Tutorial 10 BSM Model Pricing Questions & Answers PDF

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Tutorial 10 BSM Model Pricing Questions & Answers...

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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 t is   t where  is the volatility. In this problem   0 3 and, assuming 252 trading days in one year, t  1  252  0 004 so that   t  0 3 0 004  0 019 or 1.9%. 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 rate is 10% per annum, and the volatility is 30% per annum. In this case S0  50 , K  50 , r  01,   0 3 , T  025 , and ln(50  50)  (0 1 0 09  2)0 25  0 2417 d1  0 3 025

d 2  d1  03 025  00917 The European put price is 50 N( 00917) e010 25  50 N ( 0 2417)  50  0 4634e  0 1 0 25  50  0 4045  2 37

or $2.37. 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 price 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 ,   030 and T  025 . 2

d1 

ln(52  50)  (0 12  0 3  2)0 25  0 5365 0 30 0 25

d2  d1  030 0 25  0 3865 The price of the European call is 52 N(0 5365)  50 e 0 12 0 25 N (0 3865) 0 03  52  0 7042  50e    0 6504  506 or $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 35% per annum, and the time to maturity is six months?

In this case S0  69 , K  70 , r  005 ,   035 and T  05 .

d1 

ln(69  70)  (005  0 352  2)  05  01666 035 05

d2  d1  035 05  00809 The price of the European put is 70 e0 0505 N(00809)  69 N ( 0 1666) 0 025  70e    0 5323  69  0 4338  640 or $6.40. Problem 15.18. Show that the Black–Scholes–Merton formulas for call and put options satisfy put–call parity.

The Black–Scholes–Merton formula for a European call option is c  S0 N ( d1)  Ke rT N ( d 2 ) so that c  Ke rT  S0 N( d1)  KerT N( d2)  KerT or c  Ke rT  S0 N ( d1 )  Ke rT [1  N ( d2 )] or c  Ke  rT  S 0 N (d1 )  Ke rT N ( d 2 ) The Black–Scholes–Merton formula for a European put option is p  Ke  rT N ( d 2 )  S 0 N ( d1) so that p  S 0  Ke  rT N ( d 2)  S 0 N ( d1)  S 0 or p  S 0  Ke  rT N ( d 2 )  S0[1  N ( d1 )] or p  S 0  Ke rT N (d 2 )  S 0 N (d1 ) This shows that the put–call parity result c  Ke rT  p  S0 holds. Problem 15.30. Consider an option on a non-dividend-paying stock when the stock price is $30, the exercise price is $29, the risk-free interest rate is 5% per annum, the volatility is 25% per annum, and the time to maturity is four months. a. What is the price of the option if it is a European call? b. What is the price of the option if it is an American call? c. What is the price of the option if it is a European put? d. Verify that put–call parity holds.

In this case S0  30 , K  29 , r  005 ,   025 and T  4  12

d1 

2 ln(30  29)  (0 05  0 25  2)  4  12  0 4225 025 0 3333

d2 

ln(30  29)  (0 05  0 252  2)  4  12  0 2782 025 0 3333 N(0 4225)  0 6637   N(0 2782)  0 6096

N( 0 4225)  0 3363  N( 0 2782)  0 3904

a. The European call price is 30  0 6637  29 e 0 05 4 12  0 6096  2 52 or $2.52. b. The American call price is the same as the European call price. It is $2.52. Because American call options are never exercised early when there are no dividends, they are equivalent to European call options. c. The European put price is  4 12 29 e 0 05 0 3904 30 0 3363  1 05 or $1.05. d. Put-call parity states that: p  S  c  Ke  rT rT In this case c  252 , S0  30 , K  29 , p  1 05 and e   0 9835 and it is easy to verify that the relationship is satisfied....