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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 ...


0

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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