Hi, Suppose S follows a geometric Brownian motion, i.e. dS = mu S dt + sigma S dz What is the process followed by y = exp(S). I come this far: dy = (Dy/DS) dS + 0.5 (D^2 y/ DS^2) (dS)^2 Since both (Dy/Ds) = (D^2 y/ DS^2) = exp(S) = y, it follows that: dy = y dS + 0.5 y (dS)^2 = y [mu S dt + sigma S dz] + 0.5 y sigma^2 S^2 dt. My question, how do I get rid of the S in the final dy formula - simply replace it by lny? Thanks.
Ito for y=exp(S) Yes, I'd substitute in S=ln(y) and then group the dt and dz terms together. It seems an unusual process to look at though. S(t) is already an exponential process, so y would be doubly exponential. Is there a particular reason you're looking at this function or is it just for Ito practice (which is fine - nothing wrong with that)?
