This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Inputs Stock Price= S= $38.00 Exercise price = X= $35.00 Time to expiration = Tt= 3 months = 0.25 year Risk free rate = r = 6.00% standard deviation = volatility= s = 54.00% Calculations ln S/X= 0.0822 = ln (38/35) ( r + ½ s2 ) (Tt)= 0.0515 =(0.06+0.54^2/2) x 0.25 s x square root of (Tt)= 0.2700 =0.54 x square root of 0.25 Substituting the values d1= 0.4952 =(0.0822+0.0515) / 0.27 d2= 0.2252 =0.49520.27 N is the cumulative normal distribution function N(d1) 0.6898 N(d1)=1N(d1)= 0.3102 N(d2) 0.5891 N(d2)=1N(d2)= 0.4109 X * e r(Tt) = 34.4789 =35x e ^ (0.06x0.25) Thus, Value of call= S N(d1)  X * e r(Tt) * N(d2)= $5.9009 =38 x 0.6898  34.4789 x 0.5891 Value of put = X * e r(Tt) * N(d2) S N(d1) = $2.3798 =34.4789 x 0.4109  38 x 0.3102 We can also calculate the value of Put option using put call parity Put call parity c+ Xe^(rt) = p+S or p=c + Xe^(rt)Se^(q(Tt))= $2.3798 =5.9009+34.478938 which is the same as obtained above Answer: Value of call option= $5.9009 Value of put option= $2.3798...
View
Full Document
 Spring '10
 B.Mishra
 Normal Distribution, Options, Mathematical finance

Click to edit the document details