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You can refer to Shreve's book, Volume II, Section 4.4.3 . Assume that we have a generalized geometric Brownian motion $$dX_t = \sigma_t dW_t + (\alpha_t - \frac{1}{2} \sigma_t^2) dt ,$$ where the drift coefficient and the volatility are functions of $t$ also. $(dX_t)^2 = \sigma_t^2 dt + \mathcal{O}(dt^{3/2})$ . Assume that the asset price is $$S_t = S_0 ... -2 under the REAL WORLD probability (measure), it is as you think: the log is μ△t under the RISK NEUTRAL measure, the mean is changed, so the stock price is a martingale, and the mean is (μ−1/2 σ^2)△t 1 As Quartz says it is possible to make non-linear transformations taking into account skew and kurtosis, but this is mostly is limited to univariate processes (one approach for a t distribution is to match moments). For multivariate processes, it is considerably more difficult. A more general solution is to rely on Entropy Pooling. You could take views on ... 3 It's not possible with a simple linear transformation like the one you mentioned: since scale and thus the distance between mean and median are required to change, either the mean or the median will not be preserved. Therefore you must use nonlinear transformations, which will complicate quite a bit mantaining skew and kurtosis and imho will not be ... 2 Another way of seeing it is that the -\frac12\sigma^2 is just a correction term that comes from Jensen's inequality. You need this when switching from supposedly symmetric returns (normal distribution) to the skewed price process (log-normal distribution). 3 The term 1/2 * sigma-squared arises through the application of Ito's Lemma. Keep in mind that the assumption is of a stock price that follows geometric BM with a constant drift and volatility. If you set up a delta-hedge portfolio and apply Ito calculus you will end up with an adjustment in the distribution by exactly above term. Another way of interpreting ... 6 So we have the BS-Model$$dS_t=S_t(\mu dt +\sigma dW_t)$$W.l.o.g we assume S_0=1. Itô's lemma implies that$$S_t=\exp{(\sigma W_t+(\mu-\frac{1}{2}\sigma^2)t)}$$We know that W_t is normally distributed with mean 0 and variance t. Now have a look at the r.v.$$X_t=\sigma W_t+(\mu-\frac{1}{2}\sigma^2)t $\sigma W_t$ is the random part and ...