)h-CY.J

So here we have:

u = e r-o)h+CY.J = 1.056928

d = e r-O)h-CY.J = 0.950406

If we divide the formula for u by the formula for d, we get:

=

e(r-O)h+CY.J

e(r-o)h-CY.J

e2CY.J

u

d

Here h = A (= 1 month). So we have:

/CYll = 1.056928 = 1.112080

0.950406

=? Y= 0.184

So the annualized volatilty assumed in the calculations is 18.4%.

In fact the values assumed for the other parameters were r = 5% and ö = 2.3%, but it is not possible to

separate these two values based solely on the information given.

8

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ExamMFE

Solution 15

Answer: D


The results of the calculations here illustrate a problem that tends to occur with binomial trees. As we

increase the number of steps in an attempt to improve the accuracy of the results the calculated option

prices tend to oscillate.

2-step tree

Here we have h = -t and the factors for the gold price are:

u = e(r-o)h+rY.j = e(o.oS-(-o.OOS))(1/2)+O.lS.J1/2 = 1.142897

d = e(r-o)h-ry.j = e(o.OS-(-o.OOS))(1/2)-O.lS.J1/2 = 0.924441

The 2-step gold price tree looks like this:

/ 685.74 ~

600.00 '

554.66

783.73

633.92

512.75

The risk-neutral probabilities are:

p*

e(r-o)h -d

u-d

eO.OSS(1/2) -0 924441

. =0.473508

1.142897 -0.924441

1- p* = 1-0.473508 = 0.526492

The 2-step tree of option prices (calculated using discounted expected values) looks like this:

/ 0.00

23.00 '

44.80

( 0.00

( 0.00

87.25

The option price calculated using the 2-step tree is 23.00.

3-step tree

Here we have h = t and the factors for the gold price are:

u = e(r-o)h+C.j = e(o.OS-(-o.OOS))(1/3)+O.lS.J1j3 = 1.110639

d = e(r-o)h-rY.j = e(o.OS-(-o.OOS))(1/3)-O.lS.J1j3 = 0.934009


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ExamMFE



822.00

/ 666.38 ~

600.00 ~

560.41

740.11

691.27

622.41

581.34

523.42

488.88

The risk-neutral probabilities are:

* e(r-d)h -d

p =

u-d

eO.055 1/3) -0934009

. = 0.478363

1.110639-0.934009

1- P * = 1- 0.478363 = 0.521637

The 3-step tree of option prices looks like this:

0.00

~

0.00

21.94 (

4.91

0.00

9.58

38.25

18.67

65.79

11 1.2

The option price calculated using the 3-step tree is $21.94.

4-step tree

Here we have h = t and the factors for the gold price are:

u = e(r-d)h+o-.j = e(0.05-( -0.005))(1/ 4)+0.15.J1 /4 = 1.092807

d = e(r-d)h-o-.j = e(0.05-(-0.005))(1/ 4)-0.15.J1/ 4 = 0.940588


855.71

783.04

716.54

736.52

( 655 68~

673.97

600.00

616.73

633.92

564.35 580.09

530.82

545.62

499.29

469.62

The risk-neutral probabilties are:

p'

e(r-d)h -d

u-d

=

eO.055 1/4) -0.940588

0.481259

1.092807 -0.940588

1-p* =1-0.481259=0.518741

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ExamMFE

The 4-step tree of option prices looks like this:

0.00

0.00

~

0.00

0.00

22.88

7.31

0.00

14.27

0.00

37.88

27.86

60.70

54.38

92.64

130.38

The option price calculated using the 4-step tree is $22.88.

So we have: P3( 21.94) ~ P4( 22.88) ~ P2( 23.00)

Solution 16

Answer: D


Note that, since this is a call option on a stock that does not pay dividends, it is irrelevant whether the

option is American or European. The value of the option will be the same in either case.


u = e(r-o)h+O .J = e(0.06-0.00)(0.5)+0.3.J = 1.2740

d = e(r-o)h-O .J = e(0.06-0.00)(0.5)-0.3.J = 0.8335


81.149

63.698

50.000

53.092

41.675

34.735


* e r-o)h_d

p =

u-d

e(O06-0.00)0.5 _ 0 8335

. =0.4472

1.2740 - 0.8335

The corresponding tree of call prices is:

33.149

17.117

8.613

5.092

2.210

0.000

The true probabilty of an upward movement is:

p

e(a-o)h -d

u-d

=

e(0.17 -0.00)0.5 _ 0.8335

0.5794

1.2740-0.8335


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ExamMFE


We can now solve for the continuously-compounded annual expected return, r, on the call

option:

ey/2 = 0.5794 17.117)+ 1-0.5794) 2.210)

8.613

= 2xin 0.5794 17.117)+ 1-0.5794) 2.210))

8.613

r=0.461

Solution 17

Answer: B



u = e r-5)h+a.. = e 0.06-0.00) 0.5)+0.3.J = 1.2740

d = e(r-o)h-O .. = e(0.06-0.00)(0.5)-0.3.J = 0.8335


81.149

63.698

50.000

53.092

41.675

34.735


~

p'

e(r-o)h _ d

u-d

e 0.06-0.00)0.5 _ 0 8335

. = 0.4472

1.2740-0.8335

The corresponding tree of call prices is:

33.149

17.117

8.613

5.092

2.210

0.000

The true probability of an upward movement is:

e(a-o)h -d

p=

u-d

=

e 0.17 -0.00)0.5 _ 0 8335

. = 0.5794

1.2740-0.8335

We can now solve for the continuously-compounded annual expected return, r, on the call

option, over the two time steps:

8.613eY = l33.149xp2 +5.092X2p(1-p)J

y 13.611

e=-

.613

r= 0.458

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ExamMFE

Solution 18

Answer: E

Difficulty level: .

Recall from Chapter 2 that the binomial model is arbitrage-free provided that u:; erh :; d

ie eali :; erh :; e-ali .

The Cox-Ross-Rubinstein model is more likely to give rise to arbitrage when J is small and h is large, so

Choices Band E are the best ones to check first.

The Cox-Ross-Rubinstein approach can give rise to arbitrage if:

erh :; eali

The parameters from Choice E result in the inequality being true:

eO.09 .S) :; eO.os.J

1.0460 :;1.0360

Therefore, these parameters give rise to arbitrage.

Solution 19

Answer: A


The Cox-Ross-Rubinstein model uses the following values for u and d:

u = eali = eO.3.J = 1.2363

d = e-ali = e-O.3.J = 0.8089


152.847

123.631

100.000

100.000

80.886

65.425


* e r-å)h_d

p =

u-d

e O.08-0.00)O.S _ 0.8089

1.2363 - 0.8089

0.5426

The values of the call option are shown in the tree below:

57.847

32.356

18.015

5.000

2.607

0.000


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ExamMFE


Solution 20

Answer: C

Difficulty level: .

The value of delta at the initial node is:

~(S, 0) = e-ôr C(uS,h)-C(dS,h)

uS-dS

Here we have:

ö=O

uS = 52.17, dS = 48.01

C(uS,

h) = 8.45, C(dS,

h) = 5.85

So:

~(S O)=eO 8.45-5.85 0.625

, 52.17 -48.01

Solution 21

Answer: E


The value of gamma at the end of the first week is:

( h) _ ~(uS,h)-~(dS,h)

Sh' -

uS-dS

The two deltas are calculated as:

~ uS,h) = e-ôr C uuS,2h) -C udS,2h)

uuS-udS

and ~ dS,h) = e-oh C udS,2h)-C ddS,2h)

udS-ddS

Here we have:

ö=O

uuS = 54.44, udS = 50.10, ddS = 46.10

C(uuS,h)

=9.96, C(udS,

h)

=7.02, C(ddS,

h) =4.74

So: ~(uS,h) = eO 9.96-7.02 =0.67742

54.44 - 50.10

A dS h) = eO 7.02-4.74 057Ll, 50.10-46.10.

r(S h)= 0.67742-0.57 =0.02582

h 52.17-48.01

nd

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ExamMFE

Solution 22

Answer: A


The value of theta at the initial node is:

C udS ,2h) -£L1 S, 0) -l £2qS, 0) -C S, 0)

8 5,0) = where £ = udS-S

2h

Here:

5 = 50, udS = 50.10

C udS,2h)=7.02, C S,O)=7.12

L1(S, 0) = 0.625 (from Solution 20)

qs,O) ' qSh,h) =0.02582 (from Solution 21)

1

h=-

52

So:

£ = 50.10-50 = 0.10

7.02 -0.10xO.625-lx 0.102 x 0.02582 -7.12

8(5,0) = -4.2

/52

and

Solution 23

Answer: B


If there are x up movements (equal to +$0.01) in the stock price during the year, there must be

250-x down movements (equal to -$0.01) and the stock price at the end of the year wil be:

5 =5.00+0.Olx-0.Ol(250-x)

=2.50+0.02x

The probabilty we are interested in is:

Pr(S -5.00 ~ -L) = Pr((2.50+0.02x)-5.00 ~ -L)

P L -L+ 2.50J

r x~

0.02

= Pr(x ~ -50L+ 125)

Since the up and down movements are statistically independent, x has a Binomial 250,

0.51)

distribution. We can approximate this with a normal distribution with mean 250x0.51 = 127.5

and variance 250xO.51x0.49 = 62.475.

So we require:

Pr N 127.5,62.475) ~ -50L+ 125-0.5):: 0.01

The 0.5 included here is a continuity correction to improve the accuracy when approximating a discrete

distribution with a continuous distribution.


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ExamMFE


In terms of the standard normal distribution, this is:

L -50L-3J

Pr N(O,l) ~.J :: 0.01

62.475

This wil be true if:

-50L-3 :: -2.326

~62.475

or

L? 0.3077

So the smallest value of L (in dollars and cents) satisfying this requirement is $0.31.

Without the continuity correction, the answer works out to $0.32.

Solution 24

Answer: A


If the trader exercises the put option he wil be receiving cash and handing over stock. So he wil

be more likely to exercise early if the benefits of holding cash are enhanced or the benefits of

holding the stock are diminished.

Since the question says in isolation , we should ignore any knock-on effects that each event might have.

For example, payment of the special dividend might affect the stock price.

Event (1) wil not make the trader more likely to exercise early. Payment of a special dividend

wil be beneficial to a holder of the stock, which wil make early exercise of a put option less

likely.

Event (2) wil make the trader more likely to exercise early. An increase in interest rates wil be

beneficial to a holder of cash, which wil make early exercise of a put option more likely.

Event (3) wil not make the trader more likely to exercise early. A holder of a put option has

some protection against a fall in the future stock price. By exercising the option, the trader would

be giving up this insurance, which would become more valuable if the markets became more

volatile.

Solution 25

Answer: D


The tree of stock prices in the gray boxes) and option values in italics) looks like this:

Now 6 months 1 year

The stock prices are calculated from the u and d factors eg 70ud = 73.5763 .

The option values in the final column are the payoffs at the end of the year

eg max 80 -73.5763,0) = 6.4237 .

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ExamMFE


p*= erh -d = eO.05Xl -0.890 =0.465

u - d 1.181- 0.890

The option prices (assuming no exercise) at early nodes are calculated as discounted expected

values, eg:

e -Ü.05Xi62 C 6.4237p * +24.553 1- p*) J = 15.72

At the 62.3 node, the exercise value is max(80-62.3,O) = 17.7. Since this exceeds 15.72, it is

optimal to exercise the option at this node, and the value at this node is 17.7 (not 15.72).

The option value at time 0 (rounded to the nearest 5ii) is 10.75.

Solution 26

Answer: E


The tree of stock prices (in the gray boxes) and option values (in italics) looks like this:

Now 1 year 2 years

82.21

The u and d factors are:

u = e r-t5h+CY.J = e O.05-0.05)X1+0.30.J = 1.3499

and d = e(r-O)h-CY.J = e(O.05-0.05)X1+0.30.J = 0.7408

The stock prices are then calculated from these, eg 100u2 = 182.21 .

The option values in the final column are the payoffs at the end of the year,

eg max(182.21-100, 0) = 82.21 .


e r-o)h -d e O.05-0.05)X1 -0.7408

p* = 0.4256

-d 1.3499-0.7408

The option prices (assuming no exercise) at the earlier nodes are calculated as discounted

expected values, eg:

e-Ü.05x1 C82.21x p * + OX(l-p*)J = 33.28

At the 134.99 node, the exercise value is max(134.99 -100, 0) = 34.99. Since this exceeds 33.28, it

is optimal to exercise the option at this node, and the value at this node is 34.99 not 33.28).

The option value at time 0 is $14.16.


17



ExamMFE


Solution 27

Answer: C

Difficulty level: .

Remember that American and European call options on a non-dividend-paying stock have the same value.

Using the values given for u and d, we can calculate the tree of stock prices and the final payoffs

(shown in parentheses):

32.97 10.92)

25.68

20.00

22.10 0.10)

17.21

14.82 0)

The risk-neutral probability of the stock price going up is:

* e r-o)h -d

p =

u-d

eO.05 -0.8607 = 0.4502

1.2840 - 0.8607

We can then calculate the initial value of the option as the discounted risk-neutral expectation of

the payoff, which is:

e-O.05(2) i(p* t xl0.97 +2p* (l-p* )XO.l0+0 J = 2.06

Solution 28

Answer: D


Each step in the tree is of length 3 months. So, working in years, we have h = 0.25 .

The up and down ratios for each branch are then calculated as:

u = e r-O)h+ty.J = e O.08-0.09) O.25)+O.3..O.25 = eO.1475 = 1.1589

and d = e r-O)h-ty.J = e O.08-0.09) 0.25)-O.3.. = e-O.1525 = 0.8586

~ Note that, for an option on a currency, the parameter r represents the interest rate earned by the home

currency and Ö is the interest rate earned by the foreign currency.


e r-o)h _ d e O.08-0.09) O.25) - 0.8586

p* = 0.4626

u-d 1.1589-0.8586

Since this is an American option, we need to replace the calculated value (the continuation value)

with the exercise value, ie maxt1.56-S,

01 , whenever this is higher.

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ExamMFE

The tree below shows the value of 1 British pound expressed in US dollars (shown inside the

boxes) and the value of the American put option (shown above the boxes):

0

0

2.2259

0.0939

1.9207

0

0.2255

1.6573

0.1783

1.6490

1.43

0.3473

1.4229

0.3384

1.2277

0.5059

1.2216

1.0541

0.6550

0.9050

For example, the value at the 1.4229 node is calculated as:

e-D.02 (p* XO+( 1- p* )XO.3384 J = 0.1783

At the 1.0541 node the exercise value 0.5059) exceeds the continuation value 0.4984).

So the initial value of the American option is 0.2255.

Solution 29

Answer:D


The tree of stock prices and American option prices (shown above the boxes) is as follows:

0

0

585.9375

14.46

468.75

0

39.73

375

41.00

328.125

300

90

262.5

116.25

210

153

183.75

147

197.1

102.9

At each branch the up and down ratios take the same values:

u = 375 = 1.25 and d = 210 = 0.7

300 300

The risk-neutral up probabilty is calculated as:

e r-S)h _ d e OlO-0.065) - 0 7

p* = = . = 0.6102

u-d 1.25-0.7


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ExamMFE


Since this is an American option, we need to replace the calculated value (the continuation value)

with the exercise value, ie max t 300 - 5,0 J if this is higher.

At the 147 node the exercise value (153) exceeds the continuation value (133.70).

At the 210 node the exercise value (90) exceeds the contiuation value (76.60).

So the initial value of the American option is 39.73.

Solution 30

Answer: C


The value of gamma at time 0 for this option can be approximated as:

r(s h) = ,1(uS, h) - ,1(dS, h)

h' uS-dS

q This formula is only an approximation because it is actually estimating the value of gamma at time h, and

also because the estimate is based on discrete steps.

Here uS = 375 and dS = 210, and the deltas are estimated as:

( 5 h) -5 C(uuS,2h)-C(udS,2h) -0.065 0-41.00

1 u, =e =e x

uS - udS 468.75 - 262.5

-0.1863

and ,1(dS, h) = e-5 C(udS,2h)-C(ddS,2h)

udS-udS

e-o.065 x 41.00 -153 = -0.9087

262.5-147

So:

nSh,h)

-0.1863-(-0.9087)

375-210

0.0044

So the approximate value of gamma at time 0 is 0.0044.

Solution 31

Answer: E


We need to use a 2-step tree here with 6-month steps.

For options on futures, the formula for the risk-neutral "up" probabilty is:

"

p'

1-d

u-d

q The formulas for u, d and p * for an option on a future are the same as the corresponding formulas for a

stock, except that rand ö are set to zero.

So, using the information give in the question, we have:

* 1-d 2 u 4

P =-=- and -=-

u-d 3 d 3

So: 3(1- d) = 2( u - d) and 3u = 4d

Solving these simultaneous equations gives:

6 9

=-=l.2 and d=-=0.9

5 10

20





ExamMFE

We can now construct the following tree of futures prices and option values (in parentheses) for

the European option:

115.2 30.2)

96 10.7284)

80 3.7838)

86.4 1.4)

72 0.4551)

64.8 0)

For example, the payoff at the 115.2 node is calculated as maxt115.2-85,Ol = 30.2 and the value

at the 96 node is calculated as:

e-o.05x0.5 LP* x30.2 + l-p* )Xl.4 J = 10.7284

This leads to an intial value for the European option of C¡ = 3.7838.

The payoffs at maturity are the same for the American option as for the European option.

However, at the earlier nodes, we need to examine each node and replace the calculated value

with the exercise value if this is higher. This leads to the following tree for the American option:

115.2 30.2)

96 11)

80 3.8721) 86.4 1.4)

72 0.4551)

64.8 0)

For example, the exercise value at the 96 node is 96 - 85 = 11, which exceeds the calculated value

of 10.7284, and the value at the 80 node is calculated as:

e -o.05x0.5 LP * x 11 + 1- P * ) x 0.4551 J = 3.8721

This leads to an initial value for the American option of C II = 3.8721 .

The difference in value between the American and European options is therefore:

Cii -C¡ = 3.8721-3.7838 = 0.0883

q Because any diference in the two prices is due to early exercise at the 96 node, the final answer is just the

diference between the exercise value of 11 and the continuation value of 10.7284, multiplied by the risk-

neutral probability and discounted.


21




ExamMFE


Chapter 4 Solutions

Solution 1

Answer: A

Difficulty level: .

The values of d1 and d2 are:

In se -òl ) + J2T In 50e -0.04 0.5) ) + 0.252 0.5)

Ke-rT 2 52e-0.07(0.5) 2

d1 = J-f 0.25.J

d2 = d1 - J-f = -0.04862 - 0.25.J = -0.22540

-0.04862

We use d1 and d2 rounded to 2 decimal places to find the values of N(d1) and N(d2) :

N d1) = N -0.05) = 0.4801

N(d2) = N( -0.23) = 0.4090


CEur = Se-òlN(d1)- Ke-rTN(d2)

= 50e-o.04(0.5)(0.4801)_52e-o.07(0.5)(0.4090) = 2.99

Using unrounded values for d1 and d2 would give a more accurate answer of2.93.

Solution 2

Answer: B


The dividend occurs after the option expires, so it does not affect the calculation of the put price.


( Se -òl ) (J2T

In-+-

Ke-rT 2

d1 = -f

J T

in 50 ) 0.252 0.5)

52e -0.07(0.5) + 2

0.25.J

0.0645 = 0.06 to 2 decimal places)

d2 = d1 - J-f = 0.0645-0.25.J = -0.1123 = -0.11 to 2 decimal places)

Ths gives:

N -d1) = N -0.06) = 0.4761

N( -d2) = N(O.l1) = 0.5438


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The value of the put option is:

PEur = Ke-rTN(-d2) - Se-õlN(-d1) = 52e-o.07(0.5)(0.5438)_ 50(0.4761) = 3.50

Using unrounded values for d1 and d2 would give a more accurate answer of 3.64.

Solution 3

Answer: C


In(SjK) + (r-ö+O.5a.2)t

d1 =

(jJt .

In(100/98) + (0.045 - 0.02 + lO.32)X0.25

= ~ = 0.2514 (= 0.25 to 2 decimal places)

0.3vO.25

d2 = d1 - (jJt = 0.2514 - 0.3.,0.25 = 0.1014 (= 0.10 to 2 decimal places)

N(-d1) = 1- N(0.25) = 1- 0.5987 = 0.4013

N(-d2) = 1- N(0.10) = 1- 0.5398 = 0.4602

~ P = K e-rt N( -d2) 5 e-t5t N( -d1) = 4.67

- ,

8xe -0.045x0.25 x0.4602 100xe -0.02x0.25 xO.4013

In this solution, using unrounded values for d1 and d2 would give the same answer, to 2 decimal places.

Solution 4

Answer: E


Since we are given that 5 =K (ie at-the-money ), according to the Black-Scholes formula:

In(SjK) + (r-ö+O.5~)t

d1 = r;

(jvt

In(l) + (0.06 - 0 + 0.5xO.25)x0.5

= ~ =0.2616 (=0.26

to 2 decimal

places)

O.5vO.5

~ N(d1) = N(0.26) = 0.6026

d2 = d1 - (jJt = 0.2616 - 0.5.J = -0.0919 (= -0.09 to 2 decimal places)

~ N(d2) = N(-0.09) = 0.4641

~ 8 = C = Se-t5t N(d1) - Ke-rt N(d2) ~ 5 = 52.56

'- ~

x1 xO.6026 Sxe-0.03xO.4641


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Solution 5

Answer: B


We ve given a full solution to this problem below. However, the quick way to solve it is to note that the

question does not tell us the value of the risk-free interest rate r. For the question to be valid, this means

that the value of the put option must be independent of the value of r. So we can assume any convenient

value, such as r = 0, and work out the value of the put option in the usual way The values to use for the

other parameters are: 5 = 90, K = 90, T = 2, a = .J0.25 = 0.5, Ö = 0 .

Since r is unknown we must apply a different formula than the usual Black-Scholes formula.

Notice first that we have the following:

(r-ô -0.50.2)t + a.j Z

S(t) = 5(0) e whereZisN(O,l)

This is the solution to the SD E for geometric Brownian motion. See Example 7.5 in the Course Notes.

ö=o ~ e-rt S(t) = e(0-0-0.5c?)t +a.jZ

5(0)

price at tim~ t if 5(0)=1

andr=ô=O

The value (price) of the put option is:

P = e-2r EL maxi 0, S(0)e2r - S(2)JJ

= EL maxi 0,5(0)- e-2rS(2)lJ

= 5(0) El maxi 0, 1- -;(~\2) l j

price of a 2-year EtÎropean put option

where K=l, the initial stock price is 1,

and r=ô=O

Applying the Black-Scholes formula, we have:

Eimaxi0' 1 - e-2rS(2)lJ = K x e-2r x N( -d2) - 5 x e-25 x N( -d1) = 0.2736

5 (0) -- - - .. - -

. . 1 1 0.6368 1 1 0.3632

price of a 2-year E~rolean put option

where K=l, the initia stock price is 1,

and r=ô=O

In(S / K) + (r-ö + 0.5a2)

2

d1 = M

',,2

0+(0-0+0.50'2)2

O'l2

= 0.5O l2 = 0.5 x 0.5 x.f = 0.3536 (= 0.35 to 2 decimal places)

'-

iv)=:

c?=0.25

~ N( -d1) = 1- .6368 = 0.3632

d2 = d1 - O'l2 = -0.3536 (= -0.35 to 2 decimal places)

~ N(-d2) = 0.6368


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Finally, we have:

p = S (0) El ma 1 0 , i - '-:'(~ \2) r J = 90 xO.2736 = 24.62

price of a 2-year E¿ropean put option

where K = 1, the irutial stock price is 1,

and r=8=0

Solution 6

Answer: C


FP (K) =50e-rt = 50e-0.025 = 48.766

FP (5) = 50 - 2e-o.25r = 48.025

In(FP (S)/FP (K)) +l0 2t

d1 = lt

0 t

In

(0.9848) +lxO.32 xO.5

= k = 0.0339 (= 0.03 to 2 decimal places)

.3 0.5

~ N(d1) =0.5120

d2 = d1 -O lt = 0.0339- 0.3.J = -0.1178 (=-0.18 to 2 decimal places)

~ N(d2) = 0.4286

C = FP (5) N(d1) - FP (K) N(d2) = 3.69

'- '- ~ '-

8.025 0.5120 48.766 0.4286

Solution 7

Answer: B

Difficulty level: .

The value of d1 is:

in(~)+~

d1 = r;

O vT

In (50e -0.04(0.5)) + 0.252(0.5)

52e -0.07(0.5) 2

= Ii = -0.04862 (= -0.05 to 2 decimal places)

0.25 10.5

So: N(d1) = N(-0.05) = 0.4801

The value of delta for the put option is:

~ = e-OlN(d1)- e-Ol = e-o.04(0.5)(0.4801)_e-0.04(0.5) = -0.510

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Solution 8

Answer: C


Since the futures contract expires on the same date as the call option, the European futures option has the

same value at the expiration date as the European stock option.


In Se -òl ) + c?T In 75e -0.03 ) + 0.302 1)

Ke -rT 2 80e -0.08 2

d1 = ¡; .J = 0.1015 = 0.10 to 2 decimal places)

a-vT 0.30 1

d2 =d1 -a-Jf =0.1015-0.30.J =-0.1985 =-0.20 to 2 decimal places)

So: N d1) = N 0.10) = 0.5398

N d2) = N -0.20) = 0.4207


CEur = Se-otN di)-Ke-rTN d2)

= 75e-D.ü(0.5398)-80e-0.08(0.4207) = 8.22

Solution 9

Answer: D

Difficulty level: .

The Black-Scholes

formula

for the value ofa put option is P=Ke-rTN(-d2)-Se-òlN(-d1)' Since the

delta

for a put option is il=-e-òlN(-d1), this can be written as P=Ke-rTN(-d2)+ilS. So the put

option is instantaneously equivalent to a cash position of Ke-rTN(-d2) and il units of the underlying

asset.

To replicate a European put option, the amount that must be lent at the risk-free rate of return is:

Ke-rTN(-d2) = 80e-0.08(0.75)N(_0.87) = 80e-0.08(0.75) (0.1922) = 14.48

Solution 10

Answer: E

Difficulty level: . .

We don't need to determine the dividend yield to solve this problem.

Delta for a put option can be written as:

il = -e-òlN -d1)

The Black-Scholes formula for a European put option can be rewritten in terms of delta:

PEur = Ke-rTN(-d2)-Se-òlN(-d1)

PEur = Ke-rTN -d2)+

Sil

Ths equation can be used to solve for delta:

PEur =

Ke-rTN(-d2)+

Sil

2.54 = 95e -0.08(0.25) (0.3483) + 100il

il =-0.299


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Solution 11

Answer: B


We first need to determine delta for a purchased three-month 40-strike European put from the

put-call parity relationship:

C - P = 5 e -ot _ Ke -rt ~ ac _ ap = e -ot = eO = 1

as as

ap

~ - =

as

~

put

ac -1 = 0.5824 - 1 = -0.4176

as

'-

~call

Now any Greek measure of a position is the sum of the individual Greek measures of the options

making up the position. Furthermore any Greek of a written option is the negative of the Greek

for the purchased option. So for Dave s position, the delta value is:

~position = lx(-~call) + lX(-~put) = -0.5824 +0.4176 = -0.1648

Solution 12

Answer: D

Difficulty level: .

The value of the position is:

v = (-C) + (-P)

So the approximate change in value is:

dV av x dS = _(ac + ap) x

as dS dS

'-

position = -0.1648

dS

'-

9.625-40.00

= 0.0618

Solution 13

Answer: B


Following up on Solution 11, we have:

ac ap a2c a2p

---=1~ ---

as as as2 as2

a2p a2c

:: 0 ~ -=-=0.0652

as2 as2

So the gamma for Dave's position is:

a2v (a2c) ( a2p)

-= -- + -- =(-0.0652)+(-0.0652)=-0.1304

as2 as2 a2s2

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Solution 14

Answer: C


From general theory, this ratio is the absolute value of the elasticity of the position.

To compute the elasticity, we begin by computing the value of the position:

Vposition = -C - P = -(2.7804 +1.9883) = - 4.7687

since: P = C + K e-rt _Se-ot = 2.7804 + 40e-o.02 - 40 = 1.9883

Using the value of delta calculated in Solution 11, the elasticity of the position is thus:

5 ~position

Oposition =

Vposition

40 x ( -0.1648)

- 4.7687

= 1.3823

Solution 15

Answer: E

Difficulty level: .

The volatilty of a position is the absolute elasticity of the position times the volatility of the

underlying stock:

O position = I Opositionl X O stock = 1.3823 x 0.30 = 0.4147

Solution 16

Answer:D

Difficulty level: .

The return y for this position can be determined as follows:

y= Inl x a

~ '-

.3823 0.15

+ (l-lnl)

'-

0.3823

x r

'-

0.08

= 0.1768

The value of n comes from Solution 4.15.

Solution 17

Answer: A

Difficulty level: .

The Sharp ratio for any option position based on a given stock is the same as the Sharp ratio for

the underlying stock:

y- r

Sharp ratio =

O position

_Inl(a-r)

- InlO stock

a-r

0.15 - 0.08 = 0.233

0.30

stock


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Solution 18

Answer:E


The prepaid forward prices for the stock and the strike asset are:

Ft,T(S) = 5 -PVO,T(Div) = 41_4e-D.08*0.5 = 37.1568

Ft,T K) = Ke-rT = 45e-D.08 O.75) = 42.3794


In ( F~T(S) J + (j2T

FO,T K) 2

d1 = ¡;

(j T

In (37.1568) + 0.32(0.75)

42.3794 2

- 0.3..0.75

d2 = d1 - j.f =-0.3763-0.3,ß = -0.6361 = -0.64 to 2 decimal places)

-0.3763 (= -0.38 to 2 decimal places)

So: N d1) = N -0.38) = 0.3520

N(d2) = N( -0.64) = 0.2611

The value of the call option having the same strike and time until expiration as the put option is:

CEur = Ft,T S)N d1)-Ft,T K)N d2) = 37.1568 0.3520)-42.3794 0.2611) = 2.0139

We can use put-call parity to find the value of the put option:

PEur = CEur + Ft,T K) - Ft,T S)

PEur = 2.0139 + 42.3794 - 37.1568 = 7.24

Solution 19

Answer: D


The prepaid forward price of the Euro is:

Ft,T x)=xoe-r¡T =1.05e-D.032 1) =1.0169


in Ft,T S) J+ ~T

Ft,T K) 2

d1 j.f

in 1.0169 )+ 0.12 1)

100e-O.06 2

. -/ = 0.8179 = 0.82 to 2 decimal places)

.1 1

d2 =d1 - j.f =0.8179-0.1-/ =0.7179 =0.72 to 2 decimal places)

So: N d1) = N 0.82) = 0.7939

N(d2) = N(O.72) = 0.7642





ExamMFE


P ( ) () P -0.06(1)

Eur = FO,T x N d1 - FO,T(K)N(d2) = 1.0169(0.7939) -1.00e (0.7642)

=0.088

Solution 20

Answer: B

Difficulty level: . . .

