1 The Black-Scholes Model Chapter 13

2 Pricing an European Call The Black&Scholes model Assumptions: 1.European options. 2.The underlying stock does not pay dividends during the option’s life. Time line tTtT c t Max{0, S T – K}

3 Pricing an European Call The Black&Scholes model c t = NPV[Expected cash flows] c t = NPV[ E (max{0, S T – K})]. c t = e -r(T-t) max{ E (0, S T – K)}.

4 Pricing an European Call The Black&Scholes model C t =e -r(T-t) { [0  S T  K]Pr(S T  K) + E [(S T –K)  S T >K]Pr(S T >K) } c t = e -r(T-t) [ E (S T –K)  S T >K]Prob(S T >K)

5 Pricing an European Call The Black&Scholes model c t = e -r(T-t) E [(S T )  S T >K]Prob(S T > K) – e -r(T-t) KProb(S T > K)

6 The Stock Price Assumption(p282) In a short period of time of length  t the return on the stock is normally distributed: Consider a stock whose price is S

7 The Lognormal Property It follows from this assumption that Since the logarithm of S T is normal, S T is lognormally distributed

8 The Lognormal Distribution

9 Black&Scholes formula: c t = S t N(d 1 ) – Ke -r(T-t) N(d 2 ) d 1 = [ ln(S t /K) + (r +.5  2 )(T – t) ] /  √(T – t) d 2 = d 1 -  √(T – t), N(d) is the cumulative standard normal distribution. All the parameters are annualized and continuous in time

10 p t = Ke -r(T-t) N(- d 2 ) – S t N(- d 1 ). d 1 = [ ln(S t /K) +(r +.5  2 )(T – t) ] /  √(T – t) d 2 = d 1 -  √(T – t), N(d) is the cumulative standard normal distribution. All the parameters are annualized and continuous in time

11 INTEL Thursday, September 21, 2000.S = $61.48 CALLS - LAST PUTS - LAST K OCT NOV JAN APROCT NOV JAN APR

12 INTEL Thursday, September 21, S = $61.48 S = 61.48:K = 65: JAN: T-t = 121/365 =.3315yrs; R = 4.82%  r = ln[1.0482] =.047.  = 48.28% d 1={ln(61.48/65) + [ ( )].3315} / (.4828)  (.3315) = d 2 = d 1 – (.4828)  (.3315) = N( d 1) =.4979;N( d 2) =.3885 c= 61.48(.4979) – 65e -(.047)(.3315) (.3885)= 5.75 p = e- (.047)(.3315) = 8.26

13 INTEL Thursday, September 21, S = $61.48 S = 61.48:K = 70: JAN: T-t = 121/365 =.3315yrs; R = 4.82%  r = ln[1.0482] =.047.  = 48.28%. d 1={ln(61.48/70) + [ ( )].3315} / (.4828)  (.3315) = d 2 = d 1 – (.4828)  (.3315) = Next, we follow the extrapolations suggested in the text: N(-.2718) = ? N(-.5498) = ?

14 N( d 1 ) = N(-.2718) = N(-.27) -.18[N(-.27) – N(-.28)] = [ ] = N( d 2 ) = N(-.5498) = N(-.54) -.98[N(-.54) – N(-.55)] = [ ] = c t = S t N(d 1 ) – Ke -r(T-t) N(d 2 ) c = 61.48( ) – 65e -(.047)(.3315) ( ) = 4.04 p t = Ke -r(T-t) N(- d 2 ) – S t N(- d 1 ) BUT Employing the put-call parity for European options on a non dividend paying stock, we have: p = e- (.047)(.3315) = 11.81

15 Black and Scholes prices satisfy the put-call parity for European options on a non dividend paying stock: c t – p t = S t - Ke -r(T-t). Substituting the Black&Scholes values: c t = S t N(d 1 ) – Ke -r(T-t) N(d 2 ) p t = Ke -r(T-t) N(- d 2 ) – S t N(- d 1 ). into the put-call parity yields:

16 c t – p t = S t N(d 1 ) – Ke -r(T-t) N(d 2 ) - [Ke -r(T-t) N(- d 2 ) – S t N(- d 1 )] c t - p t = S t N(d 1 ) – Ke -r(T-t) N(d 2 ) - Ke -r(T-t) [1 - N(d 2 )] + S t [1 - N(d 1 ) c t - p t = S t – Ke -r(T-t)

17 The Inputs S t = The current stock price K =The strike price T – t = The years remaining to expiration r = The annual, continuously compounded risk-free rate  =The annual SD of the returns on the underlying asset

18 The Inputs S t = The current stock price Bid price? Ask price? Usually: mid spread

19 The Inputs T – t = The years remaining to expiration Black and Scholes: continuous markets  1 year = 365 days. Real world : the markets are open for trading only 252 days.  1 year = 252 days.

20 The inputs 1. r continuously compounded 2.R simple.  r = ln[1+R]. 3.R a ; R b ; n = the number of days to the T-bill maturity.

21 The Volatility (VOL) The volatility of an asset is the standard deviation of the continuously compounded rate of return in 1 year As an approximation it is the standard deviation of the percentage change in the asset price in 1 year

22 Historical Volatility (page 286-9) 1.Take n+1observed prices S 0, S 1,..., S n at the end of the i-th time interval, i=0,1,…n. 2.  is the length of time interval in years. For example, if the time interval between i and i+1 is one week, then  = 1/52, If the time interval is one day, then  = 1/365.

23 Historical Volatility 1.Calculate the continuously compounded return in each time interval as:

24 Historical Volatility Calculate the standard deviation of the u i ´s:

25 Historical Volatility The estimate of the Historical Volatility is:

26 Implied Volatility (VOL) (p.300) The implied volatility of an option is the volatility for which the Black-Scholes price equals the market price There is a one-to-one correspondence between prices and implied volatilities Traders and brokers often quote implied volatilities rather than dollar prices

27 c t = S t N(d 1 ) – Ke -r(T-t) N(d 2 ) d 1 = [ln(S t /K) + (r +.5  2 )(T – t)]/  √(T – t), d 2 = d 1 -  √(T – t). Inputs: S t ; K; r; T-t; and c t The solution yields: Implied Vol.

28 Dividend adjustments (p.301) 1.European options When the underlying asset pays dividends the adjustment of the Black and Scholes formula depends on the information at hand. Case 1.The annual dividend payout ratio, q, is known. Use: Where S t is the current asset market price

29 Dividend adjustment 2.European options Case 2.It is known that the underlying asset will pay a series of cash dividends, D i, on dates t i during the option’s life. Use: Where S t is the current asset market price

30 Dividend adjustment 2.American options: T= the option expiration date. t n = date of the last dividend payment before T. Black’s Approximationc=Max{1,2} 1.Compute the dividend adjusted European price. 2.Compute the dividend adjusted European price with expiration at t n ; i.e., without the last dividend.