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%.
108 Solution to (44) Answer: (D)
By formula (10.5), the risk-neutral probability of an up move is
.
Remark
If the put option is European, not American, then the simplest method is to use the binomial formula [p. 358, (11.17); p. 618, (19.1)]:
er(3h)
*(1 *) (300 183.75) 0 0
2 ) 3 9 . 102 300 (
*) 1 3 (
3 3 2
p p
p
= er(3h)(1 p*)2[(1 p*) × 197.1 + 3 × p* × 116.25)]
= er(3h)(1 p*)2(197.1 + 151.65p*) = e0.1 × 3 × 0.389782 × 289.63951 = 32.5997
Solution to (45) Answer: (C)
Formula (13.16) is
d u
d u
S S
. By formula (13.15) (or (10.1)), .
) (
δ
d u S
C e h Cu d
908670 .
147 0 5 . 262
153 0002 . 41
186279 .
5 0 . 262 75 . 468
0002 . 41 0
1 065 . 0 δ
1 065 . 0 δ
S e S
P e P
S e S
P e P
dd ud
dd ud h d
ud uu
ud h uu
u
Hence,
004378 .
210 0 375
908670 .
0 186279 .
0
Remark: This is an approximation, because we wish to know gamma at time 0, not at time 1, and at the stock price S0 = 300.
61022 . 210 0
375
210
* 300
1 ) 065 . 0 1 . 0 ) (
δ ( 0 )
δ
(
e
S S
S e
S d u
d p e
d u
d h h r
r
300 (39.7263)
375
(14.46034)
210 (76.5997) 90
468.75 (0)
262.5 (41.0002)
147 (133.702) 153
585.9375 (0)
328.125 (0)
183.75 (116.25)
102.9 (197.1) Option prices in bold italic signify
that exercise is optimal at that node.
109
46. 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) u/d = 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.
(v) The continuously compounded risk-free interest rate is 5%.
Let CI 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 CII CI.
(A) 0
(B) 0.022
(C) 0.044
(D) 0.066
(E) 0.088
110 Solution to (46) Answer: (E)
By formula (10.14), the risk-neutral probability of an up move is .
Substituting p* 1/3 and u/d 4/3, we have
.
Hence, and .
The two-period binomial tree for the futures price and prices of European and American options at t 0.5 and t 1 is given below. The calculation of the European option prices at t 0.5 is given by
455145 .
0
*)]
1 ( 0
* 4 . 1 [
72841 . 10
*)]
1 ( 4 . 1
* 2 . 30 [
5 . 0 05 . 0
5 . 0 05 . 0
p p
e
p p
e
Thus, CII CI e0.050.5 (11 10.72841) p* = 0.088.
Remarks:
(i) .CI e0.050.5[10.72841p*0.455145(1 p*)]3.78378 . 87207 . 3
*)]
1 ( 455145 .
0
* 11
5[
. 0 05 .
0
e p p
CII
(ii) A futures price can be treated like a stock with = r. With this observation, we can obtain (10.14) from (10.5),
1 .
* ( ) ( )
d u
d d
u d e
d u
d
p e r h r r h
Another application is the determination of the price sensitivity of a futures option with respect to a change in the futures price. We learn from page 317 that the price sensitivity of a stock option with respect to a change in the stock price is
( )
h Cu Cd
e S u d
. Changing to r and S to F yields
( )
rh Cu Cd
e F u d
, which is the same as the expression erh given in footnote 7 on page 333.
1 /
1 / 1
* 1
d u
d d
u p d
1 3 / 4
1 / 1 3 1
d 9
.
0
d u(4 / 3) d 1.2
80
96
(10.72841) 11
72
(0.455145)
115.2 (30.2)
86.4 (1.4)
64.8 (0) An option price in bold italic signifies
that exercise is optimal at that node.
111
47
.
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 information:
(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
112 Solution to (47) Answer: (B)
Let the date several months ago be 0. Let the current date be t.
Delta-hedging at time 0 means that the investor’s cash position at time 0 was 100[C(0) C(0)S(0)].
After closing out all positions at time t, her profit is
100{[C(0) C(0)S(0)]ert – [C(t) C(0)S(t)]}.