The current value of Stock X can be found using the information about Option B:

Q= SL1

(Option value)

5.0687 = 5(0.4830)

3.8117

~ 5 =40.0008

Now that we know 5, we can find delta for Option A:

Q= SL1

(Option value)

4.3272 = 40.0008L1A

5.8868

~

L1A = 0.6368

The delta for the portfolio is just the sum of the individual deltas:

L1Portfolio = OJAL1A + OJL1ß + ltL1c = 1(0.6368)+ 1(0.4830)+ 1(-.2710) = 0.8488

The elasticity of the portfolio is:

Q _ SL1Portfolio _ 40.0008(0.8488) _ 2 82

ortfolio - ( . 1) - - .

Option va ue Portfolio 12.0426

Solution 21

Answer: A


In(85/80) + (0.06-0+0.5X0.52)xi

d1 = ¡: = 0.4912 (= 0.49 to 2 decimal places)

0.5,,1

~ N(-d1) = 1- 0.6879 =0.3121

d2 =d1 -0.5.f = -0.0088 (= -0.01 to 2 decimal places)

~ N(-d2) = 0.5040

~ P = Ke-rt N(-d2) - 5 e-Ot N(-d1) = 11.44

- - ~

0xe -0.06 xO.5040 85 -~put = 1 xO.3121

~ Q = SL1put = 85 x (-.3121) = -2.318

P 11.44

~ O put = O stock x I Q I = 0.5 x2.318 = 1.159


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Solution 22

Answer: C


For a call option all of the Greeks are positive, except for Theta and Psi. Psi is not relevant for a

non-dividend-paying stock (and is not one of the possible answers in any case).

The question says usually because it is actually possible for the theta for a European call option to be

positive for some extreme sets of parameter values.

Solution 23

Answer: B


The delta of a call option in the Black-Scholes model is N d1) and the delta of a put option is

N -d1) = N d1 )-1.

The value of d1 is:

In ( Ft,T (5) J + (J2T

lFt,T K) 2

d1 = r;

(J..T

1 87 ) 0.22 0.25)

+

0.25 0.08) 2

75e -J = 1.7342 = 1.73 to 2 decimal places)

0.2 0.25

So the deltas for the call and put options are:

N(d1) = N(1.73) = 0.9582

N(d1)-1 = 0.9582-1 = -0.0418

Adding these gives the delta of the portfolio:

0.9582+ -0.0418) = 0.9164


Solution 24

Answer: A


Without the aid of a computer, the only way to find the implied volatility in this example is by trial and

error. One approach is to try the middle value ((J = 35%) first and then use the fact that options have a

positive vega, which means that increasing the volatility will increase the price.

Try (J = 35% .


In ( F~T(S) J + (J2T

L FO,T K) 2

d1 = r;

(J..T

in 110 )+ 0.352 0.5)

120 0.5 0.12-0.06) 2

.J = -0.3491 = -0.35 to 2 decimal places)

0.35 0.5

d2 = d1 - J.f = -0.3491-0.35.J = -0.5965 = -0.60 to 2 decimal places)

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So the values of N(d1) and N(d2) are:

N(d1) = N(-0.35) = 0.3632

N(d2) = N (-0.60) = 0.2743


CEur = F6,T(S)N(d1) - Frl,T(K)N(d2)

= 110e -0.12 0.5) x 0.3632 -120e -0.06 0.5) x 0.2743

=5.68

This option value is higher than the desired 4.16. So we can eliminate Answer C and tr one of

the lower values, say (j = 30 .

Now, the values of d1 and d2 are:

In ( F6,T(S) J + (j2T

L F6,T K) 2

d1 = (j.f

1 ( 110 ) 0.302(0.5) 120e0.5(0.12-0.06) + 2

0.3-J

d2 = d1 -(j.f =-0.4455-0.30-J =-0.6577 (= -0.66 to 2 decimal places)

-0.4455 (= -0.45 to 2 decimal places)

So the values of N(d1) and N(d2) are:

N(d1) = N( -0.45) = 0.3264

N(d2) = N (-0.66) = 0.2546


CEur = F6,T(S)N(d1) - F6,T(K)N(d2)

= 110e -0.12(0.5) x 0.3264 -120e -0.06(0.5) x 0.2546

=4.16

This is the desired result.

If this answer had turned out to be too low, we would have concluded by elimination) that the answer

must be (j = 32% .

Solution 25

Answer: D

Difficulty level: .

For both call options and put options on stocks and stock indices, we observe that implied volatilities are

higher when the strike price is lower, and lower when the strike price is higher.

A and C don't really make much sense at all

B is factually incorrect. Put option prices increase as the strike price increases, which corresponds

to lower implied volatilties. So this observation is the wrong way round.

E is factually incorrect. Call option prices increase as the strike price decreases, which

corresponds to higher implied volatilities. So this observation is the wrong way round.


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Solution 26

Answer: A

Difficulty level: .

It is the continuously-compounded returns on the stock (or equivalently, the log-price) that are

normally distributed, not the stock price itself.

The stock price itselfhas a lognormal distribution.

Solution 27

Answer: C

Difficulty level: .

A calendar spread involves buying one option and selling another option that differs only in

maturity date.

So there are six possible combinations (or twelve, if we take into account which of the options is

bought and which is sold). These are 30 and 34 (with a cost of :t 4), 1 and 3 (with a cost of

:t 2) etc.

The - 3 (= + 22 - 25) would arise if we sold the July put (strike 300) and bought the June put

(strike 300). This would generate a cash flow of 3 in our favor (corresponding to a negative

cost).

Solution 28

Answer: C

Difficulty level: .

Applying the Black-Scholes formulas:

d1

100 1 2) 6

loggg+ 0.055-0.01+zx0.5 Xu

0.5ll

0.2976 (= 0.30 to 2 decimal places)

and d2 = d1 -0.5& = -0.0560 (= -0.06)

So the price of the put option is:

P = Ke -rT N( -d2) - Se -õT N( -d1)

= 98e -o.055xi6i N(0.06) -100e -o.OlxA N( -0.30)

= 95.34

x

0.5239 -99.50xO.3821

=11.93

Rounded to the nearest LOLL, this is 11.90.

Solution 29

Answer:D


The volatilty of the call option can be calculated as:

51)

(Tcall = (Tstock xl Qcalll where Qcall =c (= the elasticity of the call option)

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Using the Black-Scholes formulas:

log 85 +( 0.055-0+iX0.52 )xi

d = 80 = 0.4812 = 0.48 to 2 decimal places)

0.5.f

and d2 = d1 -0.5.f = -0.0188 (= -0.02)

Using the formula for the delta of a call option and the price of a call option:

A = N d1) = N 0.48) = 0.6844

and

C = SA-Ke-rTN(d2)

= 85xO.6844-80e-o.055N(-0.02)

= 85xO.6844-75.72x0.4920

= 20.92

So:

Q _ SA _ 85xO.6844 2.781

all - C - 20.92

Finally, the volatility of the call option is:

(jeall = (jstoek xl Qealll = 50 x 2.781 = 139

Solution 30

Answer: C

Difficulty level: .

Using the Black-Scholes formula:

log 5 0) + r-0+.1 j2)T

S O)erT 2

d1 = j.J

~ +(/-0+i(j2)T

(j.J

=i(j.J

Statement (ii) tells us that T = 10 and statement (iii) tells us that (j2 = 0.4 .

So: d1 =ix.JJI =ix.J = 1

and d2 =d1 -.JJI =-1

So the price of the call option is:

C = S(0)N(d1)-Ke-rTN(d2)

= S 0)xN d1)-S 0)/ / N d2)

= 100xN(1)-100N(-1)

= 100xO.8413-100xO.1587

= 68.26


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Solution 31

Answer: C

Difficulty level: .

To deal with the discrete dividend, we can apply the Black-Scholes formula to the stock price

with the present value of the dividend deducted. Ths equals:

5* = 50 -1.50e -0.05xA = 48.52

Applying the Black-Scholes formulas:

log 48.52 +( 0.05-0+tXO.302)x 162

50 = 0.0823 (= 0.08 to 2 decimal places)

0.30ll

and d2 = d1 -0.30 = -0.1294 (= -0.13)

d1


P = Ke -rT N( -d2) - Se -oT N( -d1)

= 50e-o.025N(0.13)-48.52N(-0.08)

= 48.77x0.5517 -48.52x0.4681

=4.19

Solution 32

Answer:

A


The investor is closing out his position today, when the stock price is 85 and the call option has

4 months left. We can use the Black-Scholes formulas to find the value of the option now:

log 85 +( 0.05-0+lXO.262)x A

d - 75

-

0.26.j

and d2 = d1 -0.26.j = 0.8698 (= 0.87)


So the price of the call option now is:

C = Se-oTN(d1)-Ke-rTN(d2)

= 85 N(1.02) -75e -o.05xA N(0.87) = 12.33

- -

.8461 0.8078

The holding profit is the accumulated value of all of the investor's cashfows:

-8e 0.05xA + 12.33 = -8.27 + 12.33 = 4.06

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Solution 33

Answer: A


The formula for delta for a put option based on the Black-Scholes model is I:put = -e -òT N (-d1 ) .

Since this is a nondividend-paying stock:

I:put = -N(-d1) = -0.4364

So: N(-d1) = 0.4364 and N(d1) =1-0.4364 = 0.5636

From the table of the normal distribution, we see that this corresponds to 0.16.

So:

5 ( 1 2)

log K + r-O+ 2o- T

d1= =0.16

o-.f

We are told that T = 1 and r = 0.012. Since it is an at-the-money option, we also know that

K=S.

So:

log%+( 0.012 -0+10-2) (1)

¡; = 0.16

0-",1

0.012+l0-2

2 =0.16

0-

Rearranging this and solving the resulting quadratic equation gives:

0.50-2 -0.160-+0.012 = 0

0.16: ~ (-0.16)2 - 4(0.5)(0.012)

0-= =0.16: 0.04

=

0.20 or 0.12

2(0.5)

Since we are given an upper limit for the option price in (i), the correct value of the volatilty

must be the value that leads to the lower option price. This wil be the lower value, ie 0.12.

You can check this by doing the calculations. You wil find that the put option price works out to 0.07345

when 0- = 0.20 and 0.04185 when 0- = 0.12. Only the latter is less than 5% of the stock price.

Solution 34

Answer: D


To deal with the discrete dividend, we can apply the Black-Scholes formula to the stock price

with the present value of the dividend deducted.

The adjusted stock price equals:

5* = 50 - 5e -o.12x 2 = 45.43


15



ExamMFE


Statement (iii) tells us that (J2 = 0.01, ie (J = 0.1. Applying the Black-Scholes formulas then gives:

45.43 ( 1 2)

log--+ 0.12-0+2xO.l xl

d1 = = 1.3451 (= 1.35 to 2 decimal places)

0.1~

and d2 = d1 -0.1~ =1.2451 = 1.25)


P = Ke-rT N(-d2)-S * N(-d1)

= 45e-0.12 N(-1.25)-45.43N(-1.35)

- -

.1056 0.0885

= 0.1941

Solution 35

Anwer: C

Difficulty level: .

The values of d1 and d2 in the Black-Scholes formula are:

5 -õT)

In ~ +l(J2T

Ke-rT 2

d1 =

(J.f

20e -0.03 0.25)) 1 2

nl25e--.05(0.25) +2(0.24) (0.25)

0.24~0.25

-1.76

and d2 = d1 - J.f = -1.76 - 0.24~0.25 = -1.88

(5) ( 1 2)

In - + r-ö+2(J T

Note that this formula for d1 is equivalent to d1 = K .f

J T

So the value of one call option is:

CEur = Se-õTN(d1 )-Ke-rTN(d2)

=20e--.03(0.25) N(_1.76)_25e--.05(0.25) N(-1.88)

- -

.0392 0.0301

= 19.85N(d1 )-24.69N(d2) = 0.0350

So the value of 100 call options wil be 100xO.0350 = 3.50.

Solution 36

Answer: D


Remember that an American call option on a nondividend-paying stock has the same value as its European

counterpart. So we can treat this as a European option and use the usual Black-Scholes formula.

The Black-Scholes formulas for the value of a call option and its delta are:

CEur = Se-õT N(d1) - Ke-rT N(d2) and 11 = e-õT N(d1)

16





ExamMFE

We are not told the interest rate r, but we can work this out from the value given for delta. Since

there are no dividends, the formula for delta is:

d = N(d1) = 0.5

=? d1 =0

=?

in f)+ r-ö+l~)T =0

(Jff

Substituting the parameter values given:

in(~)+(r+lxo.302 )(0.25)

41.5

0.30.J0.25 .

o

=? r=0.1023

We also need d2, which is:

d2 =d1 - Jff =0-0.30.J0.25 =-0.15

So the value of the call option is:

CEur = 40N(d1) _41.5e-D.1023(0.25)N(d2)

= 40 é -40.453N(-0.15)

0.5

= 20-40.453N(-0.15)

We could easily look up the value of N(-0.15), which equals 0.4404, to finish off the calculation. This

gives a value of 2.18 for the call option. However, the choices require us to express the answer in terms of

an integral.

The N function is the cumulative distribution function for the standard normal distribution.

Since all the choices have an integral with an upper limit of plus 0.15, we can write:

fÛ.15 1 -lx2

N -0.15) = 1-N 0.15) = 1- 100 .J e 2 dx

o:

So we can write the value of the call option as:

1 rO.15 -lx2 )

C Eur = 20 - 40.453 1- .J J-o e 2 dx

1 rO.15 -lx2

=40.453 r, 10 e 2 dx+20-40.453

,,2íC

fÛ.15 _2 2

= 16.138 La e 2x dx - 20.453

Solution 37

Answer: A

Difficulty level: .

Recall that the historical volatility of an asset is the annualized sample standard deviation of the past

changes in the log-asset price. If the asset price follows geometric Brownian motion, this gives an estimate

of the volatility parameter (J .

The monthly log-returns for each period, calculated using the formula In ( St+h 1St) , were:

-22.3 +22.3 -22.3 +22.3 +22.3 , -22.3 , -22.3 , -22.3


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ExamMFE


So the monthly volatilty is:

.2 L1ln(St+h/Sd12

n-l

~XL(-22.3 )2 +(22.3 )2 +...+(22.3 )2 J = ¡rX22.3 = 23.8

â h =

So the annualized volatility is:

23.8%xm =82.6%

Solution 38

Answer: C


Consider first the value of this option one year from now. Using the Black-Scholes formula with

K = 51' this wil be:

SlN(d1)-lS e-rN(d2)

=

51 L N(d1)-e-O.08N(d2)J

=51

where

in( ~ )+( 0.08-0+lxO.32 )(1)

d1 = 0.3 = 0.42

d2 = d1 -(5.J = 0.42-0.3 = 0.12

So the value of the option at time 1 is:

51 r ~_e-0.08 ~J=0.1571XS1

L 0.6628 0.5478

So, in one year's time, the option has the same value as 0.1571 units of the stock. We are told that

the one-year forward price for the stock is 100. So the fair price to pay for the option in one year's

time is 0.1571xl00 = 15.71. This corresponds to a price of 15.71e-0.08 = 14.50 now.

Solution 39

Answer: D


Method 1

The value of Investor B's portfolio is:

VB = 2Vcall -3Vput = 2x4.45-3xl.90 = 3.2

Since the delta for Investor B's portfolio is 3.4, we can calculate the elasticity:

QB = tiBxS = 3.4x45 =47.8125

VB 3.2

18




Question Answer Bank - Solutions, Chapter 4 ExamMFE

Elasticities for portfolios are calculated using the formula Qportjolio = I wiQi' where the

i

weightings are based on the relative values of the components in the portfolio.

So:

Q 2x4.45 Q 1.90 Q 50 (. )

x x = ven

2x4.45 1.90 call 2x4.45 1.90 put . gi

::

8.90 Qcall 1.90

Qput = 5.0xl0.8 = 54

and Q = 2x4.45 xQ + -3xl.90 xQ = 47.8125 (calculated above)

2x4.45-3xl.90 call ,2x4.45-3xl.90, put

2.78125 -1.78125

::

8.90 Qcall - 5.70 Qput = 47.8125 x 3.2 = 153

Subtracting these two simultaneous equations gives:

7.60 Qput = -99

99

::Qput =--=-13.03

7.60

Method

2

The value of Investor A s is:

VA = 2Vcall + Vput = 2x4.45+ 1.90 = 10.8

The elasticity of this portfolio is:

QA

'-

=5.0 (given)

L\A xS

==-=

VA

L\A x45

10.8

:: L\A = 5.0x 10.8 = 1.2

45

The delta for Investor A's portfolio is:

L\A = 2L\call + L\put

'-

=1.2

:: L\call = 0.6 - O.5L\put

The delta for Investor B s portfolio is:

~ = 2L\call -3L\put = 2(0.6-0.5L\pud-3L\put = 1.2 - 4L\put :: L\put = -0.55

=3.4 (given)

So the elasticity of the put is:

L\put xS

Qput =

Vput

-o.55x45 =-13.03

1.90


19



ExamMFE


Solution 40

Answer: A


The condition that (5(1))2 :;100 is equivalent to 5(1) :;10 . So this is a cash-or-nothing call

option.

The formula for the price of this option corresponds to the second component of the Black-

Scholes call formula:

v = 100e-rT N(d2)

So the delta of this option is:

~=av =100e-rT .iN(d )=100e-rTN'(d )ad2

as as 2 2 as

The formula for d2 can be written as:

in(~)+(r-ö-t0-2 )T

d2 =

o-Jf

In 5 -InK +(r-ö -to-2 )T

o-Jf

So:

ad2 1 1 = 0.5

as = So-Jf 10xO.2

N(d2) is the cumulative distribution function of the standard normal distribution. So N'(d2) is

its density function, which is:

N'(d2) = ~exp (-t dn

,,2íC

Here we have 5 = 10, K = 10, r = 0.02, ö = 0, 0- = 0.2 and T = 1, so that:

in(10)+( 0.02-0-tXO.22 )(1)

d - 10 0

- 0.2

and

N'(d2) = N'(O) =~

J2

So: ~=100e-rTN'(d2) ad2 =100e-O.02x ~XO.5=19.55

as ,,2íC

Rounding this to the nearest whole number gives 20 shares.

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ExamMFE


and

L50 ( 1 2)(2)j

In-+ 0.05+ixO.2 12

i1(60-strike call) = N(d1) = N 60 ll = N(-2.09) = 0.0183

0.2 12

So: i1(bull spread) = lxi1(50-strike call) -1 x

i1(60-strike call) = 0.5557 -0.0183 = 0.5374

So the change in delta is:

Change in delta = 0.5374-0.5219 = 0.0155

None of the choices given are ideal, but the nearest answer is B.

If you do this calculation exactly, without applying any rounding at all (eg using Excel), the answer works

out to 0.0187.

Solution 42

Answer: C


Consider, for example, the option that wil be held during the second quarter, between times 0.25

and 0.5. Using the Black-Scholes formula with K = 0.950.25, the value of this option at time 0.25

wil be:

l e-0.25rN(-d2)-SO.25N(-di) = 50.25 L 0.ge-o.02N(-d2)-N(-d1)J

=0.950.25

where d1

in( 50.25 )+(0.08-0+lxO.32 )(0.25)

0.95025

. =Q~

.3-J0.25

d2 =d1 -(j.. =0.91-0.3-J0.25 =0.76

So the value of this option at time 0.25 is:

50.251 0.ge-o.02 ~-~J = 0.01586xSO.25

L 0.2236 0.1814

So, at time 0.25, the option covering the period from time 0.25 to time 0.5 has the same value as

holding 0.01586 units of the stock. Since the stock doesn t pay any dividends, this means that the

value of this option now is the same as the value of 0.01586 units of the stock, which is

0.01586x45.

We can apply the same logic to the options for each of the four periods and conclude that the

total value now of the rolling insurance strategy is:

Value = 4xO.01586x45 = 2.85

22





ExamMFE

Solution 43

Answer: E Difficulty level: ..

We are told very little about the nature of the derivative. In fact, we don't even know its payoff So the

only principle we can rely on is the no-arbitrage principle.

In the Black-Scholes framework, the market is assumed to be arbitrage-free and the price C(St)

of any derivative must satisfy the Black-Scholes partial differential equation:

rc(Sd=l(j2s1rt +rStÓ-t +Bt

Here, we have:

C(St) = St-k/c?

A _ ac _ a (5 -k/(J2) __ k 5 -k/c?-l

Llt - -- t - - t

aSt aSt (j2

rt = aó- =~(_~St-k/(J2_1) =_~(_~_1)St-k/(J2-2

aSt aSt (j2 (j2 (j2

B = ac =.i(s -k/c?) = 0

t at at t

So the Black-Scholes PDE is:

5 -k/c? - 1 252 ( k)( k 1)5 -k/(J2_2 5 ( k 5 -k/(J2_1) 0

t - i(j t -- --- t +r t -- t +

(j2 (j2 (j2

~

-k/(J2 1 2 ( k) ( k ) -k/(J2 ( k -k/(J2 )

rSt = i(j - (j2 - (j2 -1St +r - (j2 St

Cancelling the St-k/(J2 factor, which appears throughout:

r =l /(-~J(-~-l)-r~

/ (j2 (j2

r =1.k(~+l)-r~

2 (j2 (j2

Multiplying through by ¿i and rearranging:

r(j2 =lk( k+(j2 )-rk

lk2 +(t(j2 -r) k-r(j2 = 0


23



ExamMFE


Solving this as a quadratic equation in k :

-(l0-2 -r):t~(l0-2 -rt -4(l)(-r0-2)

k=

2(l)

=-(l0-2 -r):t~(l0-2 _r)2 +2r0-2

=-(l0-2 -r):t~(l0-2 +rt

=- l0-2 -r):t l0-2 +r)

= 2r or _0-2

We can rule out the solution k = _0-2 because the question tells us that k is positive. So the

required solution must be the other solution, for which k = 2r = 2x 0.04 = 0.08 .

If k = _0-2, then - k/0-2 = 1 and the derivative price would be SF = St. So this derivative would just

be the stock itself

Solution 44

Answer: C


From the diagram in the question, we can see that the payoff for this claim is:

tS(l)

Payoff =

42

if 0:: 5(1) ~ 42

if S(l)? 42

We can also express this in the equivalent form:

Payoff = minrS(l),

421 = 42+minrS(1)-42,01 = 42-maxr 42-5(1),01

This shows that we can replicate the payoff from the claim using a combination of

42e-o.07 = 39.16 in cash and a short position in a 42-strike put option.

Alternatively, we can replicate the claim using a combination of shares and a short call option, based on the

equation Payoff = min r 5(1),421 = 5(1) +

min rO,42-S 1)1 = 5 1) -maxrO, 5 1) -421. However, the

portfolio we have chosen to use here is easier because the delta for a cash position is zero, which simplifes

the calculations.

We can value the put option using the Black-Scholes formula:

In (45)+( 0.07 -0.03+lxO.252) (1)

d - 42 056

- 0.25 .

d2 =d1 -o-.J =0-0.25 =0.31

24





ExamMFE

So the value of the put option is:

Vput = 42e-0.07 N -Q.31)-45e-0.03 N -Q.56)

=

2.25

~ ~

.3783 0.2877

We also need the delta for the put, which is:

ó.put = -e -òl N (-d1 ) = -e -0.03 x 0.2877 = -0.2792

So the value of the claim is:

Vclaim = Vcash - Vput = 39.16 - 2.25 = 36.91

and the delta of the claim is:

Ó.claim = Ó.cash - ó.put = 0.2792

~

0

So the elasticity of the claim is:

Q . _ SÓ.claim

claim - V

claim

45

X

0.2792

36.91

0.34

The alternative portfolio requires us to hold e 0.03 = 0.9704 shares and a short call option. The reason for

holding only 0.9704 shares is that, as we receive the dividends on this holding, we can reinvest them in the

share. If we start with 0.9704 shares, this will build up to a holding of 1 full share at the end of the year.


25




ExamMFE


Chapter 5 Solutions

Solution 1

Answer: D


Since the stock is sellng for

40 and since as = 1, a2; = 0, we have the following important

as as

inormation:

40-Strike Call

45-Strike Call

Stock

Price

2.7847

1.3584

40

Delta

0.5825 0.3285

1

Gamma

0.0651

0.0524

0

Theta

-0.0173

-0.0129

For a position with value V consisting of 100 written 40-strike calls and n shares of stock, we

have:

V = -100 C1 + n 5 , C1 = price of the 40-strike call

av (ac1) as

as = -100 as + n as = - 58.25 + n

To make this derivative equal zero we must purchase n = 58.25 shares of stock. The net cost of

this position is:

V = -100C1 + nS = -100 x 2.7847 + 58.25x40 = 2,051.53

So the amount to be borrowed at the risk free rate is 2,051.53.


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ExamMFE


Solution 2

Answer: D

Difficulty level: .

The stock increases in value by £=1.25 per share and dt is one day (i.e. 1/365 years). So using

delta, gamma and theta, the approximate option price is:

ac a2c 2 ac

C1 C1 (40,3 months) + -1dS + 0.5 -21 (dS) + -1dt

as as at

C1(40,

3 months) + L11£+ 0.5 r1 £2 + ()

= 2.7847 + 0.5825xl.25 + 0.5 x 0.0651 x 1.252 + (-0.0173)

= 3.5464

N I h. . h. d ac d h .. d d 1 ( d)

te: ntis question t eta is reporte as - x t were t is in years an t = - ,one ay.

at 365

The subscript 1 refers to the 40-strike call while the subscript 2 refers to the 45-strike call.

Solution 3

Answer: A


Using the results of Solution 1 and Solution 2, the new value of the shares and options after one

day is:

58.25 x

'-

shares

41.25 -

'-

rice/share

100 x 3.5464 = 2,048.17

'- '-

options price/option

The borrowed amount has grown in one day to:

2051.53 x (1 + 0.08X~) = 2,051.98

365

, ~ eO.OS;(1/365) ,

So there is a one-day profit of:

2,048.17 - 2,051.98 = -3.81

Solution 4

Answer:D


The stock is sellng for 40$, and we know that: as = 1, a2s = O. We have the following

as as2

important information:

40-Strike Call

45-Strike Call

Stock

Price

2.7847

1.3584

40

Delta

0.5825

0.3285 1

Gamma

0.0651

0.0524

0

2





ExamMFE

For a position with value V consisting of 100 written 40-strike calls, n1 purchased shares of

stock, and n2 purchased 45-strike calls, we have:

V = -100 C1 + n1 5 + n2 C2 , C1 = 40-strike call price, C2 = 45-strike call price

av aC1 as aC2

=? as = -100 as n1 as n2 as = -58.25 n1

0.3285

n2

'- '- '-

.5825 1 0.3285

a2v a2c1 a2s a2c2

=? -i = -100 -i n1 -- n2 -i = - 6.51 0.0524 n2

as as as as

- '- '-

.0651 0 0.0524

The second order partial derivative is zero if n2 =124.24. Substituting this result into the first

order partial derivative, and then setting it equal to zero, we have n1 = 17.44 .

Solution 5

Answer: A


For 40-strike call and put options, we have:

p= C Ke-rt_ 5 =2.7847

40xe-D.02-40

=

1.9927

ap ac

- = - -1 = 0.5825 -1 = -0.4175

as as

a2p a2c

as2 = as2 = 0.0651

For a position with value V consisting of 50 written 40-strike puts and n shares of stock, we

have:

V = - 50 P n 5 , P = price of the 40-strike put

av ap as

-=-50- n- =20.875 n

as as as

To make this derivative equal zero we must buy n =- 20.875 shares of stock (i.e. sell 20.875

shares). The net cost of this position is:

V = -50P nS = -50 x 1.9927 - 20.875x40 =-934.63

So the amount to be lent at the risk free rate is 934.63.


3



ExamMFE


Solution 6

Answer: C


Within the Black-Scholes framework we know that option price satisfies a second order partial

differential equation:

r x C =

'-

.08x2.7168

a2c

0.5 j2x S2x _

as2

0.sxO.302x 802 xO.0262

ac

+ rxSx-

as

'-

.08x80x0.3285

ac

+-~

at

ac = -9.4307

at

So the one-day change in the option price is:

ac x dt = -9.4307 x ~ = -0.0258

at 365

Solution 7

Answer: E

Difficulty level: .

With re-hedging every h years, the variance of a return Rh for a period of length h years is

proportional to h2 . Since 2 days is equal to 48 hours and since var (Rh ) = 0.62 when h is 1 hour,

it follows that var R48h) = 0.62 X 482 when re-hedgig is done every 2 days ie every 48 hours).

So the standard deviation is:

0.6x48 = 28.8

Solution 8

Answer: C


We could perhaps assume that trading is permitted 24 hours a day. But, for the sake of completeness, the

solution below derives the number of

hours that trading is permitted.

If rebalancing occurs every h / n years, the formula for the standard deviation of the rate of

return earned over a period of length his:

j Rh) = In x Ji S2 j2rh)

For daily re-hedging, we are given that the daily standard deviation of returns is 2.088:

1

h=- and n=l

365

~ s2 j2r~) = 2.088

J 365

4





For hourly re-hedgig, we are given that the hourly standard deviation of returns is 0.087. If the

number of hours in the day is m, then:

h= 1 and n= 1

365xm

-- S2 j2r 1 ) = 0.087

2 365xm

2.0882- = 0.087

m

m=24

If the market-maker re-hedges hourly, then the daily standard deviation of returns can be found

as follows:

1

h=- and n=24

365

j R 1 ) =~x-- s2 j2r.J) =~X2.088 = 0.426

365 Æ J2 365.J

Solution 9

Answer: B


The value of the option position when n shares of stock are included is:

v = -50C70 + 60C75 + nS

A delta hedge is established with the purchase of n shares of stock where:

av = _ 50 ac70 + 60 aC75 + n

as as as

= - 50 x 0.569 + 60 x 0.432 + n = 0

=? n = 2.53

The value of the delta-hedged position in calls and shares is thus:

V = -50C70 + 60C75 + nS = -50x6.060 + 60x3.990 + 2.53x 71 = 116.03

Ths is the investment required of the market maker to delta hedge the option position.

Solution 10

Answer: B

Difficulty level: .

We are given the values of the three Greeks that appear in the Black-Scholes partial diferential equation,

which suggests we might be able to work from this equation.

We can use the Black-Scholes PDE to find the value of the put option:

1 2 2

rc(Sd = -(j St rt + rSt~t + 0t

2

1.(0.252)( 402 )(0.073) + 0.18( 40)( -0.337)+ (-1.017)

C St) = 2 1.15

.18

In this formula, C( St) is used generically to indicate the option price. Therefore, it is the put price, not the

call price.


