Why GBM Produces Normally Distributed Log Returns

Summary

The note derives the distribution of continuously compounded returns under geometric Brownian motion (GBM). Starting with a stock-price process whose drift and volatility are proportional to price, Itô’s lemma gives an exponential solution. Taking the logarithm of the price ratio turns that solution into a drift term plus a scaled Brownian motion.

Because Brownian motion at a fixed time is normally distributed, the log return is normal, with mean equal to drift minus half the variance rate, multiplied by elapsed time, and variance equal to the volatility squared times elapsed time. The result concerns log returns over a specified interval; it does not say that simple returns or stock prices are normally distributed. The explanation relies on GBM’s assumptions, so it is a model result rather than evidence that observed market returns always follow this distribution.

Key ideas

• Under GBM, the stock price is exponential in a Brownian motion with a drift adjustment.
• Taking the logarithm of the price ratio yields a linear function of Brownian motion.
• Log returns over a fixed interval are normal, with variance scaling with elapsed time.
• This distributional result depends on the GBM model assumptions.

Tags

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


# 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?

## Answer by Richi Wa (score 7)

https://quant.stackexchange.com/a/17288

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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