Hw4 Mfe Au14 Solution

ACTS 4302Instructor: Natalia A. Humphreys SOLUTION TO HOMEWORK 4 Lesson 6: Binomial trees: miscellaneous topics. Lesson 7: Modeling stock prices with the lognormal distribution. Problem 1 For a 1-year American put option on a stock with 6 months left to expiry, you are given: (i) The strike price is 42. (ii) The continuous dividend rate for the stock is 0.03. (iii) 6 months before expiry, the stock price is 33. (iv) The value of a 6-month call option on the stock with a strike price of 42 is 0.155. Calculate the lowest possible continuously compounded risk-free rate so that exercising the option at 6 months before expiry is optimal. Solution. Since early exercise is optimal, the present value of interest on the strike price must exceed the present value of dividends plus the value of the implicit call option: K(1 − e−rt ) ≥ S(1 − e−δt ) + C The present value of dividends is:
33 1 − e−0.5·0.03 = 0.4913 The value of the call option is given as 0.155. The present value of interest on the strike price is
42 1 − e−0.5r Hence,
42 1 − e−0.5r ≥ 0.4913 + 0.155 = 0.6463 e−0.5r = 0.9846 r = 0.031  Problem 2 The Jarrow-Rudd binomial tree is used to price an option on a non-dividend paying stock. You are given: (i) The risk-free rate is 0.045. (ii) The annual volatility is 29%. (iii) The period of the binomial tree is 7 months. Determine the risk-neutral probability of an up move. Solution. Recall that for a Lognormal (Jarrow-Rudd) tree: u = e(r−δ−0.5σ 2 )h+σ √ h , d = e(r−δ−0.5σ 2 )h−σ √ h and p∗ = e(r−δ)h − d u−d Hence, 2 √ u = e0.5833[0.045−0.5(0.29 )]+0.29√ 0.5833 = e0.2232 = 1.2501 2 d = e0.5833[0.045−0.5(0.29 )]−0.29 0.5833 = e−0.2198 = 0.8027 e0.045·0.5833 − 0.8027 1.0266 − 0.8027 p∗ = = = 0.5005 1.2501 − 0.8027 0.4474  5%.Copyright ©Natalia A.375 d=e =e = 0. in dollars. Thus. Since δ = 0.75/2 = 0.375 − 0. to avoid arbitrage. To avoid arbitrage. u and d must satisfy: d < e(r−δ)h < u For a Cox-Ross-Rubinstein tree: u = eσ √ h .000. Determine the cost of an option.06 h > 0.3%. You are given: (i) The continuously compounded risk-free rate is 5%.4817 u−d 0. note that h = 0.2 0. we need e(r−δ)h < u ⇔ e(r−δ)h < eσ eσ √ √ h.05h ⇔ √ √ 6 0. r = rd = r$ = 0. Solution. the domestic currency is dollars and the foreign currency is pounds.000 from sales in England at the end of 9 months. Solution. (ii) σ = 0.50. The company should purchase a European put on pounds with strike price 1. The company would like to guarantee that it will get at least this rate when it receives the pounds.06 h h > erh ⇔ e > e0. √ 0.8847 p∗ = = = 0. The current exchange rate is $1. Humphreys. 2014 ACTS 4302. Relative volatility of the currencies is 0.2456 e(r−δ)h − d e(0.1303 u = eσ √ √ −σ h −0. and the risk-neutral probability of an up move. which will guarantee the current exchange rate at the end of 9 months.375 = 1.5/£. AU 2014.5183 The spot exchange rate tree: Page 2 of 7 . are √ √ h = e0. SOLUTION TO HOMEWORK 4.043 δ = rf = r£ = 0. The continuously compounded risk-free rate in pounds is 3. You are given: (i) (ii) (iii) (iv) The continuously compounded risk-free rate in dollars is 4. For this option. In a Cox-Ross-Rubinstein binomial tree the up and down movements. so that it will receive at least $675.375.2.035)0. A two-period Cox-Ross-Rubinstein binomial tree is used to determine the price of options.06.043−0. d = e−σ √ h Thus. (iii) δ = 0.8847 u − d = 0.2456 1 − p∗ = 0.05h ⇔ h < ⇔ h < 1.2 0. Determine the largest period for which this tree can be used without violating arbitrage conditions on the nodes.035 Also.44 5  Problem 4 An American company expects to receive £450. Problem 3 A Cox-Ross-Rubinstein binomial tree is used to model an option. SOLUTION TO