90
91 Solution to (37) Answer: (D) We are to find the variance of
ln G 1
3[ln S(1) ln S(2) ln S(3)].
If
t ≥ 0, then it follows from equation (20.29) (with 0) that
ln S(t) ln S(0) ( ½ )t2 Z(t), t ≥ 0.
Hence,
Var[ln G] 12
3 Var[ln S(1) ln S(2) ln S(3)]
2 9
Var[Z(1) Z(2) Z(3)].
Although Z(1), Z(2), and Z(3) are not uncorrelated random variables, the increments, Z(1) Z(0), Z(2) Z(1), and Z(3) Z(2), are independent N(0, 1) random variables (McDonald 2006, page 650). Put
Z1 Z(1) Z(0) Z(1) because Z(0) 0, Z2 Z(2) Z(1),
and
Z3 Z(3) Z(2).
Then,
Z(1) + Z(2) + Z(3) 3Z1 + 2Z2 + 1Z3. Thus,
Var[ln G] 2 9
[Var(3Z1) + Var(2Z2) + Var(Z3)]
[3 2 1 ] 9
2 2 2
2
0.06.
Remarks:
(i) Consider the more general geometric average which uses N equally spaced stock prices from 0 to T, with the first price observation at time T/N,
G = 1/
1 ( / ) N
N
j S jT N
.Then,
Var[ln G] =
2
2
1 1
Var 1 ln ( / ) Var ( / )
N N
j j
S jT N Z jT N
N N
.With the definition
Zj Z(jT/N) Z((j1)T/N), j 1, 2, ... , N, we have
d ( )
d d ( ), ( )
S t t Z t
S t
2 2
14 14 (0.2)
0.06222
9 9
92
1 1
( / ) ( 1 ) .
N N
j
j j
Z jT N N j Z
Because {Zj} are independent N(0, T/N) random variables, we obtain Var[ln G]
2 2
2 1
( 1 ) Var[ ]
N
j j
N j Z
N
2 2
2 1
2 2
2 2
( 1 )
( 1)(2 1)
6
( 1)(2 1)
6 ,
N
j
N j T
N N
N N N T
N N
N N T
N
which can be checked using formula (14.19) on page 466.
(ii) Since
1
ln 1 N ln ( / )
j
G S jT N
N
is a normal random variable, the random variable G is a lognormal random variable. The mean of ln G can be similarly derived. In fact, McDonald (2006, page 466) wrote: “Deriving these results is easier than you might guess.”(iii)As N tends to infinity, G becomes
0
exp 1 Tln ( ) dS T
.The integral of a Brownian motion, called an integrated Brownian motion, is treated in textbooks on stochastic processes.
(iv) The determination of the distribution of an arithmetic average (the above is about the distribution of a geometric average) is a very difficult problem. See footnote 3 on page 446 of McDonald (2006) and also #56 in this set of sample questions.
93
38. 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[–B(t, T)r] for some functions A(t, T) and B(t, T).
(ii) B(0, 3) 2.
Based on P(0, 3, 0.05), you use the delta-gamma approximation to estimate P(0, 3, 0.03), and denote the value as PEst(0, 3, 0.03)
Find (0,3, 0.03) (0,3,0.05) PEst
P .
(A) 1.0240
(B) 1.0408
(C) 1.0416
(D) 1.0480
(E) 1.0560
94
Solution to (38) Answer: (B)
The term “delta-gamma approximations for bonds” can be found on page 784 of McDonald (2006).
By Taylor series,
P(t, T, r0 + ) P(t, T, r0) + 1
1!Pr(t, T, r0) + 1
2!Prr(t, T, r0)2 + … , where
Pr(t, T, r) –A(t, T)B(t, T)e–B(t, T)r –B(t, T)P(t, T, r) and
Prr(t, T, r) –B(t, T)Pr(t, T, r) = [B(t, T)]2P(t, T, r).
Thus,
0 0
( , , ) ( , , ) P t T r
P t T r
1 – B(t, T) + ½[B(t, T)]2 + …
and
(0,3, 0.03) (0,3,0.05) PEst
P = 1 – (2 × –0.02) + ½(2 × –0.02)2
= 1.0408
95
39. A discrete-time model is used to model both the price of a nondividend-paying stock and the short-term (risk-free) interest rate. Each period is one year.
