• 沒有找到結果。

# Exam MFE/3F Sample Questions and Solutions

N/A
N/A
Protected

Share "Exam MFE/3F Sample Questions and Solutions"

Copied!
154
0
0

(1)

1

### Sample Questions and Solutions

April 6, 2010

(2)

2

1. 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 will 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

(3)

3 Solution to (1) Answer: (A)

The put-call parity formula (for a European call and a European put on a stock with the same strike price and maturity date) is

C  P  F0,PT( )K

  PV0,T (K)

  KerT

= S0  KerT, because the stock pays no dividends

We are given that C  P  0.15, S0  60, K  70 and T  4. Then, r  0.039.

Remark 1: If the stock pays n dividends of fixed amounts D1, D2,…, Dn at fixed times t1, t2,…, tn prior to the option maturity date, T, then the put-call parity formula for European put and call options is

C  P   KerT

 S0  PV0,T(Div)  KerT, where PV0,T(Div)

### 

n i

rti ie D

1

is the present value of all dividends up to time T. The difference, S0  PV0,T(Div), is the prepaid forward price F0P,T(S).

Remark 2: The put-call parity formula above does not hold for American put and call options. For the American case, the parity relationship becomes

S0  PV0,T(Div)  K ≤ C  P ≤ S0  KerT.

This result is given in Appendix 9A of McDonald (2006) but is not required for Exam MFE/3F. Nevertheless, you may want to try proving the inequalities as follows:

For the first inequality, consider a portfolio consisting of a European call plus an amount of cash equal to PV0,T(Div) + K.

For the second inequality, consider a portfolio of an American put option plus one share of the stock.

0,PT( )

F S

0,PT( )

F S

0,PT( )

F S

0,PT( )

F S

(4)

4

2. 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 Mary and Peter are correct.

(E) None of them is correct.

(5)

5 Solution to (2) Answer: (D)

The prices are not arbitrage-free. To show that Mary’s portfolio yields arbitrage profit, we follow the analysis in Table 9.7 on page 302 of McDonald (2006).

Time 0 Time T

ST < 40 40≤ ST < 50 50≤ ST < 55 ST  55 Buy 1 call

Strike 40 11 0 ST – 40 ST – 40 ST – 40

Sell 3 calls Strike 50

+ 18 0 0 3(ST – 50) 3(ST – 50)

Lend \$1 1 erT erT erT erT

strike 55 6 0 0 0 2(ST – 55)

Total 0 erT > 0 erT+ ST – 40

> 0 erT+ 2(55 ST)

> 0

erT > 0

Peter’s portfolio makes arbitrage profit, because:

Time-0 cash flow Time-T cash flow Buy 2 calls & sells 2 puts

Strike 55 23 + 11) = 16 2(ST  55) Buy 1 call & sell 1 put

Strike 40 11 + 3 = 8 ST 40

Lend \$2 2 2erT Sell 3 calls & buy 3 puts

Strike 50 3(6  8) = 6 3(50  ST)

Total 0 2erT

Remarks: Note that Mary’s portfolio has no put options. The call option prices are not arbitrage-free; they do not satisfy the convexity condition (9.17) on page 300 of

McDonald (2006). The time-T cash flow column in Peter’s portfolio is due to the identity max[0, S – K]  max[0, K – S] = S  K

In Loss Models, the textbook for Exam C/4, max[0, ] is denoted as +. It appears in the context of stop-loss insurance, (S – d)+, with S being the claim random variable and d the deductible. The identity above is a particular case of

x  x+  (x)+,

which says that every number is the difference between its positive part and negative part.

(6)

6

3. An insurance company sells single premium deferred annuity contracts with return linked to a stock index, the time-t value of one unit of which is denoted by S(t). The contracts offer a minimum guarantee return rate of g%. At time 0, a single premium of amount  is paid by the policyholder, and π  y% is deducted by the insurance company. Thus, at the contract maturity date, T, the insurance company will pay the policyholder

π  (1 y%)  Max[S(T)/S(0), (1 + g%)T].

You are given the following information:

(i) The contract will mature in one year.

(ii) The minimum guarantee rate of return, g%, is 3%.

(iii) Dividends are incorporated in the stock index. That is, the stock index is constructed with all stock dividends reinvested.

(iv) S(0)  100.

(v) The price of a one-year European put option, with strike price of \$103, on the stock index is \$15.21.

Determine y%, so that the insurance company does not make or lose money on this contract.

(7)

7 Solution to (3)

The payoff at the contract maturity date is

π  (1 y%)Max[S(T)/S(0), (1 + g%)T]

= π  (1 y%)Max[S(1)/S(0), (1 + g%)1] because T = 1

= [/S(0)](1 y%)Max[S(1), S(0)(1 + g%)]

= (/100)(1 y%)Max[S(1), 103] because g = 3 & S(0)=100

= (/100)(1 y%){S(1) + Max[0, 103 – S(1)]}.

Now, Max[0, 103 – S(1)] is the payoff of a one-year European put option, with strike price \$103, on the stock index; the time-0 price of this option is given to be is \$15.21.

Dividends are incorporated in the stock index (i.e.,  = 0); therefore, S(0) is the time-0 price for a time-1 payoff of amount S(1). Because of the no-arbitrage principle, the time- 0 price of the contract must be

(/100)(1 y%){S(0) + 15.21}

= (/100)(1 y%)  115.21.

Therefore, the “break-even” equation is

(/100)(1 y%)115.21, or

y% = 100  (1  1/1.1521)% = 13.202%

Remarks:

(i) Many stock indexes, such as S&P500, do not incorporate dividend reinvestments.

