A martingale can have a positive expected rate of return but a zero "drift" term (modeled as a geometric Brownian motion, e^(mu*t + sigma W_t), martingales have mu=0.)
yeah, no, that confused me too the first time, that's after a change of variables. the wiki page there says the prob distribution that has mean exp(mu*t) is exp((mu - sigma^2/2)t + sigma W_t). The drift term is *not* the same as the exponent in the mean.
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ok so there is geometric Brownian motion which is exp((mu - sigma^2/2)t + sigma W_t) and mean exp(mu*t). Mu/sigma are for the Brownian motion. Due to needing to preserve expectation a "negative drift" sigma^2/2t is introduced since exp is biased.
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Then there is your formula exp(mu*t + sigma W_t) which is geometric Brownian motion(GBM) but with an incorrect mu since it is missing the sigma^2/2 term. If you add back that term to your mu you'll have GBM with mu=mu'+sigma^2/2.
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