1. Lec 13a Black & Scholes OPM
Lec 13A: Black and Scholes OPM (Stock pays no dividends) (Hull,Ch 13)
From Binomial model to Black-Scholes
Suppose S0 = 40, and r = 24%/yr (c.c.).
Question: What is the price of a CE(K=$40, T= 1-yr) ?
One simple solution: Use the Binomial OPM (with R-N probabilities).
Case I. Break 1 year period into 1 period
Let u = 1.35, d = 0.74, then R-N prob = 0.87, T = 1
Stock Price Tree R-N Prob Call Values (K=$40)
53.99 ➀ p =
C0 = ?
29.63 ➁ 1 - p =
Then, C0 = e-0.24 [ 13.99(0.87) + 0(0.13) ] = $9.59
Lec 13a Black & Scholes OPM

2. Lec 13a Black & Scholes OPM
Case II. Break 1 year into 2 sub-periods.
Set u = 1.236, d = ; then p =
Stock Price Tree R-N Prob Values at Expiration
➀ p2 = ➀ ➀
49.45
➁ p(1-p) = ➁ C0 = ? ➁
32.35
26.17 ➂ (1-p)2 = ➂ ➂
Call Value C0 = e-0.24 [ 21.14(0.556) + 0(0.380) + 0(0.065) ] = $9.24
Lec 13a Black & Scholes OPM

3. Lec 13a Black & Scholes OPM
Case III. Break 1 year into 3 sub-periods.
u =e0.30√(1/3) =eσ√T/n = 1.189, d= 1/u i.e. d=e-0.30√(1/3) =e -σ√T/n = 0.841, and
p ={e0.24(1/3) -d}/{u-d} = { } /{ }= 0.696
Call Value C0 = e-0.24 [ 27.26(0.3372) (0.4418) + 0 ] = $9.86
Stock Price Tree
R-N Prob
Call Values
67.26
p3 =
27.26
56.56
47.56
3p2(1-p) =
7.56 40
C0 = ?
33.64
3p(1-p)2 =
28.29
23.79
(1-p)3 =
Lec 13a Black & Scholes OPM

4. Lec 13a Black & Scholes OPM
Limit Case: Black-Scholes European Call Option formula (p. 4)
as n → ∞, the binomial model becomes
C0 = S0 N(d1 ) – K e-rT N(d2 )
d1 = [ ln(S0 /K) +(r+σ2/2) T ]/(σ √T) , and d2 = [ ln(S0 /K) +(r-σ2/2) T ]/(σ √T)
T = time to expiration in years (or fraction of a year)
σ2 = annualized variance of stock returns
N(d1)= area under graph of standard normal cumulative prob ( -∞ to d1 )
N(d2)= area under graph of standard normal cumul. probability (-∞ to d2)
in Excel =NORMSDIST(d1)
r= c.c. risk-free rate
Lec 13a Black & Scholes OPM

5. Lec 13a Black & Scholes OPM
Limit Case: Black-Scholes European Call Option formula (p. 4)
C0 = S0 N(d1 ) – K e-rT N(d2 )
Example: Use same data as in cases I, II etc. to price a European Call
S0 = $40, K = $40, T = 1 (1 year), σ2 = 0.09/year ⇒ σ = 0.30/yr
d1 = [ ln(40/40) +( /2) 1 ]/(0.30 √1) = 0.950
d2 = [ ln(40/40) +( /2) 1 ]/(0.30 √1) = 0.65
N(d1) = Area up to 0.95 = , N(d2) = Area up to 0.65 =
∴ C0 = 40 (0.8289) - 40(e-0.24 (1)) ( ) = $9.81
Compare to Binomial prices:
n = 1, Binomial price = $ n = 2, Binomial price = $9.24
n = 3, Binomial price = $9.86 n = 4, Binomial price = $ NOT BAD!!
Lec 13a Black & Scholes OPM

6. Lec 13a Black & Scholes OPM
Black-Scholes model for European Puts (p. 4)
Recall the Put-Call relationship: +S +P = +C +B.
Solve for the price of the put:
P0 = S0 [1-N(d1 ) ] – K e-rT [1-N(d2 ) ]
Alternatively, use the fact that N(K) + N(-K) = 1, then
P0 = -S0 N(-d1 ) + K e-rT N(-d2 )
Example: Use same data to price a European Put
S0 = $40, K = $40, T = 1 (1 year), σ2 = 0.09/year ⇒ σ = 0.30/yr
P0 = -40( ) + 40(e-0.24 (1)) ( ) = $1.27
Lec 13a Black & Scholes OPM

7. Lec 13a Black & Scholes OPM
Thank You
(A Favara)
Lec 13a Black & Scholes OPM