In such cases, the time-0 cost for receiving S(1) at time 1 is the prepaid forward price F0,1P( )S , which is less than S(0).

(ii) The identities

Max[S(T), K]  K + Max[S(T) K, 0]  K  (S(T) K)+

and

Max[S(T), K]  S(T)  Max[0, K  S(T)]  S(T) + (K  S(T))+

can lead to a derivation of the put-call parity formula. Such identities are useful for understanding Section 14.6 Exchange Options in McDonald (2006).

(8)

8 4. 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) 0 (B) 1 (C) 2 (D) 3 (E) 4

(9)

9

First, we construct the two-period binomial tree for the stock price.

The calculations for the stock prices at various nodes are as follows:

Su  20  1.2840  25.680 Sd  20  0.8607  17.214 Suu  25.68  1.2840  32.9731 Sud  Sdu  17.214  1.2840  22.1028 Sdd  17.214  0.8607  14.8161

The risk-neutral probability for the stock price to go up is

4502 . 8607 0 . 0 2840 . 1

8607 .

* 0.05 0 

 

  e

d u

d

p erh .

Thus, the risk-neutral probability for the stock price to go down is 0.5498.

If the option is exercised at time 2, the value of the call would be Cuu  (32.9731 – 22)+  10.9731

Cud = (22.1028 – 22)+  0.1028 Cdd = (14.8161 – 22)+  0

If the option is European, then Cu  e0.05[0.4502Cuu  0.5498Cud]  4.7530 and Cd  e0.05[0.4502Cud  0.5498Cdd]  0.0440.

But since the option is American, we should compare Cu and Cd with the value of the option if it is exercised at time 1, which is 3.68 and 0, respectively. Since 3.68 < 4.7530 and 0 < 0.0440, it is not optimal to exercise the option at time 1 whether the stock is in the up or down state. Thus the value of the option at time 1 is either 4.7530 or 0.0440.

Finally, the value of the call is

C  e0.05[0.4502(4.7530)  0.5498(0.0440)]  2.0585.

20

17.214 25.680

22.1028 32.9731

Year 0 Year 1 Year 2

14.8161

(10)

10

Remark: Since the stock pays no dividends, the price of an American call is the same as that of a European call. See pages 294-295 of McDonald (2006). The European option price can be calculated using the binomial probability formula. See formula (11.17) on page 358 and formula (19.1) on page 618 of McDonald (2006). The option price is er(2h)

### [

p*2Cuu

2 2

 

 + p*(1 p*)Cud 1

2 

 

 + (1 p*)2Cdd

0 2 

 

### ]

= e0.1 [(0.4502)210.9731 + 20.45020.54980.1028 + 0]

= 2.0507

(11)

11

5. 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 volatility of the exchange rate is   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.

(12)

12 Solution to (5)

Each period is of length h = 0.25. Using the first two formulas on page 332 of McDonald (2006):

u  exp[–0.010.25  0.3 0.25]  exp(0.1475)  1.158933, d  exp[–0.010.25  0.3 0.25]  exp(0.1525)  0.858559.

Using formula (10.13), the risk-neutral probability of an up move is 4626 . 858559 0 .

0 158933 .

1

858559 .

* 0

25 . 0 01 .

0

e

p .

The risk-neutral probability of a down move is thus 0.5374. The 3-period binomial tree for the exchange rate is shown below. The numbers within parentheses are the payoffs of the put option if exercised.

The payoffs of the put at maturity (at time 3h) are Puuu  0, Puud  0, Pudd  0.3384 and Pddd  0.6550.

Now we calculate values of the put at time 2h for various states of the exchange rate.

If the put is European, then Puu = 0,

Pud  e0.02[0.4626Puud  0.5374Pudd]  0.1783, Pdd  e0.02[0. 4626Pudd  0.5374Pddd]  0.4985.

But since the option is American, we should compare Puu, Pud and Pdd with the values of the option if it is exercised at time 2h, which are 0, 0.1371 and 0.5059, respectively.

Since 0.4985 < 0.5059, it is optimal to exercise the option at time 2h if the exchange rate has gone down two times before. Thus the values of the option at time 2h are Puu = 0, Pud = 0.1783 and Pdd = 0.5059.

1.43 (0.13)

1.2277 (0.3323) 1.6573 (0)

1.4229 (0.1371)

1.2216 (0.3384) 1.9207

(0)

2.2259 (0)

Time 0 Time h Time 2h Time 3h

1.6490 (0)

1.0541 (0.5059)

0.9050 (0.6550)

(13)

13

Now we calculate values of the put at time h for various states of the exchange rate.

If the put is European, then

Pu  e0.02[0.4626Puu  0.5374Pud]  0.0939, Pd  e0.02[0.4626Pud  0.5374Pdd]  0.3474.

But since the option is American, we should compare Pu and Pd with the values of the option if it is exercised at time h, which are 0 and 0.3323, respectively. Since 0.3474 >

0.3323, it is not optimal to exercise the option at time h. Thus the values of the option at time h are Pu = 0.0939 and Pd = 0.3474.

Finally, discount and average Pu and Pd to get the time-0 price, P  e0.02[0.4626Pu  0.5374Pd]  0.2256.

Since it is greater than 0.13, it is not optimal to exercise the option at time 0 and hence the price of the put is 0.2256.

Remarks:

(i) Because

h h r h h r

h h r h r

e e

e e

) ( )

(

) ( )

( 

h h

h

e e

e

1 

e h

 1

1 , we can also calculate the risk-neutral probability p* as follows:

p*  e h

 1

1 

25 . 0 3 .

