ACTS 4302 Instructor: Natalia A. Humphreys SOLUTION TO HOMEWORK 9 Lesson 15: Monte Carlo Valuation. Sufficient work must be shown to get credit for a correct answer. Partial credit may be given for incorrect answers which have some positive work. Problem 1 For a stock, you are given: (i) The stock’s prices follow a lognormal model. (ii) Current stock price is $100. (iii) The stock’s continuously compounded rate of return is 0.15. (iv) The stock’s annual volatility is 0.3. (v) The stock pays no dividends. (vi) The continuously compounded risk-free interest rate is 0.05. The stock’s price at the end of 2 years is simulated using the following uniform [0, 1) random numbers: 0.3821 0.6844 0.4562 0.0749 0.9265 Using these numbers and the inversion method to perform five trials, one number per trial, calculate the expected stock price at the end of 2 years. (A) 110.17 (B) 116.19 (C) 134.56 (D) 153.97 (E) 173.15 Key: C Solution. The parameters of the lognormal model are √ √ m = µt = (α − 0.5σ 2 )t = (0.15 − 0.5 · 0.32 )2 = 0.105 · 2 = 0.21; v = σ t = 0.3 2 = 0.4243 The following table contains generation of lognormal random numbers from given uniform [0, 1) random numbers ui ’s: ui zi = N −1 (ui ) 0.3821 -0.30 0.0827 1.0863 0.6844 0.48 0.4137 1.5123 0.4562 -0.07 0.1633 1.1774 0.0749 -1.44 -0.4010 0.6696 0.9265 1.45 0.8253 2.2825 ni = m + vzi = xi = eni = 0.21 + 0.4243zi Thus, the average projected price is: 5 S(2) = S(0) · 1X 1 xi = 100 · (1.0863 + 1.5123 + 1.1774 + 0.6696 + 2.2825) = 5 5 i=1 = 100 · 1.3456 = 134.5629 ≈ 134.56 ⇒ C Problem 2 For a stock, you are given: (i) The stock’s prices follow a lognormal model. Copyright ©Natalia A. Humphreys, 2014 ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 9. (ii) The stock’s current price is 56.12. (iii) The stock’s continuously compounded rate of return is 0.15. (iv) The stock’s annual volatility is 0.25. (v) The stock pays no dividends. (vi) The continuously compounded risk-free interest rate is 0.04. A 3-month European call option on the stock has strike price 61. Simulate the stock’s price using the following numbers drawn from a standard normal distribution: 0.48 0.93 − 0.57 − 1.39 Using these numbers, one number per simulation, calculate the expected option payoff. Solution. Since we want the undiscounted payoff, we don’t use risk-neutral probabilities. We use α = 0.15, not r = 0.05. The parameters of the lognormal model are √ √ m = µt = (α − 0.5σ 2 )t = (0.15 − 0.5 · 0.252 )0.25 = 0.0297; v = σ t = 0.25 0.25 = 0.1250 The following table contains generation of lognormal random numbers from given standard normal random numbers zi ’s: zi = N −1 (ui ) ni = m + vzi = xi = eni = 0.0297 + 0.1250zi Si = 56.12xi max(0, Si − 61) 0.48 0.0897 1.0938 61.3859 0.3859 0.93 0.1459 1.1571 64.9378 3.9378 -0.57 -0.0416 0.9593 53.8353 0 -1.39 -0.1441 0.8658 48.5906 0 Thus, the average payoff is: 1 (0.3859 + 3.9378) = 1.0809 4 Problem 3 For a stock, you are given: (i) The stock’s prices follow a lognormal model. (ii) The stock’s current price is 95. (iii) The stock’s continuously compounded rate of return is α = 0.15. (iv) The stock’s annual volatility is 0.25. (v) The stock pays quarterly dividends of 3. The next dividend will be paid 3 months from now. (vi) The continuously compounded risk-free interest rate is 0.05. A 1-year European call option on the stock has strike price 82. The payoff on the option will be based on the ex-dividend price. Simulate the stock’s price using the following standard normal random numbers in the order given to simulate the change in price for each quarter. −1.5 1.9 0.2 − 0.6 Calculate the simulated present value of the option in this run. Solution. First we calculate the quarterly price multipliers. Since we want a present value of an option we don’t use true probabilities. We