124
125 Solution to (52) Answer: (C)
This problem is a modification of #19 in the May 2007 Exam C.
U Uniform (0, 1)
N1(U) N(0, 1)
a + bN1(U) N(a, b2)
The random variable ln( / 50)S2 has a normal distribution with mean
(0.15 ½ 0.3 ) 2 0.21 2 and variance 0.32 × 2 = 0.18, and thus a standard deviation of 0.4243.
The three uniform random numbers become the following three values from the standard normal: 2.12, 1.77, 0.77. Upon multiplying each by the standard deviation of 0.4243 and adding the mean of 0.21, the resulting normal values are 1.109, 0.541, and 0.537.
The simulated stock prices are obtained by exponentiating these numbers and multiplying by 50. This yields 151.57, 29.11, and 85.54. The average of these three numbers is 88.74.
126
53. Assume the Black-Scholes framework. For a European put option and a European gap call option on a stock, you are given:
(i) The expiry 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-0 gamma of the put option is 0.07.
(v) The time-0 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-0 gamma of the cash-or-nothing call option.
(A) 5
(B) 2
(C) 2
(D) 5
(E) 8
127
Solution to (53) Answer: (B)
Let I[.] be the indicator function, i.e., I[A] = 1 if the event A is true, and I[A] = 0 if the event A is false. Let K1 be the strike price and K2 be the payment trigger of the gap call option. The payoff of the gap call option is
[S(T) – K1] × I[S(T) K2] [S(T) – K2] × I[S(T) K2] (K2 – K1) × I[S(T) K2].
Because differentiation is a linear operation, each Greek (except for omega or elasticity) of a portfolio is the sum of the corresponding Greeks for the components of the portfolio (McDonald 2006, page 395). Thus,
Gap call gamma Call gamma (K2 – K1) Cash-or-nothing call gamma As pointed out on line 12 of page 384 of McDonald (2006), call gamma equals put gamma. (To see this, differentiate the put-call parity formula twice with respect to S.) Because K2 K1 40 – 45 –5, call gamma put gamma = 0.07, and
gap call gamma 0.08, we have
Cash-or-nothing call gamma
5
07 . 0 08 .
0 0.002
Hence the answer is 1000 (–0.002) 2.
Remark: Another decomposition of the payoff of the gap call option is the following:
[S(T) – K1] × I[S(T) K2] S(T) × I[S(T) K2] K1 × I[S(T) K2].
See page 707 of McDonald (2006). Such a decomposition, however, is not useful here.
(K2 – K1) times the payoff of a cash-or-nothing call that pays $1 if S(T) K2
payoff of a K2-strike call
K1 times the payoff of a cash-or-nothing call that pays $1 if S(T) K2
payoff of an asset-or-nothing call
128
54. Assume the Black-Scholes framework. Consider two nondividend-paying stocks whose time-t prices are denoted by S1(t) and S2(t), respectively.
You are given:
(i) S1(0) 10 and S2(0) 20.
(ii) Stock 1’s volatility 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.
(v) The continuously compounded risk-free interest rate is 5%.
(vi) A one-year European option with payoff max{min[2S1(1), S2(1)] 17, 0} has a current (time-0) 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-0) price of this option.
(A) 0.66
(B) 1.12
(C) 1.49
(D) 5.18
(E) 7.86
129
Solution to (54) Answer: (A)
At the option-exercise date, the option holder will sell two shares of Stock 1 or one share of Stock 2, depending on which trade is of lower cost. Thus, the time-1 payoff of the option is
max{17 min[2S1(1), S2(1)], 0},
which is the payoff of a 17-strike put on min[2S1(1), S2(1)]. Define M(T) min[2S1(T), S2(T)].
Consider put-call parity with respect to M(T):
c(K, T) p(K, T) F0P,T(M)KerT.
Here, K = 17 and T = 1. It is given in (vi) that c(17, 1) 1.632. is the time-0 price of the security with time-1 payoff
M(1) min[2S1(1), S2(1)] 2S1(1) max[2S1(1) S2(1), 0].
Since max[2S1(1) S2(1), 0] is the payoff of an exchange option, its price can be obtained using (14.16) and (14.17):
3618 . 0 ) 25 . 0 )(
18 . 0 )(
4 . 0 ( 2 25 . 0 18 .