1 0

1

e0.15 1

1

e  0.46257.

(ii) 1  p*  1  e h

 1

1 

h h

e e

1 

e h

 1

1 .

(iii) Because   0, we have the inequalities

p*  ½  1 – p*.

(14)

14

6. 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 selling 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%.

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

Calculate the price of the block of 100 options.

(A) 0.04 (B) 1.93 (C) 3.50 (D) 4.20 (E) 5.09

(15)

15

) ( )

( )

, , , , ,

(S K r T Se N d1 Ke N d2

C    TrT (12.1)

with

T

T r

K S

d

 )

2 ( 1

) /

ln( 2

1

  (12.2a)

T d

d21 (12.2b)

Because S = \$20, K = \$25, = 0.24, r = 0.05, T = 3/12 = 0.25, and  = 0.03, we have

25 . 0 24 . 0

25 . 0 ) 24 . 20 03 1 . 0 05 . 0 ( ) 25 / 20

ln( 2

1

 

d = 1.75786

and

d2 = 1.75786 0.24 0.25 = 1.87786

Because d1 and d2 are negative, use N(d1)1N(d1) and N(d2)1N(d2).

In Exam MFE/3F, round –d1 to 1.76 before looking up the standard normal distribution table. Thus, N(d1) is 10.96080.0392. Similarly, round –d2 to 1.88, and N(d2) is thus

0301 . 0 9699 . 0

1  .

Formula (12.1) becomes

0350 . 0 ) 0301 . 0 ( 25

) 0392 . 0 (

20 (0.03)(0.25)(0.05)(0.25)

e e

C

Cost of the block of 100 options = 100 × 0.0350 = \$3.50.

(16)

16

7. Company A is a U.S. international company, and Company B is a Japanese local company. Company A is negotiating with Company B to sell its operation in Tokyo to Company B. The deal will be settled in Japanese yen. To avoid a loss at the time when the deal is closed due to a sudden devaluation of yen relative to dollar, Company A has decided to buy at-the-money dollar-denominated yen put of the European type to hedge this risk.

You are given the following information:

(i) The deal will be closed 3 months from now.

(ii) The sale price of the Tokyo operation has been settled at 120 billion Japanese yen.

(iii) The continuously compounded risk-free interest rate in the U.S. is 3.5%.

(iv) The continuously compounded risk-free interest rate in Japan is 1.5%.

(v) The current exchange rate is 1 U.S. dollar = 120 Japanese yen.

(vi) The natural logarithm of the yen per dollar exchange rate is an arithmetic Brownian motion with daily volatility 0.261712%.

(vii) 1 year = 365 days; 3 months = ¼ year.

Calculate Company A’s option cost.

(17)

17 Solution to (7)

Let X(t) be the exchange rate of U.S. dollar per Japanese yen at time t. That is, at time t,

¥1 = \$X(t).

We are given that X(0) = 1/120.

At time ¼, Company A will receive ¥ 120 billion, which is exchanged to

\$[120 billion  X(¼)]. However, Company A would like to have

\$ Max[1 billion, 120 billion  X(¼)], which can be decomposed as

\$120 billion  X(¼) + \$ Max[1 billion – 120 billion  X(¼), 0], or

\$120 billion  {X(¼) + Max[1201 – X(¼), 0]}.

Thus, Company A purchases 120 billion units of a put option whose payoff three months from now is

\$ Max[1201 – X(¼), 0].

The exchange rate can be viewed as the price, in US dollar, of a traded asset, which is the Japanese yen. The continuously compounded risk-free interest rate in Japan can be interpreted as  the dividend yield of the asset. See also page 381 of McDonald (2006) for the Garman-Kohlhagen model. Then, we have

r = 0.035,  = 0.015, S = X(0) = 1/120, K = 1/120, T = ¼.

Because the logarithm of the exchange rate of yen per dollar is an arithmetic Brownian motion, its negative, which is the logarithm of the exchange rate of dollar per yen, is also an arithmetic Brownian motion and has the SAME volatility. Therefore, {X(t)} is a geometric Brownian motion, and the put option can be priced using the Black-Scholes formula for European put options. It remains to determine the value of , which is given by the equation

 365

1 = 0.261712 %.

Hence,

 = 0.05.

Therefore, d1 =

T T r

 /2)

( 2

=

4 / 1 05 . 0

4 / ) 2 / 05 . 0 015 . 0 035 . 0

(   2

= 0.2125 and

d2 = d1  T = 0.2125  = 0.1875.

By (12.3) of McDonald (2006), the time-0 price of 120 billion units of the put option is

\$120 billion  [KerTN(d2)  X(0)eTN(d1)]

= \$ [erTN(d2)  eTN(d1)] billion because K = X(0) = 1/120

= \$ {erT[1  N(d2)]  eT[1  N(d1)]} billion

(18)

18

In Exam MFE/3F, you will be given a standard normal distribution table. Use the value of N(0.21) for N(d1), and N(0.19) for N(d1).

Because N(0.21) = 0.5832, N(0.19) = 0.5753, erT = e

0.9963 and eT = e

0.9963, Company A’s option cost is

0.9963 0.4247 0.9963 0.4168 = 0.005747 billion  \$5.75 million.

Remarks:

(i) Suppose that the problem is to be solved using options on the exchange rate of Japanese yen per US dollar, i.e., using yen-denominated options. Let

\$1 = ¥U(t) at time t, i.e., U(t) = 1/X(t).