At time 0, the stock price is S0 100 and the effective annual interest rate is r0 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.
Determine P(108) – C(108).
(A) 2.85 (B) 2.34 (C) 2.11 (D) 1.95 (E) 1.08
96
Solution to (39) Answer: (B)
We are given that the securities model is a discrete-time model, with each period being one year. Even though there are only two states of the world at time 1, we cannot assume that the model is binomial after time 1. However, the difference, P(K) – C(K), suggests put-call parity.
From the identity
x+ (x)+ x, we have
[K – S(T)]+ [S(T) – K]+ K – S(T), which yields
P(K) – C(K) F0,2P ( )K F0,2P ( )S
PV0,2(K) S(0)
K×P(0, 2) S(0).
Thus, the problem is to find P(0, 2), the price of the 2-year zero-coupon bond:
P(0, 2)
0
1
1 r
p*P(1, 2, ) (1u p*)P(1, 2, )d
0
1 * 1 *
=1 1 u 1 d
p p
r r r
.
To find the risk-neutral probability p*, we use S0
0
1 1 r or
100 1
1.05
p* 110 (1 p*) 95
.This yields p* , with which we obtain P(0, 2) = 1 2 / 3 1/ 3
1.05 1.06 1.04
0.904232.
Hence,
P(108) – C(108) 108 × 0.904232 100 2.34294.
p* Su (1 p*)Sd
105 95 2 110 95 3
97
40. The following four charts are profit diagrams for four option strategies: Bull Spread, Collar, Straddle, and Strangle. Each strategy is constructed with the purchase or sale of two 1-year European options.
Portfolio I
-15 -10 -5 0 5 10 15
30 35 40 45 50 55 60
Stock Price
Profit
One Year Six Months Three Months Expiration
Portfolio II
-15 -10 -5 0 5 10 15
30 35 40 45 50 55 60
Stock Price
Profit
One Year Six Months Three Months Expiration
Portfolio III
-8 -6 -4 -2 0 2 4 6 8 10
30 35 40 45 50 55 60
Stock Price
Profit
One Year Six Months Three Months Expiration
Portfolio IV
-4 -2 0 2 4 6 8 10
30 35 40 45 50 55 60
Stock Price
Profit
One Year Six Months Three Months Expiration
Match the charts with the option strategies.
Bull Spread Straddle Strangle Collar
(A) I II III IV
(B) I III II IV
(C) III IV I II
(D) IV II III I
(E) IV III II I
98 Solution to (40) Answer: (D)
Profit diagrams are discussed Section 12.4 of McDonald (2006). Definitions of the option strategies can be found in the Glossary near the end of the textbook. See also Figure 3.17 on page 87.
The payoff function of a straddle is
(s) = (K – s)+ + (s – K)+ = |s – K| . The payoff function of a strangle is
(s) = (K1 – s)+ + (s – K2)+
where K1 < K2.
The payoff function of a collar is
(s) = (K1 – s)+ (s – K2)+
where K1 < K2.
The payoff function of a bull spread is
(s) = (s – K1)+ (s – K2)+
where K1 < K2. Because x+ = (x)+ + x, we have
(s) = (K1 – s)+ (K2 – s)+ + K2 – K1 . The payoff function of a bear spread is
(s) = (s – K2)+ (s – K1)+
where K1 K2.
99
41. Assume the Black-Scholes framework. Consider a 1-year European contingent claim on a stock.
You are given:
(i) The time-0 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%.
(iv) The continuously compounded risk-free interest rate is 7%.
(v) The time-1 payoff of the contingent claim is as follows:
Calculate the time-0 contingent-claim elasticity.
(A) 0.24
(B) 0.29
(C) 0.34 (D) 0.39
(E) 0.44
S(1) 42
payoff
42
100 Solution to (41) Answer: (C) The payoff function of the contingent claim is
(s) min(42, s) 42 min(0, s – 42) 42 max(0, 42 – s) 42 (42 – s)+
The time-0 price of the contingent claim is V(0) F0,1P[ ( (1))] S
PV(42) F0,1P[(42S(1)) ]
42e0.07 P(45, 42, 0.25, 0.07, 1, 0.03).
We have d1 0.56 and d2 0.31, giving N(d1) 0.2877 and N(d2) 0.3783.
