FINANCIAL OPTION ANALYSIS HANDOUTS 1
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FAIR PRICING There is a market for an object called S. The prevailing price today is S 0 = 100. At this price the object S can be bought or sold by anyone for any number. A person buying the object S today must pay 100 now, and a person selling the object S will receive 100 now. It is common knowledge that the prevailing price tomorrow S 1 will either be S 1 (u) = 125 if the market for S is up or S 1 (d) = 80 if the market for S is down. A person who purchases a unit of S today will receive either 125 or 80 tomorrow (depending on the outcome), and the person who sells a unit of S today is obligated to buy back a unit of S at the new prevailing market price (125 or 80). There is also a market for an object called C. It may be bought or sold by anyone for any number. A person who buys the object C today must pay $C 0 now, and a person who sells the object C will receive $C 0 now. A person who buys C today has the option but not the obligation to buy 1 unit of the object S tomorrow at today s market price of 100. A person who sells C today has the obligation of supplying a unit of S tomorrow for the price of 100 if a holder of a C wishes to exercise his option. There is also a market for dollars, the so-called money market M. A person who buys m dollars today receives m dollars today and owes (1.10)m tomorrow; that is, the person is taking out a loan. A person who sells m dollars today must give m dollars today (to the person buying), and will receive (1.10)m the next period; that is, when a person sells dollars today, they are acting like a banker in that they are loaning money to the person buying. QUESTION: How much is a unit of C worth to you? How do you assess its value? 3
OPTION EXAMPLE 1 In each period the stock price may go up by a factor of U = 1.25 or down by a factor of D = 0.80. The stock price at time 0 is 100. Risk-free rate each period is a constant 10%. Price a one-period call option with strike price 100. Determine the self-financed replicating portfolio. 2/3 125 25 100 15.15 (0.555S, 40.40M) 1/3 80 0 NOTES 1. 0.5555 = ΔC/ΔS = [25 0] [125 80] 2. The value of the call option at time 0 = (1/1.11)[ 2/3*25 + 1/3*0] = 15.15 3. The money market amount = 15.15 0.5555[100] = -40.40 4. State Equations: h*s 1 (u) + m*r f = C 1 (u) 125h + 1.1m = 25 h*s 1 (d) + m*r f = C 1 (d) 80h + 1.1m = 0 4
OPTION EXAMPLE 2 In each period the stock price may go up by a factor of U = 1.25 or down by a factor of D = 0.80. The stock price at time 0 is 100. Risk-free rate each period is a constant 10%. Price a one-period put option with strike price 100. Determine the self-financed replicating portfolio. 2/3 125 0 100 6.06 (-0.444S, 50.50M) 1/3 80 20 NOTES 5. -0.444 = ΔC/ΔS = [0 20] [125 80] 6. The value of the put option at time 0 = (1/1.11)[ 2/3*0 + 1/3*20] = 6.06 7. The money market amount = 6.06-0.444*[100] = 50.50 8. State Equations: h*s 1 (u) + m*r f = P 1 (u) 125h + 1.1m = 0 h*s 1 (d) + m*r f = P 1 (d) 80h + 1.1m = 20 5
OPTION EXAMPLE 3 In each period the stock price may go up by a factor of U = 1.25 or down by a factor of D = 0.80. The stock price at time 0 is 100. Risk-free rate each period is a constant 10%. Price a 2-period call option with strike price 100. Determine the self-financed replicating portfolio. 2/3 156.25 56.25 125 34.09 2/3 (1.00S, 90.90M) 100 1/3 100 20.666 0 (0.7575S, 55.10M) 2/3 1/3 80 0 1/3 64 0 NOTES 9. a. 56.25 = max(156.25 100, 0). b. 0 = max(100 100, 0) c. 0 = max(64 100, 0) d. 34.09 = (1/1.1)[2/3*56.25 + 1/3*0] e. 1.00 = ΔC/ΔS = [56.25 0] [156.25 100] f. -90.90 = 34.09 1.00*125 g. 20.666 = (1/1.1)[2/3*34.09 + 1/3*0] h. 0.7575 = ΔC/ΔS = [34.09 0] [125 80] i. -55.10 = 20.66 0.7575*100 10. The value of the call option at time 0 = (1/1.21)[ 4/9*56.25 + 4/9*0 + 1/9*0 ] 6
OPTION EXAMPLE 4 In each period the stock price may go up by a factor of U = 1.25 or down by a factor of D = 0.80. The stock price at time 0 is 100. Risk-free rate each period is a constant 10%. Price a 2-period put option with strike price 100. Determine the self-financed replicating portfolio. 2/3 156.25 0 2/3 125 0 100 1/3 100 3.306 0 (-0.242S, 27.55M) 2/3 1/3 80 10.90 (-1.00S, 90.90M) 1/3 64 36 NOTES 11. a. 0 = max(100 156.25, 0). b. 0 = max(100 100, 0) c. 36 = max(100 64, 0) d. 10.90 = (1/1.1)[2/3*0 + 1/3*36] e. -1.00 = ΔC/ΔS = [0 36] [100 64] f. 90.90 = 10.90-1.00*80 g 3.306 = (1/1.1)[2/3*0 + 1/3*10.90] h 0.2424 = ΔC/ΔS = [0 10.90] [125 80] i 27.55 = 3.306-0.2424*100 12. The value of the put option at time 0 = (1/1.21)[ 4/9*0 + 4/9*0 + 1/9*36 ] 13. European Put-Call Parity: S T + P T C T = K at time T, which implies that S 0 + P 0 C 0 = K/(1.1) 2. Verification: 100 + 3.306 20.666 = 100/1.21 = 82.645. 7
OPTION EXAMPLE 5 In each period the stock price may go up by a factor of U = 1.25 or down by a factor of D = 0.80. The stock price at time 0 is 100. Risk-free rate each period is a constant 10%. Price a 2-period call option with strike price 80. Determine the self-financed replicating portfolio. 2/3 156.25 76.25 125 52.27 2/3 (1.00S, 72.72M) 100 1/3 100 35.36 20 (0.8922S, 53.87M) 2/3 1/3 80 12.12 (0.555S, 32.32M) 1/3 64 0 NOTES 14. a. 76.25 = max(156.25 80, 0). b. 20 = max(100 80, 0) c. 0 = max(64 80, 0) d. 52.27 = (1/1.1)[2/3*76.25 + 1/3*20] e. 1.00 = ΔC/ΔS = [76.25 20] [156.25 100] f. -72.72 = 52.27 1.00*125 g. 12.12 = (1/1.1)[2/3*20 + 1/3*0] h. 0.5556 = ΔC/ΔS = [20 0] [100 64] i. -32.32 = 12.12 0.5556*80 j. 35.36 = (1/1.1)[2/3*52.27 + 1/3*12.12] k. 0.8922 = ΔC/ΔS = [52.27 12.12] [125 80] l. -53.87 = 35.36 0.8922*100 15. The value of the call option at time 0 = (1/1.21)[ 4/9*76.25 + 4/9*20 + 1/9*0 ] 8