Because Company A is worried that the dollar may increase in value with respect to the yen, it buys 1 billion units of a 3-month yen-denominated European call option, with exercise price ¥120. The payoff of the option at time ¼ is

¥ Max[U(¼)  120, 0].

To apply the Black-Scholes call option formula (12.1) to determine the time-0 price in yen, use

r = 0.015,  = 0.035, S = U(0) = 120, K = 120, T = ¼, and   0.05.

Then, divide this price by 120 to get the time-0 option price in dollars. We get the same price as above, because d1 here is –d2 of above.

The above is a special case of formula (9.7) on page 292 of McDonald (2006).

(ii) There is a cheaper solution for Company A. At time 0, borrow ¥ 120exp(¼ r¥) billion,

and immediately convert this amount to US dollars. The loan is repaid with interest at time ¼ when the deal is closed.

On the other hand, with the option purchase, Company A will benefit if the yen increases in value with respect to the dollar.

(19)

19

8. 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 volatility is 30%.

(iv) The current call option delta is 0.5.

Determine the current price of the option.

(A) 20 – 20.453

0.15ex2/2dx

(B) 20 – 16.138

### 

0.15ex2/2dx (C) 20 – 40.453

0.15ex2/2dx

(D) 16.138

### 

0.15ex2/2dx20.453 (E) 40.453

### 

0.15ex2/2dx– 20.453

(20)

20 Solution to (8) Answer: (D)

Since it is never optimal to exercise an American call option before maturity if the stock pays no dividends, we can price the call option using the European call option formula

) ( )

(d1 Ke N d2 SN

C   rT ,

where

T

T r

K S

d

 ) 2 ( 1 ) /

ln( 2

1

  and d2d1  T .

Because the call option delta is N(d1) and it is given to be 0.5, we have d1 = 0.

Hence,

d2 = –0.3 0.25= –0.15 .

To find the continuously compounded risk-free interest rate, use the equation 25 0

. 0 3 . 0

25 . 0 ) 3 . 2 0 ( 1 ) 5 . 41 / 40

ln( 2

1     

r

d ,

which gives r = 0.1023.

Thus,

C = 40N(0) – 41.5e–0.1023 × 0.25N(–0.15) = 20 – 40.453[1 – N(0.15)]

= 40.453N(0.15) – 20.453 =

### 

15 .

0 /2d

2 453 .

40 2

x

e x – 20.453 =16.138

### 

0.15ex2/2dx20.453

(21)

21

9. 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 continuously 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

(22)

22 Solution to (9)

According to the first paragraph on page 429 of McDonald (2006), such a stock price move is given by plus or minus of

S(0) h,

where h  1/365 and S(0)  50. It remains to find .

Because the stock pays no dividends (i.e.,   0), it follows from the bottom of page 383 that   N(d1). By the condition N(d1) = 0.6179, we get d1  0.3. Because S  K and

  0, formula (12.2a) is

d1

T T r

 /2) (  2 or

½2T

d1 + r  0.

With d1  0.3, r  0.1, and T  1/4, the quadratic equation becomes

½2 – 0.6 + 0.1  0,

whose roots can be found by using the quadratic formula or by factorization,

½( 1)( 0.2) = 0.

We reject  = 1 because such a volatility seems too large (and none of the five answers fit). Hence,

S(0) h = 0.2  50  0.052342  0.52.

Remarks: The Itô’s Lemma in Chapter 20 of McDonald (2006) can help us understand Section 13.4. Let C(S, t) be the price of the call option at time t if the stock price is S at that time. We use the following notation

CS(S, t) = C( tS, )

S

, CSS(S, t) = 22C(S,t)

S

, Ct(S, t) = C( tS, )

t

 ,

t = CS(S(t), t), t = CSS(S(t), t), t = Ct(S(t), t).

At time t, the so-called market-maker sells one call option, and he delta-hedges, i.e., he buys delta, t, shares of the stock. At time t + dt, the stock price moves to S(t + dt), and option price becomes C(S(t + dt), t + dt). The interest expense for his position is

[tS(t)  C(S(t), t)](rdt).

Thus, his profit at time t + dt is

t[S(t + dt)  S(t)]  [C(S(t + dt), t + dt)  C(S(t), t)]  [tS(t)  C(S(t), t)](rdt) = tdS(t)  dC(S(t), t)  [tS(t)  C(S(t), t)](rdt). (*) We learn from Section 20.6 that

dC(S(t), t) = CS(S(t), t)dS(t) + ½CSS(S(t), t)[dS(t)]2 + Ct(S(t), t)dt (20.28)

= tdS(t) + ½t [dS(t)]2 + t dt. (**)

(23)

23

Because dS(t) = S(t)[ dt + dZ(t)], it follows from the multiplication rules (20.17) that

[dS(t)]2 = [S(t)]2 2 dt, (***)

which should be compared with (13.8). Substituting (***) in (**) yields dC(S(t), t) = t dS(t) + ½t [S(t)]22 dt + t dt,

application of which to (*) shows that the market-maker’s profit at time t + dt is

{½t [S(t)]22 dt + t dt}  [tS(t)  C(S(t), t)](rdt)

= {½t [S(t)]22 + t  [tS(t)  C(S(t), t)]r}dt, (****) which is the same as (13.9) if dt can be h.

