ACTS 4302 Instructor: Natalia A. Humphreys SOLUTION TO HOMEWORK 5 Lesson 8: Fitting stock prices to a lognormal distribution. Lesson 9: The Black-Scholes formula. Lesson 10: The Black-Scholes formula: Greeks. Problem 1 You are given the following monthly stock prices: t St 1 38 2 42 3 59 4 45 5 37 6 45 Estimate the continuously compounded expected rate of return on the stock. Solution. We calculate xi = ln(St /St−1 ) and corresponding x2i : xi x2i 2 42 ln(42/38) = 0.1001 0.01 3 59 ln(59/42) = 0.3399 0.1155 t St 1 38 4 45 ln(45/59) = −0.2709 0.0734 5 37 ln(37/45) = −0.1957 0.0383 6 45 ln(45/37) = 0.1957 0.0383 We need to estimate α ˆ α ˆ=µ ˆ + 0.5ˆ σ 2 , where µ ˆ = Nx ¯ s P 2 √ xi n 2 −x ¯ σ ˆ= N n−1 n N is the number of periods per year, n is the number one less than the number of observations of stock price. Since there are 6 observations, n = 5. Calculating µ ˆ: 1 45 x ¯ = ln = 0.0338 5 38 µ ˆ =N ∗x ¯ = 12 · 0.0338 = 0.4057 Copyright ©Natalia A. Humphreys, 2014 ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 5. Calculating σ ˆ2: X x2i = 0.2755 5 1 2 2 σ ˆ = 12 · · 0.2755 − 0.0338 = 0.8094 4 5 Therefore, α ˆ = 0.4057 + 0.5 · 0.8094 = 0.8104 Problem 2 You are given the following statistics for weekly closing prices of a stock St : 20 X St = 0.03489 St−1 t=1 2 20 X St (ii) ln = 0.1796 St−1 (i) ln t=1 Estimate the continuously compounded expected rate of return on the stock. (A) 0.01 (B) 0.09 (C) 0.34 (D) 0.49 (E) 0.58 Solution. By definition, if xi are observed stock prices adjusted to remove the effect of dividends, the estimate for the continuously compounded annual return is: α ˆ=µ ˆ + 0.5ˆ σ 2 , where µ ˆ = Nx ¯ and s P 2 √ xi n 2 −x ¯ σ ˆ= N n−1 n N is the number of periods per year, n is the number one less than the number of observations of stock price, xi = ln (Si /Si−1 ). The sample mean is m ˆ = 0.03489/20 = 0.001745. The sample variance is 20 s = 19 2 0.1796 − 0.0017452 20 = 0.00945 The estimated annual return is α ˆ = 52 (0.001745 + 0.5(0.00945)) = 0.3364 . Problem 3 For a 9-month European put option on a stock, you are given: (i) The stock’s price is 50. (ii) The strike price is 45. (iii) The continuous dividend rate for the stock is 2%. (iv) The stock’s annual volatility is 15% (v) The continuously compounded risk-free interest rate is 3%. Determine the Black-Scholes premium for the option. Page 2 of 8 ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 5. Copyright ©Natalia A. Humphreys, 2014 Solution. We are given: Eur. put, t = 0.75, S = 50, K = 45, δ = 0.02, σ = 0.15, r = 0.03. P = Ke−rt N (−d2 ) − Se−δt N (−d1 ), where ln 50 + 0.03 − 0.02 + 0.5 · 0.152 0.75 ln (S/K) + (r − δ + 12 σ 2 )t 0.1213 45 √ √ = d1 = = = 0.9338 ≈ 0.93 0.1299 σ t 0.15 0.75 N (−d1 ) = N (−0.9338) ≈ 1 − N (0.93) = 1 − 0.8238 = 0.1762 √ d2 = d1 − σ t = 0.9338 − 0.1299 = 0.8039 ≈ 0.8 N (−d2 ) = N (−0.8039) ≈ 1 − N (0.8) = 1 − 0.7881 = 0.2119 Ke−rt = 45e−0.03·0.75 = 43.9988 Se−δt = 50e−0.02·0.75 = 49.2556 P = 43.9988 · 0.2119 − 49.2556 · 0.1762 = 9.3233 − 8.6788 = 0.6445 Problem 4 For a 9-month European call option on a stock, you are given: (i) The stock’s price is 60. (ii) The strike price is 70. (iii) σ = 0.4. (iv) The continuously compounded risk-free interest rate is 5%. (v) The stock pays no dividend. Determine the change in the Black-Scholes premium for the option if the stock pays a quarterly dividend of 1 with the first dividend payable 3 months after the option is written, and the expiry occurs after the 3rd dividend. Solution. We are given: Eur. call, t = 0.75, S = 60, K = 70, σ = 0.4, r = 0.05. If δ = 0, then C = Se−δt N (d1 ) − Ke−rt N (d2 ), where ln 60 + 0.05 − 0 + 0.5 · 0.42 0.75 ln (S/K) + (r − δ + 12 σ 2 )t 0.0567 70 √ √ d1 = = =− = −0.1635 ≈ −0.16 0.3464 σ t 0.4 0.75 N (d1 ) = N (−0.1635) ≈ 1 − N (0.16) = 1 − 0.5636 = 0.4364 √ d2 = d1 − σ t = −0.1635 − 0.3464 = −0.5099 ≈ −0.51 N (d2 ) = N (−0.5099) ≈ 1 − N (0.51) = 1 − 0.695 = 0.305 Ke−rt = 70e−0.05·0.75 = 67.4236 Se−δt = 60 C = 60 · 0.4364 − 67.4236 · 0.305 = 26.184 − 20.5642 = 5.6198 If the dividend of 1 is paid quarterly, then the pre-paid forward price of the stock is: F P (S) = 60 − 1 · e−0.05·0.25 − 1 · e−0.05·0.5 − 1 · e−0.05·0.75 = 60 − 2.9261 = 57.0739 Page 3 of 8 ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 5. Copyright ©Natalia A. Humphreys, 2014 Using F P (S) for S and δ = 0 in the formula for C above, we obtain: C = Se−δt N (d1 ) − Ke−rt N (d2 ), where ln 57.0739 + 0.05 − 0 + 0.5 · 0.42 0.75 ln (S/K) + (r − δ + 12 σ 2 )t 0.1066 70 √ √ d1 = = =− = −0.3079 ≈ −0.31 0.3464 σ t 0.4 0.75 N (d1 ) = N (−0.3079) ≈ 1 − N (0.31) = 1 − 0.6217 = 0.3783 √ d2 = d1 − σ t = −0.3079 − 0.3464 = −0.6543 ≈ −0.65 N (d2 ) = N (−0.6543) ≈ 1 − N (0.65) = 1 − 0.7422 = 0.2578 Ke−rt = 70e−0.05·0.75 = 67.4236 Se−δt = 57.0739 C = 57.0739 · 0.3783 − 67.4236 · 0.2578 = 21.5911 − 17.3818 = 4.2093 Thus, the difference between these two options is ∆ = 5.6198 − 4.2093 = 1.4105 Problem 5 For a yen-dollar exchange rate, you are given: (i) The spot exchange rate is 120U/$. (ii) The annual volatility of the exchange rate is 20%. (iii) The continuously compounded risk-free rate for dollars is 0.06. (iv) The continuously compounded risk-free rate for yen is 0.02. Determine the Garman-Kohlhagen premium for a yen-denominated 6-month European put option on dollars with a strike price of 115U/$. Solution. We are given: Eur. put, t = 0.5, x = 120, K = 115, rf = r$ = δ = 0.06, σ = 0.2, rd = rU = r = 0.02. P = Ke−rd t N (−d2 ) − xe−rf t N (−d1 ), where 120 + 0.02 − 0.06 + 0.5 · 0.22 0.5 ln 115 ln (x/K) + (rd − rf + 12 σ 2 )t 0.0326 √ √ d1 = = = = 0.2302 0.1414 σ t 0.2 0.5 N (−d1 ) = N (−0.23) = 1 − 0.591 = 0.409 √ d2 = d1 − σ t = 0.2302 − 0.1414 = 0.0888 N (−d2 ) = N (−0.09) = 1 − 0.5359 = 0.4641 Ke−rd t = 115e−0.02·0.5 = 113.8557 xe−rf t = 120e−0.06·0.5 = 116.4535 P = 113.8557 · 0.4641 − 116.4535 · 0.409 = 52.8404 − 47.6295 = 5.211 Problem 6 You are given: (i) The 1-year futures price for gold is 550. (ii) The annual volatility of the price of gold is 0.25. (iii) The continuously compounded risk-free interest rate is 3%. Determine the Black premium for a 1-year European call option on the futures contract with a strike price of 