Hence, the time-0 put price is
P(45, 42, 0.25, 0.07, 1, 0.03) 42e0.07(0.3783) 45e0.03(0.2877) 2.2506, which implies
V(0) 42e0.07 2.2506 36.9099.
Elasticity = ln ln V
S
= V S
S V
= V S
V
= Put S
V .
Time-0 elasticity = 1 (0)
( )
(0)
T S
e N d
V
= 0.03 0.2877 45 36.9099
e
= 0.34.
Remark: We can also work with (s) s – (s – 42)+; then V(0) 45e0.03 C(45, 42, 0.25, 0.07, 1, 0.03) and
call ( )1 ( 1).
T T T T
V e e e N d e N d
S
101
42. 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(0) 50.
H Price of up-and-out call
60 0 70 0.1294 80 0.7583 90 1.6616
4.0861
Consider a special 6-month 60-strike European “knock-in, partial knock-out” call option that knocks in at H1 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:
H1 Not Hit H1 Hit
H2 Not Hit H2 Hit 0 2 max[S(0.5) – 60, 0] max[S(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.
102 Solution to (42) Answer: (D)
The “knock-in, knock-out” call can be thought of as a portfolio of – buying 2 ordinary up-and-in call with strike 60 and barrier H1, – writing 1 ordinary up-and-in call with strike 60 and barrier H2. Recall also that “up-and-in” call + “up-and-out” call = ordinary call.
Let the price of the ordinary call with strike 60 be p (actually it is 4.0861), then the price of the UIC (H1 = 70) is p – 0.1294
and the price of the UIC (H1 = 80) is p – 0.7583.
The price of the “knock-in, knock out” call is 2(p – 0.1294) – (p – 0.7583) 4.5856 .
Alternative Solution:
Let M(T)
0max ( )
t TS t
be the running maximum of the stock price up to time T.
Let I[.] denote the indicator function.
For various H, the first table gives the time-0 price of payoff of the form
.
The payoff described by the second table is
Thus, the time-0 price of this payoff is 4.0861 2 0.1294 0.7583 4.5856 . [ (½)] [ (½) 60]
I H M S
[70 (½)] 2 [80 (½)] [80 (½)] [ (½) 60]
1 [70 (½)] 1 [80 (½)] [ (½) 60]
1 [70 (½)] [80 (½)] [70 (½)] [80 (½)] [ (½) 60]
1 2 [70 (½)] [80 (½)] [ (½) 60]
[ (½)] 2 [70 (½)] [80 (½)]
I M I M I M S
I M I M S
I M I M I M I M S
I M I M S
I M I M I M
[ (½) 60]S
103
43. 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
d ( )
( ) x t
x t (r – r€)dt dZ(t),
where {Z(t)} is a standard Brownian motion and is a constant.
Let y(t) be the euro/dollar exchange rate at time t. Thus, y(t) 1/x(t).
Which of the following equation is true?
(A) d ( ) ( ) y t
y t (r€ r)dt dZ(t) (B) d ( )
( ) y t
y t (r€ r)dt dZ(t) (C) d ( )
( ) y t
y t (r€ r ½2)dt dZ(t) (D) d ( )
( ) y t
y t (r€ r ½2)dt dZ(t) (E) d ( )
( ) y t
y t (r€ r 2)dt – dZ(t)
104 Solution to (43) Answer: (E)
Consider the function f(x, t) 1/x. Then, ft 0, fx x2, fxx 2x3. By Itô’s Lemma,
dy(t) df(x(t), t)
ftdt fxdx(t) ½fxx[dx(t)]2
0 [x(t)2]dx(t) ½[2x(t)3][dx(t)]2
x(t)1[dx(t)/x(t)] x(t)1[dx(t)/x(t)]2
y(t)[(r – r€)dt dZ(t)] y(t)[(r – r€)dt + dZ(t)]2
y(t)[(r – r€)dt dZ(t)] y(t)[2dt], rearrangement of which yields
d ( ) ( ) y t
y t (r€ r + 2)dt – dZ(t), which is (E).
Alternative Solution Here, we use the correspondence between
and
W(t) W(0)exp[( – ½)t + Z(t)].
Thus, the condition given is
x(t) x(0)exp[(r r€ – ½2)t + Z(t)].