Now, at time t, the value of stock price, S(t), is known. Hence, expression (****), the market-maker’s profit at time t+dt, is not stochastic. If there are no riskless arbitrages, then quantity within the braces in (****) must be zero,

Ct(S, t) + ½2S2CSS(S, t) + rSCS(S, t)  rC(S, t) = 0,

which is the celebrated Black-Scholes equation (13.10) for the price of an option on a nondividend-paying stock. Equation (21.11) in McDonald (2006) generalizes (13.10) to the case where the stock pays dividends continuously and proportional to its price.

Let us consider the substitutions

dt  h

dS(t) = S(t + dt)  S(t)  S(t + h)  S(t), dC(S(t), t) = C(S(t + dt), t + dt)  C(S(t), t)  C(S(t + h), t + h)  C(S(t), t).

Then, equation (**) leads to the approximation formula

C(S(t + h), t + h)  C(S(t), t)  tS(t + h)  S(t) + ½t[S(t + h)  S(t)2 + t h, which is given near the top of page 665. Figure 13.3 on page 426 is an illustration of this approximation. Note that in formula (13.6) on page 426, the equal sign, =, should be replaced by an approximately equal sign such as .

Although (***) holds because {S(t)} is a geometric Brownian motion, the analogous equation,

[S(t + h)  S(t)2 = [S(t)2h, h > 0,

which should be compared with (13.8) on page 429, almost never holds. If it turns out that it holds, then the market maker’s profit is approximated by the right-hand side of (13.9). The expression is zero because of the Black-Scholes partial differential equation.

(24)

24

10. Consider the Black-Scholes framework. Let S(t) be the stock price at time t, t  0.

Define X(t)  ln[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)]  2 h, t  0, h > 0.

(iii)

### 

n  

n j X jT n X j T n

1

)]2

/ ) 1 ((

) / ( [

lim = 2 T.

(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

(25)

25 Solution to (10) Answer: (E)

(i) is true. That {S(t)} is a geometric Brownian motion means exactly that its logarithm is an arithmetic Brownian motion. (Also see the solution to problem (11).)

(ii) is true. Because {X(t)} is an arithmetic Brownian motion, the increment, X(t + h)  X(t), is a normal random variable with variance 2 h. This result can also be found at the bottom of page 605.

(iii) is true. Because X(t) = ln S(t), we have

X(t + h)  X(t) = h + [Z(t + h)  Z(t)],

where {Z(t)} is a (standard) Brownian motion and  =  –  ½. (Here, we assume the stock price follows (20.25), but the actual value of  is not important.) Then,

[X(t + h)  X(t)]2  2h 2 + 2h[Z(t + h)  Z(t)] + [Z(t + h)  Z(t)]2. With h = T/n,

### 

n

j

n T j X n jT X

1

)]2

/ ) 1 ((

) / ( [

 2T 2 / n + 2T/n) [Z(T)  Z(0)] + 

### 

n

j

n T j Z n jT Z

1

)]2

/ ) 1 ((

) / (

[ .

As n  , the first two terms on the last line become 0, and the sum becomes T according to formula (20.6) on page 653.

Remarks: What is called “arithmetic Brownian motion” is the textbook is called

“Brownian motion” by many other authors. What is called “Brownian motion” is the textbook is called “standard Brownian motion” by others.

Statement (iii) is a non-trivial result: The limit of sums of stochastic terms turns out to be deterministic. A consequence is that, if we can observe the prices of a stock over a time interval, no matter how short the interval is, we can determine the value of  by evaluating the quadratic variation of the natural logarithm of the stock prices. Of course, this is under the assumption that the stock price follows a geometric Brownian

(26)

26

motion. This result is a reason why the true stock price process (20.25) and the risk- neutral stock price process (20.26) must have the same . A discussion on realized quadratic variation can be found on page 755 of McDonald (2006).

A quick “proof” of the quadratic variation formula (20.6) can be obtained using the multiplication rule (20.17c). The left-hand side of (20.6) can be seen as

### 

0T[dZ(t)]2 .

Formula (20.17c) states that [dZ(t)]2= dt. Thus,

0T Z t

)]2

( d

[ 

### 

0Tdt  T.

(27)

27

11. 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[ln S(t + h) | S(t)]  2 h, h  0.

(ii) Var

 

 ( )

) (

) (

d S t

t S

t

S   2 dt

(iii)Var[S(t + dt) | S(t)]  S(t)22 dt

(A) Only (i) is true (B) Only (ii) is true

(C) Only (i) and (ii) are true (D) Only (ii) and (iii) are true (E) (i), (ii) and (iii) are true

(28)

28

Here are some facts about geometric Brownian motion. The solution of the stochastic differential equation

) (

) ( d

t S

t

S   dt + dZ(t) (20.1)

is

S(t)  S(0) exp[( – ½2)t + Z(t)]. (*) Formula (*), which can be verified to satisfy (20.1) by using Itô’s Lemma, is equivalent to formula (20.29), which is the solution of the stochastic differential equation (20.25). It follows from (*) that

S(t + h) = S(t) exp[( – ½2)h  Zt + h) Z(t)]], h  0. (**) From page 650, we know that the random variable Zt + h)  Z(t)] has the same distribution as Z(h), i.e., it is normal with mean 0 and variance h.

(i) is true: The logarithm of equation (**) shows that given the value of S(t), ln[S(t + h)]

is a normal random variable with mean (ln[S(t)] + ( – ½2)h) and variance 2h. See also the top paragraph on page 650 of McDonald (2006).

(ii) is true: Var

 

 ( )

) (

) (

d S t

t S

t

S = Var[dt + dZ(t)|S(t)]

= Var[dt + dZ(t)|Z(t)],

because it follows from (*) that knowing the value of S(t) is equivalent to knowing the value of Z(t). Now,

Var[dt + dZ(t)|Z(t)] = Var[ dZ(t)|Z(t)]

= Var[dZ(t)|Z(t)]

= Var[dZ(t)] independent increments

= 2 dt.