580. Page 4 of 8 Copyright ©Natalia A. Humphreys, 2014 ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 5. Solution. We are given: Eur. call, t = 1, F = 550, K = 580, σ = 0.25, r = 0.03. C = C = F e−rt N (d1 ) − Ke−rt N (d2 ), where ln (F/K) + 12 σ 2 t ln 550 + 0.5 · 0.252 √ = 580 = −0.0874 0.25 σ t N (d1 ) = 1 − N (0.09) = 1 − 0.5359 = 0.4641 √ d2 = d1 − σ t = −0.0874 − 0.25 = −0.3374 N (d2 ) = 1 − N (0.34) = 1 − 0.6331 = 0.3669 d1 = Ke−rt = 580e−0.03 = 562.8584 F e−rt = 550e−0.03 = 533.745 C = 533.745 · 0.4641 − 562.8584 · 0.3669 = 247.7111 − 206.5128 = 41.1983 Problem 7 You own the following portfolio of options on a stock whose price is 52: Number of options Option price Elasticity 20 2.35 3.30 40 1.85 -2.50 50 0.70 1.90 Calculate the elasticity of this portfolio. Solution. To calculate the value of a portfolio, take a weighted average of the elasticity of the instruments in it, rather than a sum (as you would do with option Greeks). The value of the portfolio is: W = 20 · 2.35 + 40 · 1.85 + 50 · 0.70 = 156 Each weight is Ni · Ci Ni · Ci = , where W 156 Ni is the number of shares and Ci is the value of each option. Hence, the elasticity of the portfolio is: wi = 20 · 2.35 40 · 1.85 50 · 0.70 − 2.50 · + 1.90 · = 156 156 156 = 3.30 · 0.3013 − 2.50 · 0.4744 + 1.90 · 0.2244 = 0.99423 − 1.1859 + 0.4263 = 0.2346 Ω = 3.30 · Problem 8 For a 6-month European put option on a stock, you are given: (i) (ii) (iii) (iv) (v) The The The The The stock’s price is 40. stock’s continuous dividend rate is 0.03 stock’s volatility is 0.4. strike price is 44. risk free rate is 0.06. Determine the elasticity of the put option. Page 5 of 8 ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 5. Copyright ©Natalia A. Humphreys, 2014 Solution. We are given: Eur. put, t = 0.5, S = 40, K = 44, δ = 0.03, σ = 0.4, r = 0.06. S∆put P = −e−δt N (−d1 ) Ωput = ∆put P = Ke−rt N (−d2 ) − Se−δt N (−d1 ), where 40 ln 44 + 0.06 − 0.03 + 0.5 · 0.42 0.5 ln (S/K) + (r − δ + 12 σ 2 )t 0.0403 √ √ d1 = = =− = −0.1425 0.2828 σ t 0.4 0.5 N (−d1 ) = N (0.14) = 0.5557 ∆put = −e−0.03·0.5 · 0.5557 = −0.5473 √ d2 = d1 − σ t = −0.1425 − 0.2828 = −0.4254 N (−d2 ) = N (0.43) = 0.6664 Ke−rt = 44e−0.06·0.5 = 42.6996 Se−δt = 40e−0.03·0.5 = 39.4045 P = 42.6996 · 0.6664 − 39.4045 · 0.5557 = 28.455 − 21.8919 = 6.5631 40 · 0.5473 Ωput = − = −3.3356 6.5631 Problem 9 For a stock: (i) (ii) (iii) (iv) The The The The price is 65. continuous dividend rate is 0.015. volatility of the stock is 25%. continuously compounded risk-free rate is 3%. A portfolio has 3 European put options on this stock. • The first allows sale of 200 shares of the stock at the end of 6 months at strike price 60. • The second allows sale of 300 shares of the stock at the end of 9 months at strike price 65. • The third allows sale of 400 shares of the stock at the end of one year at strike price 70. Calculate the delta for this portfolio of puts. Solution. The formula for ∆ for each put is: ∆put = −e−δt N (−d1 ). We’ll calculate this for each option. It helps a little that r − δ + 21 σ 2 = 0.03 − 0.015 + 0.5 · 0.252 = 0.04625 