期权期货与其他衍生产品第九版课后习题与答案Chapter (15).docx

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1、CHAPTER 15The Black-Scholes-Merton ModelPractice QuestionsProblem 15.1.What does the BlackScholesMerton stock option pricing model assume about the probability distribution of the stock price in one year? What does it assume about the probability distribution of the continuously compounded rate of r

2、eturn on the stock during the year?The BlackScholesMerton option pricing model assumes that the probability distribution of the stock price in 1 year (or at any other future time) is lognormal. It assumes that the continuously compounded rate of return on the stock during the year is normally distri

3、buted.Problem 15.2.The volatility of a stock price is 30% per annum. What is the standard deviation of the percentage price change in one trading day?The standard deviation of the percentage price change in time Dt is swhere sisDtthe volatility. In this problem s = 0.3 and, assuming 252 trading days

4、 in one year,DtDt = 1/ 252 = 0.004 so that s= 0.3 0.004 = 0.019 or 1.9%.Problem 15.3.Explain the principle of risk-neutral valuation.The price of an option or other derivative when expressed in terms of the price of the underlying stock is independent of risk preferences. Options therefore have the

5、same value in a risk-neutral world as they do in the real world. We may therefore assume that the world is risk neutral for the purposes of valuing options. This simplifies the analysis. In a risk-neutral world all securities have an expected return equal to risk-free interest rate. Also, in arisk-n

6、eutral world, the appropriate discount rate to use for expected future cash flows is the risk-free interest rate.Problem 15.4.Calculate the price of a three-month European put option on a non-dividend-paying stock with a strike price of $50 when the current stock price is $50, the risk-free interest

7、 rate is 10% per annum, and the volatility is 30% per annum.In this case S0= 50 , K = 50 , r = 0.1, s = 0.3, T = 0.25 , andd = ln(50 / 50) + (0.1+ 0.09 / 2)0.25 = 0.241710.3 0.25d2The European put price is= d - 0.3 0.25 = 0.0917150N (-0.0917)e-0.10.25 - 50N (-0.2417)= 50 0.4634e-0.10.25 - 50 0.4045

8、= 2.37or $2.37.Problem 15.5.What difference does it make to your calculations in Problem 15.4 if a dividend of $1.50 is expected in two months?In this case we must subtract the present value of the dividend from the stock price beforeusing BlackScholes-Merton. Hence the appropriate value of Sis0S =

9、50 -1.50e-0.16670.1 = 48.520As before K = 50 , r = 0.1, s = 0.3, and T = 0.25 . In this cased = ln(48.52 / 50) + (0.1+ 0.09 / 2)0.25 = 0.041410.3 0.25d = d21- 0.3 0.25 = -0.1086The European put price isor $3.03.50N (0.1086)e-0.10.25 - 48.52N (-0.0414)= 50 0.5432e-0.10.25 - 48.52 0.4835 = 3.03Problem

10、 15.6.What is implied volatility? How can it be calculated?The implied volatility is the volatility that makes the BlackScholes-Merton price of an option equal to its market price. The implied volatility is calculated using an iterative procedure. A simple approach is the following. Suppose we have

11、two volatilities one too high (i.e., giving an option price greater than the market price) and the other too low (i.e., giving an option price lower than the market price). By testing the volatility that is half way between the two, we get a new too-high volatility or a new too-low volatility. If we

12、 search initially for two volatilities, one too high and the other too low we can use this procedure repeatedly to bisect the range and converge on the correct implied volatility. Other more sophisticated approaches (e.g., involving the Newton-Raphson procedure) are used in practice.Problem 15.7.A s

13、tock price is currently $40. Assume that the expected return from the stock is 15% and its volatility is 25%. What is the probability distribution for the rate of return (with continuous compounding) earned over a two-year period?In this case m = 0.15 and s = 0.25 . From equation (15.7) the probabil

14、ity distribution for the rate of return over a two-year period with continuous compounding is:0.252 0.252 22j 0.15 -,i.e.,j(0.11875,0.03125)The expected value of the return is 11.875% per annum and the standard deviation is 17.7% per annum.Problem 15.8.A stock price follows geometric Brownian motion

15、 with an expected return of 16% and a volatility of 35%. The current price is $38.a) What is the probability that a European call option on the stock with an exercise price of $40 and a maturity date in six months will be exercised?b) What is the probability that a European put option on the stock w

16、ith the same exercise price and maturity will be exercised?a) The required probability is the probability of the stock price being above $40 in six months time. Suppose that the stock price in six months is ST0.352 ln ST j ln 38 + 0.16 -20.5, 0.352 0.5i.e.,ln ST() j 3.687, 0.2472Since ln 40 = 3.689

17、, we require the probability of ln(ST)3.689. This is1- N 3.689 - 3.687 = 1- N (0.008)0.247Since N(0.008) = 0.5032, the required probability is 0.4968.b) In this case the required probability is the probability of the stock price being less than$40 in six months time. It is1- 0.4968 = 0.5032Problem 1

18、5.9.Usingthe notation in the chapter, prove that a 95% confidence interval for STis betweenS e( m-s 2 / 2)T -1.96s TandS e( m-s 2 / 2) T +1.96s T00From equation (15.3):s 2 ln S jln S + m -T02 T ,s 2T 95% confidence intervals for ln STare thereforeandln S0ln S0+ (m - s 2 )T -1.96sT2T+ (m - s 2 )T +1.