Remark: The unconditional variance also has the same answer: Var

 

 ) (

) ( d

t S

t

S  2 dt.

(29)

29 (iii) is true because (ii) is the same as

Var[dS(t) | S(t)] = S(t)2 2 dt, and

Var[dS(t) | S(t)] = Var[S(t + dt) S(t) | S(t)]

= Var[S(t + dt) | S(t)].

A direct derivation for (iii):

Var[S(t + dt) | S(t)] = Var[S(t + dt) S(t) | S(t)]

= Var[dS(t) | S(t)]

= Var[S(t)dt + S(t)dZ(t) | S(t)]

= Var[S(t) dZ(t) | S(t)]

= [S(t)]2 Var[dZ(t) | S(t)]

= [S(t)]2 Var[Z(t + dt) Z(t) | S(t)]

= [S(t)]2 Var[Z(dt)]

= [S(t)]2 dt

We can also show that (iii) is true by means of the formula for the variance of a

lognormal random variable (McDonald 2006, eq. 18.14): It follows from formula (**) on the last page that

Var[S(t + h) | S(t)] = Var[S(t) exp[( – ½2)h + Zt + h) Z(t)]] | S(t)]

= [S(t)]2 exp[2( – ½2)h] Var[exp[Zt + h) Z(t)]] | S(t)]

= [S(t)]2 exp[2( – ½2)h] Var[exp[Z(h)]]

= [S(t)]2 exp[2( – ½2)h] eh2(eh2  1)

= [S(t)]2 exp[2( – ½ 2)h] eh2(h  . 2 ) Thus,

Var[S(t + dt) | S(t)] = [S(t)]2 × 1 × 1 × (dt × 2), which is (iii).

(30)

30

12. 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

d ( ) ( ) X t

X t = 0.07dt  0.12dZ(t) and

d ( ) ( ) Y t

Y t = Adt + BdZ(t), where A and B are constants.

You are also given:

(i) d[ln Y(t)]  μdt + 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

(31)

31 Solution to (12) Answer: (B)

If f(x) is a twice-differentiable function of one variable, then Itô’s Lemma (page 664) simplifies as

df(Y(t))  f ′(Y(t))dY(t) + ½ f ″(Y(t))[dY(t)]2, because f(x)

t

 = 0.

If f(x)  ln x, then f ′(x)  1/x and f ″(x)  1/x2. Hence, d[ln Y(t)] =

) ( 1 t

Y dY(t)  2 [d ( )]2 )]

( [

1 2

1 Y t

t Y 



 . (1)

We are given that

dY(t) = Y(t)[Adt + BdZ(t)]. (2)

Thus,

[dY(t)]2 = {Y(t)[Adt + BdZ(t)]}2 = [Y(t)]2 B2 dt, (3) by applying the multiplication rules (20.17) on pages 658 and 659. Substituting (2) and (3) in (1) and simplifying yields

d [ln Y(t)]  (A  2 B2

)dt  BdZ(t).

It thus follows from condition (i) that B = 0.085.

It is pointed out in Section 20.4 that two assets having the same source of randomness must have the same Sharpe ratio. Thus,

(0.07 – 0.04)/0.12 = (A – 0.04)/B = (A – 0.04)/0.085 Therefore, A = 0.04 + 0.085(0.25) = 0.06125  0.0613

(32)

32

13. Let {Z(t)} 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) 2

### 

0tsZ(s)ds

Which of the processes defined above has / have zero drift?

(A) {V(t)} only (B) {W(t)} only

(C) {U(t)} and {V(t)} only (D) {V(t)} and {W(t)} only

(E) All three processes have zero drift.

(33)

33

Apply Itô’s Lemma.

(i) dU(t) = 2dZ(t)  0 = 0dt + 2dZ(t).

Thus, the stochastic process {U(t)} has zero drift.

(ii) dV(t) = d[Z(t)]2  dt.

d[Z(t)]2 = 2Z(t)dZ(t) + 2

2 [dZ(t)]2

= 2Z(t)dZ(t) + dt

by the multiplication rule (20.17c) on page 659. Thus, dV(t) = 2Z(t)dZ(t).

The stochastic process {V(t)} has zero drift.

(iii) dW(t) = d[t2 Z(t)]  2t Z(t)dt Because

d[t2 Z(t)] = t2dZ(t) + 2tZ(t)dt, we have

dW(t) = t2dZ(t).

The process {W(t)} has zero drift.

(34)

34

14. You are using the Vasicek one-factor interest-rate model with the short-rate process calibrated as

dr(t)  0.6[b  r(t)]dt  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,

] , ), ( [

] , ), ( [

T t t r P

T t t r

dP  [r(t), t, T] dt  q[r(t), t, T] dZ(t), t  T.

You are given that (0.04, 0, 2) = 0.04139761.

Find (0.05, 1, 4).

(35)

35 Solution to (14)

For t < T, (r, t, T) is the time-t continuously compounded expected rate of return on the zero-coupon bond that matures for 1 at time T, with the short-rate at time t being r.

Because all bond prices are driven by a single source of uncertainties, {Z(t)}, the no-arbitrage condition implies that the ratio,

) , , (

) , , (

T t r q

r T t

r

, does not depend on T. See

(24.16) on page 782 and (20.24) on page 660 of McDonald (2006).

