# 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. – Ric Apr 7 '15 at 15:02
• @Richard I would rather call it continuously compounded rate of return instead of log return. – ZHI Apr 7 '15 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. – Ric Apr 8 '15 at 8:02

## 1 Answer

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. – vonjd Apr 7 '15 at 16:10
• Your comment gives additional input. One could say so many things ... I will add one more comment.. – Ric Apr 8 '15 at 7:06