5



ExamMFE


Solution 11

Answer: D


The market-maker's profit for one put option can be estimated using the following formula:

Market-maker profit -.:ë2rt - Bth- rh( i1tSt -C(St))

2

=-.:12 0.013)- -1.346)2-- 0.06 -0.048) 50) -0.158)

2 365 365

= -0.00650 + 0.00369 + .00042

=-0.00239

Since the market-maker wrote 100 options, the profit is:

100 x -0.00239 = -0.239

An alternative approach is to find the market-maker's initial hedged portfolio, which consists of -100 puts,

-4.8 shares and 255.80 cash. Initially this has a total value of zero. After 1 day the value of each put wil

be 0.158-0.048x(-1)+t(0.ü3)X12 _1.346 =0.20881. So the value of the portfolio the next day wil be

365

-100xO.20881-4.8x49 + 255.80i.06/365 = -0.239.

Solution 12

Answer: D

Difficulty level: .

We can use the delta-gamma-theta approximation:

1 2

C(St+h) C(St)+~t+-ë rt+hBt

2

= 3.35 + 3) 0.569) +.: 32 ) 0.052) + -.) -7.409) = 5.15

2 365

In fact, we didn't need to use the risk-free rate in this calculation.

Solution 13

Answer: C

Difficulty level: .

A call bull spread consists of

buying the low-strike call and selling the high-strike call. We met these in the

proof for Theorem 1.1 in Chap ter 1.

The cost of the call bull spread is:

Cost of bull spread = 6.06-3.99 = 2.07

The delta of the call bull spread is:

i1Call bull spread = 0.569 - 0.423 = 0.137

Ths is hedged by sellng 0.137 shares for:

0.137x70 = 9.59

The required investment is negative in this case because the amount received from the sale of the

shares exceeds the cost of buying the bull spread:

Investment = -9.59+ 2.07 = -7.52

6

(g BPP Professional Education




Solution 14

Answer B


Step 1 is to determine the gamma of the position to be hedged:

n

rposition = LWiri =-lxO.0235

=-0.0235

i=l

Step 2 is to purchase enough of the 105-strike option to bring the portfolio gamma to zero:

Number of 105-strike options to purchase = r Position = 0.0235 = 1.2434

r105-stike 0.0189

Notice that the gamma of the portfolio is now zero:

r' = -1(0.0235) + 1.2434(0.0189) = 0

Step 3 is to determine the delta of the new, gamma-neutral portfolio:

f.' = -1(0.6917)+ 1.2434(0.4963) = -0.0746

Step 4 is to sell a number of shares of stock equal to the delta found in step 3. In this case, that

delta is negative, so this is equivalent to buying 0.0746 shares.

Step 5 is to determine the cost of establishing this position:

Cost = Cost to establish original position

+ Cost to bring gamma to zero + Cost to bring delta to zero

Cost = -1 9.39)+ 1.2434 7.33)+0.0746 100) = 7.18

The investment is $7.18.

Solution 15

Answer: E


Step 1 is to determine the gamma and rho of the position to be hedged:

n

rpositzon = Lwzrz = -1

x

0.0235 = -0.0235

z=l

n

PPosztion = LWzPi = -lxO.1490 = -0.1490

z=l

Step 2 is to determine how much of the 105-strike call and the 100-strike put must be purchased

to bring the portfolio gamma and rho to zero.

Let:

x = amount of the 105-strike call to purchase

y = amount of the 100-strike put to purchase

Then the system of equations to solve is

0.0189x + 0.0462y = 0.0235

0.2086x-0.0408y = 0.1490


7




ExamMFE

The interest cost is:

12.44 eo.06/365 -1) = 0.00205

The profit can be calculated as the gain on the stock plus the gain on the put bear spread minus

the interest cost:

0.274 - 0.27 - 0.00205 = 0.00195

Since the put bear spread was purchased on 1,000 shares, the investor s gain is:

0.00195xl,OOO = 1.95

Solution 18

Answer:

A


How can we match 3 variables (delta, gamma, and vega) using only 2 hedge assets (the stock and the

100-strike call)? At first blush, it appears that we do not have enough assets. But the gamma and vega of

the 95-strike call are proportional to the gamma and vega of the 100-strike call, so if we hedge gamma, we

have also hedged vega. (If two options on the same underlying asset have the same term and implied

volatility, the ratio vega/gamma based on the Black-Scholes formula is 52 (j(T - t), which is independent of

the strike price.)

The best way to tackle this kind of problem is to go ahead and solve for the number of options needed to

create a gamma-neutral portfolio. If that number of options also creates a vega-hedged portfolio, then we

can solve the problem.

Step 1 is to determine the gamma and vega of the position to be hedged:

n

rposition = ¿aJiri =-lxO.0170=-0.0170

i=l

n

Vegaposition = ¿aJiVegai =-lxO.2510=-0.2510

i=l

Step 2 is to determie how much of the 100-strike call must be purchased to bring the portfolio

gamma and vega to zero.

Let: x = amount of the 100-strike call to purchase

Then the system of equations to solve is:

0.0187x = 0.0170

0.2761x = 0.2510

The solution (using either equation) is:

x=0.90909

Since the solution is the same using either equation, the system of equations has a solution, and

therefore we can gamma- and vega-hedge using just the 100-strike call option.

Step 3 is to determine the delta of the new, gamma- and vega-neutral portfolio:

/1 = -1(0.6793) + 0.90909(0.5880) = -0.14475

Step 4 is to sell a number of shares of stock equal to the delta found in step 3. Therefore, the

market-maker buys 0.14475 shares.


9



ExamMFE


Step 5 is to determine the cost of establishing this position:

Cost = Cost to establish original position

+ Cost to bring gamma and vega to zero + Cost to bring delta to zero

Cost = -1(12.26)+ (0.90909(9.56))+ 0.1448(100) = 10.91

The investment is $10.91.

Solution 19

Answer: A

Difficulty level: .

Note that all the possible answers involve purchasing or selling put options and adjusting the position in

the stock.

The gamma of a put option is the same as the gamma of the corresponding call option, and the

gamma of a stock is zero. So we must sell 100 put options to create a gamma-hedged position.

The portfolio is now long 100 calls and short 100 puts. Since the delta of a call less the delta of a

put is 1, these options have a delta of 100. We are also short 500 shares (each with a delta of 1),

making an overall delta of -400.

Since the gamma of a stock is zero, we can adjust the holding of stock to delta-hedge the portfolio

without altering its gamma. To do this, we need to purchase 400 shares.

Solution 20

Answer: D

Difficulty level: .

Since the transaction generates 1,800 in cash, the number of units of stock sold, x, and the

number of options sold y must satisfy:

50x+5y=l,800

We also know the delta of this sold portfolio must be 100, so that:

xxl+yx0.5=100

So this becomes an exercise in simultaneous equations, and the solution is:

x=20, y=160

Alternatively, in a multiple choice question, it may be quicker to check which of the combinations satisfies

both equations by direct evaluation. A, D and E satisfy the first equation. Band D satisfy the second

equation. Only D satisfies both equations.

Solution 21

Answer: E

Difficulty level: .

Options A to D are described in the notes as good strategies to reduce the risk of extreme price

movements causing losses. Option E is not necessarily going to help because:

. the new portfolio could be independent of the existing one

. the new portfolio may not even be delta-neutral.

10 ~ BPP Professional Education




ExamMFE

Solution 22

Answer: A


If the portfolio is rebalanced x times (at regular intervals) during a period of n weeks, the number

of weeks between rebalancing is t = . So, the average loss each time as a result of not

x

rebalancing is:

L(t)=t(3t+5)=~(3~+5 )

We need to multiply this by x, the number of times we incur this cost in n weeks. Also, the cost

of rebalancing x times is 20x. So the total loss over n weeks is:

L = xL (t) + 20x

= x: (3~+5 )+20X

3n2

=-+5n+20x

x

We want to find the value of x that minimizes this loss. So we differentiate with respect to x and

set to zero:

2

L = -3n + 20 = 0

x2

Ç:

n.,

X=-

J2

Techncally, we should also check that this is a minmum by looking at the second derivative:

2

L" - 6n 0

-:;

x3

Solution 23

Answer: B


If we buy x units of Call-II and y units of the stock, our portfolio wil have a value of:

v = -1000Cii +xC¡ +yS

The gamma for this portfolio is:

rportfolio =-1000rCn +xrCi +yrs

= -1000 x 0.0651 + xx 0.0746 + y xO

= -65.1 +0.0746x

Equating this to zero (to obtain a gamma-hedged portfolio) gives:

x = ~ = 872.654

0.0746


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ExamMFE


The delta for the portfolio is:

/:portfolio =-1000/:Cn +x/:c¡ +y/:s

= -1000x 0.5825 + 872.654

x 0.7773 + yx 1

= 95.814+y

Equating this to zero (to obtain a delta-hedged portfolio) gives:

y =-95.814

So we need to buy 872.7 units of Call-II and sell 95.8 units of stock.

Solution 24

Answer: D

Difficulty level: .

Using the delta-gamma approximation, the new option price when the stock price increases by

e = 31.50-30.00 = 1.50 wil be approximately:

C(St+h) c(Sd+

e/:t +te2rt

= 4.00+ 1.50x(-0.28)+txl.502 xO.l0

=4.00-0.42+0.1125

=3.6925

The price (rounded to the nearest ioii) is 3.70.

Solution 25

Answer: B

Diffculty level: ...

We wil assume here that the derivative has a fixed maturity time T.

The price at time t of the derivative can be calculated as the discounted risk-neutral expectation:

V(t) = e-r(T -t) E * (V(T) I S(t)J

From the inormation given in (iii), we know that V(t) = ert In S(t) and V(T) = erT In S(T).

So: ert In S(t) = e -r(T -t) E * (erT In S(T) I S(t) J

The exponential factors cancel here, giving:

lnS(t) = E*(1nS(T) I S(t)J

Since it is a Black-Scholes framework, S(t) follows geometric Brownian motion, with SDE

dS(t) = (r - ö)S(t)dt + as(t)dZ(t) under the risk-neutral measure. The solution to this equation,

based on the time interval (t, T) is:

S(T) = S(t) exp ((r -ö -to-2)(T - t) + o-fZ(T)-Z(tn J

:: In S(T) = In S(t)+(r -ö -to-2 )(T -t) +o-fZ(T)- Z(tn





ExamMFE

So this tells us that:

In S(t) = E* LIn s(t)+ (r -ö -to-2)(T -t) + o-rZ(T) -Z(tn 1St J

= In S t)+ r -ö -to-2 ) T -t)

since Z(T) - Z(t) ~ N(O, T - t).

Canceling the In S(t) s gives:

0= (r-ö -to-2 )(T -t)

For this equation to hold at times t ~ T , we must have:

r-ö-t0-2 =0

So: ö=r-t0-2 =0.055-t 0.30)2 =0.01

An alternative approach that can be used here is to start from the Black-Scholes partial

differential equation:

1 2 2

rc(Sd =-0- St rt +rStdt +Bt

2

Recall that this equation is derived by looking at the sources of profit on a delta-hedged position

over a short time interval of length h. To allow for the fact that the stock in this question pays

dividends (at rate ö), we need to make an extra adjustment for the dividend payment of öStdth

that wil be received during this period. Ths leads to the modified equation:

1 2 2

rc(Sd =-0- St rt +(r-ö)Stdt +Bt

2

If we evaluate the Greeks using the formula given for the derivative value, namely

C(St) = ert In St , we get:

ac rt 1 r __ a2c __ ert x -1 ac rt 1

d=-=e x- B=-=re nS

as 5 as2 52 at

Substituting these into the Black-Scholes PDE then gives:

( ) 1 2 2 rt -1 ( 5:) rt 1 rt

C St =-0- St xe x2+ r-u Stxe -+re lnSt

2 5 St -

=C St)

=?

~ =_~¿iert +(r_ö)ert + ~

2

=?

1 2

0=--0- +(r-ö)

2


13



ExamMFE


Substituting the numerical values gives ö = -t 52 + r = -tx 0.32 + 0.055 = 0.01

Solution 26

Answer: C


Using the delta-gamma approximation, the new option price when the stock price increases by

£ = 86.00 - 5(0) wil be approximately:

C(St+h) ' c(Sd+£llt +t£2rt

Substituting the values given in the question, this becomes:

2.21 = 2.34+£x -o.181)+t£2 X

0.035

0.0175£2 -0.181£+0.13 = 0

e

The solutions to this quadratic equation are:

0.181i~ -0.181)2 -4 0.0175) 0.13)

£=

2(0.0175)

0.181 iO.154 = 9.57 or 0.78

0.035

The original stock price is 5(0) = 86.00 - £. Since we are told that 5(0):; 80, the only permissible

value of £ is 0.78, which gives 5(0) = 86.00-0.78 = 85.22. Rounded to the nearest 10ii, this is

85.20.

Solution 27

Answer: B Difficulty level: .. .

Recall that the prepaid forward price is the price that would be payable immediately to fulfll a derivative

contract. So here, it is basically just the same as the derivative price.

Method 1

In the Black-Scholes framework, the market is assumed to be arbitrage-free and the price C(St)

of any derivative must satisfy the Black-Scholes partial differential equation:

rC(St) = t (52s¡rt + rStllt + Bt

Here, the price is given by the prepaid forward price, which we can use to calculate the Greeks:

C St) = 5/

A _ ac _ a (5 x) _ 5 x-I

Ut---- t -x t

aSt aSt

r - aii_ a SX-1)_ 1)Sx-2

t ---- X t -x X- t

aSt aSt

Bt = ~~ = :t ( 5/ ) = 0

14





ExamMFE

So the Black-Scholes PDE is:

rS/ =l0-2Slx(x-l)S/-2 +rSt (xS/-1 )+0

~ rS/ =l0-2x x-l)S/ +r xS/)

Cancellng the St x factor, which appears throughout:

r =l0-2x(x-l)+rx

Rearranging:

l0-2X(x-l)+rx-r = 0

l0-2X (x-l)+r(x-l) = 0

Ll0-2x+r J (x-l) = 0

One solution to this equation (as we are told) is x = 1 (ie the stock itself). The other corresponds

to:

l0-2x+r = 0

2

2r

~X=--=

0-2

2xO.04 =-2

0.202

Method 2

An alternative approach is to use the following result derived in Example 7.9 of the text:

S(Tt = S(O)n eXPL n(r-ö-l0-2)T +no-Z(T)* J

If, instead, the power is x, the time interval is (t, T) and there are no dividends (ö = 0), this

becomes:

S(Tf =S(t)X eXPLx(r-l0-2)(T-t)+xO-rZ(T)*-Z(t)*JJ

The prepaid forward price is the value of this payoff, which equals the discounted risk-neutral

expectation:

F/ T L S(T)X J = e -r(T -t) E * L S(Tf I S(t) J

= e-r(T -t)E * (S(t)X eXPL x(r -l0-2 )(T - t) +xo-rZ(T) * -Z(t) *1 J i S(t) J

= S(t)X e-r(T -t) eXPL x(r -l0-2 )(T -t) J E * L exp L xo-rZ(T) * -Z(t) *nJ

Since Z T) * -Z t)* ~ N O, T - t), the expectation can be evaluated using the moment generating

formula E L éX J = exp ( klJ + i¡zk2 0-2) for the normal distribution with k = xo- , which gives:

E * L exp L xo-rZ(T) * -Z(t) *nJ = eXPL lx20-2 (T - t) J


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ExamMFE

So interest wil have accumulated over this period in the ratio

Ke-rT 35.84

-=-,

Ke-r 32.75

ie Ker(l-T) = 1.09435, and her overall profit, allowing for interest, wil have been:

-2, 288x

1.09435 + 2,528 = +24

Solution 29

Answer: B


~ To solve this problem, we need to use the result stated at the very end of Section 5.4 of Chapter 4 of the

Course Notes. This states that a trader who is delta-hedging using the Black-Scholes model will make a

profit of zero when the stock price moves by exactly one standard deviation.

When delta-hedging using the Black-Scholes model, the profit wil be zero when e, the

movement in the underlying asset price, is exactly one standard deviation:

e= : O"St.J

We know that St = 50 and h = -2. So we need to find the volatility 0" .

365

Equating the formula for the delta of a call option on a nondividend-paying stock to the value

given:

Ll = N(d1) = 0.6179

From the normal tables, we conclude that d1 = 0.30. Since it is an at-the-money option, we have

5 = K. So the formula for d1 tells us that:

d1

In (f )+(r -ö +10 2 )T

O -f

In 1 + (0.10+10"2 )(0.25)

0 ..0.25

0.10+.10 2

2 =0.30

20

Rearranging the last equality leads to the quadratic equation:

0.50 2 -0.600 +0.10=0

Multiplying through by 10 and factoring gives:

5~ -60 +1=0

(50 -1)(0 -1)=0

So 0" = 0.2 or 0" = 1. These give values for the absolute stock price movement of:

lei = O St.J = 50~ 1 0 = 0.52 or 2.62

365

Ony the first answer is given in the choices. So this must be the required answer.


17



ExamMFE


Solution 30

Answer: A


The payoff for this gap option is:

Payoff = 5 -130) I S ?100)

Delta-hedging an option that has been sold involves purchasing Ll shares for each option sold.

So we need to calculate the option's delta.

Since the stock does not pay dividends and the risk-free interest rate is zero, the Black-Scholes

formula for the value of this option is: .

C = SN(d1) -130N(d2)

where d1 and d2 are calculated based on the trigger price K = 100 .

The delta for the option is therefore:

ac a

Ll =-=-rSN d1)-130N d2)Ì

as as

Differentiating the first term using the product rule for differentiation, we get:

Ll = r N d1)+ 5 -Z N d1) L-130-Z N d2)

L as f as

= N(d ) + SN'(d ) ad1 -130N'(d ) ad2

lIas 2 as

Since N(x) is the cumulative distribution function of the N(O,l) distribution, N'(x) is the

d. d . f . h. h. Ø() 1 _x2/2orrespon mg ensity unction, w ic is x = r; e .

,,2íC

The formula for d1 in this case is:

In ~) +ter2T

d1 =

er-l

In 5 -InK +ter2T

er-l

So:

ad1 1

as = Ser-l

ad2 =-Z(d -er-l)

= ad1 -O=~

as as 1 as Ser-l

lso:

So the formula for Ll becomes:

1 1

l = N(di)

+ Ø d1) r. -130Ø d2)--

er T Ser T

The numerical values of d1 and d2 in this case are:

in( 100)+tX12 xl

d1 = 100.. 0.5

1 1

18





ExamMFE

and d2 = d1 -eY., = 0.5 -1Ji = -0.5

Ths gives:

A = N O.5)+

Ø O.5)- 130 Ø -o.5)

100

From the Tables:

N(O.5) = 0.6915

Ø O.5) = Ø -o.5) = ~e-D.s2/2 = 0.3521

..2íC

Also:

So:

130

A = 0.6915+0.3521--xO.3521 = 0.5859

100

So the holding of shares required to hedge 1,000 option is:

1, OOOA = 1, 000xO.5859 = 586

Solution 31

Answer: C


~ The key to solving this question is to note that, if we short sell a derivative and at the same time buy A

shares of the stock, then the resulting portfolio will be instantaneously risk-free (delta hedged). So this

portfolio has zero volatility and it earns the risk-free interest rate.

Consider a portfolio consistig of a short position of one call option and a long position in A

shares of the stock. Over the next instant of time (t, t+dt) the change in the value of this

portfolio can be written (using abbreviated notation) as:

d -e+AS) =-dC+AdS

= -C rdt+ eYedZJ+AS O.ldt+ eYdZ)

= -Cr+0.1AS)dt+C -eYeC +eYASJ dZ

But, since this is a delta-hedged position, the volatilty term must be zero, so that:

eYeC =eYAS

Item iii) in the question tells us that C = 6 and item iv) tells us that AS = 9 .

So: 6eYe = 9eY

~ eYe =1.5eY

We also know that the Sharpe ratios, calculated as a - r , for the stock and the call option must be

eY

the same.

So:

O.l-r r-r

-=-

Y eYe


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ExamMFE


::

0.1-0.04 y-0.04

0 1.50

:: y=0.04+1.5(0.1-0.04)

=0.13

An alternative approach is to note that the same portfolio, because it is risk-free, must earn the

risk-free interest rate. Ths means that:

d (-C + liS) = r( -C + LiS)dt

ie -dC + lidS = 0.04(-6 + 9)dt = 0.12dt

If we now substitute the two SDEs given in the question, this becomes:

-C (ydt+ O cdZ) + liS (O.ldt + O dZ) = 0.12dt

ie (-rC + O.lLiS) dt+ L -O cC + O LiSJ dZ = 0.12dt

So equating the dt terms on each side, we get:

-y C + 0.1 liS = 0.12

Y -

6 =9

Rearranging this gives:

0.lx9-0.12

y= 0.13

6

~ Equating the dZ terms also confirms the relationship between the volatilities that we noted above.

20





ExamMFE


Chapter 6 Solutions

Solution 1

Answer: E


From the information in (v), we have:

u = Su = 48.5975 = 1.1853 d = Sd = 36.0019 0.8781

5 41 5 41

Alternatively, we could have calculated these numbers and the risk-neutral up-probability as

follows:

u = e(r-o)/i + O $í = eO.08xO.25 + O.3x.J = eO.17 = 1.1853

d = e(r-o)/i - O $í = eO.08xO.25 - 0.3x.J = e-D.13 = 0.8781

(r-o)/i d

e -

? p* =

u -d

eO.02 - 0.8781 = 0.4626

1.1853 - 0.8781

The price is 42.6732 in six months if the path followed by the stock price is up then down or

down then up. So the associated probabilty is:

2p*(1-p*) = 2x0.4626xO.5374 = 0.4972

Solution 2

Answer: C

Difficulty level: .

The price path is:

3months: down =? 50.25 = So d = 36.0019

6months: up =? 50.50 = 50.25 u = 42.6732

So the payoff to the average price 40-strike put option is:

Payoff= maxi 0,40 - A

(0.5)l = 0.66

where: A(O.5) = 50.25 +50.50 = 36.0019 + 42.6732 = 39.34

2 2


1



ExamMFE


Solution 3

Answer: B


The diagram below contains all the necessary inormation:

~

S = 57.6029

Avg = 53.1002

Payoff = 13.1002

S = 41

S = 42.6732

Avg = 45.6354

_ _ _ _ _ _ ~_ ~y~ f_~ S~6)§:i_ _ _ ___

~S=42.6732

Avg = 39.3376

Payoff = 0.00

S = 36.0019..

S = 31.6131

,

Avg = 33.8075

Payoff = 0.00

The average payoff must be computed for each of the 4 paths as the arithmetic average of the 3-

month and 6-month prices.

The price of the Asian option is the expected discounted payoff (i.e. use the risk-free rate for

discounting and the risk-neutral probabilties obtained in Solution 6.1 :

Price = 13.1002x(p*)2x e-O.5r + 5.6354x(p*)(1-p*)x e-u.5r

= 13.1002x 0.2140x 0.9608 5.6354xO.2486x 0.9608 = 4.0392

Solution 4

Answer: A


The diagram below contains all the necessary inormation:

~

S = 57.6029

Avg = 52.9089

Payoff = 12.9089

S = 41

S = 42.6732

Avg = 45.5391

_ _ _ _ _ _ ~_ ~y~ f_ ~ S~5)Jl _ _ _ _ _ _

~S=42.6732

Avg = 39.1959

Payoff = 0.00

S = 36.0019..

S = 31.6131

,

A vg = 33.7362

Payoff = 0.00

2





ExamMFE

The average payoff must be computed for each of the 4 paths as the geometric average of the 3-

month and 6-month prices i.e. avg = .JSO.25 x

50.50 ).

The price of the Asian option is the expected discounted payoff (i.e. use the risk-free rate for

discounting and the risk-neutral probabilities obtained in Solution 1):

Price

= 12.9089x(p*)2x e-Q.5r + 5.5391x(p*)(1-p*)x e-0.5r

= 12.9089x 0.2140x 0.9608 + 5.5391xO.2486x 0.9608 = 3.9768

Solution 5

Answer: C Difficulty level: .

The 250 used in the names of some of these options is based on the assumption that there are

approximately 250 trading days in a typical calendar year.

AP12 and AP250 have payoff functions that we could write as max 0, K - Al2 T) J and

max(O,K -A250(T)J. Since AP250 is sampled more frequently, the value of A250(T) wil be less

volatile than Al2 (T). As a result, AP250 wil have a lower price.

So: AP12 :; AP250

AP12 and GP12 have payoff functions max 0, K - A12 T) J and max 0, K - G12 T) J. We know

from the arithmetic-geometric mean inequality that Al2 (T) ? G12 (T). So AP12 wil have a lower

payoff, and hence a lower price, than GP12.

So: AP12 ~ GP12

If the underlying asset price happened to be equal at each of the 12 sampling points, A12 T) and G12 T)

could actually be equal. However, the model used for pricing the options wil allow for some stochastic

variation, which wil ensure that the price of AP12 is strictly lower than the price of GP12.

GS12 and GS250 have payoff functions max(O,G12(T)-ST J and max(O,G250(T)-ST J. ST is the

asset price on one particular day the year end). G12 T) is an average involving the value of ST

and 11 other values the asset prices at the end of each of the other 11 months). G250 T) is an

average involving the same values in Gi2 (T) and 238 other values (the asset prices for days that

were not at the end of any month). So the values obtained for G250 T) wil be less strongly

inuenced by ST than wil the values of G12 T). As a result, G250 T) is less strongly correlated

with ST than G12 T) is.

There is also an opposite factor at work here, namely that G250 T) has a lower volatility than G12 T) .

However, the effect described above dominates.

So: GS12 ~ GS250


3



ExamMFE


Solution 6

Answer: D


Because of the averaging feature, calculations involving Asian options inevitably involve a lot of number

crunching .

The factors for the stock price are:

u = e r-O)h+ j.J = e O.06-0) 1/12)+O.3-/1/12 = 1.09593

d = e r-O)h- j.J = e O.06-0) 1/12)-O.3-/1/12 = 0.92164

The (recombining) tree of stock prices is shown below.

65.8139 60.2211)

60.0530

55.3472 56.7322,53.5487,50.6439)

/54.7965

O~

46.0819

50.5025

46.5450 50.6147,47.7098,45.0326)

42.4708

39.1428 42.5652)

The numbers shown in parentheses are the possible averages that might apply at time 3. For

example, at the 55.3472 node, there are three possible average stock price values:

t

(54.7965

+ 60.0530+55.3472) = 56.7322

t(54.7965

+

50.5025+55.3472) = 53.5487

t(46.0819+

50.5025+

55.3472) = 50.6439

The formula for the payoff for an average strike Asian call option is max(O,ST -A(T)). This has

a nonzero value at 4 of the final nodes in the tree.

For average strike options, it is easy to confuse the payoff functions for call and put options. One way to

remember which way round they work is to spot that the ST appears in the same place in the formula for

the Asian option as for a vanila option, which would be max(O,ST - KJ .

The risk-neutral probabilty of each upward movement is:

. e r-o)h_d

p~ =

u-d

e O06-0) 1/12) -0 92164

. = 0.4784

1.09593 - 0.92164

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ExamMFE

So the value of the Asian option is:

C . = e-o.oS

Asian

(p*)3 X (65.8139 - 60.2211)

+ (p*)2 (1- p*) X (55.3472 - 53.5487)

+ (p*)2 (1- p*)x (55.3472 - 50.6439)

+ (p*)(l- p*)2 X (46.5450 - 45.0326)

= 1.56

Solution 7

Answer: C

Difficulty level: .

Option A was knocked out when the asset price dropped to 5, and so has no payoff.

Since the minimum asset price was 4, we know that it must have dropped below 5 at some point during

the period. Even if the asset price was never exactly 5, it wil still have been knocked out.

Option B was knocked in when the asset price reached 10. Since it is a put option with a strike

price of 15, and the final asset price was 8, the payoff from 1 unit is

max (0, K - ST ) = max (0,15 - 8) = 7 . So the payoff for a short holding of 5000 units is

-5000 X 7 = - 35,000 .

Option C triggered a payment when the asset price dropped to 5. Since it is a deferred rebate

option with a strike price of 20, this option makes a payment of 20 when the option expires,

irrespective of the final asset price. So the payoff for a holding of 5000 units is

5000x20 = 100,000.

The overall payoff for the barrier options in the portfolio is:

0-35,000+100,000 =+ 65,000

Solution 8

Answer: A

Difficulty level: .

During the year, the asset price wil either have risen to 15 or it wil not have. So exactly one of

Options Y and Z wil stil be in existence on December 31, and wil have an identical payoff to

OptionX.

So the two portfolios wil have identical values.

Note that this conclusion does not depend on the no-arbitrage assumption. It is based purely on the

definitions of how the payoffs are calculated. However, if we wanted to deduce that the values of the two

traders' portfolios were equal on some earlier date, such as July 1, then we would require the no-arbitrage

assumption.


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ExamMFE


Solution 9

Answer: B

Difficulty level: .

From the given six-month high and low, and from the fact that the graph of 5 (t) versus t is

continuous, we see that the barrier of 90 is never reached (current price 80, highest point

$88.25). So the purchased up and in option does not knock in. The written up and out call option

does not knock out. So Bob's profit at maturity is:

Profit = Payoffup and in - Payoffup and out - (investment) ert

= 10xO -10 x maxiO,87.75-85J - 45x eO.06x0.5 = -73.87

Solution 10

Answer: C


Compound options are always confusing because there are several dates involved. Here we're talking now

(on April 1) about a 3-month (compound) option expiring on July 1 to buy/sell what wil be a 3-month call

option expiring on October 1. So it is 6 months from start to finish

If the underlying 3-month call option was a 3-month option now, it wouldn't make much sense, since it

would expire on the same date as the compound option matures. (This would be like buying a bond on its

redemption date or taking out a mortgage that has to be repaid tomorrow.)

The option in (v) is a call-on-call. We are required to compute the price of the corresponding put-

on-calL. As a result, we need the parity relationship between these two options and the Black-

Scholes formula for an ordinary European call option:

Call-on-call - Put-on-call +

'- '-

iven: 2.89 to be determined

x e-rti

~

.50 e -O.06x0.25

Ordinary 6-month European call,

s e-otN(di) - K'e-rt N(d2)= 5.62

in(80) + (0.06 - 0 +0.5XO.32)0.5

d1 = 85 lf = -0.0383 (= -0.04 to 2 decimal places)

0.3 0.5

:= N ( d1) = 0.4840

d2 = d1 - (J.J = -0.0383 - 0.3lf = -0.2504 (=-0.25 to 2 decimal places)

:= N ( d2) = 0.4013

:= C = 5 e -ot N ( d1 ) - K e -rt N ( d2 )

= 80xlx0.4840 - 85 x e-O.06x0.5x 0.4013 = 5.62

Put-on-call = 1.70

Using unrounded values for d1 and d2 would give a more accurate answer ofl.63.

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Solution 11

Answer: D

Difficulty level: .

Initially the trader paid 15,000 to purchase the compound option.

After 3 months, he had the option to exercise the compound option, which would allow him to

purchase vanilla options on 1000 ounces of gold for 75,000. The market price of these options at

that time was 1000x 100 = 100,000. Since he had an opportunity to purchase the vanila

options for less than their market value, he wil have chosen to exercise the compound option.

Ths wil have required a further payment of 75,000, and he now owns a vanila option on 1000

ounces of gold.