In the Vasicek model, the ratio is set to be , a constant. Thus, we have

) 2 , 0 , 04 . 0 (

04 . 0 ) 2 , 0 , 04 . 0 ( )

4 , 1 , 05 . 0 (

05 . 0 ) 4 , 1 , 05 . 0 (

q q



 . (*)

To finish the problem, we need to know q, which is the coefficient of −dZ(t) in ]

, ), ( [

] , ), ( [

T t t r P

T t t r

dP . To evaluate the numerator, we apply Itô’s Lemma:

dP[r(t), t, T]  Pt[r(t), t, T]dt  Pr[r(t), t, T]dr(t)  ½Prr[r(t), t, T][dr(t)]2, which is a portion of (20.10). Because dr(t)  a[b  r(t)]dt  dZ(t), we have [dr(t)]2 = 2dt, which has no dZ term. Thus, we see that

q(r, t, T) = Pr(r, t, T)/P(r, t, T) which is a special case of (24.12)

= 

r

 lnP(r, t, T)].

In the Vasicek model and in the Cox-Ingersoll-Ross model, the zero-coupon bond price is of the form

P(r, t, T)  A(t, T) eB(t, T)r; hence,

q(r, t, T) = 

r

 lnP(r, t, T)] = B(t, T).

In fact, both A(t, T) and B(t, T) are functions of the time to maturity, T – t. In the Vasicek model, B(t, T)  [1  ea(T t)]/a. Thus, equation (*) becomes

) 0 2 ( )

1 4

( 1

04 . 0 ) 2 , 0 , 04 . 0 ( 1

05 . 0 ) 4 , 1 , 05 . 0 (

 

a

a e

e .

Because a = 0.6 and (0.04, 0, 2) = 0.04139761, we get (0.05, 1, 4) = 0.05167.

(36)

36 Remarks:

(i) The second equation in the problem is equation (24.1) [or (24.13)] of MacDonald (2006). In its first printing, the minus sign on the right-hand side is a plus sign.

(ii) Unfortunately, zero-coupon bond prices are denoted as P(r, t, T) and also as P(t, T, r) in McDonald (2006).

(iii)One can remember the formula,

B(t, T)  [1  ea(T t)]/a,

in the Vasicek model as aT t|force of interest = a, the present value of a continuous annuity-certain of rate 1, payable for T  t years, and evaluated at force of interest a, where a is the “speed of mean reversion” for the associated short-rate process.

(iv) If the zero-coupon bond prices are of the so-called affine form, P(r , t, T) A(t, T) eB(t, T)r ,

where A(t, T) and B(t, T) are independent of r, then (24.12) becomes q(r, t, T)  σ(r)B(t, T).

Thus, (24.17) is

(r, t) 

) , , (

) , , (

T t r q

r T t

r

 =

) , ( ) (

) , , (

T t B r

r T t r

 ,

from which we obtain

(r, t, T) = r  (r, t)(r) B(t, T).

In the Vasicek model, σ(r) σ, (r, t)  , and

(r, t, T) = r + σB(t, T).

In the CIR model, σ(r)  σ r , (r, t) 

 r , and

(r, t, T) = r + rB( Tt , ).

In either model, A(t, T) and B(t, T) depend on the variables t and T by means of their difference T – t, which is the time to maturity.

(v) Formula (24.20) on page 783 of McDonald (2006) is P(r, t, T) = E*[exp(

### 

tTr )(s ds) | r(t) = r],

where the asterisk signifies that the expectation is taken with respect to the risk- neutral probability measure. Under the risk-neutral probability measure, the expected rate of return on each asset is the risk-free interest rate. Now, (24.13) is

(37)

37 ]

, ), ( [

] , ), ( [

T t t r P

T t t r

dP  [r(t), t, T] dt  q[r(t), t, T] dZ(t)

 r(t) dt  q[r(t), t, T] dZ(t) + {[r(t), t, T]  r(t)}dt

 r(t) dt  q[r(t), t, T]

dZ(t) 

] , ), ( [

) ( ] , ), ( [

T t t r q

t r T t t

r

dt

### }

 r(t) dt  q[r(t), t, T]{dZ(t)  [r(t), t]dt}. (**) Let us define the stochastic process {Z t( )} by

)

~( t

Z = Z(t) 

### 

0t [r(s), s]ds.

Then, applying )

~( t

Z = dZ(t)  [r(t), t]dt (***)

to (**) yields

] , ), ( [

] , ), ( [

T t t r P

T t t r

dP  r(t)dt  q[r(t), t, T]d~( ) t Z ,

which is analogous to (20.26) on page 661. The risk-neutral probability measure is such that Z~(t) is a standard Brownian motion.

Applying (***) to equation (24.2) yields dr(t) a[r(t)]dt  σ[r(t)]dZ(t)

a[r(t)]dt  σ[r(t)]{d~( ) t

Z  [r(t), t]dt}

{a[r(t)]  σ[r(t)][r(t), t]}dt  σ[r(t)]d~( ) t Z , which is (24.19) on page 783 of McDonald (2006).

(38)

38

15. You are given the following incomplete Black-Derman-Toy interest rate tree model for the effective annual interest rates:

Calculate the price of a year-4 caplet for the notional amount of \$100. The cap rate is 10.5%.

9%

9.3%

12.6%

13.5%

11%

17.2%

Year 0 Year 1 Year 2 Year 3

16.8%

(39)

39 Solution to (15)

First, let us fill in the three missing interest rates in the B-D-T binomial tree. In terms of the notation in Figure 24.4 of McDonald (2006), the missing interest rates are rd, rddd, and ruud. We can find these interest rates, because in each period, the interest rates in

different states are terms of a geometric progression.