OPTION EXAMPLE 6 In each period the stock price may go up by a factor of U = 1.25 or down by a factor of D = 0.80. The stock price at time 0 is 100. Risk-free rate each period is a constant 10%. Price a 2-period put option with strike price 80. Determine the self-financed replicating portfolio. 2/3 156.25 0 2/3 125 0 100 1/3 100 1.469 0 (-0.107S, 12.24M) 2/3 1/3 80 4.84 (-0.444S, 40.40M) 1/3 64 16 NOTES 16. a. 0 = max(80 156.25, 0). b. 0 = max(80 100, 0) c. 16 = max(80 64, 0) d. 4.84 = (1/1.1)[2/3*0 + 1/3*16] e. -0.444 = ΔC/ΔS = [0 16] [100 64] f. 40.40 = 4.84-0.444*80 g. 1.469 = (1/1.1)[2/3*0 + 1/3*4.84] h. -0.107 = ΔC/ΔS = [0 4.84] [125 80] i. 12.24 = 1.469-0.1077*100 17. The value of the put option at time 0 = (1/1.21)[ 4/9*0 + 4/9*0 + 1/9*16 ] 18. European Put-Call Parity: S T + P T C T = K at time T, which implies that S 0 + P 0 C 0 = K/(1.1) 2. Verification: 100 + 1.469 35.36 = 80/1.21 = 66.116. 9
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CHOOSER OPTION Consider a non-dividend paying stock whose initial stock price is 62 and which has a log-volatility of σ = 0.20. The risk-free interest rate r f = 2.5% continuously compounded. Consider a 5-month option with a strike price of 60 in which after exactly 3 months the purchaser may declare this option to be a (European) call or put option. QUESTIONS: 1. Determine the value of U and D for the binomial lattice. The value for U = exp{σ(δt) 1/2 } = exp{0.20(1/12) 1/2 } = 1.05943. Note that D = 1/U = 0.94390. 2. Determine the values for the binomial lattice for 5 1-month periods. 0 1 2 3 4 5 62.00 65.68 69.59 73.72 78.11 82.75 58.52 62.00 65.68 69.59 73.72 55.24 58.52 62.00 65.68 Stock Price 52.14 55.24 58.52 49.21 52.14 46.45 3. Determine the appropriate risk-free rate. The interest rate per month R = exp(0.025*1/12) = 1.00209. 4. Determine the risk-neutral probability q of going UP. The value for q satisfies q(us 0 ) + (1-q)(DS 0 ) = RS 0, which implies that q = (R-D)/(U-D) = 0.5036. 11
5. Determine the values for the call option and put option along the lattice. 0 1 2 3 4 5 4.6686 7.0172 10.1423 13.9743 18.2315 22.7488 2.3054 3.876 6.2971 9.7137 13.7248 0.7216 1.4359 2.8571 5.6849 Call Option 0 0 0 0 0 0 0 1 2 3 4 5 2.0469 0.8344 0.1797 0 0 0 3.2858 1.5022 0.3627 0 0 5.1091 2.6646 0.7322 0 Put Option 7.6107 4.6364 1.4782 10.6604 7.8602 13.5462 6. Find the value of this Chooser Option. Compute the terminal value of the Chooser Option at t = 3 as the maximum of the call and put options at t = 3. From there we work backwards in the usual manner. 0 1 2 3 4 5 6.0483 7.3187 10.1423 13.9743 4.7847 4.4847 6.2971 5.1091 2.6646 Chooser Option 7.6107 12
CHOOSER OPTION Consider a non-dividend paying stock whose initial stock price is 62 and which has a log-volatility of σ = 0.20. The interest rate r = 2.5% continuously compounded. Consider a 5-month option with a strike price of 60 in which after exactly 3 months the purchaser may declare this option to be a (European) call or put option. QUESTIONS: 1. Determine the value of U and D for the binomial lattice. The value for U = exp{σ(δt) 1/2 } = exp{0.20(1/12) 1/2 } = 1.05943. Note that D = 1/U = 0.94390. 2. Determine the values for the binomial lattice for 5 1-month periods. 0 1 2 3 4 5 62.00 65.68 69.59 73.72 78.11 82.75 58.52 62.00 65.68 69.59 73.72 55.24 58.52 62.00 65.68 Stock Price 52.14 55.24 58.52 49.21 52.14 46.45 3. Determine the appropriate risk-free rate. The interest rate per month R = exp(0.025*1/12) = 1.00209. 4. Determine the risk-neutral probability p of going UP. The value for p satisfies p(us 0 ) + (1-p)(D S 0 ) = RS 0, which implies that p = (R-D)/(U-D) = 0.5036. 13
5. Determine the values for the call option and put option along the lattice. 0 1 2 3 4 5 4.6686 7.0172 10.1423 13.9743 18.2315 22.7488 2.3054 3.876 6.2971 9.7137 13.7248 0.7216 1.4359 2.8571 5.6849 Call Option 0 0 0 0 0 0 0 1 2 3 4 5 2.0469 0.8344 0.1797 0 0 0 2.0469 1.5022 0.3627 0 0 5.1091 2.6646 0.7322 0 Put Option 7.6107 4.6364 1.4782 10.6604 7.8602 13.5462 6. Find the value of this Chooser Option. Compute the terminal value of the Chooser Option at t = 3 as the maximum of the call and put options at t = 3. From there we work backwards in the usual manner. 0 1 2 3 4 5 6.0483 7.3187 10.1423 13.9743 4.7847 4.4847 6.2971 5.1091 2.6646 Chooser Option 7.6107 14
FINANCIAL OPTION ANALYSIS PROBLEMS A. We consider a single period binomial lattice with S 0 = 50, U = 1.20 (D = 1/1.20) and R f = 1.04. 1. a. What is the objective probability p of an upward movement in the stock price if the market s required expected return on the stock, R S, is 8%? [p = 0.6727] b. Suppose the objective probability p of an upward movement in the stock price is 0.70. What is the expected return, R S, on the stock? [R S = 9%] 2. A derivative security V has final payoffs given by V 1 = (max [S 1 50, 0]) 2, where S 1 is the final stock price. Assume the objective probability of an upward movement in the stock price p is 0.70. a. Determine the no-arbitrage value V 0 for V. [54.1958] b. Determine the market s expected return on V (using objective probabilities). [29.16%] c. Determine the replicating portfolio (hs, mm). [(5.4545S, -218.5315M)] d. Determine the portfolio weights, w S and w M, on the stock and the money market in the replicating portfolio. [w S = 5.032, w M = -4.032] e. Determine the portfolio s expected return using the portfolio weights and the expected returns on the stock and bond. [29.16%] f. Compare your answers to (b) and (e). 15