0 2 2
1 2 2
1
ln[2 (0) / (0)] ½
½ 0.1809 0.18
S S T
d T
T
, N(d1) 0.5714
2 1 ½ 0.18
d d T T , N(d2) 1 – 0.5714 0.4286
Price of the exchange option 2S1(0)N(d1) S2(0)N(d2) 20N(d1) 20N(d2) 2.856 Thus,
0,1P( ) 2 0,1 1P( ) 2.856 2 10 2.856 17.144
F M F S
and
p(17, 1) 1.632 17.144 17e0.05 0.6589.
Remarks: (i) The exchange option above is an “at-the-money” exchange option because 2S1(0) = S2(0). See also Example 14.3 of McDonald (2006).
(ii) Further discussion on exchange options can be found in Section 22.6, which is not part of the MFE/3F syllabus. Q and S in Section 22.6 correspond to 2S1 and S2 in this problem.
)
1(
,
0 M
FP
130
55. Assume the Black-Scholes framework. Consider a 9-month at-the-money European put option on a futures contract. You are given:
(i) The continuously compounded risk-free interest rate is 10%.
(ii) The strike price of the option is 20.
(iii) The price of the put option is 1.625.
If three months later the futures price is 17.7, what is the price of the put option at that time?
(A) 2.09
(B) 2.25
(C) 2.45
(D) 2.66
(E) 2.83
131
Solution to (55) Answer: (D)
By (12.7), the price of the put option is
)], ( ) (
[KN d2 FN d1 e
P rT
where
2 1 ln( / ) ½F K T
d T
, and d2 d1 T . With F K, we have ln(F / K) 0, 1 ½ 2
T ½
d T
T
, d2 ½ T , and
[ (½ ) ( ½ )] [2 (½ ) 1]
rT rT
P Fe N T N T Fe N T .
Putting P = 1.6, r = 0.1, T = 0.75, and F = 20, we get
0.1 0.75
1.625 20 [2 (½ 0.75) 1]
(½ 0.75) 0.5438
½ 0.75 0.11 0.254
e N
N
After 3 months, we have F = 17.7 and T = 0.5; hence
2 2
1 ln( / ) ½ ln(17.7 / 20) ½ 0.254 0.5
0.5904 0.59 0.254 0.5
F K T
d T
N(d1) 0.7224
7700 . 0 5 . 0 254 . 0 5904 .
1 0
2 d T
d
N(d2) 0.7794
The put price at that time is
P = erT [KN(d2) FN(d1)]
e0.1 0.5 [20 0.7794 17.7 0.7224]
2.66489
132 Remarks:
(i) A somewhat related problem is #8 in the May 2007 MFE exam. Also see the box on page 299 and the one on page 603 of McDonald (2006).
(ii) For European call and put options on a futures contract with the same exercise date, the call price and put price are the same if and only if both are at-the-money
options. The result follows from put-call parity. See the first equation in Table 9.9 on page 305 of McDonald (2006).
(iii) The point above can be generalized. It follows from the identity [S1(T) S2(T)]+ + S2(T) = [S2(T) S1(T)]+ + S1(T) that
F0,PT((S1S2) ) + F0,PT( )S = 2 F0,PT((S2S1) ) + F0,PT( )S . 1 (See also formula 9.6 on page 287.) Note that F0,PT((S1S2) ) and
0,PT(( 2 1) )
F S S are time-0 prices of exchange options. The two exchange options have the same price if and only if the two prepaid forward prices, F0,PT( )S and 1
0,PT( )2
F S , are the same.
133
56. Assume the Black-Scholes framework. For a stock that pays dividends continuously at a rate proportional to its price, you are given:
(i) The current stock price is 5.
(ii) The stock’s volatility 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 )]
2 ( ) 1 ( 2[ ) 1 2
( S S
A .
Calculate Var[A(2)].
(A) 1.51 (B) 5.57 (C) 10.29 (D) 22.29
(E) 30.57
134 Solution to (56) Answer: (A)
Var[A(2)] = 1 2 2
{E[(S(1) + S(2))2] (E[S(1) + S(2)])2}.
