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7._kafli_C_Z - 7 OPTION PRICING 1 2 In a two-period CRR...

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Unformatted text preview: 7 OPTION PRICING 1' 2. In a two-period 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 risk-neutral pricing. For that, we need to compute the risk-neutral 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 risk-neutral probability of this event, because the other two states have zero payoffs. 1' 4. Consider a single-period 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 risk-free security that pays $1 after one period regardless of what happens? We assume, as usual, that the interest rate 7' per period is constant. Solution: In order to compute the price of the risk-free security we need the interest rate. It is important to observe that the call option only pays if the stock goes up all five times. In that case, the price of the stock is 100(1.152) = 132.25, and the payoff of the option 132.25 — 115 = 17.25. If the stock goes down a single time, the option will be out- of-the-money and its payoff will be zero. Therefore, using risk-neutral pricing, for unknown risk-neutral probability p“ and interest rate per period 1", we need to have 1 (1+ 7')2 5.424 = (129217.25 or, simplifying, 1 + r = 1.723p' 33 We have two unknowns. We have another equation for the same unknowns from the formula for the risk-neutral probability, *_(1+r)—d_(1+T)—0.95 19‘ u—d _ 1.15—0.95 The solution to the two equations is p* = 0.6, 7' = 0.07. The price of the risk—free security is 1 —— = 0.9346 1.07 T 6. Suppose that the stock price today is S(t) : 2.00, the interest rate is 7" = 0%, and the time to maturity is 3 months. Consider an option whose Black—Scholes price is given by the function V(t, s) = 8262(T_t) where the time is in annual terms. What is the option price today? What is the volatility of the stock equal to? Solution: The price is V(t, 8) : 8262(T—t) = 460.5 2 6.5949 . The function V(t, 3) has to satisfy the Black—Scholes equation. We have Vt : —28262(T_t), V35 2 262(T”t), which means that 1 Vt+§-252V33:0 from which we recognize that 02 = 2, thus 0 = fl. T 10. Verify that the Black~Scholes formula for the European put option can be obtained from the formula for the call option using put—call parity. (Hint: You can use the fact that 1 — N(:c) = N(—$) 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 two-period 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 | | 50e0-05/ 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 06502-109524 — 62.3594 = 8.8534 Similarly, in node [I we have, 6060-05/12 + 61(94081 + 5e0~05/24) = 1.8097 6060.05/12 +61(76-975+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 . «we-r9. «1w.» fame-WWW") , 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 continuous-time model. Solution: The price is given by 106‘3r/12E*[1{5(T)3K1 or S(T)2K2}l = 9-8758(P*lS(T) S K1l+ P*lS(T) Z KZl) 1' 30. Show that, if S is modeled by the Merton—Black-Scholes 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 yen-dollar exchange rate to be 15%. A three-month pure-discount yen—denominated risk—free bond trades at 993 yen (nominal 1,000). Solution: We compute the domestic interest rate from the domestic bond: 10064-2“ = 98.45 with solution 7“ = 6.2486%. Similarly, we compute the Japanese rate, 1000e‘0-25W = 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 0281-0 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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