Because y(t) 1/x(t), we have
y(t) y(0)exp{[(r r€ – ½2)t + Z(t)]}
y(0)exp[(r€ r + 2 – ½()2)t + (–)Z(t)], which is (E).
Remarks:
The equation
d ( ) ( ) x t
x t (r – r€)dt + dZ(t)
can be understood in the following way. Suppose that, at time t, an investor pays $x(t) to purchase €1. Then, his instantaneous profit is the sum of two quantities:
(1) instantaneous change in the exchange rate, $[x(t+dt) – x(t)], or $ dx(t), (2) € r€dt, which is the instantaneous interest on €1.
Hence, in US dollars, his instantaneous profit is dx(t) + r€dt × x(t+dt)
dx(t) + r€dt × [x(t) + dx(t)]
dx(t) + x(t)r€dt, if dt × dx(t) 0.
d ( )
d d ( ) ( )
W t t Z t
W t
105
Under the risk-neutral probability measure, the expectation of the instantaneous rate of return is the risk-free interest rate. Hence,
E[dx(t) x(t)r€dt | x(t)] x(t) × (rdt), from which we obtain
r€)dt.
Furthermore, we now see that {Z(t)} is a (standard) Brownian motion under the dollar-investor’s risk-neutral probability measure.
By similar reasoning, we would expect d ( )
( ) y t
y t (r€ r)dt + ωdZ€(t),
where {Z€(t)} is a (standard) Brownian motion under the euro-investor’s risk-neutral probability measure and is a constant. It follows from (E) that ω and
Z€(t) Z(t) t.
Let W be a contingent claim in dollars payable at time t. Then, its time-0 price in dollars is
E[ert W],
where the expectation is taken with respect to the dollar-investor’s risk-neutral
probability measure. Alternatively, let us calculate the price by the following four steps:
Step 1: We convert the time-t payoff to euros, y(t)W.
Step 2: We discount the amount back to time 0 using the euro-denominated risk-free interest rate,
exp(r€t) y(t)W.
Step 3: We take expectation with respect to the euro-investor’s risk-neutral probability measure to obtain the contingent claim’s time-0 price in euros,
E€[exp(r€t) y(t)W].
Here, E€ signifies that the expectation is taken with respect to the euro-investor’s risk-neutral probability measure.
Step 4: We convert the price in euros to a price in dollars using the time-0 exchange rate x(0).
We now verify that both methods give the same price, i.e., we check that the following formula holds:
x(0)E€[exp(r€t) y(t)W] E[ert W].
This we do by using Girsanov’s Theorem (McDonald 2006, p. 662). It follows from Z€(t) Z(t) t and footnote 9 on page 662 that
E€[y(t)W] E[(t)y(t)W], d ( )
E ( ) (
( )
x t x t r x t
106 where
(t) exp[)Z(t) – ½()2t] exp[Z(t) – ½2t].
Because
y(t) y(0)exp[(r€ r + ½2)t – Z(t)], we see that
exp(r€t)y(t)(t) y(0)exp(rt).
Since x(0)y(0) 1, we indeed have the identity
x(0)E€[exp(r€t) y(t)W] E[ert W].
If W is the payoff of a call option on euros, W [x(t) – K]+, then
x(0)E€[exp(r€t) y(t)W] E[ert W]
is a special case of identity (9.7) on page 292. A derivation of (9.7) is as follows. It is not necessary to assume that the exchange rate follows a geometric Brownian motion.
Also, both risk-free interest rates can be stochastic.
The payoff of a t-year K-strike dollar-dominated call option on euros is $[x(t) – K]+
[$x(t) – $K]+
[€1 $K]+
[€1 €y(t)K]+
K × €[1/K y(t)]+,
which is K times the payoff of a t-year (1/K)-strike euro-dominated put option on dollars.
Let C$(x(0), K, t) denote the time-0 price of a t-year K-strike dollar-dominated call option on euros, and let P€(y(0), H, t) denote the time-0 price of a t-year H-strike
euro-dominated put option on dollars. It follows from the time-t identity $[x(t) – K]+ K × €[1/K y(t)]+
that we have the time-0 identity
$ C$(x(0), K, t) K × € P€(y(0), 1/K, t) $ x(0) × K × P€(y(0), 1/K, t) $ x(0) × K × P€(1/x(0), 1/K, t), which is formula (9.7) on page 292.
107
For Questions 44 and 45, 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%.