To find the accumulation factor ert, we can use put-call parity:
C(0) – P(0) S(0) – KerT, C(t) – P(t) S(t) – Ker(Tt), where T is the option expiration date. Then,
ert ( ) ( ) ( ) (0) (0) (0) S t C t P t
S C P
= 50 14.42 0.26 40 8.88 1.63
= 35.84
32.75 = 1.0943511.
Thus, her profit is
100{[C(0) C(0)S(0)]ert – [C(t) C(0)S(t)]}
100{[8.88 0.794 × 40] × 1.09435 – [14.42 0.794 × 50]}
24.13 24
Alternative Solution: Consider profit as the sum of (i) capital gain and (ii) interest:
(i) capital gain 100{[C(0) C(t)] C(0)[S(0) – S(t)]}
(ii) interest 100[C(0) C(0)S(0)](ert – 1).
Now,
capital gain 100{[C(0) C(t)] C(0)[S(0) – S(t)]}
100{[8.88 14.42] [40 – 50]}
100{5.54 + 7.94} 240.00.
To determine the amount of interest, we first calculate her cash position at time 0:
100[C(0) C(0)S(0)] 100[8.88 400.794]
100[8.88 31.76] = 2288.00.
Hence,
interest = 2288(1.09435 – 1) = 215.87.
Thus, the investor’s profit is 240.00 – 215.87 = 24.13 24.
Third Solution: Use the table format in Section 13.3 of McDonald (2006).
Position Cost at time 0 Value at time t Short 100 calls 100 8.88 = –888 –100 14.42 = 1442 100 shares of stock 100 0.794 40 = 3176 100 0.794 50 = 3970
Borrowing 3176 888 = 2288 2288ert = 2503.8753
Overall 0 24.13
113
Remark: The problem can still be solved if the short-rate is deterministic (but not necessarily constant). Then, the accumulation factor ert is replaced by
exp[
0tr s s( )d ], which can be determined using the put-call parity formulasC(0) – P(0) = S(0) – K
exp[
0Tr s s( )d ], C(t) – P(t) = S(t) – Kexp[ T ( )d ]t r s s
.If interest rates are stochastic, the problem as stated cannot be solved.
114
48. The prices of two nondividend-paying stocks are governed by the following stochastic differential equations:
1 1
d ( )
0.06d 0.02d ( ), ( )
S t t Z t
S t
2 2
d ( )
0.03d d ( ), ( )
S t t k Z t
S t
where Z(t) is a standard Brownian motion and k is a constant.
The current stock prices are S1(0) 100 and S2(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
115 Solution to (48) Answer: (E)
The problem is a variation of Exercise 20.12 where one asset is perfectly negatively correlated with another.
The no-arbitrage argument in Section 20.4 “The Sharpe Ratio” shows that 02
. 0
04 . 0 06 .
0
= k
04 . 0 03 .
0
or k = 0.01, and that the current number of shares of Stock 2 in the hedged portfolio is
(0)
) 0 (
2 1 1
S k
S
=
50 ) 01 . 0 (
100 02 . 0
= 4,
which means buying four shares of Stock 2.
Alternative Solution: Construct the zero-investment, risk-free portfolio by following formula (21.7) or formula (24.4):
I(t) S1(t) N(t)S2(t) W(t),
where N(t) is the number of shares of Stock 2 in the portfolio at time t and W(t) is the amount of short-term bonds so that I(t) 0, i.e.,
W(t) [S1(t) + N(t)S2(t)].
Our goal is to find N(0). Now, the instantaneous change in the portfolio value is dI(t) dS1(t) N(t)dS2(t) W(t)rdt
S1(t)[0.06dt + 0.02dZ(t)] N(t)S2(t)[0.03dt kdZ(t)] 0.04W(t)dt
(t)dt (t)dZ(t), where
(t) 0.06S1(t) 0.03N(t)S2(t) 0.04[S1(t) N(t)S2(t)]
0.02S1(t) 0.01N(t)S2(t), and
(t) 0.02S1(t) kN(t)S2(t).
The portfolio is risk-free means that N(t) is such that (t) = 0. Since I(t) 0, the no-arbitrage condition and the risk-free condition mean that we must also have (t) 0, or
) ( 01 . 0
) ( 02 . ) 0 (
2 1
t S
t t S
N .