The price of gold is now 500 per ounce. Since the vanila option is a put option with a strike

price of 600, it is in-the-money on the expiration date. It has a payoff of 100 per ounce, making

a total of 1000x 100 = 100,000.

So the trader's overall profit is:

Profit = -15,000-75,000+ 100,000 = + 10,000

Solution 12

Answer: D


Luckily, we don't need to use the Black-Scholes model to find the value of the compound put option

directly. This would involve bivariate normal calculations. The approach we can use is to use the Black-

Scholes model to value the vanila call option (Option A), then use put-call parity to find the value of

Option C from the value of Option B.

The value of Option A is:

VA = Se-ôtN(d1)-Ke-rTN(d2)

= 50e-o.04N(d1)-50e-0.06N(d2)


( Se -õl) a-2T ( 50e -0.04 ) 0.252 (1)

In +- In +

K2e-rT 2 50e-0.06 2

d1 = a-ff 0.25 0.205 (= 0.21 to 2 decimal places)

d2 = d1 - a-ff = 0.205 -0.25 = -0.045 (= -0.05 to 2 decimal places)

So: N(d1) = N(0.21) = 0.5832

N(d2) = N(-0.05) = 0.4801

So: VA = 50e-o.04(0.5832) -50e-0.06(0.4801) = 5.41

The price of Options Band C satisfy the put-call parity relationship for compound options:

(Vaiue of J ( Value of J ( Value of J -rt

ompound Call Option - Compound Put Option = Underlying Option -xe 1

0.90 - Vc = 5.41-10e -0.06(0.5)

The price of Option C is 5.19.


=? Vc =5.19


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ExamMFE


Solution 13

Answer: B

Difficulty level: .

The risk-neutral probabilty of the stock moving up in price is:

* e r-d)h -d

p =

u-d

Se r-d)h -Sd

Su-Sd

60eO.07 -55 0.4675

75-55

Since both of the possibilities for the final stock price are below the trigger price, there wil be a

payoff (equal to 65-51) in both cases. The payoff at the upper node is -10 and the payoff at the

lower node is +10.

Recall that a gap option must make a payment if the payoff is triggered, even if the amount is negative.

So the current value of the gap put option is:

pgap _ p*(-10)+(1-p*)(10)

Eur - eO.07

_ 0.4675 -10) + 1- 0.4675) 10)

- eO.07

0.61

Solution 14

Answer: D


The value of the European gap call option, which we are told is equal to zero, is:

cgap =Se-dtN(d )-K e-rTN(d )

Eur 1 1 2


se-ôT) a-2T

In +-

K2e-rT 2

d1 =

a-JT 0.25-J

d2 = d1 -a-JT = 0.1732-0.25-J = -0.0035 = -0.00 to 2 decimals)

In 50e -0.04 0.5) ) + 0.252 0.5)

50e-O.07 0.5) 2

0.1732 = 0.17 to 2 decimals)

Remember that d1 and d2 are calculated based on the trigger price K2.

So: N d1) = N 0.17) = 0.5675

N d2) = N -O.OO) = 0.5000

To find K1, we need to solve the following equation:

Se-dtN(d1)-K1e-rTN(d2) = 0

50e -0.04(0.5)(0.5675) - K1 e -0.07(0.5)(0.5000) = 0

50e -0.04(0.5) (0 5675)

K1 . 57.61

e -0.07(0.5) (0.5000)

The stock price in 6 months exceeds the trigger price. So the payoff wil be:

Payoff = 55 - 57.61 = -2.61

Using unrounded values for d1 and d2 would give a more accurate answer of -2.90.

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ExamMFE

Solution 15

Answer: A

Difficulty level: .

The call and put deltas are related:

C P - 5 -ot K -rt ac ap - -ot - 1 if s: - 0

- e - e =? ----e - v-

as as

So we

have:

ac ap

- = 1+ -=1-0.8=0.2

as as

Solution 16

Answer: D

Difficulty level: .

According to the Black-Scholes formula adapted to gap options, we have:

Cgap=S e-otN(d1) -K1 e-rtN(d2) (hereK=100,K1=95)

'- '-

uropean call delta value today of 1

at time t if S(t);'K

= 80xO.20 - 95xO.12 =4.60

Solution 17

Answer: E

Difficulty level: .

It is important to make sure you get the exchange the right way round. If we exercise this option, we wil

receive Stock B and hand over Stock A.

The Black-Scholes formula for the value of the exchange option is:

CExchange = F6,T(B)N(d1)-Ft,T(A)N(d2)

where the prepaid forward prices of the stocks are:

FO~T A)=SAe-oAT =50e-D.05 O.5) =48.765

and Ft,T(B) = SBe-OBT = 50e-O.03(O.5) = 49.256

We need to calculate the relative volatilty between the two asset prices, which is:

CJ = ~CJÃ + CJã - 2pCJ A CJB = ~(0.2)2 + (0.3)2 - 2(0.5)(0.2)(0.3) = ,J0.07


in F6,T B) J+ ~T

Ft,T(A) 2

d1 =

CJ.J

1 (49.256) 0.07(0.5)

-+

8~.J 2 =0.1471 (=0.15 to

2

decimal places)

0.07 0.5

d2 = d1 -CJ.J =0.1471-,J0.07.J = -0.0400 = -0.04 to 2 decimal places)

So: N(d1) = N(0.15) = 0.5596

N(d2) = N(-0.04) = 0.4840


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ExamMFE


The value of the exchange option is:

CExchange = Ft,T(B)N(d1) - Ft,T(A)N(d2)

= 49.256 0.5596) - 48.765 0.4840) = 3.96


Solution 18

Answer: C


We need the Black-Scholes formula for exchange options:

FP (51) = 51 (O)e-óìt = 120 x e-O.05x0.5 = 117.0372

FP (52) = 52 (O)e-oit = 135 x e-OxO.5 = 135.00

(J = ~(Jl + (J5: - 2p(J1(J2 = ~0.16 + 0.09 -2xO.7x0.4xO.3 = 0.2864

(FP(S )J

In 1 + 0.5(J2t

d1 = FP(S2) = -0.1428+0.5xO.28642x0.5 =-0.6039 (=-0.60

to 2

decimals)

(J-J 0.2864v

=? N(d1) =N(-0.60) = 0.2743

d2 = d1 -(J-J = -0.6039 - 0.2864v = -0.8064 (= -0.81to 2 decimals)

=? N(d2) =

N(-0.81)

= 0.2090

=? C = FP (Sl)N(d1) - FP (S2)N(d2) = 3.89


Solution 19

Answer: A


We want to relate the claim here to the payoff in the exchange option in the previous question:

51 (0.5) + 52 (0.5) = miniS1 (.5),52 (0.5)) + maxiS1 (.5),52 (0.5))

maxi 51 (.5),52 (0.5)1 = 52 (0.5) + maxi 0,51 (0.5) - 52 (0.5)1

=? mini 51 (.5), 52 (0.5)1 = 51 (0.5) - ~axi 0, 51 (0.5) - 52 (0.5)r

exchange option payoff

By the law of one price, we must have:

Price for claim min 151 0.5)52 0.5)1 = FP 51) - Exchange option price

= 117.04 - 3.60 = 113.44

'-

olution 18

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ExamMFE

Solution 20

Answer: E

Difficulty level: .

Option 1 has the same payoff as a vanila call option, provided that the asset price stays below the

barrier throughout the period:

Payoff1 = max O,ST -K)xI maxSt ~ Hi)

Similarly for Option 2:

Payoffi = max(O,ST -K)xI(maxSt ~ H2)

So the payoff for a portfolio consisting of a long position in Option 1 and a short position in

Option 2 is:

Total payoff = max(O,ST -K)xI( maxSt ~ H1)-max(O,ST -K)xI( maxSt ~ H2)

= max O,ST -K)x I maxSt ~ H1)-I maxSt ~ H2))

We can simplify the expression in parentheses, i max St ~ Hi) - I max St ~ H 2)) , by considering

the possible ranges in which ST might lie:

If maxSt ~ Hi ~ H2, this equals 1-1 = O.

If Hi ~ maxSt ~ H2, this equals 0-1 = -1.

If Hi ~ H2 ~ maxSt, this equals 0-0 = O.

So the expression in parentheses is the same as -I(H1 ~ maxSt ~ H2).

So the total payoff equals:

Total payoff = -max(O,ST -K)xI(H1 ~ maxSt ~ H2)

= min O,K -ST )xI H1 ~ maxSt ~ H2)

The last line is derived using the identity -max(a, b) = min( -a,-b) .

Solution 21

Answer: E

Difficulty level: .

To calculate the arithmetic average we need to add up the 12 prices and divide by 12:

A(T) = 20+ 13+ 19+26+26+29+28+29+32+34+37 +41 = 27.83

12

For the geometric average we multiply together the 12 prices and take the 12th root:

G(T) = (20x13x19x26x26x29x28x29x32x34x37x41)1/12 = 26.67

The excess of the arithmetic average over the geometric average is:

A(T)-G(T) = 27.83-26.67 = 1.16


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ExamMFE


Solution 22

Answer: E


For the geometric average we multiply together the 12 prices and take the 12th root:

G(T) = (42x50x43x32x27X35x32x38x49x56x57x48)1f12 = 41.34

The payoff to the holder of an average price put option based on this average is:

max(O,K -G(T)) = max(O,50-41.34J = $8.66

To calculate the arithmetic average we need to add up the 12 prices and divide by 12:

A(T)

42 + 50 + 43 + 32 + 27 + 35 + 32 + 38 + 49 + 56 + 57 + 48 = 42.42

12

The payoff to the holder of an average strike call option based on this average and the final stock

price of $48 is:

max(O,ST -A(T)) = max(O,48-42.42J = $5.58

Since the investor has bought option A and sold option B, his payoff is:

P = $8.66 - $5.58 = $3.08

Solution 23

Answer: D


An up-and-out option wil cease to exist if the stock price 5 exceeds the barrier level H at any

time during the four months. If the barrier level becomes very large, this wil become extremely

unlikely, so that the option wil never be knocked out . This means we are effectively valuing a

vanila European put option and, since the stock price follows geometric Brownian motion, we

can use the Black-Scholes formula.


In( Ft,T(S) J+ (j2T

L Ft,T K) 2

d1 =

(jff

1 40 ) 0.252 4/12)

45e-D.06(4/12) + 2

0.25.J4/12

-0.6053 = -0.61 to 2 decimals)

d2 = d1 -(jff = -0.6053-0.25.J4/12 = -0.7496 (= -0.75 to 2 decimals)

So the values of N(-d1) and N(-d2) are:

N -d1) = N 0.61) = 0.7291

N -d2) = N 0.75) = 0.7734

The value of the put option is:

PEur = Ft,T(K)N( -d2) - Ft,T(K)N(-d1) = 45e -O.06( 4/12) (0.7734) -40(0.7291)

=$4.95

Using unrounded values for d1 and d2 would give a more accurate answer of$5.01.

12





ExamMFE

Solution 24

Answer: B

Difficulty level: .

A compound option has 4 different combinations. A compound-compound option could either

be a call option on a compound option or a put option on a compound option. This gives

4+ 4 = 8 possible combinations altogether.

Alternatively, you can calculate the answer as 23 = 8, since there are three levels involved, each of

which could be a call or a put.

Solution 25

Answer: B

Difficulty level: .

The underlying asset of the underlying option (which could, for example, be a stock) doesn't

necessarily have a term. All the other inormation is needed to price a compound option.

Solution 26

Answer: E


We can calculate the prices of the put options using put-call parity.

If K = 70, the value of the put option assuming that we do not exercise it immediately is:

P =C+Ke-rT -Se-õT = 0.00+70e-0.04 _50e-0.08 = 21.10

The exercise value is:

K - 5 = 70 - 50 = 20

Since 21.10:; 20, it would not be optimal to exercise the 70-strike option early.

If K = 60, the value of the put option assuming that we do not exercise it immediately is:

p = C + Ke -rT - Se -õT = 0.71 + 60e -0.04 - 50e -0.08 = 12.20

The exercise value is:

K 5 =60 50 = 10

Since 12.20 :;10, it would not be optimal to exercise the 60-strike option early.

The put options with strike prices 40 and 50 are out-of-the-money. So it would not be optimal to

exercise them whatever their European values are.

So it is not optimal to exercise any of the put options early.

Solution 27

Answer: B


This question tacitly assumes that we are in a Black-Scholes world.

The value of the vanilla call option is:

Cvanila = Se-õTN(d1)-Ke-rTN(d2)

= 80e-õTN(d1)-100e-rTN(d2)

Here d1 and d2 are calculated based on a trigger price of K = 100 .


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ExamMFE


Since the delta for a vanilla call option is /1 = e -òl N d1), and we are told that the price of this

option is 4, we can conclude that:

80/1-100e-rTN(d2) = 4

ie 80(0.2)-100e-rTN(d2)=4

~ e-rTN(d2) = 0.12

The gap call option pays S(T) - 90 if S(T):; 100, and zero otherwise. So it is equivalent to a

combination of:

1. a long position in an asset-or-nothing option with a payoff of 1 unit of stock if S(T) :;100

2. a short position in a cash-or-nothing option with a payoff of 90 if S(T) :; 100 .

Using the Black-Scholes formula for these digital options where again d1 and d2 are calculated

based on a trigger price of K = 100), we have:

Cgap = Value(asset-or-nothing option)- Value(cash-or-nothing option)

= Se-òlN(d1) - 90e-rTN(d2)

= 80/1-90e-rTN d2)

= 80 0.2) - 90 0.12)

=5.2

Solution 28

Answer: B

Difficulty level: .

The arithmetic average of the stock price is:

5* = A (105+ 120+...+ 110+115) = A xl,320 = 110

So the payoff from the Asian call option in (i) is max(110 -100,0) = 10 .

The stock price hits the barrier of 125 (eg on October 31), so the up-and-out option in (ii) is

knocked out and hence has a zero payoff.

The stock price hits the barrier of 120 (eg on February 29), so the up-and-in option in (iii) is

knocked in. Its payoff, based on the final stock price of 115 and a strike price of 110 is

max(115 -110, 0) = 5.

So the largest payoff is 10 for (i) and the smallest is 0 for (ii). Therefore the difference is 10.

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ExamMFE

Solution 29

Answer: D

Difficulty level: .

The option wil make a payment if the index is below the trigger price of 1, OOOx (1- 40 ) = 600 .

Using the Black-Scholes formulas:

log 1,000 +( 0.025-0.02+lX0.202 )xi

d1 = 600 0.20.J 2.6791 (= 2.68 to 2 decimal places)

So the price of an asset-or-nothing put option is:

P = Se-olN(-d1)

= 1, OOOe -0.02x1 N -2.68)

= 980.20xO.0037

=3.63

So the price for one milion of these options would be 3.63 milion.

Solution 30

Answer: A


The stock price follows geometric Brownian motion with SDE:

dS(t) = 0.10dt+0.30dZ(t)

S(t)

The solution to this equation is:

S(t) = 5(0) eXPL( 0.lo-lxO.302 ) t+ 0.30Z(t) J = 100exp (0.055t+ 0.30Z(t))

The call option wil be exercised if 5 ( ;2 ) :; 125. The probability of this happenig is:

PrLS ;2) :;125 J =prLl00exPL 0.055

(;2)

+0.30Z

;2)J :;125 J

=pr Z ;2):; O.~O tIn ~~~ -0.055 ;2) J J

= Pr (Z(0.75) :; 0.6063)

= Pr (N(O, 0.75):; 0.6063)

=pJ N(O,l):; 0.6063-0J

L .J0.75

=1-N(0.70)

'-

.7580

=0.242


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ExamMFE


The situation described here is consistent with the Black-Scholes framework. So an alternative

approach is to use the result that the risk-neutral probabilty that ST :; K ie when the expected

rate of return on the stock price is the risk-free rate r) is N d2)

, where:

d2

in f)+ r-ö-t0-2 )T

o-.f

If the expected rate of return is some other value, a say, the probabilty that ST:; K can be

found by replacing r with a, ie it wil be N(d2), where:

in ~ )+ a-ö -to-2) T

d2 =

o-.f

Here, we have a = 0.10, so that:

in ~ )+ a-ö -to-2)T

d2 =

o-.f

in( 100)+( 0.10-0-tXO.302 )(0.75)

125 =-0.70

0.30-,0.75

So: Pr Sr :; KJ =N d2) = N -o.70) = 0.242

Solution 31

Answer: B


Consider the position at time t = 1. The chooser allows us to choose between the call option and

the put option at that time. So its value equals the greater of the two:

Value of chooser (t = 1) = max1Value of call (t = 1), Value of put (t = 1)1

But we know from put-call parity that:

Value of call (t = 1) - Value of put (t = 1) = Value of stock (t = 1) -100e-2r

Since the interest rate is zero, this means that:

Value of put t = 1) = Value of call t = 1)+ 100- Value of stock t = 1)

So the value of the chooser at time 1 can be expressed as:

Value of chooser (t = 1) = max1Value of call (t = 1), Value of call (t = 1) + 100 - Value of stock (t = 1)1

= Value of call t = 1)+ max10,100- Value of stock t = 1)1

Note that the max term in this last expression is just the payoff from a put option that expires at

time 1. So we can replicate the chooser at time 1 using the 3-year call option and a put option that

expires at time 1.

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ExamMFE

Since all three components in this equation are options, which do not generate any cashfows, it

follows that we can also replicate the chooser option at time 0 using the same two options.

So: .value of chooser t = O~ = .value of 3-year call t = O~+ Value of 1-year put t = 0)

=20 (from question) =c'(3)

Again, we can use put-call parity to find the value of the 1-year put option:

Value of 1-year put t = 0) = .value of 1-year call t = O~+ 100- .value of stock t = O~ = 9

~ =%

So: C 3) = 20 - .value of 1-year put t = O~ = 11

;9

Solution 32

Answer: D


All the options in this question pay a multiple (sometimes zero) of max(S -60,0).

Note that the option with a barrier of infinity is equivalent to a vanilla call option.

The barriers involved in the special option are 70 and 80. So we can try to replicate the special

option using a combination of:

. x vanila options (barrier = 00 )

. y options with a barrier of 80

. z options with a barrier of 70.

One way to find the correct values of x, y and z is to match up the payoff multiple that would

be paid from the portfolio above with the payoff required for the special option when the

maximum stock price equals each of the critical values 85,75 and 65.

Maximum = 85:

x=l

Maximum = 75:

x+y=2

Maximum = 65:

x+y+z=O

For example, if

the maximum stock price is 75, the vanila option wil pay lxmax(S-60,O) and the 80-

barrier option wil also pay 1 x max( 5 - 60,0). So we match this to the payoff from the special option,

which is 2xmax S-60,O).

We can easily solve these equations to find that x = 1, Y = 1 and z = -2. So we can replicate the

special option using:

. 1 vanila optionl (barrier = 00 )

. 1 options with a barrier of 80

. -2 options (ie a short position in 2 options) with a barrier of 70.


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ExamMFE


The value of this portfolio is:

lx4.0861 + lxO.7583-2xO.1294 = 4.5856

Solution 33

Answer: B


The gap call option pays 5 - 45 if S? 40 and nothing otherwise. The payoff from 0.005 cash-or-

nothig options would be 5 if S? 40 and nothing otherwise. So these two derivatives together

would have a payoff of 5 - 40 if S? 40 and nothig otherwise. Ths is the same as a vanila call

option with strike price 40.

So: 1 Gap call + 0.005 All-or-nothing calls = 1 Vanilla call

We can rearrange this to get:

1 All-or-nothing call = 200 Vanila calls - 200 Gap calls

We are given the gamma for the gap call option. To find the gamma for the vanila call option,

we can consider the put-call parity equation:

CEur (K, T) - PEur (K, T) = 5 - Ke-rT

If we differentiate the right-hand side twice with respect to 5 (to find gamma), we see that the

gamma for the right-hand side equals zero. So the gamma for the left-hand side must also be

zero, which means that the gamma for a vanilla call option is the same as the gamma for the

corresponding put option.

So: r (All-or-nothig call) = 200 r (Vanilla put) - 200 r (Gap call) = -2

- ~

0.07 =0.08

Solution 34

Answer: A


Note that 5(1 and 5(2) are not independent, so we can't work out the variances of 5(1 and 5(2) separately

and just add them together . We would need to include a covariance term, as well. It is possible to do

this, but this method is quite complicated.

One way to work out the variance of A(2) is to work in terms of expectations, rather than

variances, by using the formula:

var(A(2)) = EL A(2)2 J-(E( A(2)))2

Since we are told that the expected rate of stock appreciation is 5 per annum, we know that

E (S(t)) = S(0)eO.05t. So we can easily work out E (A(2)) as:

E( A(2)) = ELt( 5(1) + 5(2)) J = t( E (5(1))+ E (5(2))) =t( 5eO.05 +5e2XO.05) = 5.3911

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ExamMFE

We also know that the Black-Scholes model assumes that the stock price follows geometric

dS(t)

Brownian motion. In this case, the SDE for this process is - = 0.05dt+ O.2dZ(t) and the

S(t)

solution to this SDE is S(t) = S(O)exPL (0.05- 1J2XO.22) t+ 0.2Z(t) J = 5exp(0.03t+ O.2Z(t)). This

allows us to express A(2) in terms of the values of the standard Brownian motion:

A(2) = t( 5(1)+5(2)) =tL 5eO.03+0.2Z(1) +5eO.06+0.2Z(2) J

We want to work out the expectation of the square of this quantity. But again, we have the

problem that Z(l) and Z(2) are not independent. However, by taking out a factor, we can write

this in terms of the independent increments Z(l) and Z(2)-Z(1) as:

A(2) = tx 5eO.03+0.2Z(1) II + eO.03+0.2rZ(2)-Z(1)J J

We now

have:

E L A(2)2 J ='¡x25E L eO.06+0.4Z(1) J E ¡ii + eO.03+0.2rZ(2)-Z(1)J r J

=.¡x 25E L eO.06+0.4Z(1) JEll + 2eO.03+0.2rZ(2)-Z(1)J + eO.06+0.4rZ(2)-Z(1)J J

Since Z(l) and Z(2)- Z(l) each have a N(O,l) distribution, we can simplify this using the result

that, if X ~ N(O, 1) ,then E L ea+bX J = ea+l/b2 .

One way to see this is to note that, if X~N(O,l), then a+bX~N(a,b2) and

ea+bX ~ LogNormal a, b2). So E L ea+bX J is the mean of the LogNormal a, b2) distribution, which is

ea+l/b2

So we get:

E L A(2)2 J =.¡x 25 ~L eO.06~O.4Z(1) J 11 + 2 ~l eO.03+0.2rZ(2)-Z(1)J J+ ~l eO.06+0.4rZ(2)-Z(1)J Jl

eO.14 0 05 0 14

. e .

=30.5744

So, finally:

var(A(2))=EL A(2)2J_(E(A(2)))2 =30.5744-(5.3911)2 =1.51


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ExamMFE


Solution 35

Answer: A


The owner of the option whose value we are tring to find wil sell either two shares of stock 1 or

one share of stock 2, whichever has the lowest value, for the strike price of 17 if this is higher. So

the payoff at time 1 for this option is:

maxU7 - minf2S1 (1),52 (1)), OJ

If we write M(l) = minf2S1 (1),52 (1)) to denote the minimum expression, this is

maxt17-M(1),Oj. But we are told that the value of an option with payoff maxtM(1)-17,Oj is

1.632. So these two options are a put and a call option on M(l) with a strike price 17. Their

values at time 0 are therefore related by the put-call parity equation:

CEur (K, T) - PEur (K, T) = F6,T - Ke-rT

ie CEur

(17

,l)-PEur

(17,1)

=

F61 _17e-o.05

- -

.632 16.17

In this equation, F6,l denotes the prepaid forward price at time 0 of M(l) .

~ At first sight you might think that F6,l is just equal to the current value M(O), which equals

min(2S1 (0),52 (0)) = min(2x 10,20) = 20. However, in this situation the value of M(l) depends on the

value of each of the two individual assets and it is not possible to replicate this payoff at time 1 with

certainty by simply investing in a portfolio consisting of a combination of the two assets. Instead, we

would need to buy an exchange option that would guarantee paying the smaller value of the two assets,

however the individual prices might move.

~ To highlight this point, you can see that there would be a big diference between the situation given here,

where 52 0) = 20, and an alternative scenario where 52 0) = 100, even though the current value of the

minimum would be 20 in both cases.

This is the prepaid value of a payoff equal to M(l) = minf2S1 (1),52 (1)). Ths is similar to the

exchange options described in the Course Notes, except that it involves a minimum rather than a

maximum. However, we can write the payoff in an equivalent form that involves a maximum:

M(l) = 251 (1) -maxfO,

251 (1) -52 (1))

~ The logic used to derive this identity was to treat 251 (1) as the starting point for the minimum, but then

to subtract the diference between 251 1) and 52 1) if the price of 52 1) is lower.

The prepaid value of 251 (1) is just 251 (0), which equals 2xl0 = 20. The prepaid value of

maxfO,

251

(1)-52 (1)) is the value of an exchange option whose value can be found using the

Black-Scholes model based on a volatilty of:

2 2 2

CJ = CJs + CJK - 2pCJSCJK

=0.182 +0.252 -2x-0.40xO.18xO.25

= 0.1309 = (0.3618)2

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The value of the exchange option is then:

C~x::ange = Se-JsTN(d1 )-Ke-JKTN(d2)

Here, 5 corresponds to the asset we wil receive, ie 251, and K corresponds to the asset we must

hand over, ie 52. Since neither stock pays dividends, this becomes:

c~x::ange = 251 (0)N(d1 )-52 (0)N(d2)

Here the formulas for d1 and d2 are:

in(2S1 (0)J+ler2T

52 (0) 2

d1 =

er-

0+lxO.1309(1)

r; 0.18

0.3618",1

d2 =d1 -er- =0.18-0.3618=-0.18

So: C~:c:ange = 2xl0xN(0.18)-20xN(-0.18) = 2.856

- -

.5714 0.4286

So the prepaid value we require is:

Frl1 = 20-2.856 = 17.144

The put-call parity equation then becomes:

1.632 - PEur (17,1) = 17.144-16.17

Rearrangig this gives:

PEur (17,1) = 1.632-17.144+ 16.17 = 0.658


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ExamMFE


Chapter 7 Solutions

Solution 1

Answer: B


We can write this probability in terms of the increments of the Brownian motion:

P (Z(10) :; 11 Z(5) = -1) = P (2(10) - Z(O) :; 11 Z(5) - 2(0) = -1)

We can split up the increment Z 10) - 2 0) so that we have non-overlapping time periods:

p Z 10) :;11 Z 5) = -1) = p 1 Z 5) - Z O)J +12 10) - Z 5)J :;11 Z 5) - 2 0) = -1 J

= p( -1 +f2(10)-Z(5)J :;11 Z(5)-Z(0) = -1 J

= P 2 1O)- 2 5) :; 21 2 5) - Z O) = -1)

= P(Z(10)-Z(5):; 2)

The distribution of this increment is normal with mean zero and variance 5.

So: P(Z(10)-2(5):; 2) = 1-N( 2 0) = 1-N(0.89) = 1-0.8133 = 0.1867

So: P Z 10) :;11 2 5) = -1) = 0.1867

In this question we used the following results and tricks, which are often required in calculations involving

Brownian motion:

1. We can split a time interval into two shorter ones to create non-overlapping time intervals, so that

the increments wil be statistically independent.

2. Information specified in the conditional clause of a conditional probability statement can be treated

as known (eg we used the fact that Z(5) - Z(O) = -1 to simplif the first part).

3. If the main clause in a conditional probability statement is independent of the conditional clause,

the conditional clause can be dropped.

4. The increment, X(t+s)-X(t), of

Brownian motion has a normal distribution with mean zero and

variance s.

Solution 2

Answer: D

Difficulty level: .

To pick the right answer for this question, you have to read the wording carefully and be clear about the

definition of a martingale. In particular, you need to distinguish between Brownian motion itself and the

increments of Brownian motion, which also form a stochastic process (assuming you keep the value of s

fixed). The first of these is a martingale, but the second is not.

A stochastic process, X(t), is a martingale if its expected future value (at time t + u, say) is

always equal to its actual current value:

E(X(t+ u) 1 X(t)) = X(t)


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ExamMFE


We've used" u ", rather than" s " here, because there's already an " s " in the definition of the increment

given in the question.

If X(t) = Z(t+ s) - Z(t) is an increment of Brownian motion, then:

E X t+u) I X t)) = E Z t+u+s)-Z t+u) I Z t+s)-Z t))

If u:; s, the time intervals (t,t+s) and (t+u,t+u+s) do not overlap, and this simplifies to:

E X t+u) I X t)) = E Z t+u+s)-Z t+u)) = 0

But for X(t) to be a martigale, we would need this to equal X(t), ie Z(t+s)-Z(t). So the

increments of Brownian motion do not form a martingale.

If X(t) refers to the Brownian motion itself rather than its increments, so that X(t) = Z(t), then we have:

E(X(t+u) i X(t)) = E(Z(t+u) I Z(t)) =E(Z(t)+tZ(t+u)-Z(t)ll Z(t)) = Z(t)+O = X(t)

So Brownian motion itself i§ a martingale.

Solution 3

Answer: C


The processes in B through E are very similar to the process in A. So a good approach here is to start by

working out the expectation required to decide whether A is a martingale. We can then apply simple

adjustments to find the expectations for the other processes.

If A is a martingale, then the following equation would have to be true:

?

EL exp(-Z(t+s)) I Z(t)J ; exp(-Z(t))

The expectation on the left-hand side is:

EL exp(-Z(t+ s)) I Z(t) J = El eXPL -Z(t) -iZ(t+ s) - Z(t)JJ I Z(t) J

= exp(-Z(t)) El eXPL -iZ(t+s) - Z(t)JJJ

Since Z(t+s)-Z(t) is an increment of Brownian motion, it has a normal distribution with mean 0

and variance s. From statistical theory, it follows that -iZ(t+s)-Z(t)J also has a normal

distribution with mean 0 and variance s, and eXPL -iZ(t+s)-Z(t)JJ has a lQormal

distribution with parameters f1 = 0 and (j2 = s .

So the expectation El eXPL -iZ(t+s)-Z(t)JJJ is just the mean of this lognormal distribution,

which we can evaluate using the formula in the Tables:

El eXPL -iZ(t+s)-Z(t)JJJ = eXPLf1+l(j2 J = eXPLls J

An alternative way to evaluate this expectation is to spot that it has the same form as the moment

generating function of a normal distribution, evaluated at -1. Failing that, you could go back to first

principles and express the expectation as an integral, which you can evaluate by completing the square

So: EL exp( -Z(t+s)) I Z(t) J = exp( -Z(t))

exp

lls J

Since the right-hand side contains an "extra" eXPLls J factor, this shows that the process in A is

not a martingale.

