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Unformatted text preview: 7 OPTION PRICING 1' 2. In a twoperiod CRR model with r = 1% per period, 5(0) = 100, u = 1.02, and
d = 0.98, consider an option that expires after two periods, and pays the value of the squared
stock price, 32(t), if the stock price S (t) is higher than $100.00 when the option is exercised.
Otherwise (when S(t) is less or equal to 100), the option pays zero. Find the price of the
European version of this option. Solution: After two periods, the stock can take three possible values, 3,“,(2) = 104.04,
Sud(2) = 99.96, 5.“,(2) = 96.04. The payoff of the option will then be 0....(2) = 10,824.32
if the stock goes up in price twice, and zero in the other two possible states. In order to
price this derivative, we can use riskneutral pricing. For that, we need to compute the
riskneutral probability p‘ as ,_(1+r)—d_1.01—0.98 1” u — d ‘ 1.02 — 0.98 = 0'75 The price of the derivative is 1
1.012 where we only needed to take into account the state resulting if the stock goes up twice and 0(0) = (0.75‘2  10, 824.32) = 5, 968.71 the riskneutral probability of this event, because the other two states have zero payoffs. 1' 4. Consider a singleperiod binomial model with two periods where the stock has an
initial price of $100 and can go up 15% or down 5% in each period. The price of the European
call option on this stock with strike price $115 and maturity in two periods is $5.424. What
should be the price of the riskfree security that pays $1 after one period regardless of what
happens? We assume, as usual, that the interest rate 7' per period is constant. Solution: In order to compute the price of the riskfree security we need the interest
rate. It is important to observe that the call option only pays if the stock goes up all five
times. In that case, the price of the stock is 100(1.152) = 132.25, and the payoff of the
option 132.25 — 115 = 17.25. If the stock goes down a single time, the option will be out
ofthemoney and its payoff will be zero. Therefore, using riskneutral pricing, for unknown
riskneutral probability p“ and interest rate per period 1", we need to have 1
(1+ 7')2 5.424 = (129217.25 or, simplifying,
1 + r = 1.723p' 33 We have two unknowns. We have another equation for the same unknowns from the formula for the riskneutral probability, *_(1+r)—d_(1+T)—0.95
19‘ u—d _ 1.15—0.95 The solution to the two equations is p* = 0.6, 7' = 0.07. The price of the risk—free security is 1
—— = 0.9346
1.07 T 6. Suppose that the stock price today is S(t) : 2.00, the interest rate is 7" = 0%, and
the time to maturity is 3 months. Consider an option whose Black—Scholes price is given by
the function V(t, s) = 8262(T_t) where the time is in annual terms. What is the option price today? What is the volatility
of the stock equal to? Solution: The price is
V(t, 8) : 8262(T—t) = 460.5 2 6.5949 . The function V(t, 3) has to satisfy the Black—Scholes equation. We have Vt : —28262(T_t),
V35 2 262(T”t), which means that 1
Vt+§252V33:0 from which we recognize that 02 = 2, thus 0 = ﬂ. T 10. Verify that the Black~Scholes formula for the European put option can be obtained
from the formula for the call option using put—call parity. (Hint: You can use the fact that
1 — N(:c) = N(—$) for the normal distribution function.) Solution: We have
19(0) 2 c(0)+Ke_TT—S(0) = S(0)(N(d1)—1)+Ke—TT(1—N(d2)) = Ke"TTN(—d2)—S(0)N(—d1). T 12. In order to avoid the problem of implied volatilities being diﬁerent for different
strike prices and maturities, a student of the Black—Scholes theory suggests making the
stocks volatility a a function of K and T, 0(K, T). What is wrong with this suggestion, at
least from the theoretical/ modeling point of View? (In practice, though, traders might use
different volatilities for pricing options with different maturities and strike prices.) Solution: The stock’s volatility a in the Merton—Black—Scholes model is a constant
associated with the given stock. Thus, theoretically, it cannot change with the strike price and the maturity of options written on the stock. 34 T 14. In a twoperiod CRR model with r = 1% per period, 3(0) = 100, u = 1.02
and d = 0.98, consider an option that expires after two periods, and pays the value of the
squared stock price, S20), if the stock price S(t) is higher than $100.00 when the option
is exercised. Otherwise (when S (t) is less or equal to 100), the option pays zero. Find the
price of the American version of this option. Solution: We present the corresponding tree for the stock price. I 104.04
102 [10824.32]
[10404]
100 H 99.96
98 [0]
0
[ ] 96.04 [0] In brackets we record the payoff of the American option if we exercise it in the corre—
sponding node. In node I I the payoff and value of un—exercised option is zero. In node I, if
we exercise the option, the payoff is 1022 = 10. 404. We have to compare this payoff with the
value of the un—exercised option. For that, we need to compute the risk—neutral probability
p* as * (1+r)—d 1.01—0.98
p : u—d 2102—098 2
The value of the un—exercised option at that node is 0.75 1
—— 0.75  10,824.32 2 .037. 6.
