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.
(iv) The continuously compounded risk-free interest rate is 8%.
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
48 Solution to (19) Answer: (C)
This problem is based on Exercise 14.21 on page 465 of McDonald (2006).
Let S1 denote the stock price at the end of one year. Apply the Black-Scholes formula to calculate the price of the at-the-money call one year from today, conditioning on S1. d1 [ln (S1/S1) + (r + σ2/2)T]/( T ) (r + 2/2)/ 0.417, which turns out to be independent of S1.
d2 d1 T d1 0.117
The value of the forward start option at time 1 is C(S1) S1N(d1) S1er N(d2)
S1[N(0.417) e0.08 N(0.117)]
S1[N(0.42) e0.08 N(0.12)]
S1[0.6628 e-0.08 0.5438]
0.157S1.
(Note that, when viewed from time 0, S1 is a random variable.)
Thus, the time-0 price of the forward start option must be 0.157 multiplied by the time-0 price of a security that gives S1 as payoff at time 1, i.e., multiplied by the prepaid forward price F0P,1(S). Hence, the time-0 price of the forward start option is
0.157F0P,1(S) = 0.157e0.08F0,1(S) = 0.157e0.08100 14.5
Remark: A key to pricing the forward start option is that d1 and d2 turn out to be independent of the stock price. This is the case if the strike price of the call option will be set as a fixed percentage of the stock price at the issue date of the call option.
49
20. 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
50 Solution to (20) Answer: (D) Applying the formula
portfolio
S
portfolio value to Investor B’s portfolio yields
3.4 2C – 3P. (1)
Applying the elasticity formula
portfolio S
ln
ln[portfolio value]
value portfolio
S
S
portfolio value to Investor A’s portfolio yields
5.0
P C
S
2 (2C + P) =
9 . 1 9 . 8
45
(2C + P), or
1.2 = 2C + P. (2)
Now, (2) (1) 2.2 = 4P. Hence, time-0 put option elasticity = P =
P
S P =
4 2 . 2 9 .
145 = 13.03, which is (D).
Remarks:
(i) If the stock pays no dividends, and if the European call and put options have the same expiration date and strike price, then C P = 1. In this problem, the put and call do not have the same expiration date and strike price; so this relationship does not hold.
(ii) If your copy of McDonald (2006) was printed before 2008, then you need to replace the last paragraph of Section 12.3 on page 395 by
http://www.kellogg.northwestern.edu/faculty/mcdonald/htm/erratum395.pdf The ni in the new paragraph corresponds to the i on page 389.
(iii) The statement on page 395 in McDonald (2006) that “[t]he elasticity of a portfolio is the weighted average of the elasticities of the portfolio components” may remind students, who are familiar with fixed income mathematics, the concept of duration.
Formula (3.5.8) on page 101 of Financial Economics: With Applications to
Investments, Insurance and Pensions (edited by H.H. Panjer and published by The Actuarial Foundation in 1998) shows that the so-called Macaulay duration is an elasticity.
(iv) In the Black-Scholes framework, the hedge ratio or delta of a portfolio is the partial derivative of the portfolio price with respect to the stock price. In other continuous-time frameworks (which are not in the syllabus of Exam MFE/3F), the hedge ratio may not be given by a partial derivative; for an example, see formula (10.5.7) on page 478 of Financial Economics: With Applications to Investments, Insurance and Pensions.
51
21. The Cox-Ingersoll-Ross (CIR) interest-rate model has the short-rate process:
d ( )r t a b r t[ ( )]dt r t( ) d ( )Z t , where {Z(t)} 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:
d [ ( ), , ]
[ ( ), , ]d [ ( ), , ]d ( ) [ ( ), , ]
P r t t T
r t t T t q r t t T Z t
P r t t T , t T .
You are given (0.05, 7, 9) 0.06.
Calculate (0.04, 11, 13).
(A) 0.042 (B) 0.045 (C) 0.048 (D) 0.050 (E) 0.052
52 Solution to (21) Answer: (C)
As pointed out on pages 782 and 783 of McDonald (2006), the condition of no riskless arbitrages implies that the Sharpe ratio does not depend on T,
( , , )
( , ).
( , , ) r t T r q r t T r t
(24.17)
(Also see Section 20.4.) This result may not seem applicable because we are given an for t = 7 while asked to find an for t = 11.