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ExamMFE

However, for the process in C, we have:

E eXPL-Z t+ s) -t t+s) J I Z t) J = eXPL -t t+s) JE exp -Z t+s)) I Z t) J

= eXPL -t(t+s) Jexp(-Z(t))exPLts J

=exPL -Z(t)-ttJ

So the expected value of process C at time t + s is always equal to its value at time t, which

means that it is a martingale.

Solution 4

Answer: D Difficulty level: .

We

have:

5 = 55 eO.045t + O.5z(t)

as as a2s

=? - = 0.0455, - = 0.55 , -i = 0.255

at az az

as as 1 a2s 2

=? dS(t) = at dt + az dZ(t) + 2 az2 (dZ(t))

= 0.045S(t)dt + O.5S(t)dZ(t) +0.125S(t)(dZ(t))2

'-

dt

= 0.17S(t)dt + O.5S(t)dZ(t)

Solution 5

Answer: A Difficulty level: .

We are given:

X(S,t) = 3é(t) + t X Set)

Using Ito's Lemma:

dX S,t) = XsdS+tXss dS)2 + Xtdt

The partial derivatives are given by:

Xs = 3eS(t) + t, Xss = 3eS(t), Xt = S(t)

Substituting these into the equation for dX(S,t) gives:

dX(S, t) = (3é(t) + t )dS +t( 3eS(t) )(dS)2 + S(t)dt

= (3é(t) + t)( as(t)dt + o-S(t)dZ(t)) +t( 3é(t)) ifS(t)2 dt + S(t)dt

= (( 3é(t) + t) as(t) +t( 3é(t)) ifS(t)2 + S(t) Jdt +( (3é(t) + t) o-S(t) JdZ(t)

The drift is given by the coeffcient of dt , which simplifies to:

Drift = (3é(t) + t) as(t) +t( 3é(t)) o-2S(t)2 + S(t)

= S(t)( (3a+tifS(t) )é(t) + at+ 1 J


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ExamMFE


Solution 6

Answer:B

Difficulty level: .

The stochastic differential equation for X (t) describes Brownian motion with drift:

dX(t) = 0.09dt + O.4dZ(t) ~ X(t) =0.09t + O.4Z(t)

So in general we have:

var(X(t)) = var(0.09t + O.4Z(t)) = var(O.4Z(t))

= 0.42 var(Z(t)) = O.16t = 0.32 when t = 2

Solution 7

Answer: A

Difficulty level: .

From the previous solution, we have:

X(t) =0.09t + O.4Z(t) ~ X(2) = 0.18 + 0.4 Z(2) where Z(2)-N(O,2)

~ Pr(X(2) :;0.15) = Pr(0.18 + 0.4 Z(2) :; 0.15)

~ Pr(Z(2) :; -0.075)

( -0075 - 0)

~ Pr N(O,l):;.-l =1-N(-0.05)=N(0.05)=0.5199

Solution 8

Answer: E

Difficulty level: .

The original process is:

dr(t) = (0.06-r(t))dt+0.01dZ(t)

We are creating a new process V(t) by

applying the function exp (-tr(t)) to this process. So we can find

the SD E for the new process by applying Itô's lemma.

The partial derivatives required for Itô s lemma are:

av a

-=-exPL -tr J = -texPL -tr J =-tV

ar ar

a2v a

-- =-l-texPL -tr JJ = t2 eXPL -tr J = t2V

ar ar

av a

-=-exPL -tr J = -rexPL -tr J =-rV

at at

We have abbreviated the r(t) s and V(t) to r s and v s here to make it clearer which variables are

involved in calculating the derivatives. To apply Itô s lemma correctly, we need to think of the new process

value, V, as being a function of the original process value, r, and the time t.

Itô s lemma then tells us that:

av 1 a2v 2 av

dV(t) =-dr+--(dr) +-dt

ar 2 ar2 at

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ExamMFE

Substituting the partial derivatives gives:

dV(t) = -tVdr+tt2V(dr)2 -rVdt

= V i-tdr+tt2 dr)2 -rdtl

The SDE for r(t) is:

dr(t) = (0.06-r(t))dt+0.OldZ(t)

So: dr t))2 =UO.06-r t)) dt+ 0.01dZ t)ì2 = 0.01)2 dZ t))2 = O.OOO1dt

Substituting these into the SDE for V(t) gives:

dV(t) = V(t)i -t( 0.06 - r(t)) dt - O.OltdZ(t) +tt2 (O.OOOldt) - r(t)dtl

= V t)tL -D.06t+

0.00005t2 + (t-l)r(t) J dt -O.OltdZ(t)l

or dV t) = r -0.06t+

0.00005t2 + t -l)r t)J dt- O.Olt dZ t)

V t) L

Solution 9

Answer: A

Difficulty level: .

In general, we know that the equation in i) describes geometric Brownian motion:

a- O.5o.2)t + CTZ t)

dX t) = aX t)dt + o-X t)dZ t) =? X t) = X O)e

So here we have:

a = 0.09 , 0- = 0.4 , X 0) = 55 =? X t) = 55eo.Olt + 0.4Z t)

Solution 10

Answer: E

Difficulty level: .

From the previous solution, we have:

X(t)

=

55eo.Olt+OAZ t) =? X 2)

=

55eo.02

+

0.4Z(2) =?

Pr X 2) :; 65) = pr 55eo.02 + 0.4Z 2) :; 65)

=pr Z 2):; in 65/55)-0.02J

0.4

= Pr(Z(2) :; 0.3676)

= pr(N(O 1) :; 0.3676)

, .J

= 1 - N 0.26) = 0.3974


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ExamMFE


Solution 11

Answer: A


The SDE for the gold price is:

dG(t) = 0.06dt+0.ldZ(t)

G(t)

This process is geometric Brownian motion. The solution to this equation is:

G(T) = G(0)e(0.06-tXO.12)T+O.1Z(T)

= G(O) e 0.055T +O.lZ(T)

The notes give the SD E for geometric Brownian motion and its solution in the form:

dS(t)

-= adt+a-dZ(t)

S(t)

~

S T) = S 0)e a-ta2)T+aZ T)

Here we have a = 0.06 and J = 0.1 .

The probabilty that the gold price at time 5 wil exceed 1000 is:

Pr(G(5) :;1000 I G(O) = 500) = Pr( G(0)eO.055(5)+0.lZ(5) :;1000 I G(O) = 500 J

= Pr( 500eO.275+0.1Z(5) :;1000 J

_ P (0.275+0.1Z(5) 1000J

r e :;-

00

= Pr(0.275+0.1Z(5):; ln2)

= pr Z 5):; in2-0.275J

0.1

= Pr(N(O,5):; 4.1815)

=pri N O,l):; 4.~5J

=1-N(1.87)

=1-0.9693

= 0.031

So the required probability is 3.1 %.

Solution 12

Answer: D


If we let X(t) = Z(t)4 - 6Z(t)2 , we can use Itô s lemma to find the SDE for X(t)

, and hence find its

drift.

The original process here is standard Brownian motion, whose SDE is just:

dZ t) = dZ t)

The process X(t) is calculated from Z(t) by applying the function:

X =Z4 -6Z2

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ExamMFE

The partial derivatives required for Itô's lemma to find the SDE for X(t) are:

ax =~ z4 -6Z2)=4Z3 -12Z

az az

a2 x = ~(4Z3 -12Z) = 12Z2 -12

az2 az

ax =0

at


dX(t) = ax dZ+~ a2x (dZ)2 + ax dt

az 2 az2 at


dX(t) = (4Z3 -12Z)dZ +~(12Z2 -12)(dZ)2 +0

2

Since (dZ(t))2 = dt, this simplifies to:

dX(t) = l6Z(t)2 - 6 J dt + l4Z(t)3 -12Z(t) J dZ(t)

So the drift of the process X(t) is:

6Z(t)2 -6

This wil have a positive value when Z(t)2 :;1, ie when I Z(t) I :;1, but not otherwise.

Solution 13

Answer: C


A process S(t) is geometric Brownian motion if its SDE is of the form:

dS t) = adt+ CJdZ t) or dS t) = as t)dt+ CJS t)dZ t)

S(t)


S(T) = 5(0) e (a-l(T2)T +CZ(T)

We can decide whether the other processes are also geometric Brownian motion by checking

whether either their SDEs or their solutions have the same form as these equations.

(The parameter values need not be the same as for S(t).)

The relationship U(t) = S(t) + 100 tells us that:

dU(t) = dS(t)

Using the SDE we are assuming for S(t), this tells us that:

dU t) = dS t) = as t)dt + CJS t)dZ t)

But since, S(t) = U(t) -100, this is:

dU t) = a U t) -100) dt+ CJ U t) -100) dZ t)


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ExamMFE


This SDE doesn't have the correct form for U(t) to be geometric Brownian motion because

neither the drift nor the volatility is proportional to U(t).

The relationship V(t) = 10S(t) tells us that:

dV(t) = 10dS(t)

Using the SDE we are assuming for S(t), this is:

dV(t) = lOtaS(t)dt+O S(t)dZ(t))

= 10S(t)1 adt+ O dZ(t)J

= V(t)1 adt+ O dZ(t))

dV(t)

-=adt+O dZ(t)

V(t)

or

This does have the correct form for V(t) to be geometric Brownian motion.

For W(t)

, it is probably easier to work from the solutions, rather than from the SDEs.

The solution for S(T) is:

S(T) = 5(0) e (a-ta2)T +aZ(T)

Note that this formula applies at il time T , and so it does tell us about the future behavior of the process,

not just the value at one time point.

So: W(T) = S(T)2

= L S(0)e(a-ta2)T+aZ(T) r

2

= 5(0)2 e(2a-a )T +2aZ(T)

We know that W(O) = 5(0)2. If we let O w = 20 and aw = 2a+ 0 2, this can be written as:

W(T) = W(O) e(aw-taiv)T +awZ(T)

This has the correct form for W (t) to be geometric Brownian motion.

Alternatively, you could use Itô's lemma to find the SDE for W(t), which is:

dW(t) = (2a+l0 2)S(t)2 dt+ 20 S(t)2 dZ(t)

= (2a+l0 2)W(t)dt+ 20 W(t)dZ(t)

This has the correct form for W(t) to be geometric Brownian motion.

Solution 14

Answer: B


In general, we know that the equation in (iii) describes geometric Brownian motion:

dX(t) (a - O.5(2)t + aZ(t)

X(t) = at + 0 dZ(t) =? X(t) = X(O)e

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ExamMFE


So:

a=-O.Ol, 0 =0.20, dG(t) = -0.01

dt+ 0.20

dZ(t)

, G(t) ,

Soluti'on 7.7

=? G(t) = G(O) e(-O.Ol- 0.5xO.04)t + 0.20Z(t) = G(O) e-o.03t + 0.20Z(t)

Now let's determine G(O) when T=l year:

G(t) = X(t)i.02(T-t) =? G(O) = 0.85eo.02(1-0) = 0.8672

Therefore:

G(t) = 0.8672 e-O.03t+

0.20Z(t)

Now we can determine the probability sought:

Pr( G(0.5) :; 0.88) = pr( 0.8672 e -0.03x0.5 + 0.20Z(0.5):; 0.88)

( ( ) in(0.88/0.8672)+0.015J

= Pr Z 0.5 :;

0.20

= Pr(N(O,O.5) :;0.1484)

= 1- N(0.1484)

-J

= 1 - N(0.21) = 0.4168

Solution 17

Answer: E


The standard SDE for stock price following geometric Brownian motion is:

dS(t)

- = (a-ö)dt + O dZ(t)

5 (t)

From the first SDE in (iv), we see that:

0 1 = 0.25 and 0.09 = a1 -Si = a1 -0.02 =? a1 = 0.11

From the second SDE in (iv), we have:

0 2 = band 0.13 = a2 -ö2 = a2 -0 =? a2 = 0.13

Now notice that the stock prices are perfectly correlated since the same Brownian motion drives

them both. So they must have identical Sharpe ratios:

a1 -r a2-r

-=-=?

0 1 0 2

0.11 - 0.045 = 0.13 - 0.045 =? b = 0.3269

0.25 b

10





ExamMFE

Solution 18

Answer: E Difficulty level: .

Ths is the mean revertig Ornstein-Uhlenbeck process. In general, we have:

dX t) = Â. a - X t)) dt + a-dZ t)

Although this process mean-reverts to the value a, the random component does not tend to

zero. So the limit lim X (t) = a does not exist.

t-'oo

Solution 19

Answer: D Difficulty level: .

Again, this process mean-reverts to the value a = 0.06. When t is large, the random component

is proportionalto Z (t) , which has a mean of zero. So the limit lim E LX (t) J = a = 0.06 .

t-'oo

Solution

20

Answer: E


The original process is:

dX t) = Â. a- X t)) dt + a-dZ t)

The process N(t) is calculated from X(t) by applying the function:

N = eÅt (X -aJ

The partial derivatives required for Itô's lemma to find the SDE for N(t) are:

aN =.if eÅt X -aJl = eÅt

ax ax 1 f

a2N _ a eÅt-O

ax2 - ax -

aN = _~.J eÅt X -aJl = Â.eÅt X -aJ

at at 1 f


dN(t)

= aN dX+2. a2N (dX)2+ aN dt

ax 2 ax2 at


dN(t) = eÅt dX +0+ Â.eÅt (X -aJ dt

The SDE for X(t) is:

dX(t) = Â.( a - X(t)) dt + a-dZ(t)


11



ExamMFE


Substituting this into the SDE for N(t) gives:

dN(t) = eÂt tÂ(a-X(t))dt+o-dZ(t)l+ÂeÂt (X(t)-a)dt

= ÂeÂt a- X t))dt+ o-eÂt dZ t)

+ ÂeÂt X t)-a)dt

= Odt+ o-eÂtdZ(t)

SO liN t), the drift of the process N t)

, is O.

Since the drift is zero, the process N(t) is actually a martingale.

The process S(t) is calculated from N(t) by applying the function:

S=N2

The partial derivatives required for Itô's lemma to find the SDE for S(t) are:

as =~N2 =2N

aN aN

a2s =~(2N)=2

aN2 aN

as =~N2 =0

at at


dS(t) = as dN +~ a2s (dN)2 + as dt

aN 2 aN2 at


dS(t) = 2NdN +~(2)(dN)2 +0

2

The SDE for N(t) is:

dN(t) = o-eÂt dZ(t)

and the equation for N(t) itself (given in the question) is:

N(t) = eÂt (X(t)-a)

Substituting this into the SDE for S(t) gives:

dS(t) = 2eÂt (X(t)-a)aeÂtdZ(t)+L o-eÂtdZ(t)J2

= 2eÂt X t) -a) o-eÂt dZ t) + 0-2e2Ât dt

= 0-2e2Ât dt + 20-e2Ât X t) -a) dZ t)

So lis t) , the drift of the process S t), is 0-2 e2Ât .

So: liN (t) + lis (t) = 0 + 0-2 e2Ât = 0-2 e2Ât





ExamMFE

Solution 21

Answer: C


To illustrate the calculations required for this question, if aWi (t) + bW2 (t) has the same variance as

Z2 t), then var aWi t)+ bW2 t)J = var Z2 t)J, so that a2t+ b2t = t and a2 +b2 = 1.

Similarly, if aWi t) + bW2 t) and cW1 t) + dW2 t) + eW3 t) have the same covariance as Z2 t) and

Z3(t) , then covL aWi t)+ bW2 t),cW1 t)+ dW2 t)+ eW3 t)J = COV Z2 t),Z3 t)J, so that

acvar W1 t)J + bdvar W2 t)J = COV Z2 t),Z3 t)J. SO act + bdt = P23JtJt and ac+ bd = 0.5 .

We need to ensure that the Z 's have the correct variances. Zl already has the correct variance,

since it equals Wi' which is Brownian motion. To get the correct variances for Z2 and Z3' we

need:

a2 +b2 =1

c2 +d2 +e2 =1

We also need to have the correct covariances, which leads to the equations:

a=0.8

c = 0.4

ac+bd =0.5

We can now solve to find all the constants:

b = ~1- a2 = ~1- 0.8)2 = ..0.36 = 0.6

d= O.5-ac 0.5-0.8 0.4)

0.3

b 0.6

e = ~1- c2 - d2 = ~1- 0.4)2 - 0.3)2 = ..0.75

Solution 22

Answer: D

Difficulty level: .

A is true, since each component has zero drift, so that the combination wil also have zero drift.

B follows directly from A, since this is a linear combination with coefficients of 1 and -1.

C is a martingale because we are removing the drift.

D is not usually correct. For example, if Z t) is Brownian motion, then Z t) is a martigale.

But Z(t)xZ(t) = Z(t)2 has a drift of t (since we know that Z(t)2 -t is a martingale).

E is correct, since Z(t) has a standard normal distribution, which is symmetrical about zero.

We can prove this in detail as follows. Since Z (t) has a standard normal distribution, so also does -Z( t) .

It follows that L Z (t) rand L -Z (t) r have the same statistical distributions. Therefore they have the

same expectations:

E1LZ t)rJ = E1L -Z t)rJ


13



ExamMFE


:: E1LZ t)TJ = -It E1LZ t)TJ

Since n is a positive odd number, (_l)n = -1, and we have:

E1LZ(t)TJ =-E1LZ(t)TJ

:: 2E1LZ t)TJ =0

:: E1LZ t)TJ = 0

Solution 23

Answer: E


The process in A is a linear combination of two more recognizable martingales, namely minus 3

times Z(t) and mius 12 times Z(t)2 -t:

12t-3l Z(t)+4Z(t)2J=-3Z(t)-12L Z(t)2_tJ

The process in B is an example of the martingale eÂZ t)-0.5Â2t with a value of À = -4.

The processes in C and E both involve Z (t)3 , so let's construct a martingale from Z (t)3 :

El Z(t)3IZ(s)J = ELfZ(t)-Z(s)+z(s)l3Iz(s)J

= ELf Z( t) - Z(s )ì3 J +3Z(s )EL fZ( t) - Z(s )ì2 J

+3Z (s)2 EL Z( t) -Z (s) J +Z(s)3

= 3Z(s)(t-s)+Z(s)3

Notice that two of the terms disappear because we know that the odd moments of Z(t)-Z(s)

are zero (from the previous question). Now we can use the fact that Brownian motion itself is a

martingale to finish off:

El Z( t)3IZ(s)J = 3Z(s)t-3Z(s)s+Z(s)3

= 3tEL Z (t)IZ( s) J - 3Z( s)s + Z (s)3

Rearranging then gives:

El Z(t)3 -3tZ(t)lz(s)J = Z(s)3 -3sZ(s)

The process in C however is mius 2 times this martingale, and hence a martingale itself.

The process in D may look confusing because it has e and TC in it, but these are both constants.

When we multiply the values of a process with no drift by a constant (eg the constant e ) and/ or

rescale the time argument (eg by a factor of TC), it remains a process with no drift.

These types of transformations are equivalent to stretching or shrinking the process along the vertical

and/or horizontal axes.

14





ExamMFE

We can check whether the process in E is a martingale by expressing it in terms of the martingale

derived in C and the familiar martingales, Z(t) and Z(t)2 -t:

Z t)3 -3tZ t)2 =iZ t)3 -3tZ t)1+3trZ tn-3ttZ t)2 -tJ-3t2

So: EL Z t)3 -3tZ t)21 Z s)J =iZ s)3 -3sZ s)1+3trZ sn-3ttZ s)2 -sJ-3t2

The right-hand side of this equation does not even simplify to a function of s. (The ts don't

cancel). So it cannot have the correct form for a martingale.

Solution 24

Answer: A


Apply Taylor's formula to V(t) = f(X(t)) = e-2X(t):

dV (t) = df (X (t))

= af dX(t)+. a2 ~ L dX (t) J2

ax 2ax

= _2e-2X(t) L Y (t)dt+~Y( t)dZ( t) J+~L 4e-2X(t) JL Y( t )dt+~Y(t)dZ( t) r

= _2e-2X(t)y( t)dt - 2e -2X(t) ~Y(t)dZ(t)+ 2e -2X(t)y (t)dt

= Odt- 2e -2X(t) ~Y (t)dZ (t)

In the last line the drift ( dt ) terms have canceled, so that V (t) is indeed a martingale.

Solution 25

Answer: B


Apply Taylor's formula to X(t) = f(Z(t)) = Z(t)e-rZ(t):

dX t) = df Z t))

= af dZ( t)+. a2 ~ L dZ(t) J2

az 2 az

= e -rZ(t) L 1- rZ( t) J dZ( t) +~L e -rZ(t) (-r) -re -rZ(t) U -rZ(tn Jdt

= re -rZ(t) L 0.5rZ (t) -lJ dt+ L 1-rZ(t) Je -rZ(t)dZ (t)


15



ExamMFE


Solution 26

Answer: C

Difficulty level: .

D is the definition of the Sharpe ratio. It equals the risk premium of an asset divided by the

volatility of that asset:

a-r

(J

where a is the continuously-compounded expected rate of return for the asset

(J is the volatility of the asset and

r is the continuously-compounded risk-free interest rate.

So, the Sharpe ratio wil:

. increase if the risk free rate decreases

. decrease if the expected rate of return decreases

. decrease if the volatilty of the asset increases.

So the statements in A, Band E are all correct. This just leaves C. It is true to say that assets

whose prices are perfectly correlated have the same Sharpe ratio, but that doesn t necessarily

mean that the converse is true. Consider two independent assets that coincidentally have a

Sharpe ratio of 0.25 say.

Solution 27

Answer: C


The payoff of this derivative can be written as:

max ( 51 (3), 52 (3)) = 52 (3) + max (51 (3) - 52 (3), 0)

Statement (iv) tells us that dividends for Stock 2 are paid continuously at a rate of 10 per

annum. So, if we purchase initially e-O.3 units of Stock 2 and reinvest the dividends, at time 3 we

wil have 1 unit of stock, and its value wil be 52 (3) .

If we purchase one of the exchange options in statement (v), at time 3 its payoff wil be

max(Sl(3)-S2(3),O) .

So a portfolio of e-O.3 units of Stock 2 and one exchange option wil replicate the payoff of the

claim. So the initial value of the claim is:

e-O.3 x200+ 10 = 158

Solution 28

Answer: A


Applying Itô's lemma to the function C = Se(r-r*)(T -t) , we get:

dC = ac dS +l a2c (dS)2 + ac dt

as 2 as2 at

16





ExamMFE

The partial derivatives are:

aG = e(r-r*)(T -f) a2G = 0

as as2

d aG _ ( *)5 (r-r*)(T -f)

n --- r-r e

at

So: dG = e(r-r*)(T -f)dS + 0 -(r _r*)Se(r-r*)(T -f)dt

= G dS-(r-r*)Gdt

5

= G (O.ldt + O.4dZJ + 0.02Gdt

= G(O.12dt+0.4dZ)

Solution 29

Answer: A


Applying Itô s lemma to the function L = In Y , we get:

d(1n YJ = dL =~dY +l a2L (dy)2 + aL dt

ay 2 ay2 at

The partial derivatives are:

So:

aL 1 a2L 1 aL =0

- - and

ay -y ay2 -- y2 at

dL = .:dY _~(dy)2

Y 2y2

= dY _ (dy)2

Y 2 Y

= (Gdt + Hdz)-l(Gdt+ HdZf

=

(Gdt+HdZ)-lH2dt

=( G-lH2 )dt+HdZ

Comparing the drift and the volatility coeffcients with the equation in statement (i), we see that:

G-lH2 = 0.06 and H = (J

Since Assets X and Yare both driven by the same Brownian motion, they must have the same

Sharpe ratio, which gives us another relationship between G and H :

G-r

-=

H

0.07-r

0.12

ie

G-0.04

H

0.07 -0.04 1

0.12 4

~ H=4G-0.16

Substituting this expression for H into the earlier equation gives:

G-l(4G-0.16)2 = 0.06

Expanding and rearranging gives:

8G2 -1.64G+0.0728=0


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ExamMFE


Solving this using the quadratic formula, we get:

1.641:~1.642 -4 8) 0.0728) 1.641:0.6

G = = = 0.14 or 0.065

16 16

The corresponding values of Hare:

H = 4G-0.16 = 0.4 or 0.1

Since H = (J and statement (iii) tells us that (J ~ 0.25, we can eliminate the first possibilty and

conclude that H = 0.1 and G = 0.065 .

Solution 30

Answer: D


Applying Itô s lemma to the function Y = ~ = X-I, we get:

X

dY = ay dX +l a2y (dX)2 + ay dt

ax 2 ax2 at

=_y2 f2 4- y-1ldt+8dZl+ y3 x64dt

= f(-8y2 +2Yldt-8y2dZl+ y3 x64dt

= (2Y _8y2 +64y3ldt-8y2dZ

The function a(y) corresponds to the drift term, ie a(y) = 2y _8y2 + 64y3 .

When y = l ' the value of this function is:

2 3

a l) = 2 l)-8 t) +64 l) = 1-2+8 = 7

Solution 31

Answer: C


Suppose the investor purchases n1 units of Asset 1 and n2 units of Asset 2. The value of the

resulting portfolio wil be:

P t) = niSi t) + n2S2 t)

The change in the value of this portfolio over the infinitesimal time interval t,t+dt) wil be:

dP(t) = n1 dS1 (t) + n2 dS2 (t)

18





ExamMFE

Using the SDEs given, this is:

dP t) = niSi t) 0.08dt+ 0.2dZ t))+

n2dS2 (t) (0.0925dt-0.25dZ(t))

= (0.08n1 51 (t) + 0.0925n2S2 (t)) dt+ (0.2n1 51 (t) - 0.25n2S2 (t)) dZ(t)

In order for this to behave in the same way as a risk-free asset, it must have zero volatility. So we

must have:

0.2n1S1 (t) - 0.25n2 52 (t) = 0 ~ n2S2 (t) = 0.2 niSi (t) = 0.8n1 51 (t)

0.25

Since the total amount allocated is 1,000, we also know that:

P t) =

niSi (t)+n2S2 (t) = 1,000

So:

1,000

niSi (t)+0.8n1S1 (t) = 1,000 ~ niSi (t) =-= 555.56

1.8

An alternative approach is to note that, since the prices of the two assets both depend on the

same source of randomness Z(t), the Sharpe ratio must be the same in each case, so that:

0.08-r

0.2

0.0925-r

-0.25

Solving this gives:

r = 0.0385 = 0.08556

0.45

We can see from the SDEs that asset 1 earns 8 (=0.08) on average and asset 2 earns 9.25

(=0.0925). So, if the investor invests a proportion íC1 of the 1,000 in asset 1 and a proportion

1- íC1 in asset 2, the overall average rate of return wil be 0.08íC1 + 0.0925 1- íC1 ) . If this

combination is to synthesize the risk-free asset, this rate of return must equal r.

So: 0.08íC1 + 0.0925(1- íC1 ) = 0.08556

~ íC = 0.0925 - 0.08556 0.55556

1 0.0925 - 0.08

So the amount that must be invested in asset 1 is 1, OOOx 0.55556 = 555.56.

Solution 32

Answer: C


From (ii) we know that, under the real-world probabilty measure, the SDE for S(t) is:

dS t) = 0.05dt + 0.2dZ t) or dS t) = 0.05S t)dt + O.2S t)dZ t)

S(t)

Note that this equation must be based on the real-world probabilities because the drift (0.05) is not equal to

the risk-free interest rate (0.03).


19



ExamMFE


The solution to this geometric Brownian motion equation is:

S(t) = 5(0) expi( 0.05-lXO.22) t+ 0.2Z(t) J = 0.5 exp (0.03t+ O.2Z(t)J

Here we have used the standard result that the solution to the SDE dX(t) = aX(t)dt+ oX(t)dZ(t) is

X t) = X O) exp L a-lo.2 ) t+ oZ t) J .

When t = 1, we have:

5 1) = 0.5

exp(0.03+

0.2Z(1)J

and EL S lt J = Elf 0.5

exp

(0.03

+0.2Z(1)Jr J

= O.5a eO.03a E L exp 0.2aZ 1) JJ

But, using the properties of standard Brownian motion, we know that Z(l) - N(O,l). It follows

that 0.2aZ 1)-N O,O.04a2) and exp 0.2aZ 1)J-LogNormal O,O.04a2). Using the formula

exp (.u +l 0.2) for the mean of the lognormal distribution, this tells us that:

E L exp O.2aZ l)J = exp 0+lxO.04a2) = exp 0.02a2)

So: E L S l)a J = 0.5a eO.03a eO.02a2 = 0.5a eO.03a+O.02a2

Using the information in iii), this gives:

2

1.4 = O.5a eO.03a+O.02a

We can now find the value of a by taking logs to obtain a quadratic equation:

In 1.4 = aln 0.5+ 0.03a+ 0.02a2

0.02a2 + (0.03 + In O.5)a - In 1.4 = 0

0.02a2 - 0.6632a - 0.3365 = 0

Since we are told that a is negative:

0.6632:: ~ -0.6632)2 - 4 0.02) -0.3365) 0.6632:: 0.6831

a = = -0.5 very nearly)

0.02) 0.04

20





ExamMFE

If we now go back to the original SDE for S(t), Girsanov's theorem tells us that the

corresponding SDE under the risk-neutral probability measure is obtained by changing the drift

so that it equals the risk-free interest rate:

dS t) =

O.03dt+ 0.2dZ(t)

*

S(t)

Again, we can solve this to get:

S(t) = 5(0) exp L( 0.03-lXO.22) t+O.2Z(t) * J = O.5exp (0.01t+ 0.2Z(t) * J

and when t = 1, we have:

5 1) = 0.5 exp O.Ol +0.2Z 1) * J

Also: E L S l)a J = O.5a eO.01a+O.02a2

The price at time 0 of the contingent claim (derivative payoff) is the discounted expected payoff

calculated using risk-neutral probabilities:

e-r E * L S l)a J = e-O.03 O.5a eO.01a+O.02a2 = e-O.03 0.5-0.5 eO.01 -o.5)+O.02 -0.5)2 = 1.372

Solution 33

Answer: A


We are given that:

51 t) = 51 0) exp 0.1t+0.2Z t)J and 52 t) = 52 0) exp 0.125t+

0.3Z(t)J

The corresponding SDEs are:

dS1 (t) = (0.1+lXO.22 )dt+0.2dZ(t) = 0.12dt+0.2dZ(t)

51 (t)

ie dS1 t) = 0.1251 t)dt+ 0.251 t)dZ t)

and dS2 t) = 0.125+lxO.32) dt+ 0.3dZ t) = 0.17 dt+0.3dZ t)

52 (t)

ie dS2 t) = 0.1752 t)dt + 0.352 t)dZ t)

Since the prices of these two assets are both determied by the same single source of randomness,

ie Z t)

, their Sharpe ratios, calculated as a - r , must be the same.

(7

So:

0.12-r

0.2

0.17 -r

0.3


21



ExamMFE


0.3(0.12 - r) = 0.2(0.17 - r) := r = 0.002 = 0.02 ie 2%

0.1

Solution 34

Answer: E Difficulty level: .

Remember that the SDE for a process that is arithmetic Brownian motion has the form

dX t) = adt + o-dZ t) .