B. A non-dividend paying stock has an initial price = 100. Its price path is modeled as a binomial lattice with U = 1.3 and D = 1/1.3. Period length = 1 year. The risk-free rate is 6% per year. 1. Determine the value of a 2 year European put option with strike price K = 100. [7.43] 2. An American put permits the holder to exercise at any date t, and receive the intrinsic value max (0, K S t ). Hence, at each time t, the holder can take one of two possible actions: no exercise, as in a European option, or exercise. Determine the value of a 2 year American put option with strike price K = 100. [9.84] 16
C. Smith knows that the value of a six month European call option with strike price K = 24 on a nondividend paying stock is 6.80. The current value of the stock is 26. However, he wishes to price a European put option on the same stock with the same strike price and maturity. He knows that the risk-free rate is 2.50% per year, but he is stuck because he does not know the log-volatility σ upon which he would calculate the value of U. He comes to you for help. What say you? [4.50] 17
D. Do the following markets exhibit arbitrage? If so, demonstrate. If not, state why not. 1. 2. Securities S1 S2 price at time 0 5 10 up state at time 1 20 60 down state at time 1 10 30 Securities S1 S2 S3 price at time 0 60 115 1 up state at time 1 90 100 1.05 down state at time 1 40 150 1.05 3. Securities S1 S2 S3 S4 S5 S6 S7 S8 S9 Price at time 0 20 1 100 100 100 100 200 200 305 Up state at time 1 28 1.05 100 110 140 50 260 160 385 Down state at time 1 14 1.05 110 100 70 160 160 260 245 18
E. The evolution of the stock price over 2 periods is shown in the figure below. Let S 2 denote the (random) value of the stock price at t = 2. The appropriate risk-adjusted rate of return (cost of capital) is 20% per period. The risk-free rate is 4% per period. In this problem we shall consider pricing a European square root derivative security with strike price 140 that pays off ( S 2 140 ) 1/2 at time t =2. 220 150 115 120 4.733 90 80 1. Determine the objective probabilities along the lattice. 2. Determine the final period payoffs for the square root option. (Enter them in the figure.) 3. Determine the Decision-Tree value of the square root option by (1) first computing the expectation of the final period payoffs using the objective probabilities, and then (2) discounting using the risk-adjusted rate of return. 4. Fill out the above figure by placing the correct value of the square root option as the 2 nd entry, and recording the self-financed replicating portfolio as the 3 rd entry. Determine the risk-neutral probabilities along the lattice and mark them with an asterisk. 5. Suppose the current market value of the square root option is the Decision-Tree value. Conceptually explain how you would use your answer to (d) to obtain a risk-free profit from the incorrect pricing. Be specific about your dynamic trading strategy. (No further calculations are required.) 19
Sample Worksheet 1: 220 150 115 120 5.881 90 80 Sample Worksheet 2: 220 150 115 120 90 80 20
Asset Price Dynamics Example Data: σ = 0.30, risk-free rate = 0.08. S(0) = 36. 5-month horizon. U = exp(σ Δ t ). One-month intervals: 0 1 2 3 4 5 0 1 2 3 4 5 36.00 39.26 42.61 46.68 50.90 55.51 33.01 36.00 39.26 42.81 46.68 30.27 33.01 36.00 39.26 27.76 30.27 33.01 q = 0.51680 (0.51693) 25.46 27.76 23.35 4.74 2.69 1.10 0.17 0.00 0.00 4.36 2.51 1.04 0.17 0.00 0.00 6.99 4.43 2.10 0.36 0.00 6.40 4.11 1.98 0.36 0.00 9.73 6.99 4.00 0.74 8.94 6.46 3.74 0.74 12.24 9.73 6.99 11.71 9.46 6.99 American Put Option 14.54 12.24 European Put Option 14.28 12.24 16.65 16.65 Half-month intervals: 36.00 38.27 40.69 43.26 45.99 48.90 51.98 55.27 58.76 62.47 66.41 33.86 36.00 38.27 40.69 43.26 45.99 48.90 51.98 55.27 58.76 31.85 33.86 36.00 38.27 40.69 43.26 45.99 48.90 51.98 29.96 31.85 33.86 36.00 38.27 40.69 43.26 45.99 28.18 29.96 31.85 33.86 36.00 38.27 40.69 26.50 28.18 29.96 31.85 33.86 36.00 24.93 26.50 28.18 29.96 31.85 23.45 24.93 26.50 28.18 q = 0.51190 (0.51194) 22.06 23.45 24.93 20.75 22.06 19.51 4.73 3.26 2.03 1.08 0.45 0.11 0.00 0.00 0.00 0.00 0.00 6.30 4.57 3.04 1.76 0.80 0.22 0.00 0.00 0.00 0.00 8.15 6.21 4.40 2.78 1.42 0.46 0.00 0.00 0.00 10.04 8.15 6.14 4.23 2.43 0.95 0.00 0.00 11.82 10.04 8.15 6.14 4.00 1.95 0.00 13.50 11.82 10.04 8.15 6.14 4.00 15.07 13.50 11.82 10.04 8.15 16.55 15.07 13.50 11.82 American Put option 17.94 16.55 15.07 19.25 17.94 20.49 4.41 3.094 1.956 1.058 0.442 0.109 0.00 0.00 0.00 0.00 0.00 5.82 4.309 2.912 1.711 0.794 0.224 0.00 0.00 0.00 0.00 7.444 5.804 4.191 2.685 1.396 0.461 0.00 0.00 0.00 9.216 7.534 5.799 4.056 2.387 0.947 0.00 0.00 11.04 9.405 7.6668 5.833 3.914 1.946 0.00 12.83 11.29 9.644 7.884 6.006 4.00 14.54 13.10 11.56 9.909 8.15 16.15 14.8 13.36 11.82 European Put option 17.68 16.42 15.07 19.12 17.94 20.49 21