Now, equation (24.12) in McDonald (2006) is
( , , ) ( ) ( , , ) / ( , , )r ( ) ln[ ( , , )], q r t T r P r t T P r t T r P r t T
r
the substitution of which in (24.17) yields
( , , )r t T r ( , ) ( )r t r ln[ ( , , )]P r t T
r
.
In the CIR model (McDonald 2006, p. 787), ( )r r , ( , ) r t r
with being a constant, and ln[ ( , , )]P r t T B t T( , ).
r
Thus,
( , , )r t T r
= ( , ) ( )r t r ln[ ( , , )]P r t T
r
= r
× r ×[B t T( , )]
= rB t T( , ), or
( , , )
1 ( , ) r t T
B t T r
.
Because ( , )B t T depends on t and T through the difference T t we have, for ,
1 1 2 2,
T t T t
1 1 1 2 2 2
1 2
( , , ) ( , , ) r t T r t T .
r r
Hence,
(0.04, 11, 13) 0.04 (0.05, 7, 9) 0.8 0.06 0.048.
0.05
Remarks: (i) In earlier printings of McDonald (2006), the minus sign in (24.1) was given as a plus sign. Hence, there was no minus sign in (24.12) and would be a negative constant. However, these changes would not affect the answer to this question.
(ii) What McDonald calls Brownian motion is usually called standard Brownian motion by other authors.
53 22. You are given:
(i) The true stochastic process of the short-rate is given by
d ( )r t 0.09 0.5 ( ) d r t t0.3d ( )Z t ,
where {Z(t)} is a standard Brownian motion under the true probability measure.
(ii) The risk-neutral process of the short-rate is given by
d ( )r t 0.15 0.5 ( ) d r t t( ( ))d ( )r t Z t , where ~( )}
{ tZ is a standard Brownian motion under the risk-neutral probability 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-short-rate derivative does not pay any
dividend or interest.
(iv) g(r(t), t) satisfies
dg(r(t), t) (r(t), g(r(t), t))dt 0.4g(r(t), t)dZ(t).
Determine (r, g).
(A) (r 0.09)g (B) (r 0.08)g (C) (r 0.03)g (D) (r + 0.08)g (E) (r + 0.09)g
54
Solution to (22) Answer: (D)
Formula (24.2) of McDonald (2006),
dr(t) a(r(t))dt (r(t))dZ(t),
is the stochastic differential equation for r(t) under the true probability measure, while formula (24.19),
,
is the stochastic differential equation for r(t) under the risk-neutral probability measure, where (r, t) is the Sharpe ratio. Hence,
(r) = 0.3, and
0.15 – 0.5r = a(r)r)(r, t)
= [0.09 – 0.5r(r)(r, t)
= [0.09 – 0.5r](r, t).
Thus, (r, t) = 0.2.
Now, for the model to be arbitrage free, the Sharpe ratio of the interest-rate derivative should also be given by (r, t). Rewriting (iv) as
d ( ( ), ) μ( ( ), ( ( ), ))
d 0.4d ( ) ( ( ), ) ( ( ), )
g r t t r t g r t t
t Z t
g r t t g r t t [cf. equation (24.13)]
and because there are no dividend or interest payments, we have
4 . 0
) , (
)) , ( ,
( r
t r g
t r g
r
= (r, t) [cf. equation (24.17)]
= 0.2.
Thus,
(r, g) = (r + 0.08)g.
Remark: d ( ) d ( )Z t Z t ( ( ), )dr t t t. This should be compared with the formula on page 662:
d ( ) d ( ) d d ( ) rd Z t Z t t Z t t
.
Note that the signs in front of the Sharpe ratios are different. The minus sign in front of
(r(t), t)) is due the minus sign in front of q(r(t), t)) in (24.1). [If your copy of McDonald (2006) has a plus sign in (24.1), then you have an earlier printing of the book.]
d ( )r t a r t( ( ))( ( )) ( ( ), ) dr t r t t t( ( ))d ( )r t Z t
55
23. 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 0. For
0 let C(s, t) be the price of one unit of the call option at time t, if the stock t T, price is s at that time. You are given:
(i) d ( )
0.1d d ( ) ( )
S t t Z t
S t , where is a positive constant and {Z(t)} is a Brownian motion.
(ii)
(iii) C(S(0), 0) 6.
(iv) At time t 0, the cost of shares required to delta-hedge one unit of the call option is 9.
(v) The continuously compounded risk-free interest rate is 4%.
Determine (S(0), 0).
