Start with the log return of a process, ln(S_T/S_t)

Assume that S follows the process dS_t = µdt + σdW_t, where µ is drift, σ is volatility, and W_t is a random process that is N~(0,1)

Find the stochastic derivative of the process ln(S) using the Ito formula, df_{t =} f_t  + f_s + 1/2  f_ss :

             dln(S) = 1/S(dS) – (1/2)(1/S²)(dS²)

             dln(S) = (µ – ½ σ²)dt + σdW_t

This says that the logarithm of a random process that is normally distributed is concave and has a mean of (µ – ½ σ²) and that

             ln(S_T/S_t) = ln(S1) – ln(S0) = (µ – ½ σ²)(T-t) + σ(W(T) – W(t))

             ln(S_T) = ln(S_t) + (µ – ½ σ²)(T-t) + σ(W(T) – W(t))

Assume that S_t is lognormally distributed, meaning it is an exponentiated normal process.  Take the exponential of both sides of the above equation:

             S_T = S_te^{(µ – 1/2σ^2)dt + σdWt}

The lognormal process Se^X, X ~ N(µ, σ), is convex and has a mean of (µ + ½ σ²), so the above equation has an expectation of

             S_T = S_te^µt

since the- ½ σ²  concavity of the log of a normal and the + ½ σ² implicit convexity in the exponentiated normal cancel each other and the expectation of dW_t is 0.  This is necessary for the process S_t to be a martingale when µ = 0.