19、96s295% confidence intervals for STare thereforei.e.eln S0 +(m-s 2 / 2)T -1.96s Tandeln S0 +(m-s 2 / 2)T +1.96s TProblem 15.10.S e( m-s 2 / 2)T -1.96s T0andS e( m-s 2 / 2) T +1.96s T0A portfolio manager announces that the average of the returns realized in each of the last 10 years is 20% per annum.

20、 In what respect is this statement misleading?This problem relates to the material in Section 15.3. The statement is misleading in that a certain sum of money, say $1000, when invested for 10 years in the fund would have realized a return (with annual compounding) of less than 20% per annum.The aver

21、age of the returns realized in each year is always greater than the return per annum (with annual compounding) realized over 10 years. The first is an arithmetic average of the returns in each year; the second is a geometric average of these returns.Problem 15.11.Assume that a non-dividend-paying st

22、ock has an expected return of mand a volatility of s . An innovative financial institution has just announced that it will trade a derivative that paysToff a dollar amount equal to ln Sat time T where STdenotes the values of the stockprice at time T .a) Use risk-neutral valuation to calculate the pr

23、ice of the derivative at time t in term of the stock price, S, at time tb) Confirm that your price satisfies the differential equation (15.16)a) At time t, the expected value of ln STis from equation (15.3) ln S + (m -s 2 / 2)(T - t)In a risk-neutral world the expected value ofln ST is therefore ln

24、S + (r -s 2 / 2)(T - t)Using risk-neutral valuation the value of the derivative at time t ise-r (T -t )ln S + (r -s 2 / 2)(T - t)b) Ifthenf = e- r (T -t )ln S + (r -s 2 / 2)(T - t)f()t = re- r (T -t )ln S + (r -s 2 / 2)(T - t) - e-r (T -t ) r -s 2 / 2f = e- r (T -t )SS2 fS 2= - e- r (T -t )S 2The le

25、ft-hand side of the Black-Scholes-Merton differential equation ise- r (T -t ) r ln S + r(r -s 2 / 2)(T - t) - (r -s 2 / 2) + r -s 2 / 2= e- r (T -t ) r ln S + r(r -s 2 / 2)(T - t)= rfHence the differential equation is satisfied.Problem 15.12.Consider a derivative that pays off S n at time T where Si

26、s the stock price at that time.TTWhen the stock pays no dividends and its price follows geometric Brownian motion, it can beshown that its price at time t (t T ) has the formh(t,T )Snwhere S is the stock price at time t and h is a function only of t and T .(a) By substituting into the BlackScholesMe

27、rton partial differential equation derive an ordinary differential equation satisfied by h(t,T ) .(b) What is the boundary condition for the differential equation for h(t,T ) ?(c) Show thath(t,T ) = e0.5s 2n(n-1)+r (n-1)(T -t )where r is the risk-free interest rate and sis the stock price volatility

28、.If G(S, t) = h(t,T )S n then G / t = h S n , G / S = hnS n-1 , and 2G / S 2 = hn(n -1)Sn-2twhere ht= h / t . Substituting into the BlackScholesMerton differential equation weobtain1h + rhn +s 2 hn(n -1) = rht2The derivative is worth S n when t = T . The boundary condition for this differential equa

29、tion is therefore h(T ,T ) = 1The equationh(t,T ) = e0.5s 2n(n-1)+r (n-1)(T -t )satisfies the boundary condition since it collapses to h = 1 when t = T . It can also be shown that it satisfies the differential equation in (a). Alternatively we can solve the differential equation in (a) directly. The

30、 differential equation can be written1th = -r(n -1) -s 2 n(n -1)The solution to this isorh2ln h = r(n -1) + 1 s 2n(n -1)(T - t) 2h(t,T ) = e0.5s 2n(n-1)+r (n-1)(T -t )Problem 15.13.What is the price of a European call option on a non-dividend-paying stock when the stock price is $52, the strike pric

31、e is $50, the risk-free interest rate is 12% per annum, the volatility is 30% per annum, and the time to maturity is three months?In this case S0= 52 , K = 50 , r = 0.12 , s = 0.30 and T = 0.25 .d = ln(52 / 50) + (0.12 + 0.32 / 2)0.25 = 0.536510.30 0.25d = d21- 0.30 0.25 = 0.3865The price of the Eur

32、opean call is52N (0.5365) - 50e-0.120.25 N (0.3865)= 52 0.7042 - 50e-0.03 0.6504= 5.06or $5.06.Problem 15.14.What is the price of a European put option on a non-dividend-paying stock when the stock price is $69, the strike price is $70, the risk-free interest rate is 5% per annum, the volatility is