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Asset Price Dynamics Example Data: σ = 0.30, risk-free rate = 0.08. S(0) = 36. 5-month horizon. Period length = 0.5 months. U = exp(σ Δ t ). 0 1 2 3 4 5 6 7 8 9 10 Prob S 10 /S 0 ln [S 10 /S 0 ] (q = 0.5119) 36.00 38.27 40.69 43.26 45.99 48.90 51.98 55.27 58.76 62.47 66.41 0.001235524 1.844802889 0.612372436 33.86 36.00 38.27 40.69 43.26 45.99 48.90 51.98 55.27 58.76 0.011780805 1.632149650 0.489897949 31.85 33.86 36.00 38.27 40.69 43.26 45.99 48.90 51.98 0.050548837 1.444009273 0.367423461 29.96 31.85 33.86 36.00 38.27 40.69 43.26 45.99 0.128529724 1.277556123 0.244948974 28.18 29.96 31.85 33.86 36.00 38.27 40.69 0.214469383 1.130290283 0.122474487 26.50 28.18 29.96 31.85 33.86 36.00 0.245397552 1.000000000 0.000000000 24.93 26.50 28.18 29.96 31.85 0.194990143 0.884728476-0.122474487 23.45 24.93 26.50 28.18 0.106242502 0.782744477-0.244948974 22.06 23.45 24.93 0.037988595 0.692516329-0.367423461 20.75 22.06 0.008049416 0.612688916-0.489897949 19.51 0.000767517 0.542063331-0.612372436 0.014574500 E[S 10 /S 0 ] = 1.03384409 E(ln [S 10 /S 0 ]) = 0.014574500 NOTES: 1. Δt = 1/24. 2. U = exp(0.30 1 / 24 ) = exp(0.061237244) = 1.063115111. 3. υ = (r σ 2 /2) = (0.08 (0.30) 2 /2) = 0.035. 4. q = 0.5(1 + 0.035/0.30 1 / 24 ) = 0.511907242. q = (1.003383780 0.940600062)/(1.06315111 0.940600062) = 0.512303395. 5. (1 + 0.08/24) 10 = 1.033837804. 6. exp(0.08(5/12)) = 1.033895114. 7. E (ln [S 1 /S 0 ]) = 0.5119(0.061237244) + 0.4881(-0.061237244) = 0.001445199. 8. E (ln [S 10 /S 0 ]) = 10(0.001445199) = 0.01445199. 9. υ(5/12) = 0.014583333. 10. P(S 10 38) = 0.593435725. 11. P(S(5/12) 38) = P{ [ln S(5/12)/S(0) - (0.035)(5/12)]/[0.30 5 / 12 ] [ln 38/36-0.01458333]/[0.193649167] } = Φ(0.203893921) = 0.580782. 23
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Table 1: A stochastic volatility, random interest rate model t = 0 t = 1 t = 2 S 0 = 4, r 0 = 25% S 1 (U) = 8, r 1 (U) = 25% S 2 (UU) = 12 S 1 (D) = 2, r 1 (D) = 50% S 2 (UD) = S 2 (DU) = 8 S 2 (DD) = 2 Option Analysis with Stochastic Interest Rates In this problem we consider a two-period, stochastic volatility, random interest rate model. The stock prices and interest rates are provided in Table 1. Consider the European option whose final payoffs are V 2 = max{s 2 7, 0}. Determine the value of this option at times 0 and 1. 25
Table 2: Solution to the stochastic volatility, random interest rate model t = 0 t = 1 t = 2 1.00 4 = 1 1.25 [0.5(2.4) + 0.5(0. 1)] 2.4 = 1 1.25 [0.5(5) + 0.5(1)] 5 0. 1 = 1 1.5 [(1/6)(1) + (5/6)(0)] 1 0 The risk-neutral q changes along the tree since the risk-free rate is stochastic. Otherwise, all the calculations are the same, since at each node the replicating portfolio and corresponding risk-neutral discounted expectation ideas still apply. The solution is provided in Table 2. 26
Asian Option Consider a non-dividend paying stock S whose price process follows a binomial lattice with U = 2 and D = 0.5. R = 1.25 and S 0 = 4. Define Y t := t S k, t = 0, 1, 2, 3 k=0 to be the sum of the stock prices between times zero and t. 1. Consider a (European) Asian call option that expires at time three and has a strike price K = 4; that is, its payoff at time three is { Y3 } max 4 4, 0. This is like a European call option, except the payoff of the option is based on the average stock price rather than the final stock price. Let V t (s, y) denote the price of this option at time n if S t = s and Y t = y. In particular, { y } V 3 (s, y) = max 4 4, 0. (a) Develop an algorithm for computing V t recursively. In particular, write a formula for V t in terms of V t+1. (b) Apply the algorithm developed in (a) to compute V 0 (4, 4), the price of the Asian option at time zero. (c) Provide a formula for δ t (s, y), the number of shares of stock that should be held by the replicating portfolio at time t if S t = s and Y t = y. 2. What is the value of a (European) Asian put option that expires at time three and has a strike price K = 4; that is, its payoff at time three is { max 4 Y } 3 4, 0. 27
1. Risk-neutral q = 0.5. ( ) (a) V t (s, y) = (1/1.25) 0.5V t+1 (us, y + us) + 0.5V t+1 (ds, y + ds). (b) V 3 (32, 60) = 11; V 3 (8, 36) = 5; V 3 (8, 24) = 2; V 3 (2, 18) = 0.5; V 3 (8, 18) = 0.5; V 3 (2, 12) = V 3 (2, 9) = V 3 (0.5, 7.5) = 0. V 2 (16, 28) = (1/1.25)[0.5(11) + 0.5(5)] = 6.40. V 2 (4, 16) = (1/1.25)[0.5(2) + 0.5(0.5)] = 1.0. V 2 (4, 10) = (1/1.25)[0.5(0.5) + 0.5(0)] = 0.20. V 2 (1, 7) = (1/1.25)[0.5(0) + 0.5(0)] = 0. V 1 (8, 12) = (1/1.25)[0.5(6.4) + 0.5(1.0)] = 2.96. V 1 (2, 6) = (1/1.25)[0.5(0.2) + 0.5(0)] = 0.08. (c) V 0 (4, 4) = (1/1.25)[0.5(2.96) + 0.5(0.08)] = 1.216. Remark. The value of this Asian option, 1.216, equals the discounted expectation of the final payoffs using the risk-free rate and the risk-neutral probability, i.e., 1 11 + 5 + 2 + 0.5 + 0.5 + 0 + 0 + 0 (1.25) 3. 8 Simulation is an especially useful computational approach for valuing pathdependent, European-style derivative securities, since their value can be obtained as a (discounted) sample average of the value along a sample path. δ t (s, y) = V t+1(us, y + us) V t+1 (ds, y + ds). (u d)s 28
No Arbitrage Bounds Consider a family of call options on a non-dividend paying stock, each option being identical except for its strike price. The value of the call with strike price K is denoted by C(K). Prove the following two general relations using arbitrage arguments: 1. If K 2 > K 1, then K 2 K 1 C(K 1 ) C(K 2 ). 2. If K 3 > K 2 > K 1, then C(K 2 ) ( K3 K ) ( 2 K2 K ) 1 C(K 1 ) + C(K 3 ). K 3 K 1 K 3 K 1 Hint: For both parts find a portfolio that is guaranteed to have no negative but sometimes positive final payoffs. If there is to be no arbitrage, such a portfolio must cost something to acquire today. In each part plot the final payoffs for each portfolio, otherwise known as the payoff diagram. 29
