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The solution to the above SDE is (this is will known and can be seen by applying Ito's lemma) $$ S_t = S_0 \exp\left( (u-\sigma^2/2) t + \sigma B_t \right), $$ Thus the log-return is given by $$ \log(S_t/S_0) = (u-\sigma^2/2) t + \sigma B_t $$ and is normally distributed as $B_t$, Brownian motion at time $t$, is normally distributed. In fact the distribution ...


The average of the exponentials is not the exponential of the average. It is always higher due to convexity (Jensen inequality). So there is no contradiction between the average of $X_T$ being negative and the average of $S_T$ being $S_0$. So the question is: are your results really significantly different from what you would expect? Have you tried ...

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