# Why is rate of return on the stock normally distributed under GBM?

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?

• I edited the question a bit to make it clearer - I hope, I got it right. Apr 7, 2015 at 15:02
• @Richard I would rather call it continuously compounded rate of return instead of log return.
– ZHI
Apr 7, 2015 at 15:07
• 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. Apr 8, 2015 at 8:02

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.

• +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. Apr 7, 2015 at 16:10
• Your comment gives additional input. One could say so many things ... I will add one more comment.. Apr 8, 2015 at 7:06