In particular,
5 4 . 0
2 ) 0 ( 01 . 0
) 0 ( 02 . ) 0 0 (
2
1
S
N S .
Remark: Equation (21.20) on page 687 of McDonald (2006) should be the same as (12.9) on page 393,
option || × .
Thus, (21.21) should be changed to sign() ×
option option
r
.
r
116
49
.
You use the usual method in McDonald and the following information 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).(i) The period is 3 months.
(ii) The initial stock price is $100.
(iii) The stock’s volatility is 30%.
(iv) The continuously compounded risk-free interest rate is 4%.
At the beginning 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 will exercise the put option at the beginning of the period.
(A) 114 (B) 115 (C) 116 (D) 117 (E) 118
117
Solution to (49) Answer: (B)
1.173511 0.869358 S initial stock price 100
The problem is to find the smallest integer K satisfying
K S erh[p* Max(K Su, 0) + (1 p*) Max(K Sd, 0)]. (1) Because the RHS of (1) is nonnegative (the payoff of an option is nonnegative), we have the condition
K S 0. (2)
As d 1, it follows from condition (2) that Max(K Sd, 0) K Sd, and inequality (1) becomes
K S erh[p* Max(K Su, 0) + (1 p*) (K Sd)]. (3) If K ≥Su, the right-hand side of (3) is
erh[p* (K Su) + (1 p*) (K Sd)]
erhK ehS
erhK S,
because the stock pays no dividends. Thus, if K ≥Su, inequality (3) always holds, and the put option is exercised early.
We now investigate whether there is any K, S K Su, such that inequality (3) holds. If Su K, then Max(K Su, 0) 0 and inequality (3) simplifies as
K S erh × (1 p*) (K Sd), or
K
)
* 1 ( 1
)
* 1 ( 1
p e
d p e
rh rh
S. (4)
The fraction
)
* 1 ( 1
)
* 1 ( 1
p e
d p e
rh rh
can be simplified as follows, but this step is not necessary. In McDonald’s forward-tree model,
1 p* p*×e h, from which we obtain
1 p*
e h
1
1 .
(r )h h rh h (0.04 / 4) (0.3/ 2) 0.16
u e e e e
(r )h h rh h (0.04 / 4) (0.3/ 2) 0.14
d e e e e
118 Hence,
)
* 1 ( 1
)
* 1 ( 1
p e
d p e
rh rh
rh h
rh h
e e
d e e
1 1
h rh
h h
e e
e e
1
1 because 0
1e h erh
1 .
Therefore, inequality (4) becomes K
rh
h e
e
1
1 S
0.15 0.01 1
1
e e S 1.148556×100 114.8556.
Thus, the answer to the problem is 114.8556 115, which is (B).
Alternative Solution:
1.173511 0.869358 S initial stock price = 100
p* = = 0.46257.
Then, inequality (1) is
K 100 e0.01[0.4626 × (K 117.35)+ 0.5374 × (K 86.94)+], (5) and we check three cases: K ≤ 86.94, K ≥ 117.35, and 86.94 K 117.35.
For K ≤ 86.94, inequality (5) cannot hold, because its LHS 0 and its RHS 0.
For K ≥ 117.35, (5) always holds, because its LHS K 100 while its RHS e0.01K 100.
For 86.94 K 117.35, inequality (5) becomes
K 100 e0.01 × 0.5374 × (K 86.94), or
K 100 0.010.010.5374 86.94
1 0.5374
e e
114.85.
Third Solution: Use the method of trial and error. For K 114, 115, … , check whether inequality (5) holds.
Remark: An American call option on a nondividend-paying stock is never exercised early. This problem shows that the corresponding statement for American puts is not true.
(r )h h rh h (0.04 / 4) (0.3/ 2) 0.16
u e e e e
(r )h h rh h (0.04 / 4) (0.3/ 2) 0.14
d e e e e
0.3/ 2 0.15
1 1 1 1
1+1.1618
1 1
1 e h e e
119 50. Assume the Black-Scholes framework.
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
120 Solution to (50) Answer: (A)
This problem is a modification of #4 in the May 2007 Exam C.
The conditions given are:
(i) S0 = 0.25, (ii) = 0.35, (iii) = 0.15.
We are to seek the number S such that 0.5U Pr(S0.5 S0.5U ) = 0.95.
