Let us assume the geometric Brownian motion, and we have $$dS_t= uS_tdt+\sigma S_tdz,$$ and $S_t$ follows a log-normal distribution, but why is $r_t$, the continuously compounded rate of return, normally distributed?
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$\begingroup$ I edited the question a bit to make it clearer - I hope, I got it right. $\endgroup$– Richi WaApr 7, 2015 at 15:02
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$\begingroup$ @Richard I would rather call it continuously compounded rate of return instead of log return. $\endgroup$– ZHIApr 7, 2015 at 15:07
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$\begingroup$ It is absolutely usual to call it log-return as far as I know ... it was simply necessary to say what $r_t$ could mean ... please formulate your question clearly in the future. $\endgroup$– Richi WaApr 8, 2015 at 8:02
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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 of the expression above is $N( (u-\sigma^2/2) t, t \sigma^2)$. If we assume that $t=1$ (one day or one year) then we get $N( (u-\sigma^2/2),\sigma^2)$ and have the interpretation of the parameters in this frequency.
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1$\begingroup$ +1, perhaps you should add that the log of a log-normally distributed random variable is normally distributed, that log of a division is equal to the difference between the logs of both values and that the difference of two normally distributed random variables is again normally distributed. $\endgroup$– vonjdApr 7, 2015 at 16:10
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1$\begingroup$ Your comment gives additional input. One could say so many things ... I will add one more comment.. $\endgroup$– Richi WaApr 8, 2015 at 7:06