% 6 . 135 10

. 0

172 . 0 135 .

0   dd

dd

r r

% 6 . 168 13

. 0 11 .

0   uud

uud

uud r

r r

% 9 . 11 8

. 0

168 . 0 11 .

0 2   

 

ddd

ddd

r r

The payment of a year-4 caplet is made at year 4 (time 4), and we consider its discounted value at year 3 (time 3). At year 3 (time 3), the binomial model has four nodes; at that time, a year-4 caplet has one of four values:

, 394 . 168 5

. 1

5 . 10 8 .

16  

, 729 . 136 2

. 1

5 . 10 6 .

13  

, 450 . 11 0 . 1

5 . 10 11 

and 0 because rddd = 8.9%

which is less than 10.5%.

For the Black-Derman-Toy model, the risk-neutral probability for an up move is ½.

We now calculate the caplet’s value in each of the three nodes at time 2:

4654 . 172 3

. 1

2 / ) 729 . 2 394 . 5

(  

, 1.4004

135 . 1

2 / ) 450 . 0 729 . 2

(  

, 0.2034

106 . 1

2 / ) 0 450 . 0

(  

. Then, we calculate the caplet’s value in each of the two nodes at time 1:

2.1607

126 . 1

2 / ) 4004 . 1 4654 . 3

(  

, 0.7337

093 . 1

2 / ) 2034 . 0 40044 . 1

(  

. Finally, the time-0 price of the year-4 caplet is 1.3277

09 . 1

2 / ) 7337 . 0 1607 . 2

(  

.

Remarks:

(i) The discussion on caps and caplets on page 805 of McDonald (2006) involves a loan.

This is not necessary.

(ii) If your copy of McDonald was printed before 2008, then you need to correct the typographical errors on page 805; see

http://www.kellogg.northwestern.edu/faculty/mcdonald/htm/typos2e_01.html

(iii)In the earlier version of this problem, we mistakenly used the term “year-3 caplet” for

“year-4 caplet.”

(40)

40

Alternative Solution: The payoff of the year-4 caplet is made at year 4 (at time 4). In a binomial lattice, there are 16 paths from time 0 to time 4.

For the uuuu path, the payoff is (16.8 – 10.5)+

For the uuud path, the payoff is also (16.8 – 10.5)+

For the uudu path, the payoff is (13.6 – 10.5)+

For the uudd path, the payoff is also (13.6 – 10.5)+

: :

We discount these payoffs by the one-period interest rates (annual interest rates) along interest-rate paths, and then calculate their average with respect to the risk-neutral probabilities. In the Black-Derman-Toy model, the risk-neutral probability for each interest-rate path is the same. Thus, the time-0 price of the caplet is

16 1

## {

168 . 1 172 . 1 126 . 1 09 . 1

) 5 . 10 8 . 16 (

+ 1.09 1.126 1.172 1.168 )

5 . 10 8 . 16 (

+

136 . 1 172 . 1 126 . 1 09 . 1

) 5 . 10 6 . 13 (

+

136 . 1 172 . 1 126 . 1 09 . 1

) 5 . 10 6 . 13 (

+ ………

= 8

1

## {

1.091(16.126.8110.172.5)1.168

+ 1.09 1.126 1.172 1.136 )

5 . 10 6 . 13 (

+ 1.09 1.126 1.135 1.136 )

5 . 10 6 . 13 (

+

136 . 1 135 . 1 093 . 1 09 . 1

) 5 . 10 6 . 13 (

+ 1.09 1.126 1.135 1.11 )

5 . 10 11 (

+

11 . 1 135 . 1 093 . 1 09 . 1

) 5 . 10 11 (

+

11 . 1 106 . 1 093 . 1 09 . 1

) 5 . 10 11 (

+ 1.09 1.093 1.106 1.09 )

5 . 10 9 (

## }

= 1.326829.

Remark: In this problem, the payoffs are path-independent. The “backward induction”

method in the earlier solution is more efficient. However, if the payoffs are path- dependent, then the price will need to be calculated by the “path-by-path” method illustrated in this alternative solution.

(41)

41

16. Assume that the Black-Scholes framework holds. Let S(t) be the price of a nondividend-paying stock at time t, t ≥ 0. 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

, [ ( ) ] ( )

P x x

Ft T S TS t ,

where the left-hand side is the prepaid forward price at time t of a contingent claim that pays S T( )x 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

• Similar to futures options except that what is delivered is a forward contract with a delivery price equal to the option’s strike price. – Exercising a call forward option results

• Similar to futures options except that what is delivered is a forward contract with a delivery price equal to the option’s strike price.. – Exercising a call forward option results

• A call gives its holder the right to buy a number of the underlying asset by paying a strike price.. • A put gives its holder the right to sell a number of the underlying asset

• A put gives its holder the right to sell a number of the underlying asset for the strike price.. • An embedded option has to be traded along with the

• Similar to futures options except that what is delivered is a forward contract with a delivery price equal to the option’s strike price.. – Exercising a call forward option results

• Similar to futures options except that what is delivered is a forward contract with a delivery price equal to the option’s strike price.. – Exercising a call forward option results

• It works as if the call writer delivered a futures contract to the option holder and paid the holder the prevailing futures price minus the strike price.. • It works as if the

• The XYZ.com bonds are equivalent to a default-free zero-coupon bond with \$X par value plus n written European puts on Merck at a strike price of \$30. – By the