(A) 0.10 (B) 0.12 (C) 0.13 (D) 0.15 (E) 0.16
d ( ( ), )
( ( ), )d ( ( ), )d ( ), 0
( ( ), ) C
C S t t
S t t t S t t Z t t T
C S t t
56 Solution to (23) Answer: (C) Equation (21.22) of McDonald (2006) is
option SVS ( )
r r
V ,
which, for this problem, translates to
( ) ( ( ), )
( ( ), ) 0.04 (0.1 0.04).
( ( ), ) S t S t t S t t
C S t t
Because
(0) ( (0),0) 9 ( (0),0) 6 1.5
S S
C S
,
we have
(S(0), 0) 0.04 + 1.5 × (0.1 0.04) 0.13
(which is the time-0 continuously compounded expected rate of return on the option).
Remark: Equation (21.20) on page 687 of McDonald (2006) should be the same as (12.9) on page 393,
option = || × , and (21.21) should be changed to
r = sign() ×
option option
r
. Note that , option, and option are functions of t.
57 24. Consider the stochastic differential equation:
dX(t) = [ – X(t)]dt dZ(t), t ≥ 0,
where and are positive constants, and {Z(t)} is a standard Brownian motion.
The value of X(0) is known.
Find a solution.
(A) X(t) X(0) et (1–et) (B) X(t) X(0) +
0tds
0tdZ(s)(C) X(t) X(0) +
0tX(s)ds
0tX(s)dZ(s)(D) X(t) X(0) (et – 1) d ( )
0t e s Z s
(E) X(t) X(0) et (1–et)
0te(ts)dZ(s)58 Solution to (24) Answer: (E)
The given stochastic differential equation is (20.9) in McDonald (2006).
Rewrite the equation as
dX(t) X(t)dt = dt dZ(t).
If this were an ordinary differential equation, we would solve it by the method of integrating factors. (Students of life contingencies have seen the method of integrating factors in Exercise 4.22 on page 129 and Exercise 5.5 on page 158 of Actuarial
Mathematics, 2nd edition.) Let us give this a try. Multiply the equation by the integrating factor et, we have
et dX(t) + etX(t)dt etdt etdZ(t). (*) We hope that the left-hand side is exactly d[etX(t)]. To check this, consider f(x, t) = etx,
whose relevant derivatives are fx(x, t) = et, fxx(x, t) = 0, and ft(x, t) = etx. By Itô’s Lemma,
df(X(t), t) et dX(t) + 0 + et X(t)dt,
which is indeed the left-hand side of (*). Now, (*) can be written as d[esX(s)] esds esdZ(s).
Integrating both sides from s = 0 to s = t, we have
) ( d )
1 ( ) ( d d
) 0 ( )
( 0 0 0
0X e s e Z s e e Z s
e t X
et
t s
t s t
t s ,or
etX(t) X(0) (et – 1) d ( )
0te s Z s
.
Multiplying both sides by et and rearranging yields
X(t) X(0)et (1–et) d ( )
0e Z s et
t s
X(0)et (1–et) d ( )
0
)
( Z s
te t s
,
which is (E).
59
Remarks: This question is the same as Exercise 20.9 on page 674. In the above, the solution is derived by solving the stochastic differential equation, while in Exercise 20.9, you are asked to use Itô’s Lemma to verify that (E) satisfies the stochastic differential equation.
If t 0, then the right-hand side of (E) is X(0).
If t > 0, we differentiate (E). The first and second terms on the right-hand side are not random and have derivatives X(0)et and et, respectively. To differentiate the stochastic integral in (E), we write
) (
0 d
)
( Z s
t e t s
= et
0tesdZ(s),which is a product of a deterministic factor and a stochastic factor. Then,
).
( d d ) ( d
)]
( d [ )
( d )
d (
) ( d d
) ( d )
d ( ) ( d d
0 0
0 0
0
t Z t s Z e e
t Z e e s Z e e
s Z e e
s Z e e
s Z e e
t s
t
t t s t
t
t s
t s t t s t
t
Thus,
), ( d d ] ) ( [
) ( d d d
)]
1 ( ) ( [
) ( d d d
) ( d )
0 (
) ( d d ) ( d d
d ) 0 ( )
( d
0
0
t Z t t
X
t Z t e t e t
X
t Z t e t s Z e e e
X
t Z t s Z e e t
e t e X t
X
t t
t s t t t
t s
t t
t
which is the same as the given stochastic differential equation.
60