1.01( ) 8' 8 Therefore, it is optimal to exercise early at this node. We now compute the price of the
option at the initial moment as
1
A(0) 2 1——0—1(0.75' 10, 404) : 7725.743. 1‘ 16. Find the price of a 3—month European call option with K = 100, r = 005, 5(0) =
100,u = 1.1 and d = 0.9 in the binomial model, if a dividend amount of D = $5 is to be
paid at time 7‘ = 1.5 months. Use the binomial tree with time step At = 1/12 years to
model the process Sg(t) = S(t) — e_’(7_t)D for t < 7'. Solution: As usual, we assume that 7" = 0.05 is the annual interest rate. In the following graph we represent the tree for the stock S and for its value SC; net of dividends, recorded
in the brackets: 35 Payoff 126.486 26.4865
111 /
/ 114.987\
I
109.524 103.488 3.488
[10453] \ /
IV
100 / 94.08
[95.031]\ /
H \
90.518 84.673 0
[85.528] \
V /
76.97 The initial price of the stock net of the dividend is 59(0) 2 8(0) — e’TTD : 100 —
6’0‘05'1‘5/12  5 = 95.0312. This part of the price of the stock can go up or down with factors
u or d. The total price S(t) of the stock, including dividend, is found from S(t) 2 Soft) +
6‘0‘05(T‘t)5. After the second period the dividend has already been paid and the price of the
stock net of the dividend is the total price of the stock, so we keep using u and d to track
the possible values of the stock, but do not have to add the dividend. We now compute the price of the option using backward induction. In node V the price
is trivially zero. In nodes III and IV we can use risk—neutral probability, since the dividend
has already been paid, or the replicating portfolio method. Since there are dividends, we
cannot use the usual risk—neutral probability and we will have to ﬁnd the replicating portfolio
in the previous nodes. Thus, we decide to use this procedure for the whole tree. As usual,
we denote 60 the investment at the risk—free rate and 61 the number of shares to hold. In node III, we have 26.4865
3.4889   50e005/ 12 + 611264865
50e0~05/12 + 611034889 with solution 60 = *995842; 61 = 1. The value of the option at node III is, then,
0111(2) = 114.9877 — 99.5842 2 15.4035 For node IV, H 60610“ 12 + 611034899 3.4899 6060'05/12 + 6184.673 H
o 36 with solution 60 : —15.6349; 61 = 0.1854. The price of the option at node IV is, then,
sz(2) : 0185494081 — 15.6349 2 1.8097 N ow, we have to take into consideration that between t = 1 and t = 2 the dividend is paid.
An investor that buys one share of stock at moment t = 1, at moment t = 2 will have one
share of the stock plus the future value of the dividend, 560‘05/24. Thus, we solve for the replicating portfolio at node I from 6060.05/12 + 61(1149877 _+_ 560.05/24) : 154305
50e°~°5/ 12 + 61(94081 + 51900504) 18097 with solution 60 = —62.3594; 61 = 0.6502. The price of the option at node I is, then, (71(1) 2 06502109524 — 62.3594 = 8.8534
Similarly, in node [I we have, 606005/12 + 61(94081 + 5e0~05/24) = 1.8097
6060.05/12 +61(76975+5€U.O5/24) : O with solution 60 = —8.6379; 61 = 0.1058. The price of the option at node [I is, then,
011(1) 2 0.1058  90.518 — 8.6379 2 0.9387 Finally. at the initial point we have 5060~05/12+61109.524 = 8.8534
60e0‘05/12+6190.518 = 0.9387 with solution (50 = —36.6024; 61 = 0.4164. The price of the option at the initial time is, then, 0(0) = 0.4164 . 100 — 36.6024 = 5.0403 1' 18. Consider the following two—period setting: the price of a stock is $50. Interest rate
per period is 2%. After one period the price of the stock can go up to $55 or drop to $47 and
it will pay (in both cases) a dividend of $3. If it goes up the ﬁrst period, the second period
it can go up to $57 or down to $48. If it goes down the ﬁrst period, the second period it
can go up to $48 or down to $41. Compute the price of an American put option with strike
price K = 45 that matures at the end of the second period. Solution: In the following graph we present the tree corresponding to the stock and put
payoff: 37 . «wer9. «1w.» fameWWW") , 41 4 In parenthesis we present the stock prices after dividends are paid. \Ne solve recursively.