use r = 0.05, not α = 0.15. The parameters of the lognormal model are √ √ m = µt = (r − 0.5σ 2 )t = (0.05 − 0.5 · 0.22 )0.25 = 0.0047; v = σ t = 0.2 0.25 = 0.1250 Page 2 of 9 Copyright ©Natalia A. Humphreys, 2014 ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 9. The stock price is calculated recursively, multiplying the previous price by the multiplier and subtracting the dividend. The following table contains generation of lognormal random numbers from given standard normal random numbers zi ’s: zi = N −1 (ui ) ni = m + vzi = xi = eni = 0.0047 + 0.1250zi S0.25t = S0.25(t−1) xi − 3 -1.5 -0.1828 0.8329 76.1278 1.9 0.2422 1.2740 93.9893 0.2 0.0297 1.0301 93.8215 -0.6 -0.0703 0.9321 84.4512 Thus, the value of the call option is: e−0.05 (84.4512 − 82) = 2.3317 Problem 4 The current exchange rate between dollars and pounds is $1.30/£. You are given: (i) (ii) (iii) (iv) The The The The exchange rate follows the Black-Scholes framework. continuously compounded risk-free rate in dollars is 4%. continuously compounded risk-free rate in pounds is 2%. volatility of the exchange rate is 0.2. A 3-month dollar-denominated European call option on pounds has a strike price of $1.25/£. The value of this option is estimated using simulation. One random number is used for each trial. You are given the following standard normal random numbers to be used for the simulation: 0.35 0.74 − 0.25 1.72 Determine the resulting price for the option. Solution. Since we want a present value of an option we don’t use true probabilities. We use r = r$ = 0.04, and δ = r£ = 0.02. The parameters of the lognormal model are √ √ m = µt = (r − δ − 0.5σ 2 )t = (0.04 − 0.02 − 0.5 · 0.22 )0.25 = 0; v = σ t = 0.2 0.25 = 0.1 The following table contains generation of lognormal random numbers from given standard normal random numbers zi ’s: zi = N −1 (ui ) ni = m + vzi = xi = eni = 0.1zi Si = 1.3xi Payoff=Si − 1.25 0.3500 0.0350 1.0356 1.3463 0.0963 0.7400 0.0740 1.0768 1.3998 0.1498 -0.2500 -0.0250 0.9753 1.2679 0.0179 1.7200 0.1720 1.1877 1.5440 0.2940 Page 3 of 9 Copyright ©Natalia A. Humphreys, 2014 ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 9. Thus, the value of the call option is: 1 e−0.04·0.25 · (0.0963 + 0.1498 + 0.0179 + 0.2940) = 0.99 · 0.1395 = 0.1381 4 Problem 5 For a stock, you are given: (i) The stock’s prices follow a lognormal model. (ii) The stock’s current price is 37.60. (iii) The stock’s continuously compounded rate of return is α = 0.25. (iv) The stock’s annual volatility is 0.35. (v) The stock pays no dividends. (vi) The continuously compounded risk-free interest rate is 0.05. Simulate the stock’s price for 6 months using the following standard normal random numbers in the order given to simulate the change in price for each month. −1.3 1.7 0 − 0.4 0.3 0.6 Calculate the simulated present value of a 6-month Asian arithmetic average strike call option on the stock. Solution. First we calculate the monthly price multipliers. Since we want a present value of an option we don’t use true probabilities. We use r = 0.05, not α = 0.15. The parameters of the lognormal model are r √ 1 2 2 1 = 0.1010 m = µt = (r − 0.5σ )t = (0.05 − 0.5 · 0.35 ) = −0.0009; v = σ t = 0.35 12 12 The stock price is calculated recursively starting with 45.50 and calculating each month’s price as the previous month’s price times the multiplier. The following table contains generation of lognormal random numbers from given standard normal random numbers zi ’s: zi = N −1 (ui ) ni = m + vzi = xi = eni = −0.0009 + 0.1010zi Si = Si−1 xi -1.3000 -0.1323 0.8761 32.9410 1.7000 0.1708 1.1863 39.0774 0.0000 -0.0009 0.9991 39.0408 -0.4000 -0.0414 0.9595 37.4593 0.3000 0.0294 1.0298 38.5759 0.6000 0.0597 1.0615 40.9484 The strike price, the average of the stock prices for 6 months, is 