The Black-Scholes model assumes that the stock price S(t) is geometric Brownian motion with an

SDE of the form:

dS(t)

- = adt+ o-dZ t)

S(t)


S t) = 5 0)

exp

L a-t0-2) t+o-Z t)J

:= lnS t) = lnS O)+ a-t0-2 )t+O-Z t)

'-

X(t)

The increments of this log process are:

dX t) = a-t0-2) dt+ o-dZ t)

Alternatively, we could derive this SDE by applying Itô's formula to the function X(t) = In S(t) .

Since the coefficients in the SDE for X t), ie the drift a-t0-2 and the volatilty 0-, are both

constant, X(t) is arithmetic Brownian motion. So statement (i) is correct.

From the equation above for X t)

, we find that:

X t+h)-X t) = a-t0-2 )h+O- Z t+h)-Z t))

:= var(X(t+h)- X(t)) = 0+0-2 var (Z(t+h) - Z(t)) = 0-2h

'-

N(O,h)

So statement ii) is also correct.

As n ~ 00, the limit on the left-hand side of statement (iii) becomes 1 (dX(t)f ' which we can

evaluate using the SDE above:

1 dX t))2 = 1L a-t0-2)dt+o-dZ t)r = 1 0-2dt=0-2 1 dt=0-2T

22





ExamMFE

Here, we've squared the expression in brackets and used the usual results that (dt)2 =0, dtxdZ(t)=O

and (dZ(t))2 = dt .

So statement iii) is also correct.

Solution 35

Answer: E

Difficulty level: .

The Black-Scholes model assumes that the stock price S(t) satisfies an SDE of the form:

dS(t)

-=adt+CTdZ(t)

S(t)


S t) = 5 0) exp L a -t CT2 ) t + CTZ t) J

~ In S t) = In 5 0) + a -t CT2 ) t + CTZ t)

The increments of this log process over a finite time interval (t, t + h) are:

In S t +h)- In S t) = a-tCT2 ) h + CT Z t+ h) - Z t))

So: var lnS t+h) I S t)) = var InS t+h) IlnS t))

= var In S t+ h)- In S t) I In S t))

= var1( a-tCT2 )h+CT(Z(t+h)-Z(t)J1

= CT2 var Z t+h)-Z t)) = CT2h

~

N(O,h)

Here, we've used the fact that the Brownian increment Z(t + h) - Z(t) is independent of the past history of

the process up to time t.

So statement i) is correct.

The variance of the proportional change in the stock price over the infinitesimal time interval

(t,t+dt) is:

varr dS t) i

S(t)J

=

var(adt+CTdZ(t)

IS(t))

=

CT2 var dZ t)) =CT2dt

L S t) ~

N(O,dt)

So statement ii) is also correct.


23



ExamMFE


The change in the stock price over the ininitesimal time interval (t, t + dt) is:

S(t + dt) - S(t) = dS(t)

So: var S t+dt) I S t)) = var S t+dt)-S t) I S t))

= var dS t) I S t))

= var L S(tH adt+ o-dZ(t)J I S(t) J

=S t)2 var adt+o-dZ t)I

S(t))

=S t)20-2 var dZ t))

= S(t)2 0-2dt

So statement iii) is also correct.

Solution 36

Answer: B

Difficulty level: .

Since there is a single source of randomness, the Sharpe ratio, calculated as a - r wil be the

0-

same for both assets. So, using the SDEs for the two asset prices and the value we are given for

the continuously-compounded interest rate r, we have:

0.07 -0.04

0.12

A-0.04

B

=? A = 0.25B + 0.04

The volatilty for a geometric Brownian motion process Y(t) and the corresponding log process

In Y(t) are the same, apart from the proportionality factor Y(t). So, using the equation given in

i), we see that B = 0.085 .

So: A = 0.25 0.085) + 0.04 = 0.06125

Solution 37

Answer: E


We can determie which of these processes has zero drift by finding their SDEs.

For U t)

, we have:

dU t) = U t+ dt) - U t) = PZ t+ dt)- 2) -PZ t)- 2) = 21 Z t+ dt) - Z tn = 2dZ t)

Since there is no II dt II term in this equation, the process U t) has zero drift.

24





ExamMFE

The SDE for V(t) is:

dV(t) = dt(Z(t)J2J-dt

= 2Z t)dZ t)+l 2) dZ t)J2 -dt

'-

t

= 2Z(t)dZ(t)

We ve used the Itô / Taylor series method to evaluate d t (Z(t) f J here.

Again, there is no dt term in this equation, so the process V(t) also has zero drift.

In fact, Z t)f -t is a well-known martingale based on standard Brownian motion. So you might have

recognized straight away that V(t) has no drift.

The SDE for W(t) is:

dW t) = dtt2Z t)j-2d1.b SZ S)dSl

We can evaluate dtt2 Z(t)j using the Taylor series method, remembering to include the extra t term.

The first d term on the right-hand side is:

dtt2Z(t)j= a~ tt2ZjdZ(t)+l a~2 tt2zjX(dZ(t)J2 + :t tt2zjdt

= t2 dZ(t) + ~ + 2tZ(t)dt

d1.b sZ(s)ds llOOkS more difcult to evaluate because it involves an integral. However, if we go back to

first principles, it is actually quite easy.

The second d term on the right-hand side is:

f rt 1 rt+dt rt rHdt

p/Z s)dsJ= J: sZ s)ds- J:sZ s)ds= Jt sZ s)ds=

tZ(t)dt

If you think of the integral as the sum of a sequence of very small elements, this last equation should be

obvious, since the only diference between the integral going up to t and the integral going up to t + dt is

that

we have added the

final element tZ t)dt. In general, d1.b f S,Z S))dsl = f t,Z t)).

Combining these gives:

dW t) = t2dZ t)+ Jß - Jß = t2dZ t)


25



ExamMFE


Again, there is no II dt II term in this equation, so the process W(t) also has zero drift.

Solution 38

Answer: E


This is the SDE for the Omstein-Uhlenbeck process. The solution to this equation is a standard result. It

can be derived using the method given below.

To solve this SDE, let Y(t) = eítt X(t).

We wil see that the SDE for Y(t) is easier to solve than the SDE for the original process X(t).

We can then find the SDE for Y(t) using the Itô/Taylor series method:

dY(t) = d(eíttX(t)J

= a~ ieíttXidX(t)+l a:2 ieíttXix(dX(t))2 + :t ieíttXidt

= eítt dX(t) + ~ + Â.eítt X(t)dt

=eítt 1Â.(a-X(t))dt+O dZ(t)l+ ~ +Â.eíttX(t)dt

= eítt i Â.adt + O dZ(t) J

= aÂ.eítt dt + O eítt dZ(t)

This SDE now has a simpler form, as there are no X(t) 's or Y(t) 's on the right-hand side.

If we rename the time variable as II s II and integrate over the time period (0, t), we get:

1 dY(s) = a 1 Â.eíts ds + 0 1 eM dZ(s)

'-

Y(t)-Y(O)

The first integral on the right-hand side is:

.b Â.eíts ds = a eíts I ~ = a ( eítt - 1 )

Since Y(t) = eítt X(t) and Y(O) = eO X(O) = X(O) , we now have:

So: eíttX(t)-X(O)

=

a

(eítt -1)+0 .beMdZ(S)

We can rearrange this equation to make X(t) the subject by taking the X(O) over to the right-

hand side, then dividing through by eítt:

eítt X(t) = X(O) + a( eítt -1) + 0 1 eM dZ(s)

26





ExamMFE

X(t) = e-Ât ix(o)+a( eÂt -1) + 0 1 eÂS dZ(s) 1

= X(O)e-Ât +a( 1- e-Ât) + 0 .b e-Â(t-s) dZ(s)

Solution 39

Answer: E


Remember that the solution to the SDE d~~~) = adt+ O dZ(t) is S(t) = 5(0) eXPL (a-t0 2) t+ O Z(t) L

Jane's portfolio at time t consists of two components:

. an amount lpW(t) held in stock

. an amount (l-lp) W(t) held in the risk-free asset.

The component invested in stock wil change in value in the same way as the stock price S(t):

d(lpW(t)J =adt+O dZ(t)

lpW(t)

~ d(lpW(t)J = lpW(t)1 adt+O dZ(t)J

The component invested in the risk-free asset wil earn interest at the risk-free rate:

dC(l-lp)W(t)J rdt

(l-lp) W(t)

~ d l-lp)W t)J= l-lp)rW t)dt

So the SDE for the whole portfolio wil be:

dW(t) = dClpW(t) J + d( (l-lp) W(t)J

= lpW(t)1 adt+O dZ(t)J+(l-lp)rW(t) dt

= tlpa W(t) + (l-lp)r W(t)r dt + lpO W(t)dZ(t)

or

dW(t) = tlpa+ (l-lp)rr dt + lpO dZ(t)

W(t)

This almost matches A, except the lp in the lpO is missing. So we can eliminate A and consider the other

choices, which are all equations for W(t) .

Since lp, a, rand 0 are all constants, this SDE is for a geometric Brownian motion process.

Its solution is:

W(t) = W(O) exp L llpa+(l-lp)r -t(lpO )2 J t+ rp Z(t) J


27



ExamMFE


This doesn't match B because the Yi term is missing and it doesn't match C because the rp isn't squared.

The stock price satisfies the SDE:

dS(t)

-=adt+O'dZ(t)

S(t)


S(t) = s(o)exp(( a-f0'2) t+O'Z(t)J

So (S(t)jS(O)Jai, which appears in the remaining answers D and E, is equal to:

(S(t)jS(O)Jai =fexp((a-f0'2 )t+O'Z(t)Jr

= exp (rp( a-f0 2 ) t+rpO Z(t) J

The equation we derived earlier for W(t) can now be written as:

W(t) = W(O) exp (1 rpa+(l- rp)r-f(rpO')2j t+ rpO' Z(t) J

= W(O) exp (1 rpa+(l- rp)r-f(rpO )2j t+ rpO Z(t) J

x~ S(t)jS(O) Jai exp ( -rp( a-f 0 2) t - rpO Z(t) J

~1

We can cancel the a terms and the Z(t) terms to simplify this:

W(t) = W(O) (S(t)jS(O) Jai exp U (1- rp)r-f(rpO')2j t J exp ( -rp( _f0'2 ) t J

= W(O) (S(t)jS(O)Jai exp U (1- rp)r -f(rpO'f +frp0'2 l t J

= W(O) (S(t)jS(O) Jai exp (1 (1- rp)r +f rp(l- rp)0 2 l t J

= W(O) (S(t)jS(O) Jai exp ((1- rp)1 r+frp0'2 l t J

Ths matches E.

Solution 40

Answer: A Difficulty level: .

As n -? 00, the limit given in the question becomes 1 (dZ(t) f ' which we can evaluate as

follows:

1 (dZ(t)J3 = 1 f( dZ(t)f dZ(t)l = 11 dt dZ(tH = 10= 0

28





ExamMFE

We've used the

usual results

that (dt)2 =0, dtxdZ(t)=0 and (dZ(t))2 =dt here.

So this is a determiistic quantity with a fixed value of zero. Ths corresponds to the degenerate

normal distribution N(O, 0), which has mean zero and variance zero.

Solution 41

Answer: B


Applying the Itô/Taylor method to X(t) = (R(t))2 tells us that:

dX t) = 2R t)dR t)+l 2) dR t))2 = 2R t)dR t)+ dR t))2

To simplify this further, we need to find the SDE for R(t).

This question would be straightforward ifwe were given the SDEfor R(t). But, since we are not, we need

to use a little ingenuity. Here's one method that can be used.

We are given the following equation for R(t):

R t) = R O)e-t +0.05 1-e-t )+0.11 é-t .JR s)dZ s)

This takes a slightly simpler form if we multiply through by et:

R t)et =R 0)+0.05 et -1)+0.11es .JR s)dZ s)

The increments of this equation over the time interval (t, t+ dt) are:

4 R(t)et J = 0+0.05~+0.let .JR(t)dZ(t)

=etdt

To derive the increment of the integral term, we've thought of the integral as the sum of a sequence of very

small elements. The only diference between the value of an integral going up to t and the value of an

integral going up to t+ dt is the value of the final element, which is et .JR t)dZ t).

If we let Y(t) = R(t)et and apply the Itô/Taylor series method, the left-hand side of this equation

becomes:

drR t)etJ=dY t)=ay dR t)+la2Y2 dR t))2+ay dt

L aR aR at

=etdR t)+~ +R t)etdt

= et dR(t) + R(t)et dt


29



ExamMFE


So, substituting this into the previous equation, we get:

et dR(t) + R(t)et dt = 0.05et dt + O.let .JR(t)dZ(t)

Rearranging this gives:

et dR(t) = -R(t)et dt + 0.05et dt + O.let .JR(t)dZ(t)

= (0.05 - R(t)J et dt + O.let .JR(t)dZ(t)

~ dR t) = 0.05- R t)J dt+ O.l.JR t)dZ t)

So we see that R t) is in fact a CIR process, as you might have guessed from the presence of the square

root.

If we now substitute this into our equation for dX(t) at the start of the solution, we get:

dX(t) = 2R(t)dR(t) + (dR(t) J2

2

= 2R(t*0.05- R(t)J dt+ 0.1.JR(t)dZ(t)J+l(0.05 - R(t)J dt+ O.iJR(t)dZ(t)J ,

=O.OlR(t)dt

= i O.lR t) -2R t)2 J dt

+ 0.2R(t).JR(t)dZ(t)l

+

O.OlR(t)dt

= ( O.l1R(t) - 2R(t)2 J dt+ 0.2R(t).JR(t)dZ(t)

Since X(t) = (R(t)J2, we can write R(t) = .JX(t) . So this becomes:

dX(t) = L O.l1.JX(t) - 2X(t) J dt+0.2 (X(t)f/4 dZ(t)

This matches answer B.

Solution 42

Answer: D


Remember that the solution to the SDE d~~;) = adt+ adZ t) is S t) = 5 0) exp a--r 2) t+aZ t) L

The log of the geometric average is:

In G = In r (S(1)XS(2)XS(3)Jl/3 J =t(1n 5(1)+ In 5(2) + In S(3)J

The solution to the SDE given for the stock price is:

S t) = s o)exp 0.03--rXO.22) t+0.2Z t)J = S 0)exp 0.Olt+0.2Z t)J

~ In S t) = In 5 0) + O.Olt + 0.2Z t)





ExamMFE

So we can express In G in terms of the log of the underlying standard Brownian motion:

In G = l(1n 5(1)+ In 5(2)+ In 5(3))

Lin 5(0) + 0.01 + 0.2Z(1) :

=l +lnS(0)+0.02+0.2Z(2)

+ In 5(0) + 0.03 + 0.2Z(3)

= lnS(0)+0.02+ 0.2 x(Z(1)+Z(2)+Z(3))

3

So, to work out the variance of In G, we need to work out the variance of Z(l) + Z(2) + Z(3)

,

which we can do by first expressing it in terms of independent (non-overlapping) increments:

Z(l) + Z(2)+ Z(3) =i Z(3) -Z(2)) + 2i Z(2) -Z(l)) +3iZ(1) - Z(O))

One way to get the correct expression here is to work backwards. Our expression wil involve the

increments Z(3) - Z(2), Z(2) - Z(l) and Z(l) - Z(O). Since we need to have Z(3) with a coeffcient of 1,

our expression will contain Z(3) - Z(2). But this expression implies a coeffcient of -1 for Z(2). So we

need to multiply the increment Z(2) - Z(l) by 2 to fix this. But this now gives a coeffcient of -2 for the

Z(l) term, so we need to multiply the Z(l) - Z(O) increment by 3. And since Z(O) = 0 , we don t have to

worry about this term.

We then have:

var Z l)+ Z 2)+ Z 3)) = var Li Z 3) - Z 2)) + 2i Z 2) - Z l)) +3iZ 1) -Z O)JJ

= var (Z(3)- Z(2))+ varL 2iZ(2)- Z(l)lJ + varL 3iZ(1)- Z(O))J

= var (Z(3)- Z(2))+ 22 var (Z(2) - Z(l)) +32 var (Z(l) -Z(O))

- - -

(O,l) N(O,l) N(O,l)

=1+4+9=14

So, finally, we get:

(02)2

var

(1n

G) = ~ x14 = 0.062

Solution 43

Answer: E

Difficulty level: .

Applying the Itô/Taylor method to y(t) = (x(t)r1 tells us that:

dy(t)=-(x(t)r2 dx(t)+-tX2(x(t)r3 (dx(t))2

=-(x(t)r1 dx(t) +lX2(x(t)r1 (dX(t)J2

x t) 2 x t)


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ExamMFE


dx(t) -1

ince -=(r-r€)dt+O dZ(t) and y(t)

=

(x(t)) ,this

becomes:

x(t)

dy(t) = -( x(t) r1 ((r - r€ )dt + O dZ(t)) +lx 2 (x(t) r1 ((r - r€ )dt+ O dZ(t))2

= -y(t) ((r -r€ )dt + O dZ(t))+ y(t) 0 2 dt

=?

dy(t) = _ ((r _ r€ )dt+ O dZ(t)) + ~ dt = (r€ -r + 0 2 )dt - O dZ(t)

y(t)

Solution 44

Answer: E


Since there is a single source of randomness (the Brownian motion Z(t)), the Sharpe ratio,

a-r

calculated as - wil be the same for both assets. So, using the value we are given for the

0'

continuously-compounded interest rate r, we have:

0.06-0.04

0.02

0.03-0.04

k

=? k= -0.01

The value we have calculated for the volatility k of stock 2 is negative. This means that the movements of

the two stock prices are negatively correlated. When the price of stock 1 goes up, the price of stock 2 tends

to go down, and vice versa.

If the portfolio being set up contains 1 unit of stock 1, x units of stock 2 and y units of the risk-

free bond, then its value at time t wil be:

P(t) = 51 (t)+XS2 (t)+yB(t)

The change in the portfolio value over the next time instant (t, t+ dt) wil be:

dP(t) = dS1 (t) + xdS2 (t) + ydB(t)

= 51 (t) (0.06dt+0.02dZ(t)) +xS2 (t) (0.03dt -O.OldZ(t))+yB(t) (0.04dtJ

= (0.0651 (t) + 0.03xS2 (t)+ 0.04yB(t)J dt+ (0.0251 (t) -0.OlxS2 (t)) dZ(t)

For this to be (instantaneously) risk-free, we need the volatilty term to equal zero. Based on the

current prices, this requires:

0.0251 (t) - 0.OlxS2 (t) = 0

=? 0.02xl00-0.0lxx50 = 0

2

=?x=-=4

0.5

As a shortcut in this question, once we had established the value k = -0.01, we could note that Stock 2 is

half as volatile as Stock 1 (by comparing the coeffcients 0.02 and -0.01 in the SDEs). So to achieve a risk-

free portfolio, we wil need a holding of Stock 2 that is worth twice as much as our holding of Stock 1. But

since the price of Stock 2 (50) is half the price of Stock 1 (100), this means that we need to hold 4 times the

number of units (shares) of Stock 2.

32





ExamMFE

Solution 45

Answer: A

Difficulty level: .

If a denotes the expected anualized continuously-compounded expected rate of stock price

appreciation, the future price of the stock at time T, according to the Black-Scholes model, wil

be:

S(T) = 5(0)

exp

L(a-to.2)T +O Z(T)J

In this formula, the effect of any dividends that are payable is already incorporated in the parameter a .

Since Z(T) ~ N(O, T) =.J xN(O,l), the upper confidence limit for Z(T)

, allowing for 5% in each

tail, is 1.645.J .

We are assuming here that a symmetrical confidence interval based on the underlying normal distribution

is intended here.

So the upper confidence limit for S(T) is:

S(T) = S(0)exPL(a-to-2)T + 1.645o-.JJ

= 0.25 eXPL (0.15 -tXO.352 )(0.5)+ 1.645xO.35~J

=0.393

Solution 46

Answer: D


Girsanov s theorem tells us that, when we are working with Brownian motion, the Brownian

increments under the risk-neutral and real-world probability measures are related by adjusting

the drift by the Sharpe ratio:

dZ(t)*=dZ(t)+( a~r)dt

If we add up (integrate) these increments over the period 0:: t :: 0.5, we get:

Z(O.5)* = Z(O.5) + ( a ~ r ) (0.5)

Taking expectations based on risk-neutral probabilties gives:

E * (Z(O.5) *) = E* (Z(0.5))+( a~r) (0.5)

We know that, under the risk-neutral measure, Z(t)* ~ N(O, t), so that E * (Z(O.5) *) = 0 . The

question also tells us that E * (Z(O.5)) = -0.03 .

So:

(a-r)

= -0.03 + -; (0.5)


33



ExamMFE


~

a-r = 0.03 = 0.06

6 0.5

In this equation, 6 is the volatility of the stock and a is the real-world expected return on the

stock (including reinvestment of dividends). We can see from the SDE given in the question that

6 = 0.25. We can also see from the SDE that the expected real-world rate of return on the stock is

0.05 (ie 5 ). However, this is just the rate of share price growth without reinvestment of the

dividends received. Since the dividend rate is 1 , the total rate of return is a = 0.05+ 0.01 = 0.06.

So: 0.06 - r = 0.06

0.25

~ r

=0.06-0.06x0.25=

0.045

Solution 47

Answer: A


We wil assume here that the derivative has a fixed maturity time T. The quantity Ft,T (52) given in the

question represents the fair price at time t that a trader should agree to pay at time T to receive a cash

payment equal to S(T)2.

The price of the derivative at time t can be calculated as the forward price discounted at the risk-

free interest rate (which is the same as the prepaid forward price). Ths equals:

e-r(T -t) Ft,T (52) = r S(tn 2 exp ((0.18 - r)(T - t) J

Method 1

The price of this derivative at time t should equal the discounted risk-neutral expectation of the

payoff. So we should have:

r S(tn 2 exp ((0.18 - r)(T - t) J = e -reT -t) E * L S(T)2 I S(t) J

We can rearrange this to get:

E*L S T)21 S t)J =rS tn2 exp 0.18 T -t)J ... 1)

The SDE given for S(t) in the question is geometric Brownian motion (as it must be for a Black-

Scholes framework) and the solution to this SDE over the time period (t, T) is:

S T) = S t) exp L lL -t 2) T - t) + 6 Z T) - Z t)) J

The square of this is:

S T)2 = S t)2 eXPL 2 lL-t62 ) T -t)+26 Z T)-Z t))J

= S t)2 exp L 2lL - 2) T - t) + 26 Z T) - Z t)) J

34





ExamMFE

We can use this to evaluate the expectation in equation (1) above directly:

E* S T)2 i S t)J = E*l S t)2 eXPL 2ll- j2 ) T-t)+2 j Z T)-Z t) )JJ

= S(t)2 exp L (2ll- (j2 ) (T -t) J E * (exp L 2(j( Z(T) - Z(t)) JJ

To simplify this further, we can use the fact that Z(T)-Z(t)~N(O,T-t), and is independent of

S t), to get:

E* S T)2 i S t)J = S t)2 eXPL 2ll- j2 ) T -t)Jexp 0+lX 2 j)2 T -t)J

= S( t)2 exp L ( 2ll + (j2 ) (T - t) J

Here we ve used the result that, if X ~ N (ll, (j2 ), then E ( er X J = exp (r ll + l r2 (j2 ) .

If we compare this with equation (1) above, we see that:

2ll + (j2 = 0.18

So: ll =l(0.18-(j2) =l(0.18-0.42) = 0.01

Method 2

Since we are assuming a Black-Scholes framework, this price should satisfy the Black-Scholes

partial differential equation. The basic form of the Black-Scholes PDE, when the underlying asset

is a nondividend-paying stock, is:

1 2 2

rC t, St) = - j St rt + rStl1t + 0t

2

Recall that this equation is derived by looking at the sources of profit on a delta-hedged position

over a short time interval of length h. To allow for the fact that the stock in this question pays

dividends (at rate ö), we need to make an extra adjustment for the dividend payment of öStl1th

that wil be received during this period. This leads to the following modified equation, which

correctly takes into account the dividends:

1 2 2

rC t,St) =- j St rt + r-ö)Stl1t +Ot

2

If we evaluate the Greeks using the formula for the derivative value, namely

C t, St) = r S t)J2 exp 0.18 -r) T - t)) , we get:

11= ac =2Sexp 0.18-r) T-t))=2 C t,Sd

as St

r= a2c =2exp((0.18-r)(T-t))=2 C(t,St)

as2 Sf


35



ExamMFE


ac 2

=- = -(0.18-r)S exp(0.18(T -t)) = -(0.18-r)C

at

Substituting these into the Black-Scholes PDE then gives:

rC(t 5 )=2.(J2S2x2 C(t,Sd +(r-ö)S x2 C(t,Sd (0.18-r)C(t,St)

t 2 t 52 t 5

t

Canceling the St' s and dividing through by C (t, St ) , this becomes:

r = (J2 +2(r-ö)-(0.18-r)

or 0 = (J2 +2(r-ö)-0.18

The SDE given in the question tells us that the instantaneous risk-free rate of drift of the share

price without dividends reinvested) is l1. So the overall rate of drift with dividends reinvested)

is l1 + Ö , which must equal the risk-free interest rate if we are using risk-neutral probabilties. So

r = l1 + ö. We can also see from this equation that (J = 0.4 .

Substituting these values gives:

0=0.42 +2l1-0.18

=: l1 = 0.01

Solution 48

Answer: A


The quadratic variation of a process X(t) over the time interval (0, TJ is defined to be

lim iJ x(jT)_x(U-l)T)J2. This corresponds to a sum of

the

form lim I (X(t+dt)-X(t))2

n-700 j=ll n n OgS;T

, where the limit involves dividing the time interval into smaller and smaller pieces, or the integral

1(dX(t))2.

For the process W(t)

, the quadratic variation over the interval 0, TJ is:

Vf(W) = ( dW(t))2

But, from ordinary calculus, we know that:

dW(t) = d(t2 J = 2tdt

So:

(dW(t))2 =

(2tdtf =4t2 (dt)2 =0

'-

0

=: Vf4 (W) = 0

36





ExamMFE

The graph of the process X(t) as a function of t is a step function that is flat apart from at the

integers. So, over the range (0,2.4), the increments xC~)-X( (j-nl)T) wil all be zero except

at the times corresponding to t = 1 and t = 2. So the quadratic variation over the interval 0,2.4)

is:

2 ) 2 2

V2.4 X = 1 + 1 = 2

For the process Y t)

, the quadratic variation over the interval 0, T) is:

Vf(Y) = 1 (dY(t))2

Using stochastic calculus, we know that:

dY t))2 =

(2dt+0.9dZ(t))2 =0.81dt

So: Vf Y)

= 1

dY t))2= 10.81dt=

0.81T ~Vf4 Y)=0.81x2.4=1.944

Arrangig these in order, we get:

Vf4 W) ~ Vf4 Y) ~ Vf4 X)

'- '- '-

0 =1.944 =2

Solution 49

Answer: C


Since Y t) = S t) f and the stock price S t) must be positive, we know that S t) = Y t)r2 .

Applying the Itô/Taylor method to this function tells us that:

dS(t)=~dY(t)+l a2s (dY(t))2

ay 2 ay2

= t Y t)rl/ dY t)+txt -t ) Y t)r1l/ dY t))2

If we now adjust the Y(t) factors and use the SDE for dY(t) given in the question, we get:

Y(t)

dS(t) =l(Y(t)r2 dY(t) _l(Y(t)r21 dY(t)J2

2 Y t) 8 L Y t)

= tS t) 1.2dt - O.5dZ t)) -tS t) r 1.2dt - O.5dZ t) )2,

=0.25dt

= S t) 0.56875dt-0.25dZ t))

So S(t) is geometric Brownian motion. The solution to this equation is:

S t) = 5 0) eXPL 0.56875 -tXO.252 ) t-0.25Z t) J

= 5 0) exp 0.5375t - 0.25Z t))


37



ExamMFE


Since Z(t) is standard Brownian motion, the distribution of Z(2) is N(O,2). So a 90

(symmetrical) confidence interval for the value of Z(2) is :t.645v'.

Using the previous equation, the corresponding values of 5(2) are:

5(2) = 5(0) exp L 0.5375(2) - 0.25(: 1.645v') J

Since Y O) = 64, we know that 5 0) = 8. So the confidence interval for 5 2) is:

5(2) = 8exPL 0.5375(2) - 0.25(: .645v') J

The upper limit U is obtained by taking the negative sign in this expression, which gives 41.93.

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ExamMFE


Chapter 8 Solutions

Solution 1

Answer: A


The proposed model is:

dP(r, t, T) = Lr(t)+ ß~r(t)(T - t) J P(r, t, T)dt+ o-~T - tP(r, t, T)dZ(t)

which is of the form:

dP r, t, T) = a r, t, T)dt + q r, t, T)dZ t)

P(r,t,T)

where: a r,t,T) = r t)+ ß~r t) T -t)

and q r, t, T) = o-~T - t

The Sharpe ratio is equal to:

a(r,t,T)-r(t)

q(r,t,T)

r(t) + ß~r(t)(T - t) - r(t)

o ~T t

= ß ~r(t)

0-

Ths ratio depends on r and so is not constant. So Statement 1 is factually correct. However, it is

quite reasonable for the risk premium on a bond to depend on the level of interest rates, so this

isn't a problem. So this is not a valid objection for a bond price modeL.

The Sharpe ratio above is independent of T, so Statement 2 is not factually correct.

Ø r, t, T)

If the Sharpe ratio had been dependent on T then this would have been a valid objection, since we require

the Sharpe ratio for bonds of diferent maturities to be equal.

The bond volatility is o-~T - tP(r, t, T), which does decrease as maturity approaches, ie as

T - t ~ 0 , so Statement 3 is not factually correct.

We would expect the volatility to decrease as maturity approaches, since the zero-coupon bond price must

approach the bond's redemption value, which has a fixed value ofl.

Solution 2

Answer: E

Difficulty level: .

The Vasicek model for the short rate is:

dr t) = a b-r t)Jdt+o-dZ

where a, band 0- are positive constants and Z is Brownian motion.


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ExamMFE


Since dZ is normally distributed, O dZ can be arbitrarily large and negative, causing r to

become negative.

So Statement 1 is correct.

When r(t) ~ b the drift is positive, but when r(t):; b the drift is negative, so the process is mean-

revertig to b .


The Sharpe ratio for the Vasicek model is normally assumed to be constant and is typically

denoted by ø.


Solution 3

Answer: B

Difficulty level: .

The Cox-Ingersoll-Ross model for the short rate is:

dr(t) = arb - r(t)) dt + O ~r(t)dZ

where a, band 0 are positive constants and Z is Brownian motion.

Since dZ is normally distributed, it can take arbitrarily large and negative values. However, as

r approaches zero, the drift increases and the ~r(t) in the volatilty term causes the volatility to

decrease, thus preventing r from actually reducing to zero or below.

So Statement 1 is not correct.

When r(t) ~ b the drift is positive, but when r(t):; b the drift is negative, so the process is mean-

reverting to b .


The Vasicek model, and not the Cox-Ingersoll-Ross model, assumes the short rate follows an

Ornstein-Uhlenbeck process.