33、35% per annum, and the time to maturity is six months?In this case S0= 69 , K = 70 , r = 0.05 , s = 0.35 and T = 0.5 .d = ln(69 / 70) + (0.05 + 0.352 / 2) 0.5 = 0.166610.35 0.5d = d21- 0.35 0.5 = -0.0809The price of the European put is70e-0.050.5 N (0.0809) - 69N (-0.1666)= 70e-0.025 0.5323 - 69 0.4

34、338= 6.40or $6.40.Problem 15.15.Consider an American call option on a stock. The stock price is $70, the time to maturity is eight months, the risk-free rate of interest is 10% per annum, the exercise price is $65, and the volatility is 32%. A dividend of $1 is expected after three months and again

35、after six months. Show that it can never be optimal to exercise the option on either of the two dividend dates. Use DerivaGem to calculate the price of the option.Using the notation of Section 15.12, D = D212= 1, K (1- e- r (T -t ) ) = 65(1- e-0.10.1667 ) = 1.07 ,and K (1- e- r (t -t ) ) = 65(1- e-0

36、.10.25 ) = 1.60 . Since2 1andD K (1- e- r (T -t2) ) 1D K (1- e- r (t -t ) )2 12It is never optimal to exercise the call option early. DerivaGem shows that the value of the option is 10.94 .Problem 15.16.A call option on a non-dividend-paying stock has a market price of $2.50. The stock price is$15,

37、the exercise price is $13, the time to maturity is three months, and the risk-free interest rate is 5% per annum. What is the implied volatility?In the case c = 2.5 , S0= 15 , K = 13 , T = 0.25 , r = 0.05 . The implied volatility must becalculated using an iterative procedure.A volatility of 0.2 (or

38、 20% per annum) gives c = 2.20 . A volatility of 0.3 gives c = 2.32 . A volatility of 0.4 gives c = 2.507 . A volatility of 0.39 gives c = 2.487 . By interpolation the implied volatility is about 0.396 or 39.6% per annum.The implied volatility can also be calculated using DerivaGem. Select equity as

39、 the Underlying Type in the first worksheet. Select Black-Scholes European as the Option Type. Input stock price as 15, the risk-free rate as 5%, time to exercise as 0.25, and exercise price as13. Leave the dividend table blank because we are assuming no dividends. Select the button corresponding to

40、 call. Select the implied volatility button. Input the Price as 2.5 in the second half of the option data table. Hit the Enter key and click on calculate. DerivaGem will show the volatility of the option as 39.64%.Problem 15.17.With the notation used in this chapter(a) What is N (x) ?(b) Show that S

41、N (d ) = Ke- r (T -t ) N (d12) , where S is the stock price at time to T - td = ln(S / K ) + (r +s 2 / 2)(T - t)1d = ln(S / K ) + (r -s 2 / 2)(T - t)2s T - t(c) Calculate d1/ S and d2/ S .(d) Show that whenc = SN (d ) - Ke- r (T -t ) N (d )12cs= -rKe-r (T -t ) N (d ) - SN (d )t21 2 T - twhere c is t

42、he price of a call option on a non-dividend-paying stock.(e) Show that c / S = N (d ) .1(f) Show that the c satisfies the BlackScholesMerton differential equation.(g) Show that c satisfies the boundary condition for a European call option, i.e., thatc = max(S - K , 0) as t tends to T.(a) Since N (x)

43、 is the cumulative probability that a variable with a standardized normal distribution will be less than x , N (x) is the probability density function for a standardized normal distribution, that is,2pN (x) =1e- x2 2(b) N (d ) = N (d12p12 d 2+sT - t )1T - t=exp - 2 -s d- 2 s 2 (T - t)22T - t1Because

44、= N (d2 )exp -s d2- 2 s (T - t)2it follows thatdln(S K ) (r2 2)(Tt)Tt2expd12 (Tt)Ke r(T t)Tt22SAs a resultSN (d )Ke r(T t)N (d )12which is the required result.(c) (c)K2Ttdln S 1(r2 )(Tt)2TtlnSlnK(r2 )(Tt)Henced1STt1SSimilarlyanddlnSlnK(r2Tt22)(Tt)d1STt2STherefore:dd12SS(d) (d)cSN (d ) Ke r(T 1cdt)N

45、(d )2dtFrom (b):SN (d )11trKe r(T t)N (d2) Ker(T t)N (d )22tHenceSN (d )Ke r(T t)N (d )12c ddrKe r(T t)N (d ) SN (d )12t21ttSinceTtdd12(Tt)dd12ttt2 T - tHence= -scs= -rKe-r (T -t ) N (d ) - SN (d )t21 2 T - t(e) From differentiating the BlackScholesMerton formula for a call price we obtaincdd= N (d ) + SN (d )1 - Ke- r (T -t