32.9410 + 39.0774 + 39.0408 + 37.4593 + 38.5759 + 40.9484 = 38.0071 6 Thus, the value of the call option is: e−0.05·0.5 (40.9484 − 38.0071) = 2.8686 Page 4 of 9 Copyright ©Natalia A. Humphreys, 2014 ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 9. Problem 6 You simulate option A. To lower the variance of the estimate, you simulate option B, which is highly correlated with option A and for which you have an exact value, using the same random numbers, and use the control variate method. The results of the trials are: Trial Value of option A Value of option B 1 1.57 1.88 2 3.07 2.19 3 4.82 1.07 4 2.95 2.32 5 3.82 3.74 The exact value for option B is 2.62. Determine the simulated value of option A. (A) 1.61 (B) 2.24 (C) 2.87 (D) 3.25 (E) 3.63 Key: E Solution. For the control variate method, ¯ A∗ = A¯ + E[B] − B The means of the simulated values are: 16.23 ¯ = 11.2 = 2.24 A¯ = = 3.246, B 5 5 Therefore, the simulated result is A∗ = 3.246 + 2.62 − 2.24 = 3.626 ≈ 3.63 Problem 7 A variable X is simulated. A control variate Y is used to make the simulation more efficient. After 100 trials, 100 random values xi for X and yi for Y are generated. You are given: (i) 100 X xi = 91.88 i=1 (ii) 100 X yi = 91.34 i=1 (iii) 100 X x2i = 87.15 i=1 (iv) 100 X yi2 = 85.55 i=1 Page 5 of 9 Copyright ©Natalia A. Humphreys, 2014 ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 9. (v) 100 X xi yi = 86.24 i=1 Let s21 be the variance of the naive Monte Carlo estimate of X and s22 the variance of the control variate estimate of Y . Calculate the proportional reduction in variance from using the control variate method, 1 − s22 /s21 (A) −11.58 (B) −2.47 (C) 0.08 (D) 0.92 (E) 1.00 Key: D Solution. The variance of the naive Monte Carlo estimate of X is 1 100 87.15 1 2 = 0.000276 s21 = · − · 91.88 100 99 100 1002 The variance of the control variate estimate of X is ¯ + V ar(Y¯ ) − 2Cov(X, ¯ Y¯ ) s22 = V ar(X ∗ ) = V ar(X) ¯ is calculated above. V ar(X) 1 100 85.55 1 2 ¯ V ar(Y ) = · − · 91.34 = 0.000214 100 99 100 1002 1 100 86.24 ¯ ¯ Cov(X, Y ) = · − 0.9188 · 0.9134 = 0.000234 100 99 100 Hence, s22 = 0.000276 + 0.000214 − 2 · 0.000234 = 0.000022 And the reduction in variance is 0.000022 1 − s22 /s21 = 1 − = 0.920501 ≈ 0.92 0.000276 Problem 8 You simulate a 3-month Asian arithmetic average price put option. To reduce the variance of the estimate, you use the control variate method, with a 3-month Asian geometric average price put option as the control variable. You are given: (i) The stock price is 40. (ii) The stock pays no dividends. (iii) The strike price is 39.5. (iv) The continuously compounded risk-free interest rate is 0.03. (v) The average price is based on an average of 3 monthly prices. (vi) The stock’s annual volatility is 0.15. (vii) The exact value of the geometric option is 0.642 Use the following random numbers from the standard normal distribution to perform one trial: −1.25 0.64 − 0.8 Determine the resulting simulated price for the arithmetic option. (A) 0.571 (B) 0.638 (C) 0.982 (D) 1.211 (E) 1.785 Key: B Solution. For the control variate method, ¯ + E[Y ] − Y¯ X∗ = X Page 6 of 9 Copyright ©Natalia A. Humphreys, 2014 ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 9. First we calculate the monthly price multipliers. Since we want a present value of an option we don’t use true probabilities. We use r = 0.03. The parameters of the lognormal model are r √ 1 2 2 1 m = µt = (r − 0.5σ )t = (0.03 − 0.5 · 0.15 ) = 0.0016; v = σ t = 0.15 = 0.0433 12 12 The stock price is calculated recursively starting with 40 and calculating each month’s price as the previous month’s price times the multiplier. The following table contains generation of lognormal random numbers from given standard normal random numbers zi ’s: zi = N −1 (ui ) ni = m + vzi = xi = eni = 0.0016 + 0.0433zi Si = Si−1 xi -1.25 -0.0526 0.9488 37.9517 0.64 0.0293 1.0297 39.0792 -0.8 -0.0331 0.9675 37.8077 The arithmetic average of the stock prices for 3 months, is 37.9517 + 39.0792 + 37.8077 = 38.2795 3 The payoff after 3 months is: 39.5 − 38.2795 = 1.2205 Thus, the simulated value of the 3-month Asian arithmetic average price put option is: e−0.03·0.25 1.2205 = 1.2113 The geometric average of the stock prices for 3 months, is 1 (37.9517 · 39.0792 · 37.8077) 3 = 38.2753 The payoff after 3 months is: 39.5 − 38.2753 = 1.2247 Thus, the simulated value of the 3-month Asian geometric average price put option is: e−0.03·0.25 1.2247 = 1.2155 The simulated value in this run of the arithmetic average price put option using the control variate method is: X ∗ = 1.2113 + (0.642 − 1.2155) = 0.6378 ≈ 0.638 Problem 9 The time-t price of a nondividend paying stock following the Black-Scholes framework is S(t). You are given: (i) The current price of the stock is 55. (ii) The continuously compounded risk-free interest rate is 0.04. (iii) The volatility of the stock is 0.35. A European put option on the stock with strike price 45 expires in one year. To estimate its price, simulation is used. Stratified sampling is used with four strata: [0, 0.25), [0.25, 0.50), [0.50, 0.75), [0.75, 1). Uniform random numbers u4k+i are sent to stratum i for k ≥ 0 and 1 ≤ i ≤ 4. Use the following four uniform random numbers u1 . . . u4 for the first four trials: 0.66 0.42 0.104 0.29 Page 7 of 9 Copyright ©Natalia A. Humphreys, 2014 ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 9. Determine the resulting estimate of the option’s price. Solution. The parameters of the lognormal model are √ m = µt = (r − 0.5σ 2 )t = (0.04 − 0.5 · 0.352 ) = −0.0213, v = σ t = 0.35 Let us stratify the numbers by mapping the to the appropriate quartile: 0.66 7−→ 0.66 · 0.25 = 0.1650 0.42 7−→ 0.42 · 0.25 + 0.25 = 0.3550 0.104 7−→ 0.104 · 0.25 + 0.5 = 0.5260 0.29 7−→ 0.29 · 0.25 + 0.75 = 0.8225 The following table contains generation of lognormal random numbers from given uniform [0, 1) random numbers ui ’s: ui zi = N −1 (ui ) 0.1650 -0.9700 -0.3608 0.6972 38.3434 6.6566 0.3550 -0.3750 -0.1525 0.8586 47.2207 0 0.5260 0.6500 0.2063 1.2291 67.5983 0 0.8225 0.9250 0.3025 1.3532 74.4281 0 ni = m + vzi = xi = eni = −0.0213 + 0.35zi Si = 55xi Payoff=45 − Si Thus, the price of the put option is: e−0.04 · 1.6641 = 1.5989 Problem 10 Stock prices are lognormally distributed with annual return parameters µ = 0.25 and σ = 0.10. The stock price after 2 years is simulated using the antithetic variate method. The following random numbers from the uniform distribution on [0, 1] are used: 0.2709 0.834 0.1515 Six trials are run. Determine the simulated average ratio of the stock price at the end of two years to the current stock price. Solution. We use the three random numbers ui and their compliments 1 − ui . By the symmetry of the normal distribution, the generated standard normal numbers of the complements will be negative the standard normal numbers generated from the originals. The parameters of the lognormal model are √ √ m = µt = 0.25 · 2 = 0.5; v = σ t = 0.1 2 = 0.1414 The following table contains generation of lognormal random numbers from given uniform [0, 1) random numbers ui ’s: Page 8 of 9 Copyright ©Natalia A. Humphreys, 2014 ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 9. ui zi = N −1 (ui ) 0.2709 -0.6100 0.4137 1.5125 0.8340 0.9700 0.6372 1.8911 0.1515 -1.0300 0.3543 1.4252 0.7291 0.6100 0.5863 1.7972 0.1660 -0.9700 0.3628 1.4374 0.8485 1.0300 0.6457 1.9073 ni = m + vzi = xi = eni = 0.5 + 0.1414zi Thus, the average change is: 6 1X S(2) 1 = xi = (1.5125 + 1.8911 + 1.4252 + 1.7972 + 1.4374 + 1.9073) = 1.6618 S(0) 6 6 i=1 Page 9 of 9
© Copyright 2024 ExpyDoc