The second expectation is easier to evaluate. By (20.29) on page 665 of McDonald (2006),
S(t) S(0)exp[( ½2)t Z(t)].
Thus,
E[S(t)] S(0)exp[( ½2)t]×E[eZ(t)]
S(0)exp[( )t]
by (18.13). (See also formulas 18.21 and 18.22.) Consequently, E[S(1) S(2)] = E[S(1)] E[S(2)]
= 5(e0.05 e0.1), because condition (iii) means that
We now evaluate the first expectation, E[(S(1) + S(2))2]. Because ( 1) 2
exp{( δ ½ ) [ ( 1) ( )]}
( )
S t Z t Z t
S t
and because {Z(t 1) Z(t), t 0, 1, 2, ...} are i.i.d. N(0, 1) random variables (the second and third points at the bottom of page 650), we see that
( 1)
, 0, 1, 2, ( )
S t t
S t
is a sequence of i.i.d. random variables. Thus, E[(S(1) + S(2))2] =
2 (2) 2
E (1) 1 (1) S S
S
=
2 2
2 (1) (2)
(0) E E 1
(0) (1)
S S
S S S
=
2 2
2 (1) (1)
(0) E E 1
(0) (0)
S S
S S S
.
By the last equation on page 667, we have E ( )
(0) S t a
S
e[ (a δ) ½ ( 1)a a 2]t.
(This formula can also be obtained from (18.18) and (18.13). A formula equivalent to (18.13) will be provided to candidates writing Exam MFE/3F. See the last formula on
135
the first page in http://www.soa.org/files/pdf/edu-2009-fall-mfe-table.pdf ) With a 2 and t = 1, the formula becomes
2
) 0 (
) 1 E (
S
S exp[2×0.05 0.22] e0.14.
Furthermore,
2 2
0.05 0.14
(1) (1) (1)
E 1 1 2E E 1 2
(0) (0) (0)
S S S
e e
S S S
.
Hence,
E[(S(1) + S(2))2] 52 × e0.14 × (1 2e0.05 e0.14) 122.29757.
Finally,
Var[A(2)] ¼×{122.29757 [5(e0.05 e0.1)]2} 1.51038.
Alternative Solution:
Var[S(1) S(2)] = Var[S(1)] + Var[S(2)] + 2Cov[S(1), S(2)].
Because S(t) is a lognormal random variable, the two variances can be evaluated using the following formula, which is a consequence of (18.14) on page 595.
Var[S(t)] = Var[S(0)exp[( ½2)t Z(t)]
= S2(0)exp[2( ½2)t]Var[eZ(t)]
= S2(0)exp[2( ½2)t]exp(2t)[exp(2t) 1]
= S2(0) e2( )t [exp(2t) 1].
(As a check, we can use the well-known formula for the square of the coefficient of variation of a lognormal random variable. In this case, it takes the form
2
Var[ ( )]
{E[ ( )]}
S t
S t = e2t 1.
This matches with the results above. The coefficient of variation is in the syllabus of Exam C/4.)
136 To evaluate the covariance, we can use the formula Cov(X, Y) = E[XY] E[X]E[Y].
In this case, however, there is a better covariance formula:
Cov(X, Y) = Cov[X, E(Y | X)].
Thus,
Cov[S(1), S(2)] = Cov[S(1), E[S(2)|S(1)]]
= Cov[S(1), S(1)E[S(2)/S(1)|S(1)]]
= Cov[S(1), S(1)E[S(1)/S(0)]]
= E[S(1)/S(0)]Cov[S(1), S(1)]
= e Var[S(1)].
Hence,
Var[S(1) S(2)] = (1 + 2e )Var[S(1)] + Var[S(2)]
= [S(0)]2[(1 + 2e )e2( )(e2 1) + e4( )( 1)]
= 25[(1 + 2e0.05)e0.1(e0.04 1) + e0.2(e0.08 1)]
= 6.041516, and
Var[A(2)] = Var[S(1) S(2)]/4
= 6.041516 / 4
= 1.510379.
Remark: #37 in this set of sample questions is on determining the variance of a geometric average. It is an easier problem.
2 2
e
137