is the same for all 3 options. For the first option we have: ln (S/K) + (r − δ + 12 σ 2 )t ln 65 + 0.04625 · 0.5 √ √ = 60 = 0.5836 σ t 0.25 0.5 N (−d1 ) = N (−0.58) = 1 − N (0.58) = 1 − 0.719 = 0.281 d1 = ∆1 = −e−0.015·0.5 · 0.281 = −0.2789 For the second option we have: ln 65 · 0.75 65 + 0.04625 √ = 0.1602 0.25 0.75 N (−d1 ) = N (−0.16) = 1 − N (0.16) = 1 − 0.5636 = 0.4364 d1 = ∆2 = −e−0.015·0.75 · 0.4364 = −0.4315 Page 6 of 8 Copyright ©Natalia A. Humphreys, 2014 ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 5. For the third option we have: ln 65 70 + 0.04625 = −0.1114 0.25 N (−d1 ) = N (0.11) = 0.5438 d1 = ∆3 = −e−0.015 · 0.5438 = −0.5357 Delta for the portfolio is the weighted sum: ∆portf olio = 200∆1 + 300∆2 + 400∆3 = = 200 · (−0.2789) + 300 · (−0.4315) + 400 · (−0.5357) = = −55.7801 − 129.4554 − 214.2815 = −399.51 Problem 10 Five observations of the logged ratios of stock prices, in order, are: −0.06, 0.03, 0.07, 0.11, 0.17 a) Determine the y-axis label of the point on the normal probability plot corresponding to 0.03. b) Determine the smoothed 60th percentile. Solution. Note that since the data is given to us in ascending order, it’s called order statistics. a) Let us determine the values of Fn (xi ) and corresponding values of the inverse normal cumulative probability function (quantiles) for these observation. Since n = 5 and Fn (xi ) = (2i − 1)/2n, i = 1, . . . n, Therefore, the y-axis label of the point on the normal probability plot corresponding to 0 is xi yi = F5 (xi ) N −1 (yi ) −0.06 1 10 = 0.1 -1.28 0.03 3 10 = 0.3 -0.525 0.07 5 10 = 0.5 0 0.11 7 10 = 0.7 0.525 0.17 9 10 = 0.9 1.285 F5 (0) = 0.3. th b) The ith observation represents 2i−1 percentile, so the third observation is the 50th percentile and 2n the forth observation is the 70th percentile. Note that the 60th percentile should lie half way between the third and the forth observation. Hence, the 60th percentile is: 0.07 · 0.5 + 0.11 · 0.5 = 0.09 Problem 11 You are given the following information on the price of a stock: Page 7 of 8 Copyright ©Natalia A. Humphreys, 2014 ACTS 4302. AU 2014. SOLUTION TO HOMEWORK 5. Date Stock price Jul. 1, 2007 35.30 Aug. 1, 2007 33.90 Sep. 1, 2007 41.20 Oct. 1, 2007 31.95 Nov. 1, 2007 38.25 Dec. 1, 2007 46.18 Estimate the annual volatility of continuously compounded return on the stock. (A) 0.20 (B) 0.62 (C) 0.68 (D) 0.69 (E) 0.75 Key: D Solution. To estimate volatility from historical data: s P P 2 √ n xt St xt 2 σ ˆ= N −x ¯ , where xt = ln , x ¯= n−1 n St−1 n N is the number of periods per year, n is the number one less than the number of observations of stock price. In our problem N=12 and n=5. We calculate xt = ln(St /St−1 ) and x2t : St xt x2t 35.30 33.90 -0.0405 0.0016 41.20 0.1950 0.0380 31.95 -0.2543 0.0647 38.25 0.1800 0.0324 46.18 0.1884 0.0355 Summing up the third column and its squares, 5 X 5 X 0.2687 = 0.05373, x2t = 0.1722 5 t=1 t=1 √ 5 0.1722 s2 = − 0.053732 = 0.03944, s = 0.03944 = 0.1986 4 5 That is the monthly volatility. The annual volatility is √ 0.1986 12 = 0.688 ≈ 0.69 xt = 0.2687, x ¯= Page 8 of 8