The random variable ln(S0.5/ 0.25) is normally distributed with mean (0.15 ½ 0.35 ) 0.5 0.044375,2
standard deviation 0.35 0.5 0.24749.
Because N−1(0.95) 1.645, we have
0.044375 0.24749 N1(0.95) 0.4515 . Thus,
0.5
S = U 0.25e0.45150.3927.
Remark The term “confidence interval” as used in Section 18.4 seems incorrect, because St is a random variable, not an unknown, but constant, parameter. The expression
Pr(StL St StU) 1 p
gives the probability that the random variable St is between StL and S , not the tU
“confidence” for St to be between StL and S . tU
121 51. Assume the Black-Scholes framework.
The price of a nondividend-paying stock in seven consecutive months is:
Month Price 1 54 2 56 3 48 4 55 5 60 6 58 7 62
Estimate the continuously compounded expected rate of return on the stock.
(A) Less than 0.28
(B) At least 0.28, but less than 0.29 (C) At least 0.29, but less than 0.30 (D) At least 0.30, but less than 0.31 (E) At least 0.31
122 Solution to (51) Answer: (E)
This problem is a modification of #34 in the May 2007 Exam C. Note that you are given monthly prices, but you are asked to find an annual rate.
It is assumed that the stock price process is given by d ( )
( ) S t
S t dt dZ(t), t 0.
We are to estimate , using observed values of S(jh), j 0, 1, 2, .. , n, where h 1/12 and n 6. The solution to the stochastic differential equation is
S(t) S(0)exp[(½t Z(t)].
Thus, ln[S((j+1)h)/S(jh)], j 0, 1, 2, …, are i.i.d. normal random variables with mean (½)h and variance h.
Let {rj} denote the observed continuously compounded monthly returns:
r1 = ln(56/54) = 0.03637, r2 = ln(48/56) = 0.15415, r3 = ln(55/48) = 0.13613, r4 = ln(60/55) = 0.08701, r5 = ln(58/60) = 0.03390, r6 = ln(62/58) = 0.06669.
The sample mean is r =
n j
rj
n 1
1 =
n 1
0
( ) ln ( )
S tnh
S t = 1 6
ln62
54 = 0.023025.
The (unbiased) sample variance is
n j
j r n 1 r
)2
1 (
1 = 2 2
1
1 ( )
1
n j j
r nr
n
= 6 2 21
1 ( ) 6
5 j r j r
= 0.01071.Thus, (½) + ½ is estimated by
(0.023025 + ½ × 0.01071) × 12 0.3405.
123 Remarks:
(i) Let T = nh. Then the estimator of ½ is r
h = 1 nh
ln ( ) (0) S T
S = ln[ ( )] ln[ (0)]
0
S T S
T
.
This is a special case of the result that the drift of an arithmetic Brownian motion is estimated by the slope of the straight line joining its first and last observed values.
Observed values of the arithmetic Brownian motion in between are not used.
(ii) An (unbiased) estimator of 2 is
h
1 2 2
1
1 ( )
1
n j j
r nr
n
= T1 2 21
1 ( )
( ) ln
1 1 (0)
n j j
n S T
n r n S
≈
T 1
1 n n
2 1
( )
n j j
r
for large n (small h)=
T 1
1 n
n n1{ln[ ( / ) / (( 1) / )]}2
j
S jT n S j T n
,which can be found in footnote 9 on page 756 of McDonald (2006). It is equivalent to formula (23.2) on page 744 of McDonald (2006), which is
= 1
h 1
1
n n1{ln[ ( / ) / (( 1) / )]}2
j
S jT n S j T n
.(iii) An important result (McDonald 2006, p. 653, p. 755) is: With probability 1,
lim
n
2 1
{ln[ ( / ) / (( 1) / )]}
n j
S jT n S j T n
= 2T,showing that the exact value of can be obtained by means of a single sample path of the stock price. Here is an implication of this result. Suppose that an actuary uses a so-called regime-switching model to model the price of a stock (or stock index), with each regime being characterized by a different . In such a model, the current regime can be determined by this formula. If the price of the stock can be observed over a time interval, no matter how short the time interval is, then is revealed immediately by determining the quadratic variation of the logarithm of the stock price.
ˆH2
124