So Statement 3 is not correct.

Solution 4

Answer: B


The probability measure assumed in the specifcation of an Ito process (whether it is the real-world or the

risk-neutral measure) affects the drif of the process. As a result, the long-term average value of the process

difers according to the probability measure assumed in the model (although, in practice, the adjustment

wil usually be small). This is why two diferent parameters, l1 and l1 *, are mentioned in the question.

In the notation implied by the question, the Vasicek model is specified by the following SDE

(under the risk-neutral probabilty measure):

drt = a(l1 * -rt ) dt + 0 0 dZt

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ExamMFE

To apply this modeL, we need to specify values for:

the mean reversion rate a, which we are told is equal to 0.2

l1 * , which is the long-term value of the short rate in a risk-neutral world

ro, the initial value of the short rate

the volatility parameter (To, which remains constant over time.

Once this model has been specifed, the statistical distributions (under the risk-neutral probability measure)

of the future values of the short rate and the integral r(s)ds (both conditional on the initial value

assumed for the short rate) are fully determined. So no further information about the future values of the

short rate (which evolve randomly ) are required.

The zero-coupon bond price at time 0 can then be calculated from the formula:

P(O,T,r(O)J = E~ Le-R(O,T) J = E~ L exp( -, r(S)ds)J

This does not require us to specif any further inputs (apart from T itself).

Solution 5

Answer: E


The CIR model of the short rate can be specified by the following SDE:

dr t) = a b-r t)Jdt+ T~r t)dz t)

In this equation, (T is a parameter whose value must be specified. The volatility of the process at

time t is (T~r(t). Since the short rate r(t) wil vary stochastically (randomly) in the future, so

also wil the value of the volatilty.

Solution 6

Answer: C


In both the CIR and Vasicek models interest rates follow diffusion models of a specific type. The

bond price formulas in these models are of the form P r , t , T) = A t , T) e - B t , T) r t). Both models

are time-homogeneous, which means that the values of the functions involved depend only on

T - t , the length of the time interval involved, not on the actual values of t and T. So we have:

T-t=T1-t1 ~A(t,T)=A(t1,T1) and B(t,T)=B(t1,T1)

So the given inormation supplies us with 2 equations in the two unkowns A(O ,3) and B(O,3):

P(0.04,O ,3) = 0.91792 = A(O,3)e -B(O,3)xO.04

P(0.05,O,3) =P(0.05,3,6) = 0.89990 = A(O,3)e-B(O,3)XO.05

~ 0.91792 = eO.01B(O,3) ~ B(O,3) = 1.98266 ~ A(O,3)= 0.99368

0.89990


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Solution 7

Answer: D

Difficulty level: .

Using results from Solution 8.5, we have:

100P 0.06,t,t+3) = 100XA O,3)e-B O,3)XO.06 = 88.22

Solution 8

Answer: E


The trick suggested in this question provides a method that can sometimes be used to find the expected

value of a process defined by a stochastic diferential equation.

The short rate r(t) follows the SDE:

dr t) = 0.2 0.05-r t))dt+0.OldZ t)

If we take expectations this becomes:

E dr t)) = 0.210.05- E r t))J dt+O.01 E dZ t))

The Brownian increment dZ(t) has a mean of zero. Also, the expected value of the increment

dr(t) wil equal the change in the expected value of r(t), which equals the change in m(t):

E (dr(t)) = E (r(t + dt) - r(t)) = E (r(t + dt)) - E (r(t)) = m(t + dt) - m(t) = dm(t)

So we deduce that:

dm t) = 0.210.05 -m t)J dt

Note that m(t) is not a stochastic quantity and that this is an ordinary, not a stochastic, diferential

equation.

We can solve this differential equation by rearranging it to separate the variables, and then

integrating from time 0 to time T (say). This gives:

rT dm(t) = rT 0.2dt

.b 0.05-m(t) .b

So: -ln10.05-m t)jl~ = 0.2T

- In 1 0.05 - m(T)J + In 10.05 - r(O)J = 0.2T

To obtain the last line, we have used the fact that m(O) = E(r(O)) = r(O) .

We can now exponentiate both sides and rearrange to find m T)

, which gives:

m(T) = r(O)e -D.2T + 0.05 (1- e -D.2T)

Replacing the T' s with t' s then gives the required formula.

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Solution 9

Answer: C


Observation 1) is correct. The model involves only 4 parameters - a, b, r O) and ). The

values for these parameters can usually be chosen (by solving a set of simultaneous equations) so

that the model reproduces the prices of any 4 or fewer) bonds, but it cannot reproduce the prices

of more than 4 bonds except by coincidence).

This is the problem referred to in one of the learning outcomes for this topic, which states: Explain why

the time-zero yield curve in the Vasicek and Cox-Ingersoll-Ross bond price models cannot be exogenously

prescribed.

Model 3 is called the Hull-White modeL. It gets round this problem by using a function in place of the

parameter b. The function can be thought of as an infinite set of parameters, one for each future time.

Observation (2) is correct. The price of a zero-coupon bond can be calculated as a risk-neutral

expectation of an exponential function involving the stochastic quantity 'r(s)ds. With Models 1

and 3, which assume a constant volatility, the distribution of this quantity is normally distributed

and the ZCB price is lognormally distributed. As a result, the ZCB price can take any positive

value, including values greater than 1.

Note that a zero-coupon bond price exceeding $100 implies that the spot rate of interest covering the same

term, which can be calculated as -~ In P t, T, r t))

, is negative.

T-t

This problem is most likely to occur if

the current rate r t) and the long-term rate b are close to zero

the parameter a has a small value so that the mean reversion is weak)

the volatility () is high

the bond has a short remaining life T - t .

In calculations with realistic parameter values, this situation doesn't usually arise.

Observation 3) is not correct. Although the future short rate evolves stochastically randomly)

under each of these models, the price of a zero-coupon bond is calculated as an expected value,

using the formula p t, T,r t)) = E; L exp - r r S)ds) L and hence has a determistic value.

Solution 10

Answer: D


Let Rt T, T +s) denote the annualized effective interest rate applying between times T and

T + s , implied by the yield curve at time t.

From the question, we know that:

RO(O,2)

=0.08

Ro(O,3) = 0.09


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ExamMFE


The rate underlying the FRA, Ro(2,3), can be found by rearranging the identity:

(1 + Ro(O,2)J2 (1 + Ro(2,3)J = (1 + Ro(O,3)J3

to give:

RO(2,3) (1+Ro(O,3)J3 -1 = 1.093 -1 =11.0279%

(1+Ro(O,2)J2 1.082

The payoff at time 3 under the FRA is given by:

FRA payoff = 10mx(Ro(2,3)-R2(2,3)J

The value of this payoff at the exercise date (time 2) would be:

10mx Ro(2,3)-R2(2,3)

1+R2(2,3)

A two-year call option on this FRA with strike K would have a payoff at time 2 given by:

Option payoff = $10mxmaJ 0, Ro(2,3)-R2(2,3) KJ

L 1+R2(2,3)

It costs nothing to enter into an FRA, so the strike for an at-the-money FRA option wil be zero.

Setting K = 0 and rearranging gives:

Option payoff = 10mxmaJ 0, RO(2'3)-R2(2'3)J

L 1+R2(2,3)

= 10mxU+R (2,3nmaJo, 1+Ro(2,3)-1-R2(2,3) J

L U+R2(2,3nu+Ro(2,3n

= 10mxU+Ro(2,3nmaJo, 1 1 J

L 1+R2(2,3) 1+

Ro(2,3)

1

This is the same as the payoff to 10mxtl+Ro(2,3n bond call options with strike ,so

1+Ro(2,3)

the FRA option can be valued using the Black formula:

c = $10mxtl + RO(2,3nx,p(O,2)(FN(d1)-KN(d2)l

Value of 1 'call option

where:

P(O,2) = 1 1

(1+RO(O,2)J2 = 1.082

1

(so that F and K cancel with the t 1 + Ro (2,3) J factor)

1+Ro(2,3)

=K

In(F/K)+t0.2x2 0' 0.10 .

d1 = M = M = M =0.0707 (=0.07 to 2 decimal places)

0' ,2 ,2 ,2

d2 = d1 -O'-J = 0.0707 -0.10-J = -0.0707 (= -0.07 to 2 decimal places)

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ExamMFE

So:

c = 10mx N(d1) -N(d2)

(1+Ro(O,2)f

= 10mx 0.5279 - 0.4721

1.082

= 0.478m

Using unrounded values for d1 and d2 would give a more accurate answer of 0.483m.

Solution 11

Answer: E


We need to use a standard trick here, which involves expressing the payoff function from the cap in a form

that matches the payoff function of an option on a zero-coupon bond.

The present value at time 1 of the payment from the caplet (ignoring the 1,000,000 factor) is:

max(R2 -0.05,0) (R2 -0.05 )

=

max ,0

1+R2 1+R2

=max((1+R2)-1.05,o)

1+R2

=max(l- 1.05 ,0)

1+R2

= 1.05 max (-- -~, 0)

1.05 1+R2

= 1.05 max (-- - PI' 0)

1.05

where PI denotes the price at time 1 of a zero-coupon bond maturing at time 2.

So the value at time 1 of the caplet matches the payoff from a put option expiring at time 1 with a

strike price of -- on the zero-coupon bond, multiplied by 1.05.

1.05

Since the log of the ZCB price has a normal distribution, the ZCB price itself wil have a

lognormal distribution. We can therefore use the Black-Scholes formula to value the put option.

The values of d1 and d2 for the Black-Scholes formula are:

in( 0.90 )+.1(0.0001)(1)

1jl.05xO.95 2

d1 = -0.5227 (= -0.52 to 2 decimal places)

0.01 1

and d2 = d1 - (j.f = -0.5227 -0.01 = -0.5327 (= -0.53 to 2 decimal places)

Note that the l-year zero-coupon bond price takes the place of the usual e -rT factor in this calculation.


1

- x 0.95N(0.53) - 0.90x N(0.52)

1.05

= --XO.95 xO.7019 - 0.90x 0.6985

1.05

=0.00640


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ExamMFE


So the value of the caplet is:

l,OOO,OOOxl.05xO.00640 = 6,720

Using unrounded values for d1 and d2 would give a more accurate answer of 6,794.

Solution 12

Answer: D


We can find the value of this interest rate cap by valuing each caplet separately and then adding the

answers together.

Payment at time 1 (frst caplet)

The amount of the payment for the first caplet (payable at time 1) is known at the outset, since we

can deduce the effective interest rate applicable over the first year from the price of the ZCB

maturing at time 1:

1

-=0.95

1+i1

So the amount of the first caplet wil be:

=? i = 0.052632

l,OOO,OOOx(0.052632-0.05) = 2,632

and its value at time 0 is:

2,632xO.95 = 2,500

Payment at time 2 (second caplet)

We have already calculated (in the previous question) the value of the caplet payable at time 2,

which was 6,720.

Payment at time 3 (third caplet)

We can find the value of the caplet payable at time 3 using the same method as in the previous

question, ie by expressing its value in terms of a European put option exercisable at time 2 on a

ZCB maturing at time 3.

The values of d1 and d2 for the Black-Scholes formula are:

in 0.85 )+1- 0.0003) 2)

1jl.05xO.90

~ M -0.3294 (=-0.33 to 2 decimal places)

,,0.0003 x,,2

and d2 = d1 -ry.J = -0.3294-,J0.0003 xJ2 = -0.3539 (= -0.35 to 2 decimal places)

d1

Here the 3-year zero-coupon bond price takes the place of the e-rT factor.


1

- x 0.90N 0.35) - 0.85 x N 0.33)

1.05

= --X 0.90 xO.6368 - 0.85x 0.6293

1.05

=0.01092

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ExamMFE

So the value of this caplet is:

l,OOO,OOOxl.05xO.Ol092 = 11,470

So the total value for the cap is:

2,500+ 6,720 + 11,470 = 20,690

Using unrounded values for d1 and d2 would give a more accurate answer of 22,310.

Solution 13

Answer: B

Difficulty level: .

Po (1,3) = Pt,3~ = 0.8163 = 0.86527 ~ l,OOOPo (1,3) = 865.27

P 0,1 0.9434

Solution 14

Answer: B


Since the option is at-the-money and the current price of a three-year zero-coupon bond maturing

for 1,000 is 816.30, the strike is K =816.30 .

According to Black's formula:

P(O,3)

F = forward bond price = 1,000 ( = 865.27

P 0,1)

K = strike price = l,OOOP(O,3) = 816.30

The rest of the calculations are as follows:

d - In(F / K) + O.5a.2T _ 0.0583 + 0.5xO.052xl

-11903 (-119 2 d . 1 1 )

- a.J - 0.05.J -. -. to ecima paces

d2 = d1 -a.J = 1.1903 - 0.05 = 1.1403 (= 1.14 to 2 decimal places)

~ P = P(O,l)(KxN(-d2) - FxN(-d1))

= 0.9434(816.30xN(-1.14) - 865.27xN(-1.19))

= 0.9434(816.30xO.1271 - 865.27xO.1170)

= 2.37

Solution 15

Answer: C


We can use the binomial interest rate tree to calculate the prices of 2-year and 3-year zero-coupon bonds.

These can be used to deduce the required forward rate.

Let P(O, T) denote the price at time 0 of a zero-coupon bond that pays 1 at time T.

Let Rt(T, T +s) denote the continuously-compounded interest rate applying between times T

and T + s , implied by the yield curve at time t.

The interest rate we need to find is then Ro (2,3) .


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ExamMFE


The value at time 0 of 1 payable at time 3 can be expressed in the following two ways:

P(O,3)

and e-Ro 2,3)p O,2)

Equating these two expressions gives:

R 23)=_ln P O,3))=ln P O,2))

, P(O,2) P(O,3)

We can construct a binomial interest rate tree from the inormation given in the question and use

this to derive the required bond prices.

12

..

0

-- 8

/

~

8

6 ~

-- 4

Time

0 Time 1

Time

2

P(O,2) = e-o.08 (fe-0.10 +-te-0.06) = 0.85231

P O,3) = e-0.08 L -te -0.10 -te-0.12 +-te -0.08) +-te-o.06 -te -0.08 +-te-0.04 ) J = 0.78741

So:

R (2 3) = in( P(O'2)) = In

(0.85231) = 7.92

o , P(O,3) 0.78741

Solution 16

Answer: D

Difficulty level: .

The rates are determined as follows:

u = ru = 0.07704 = 1.2840, d = rd 0.04673 = 0.77883

ro 0.06000 ro 0.06000

rdd = d rd = 0.03639 , rud = rdu = d ru = 0.06000 , ruu = u ru = 0.09892

=? ruu + rud + rdd = 0.19531

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ExamMFE

Solution 17

Answer: A


The value of the interest rate cap is the benefit of making an interest payment equal to the

mimum of 7.0% of $5,000 and 5,OOOr where r :;0.070 is the prevailng rate.

Here is the tree:

Year 3 Rate

Year 2 Rate

~ 1'~9.892o/~

r= 6.000

~

=4.673%

r= 6.000

Year 1 Rate

r =3.639%

We see that the 7% cap is exceeded in 2 places (boxed in above). So the value of the cap is

computed from a saving of (0.09892 - 0.07000)x5,OOO = 144.60 at the end of year 3 on the top

branch, and a saving of (0.07704 - 0.07000)x5,OOO = 35.20 at the end of year 2 on the top branch:

35.20

cap value = 0.5x

1.06x1.07704

O 52 144.60

. x

1.06xl.07704xl.09892

= 44.23

Solution 18

Answer: B

Difficulty level: .

The price of the one-year bond gives us:

0.9524 = ~ ~ ro=0.04998

1 + ro

Solution 19

Answer: A


We are given that 0 1 =0.08. In the BDT model the rates at time 1 are:

rd = R1 , ru = R1 e20 1,j = R1 e°.l6 (h = 1 year)


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ExamMFE


With this model we must duplicate the given bond price data. So we have the following equation

based on the given price of a 2-year bond:

1.8868 = 1

1 eO.16 R1

0.8985 = 0.5xO.9524 x -- 0.5xO.9524 x -- =:

1 ru 1 rd

1

1 + R1

Either by iteration or using the quadratic formula, it follows from this equation that:

rd = R1 =0.05522 .

Solution 20

Answer: E

Difficulty level: .

So: ru = R1 eO.16 = 0.0648

Solution 21

Answer: A

Difficulty level: .

At time 1 the prices for a 1-year zero-coupon bond maturing for 1,000 are:

d . 1,000

wn state: price = - =

l+rd

. 1,000

up state: price = - =

l+ru

1,000

1 + 0.05522

1,000

1 + 0.06480

= 947.67

= 939.14

Since we have the option to buy this bond for 940, the option price is:

1 1

= 0.5x - x maxi0 ,939.14 - 940J O.5x - x maxi0 ,947.67 - 940J = 3.65

1 ~ 1 ~

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ExamMFE

Solution 22

Answer: E


A 3-step Black-Derman-Toy tree has the following form:

r = R é0 2.J

~~

uu 2

~

u = R1 e20 1.J

ro =Ro

~

~

r - R e20 2.J

~

d - 2

~

rd =R1 ~

~

~

rdd = R2

~

~

Time

0

Time 1

Time 2

Time

3

where the interest rates are short rates applying over a 1-year period. Since we have the bond

prices at 1-year intervals, we set h = 1 .

The 1-year bond price gives us:

0.9346 =--

l+Ro

1

~ Ro = 0.9346 -1 = 0.06998

Next, we wish to find values for R1 and 0 1' We know the 1-year yield volatility at time 1 is 10%,

so 0 1 must satisfy the equation:

( R i0 1 )

.10 = O.5Xlni 1R1 = 0 1

~ 0 1 =0.10

Using the tree above, we can write the 2-year bond price as:

0.8495 = 0.9346( 0.5x--+0.5x 1 2 )

L 1+R1 1+R1e 0 1

This can be rearranged to give:

0.8495( 1 + R1)( 1 + R1 e20 1 ) = 0.9346L 0.5X( 1 + R1 e20 1 ) + 0.5x(1 + R1) J

~ 0.8495L 1 + R1 (1 + eO.2)+ RreO.2 J = 0.9346X0.5L 2+ R1 (1 +eO.2) J

~ L 0.8495eo.2 JRr +L(0.8495-0.9346XO.5)( 1 +eO.2) JR1 +(0.8495-0.9346) = 0

~ 1.03758xRr +

0.84902

x

R1 -0.0851 = 0


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ExamMFE


Solving this quadratic gives a positive root of 0.09027, so we get:

rd = R1 = 0.09027

and ru = R1 e20 1 = 0.09027 eO.2 = 0.11026

Solution 23

Answer: B


The Black-Derman-Toy tree has the following form:

.-

r =R é0 2

uu 2

.-

rdd = R2

~

~

~

~

~

r - R e20 1

u - 1

ro =Ro

~

~

rd =R1

~

~

~

rud = R2e20 2

Time 0

Time 1

Time

2

Time

3

We now need to find R2 and 0 2. Denoting the two possible 2-year bond prices at time 1 by

P(l,3,rd) and P(l,3,ru)' we can write the 3-year bond price at time 0 as:

0.7722 = 0.9346(0.5x P(l,3, rd)+O.5x P(l,3, ru))

=0.934JO.5X~(0.5X~+0.5X 1 2 )

L 1+R1 1+R2 1+R2e 0 2

1 (1 l)J

O.5x O.5x +O.5x

1 + R1 e20 1 1 + R2e20 2 1 + R2é0 2

Substituting in the values of R1 = 0.09027 and R1 e20 1 = 0.11026 (which were derived in Solution

22) gives:

¡ 0.5 (1 1)

0.7722=0.9346x0.5 -+ 2

1.09027 1 + R2 1 + R2e 0 2

0.5 (1 l)J

+ 1.11026 1+R2e20 2 + 1+R2é0 2

Rather than trying to solve the simultaneous equations for R2 and ()2' it will be much easier just to try

out the five possible combinations given in the question to see which one works.

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ExamMFE

The various combinations of highest interest rate at time 2 (ie R2é0 2) and 0 2 give the following:

0 2

R2é0 2

R2e20 2

R2

3-year

P(l,3,rd) P(l,3,ru)

bond

(highest)

(medium)

(lowest)

price

A

18.991%

14.237% 9.738%

6.661 % 0.8479

0.8046 0.7722

B 14.029%

13.058%

9.863%

7.450% 0.8442

0.8082

0.7722

C

18.991%

12.782% 8.743%

5.980%

0.8545

0.8134 0.7794

D 12.000%

12.591% 9.905% 7.791%

0.8427

0.8097

0.7722

E 14.029%

7.450%

5.627% 4.251%

0.8741

0.8455 0.8035

The first two columns are the values given in A to E in the question. The other columns have been

calculated based on these.

From this, we can rule out Answers C and E, since they don't reproduce the correct 3-year bond

price.

The 2-year volatilty at time 1 is given by:

0.5Xln( P(l,3,ru r1/2 -1)

L P(l,3,rdr1/2_1

The bond values for Answers A, Band D give 14.45%, 12.00% and 10.99% respectively.

So B is the correct answer.

Solution 24

Answer: A


The Black-Derman-Toy interest rates were found in the solutions to Solution 22 and Solution 23,

and are shown in the binomial tree below.

2-year bond price

Payoff

~

~13.058%

-- 9.863

~9.863%

9.027%~

~7.450%

80.824 o

11.026%

6.998%

84.423 2.423

Time

0

Time

1 Time

2

The 2-year bond prices at time 1 are found from the equation:

100 (1 1 J

P(l,3,r¡) =- O.5x-+O.5x-

l+r¡ l+r¡u l+r¡d

where i denotes either an up jump or a down jump in the first step.


15



ExamMFE


In fact, we calculated the values of P l,3, ru) and P l,3, rd) in the table in Solution 23.

Since the option is a European call option with a strike of 82, the payoff at time 1 is given by:

Payoffz = max(O,P(l,3, rz) -82)

The payoffs can then be discounted back to time 0 to give the price of the option:

1

C = 1.06988 (0.5xO+0.5x2.423) = 1.132

Solution 25

Answer: D


The Black-Derman- Toy interest rates were found in the solutions to Solution 22 and Solution 23,

and are shown in the binomial tree below.

I-year bond price

Payoff

'-13.058

88.450

3.550

~

11.026

--

.998

9.863

91.022

0.978

~

9.027 ::

9.863

91.022

0.978

7.450

93.066

0

Time 0

Time

1

Time 2

The 1-year bond prices at time 2 are found from the equation:

100

P(2,3,rzj)=-

l+r..

J

where i and j denote either up or down jumps.

Since the option is a European put option with a strike of 92, the payoff at time 2 is given by:

Payoffzj =maxLO,92-P 2,3,rzj)J

The payoffs can then be discounted back through the tree to give the price of the option:

1 L 1

P = O.5x 0.5x3.550+0.5xO.978)

1.06988 1.11026

+ 0.5X~ 0.5XO.978+ 0.5XO)J

1.0927

=1.162

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ExamMFE

Solution 26

Answer: C


The price of the put option can be found using the standard put-call parity relationship:

C+Kxdo,l = P+So

where C =

K=

T=

price of the call option at time 0, in this case 0.34

strike price, in this case 91

term of option, in this case 1

dO,l = discount factor applying from time 0 to time T, in this case 0.9346

P = price of the put option at time 0, the unknown we re tring to find

So = price of the underlying asset at time 0, in this case a bond that matures at time 2,

with market value 84.95 (given in the table of market data for Solution 22 to

Solution 26).

Although the option is on a l-year bond, we need to consider a bond that matures on a fixed date (ie time 2)

rather than a bond with a fixed term.

Substituting these values in the put-call parity relationship gives:

P=C+Kxdo,l -So

= 0.34 + 91 x

0.9346 - 84.95

= 0.44

Solution 27

Answer: E


The forward price (F ) for purchasing the bond in 1 year's time can be calculated from:

0.9434F = 0.8817 =? F = 0.9346

Using the Black formula:

log Fo +lo-2T

d _ K 2

- o--i

1 0.9346 1 0052 1

g-+-x. x

0.9259 2

0.05~


and d2 = d1 -0.05~ = 0.1620 (= 0.16)

and C = e-rT (FN(d1)-KN(d2))

= 0.9434(0.9346xN(0.21) -0.9259xN(0.16))

= 0.9434 (0.9346 x 0.5832 - 0.9259 x 0.5636)

=0.022


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ExamMFE


Solution 28

Answer: A


The tree of interest rates (in the gray boxes) and caplet payments (in italics) looks like this:

Now

1 year

2 years

3 years

2.392

At the end of the second year, the caplet wil make a payment of (7.704 -7.5 ) x 100 = 0.204 if

interest rates in 1 year's time are at the 7.704 node, which has a (risk-neutral) probabilty of -t.

The value of this caplet is equal to the discounted expected value of the payoff, which equals:

lxO.204x 1 = 0.089

2 1.0600

x

1.07704

At the end of the third year, the caplet wil make a payment of (9.892 -7.5 )xl00=2.392 if

interest rates in 2 years' time are at the 9.892 node, which has a (risk-neutral) probabilty of

-tx-t = t. The value of this caplet is:

1

lx2.392x = 0.477

4 1.0600xl.07704xl.09892

So the value of the cap is:

0.089+0.477 = 0.57

Solution 29

Answer: E


According to the Vasicek model, the price of a zero-coupon bond is:

P r t),t, TJ = A t, T)e-B t,T)r t)

Since this model is time-homogeneous, the values of the functions A and B depend only on the

time interval T - t , which equals 2 years for each of the three bonds in the question. So we can

use the first two bond prices given to find the values of A and B:

P 0.04,O,2J = Ae-O.04B = 0.9445 for Bond 1

P(0.05,l,3J = Ae-O.05B = 0.9321 for Bond 2

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ExamMFE

Dividing the first equation by the second gives:

fie-O.04B

fie-O.05B

0.9445 = 1.0133

0.9321

:: eO.01B = 1.0133

::

B = In 1.0133 = 1.3216

0.01

Using the equation for Bond 1 then gives:

Ae-D.04x1.3216 = 0.9445

:: A

=0.9958

The equation for Bond 3 is:

P(r*,2,4)

=

Ae-Br =0.8960

Substituting the values for A and B gives:

0.9958e-1.3216r = 0.8960

r*=-~ln 0.8960 =0.08

1.3216 0.9958

::

Solution 30

Answer: E


In this tree the interest rate (continuously-compounded) wil be:

. 12 in the first year

. 15 or 9 in the second year

. 18 ,12 or 6 in the thid year.

The value of the one-year bond at the end of the second year wil be:

· lx e-O.18 = 0.8353 if the steps in the second and third years are uu

. 1 x e -D.12 = 0.8869 if the steps in the second and third years are ud or du

· lxe-O.06 = 0.9418 if the steps in the second and thid years are dd.

The corresponding payoffs from the put option are:

· max(0.9 -0.8353, 0) = 0.0647

· max(0.9-0.8869,O) = 0.0131

· max(0.9-0.9418,0) = o.


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ExamMFE


Applying the appropriate interest discount factors and risk-neutral probabilties, gives the value

of the option at time 0:

Value = e-o.12e-o.15 xO.72 x

0.0647 +e-o.12e-o.15 xO.7xO.3xO.0131+e-o.12e-o.09 xO.3xO.7xO.0131 + 0

uVu 1\ uVd ,\ iu 'da

= 0.0242 0.0021 0.0022 0 = 0.0285

Solution 31

Answer: E


The general form of a Black-Derman-Tòy interest rate tree looks like this:

~

=R éCT2

uu 2

~

~

- R e2CT1

u - 1

~

o =Ro

~

~

rud = R2e2CT2

~

rd =R1

~

~

~

rdd = R2

~

~

Here we have:

~

0%

60%

~

ro

~

~

~

30%

~

~

40%

?

~

~

0%

The value of rud has been calculated as the geometric average of ruu and rdd:

rud = R2e2CT2 = ~ R2éCT2 xR2 =~ ruu xrdd = ~ 0.80xO.20 =0.40 (=40 )

We do not know the value of ro. However, we can calculate the 2-year forward price of the

bond by finding the bond price at time 0 and dividing by the 2-year discount factor.

Using the tree and remembering that, for the BDT model, the up and down risk-neutral

probabilties are all equal to 1/, we can calculate the initial bond price as the discounted value of

the expected payoff (which always equals 1):

P(O,

3) =--x.:xf 1 + 1 + 1 + 1 1. = 0.4960

l ro 4 li.6x1.8 1.6xl.4 1.3xl.4 l.3xl.2f l ro

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ExamMFE

The 2-year discount factor at time 0 (which is the same as the price at time 0 of a 2-year zero-

coupon bond) is:

P(O,2) =~x2.xL~:_+_~J = 0.6971

l+ro 2 11.6 1.3J l+ro

So the 2-year forward price of 1,000 3-year bonds is:

1, OOOx P O,

3)

= 1, 000

x 0.4960/0.6971 = 1, OOOx 0.4960 = 712

P(O,

2) l+ro l+ro 0.6971

Solution 32

Answer: C


Since the bond maturity date T is fixed, we can consider the bond price P(r(t),t,T) to be a

function of r(t) and t. Itô's lemma then tells us that:

ap a2p 2 ap

dP =-dr+t-- dr) +-dt

ar ar at

ap a2p 2 ap

-H 0.008 - O.lr t)) dt + 0.05dZ t) l+t--H 0.008 - O.lr t)) dt + 0.05dZ t)1 +- dt

ar ar v at

O.Os2dt

r ap 2 a2 P aPt ap

(0.008-0.1r(t))-+txO.05 --+- dt+0.05-dZ(t)

~ ~ ~ ~

We have been asked to find a value for the drift coefficient a, which corresponds to the

expression in braces divided by P. Ths is difficult to work out directly, but we can work it out

indirectly from the volatility term in the equation given in (iii), which is -qP, which must

ap

correspond to 0.05-, so that:

ar

1 ap

q=-o.05x--

par

The equations given in (i) and (ii) correspond to the Vasicek model, dr(t) = a (b -r(t)) dt+ adZ(t).

The first equation can be written in the form dr t)

=

0.1

(0.08-r(t))dt+0.05dZ(t)

, which tells us

that a = 0.1.

