August 8th, 2005, 2:55 pm
Ok, so under the risk neutral measuredS_t = r S_t dt + sigma S_t dWt (if you are doing this in the normal BS framework)Solving this SDE gives usS_t= S_0exp((r-0.5sigma^2)t + sigmaW_t)so So Ln(S_t/X) is normally distributed with mean = ln(S_0/X) +(r-0.5*sigma^2)t and variance sigma^2 tThe value V in the RN world isV = exp(-rt)E[Phi(S)] where Phi is the payoff functionSo, in this case V = exp(-rt)E[Ln(S_t/X) ; Ln(S_t/X>0]let Z_t = ln(S_t/X)so V = exp(-rt)E[Z_t; Z_t>0]= exp(-rt) *1/sqrt(2*Pi)*S * int_{0}^{infty}z * exp(-(z-M)^2/2*S^2 dzwhere M is the mean and S is the stdev of Zdo the integral, and substitute back in, and that is your answer.