HOMEWORK 4.3259 = 0.6954 x0 = 1.5183 · 0.3271 ddx0 = 1.1729 > 0.0848 = 38.375 · 0.0882$/£ This is the cost of a put to buy £1.375 · 0. we use 0.1729.143.000 is: 450.Copyright ©Natalia A. AU 2014.5 − 1.5183 · 0.375 · 0.684.1741 The put tree: Puu = 0 Pu Pud = 0 P Pd Pdd = 0.1729 in further calculations: P = e−0.3259 Pd = e−rh (p∗ Pud + (1 − p∗ )Pdd ) = e−0. The total price of a put on £450.000 is: 450.000 · 0.0882 = 39. Humphreys.9163 ux0 = 1.5183 · 0. we continue with this value of Pd : P = e−0.0848$/£ This is the cost of a put to buy £1.5 udx0 = 1.1662.043·0.3271 = 0.1662 = 0.1729 = 0.5 dx0 = 1. the answer would be slightly different: If exercised.74  Problem 5 You are given the following information on the price of a stock: Page 3 of 7 . 1. Since 0.1662 Since the put is the European put.000 · 0. The total price of a put on £450. 2014 ACTS 4302.043·0.61 Note that if the put were an American put.043·0. u2 x0 = 1. (A) 0.2687 = 0.90 -0.69 (E) 0. (ii) The probability that the stock’s price will be less than 98 at the end of 6 months is 0.95 Nov.90 Sep.1986 4 5 That is the monthly volatility.1722 5 t=1 t=1    √ 5 0.0380 31.0016 41.18 0.20 (B) 0.1722 s2 = − 0.0324 46.30 Aug.Copyright ©Natalia A. s = 0.75 Key: D Solution. 2014 ACTS 4302.053732 = 0.18 Estimate the annual volatility of continuously compounded return on the stock. x2t = 0. Page 4 of 7 . 1.2687.69  xt = 0. AU 2014.1986 12 = 0. 1. 2007 41. 2007 38. x ¯= σ ˆ= N n−1 n St−1 n N is the number of periods per year. 2007 31.1884 0.25 0.0355 Summing up the third column and its squares. Date Stock price Jul.2543 0.3483. 2007 33.05373. The annual volatility is √ 0. We calculate xt = ln(St /St−1 ) and x2t : St xt x2t 35.1950 0. 1.68 (D) 0. 2007 46.30 33. SOLUTION TO HOMEWORK 4. To estimate volatility from historical data: s P P 2    √ xt n St xt 2 −x ¯ . 5 X 5 X 0.1800 0. x ¯= Problem 6 A stock’s price follows a lognormal model. You are given: (i) The current price of the stock is 105.62 (C) 0.25 Dec.20 0. In our problem N=12 and n=5. n is the number one less than the number of observations of stock price. Humphreys.688 ≈ 0. 2007 35. 1. 1.03944 = 0.95 -0. where xt = ln .20 Oct.0647 38. 1.03944.0405 0. 5v = 105e−0.2758σ = −0.0468 = 105. µ = −0. Calculate the expected price of the stock at the end of one year. Solution.75µ = −0. 0.069 + 0.39 and z0.3483 P r (S0. AU 2014.7123 = 0.7123. Therefore.091 + 0. Humphreys.56.7123 = 115 The normal percentiles are: z0.5µ 0. 2014 ACTS 4302.75µ+ 0. we solve the system:   0.75) = 0.091 + 0.75).866σ σ 0.5) = ln dˆ2 (0.138 ⇔   0.75) = µ · 0.01864+0.5517σ = −0.56 Thus.069+0.75 On the other hand.75 < 115) = N (−dˆ2 (0.5 σ 0.5 < 98) = N (−dˆ2 (0.5) = −0.Copyright ©Natalia A.3483 = −0.75µ + 0.5) and dˆ2 (0.75µ 0.069  µ − 0. we obtain: 105 98 √+
ln dˆ2 (0. The probability statements result in these two equations expressing the percentiles in terms of standard normal coefficients of the 100pth percentile.2164.75µ −0.01864t. or zp : √ 105e0.866σ    0.39 ⇔   0. SOLUTION TO HOMEWORK 4.75)) = 0.7071σ = 0.56 0.069 + 0.  Page 5 of 7 .091 µ + 0.1213 to obtain: σ = 0.485σ = −0.3483 and N (z) = 0.0468t and the expected value of the stock after one year is 2 E[S1 ] = S0 em+0. v 2 = σ 2 t = 0.2758σ = −0.5µ 0.2758σ   −0. where

ln SK0 + (α − δ + 0. Hence.5σ 2 )t ln SK0 + (µ + σ 2 )t ˆ √ √ d1 = = σ t σ t

ln SK0 + µt ln SK0 + (α − δ − 0.7123.7123 Calculating dˆ2 (0.5