Solution: For both parts below one finds a portfolio that is guaranteed to have no negative but sometimes positive final payoffs. If there is to be no arbitrage, such a portfolio must cost something to acquire today. In each case below it will be instructive to plot the final payoffs for each portfolio, otherwise known as the payoff diagram. 1. Verify that the portfolio C(K 1 ) + C(K 2 ) + (K 2 K 1 ) has no negative but sometimes positive final payoffs. Hence, its cost today must be nonnegative, which establishes the result. 2. Consider the portfolio mc(k 1 ) C(K 2 ) + nc(k 3 ) with m, n > 0 and m < 1. This portfolio s final payoffs are zero if S T K 1 and will be positive on the interval K 1 S T K 2. Since m < 1 the payoffs decline on the interval K 2 S T K 3. If we set m so that m(k 3 K 1 ) = (K 3 K 2 ), then the payoffs will remain nonnegative in the interval K 2 S T K 3. In particular, the payoff when S T = K 3 will be zero. If we further set n so that m + n = 1, the final payoffs will be zero on the interval S T K 3. We conclude that the portfolio ( K3 K ) ( 2 K2 K ) 1 C(K 1 ) C(K 2 ) + C(K 3 ) K 3 K 1 K 3 K 1 has no negative but sometimes positive final payoffs. nonnegative, which establishes the result. Hence, its cost today must be 30
FINANCIAL OPTION ANALYSIS SAMPLE PROBLEMS I. Single-period 1. The price of asset S at time t = 0 is S 0 = 20. The asset s value at time t = 1 is S 1 (u) = 28 in the up state is S 1 (d) = 14 in the down state. The risk-free rate is 5%. The objective probability of the up state is 0.60. A derivative security V has final payoffs at time t = 1 of V 1 (u) = 126 in the up state and V 1 (d) = 84 in the down state. a. Determine the expected return, R S, on the asset S. b. Determine the no-arbitrage value for V. c. Determine the expected return on V under the objective probabilities. d. Determine the replicating portfolio for V. e. Determine the portfolio weights of S and M in the replicating portfolio for V. f. Show how the answer to (c) can be obtained using the expected returns under the objective probabilities of the underlying assets in the replicating portfolio. 2. The price of asset S at time t = 0 is S 0 = 40. The asset s value at time t = 1 is S 1 (u) = 50 in the up state is S 1 (d) = 30 in the down state. The risk-free rate is 2.5%. The objective probability of the up state is 0.70. A derivative security V has final payoffs at time t = 1 of V 1 (u) = 100 in the up state and V 1 (d) = 50 in the down state. a. Determine the expected return, R S, on the asset S. b. Determine the no-arbitrage value for V. c. Determine the expected return on V under the objective probabilities. d. Determine the replicating portfolio for V. e. Determine the portfolio weights of S and M in the replicating portfolio for V. f. Show how the answer to (c) can be obtained using the expected returns under the objective probabilities of the underlying assets in the replicating portfolio. II. Multi-period 3. A non-dividend paying stock has an initial price S 0 = 100. Its price path is modeled as a binomial lattice with U = 1.25 and D = 0.80. Period length is 1 year. The risk-free rate is 10% per year. a. Determine the value of a 2 year American call option with strike price K = 90. b. Determine the value of a 2 year European call option with strike price K = 90. c. Use Put-Call Parity to value a 2 year European put option with strike price K = 90. 4. A non-dividend paying stock has an initial price S 0 = 100. Its price path is modeled as a binomial lattice with U = 1.25 and D = 0.80. Period length is 1 year. The risk-free rate is 10% per year. a. Determine the value of a 2 year American put option with strike price K = 90. b. Determine the value of a 2 year European put option with strike price K = 90. c. Use Put-Call Parity to value a 2 year European call option with strike price K = 90. 5. The risk-free rate is 2.5% and the period length equals 3 months. Determine the value of the parameter U in a binomial lattice if log-volatility σ = 0.60, and determine the value of the risk-neutral probability q. 6. The risk-free rate is 4% and the period length equals 6 months. Determine the value of the parameter U in a binomial lattice if log-volatility σ = 0.80, and determine the value of the risk-neutral probability q. 31
7. The evolution of the stock price over 2 periods is shown in Figure 1 below. Let S 2 denote the (random) value of the stock price at t = 2. The appropriate risk-adjusted rate of return (cost of capital) is 20% per period. The risk-free rate is 4% per period. In this problem we wish to price a European derivative security that pays off ( S 2 64 + S 2 114 + S 2 181.50 ) at time t =2. a. For each box in the figure below place the correct value of the derivative security as the 2 nd entry and record the self-financed replicating portfolio as the 3 rd entry. 181.50 130 100 114 80 64 b. Determine the value at time t = 0 of this derivative security according to DTA. c. Suppose the current market value of the derivative security is 150. Exactly explain how you could guarantee a risk-free profit from the incorrect pricing. Be specific with numbers. 32
8. The evolution of the stock price over 2 periods is shown in Figure 1 below. Let S 2 denote the (random) value of the stock price at t = 2. The appropriate risk-adjusted rate of return (cost of capital) is 12% per period. The risk-free rate is 5% per period. In this problem we wish to price a European derivative security that pays off S 2 119 1/2 at time t =2. a. For each box in the figure below place the correct value of the derivative security as the 2 nd entry and record the self-financed replicating portfolio as the 3 rd entry. 189 150 100 119 70 49 b. Determine the value at time t = 0 of this derivative security according to DTA. c. Suppose the current market value of the derivative security is priced according to DTA. Exactly explain how you could guarantee a risk-free profit from the incorrect pricing. Be specific with numbers. 33