For the Vasicek model, the bond price has the form:

1- -a T-t)

P(r(t),t,T)=A(t,T)e-B(t,T)r(t), where B(t,T)= e

a

So: ~: = -B(t, T)xA(t, T)e-B(t,T)r(t) = -B(t, T)P

Therefore:

1 ap 1_e-O.1 S-2) J

q 0.04,2,5J =-0.05x--= 0.05 B 2,5)

=

0.05 = 0.1296

P ar 0.1


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ExamMFE


Statements (i) and (ii) tell us the rates of return that would be obtained on a cash account using

the true (real-world) probabilty measure and the risk-neutral probability measure. We can use

this information to find the Sharpe ratio:

Sh . I

Real-world drift-risk-neutral driftl

arpe ratio =

Volatilty

L 0.008-~J-L 0.013- ~J

=0.1

0.05

Since there is only once source of randomness, the Sharpe ratio for the bond wil have the same

value. So, from (iii) when t = 2, T = 5 and r(2) = 0.04 :

a (0.04,2,5) - 0.04

Sharpe ratio ( ) 0.1

q 0.04,2,5

~ a(0.04,2,5) = 0.04

+O.lq

(0.04,

2,5) = 0.04+0.1(0.1296) = 0.053

Solution 33

Answer: E


Since the bond maturity date T is fixed, we can consider the bond price P(r(t), t, T) to be a


ap 1 a2p 2 ap

dP=-dr+--(dr) +-dt

ar 2 ar2 at

ap a2p 2 ap

= -10.6 (b - r(t)) dt+ O dZ(t)J+t-i1 0.6 (b -r(t)) dt+ O dZ(t)J +-dt

ar ar , at

¡j2dt

r ap 1 2a2p aPt ap

= 0.6(b-r(t))-+-0 -+- dt+O -dZ(t)

ar 2 ar2 at ar

The SDE given in the question for the bond price can be written in the form dP = aPdt-qPdZ and we

have been asked to find a value for a. Comparing this with the equation we have just derived, we see that

a equals the expression in braces divided by P. This is diffcult to work out directly, but we can work it

out indirectly from the volatility term.

The volatilty in the SDE given for P is -qP, which corresponds to 0 ap in the equation above.

ar

So:

ap

-qP=O -

ar

1 ap

~q=-O x--

par

For the Vasicek model, the bond price has the form:

1 -aCT -t)

P(r(t),t,T)=A(t,T)e-B(t,T)r(t), where B(t,T)= -e

a

So:

~~ = -B(t, T)xA(t, T)e-B(t,T)r(t) = -B(t, T)P

1 ap

q = -O xpa; = O B

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ExamMFE

Therefore:

(1_e-O.6(4-1) J

q(0.05,l,4)

=

O B(l,4)

=

0 =1.39120

0.6

(1- e -0.6(2-0) J

and q

(0.04,0,

2)

=O B(2,5) =0 =1.16470

0.6

To get a link between the volatility and the drift terms, we need to use the Sharpe ratio.

For the Vasicek model, the Sharpe ratio takes a constant value. So we can derive an equation by

equating its value when t = 1, T = 4 and r(l) = 0.05 and when t = 0, T = 2 and r(O) = 0.04 :

. a(0.05,l,4)-0.05 a(0.04,O,2)-0.04

Sharpe ratio = =

q(0.05,l,4) q(0.04,O,2)

=?

a(0.05,l,4)-0.05

1.39120

0.04139761-0.04

1.16470

1.3912

=? a(0.05,l,4) = (0.04139761-0.04)x-+0.05 =0.0517

1.1647

Solution 34

Answer: C


Since the bond maturity date T is fixed, we can consider the bond price P(r(t), t, T) to be a


ap a2p 2 ap

dP=-dr+l-i(dr) +-dt

ar ar at

ap a2p 2 ap

=-la(b-r(t))dt+O v'r(t) dZ(t)l+l-iia(b-r(t))dt+O v'r(t) dZ(t)l +-dt

r ar, ,at

=a2;(t)dt

Lap 2 a2 P ap ~ ap

a(b-r(t))-+lO r-i+- dt+O v'r(t) -dZ(t)

ar ar at ar

The SDE given in the question for the bond price can be written in the form dP = aPdt-qPdZ and we

have been asked to find a value for a. Comparing this with the equation we have just derived, we see that

a equals the expression in braces divided by P. This is difcult to work out directly, but we can work it

out indirectly from the volatility term.

The volatilty in the SDE given for P is -qP, which corresponds to O v'r(t) ap in the equation

ar

above.

So:

ap

-qP = O v'r(t)-

ar

1 ap

=? q = -O v'r(t) x--

par

For the CIR model, the bond price has the form:

P(r(t), t, T) = A(t, T)e-B(t,T)r(t)

So: ~~ = -B(t, T)xA(t, T)e-B(t,T)r(t) = -B(t, T)P


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ExamMFE


=? q = - J~r t) x2- ap = J~r t) B

par

Therefore:

q (0.04,11,13) = (J.,0.04 B(l1,13)

and q 0.05,7,9) = J.,0.05 B 7,9)

To get a link between the volatility and the drift terms, we need to use the Sharpe ratio.

For the CIR model, the Sharpe ratio equals ø = ø .¡ / J ,where ø is a constant. So when t = 11,

T = 13 and r(l1) = 0.04 , we have:

- .,0.04 a (0.04,11,13) - 0.04

ø-=

J q (0.04,11,13)

When t = 7 , T = 9 and r(7) = 0.05 , we have:

-.,0.05 a(0.05,7,9)-0.05

ø-=

J q(0.05,7,9)

Equating the values of ø /(J from these two equations, gives:

ø a(0.04,

11,

13)-0.04

(J = .,0.04xq(0.04,l1,13)

a(0.05,7,9) - 0.05

.,0.05 xq (0.05,7,9)

Using the formulas derived above for the values of q and the value given in the question for

a 0.05,7,9), we get:

a(0.04,

11,13) -0.04 0.06-0.05

.,0.04 x

(J.,0.04 B(l1,

13) .,0.05 x

(J.,0.05 B(7,9)

Since the CIR model is a time-homogeneous model, the values of the function B(t, T) depend

only on the time interval T - t. So B(l1, 13) = B(7, 9) and these two factors cancel (as do the (J s).

So:

a(0.04, 11, 13) -0.04

0.04

0.06-0.05

0.05

=? a(0.04,

11,13) = 0.048

Solution 35

Anwer: D


rdd

16.8

9.3

17.2

9

12.6

ruud

13.5

11

rddd

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ExamMFE

According to the Black-Derman-Toy modeL, the two interest rates we are given for the thid year

(in the third column of the table above) are calculated as:

ruu = R2é0 2 = 0.172 and rud = R2e20 2 = 0.135

So we can calculate rdd as:

2

(R2e20 1 )

rdd =R2 =

R2éO l

0.1352

=

0.172

0.106 ie 10.6%

The interest rates we are given for the fourth year (in the final column of the table) are:

ruuu = R3é0 3 = 0.168, rudd = R3e20 3 = 0.11

So we can calculate ruud as:

ruud = R3é0 3 = ~R3é0 3 xR3e20 3 = ,J0.168 x

0.11 = 0.136 ie 13.6%

and

3

(R3e20 1 )

rddd = R3 =

R3e60 1

=~0.113 =0.089 ie 8.9%

0.168

The 4-year caplet has a payoff of 100maxir4 -0.105,01 at time 4, where r4 denotes the effective

interest rate in year 4. We can calculate the price of this caplet at each of the earlier nodes in the

usual way by applying the risk-neutral probabilities (equal to Yz for each move) and discounting

at the appropriate risk-free interest rates.

This leads to the following tree of values for the caplet:

6.3

4.06

2.43

3.1

1.45

1.59

0.80 0.5

0.23

0

For example, the value shown at the 16.8% node is the value of the caplet at the end of the fourth

year, which is:

100maxr0.168-0.105, OJ = 6.3

The value at the 17.2% node is calculated as:

1 1 1 1

-x-x6.3+-x-x3.1 = 4.06

2 1.168 2 1.136

The intial value of 1.45 shown in the table is the value of the caplet at the end of the first year.

So the value of the caplet at time 0 is 1.45 = 1.33 .

1.09


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ExamMFE


Solution 36

Answer: D


We are told that the SDEs for the short rate under the real-world and risk-neutral probabilty

measures are:

dr(t) = (0.09 - O.5r(t)) dt + 0.3dZ(t)

and dr(t) = (0.15 -O.5r(t)) dt + 0 (r(t)) dZ(t)

Girsanov's theorem tells us that, for models based on Brownian motion, changing between these

two probabilty measures is equivalent to changig the drift (without altering the volatilty). So

this tells us that 0 (r(t)) = 0.3. If we then equate the two equations for dr(t)

, we get:

(0.09 - O.5r(t)) dt + 0.3dZ(t) = (0.15 -0.5r(t)) dt+ 0 ( r(t)) dZ(t)

'-

0.3

=: dZ(t) = dZ(t) - 0.02dt

~ The adjustment of -0.02 to the drift that appears in this equation is the Sharpe ratio. It can be either

positive or negative, depending on the sign convention used to define the volatility terms in the SDEs.

We can write the equation for the asset value g(r(t), t) in the form:

dg(r(t), t)

g(r(t),t)

ll (r(t), g (r(t), t))

( ) dt-O.4dZ(t)

r(t),t

This equation also allows us to deduce the Sharpe ratio, using the formula:

a-r

Sharpe ratio =-

0

~ In this formula, a denotes the expected real-world rate of return on the asset.

So:

Sharpe ratio =

ll (r(t), g (r(t), t))

g(r(t),t)

0.4

r(t)

0.02

Rearranging this equation gives:

ll (r(t), g (r(t), t)) = (r(t) + 0.08) g (r(t), t)

26

(§ BPP Professional Education




Solution 37

Anwer: D


The tree of interest rates is as follows:

ro

5

6

rud

3

2

According to the Black-Derman- Toy model, the interest rates for the third year in the final

column) are calculated as R2 é0 2, R2 e20 2 and R2.

So: R2é0 2 = 0.06 and R2 = 0.02

~ rud = R2e20 2 = ~R2 xR2é0 2 = .J0.02xO.06 = 0.0346 ie 3.46%

There are two possibilties for the price of the 3-year bond at the end of the first year (which wil

then be a 2-year bond).

If the interest rate moves to 5 , the bond price wil be:

1 11 1 1 1;

- -x-+-x- =0.9095

1.05 2 1.06 2 1.0346

and the corresponding bond yield wil be:

0.9095-1/2 -1 = 0.0486

If the interest rate moves to 3 , the bond price wil be:

1 11 1 1 1;

- -x-+-x- =0.9451

1.03 2 1.0346 2 1.02

and the corresponding bond yield wil be:

0.9451-1/2 -1 = 0.0286

The volatilty of the 3-year bond price can then be calculated as:

lIn (0.9451) = 1.92%

2 0.9095

~ This doesn't match any of the choices. So the question must be referring to the volatility of the bond yield,

rather than the bond price.

The volatilty of the 3-year bond yield is:

iln(0.0486) = 26.4%

2 0.0286


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ExamMFE


Solution 38

Answer: A


The structure of the 2-period Black-Derman-Toy model looks like this:

ru = R1e2CY1

ro =Ro

~

~

rd =R1

We can calculate the value of ro, which is the effective interest rate for the first year, from the

price of the 1-year bond:

1

-=0.9434 =HO =0.06 (=6 )

l+ro

We are given the volatility in year 1, which is 0 1 = 0.10.

There are two possibilties for the interest rate in the second year. Under the BDT model using

risk-neutral probabilties, these possibilties are equally likely. So the value of the 2-year bond is:

1 11 1 1 1;

-x +-x- = 0.8850

1.06 2 1+R1e2CY1 2 1+R1

Rearrangig this gives:

1 1

02 +-=2xO.8850xl.06=1.8762

1+R1e' 1+R1

1 + R1 + 1+ R1 eO.2 = 1.8762

( 1 + R1 eO.2) (1 + R1)

This simplifies to:

2.2916Rr + 1.9464R1 -0.1238 = 0

We can solve this using the quadratic formula:

R = -1.9464-2.2188 0.0594

1 2(2.2916)

q The other root is negative, which is not possible for an interest rate.

Solution 39

Anwer: B


The delta-gamma approximation tells us how to approximate the new price of the bond when the

interest changes by ë = 0.03 -0.05 = -0.02, ie from 5 to 3 :

PEst (0,3, 0.03) = P(O,

3, 0.05)+ ëLi+lë2r (*)

ap a2p

where Li=- and r=-.

ar ar2

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ExamMFE

One way to find the Greeks here, and to deal with the exponential function in the bond price

formula, is by workig with the log of the bond price:

In P(t, T, r) = In 1 A(t, T)exp (-B(t, T)r n = In A(t, T) - B(t, T)r

Differentiating with respect to r gives:

1 ap . L1

-- = -B(t T) ie - = -B(t T)

par ' P ,

Differentiating again (using the product rule) gives:

_~(ap)2 +.. a2 p = 0

2 ar P ar2 .

~ ~=(~r

If we divide Equation *) through by P, we get:

PEst (0, 3, 0.03) L1 1 2 r

=l+e +-e

P(O,

3, 0.05) P(O,

3,

0.05) 2 P(O,

3,

0.05)

Using the formulas derived above for the Greeks, with t = 0 and T = 3 , this becomes:

PEst

(0,3,0.03)

=l-eB(O 3)+le2 (B(O 3))2

P(O,

3,

0.05) 2

= 1-(-0.02)(2)+1-(-0.02)2 (2)2

=1.0408


29




ExamMFE


Chapter 9 Solutions

Solution 1

Answer: E

Difficulty level: .

The chi-squared test statistic is:

2

L(O-E)

E

where 0 is the observed frequency and E is the expected frequency. If the values conform to a

uniorm distribution, the expected frequency for each group must be equal to 1,000 = 50 .

20

So we

have:

L 0-E)2

E

L OJ - 50)2 = ~ L O? - 2 x 50 L O. + 20 x 502 = 51,850 2,000 + 1, 000 = 37

50 50 J 50 ~ 50 50

=1,000

This is an observed value from a chi-squared distribution with 19 degrees of freedom (there are

no estimated parameters). We are given that Xl9(l ) = 36.19. So we wil reject Ho, even at the

1 leveL.

Solution 2

Answer: B

Difficulty level: .

The distribution function for this distribution is:

B)a 50 )2.5

F x)=l- - =1--

B+x 50+x

Now set u=F(x) and solve for x in terms of u:

u = 1_ ~)2.S

50+x

~

~ = 1_u)1/2.S = 1_u)0.40

50+x

~

x = 50 1- u rO.40 -1 )


1



ExamMFE


Solution 3

Answer: D

Difficulty level: .

The probabilty function for this distribution is:

Pr X =x) = GJ 0.75)X 0.25)2-x, x =0,1,2

From this expression we obtain the following probabilties:

Pr X = 0) = 0.0625, Pr X = 1) = 0.375, Pr X = 2) = 0.5625

So we assign random values from the distribution as follows:

Range of values for u¡

Assigned value in the

binomial distribution

o :: u¡ ~ 0.0625

0

0.0625 :: u¡ ~ 0.4375

1

0.4375:: u¡ :: 1

2

Note the convention that any values lying exactly on a boundary are assigned to the larger possible

outcome.

So our random sample of values from the binomial distribution is:

2

1

2

2

1

1

2

2

1

1

This sample has a sample mean of 1.5.

Solution 4

Answer: C

Difficulty level: .

The distribution function for this Pareto distribution is:

F X)=l- 400 )6

400+x

If we set this equal to u and invert the formula, we find that:

400

x= 1/6 400

(l-u)

Using this formula to obtain the sample values corresponding to the sample of U(O,l) values

given, we obtain:

74.25 137.43 5.60 223.08 22.51

33.17 142.88 34.35 209.37 4.87

The mean of this random sample is 88.75.

2





ExamMFE

Solution 5

Answer: C


The distribution function of the inverse exponential distribution is:

F x) = e-B/x

Note that this distribution is diferent from the regular exponential distribution, which has distribution

function F(x) = 1- e-x / B .

So, writing u = F(x) and rearrangig, we get:

-100

x=-

lnu

Substituting in the values for u from our sample of random numbers, we find that the random

sample from the inverse exponential distribution is:

224

103

537

574

40

106

1,378

1,199

79

38

We can estimate the median as the average of the middle two values when these are arranged in

order, which gives -t 106 + 224) = 165 .

Solution 6

Answer:B


The annual number of deaths follows a binomial distribution with n = 100 trials and q =0.03 .

So the probabilty function is:

Pr(X=x) = C~O) O.03x 0.97100-x for x=O, 1, 2, ...,100

Under the Inversion Method a random number u is transformed to the smallest x such that

F x):; u.

We can construct the following table:

x

0

1

2

...

Pr(X=x)

0.048

0.147

0.225

...

Pr X :: x)

0.048

0.195

0.420

...

Using the random numbers given, we can see that:

u1 =0.20 ~ x=2

u2 =0.03 ~ x = 0

u3 =0.09 ~ x = 1

The sample mean is the average of these, which equals 1.


3



ExamMFE


Solution 7

Answer: C


We need to

simulate values X from

the N(ll,(Y) distribution

with ll=15,OOO and (Y=2,OOO.

We can write X = ll+(YZ, where Z has a N (0,1) distribution.

We can simulate the required values of Z by looking for the random values given in the body of

the normal distribution table and reading off the corresponding Z value. The Z values can then

be converted to the corresponding X value:

u =0.5398

~ Z=0.10

~ X =15,200

~ X=12,600

~ X =9,000

~ X =16,600

u = 0.1151 (= 1-0.8849) ~ Z = -1.20

u=0.0013 (=1-0.9987) ~ Z=-3.00

u =0.7881

~ Z=0.80

Totaling the last column, we find that the insurer s simulated claim cost for the four-month

period is 53,400.

Solution 8

Answer: C


The standard deviation of X is :í.

The standard error is obtained by replacing (Y2 (which is unkown) with the sample variance,

which is:

S2 = . (f x; - 8x2 J =. ( 144,939 - 8x 133.3752) = 375.4

7 i=l 7

For the standard error to be less than or equal to 3, we require:

se(X)= In::3

S2

=? n ?-= 41.7

9

So the smallest sample size is 42, which means that we require an additional 34 simulations.

4





ExamMFE

Solution 9

Answer: A


Suppose we want to be 100 (1- a) certain that our estimator Xn is within k ll of the true value

ll. How many simulations are needed? (In other words, how big must n be?)

If you take a large sample (size n) of values from a particular distrbution (not necessarily, a

normal distribution) that has a mean of ll and a variance of ¿i, the Central Limit Theorem says

that the sample mean Xn wil have a distribution that is approximately a normal distribution

with mean ll and variance 52 In .

Xn -ll

If we standardize the sample mean, this tells us that has a distribution that is

5/ .J

approximately N(O,l).

Using this result, we have:

1 - a = Pr 1- k ) ll :: Xn :: 1 + k) ll )

(-kll X - ll kll J

= Pr 5 /.J :: 5 ï.J :: 5 / .J

pr( -kllc::N(O,l):: kllcJ

(5/..n (5/..n

Since 1-a = Pr -za/2 :: N O ,1) :: Za/2) we must have:

za/2 = (5~~ ~ n=(Za:2 r x :~

Since both ll and 52 are unkown, we replace them in the formula above with their sample

n

estimates jL=xn and s2 = ¿(Xz -xn)2 /(n-l). We cannot determie the value of n precisely, but

z=l

we can be reasonably safe in stopping the simulation when we have:

n ?(Za/2)2 x ~

k (xnl

In this case, we have k=0.02 and zO.025 =1.96 . So the inequality is:

S2

n ? 9,604 x --

(xn)

Ths is only satisfied for the data in case 1.


5



ExamMFE


Solution 10

Answer: E


Using the same method as in the previous examples, we require:

n?(1.96)2 x(j: =(1.96)2 xl.22 =2,212.8

0.05 l1 0.05

So the sample size must be at least 2,213.

The reason we are told that it is a non-negative random variable is that this ensures that the mean l1 has a

positive value. If l1 was negative, the statement that the population standard deviation is 20% larger

than the population mean would not make sense.

Solution 11

Answer: E


Firstly, what shall we use to estimate F(300)? Since F(300) is P(X:: 300), where X has the

given exponential distribution, the following procedure seems appropriate:

(1) Take sample values x1,x2,...,xn from an Exp(100) distribution.

(2) Observe how many of these n values are less than 300, let's say k of them.

(3) Our estimate of F(300) is f = k / n .

We see that k is an observed value from a binomial distribution, with parameters n (the number

of simulations) and f (the true value of F(300)). We want:

PrL 0.99 f:: j:: 1.01f J? 0.99

Using the normal approximation to the binomial distribution, we see that k is approximately

normal with parameters l1 = nf and (j2 = nf(l- f). So k / n is approximately normal with

parameters l1 = f and (j2 = f(l- f) .

n

So the probabilty becomes (standardizing in the usual way):

pri 0.99 f:: N(f, f(l; f)):: 1.01f J? 0.99

L O.Olf O.Olf j

= Pr re :: N O, 1) :: + re ? 0.99

f(l- f) f(l- f)

n n

Using the appropriate point of the standard normal distribution, we see that we need:

0.01 f = 2.5758

~f l: f)

6





ExamMFE

But we know the true value of f. Using the distribution function of the exponential distribution,

we

have:

f = 1_e-300/100 = 1-e-3 = 0.95021

So we now have the equation:

0.01 x 0.95021

= 2.5758

0.95021

x

0.04979

n

Solving this equation, we find that n = 3,477.

The situation described here is rather artifcial since we can easily work out the answer directly without

using a Monte Carlo approach.

Solution 12

Answer: A

Difficulty level: .

We want the value of k such that Pr X:: k) = 0.7:

(iog X - 6 log k - 6)

Pr(X:: k) = 0.7 :: Pr(log X:: log k) = 0.7 :: Pr Ja :: Ja = 0.7

0.5 0.5

But the quantity on the left-hand side of this inequality has a standard normal distribution. So,

using the Tables, we find that:

10gk-6

Ja

0.5244

Solving this equation, we find that:

k = é.3708 = 584.529

Solution 13

Answer: A

Difficulty level: .

If X has a lognormal distribution, then In X - f. - N(O,l). From the Tables, the 10th and 90th

(Y

percentiles of the N(O,l) distribution are : 1.282. The inormation given in the question tells us

that:

lnl00-f. -1.282 and In

2500-f. =1.282

(Y (Y

We can rearrange these to get:

etl-1.2820 = 100 etl+1.2820 = 2,500

(9 BPP Professional Education 7



ExamMFE


Since In X - l1 ~ N(O,l) and the N(O,l) distribution is symmetrical about zero, the median of the

(Y

lognormal distribution is the value where In X - l1 0, ie X = efJ .

(Y

So, multiplying the previous two equations together, we get:

e2fJ =2,500xl00=250,OOO

We now take the square root to find the median:

median = efJ = 500

Solution 14

Answer: D

Difficulty level: .

If the expected rate of return is a, then the expected stock price is:

E St ) = Soe a-S)t

So: 58 = E 50.5) = 55 e a-O)xO.5

Solving this gives a = 0.10622 .

Solution 15

Answer: B


We can use the lognormal stock price model in the following form:

St =Soe(a-S-O.5c?)t eU.jz where Z-N(O,l)

From the Tables of the normal distribution, we see that 0.8412 corresponds to Z = 1 .

So: 74 = So e a-S-0.5u2)t eU.j

Similarly, 0.1587 corresponds to Z = -1.

So: 56 = So e a-s-0.su2)t e-u.j

Dividing the two equations and substituting t = 0.5 , we have:

74 = e2u.j =: Y = 0.197

56

8





ExamMFE

Solution 16

Anwer: D


If the expected rate of return of the stock price is a , then the mean of 52 is:

110 = E (52) = Soe(a-å)x2

and the variance is:

2,420 = var( 52) = (So e(a-å)x2 )2 (ec?X2 -1)

'-

102

We can solve these to get:

0.2 = (ec?X2 -1) =? (Y2 =0.09116 =? (y = 0.30193

Solution 17

Answer: A


We can simulate the stock price using the formula:

5 -5 (r-å-O.5c?)txea.Jep-l(u)

0.5 - ° e

= 70 eO.0025 e°.21213ep-l(u)

From the standard normal probability table, we have:

U1 = 0.6293 =? -1 (U1) = 0.33

u2 = 0.1611 =? -1 (U2) = -0.99

u3 = 0.5000 =? -1 (U3) = 0

So the three simulated stock prices are:

50.5,1 = 70 eO.0025 eO.21213ep-l(Ul) = 75.26

50.5,2 = 70 eO.0025 e°.21213ep-l(U2) =56.88

50.5,3 = 70 eO.0025 eO.21213ep-l(U3) =70.18

So the average simulated price is 67.44.


9



ExamMFE


Solution 18

Answer: E


Statement I is false. Either put options or call options can be more valuable, depending on the

relationship between the stock price and the strike price. If the strike price is much higher than

the stock price, then put options wil be more valuable than call options. If the strike price is

much lower than the stock price, then call options wil be more valuable than put options.

Statement II is false. Although this statement would be true if the underlying asset did not pay

any dividends, it is not necessarily true for call options on dividend-paying stocks, for example.

Statement II is false. The Black-Scholes formula wil always give a positive theoretical option

price.

Solution 19

Answer: E


We are given monthly data, so in this case h = A . We know that EL St¡ 15tH J = 5tH ia and we

require an estimate for a , the annualized expected return on the stock.

Define ri to be:

(St. J

ri = log --

5tH

These values correspond to the observed values of h (a-O.5o.2).

56

Using the data given in the question, we have r1 = log- = 0.036368, and similarly

54

r2 = -0.154151, r3 = 0.136132, r4 = 0.087011, rs = -0.033902 and r6 = 0.066691 .

The sample mean of these values is:

1 6

, = - L ri = 0.023025

6 i=l

This is an estimate of h a - 0.50.2 ). So, to find an estimate of a, we also need an estimate of 52,

which we can find by considering the variance of the monthly returns:

hâ.2 = ~ f (ri - T)2 =. r t ri2 _,2 J =. L 0.056785 - 6 x 0.0230252 J = 0.010721

n -1 i=l 5li=1 5

Our estimate for a can now be found from the equation:

, = h(â-0.5d2) = ii â-0.5xhd2

So: â = 12(, +0.5xhd2) = 12(0.023025+ 0.5xO.010721) = 0.3406





ExamMFE

Solution 20

Answer: C


From the first two pieces of inormation given in the question, we see that iog( ~~) has a normal

distribution with parameters:

f1 = ( 0.15 - 0.~2 J (2) = 0.21

and

(72 = 0.32 x2 = 0.18

So we need to simulate values from a N(0.21,O.18) distribution.

First, we can use our U(O,l) numbers to simulate values from a N(O,l) distribution. We use the

inverse of the N(O,l) distribution function. Using the normal distribution tables, we see that

(2.12) = 0.9830, (-1.77) = 0.0384 and (0.77) = 0.7794. So our simulated N(O,l) values are

21 = 2.12, 22 = -1.77 and 23 = 0.77 .

We can now simulate values from Y ~ N(0.21, 0.18) by using the relationship y¡ = 0.21 + 2¡.J0.18 ,

where 2¡ is a simulated N(O,l) value. So our simulated values of N(0.21,O.18) are:

1.1094

-0.5409

0.5367

These are related to the simulated stock prices via the relationship y¡ = iog ;~). Inverting this

formula gives us:

s¡ =50eYi

So the values of s¡ are:

Sl = 50e1.1094 = 151.63

s2 = 50e-D.5409 = 29.11

and s3 = 50e°.5367 = 85.52

The mean of these three values is 88.75.

Solution 21

Answer: A


We know that, using the lognormal stock price model, the price of a stock at time t, St, is given

by:

5 -5 a-J-O.5~)t j-fz

t - oe e

where Z is a standard normal N(O,l) variable.


11



ExamMFE


With this model, E( Stl = Soe(a-J)t, and the question tells us that a = 0.15 and Ö = o. So, putting

in the numerical values, we find that the share price at time t = 0.5 is:

51/2 = 0.25 e(0.15 - 0.5xO.352) x 1/2 eO.35.J Z = 0.25 eO.04375 + 0.247487 Z

The upper 95 point of the standard normal is 1.6449 (the 90 confidence interval runs from the

5th percentile to the 95th percentile). So the upper limit of the confidence interval for the price is:

0.25 eO.044375 + 0.247487 x 1.6449 = 0.39265

Solution 22

Answer: A


The formula for the stock price at time t = 0.5 is:

St = So e(a-J-0.5o-2)t eo-.. Z

= 60 e(0.09-0-0.5xO.16)xO.5 eO.4xJrz

= 60eO.005 e.Jz where Z - N O,l)

The call option expires in the money if the price at time t is more than the strike price,

ie if St :; 62 . The probabilty of this event is:

r 62 j

10g--0.005

Pr 60eO.005e.Jz :;62) =Pr Z:; 60 =1- l 0.09825)

=0.4609

-J0.08

Solution 23

Answer: A


Recall that the value of this call option is E * (e -rt max(St - K, 0) J, where the expectation is calculated

using risk-neutral probabilities. If we assume that St follows the lognormal model (geometric Brownian

motion), this is precisely what the Black-Scholes call option pricing formula gives.

This situation is consistent with the assumptions of the Black-Scholes modeL. So we can use the

usual Black-Scholes formula (with ö = 0) to find the value of the call option:

In

(So / K)+(r-ö +0.50.2) t

(j.J

In So / K) + r -ö - 0.5 j2) t

d2

(j.J

0.18459 =? (l (d1 ) = 0.5732

1

-0.09825 =? (l (d2) = 1- 0.5391 = 0.4609

The price of the call option is given by the Black-Scholes formula:

P = So (l ( d1 ) - Ke -rt (l ( d2 )

= 60 x 0.5732 - 62e -û.09xO.5 x 0.4609

= 7.0736

12




Question Answer Bank - Solutions, Chapter 9 ExamMFE

Solution 24

Answer: E Difficulty level: ...

With the method described, U1 wil be transformed to a value VI from the Uniform(O,O.25) distribution,

U2 will be transformed to a value V2 from the Uniform(0.25,0.5) distribution etc. This is then repeated

for US,U6,U7,US.

To transform a value U from the Uniform O,l) distribution to a value V from the Uniform a, b)

distribution, we can apply the function V = a+(b-a)U .

So the calculations here are:

U1 = 0.4880

Convert to Uniform O,O.25)

~ VI = 0+0.25U1 = 0.1220

U2 =0.7894

Convert to Uniform 0.25,O.5)

~ V2 =0.25+0.5U2 =0.44735

U3 =0.8628


~ V3 =0.5+0.25U3 =0.7157

U4 =0.4482

Convert to Uniform 0.75,l)

~ V4 = 0.75 + 0.25U4 = 0.86205

Us =0.3172

Convert to Uniform O, 0.25)

~ Vs =0+0.25Us =0.0793

U6 =0.8944


~ V6 =

0.25 + 0.5U6 =0.7236

U7 =0.5013


~ V7 = 0.5 + 0.25U7 = 0.625325

Us = 0.3015

Convert to Uniform 0.75,l)

~ Vs = 0.75 + 0.25Us = 0.825375

In fact, because we are only asked to find the diference between the smallest and the largest values, we only

really needed to consider the values for the Uniform(O,0.25) and Uniform(0.75,l) distributions.

The smallest of the V values is Vs = 0.0793 , which wil be converted into the normal value:

25 = N-1 (0.0793) = _N-1 (1-0.0793) = -N-1(0.9207) = -1.41

The largest of the V values is V4 = 0.0793 , which wil be converted into the normal value:

Z4 = N-1 (0.86205) = 1.09

The difference between these two values is 1.09-(-1.41) = 2.50.

bpp questions spring 2011

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