I am having trouble understanding a concept from Shreve Volume II (page 148 on the bottom)The text says the following:"If alpha and theta are const, we have the usual geometric Brownian motion model, and the distribution of S(t) is log-normal. In the case where alpha and theta are constant,S(t) = S(0)*exp{ theta*W(t) + ( alpha - 1/2*theta*theta)*t }One can incorrectly argue from this formula that since Brownian motion is a martingale (ie., it has no overall tendency to rise or fall), the mean rate of return for S(t) must be (alpha - 1/2*theta*theta). The error in this argument is that although W(t) is a martingale, S(0)*exp{theta*W(t) } is no a martingale. The convexity of the function exp{ theta*x}imparts an upward drift to S(0)*exp{theta*W(t) }. In order to correct for this, one must subtract 1/2*theta*theta*d in the exponential"I do not understand why this is true? Can anyone explain it better?Thanks

Focus on the later part and notice S(t) is a process with mean rate of return alpha. The (-1/2)*(volatility^2) is to make S(0)exp{...} (not have alpha*t) is a exponential martingale.Please correct if I am wrong ^_^

Well this argument is a big pain. I would suggest something more brutal, directly evaluate the expectation that you want by simple integration

E[St]=S0e^(alpha-1/2*theta^2)t*E[e^(theta Wt)]So if you think that E[St]=S0e^(alpha-1/2*theta^2)t, you are assuming that E[e^(theta Wt)]=e^(theta E[Wt])=e^(theta*0)=1. But this assumption is wrong: in general E[f(X)] != f(E[X]). Google "graphical proof of Jensen's inequality" if you are not sure why this is.In fact E[e^(theta Wt)]=e^(1/2*theta^2)t, (use the moment generating function of a normal variable to prove this) and so E[St]=S0e^alpha*t.