III. Path Dependent 9. A non-dividend paying stock has an initial price S 0 = 100. Its price path is modeled as a binomial lattice with U = 1.25 and D = 0.80. Period length is 1 year. The risk-free rate is 10% for the first year, and 2.5% for the second year. Determine the value of a 2-year European-style path-dependent derivative security whose payoff at time t = 2 is max {S 0, S 1, S 2 } S 2, namely, the difference between the maximum stock price observed over the 2-year period and the stock price at time t = 2. 10. The price process of a non-dividend paying stock over the next 2 years is shown in the following table: t = 0 t = 1 t = 2 100 130 171 80 102 57 The risk-free rate is 5% per year. Define Y 2 = S 0 + S 1 + S 2 denote the sum of the stock prices between times zero and 2. A European Asian put option that expires at time t = 2 and has strike price K = 104 has its payoff at time t = 2 equal to max{104 - Y 2 /3, 0}. (This is like a European put option, except that the payoff of this option is based on the average stock price rather than the final stock price.) Determine the no-arbitrage value of this path-dependent option. 11. Consider a binomial lattice with S 0 = 4, U = 2 and D = 0.50. The risk-free rate if 25%. A 3-year, lookback option is a European path-dependent derivative security that pays off at time three. V 3 = max {0 t 3} S t S 3 a. Determine the no-arbitrage value of this lookback option. b. Determine the replicating portfolio at time t = 0. 12. A non-dividend paying stock has an initial price S 0 = 36. Its price path is modeled as a binomial lattice with U = 1.5 and D = 2/3. Period length is 1 year. The risk-free rate is 2.5% for the first year, and 10% for the second year. Determine the value (to the nearest penny) of a 2-year American-style path-dependent derivative security whose payoff at any time t, if exercised, is the stock price at time t less the minimum stock price observed up to this time t. 13. A non-dividend paying stock has an initial price S 0 = 36. Its price path is modeled as a binomial lattice with U = 1.5 and D = 2/3. Period length is 1 year. The risk-free rate is 2.5% for the first year, and 10% for the second year. Determine the value (to the nearest penny) of a 2-year American-style path-dependent derivative security whose payoff at any time t, if exercised, is the maximum stock price observed up to this time t less the stock price at time t. 34
IV. Arbitrage 14. Does the following market exhibit arbitrage? If so, provide a concrete example of arbitrage. If not, state why not. Securities S1 S2 S3 S4 S5 S6 S7 S8 S9 Price at time 0 20 1 100 100 100 100 200 200 305 Up state at time 1 28 1.05 100 110 140 50 260 160 385 Down state at time 1 14 1.05 110 100 70 160 160 260 245 15. Consider dividend-price data for a complete, no-arbitrage market with the following three securities: Security 1 Security 2 Security 3 Payoff Vector V Price at t = 0 100 80 1.0? Payoff in state 1 at t = 1 200 0 1.1 420 Payoff in state 2 at t = 1 240 0 1.1 460 Payoff in state 3 at t = 1 0 176 1.1 44 a. Use the replicating portfolio approach to determine the correct value of the payoff vector V at time 0. b. Use the risk-neutral approach to determine the correct value of the payoff vector V at time 0. 16. a. Does the following market exhibit arbitrage? If so, provide a concrete example of arbitrage. If not, state why not. Securities S1 S2 S3 S4 S5 Price at time 0 60 8 60 40 22 State 1 125 0 110 0 22 State 2 80 0 110 0 11 State 3 0 28 0 105 21 State 4 0 14 0 105 42 b. Determine the risk-free rate for this market. 17. Consider the following market data: Securities S 1 S 2 S 3 Price at time 0 20 10 80 State 1 at time 1 26 10 68 State 2 at time 1 16 11 Z a. Determine the state-prices y 1 and y 2. b. Determine the value of Z in the table so that this market does not exhibit arbitrage. c. Determine the risk-free rate for this market. 35
V. Asset Price Dynamics 18. Consider a non-dividend paying stock whose price follows Geometric Brownian motion, ds t = μs t dt + σ S t db t with (annual) parameters μ = 0.20, σ = 0.40, and S 0 = 100. a. Determine the expected value for the stock price after 3 months (0.25 yr). b. Determine the median value s* for the stock price after 3 months. Recall that s* is defined by the equation P[S(0.25) s*] = 0.50. c. Determine the standard deviation of the stock price after 3 months. d. Consider a 1-year binomial lattice representation of S with 1-month periods. What is the probability that a simulated stock price path will have its terminal stock price equal to 100? 19. Consider a non-dividend paying stock whose price follows Geometric Brownian motion, ds t = μs t dt + σ S t db t with (annual) parameters μ = 0.12, σ = 0.60, and S 0 = 100. Risk-free rate is 2.5%. a. Determine the expected value for the stock price after 6 months (0.50 yr). b. Determine the median value s* for the stock price after 6 months. Recall that s* is defined by the equation P[S(0.50) s*] = 0.50. c. Determine the standard deviation of the stock price after 6 months. d. Consider a 1-year binomial lattice representation of S with 2-month periods. What is the probability that a simulated stock price path will have its terminal stock price equal to 100? e. Set up the Black-Scholes equation to value a six-month call option on this stock with a strike price of 105. Be explicit but do not calculate. f. What is the probability that the stock price at the end of 6 months will be no higher than 110? 36
