# Is return normal or log normal?

I read that return is normal and stock price is log normal. But I also read that return is log normal. So I am confused about which it is.

In the 14th Chapter of Options, Futures, and other Derivatives, John C Hull says

... the expected percentage return required by investors from a stock is independent of stock's price...

A reasonable assumption is that the variability of the return in a short period of time, $$\Delta t$$, is the same regardless of the stock price.

Based on these two observations, he got the model : $$\frac{\Delta S}{S}=\mu\Delta t + \sigma\epsilon\sqrt{\Delta t}$$

where $$S$$ is price, $$\epsilon$$ has a standard normal distribution. From this model, it seems like return has a normal distribution. This model also says $$S$$ follows a geometric Brownian motion and so we can apply Ito's lemma on some function of $$S, t$$. Applying Ito's Lemma to $$lnS$$ he arrived at the equation $$lnS_T - lnS_0\sim N((\mu-\frac{\sigma^2}{2})T, \sigma^2 T)$$ His conclusion from this is that price follows a log normal distribution. But it seems that return $$r$$ is also normal since $$lnS_T-lnS_0=ln\frac{S_T}{S_0}=lnr$$ is normal.

My question is, assuming the proposed model is right, does return have a normal or log normal distribution? Or is it a matter of time span, for small $$\Delta t$$ return is normal, and for longer periods $$T$$ return is log normal?

• If returns had a lognormal distribution, prices would never go down. May 19 at 10:32

• $$S_t$$ has a lognormal distribution for any $$t$$
• $$S_t/S_0$$ has a lognormal distribution for any $$t$$
• $$S_t/S_0$$ is approximately normal for any small $$t$$.
The equation with $$\Delta$$'s is less a definition of the model, and more a guideline to its numerical implementation.
• The model is still based on expected return being independent of stock price. The equation with deltas is one way of writing out the intuition, but it is not a consistent model on its own, because carrying it out for two steps with $\Delta t =1$ gives a different result from one step with $\Delta t =2$. May 5 at 12:47