Obviously, in node I the put is worth 0. In node II, if we exercise the option we get
45 — 44 = 1 (since we would exercise after the dividend is paid). We compute the value
of the un—exercised option and take the higher of the two. In order to compute the value
of the un—exercised option we compute the value of the replicating portfolio, by ﬁnding the
number of shares 61 and the amount 60 to be deposited in the bank: 1.0260 + 4161 4 with solutions 61 = —4/7 and (50 = 26.89. The value of the put is, then,
P1(1) = 26.89 — (4/7)44 = 1.748 This is larger than one, and therefore, we will not exercise the option until maturity. The replicating portfolio at the initial node is obtained from 1.0260 + 5561
1.0260 + 4761 0
1.748 With solutions 61 = —0.2185 and 60 = 11.7812. The value of the put is, then,
P(0) = 11.7812 — 0.2185  50 = 0.8568. 1‘ 20. Consider a Merton—Black—Scholes model with 7" = 0.07, a = 0.3, T = 0.5 years,
S (0) = 100, and a call option with the strike price K = 100. Using the normal distribution
table (or an appropriate software program), ﬁnd the price of the call option, when there are
no dividends. Repeat this exercise when (a) the dividend rate is 3%; (b) the dividend of
$3.00 is paid after three months. 38 Solution: With no dividends we get
d1 = 0.2711, d2 = 0.0589, N(d1) = 0.6068, N(d2) = 0.5235, C = 10.1338 .
When the dividend rate is 3% we get
d1 = 0.2003 , d2 = —0.0118, N(d1) = 0.5794, N(d2) = 0.4953, C = 9.2506 . When the dividend of $3.00 is paid after three months, we replace S (0) = 100 by S (0) —
D(0) : 100 — 3e‘0'25T : 97.0524 and get d1 = 0.13 , d2 = —0.0821, N(d1) = 0.5517, N(d2) = 0.4673, C = 8.4253 . T 22. In the context of the previous two problems, with no dividends, compute the price
of the chooser option, for which the holder can choose at time t1 = 0.25 years whether to
hold the call or the put option. Solution: For the price of a call option maturing at t1 we get
d1 = 0.1917, d2 = 0.0417, N(d1) : 0.576, N(d2) = 0.5166, C = 6.8343 . For the price of a put option maturing at 231 and with strike price Ke‘ra—tl) = 98.2652 we get
d1 = 0.3083, d2 = —0.1583, N(d1) = 0.6211, N(d2) = 0.5629, P = 4.3149 . The price of the chooser option is C + P : 11.1492. T 24. Provide a proof for expression (7.43) for the price of a digital option. Compute
the price if the option pays $1.00 if the stock price at maturity is larger than $100.00, and
it pays $0.00 otherwise. Use the same parameters as in the previous three problems. Solution: The ﬁrst equality in (7.43) is obvious. The second equality is proved in
Appendix 7.9 in the book. The price is N(0.0589) = 0.5235. ’r 26. Let 5(0) 2 $100.00, K1 2 $92.00, K2 = $125.00, 7* = 5%. Find the Black—Scholes
formula for the option paying in three months $10.00 if S(T) S K1 or if S(T) 2 K2, and
zero otherwise, in the Black—Scholes continuoustime model. Solution: The price is given by 106‘3r/12E*[1{5(T)3K1 or S(T)2K2}l = 98758(P*lS(T) S K1l+ P*lS(T) Z KZl) 1' 30. Show that, if S is modeled by the Merton—BlackScholes model, then S and its futures price have the same volatility. 39 ”may”, Ir??‘."‘ ....__ ‘ . _ Solution: We have
dF(t) = eT(T“t)dS(t) — reT(T_t)S(t)dt = Fungi — 7")dt + adW(t)] We see that the volatility is the same. 1' 32. Compute the price of a European call on the yen. The current exchange rate is
108, the strike price is 110, maturity is three months and t he price of a three—month T—bill
is $98.45. We estimate the annual volatility of the yendollar exchange rate to be 15%.
A threemonth purediscount yen—denominated risk—free bond trades at 993 yen (nominal
1,000). Solution: We compute the domestic interest rate from the domestic bond:
100642“ = 98.45
with solution 7“ = 6.2486%. Similarly, we compute the Japanese rate,
1000e‘025W = 993 with solution rf = 2.81%. We observe that, in order to apply the Black—Scholes formula,
we need the exchange rate expressed in dollars per yen (and not the other way around, as quoted in the problem). We now compute the discounted value of the exchange rate dollars per yen. 1 —0 02810 25
‘ ' = . 91
1088 0 00 94 and we can now apply the formula with 0.009194 for the underlying and 1/110 for strike
price. We get C (0) = 000041064, representing the value of the right (in dollars) to buy one
yen at an exchange rate of 110 yens per dollar, in three months. 40 " ,1”, .~ ‘ wwmmrﬁrwwvw:'rr'ﬁvci. ,K 1 r, "Ema”W“ ~ , ~ 2* ...
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 Spring '09
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