FINANCIAL OPTION ANALYSIS SAMPLE PROBLEM SOLUTIONS 1. a. R S = [0.6(28) + 0.4(14)]/20-1 = 0.12. b. Since [0.5(28) + 0.5(14)]/1.05 = 20, q = 0.5. V 0 = [0.5(126) + 0.5(84)]/1.05 = 100. c. r V = [0.6(126) + 0.4(84)]/100-1 = 0.092 or 9.2%. e. (126-84)/(28-14) = 3. So h = 3. Thus, replicating portfolio is (3S, 40M). e. 60/100 is invested in S, so w S = 0.6 and w M = 0.4. f. 0.6(12) + 0.4(5) = 9.2. 2. a. R S = [0.7(50) + 0.3(30)]/40-1 = 0.10. b. Since [0.55(50) + 0.45(30)]/1.025 = 40, q = 0.55. V 0 = [0.55(100) + 0.45(50)]/1.025 = 75.61. c. R V = [0.7(100) + 0.3(50)]/75.61-1 = 0.1242 or 12.42%. e. (100-50)/(50-30) = 2.5. So h = 2.5. Thus, replicating portfolio is (2.5S, -24.39M). e. 100/75.61 is invested in S, so w S = 1.3226 and w M = -03226. f. 1.3226(10) - 0.3226(2.5) = 12.42. 3. a. t = 0 t = 1 t = 2 100 125 [43.182] 156.25 [66.25] 28.0073 = 80 [6.061] 100 [10] [2/3(43.182)+1/3(6.061)]/1.1 since exercise value = 0 64 [0] b. [(4/9)(66.25)+(4/9)(10)]/(1.1) 2 = 28.0073. c. 90/(1.1) 2 = S + P C = 100 + P 28.0073. P = 2.3875. 4. a. t = 0 t = 1 t = 2 100 125 [0] 156.25 [0] 3.03 = 1/3(10)/1.1 80* [10] 100 [0] since 10 > 1/3(26)/1.1 64 [26] b. (1/9)(26)/(1.1) 2 = 2.3875. c. 90/(1.1) 2 = S + P C = 100 + 2.3875 C. C = 28.0073. 5. U = exp[0.60(0.25) 1/2 ] = 1.3499. q = [exp(0.025*0.25) 0.7408]/[1.3499 0.7408] = 0.4358. 6. U = exp[0.80(0.50) 1/2 ] = 1.7607. q = [exp(0.040*0.50) 0.5680]/[1.7607 0.5680] = 0.3791. 37
7. a. Note: Objective probabilities are in parentheses. 181.50 185 0.314 (0.622) 130 133.36 0.48 (0.80) (1S, 3.36M) 100 114 132.85 0.686 (0.377) 117.5 (-0.185S, 151.3M) 0.384 (0.64) 80 0.52 (0.20) 142.60 (-1S, 222.60M) 0.616 (0.36) 64 167.5 b. [185(0.80*0.622) + 117.5(0.80*0.377 + 0.20*0.64) + 167.5(0.20*0.36)] / (1.2) 2 = 107.43. c. Derivative security is overpriced, so you would want to sell it. Collect 150 and use 132.85 of it to purchase the replicating portfolio of ( 0.1848S, 151.33M). Invest the 17.15 in the bank or buy lunch. If the price goes up next period to 130, rebalance the portfolio to (1S, 3.36M), which you can afford to do since it will cost 133.36 and this is precisely what the portfolio ( 0.1848S, 151.33M) equals when the price = 130. If the price goes down next period to 80, rebalance the portfolio to ( 1S, 222.60M), which you can afford to do by the same reasoning as before. Finally, when period 2 comes around, your updated portfolio will exactly match the final payoffs of the derivative security (185, 117.5 or 167.5) regardless of the final state and so you will be able to meet your obligations. (If you were forced to buy back the derivative security at time 1, you would have the exact money to do so.) 38
8. a. Note: Objective probabilities are in parentheses. 189 8.3667 0.55 (0.7) 150 4.3825 0.4375 (0.525) (0.1195S, -13.5459M) 100 119 4.6007 0.45 (0.3) 0 (-0.00996S, 5.5967M) 0.35 (0.42) 70 0.5625 (0.475) 5.1793 (-0.1195S, 13.5459M) 0.65 (0.58) 49 8.3667 b. 8.367[(0.525)(0.7) + (0.475)(0.58)] / (1.12) 2 = 4.289. c. Derivative security is underpriced, so you would want to buy it. Sell the RP for 4.6007, i.e., acquire the portfolio (0.00996S, -5.5967M), and buy the derivative security for 4.289. Collect 0.3117 and invest it in the bank. If the price goes up next period to 150, you owe 4.3825. Purchase the new RP of (-0.1195S, 13.5459M), collect 4.3825 and pay off the obligation. If the price goes down next period to 80, you owe 5.1793. Purchase the new RP of (0.1195S, -13.5459M), collect 5.1793 and pay off the obligation. Finally, when period 2 comes around, your updated portfolio s obligation will exactly match the final payoffs of the derivative security (8.3667, 0 or 8.3667) regardless of the final state, and so you will be able to meet your obligations. 9. There are four possible paths. The payoffs in the states uu, ud, du, dd are, respectively, 0, 25, 0, 36. Thus, the time 0 no-arbitrage value is [(2/3)(0.5)(25)]/(1.1)(1.025) + [(1/3)(0.5)(36)]/(1.1)(1.025) = 12.71. 10. There are four possible paths. The average values along the uu, ud, du, and dd paths are, respectively, 133.67, 110.67, 94, and 79. The values of the Asian put option for these paths are, respectively, 0, 0, 10, and 25. The q probability of path du is (0.5)(0.6) = 0.3, and the q probability of path dd is (0.5)(0.4) = 0.2. Therefore, the time 0 no-arbitrage value is [(0.3)(10) + (0.2)(25)]/(1.05) 2 = 7.2562. 11. a. The final payoffs of this lookback option in states uuu, uud, udu, udd, duu, dud, ddu, ddd are, respectively, 32-32=0, 16-8=8, 8-8=0, 8-2=6, 8-8=0, 4-2=2, 4-2=2, 4-0.5=3.5. Since the q probability is 0.5, the time 0 no-arbitrage value is [(0+8+0+6+0+2+2+3.5)/8]/1.25 3 = 1.376. b. To determine the RP at time t = 0, we have to find the time 1 no-arbitrage values of this lookback option. The time 1 value when S = 8 is [(0+8+0+6)/4]/1.25 2 = 2.24, and the time 1 value when S = 2 is [(0+2+2+3.5)/4]/1.25 2 = 1.20. (Notice how [0.5(2.24)+0.5(1.20)]/1.25 = 1.376, as it should.) So, the RP is (0.1733S, 0.6827M). 39