105 115 √+ = µ · 0.5µ − 0. we obtain: −dˆ2 (0. our statements could be written as: P r (S0.75µ + 0.069 0. using the standard normal probability table for N (z) = 0.091 This is the same system as in the first approach above.75 0.7071σ = −0. Recall from Lesson 7 that probabilities of payoffs of stock prices are: P r(St < K) = N (−dˆ2 ).39 −dˆ2 (0.5σz0.5µ = 0. using the theory of percentiles.01864.5µ − 0. m = µ · t = −0.75σz0.091+0.485σ = 0.6467σ = 0.5µ+ √0.5·0. (iii) The probability that the stock’s price will be less than 115 at the end of 9 months is 0.5)) = 0. and P r(St > K) = N (dˆ2 ).3483 = 98 105e0.5σ 2 )t ˆ √ √ = d2 = σ t σ t √ dˆ2 = dˆ1 − σ t Hence.5025 This problem could be solved slightly differently.485σ = 0. 645 √ and 50e2µ+1.07.4 5 = 0. the answer is (15. (ii) α = 0.2039)(1.20 = 50eµ−1.05 = 1.14 − 0.85 √ 50e2µ+1.03. 5. we get
e−1.8944.5σ 2 )t = (0.05).645 2σ Calculating: √ √ e1. 73. We know that 41. Solution. Humphreys.4909) Multiplying by the initial stock price of 95. Construct a 95% confidence interval for the price of the stock at the end of five years.8031. 2014 ACTS 4302. e1.92 e3.4049 50e2µ−1. You are given: (i) The stock’s initial price is 60.92σ = and the product is e2µ = We need 41.66. SOLUTION TO HOMEWORK 4.645 2σ =√e1. The confidence interval is −0. 521.1461 = 1.  Problem 9 A stock’s price follows a lognormal model. Page 6 of 7 .7031 = (0.20 · 73. Construct a 90% confidence interval for the price of the stock at the end of two years.20.42 )5 = −0.2039 502 √ 2σ 50e2µ−1.05 = 50eµ+1.14.Copyright ©Natalia A.05 √ The lognormal parameter v is 0.20 ln 1.645 2σ = 50(1.4049) = 84.645 2σ = 50(1. Problem 7 A stock’s price follows a lognormal model.  Problem 8 A stock’s price follows a lognormal model.8944) = (−1.7731 σ= = 0.05 = 1.07 − 0. 1.85. A 95% confidence interval for the price of the stock at the end of one year is (41. Solution.7031) Exponentiating.645 2·0.56 Thus the 90% confidence interval for the price of the stock at the end of two years is (42. (iv) σ = 0.7731 41. AU 2014. (iii) δ = 0.96σ so the quotient is: 73. You are given: (i) The initial price is 95.64). (iii) The stock’s continuously compounded dividend rate is 0. (ii) The stock’s continuously compounded rate of return is 0.1461 3.5 · 0.96(0.2039)/1.4. The current price of the stock is 50.05 ± 1.06.1648.4049 = 42. 84.56).8031 . The lognormal parameter m for the distribution of the stock price after five years is (α − δ − 0.96σ and 73. 32 N (dˆ1 ) = N (−0.3716) = N (−0.1822 1 − N (0.62) 1 − 0.03.063 (0.07 σ t 0.5 K √ √ dˆ1 = = = −0.36 0.5161  Page 7 of 7 . AU 2014.3716) 1 − N (0.8273 ≈ 1 − N (0.6255 = 0.25 = −0.362 ))0.25 σ t √ dˆ2 = dˆ1 − σ t = −0.4721 N (dˆ2 ) = N (0.3716) = 61.3217 ≈ −0.156 − 0. Humphreys. St − K)] = S0 e(α−δ)t N (dˆ1 ) − KN (dˆ2 ) Calculating dˆ1 and dˆ2 :
S0 ln + (α − δ + 0.36.3716 − 0. Calculate the expected payoff for the call option.5279 = 0.03 + 0. (iv) Volatility is 0.5(0.25.7324 E[St |St > 70] = 60e0.07) = 1 − N (0.5 √ √ dˆ2 = dˆ1 − σ t = −0. Copyright ©Natalia A. Solution. Solution. Calculate the conditional expected value of the stock after 1 year given that it is greater than 70.156.32) = 1 − 0.37) 1 − 0. We use the formula E[max(0.5 = −0. (iii) The stock’s continuously compounded dividend rate is 0.5σ 2 )t ln (42/47) + (0.32) = 1 − N (0.3745) = 3.06 − 0.03 + 0. St − K)] = 42e0.6216) 1 − N (0.5(0. You are given: (i) The stock’s initial price is 42. SOLUTION TO HOMEWORK 4.252 ) ˆ √ d1 = = = −0. A European call option on the stock expires in 6 months and has strike price 47.3716 0.8273 = 61.6443 ≈ 61.0671 ≈ −0. (ii) The stock’s continuously compounded rate of return is 0.07) = 1 − 0.0671 − 0.8273 = 82.4721) − 47(0.3745 The expected payoff for the call option is E[max(0.03  Problem 10 A stock’s price follows a lognormal model.6216)   1 − N (0.8273 = 61.6216) 1 − N (0.36 0. 2014 (iv) Volatility is 0.6216 Then N (−0. We use the formula E[St |St > K] = S0 e(α−δ)t N (dˆ1 ) N (dˆ2 ) Calculating dˆ1 and dˆ2 :
ln SK0 + (α − δ + 0.ACTS 4302.5σ 2 )t ln (60/70) + 0.