12. Stock price paths option value over time 36 54 uu: 81 12.0787 21.2727 45 ud: 36 0 24 du: 36 5.6727 12 dd: 16 0 Risk-neutral probability q 1 = 0.43 and q 2 = 52. 21.2727 = max{ [0.52(45) + 0.48(0)]/1.1, 54-36 }. 5.6727 = max{ [0.52(12) + 0.48(0)]/1.1, 24-24 }. 12.0787 = [0.43(21.2727) + 0.57(5.6727)]/1.025. 13. Stock price paths option value over time 36 54 uu: 81 9.9682 7.8545 0 ud: 36 18 24 du: 36 12* 0 dd: 16 20 Risk-neutral probability q 1 = 0.43 and q 2 = 52. 7.8545 = max{ [0.52(0) + 0.48(18)]/1.1, 54-54 }. 12 = max{ [0.52(0) + 0.48(18)]/1.1, 36-24 }. (Exercise.) 9.9682 = [0.43(7.8545) + 0.57(12)]/1.025. 14. First four securities show that q = 0.5. To be consistent, the value S9 = [0.5(385) + 0.5(245)]/1.05 = 300, which is less than 305. The portfolio (10S1, 100S2) replicates S9 and costs 300. So sell S9 for 305 and buy this replicating portfolio for 300, and pocket the difference of 5. 15. a. Since S2 does not payoff in either state 1 or 2 only S1 and S3 can be used to replicate the payoffs of the Vector V in these two states. (S3 is of course our old friend M.) We re back to the (hs1, M) : here, h = (420-460)/(200-240) = 1 and M = S3 = 200. Now use state 3 to pin down the number of units of S2 to hold: 176(S2) + 1.1(200) = 44 S2 = -1. Thus, the replicating portfolio is (1S1, 1S2, 200S3) for a cost today of 220. b. Let q = (q 1, q 2, q 3 ) denote the risk-neutral probability vector. Obviously R = 1.1. Discounted expectation using q and R applied to S2 implies that (1/1.1)176q 3 = 80 q 3 = 0.5. Thus, q 1 + q 2 = 0.5. Discounted expectation using q and R applied to S1 implies that (1/1.1)[200q 1 + 240q 2 ] = 100 q 1 =q 2 = 0.25. Risk-neutral valuation says that (1/R)E q [V] = (1/R) q T V = p for any vector V. Thus, the value of V is (1/1.1)[0.25(420) + 0.25(460) + 0.5(44)] = 220, which coincides with the answer in part (a), as it should. 40
16. a. We can use the first four assets to determine the state-prices. The equations are: 125y 1 + 80y 2 = 60. 110y 1 + 110y 2 = 60. These equations imply that y 1 = 4/11, y 2 = 2/11. 28y 3 + 14y 4 = 8. 105y 3 + 105y 4 = 40. These equations imply that y 3 = 4/21, y 4 = 4/21. Now we can use these state-prices to determine the no-arbitrage value of asset 5 as: (4/11)22 + (2/11)11 + (4/21)21 + (4/21)42 = 22. Since this IS the price of asset 5, this market does NOT exhibit arbitrage. b. Recall that the reciprocal of the sum of the state-prices equals = 1 + r f. Thus, 1+r f = 1/(4/11 + 2/11 + 4/21 + 4/21) = 231/214 = 1.0794. r f = 7.94%. Alternatively, one can easily combine assets 3 and 4 to obtain a constant payoff vector. For example, a purchase of 1 unit of asset 3 and 110/105 units of asset 4 yields a constant payoff of 110. The cost of this portfolio is 60(1) + 40(110/105) = 101.905. Thus, the total return on the risk-free security is 110/101.905 = 1.0794, same as above, as it should. 17. a. The equations are: 26y 1 + 16y 2 = 20. 10y 1 + 11y 2 = 10. These equations imply that y 1 = 10/21, y 2 = 10/21. b. 80 = 68y 1 + Zy 2 = 68(10/21) + Z(10/21). So, Z = 100. c. R = 1/(y 1 + y 2 ) = 1.05. So risk-free rate is 5% in this market. 41
18. a. E[S(0.25)] = S(0)exp(0.20*0.25) = 105.127. b. υ = 0.20 (0.40) 2 /2 = 0.12. ln[s(0.25)/s(0)] ~ Normal with mean = 0.12(0.25) and variance = (0.40) 2 (0.25). The mean and median values are the same for a normally distributed random variable, and so the median value of ln[s(0.25)/s(0)] is 0.03, which implies the median value for S(0.25) is 100*exp(0.03) = 103.045. c. ln [S(t)/S(0)] := X(t) ~ N(υt, σ 2 t), where υ = µ - 0.5*σ 2 = 0.12. S(t) = S(0)exp[X(t)]. E[S(t)] = S(0)E[exp(X(t))] = S(0)exp[υt + 0.5σ 2 t] = 100exp[0.12(0.25) + 0.5(0.16)(0.25)] (when t = 0.25) = 100exp(0.05) = 105.127. (Same answer as part a.) E[S(t) 2 ] = S(0) 2 E[exp(2X(t))] = S(0) 2 exp[2υt + 2σ 2 t] = 10000exp[0.06+0.08] = 11502.74 (when t = 0.25). Var[S(0.25)] = 11502.74 (105.127) 2 = 451.05. Stdev[S(0.25)] = 21.24. d. p = 0.5(1 + (0.12/0.40)(1/12) 1/2 ) = 0.5433. For a simulated stock price at the end of the 12 th month to have a value of 100, there must have been exactly 6 up and 6 down transitions. The probability of this occurrence is therefore ( 12 C 6 )*(0.5433) 6 (0.4567) 6 = 0.2156. 19. a. E[S(0.50)] = S(0)exp(0.12*0.50) = 106.184. b. υ = 0.12 (0.60) 2 /2 = -0.06. ln[s(0.50)/s(0)] ~ Normal with mean = (-0.06)(0.50) and variance = (0.60) 2 (0.50). The mean and median values are the same for a normally distributed random variable, and so the median value of ln[s(0.50)/s(0)] is -0.06, which implies the median value for S(0.50) is 100*exp(-0.06) = 94.176. c. ln [S(t)/S(0)] := X(t) ~ N(υt, σ 2 t), where υ = µ - 0.5*σ 2 = -0.06. S(t) = S(0)exp[X(t)]. E[S(t)] = S(0)E[exp(X(t))] = S(0)exp[υt + 0.5σ 2 t] = 100exp[-0.06(0.50) + 0.5(0.36)(0.50)] (when t = 0.50) = 100exp(0.06) = 106.184. (Same answer as part a.) E[S(t) 2 ] = S(0) 2 E[exp(2X(t))] = S(0) 2 exp[2υt + 2σ 2 t] = 10000exp[-0.06+0.36] = 13498.59 (when t = 0.50). Var[S(0.50)] = 13498.59 (106.184) 2 = 2223.55. Stdev[S(0.50)] = 47.15. d. p = 0.5(1 + (-0.06/0.60)(2/12) 1/2 ) = 0.4796. For a simulated stock price at the end of the 12 th month to have a value of 100, there must have been exactly 3 up and 3 down transitions. (Period length = 2 months.) The probability of this occurrence is therefore ( 6 C 3 )*(0.4796) 3 (0.5204) 3 = 0.3109. e. d 1 = [ln(100/105) + (0.025 + 0.6 2 /2)(0.50)]/[0.60(0.50) 1/2 ] = 0.1266. d 2 = 0.1266 0.60(0.50) 1/2 = -0.2977. Kexp(-rT) = 105exp(-0.025(0.50)) = 103.696. C = 100Φ(0.1266) 103.696Φ(-0.2977). f. P{S(0.5) 110} = P{S(0.5)/100 110/100} = P{ln[S(0.5)/100] ln[1.1]} = P{ [ ln[s(0.5)/100] (-0.06)(0.5) ]/(0.6(0.5) ½ [ ln[1.1] (-0.06)(0.5) ]/(0.6(0.5) ½ } = P{Z 0.2954} = Φ(0.2954) = 0.616 or 61.6%. 42