# What is the Kurtosis of Returns in Geometric Brownian Motion?

Suppose that $$dS_t=S_t(\mu\mathop{dt}+\sigma\mathop{dW_t})$$ which has solution $$S_t=S_0\exp\left(t\left(\mu+\frac{\sigma^2}{2}\right)+\sigma W_t\right),$$ such that $$W_t$$ is a Wiener process, $$\mu$$ is drift, and $$\sigma$$ is volatility. So can $$\begin{split} K(\delta)&=\text{kurt}\left(\frac{S_{t+\delta}-S_t}{S_t}\right),\\ &=\text{kurt}\left(e^{\frac{\delta \sigma^{2}-2\sigma W_t}{2}+\delta\mu+\sigma W_{t+\delta}}-1\right), \end{split}$$ be simplified for all $$\delta$$? In simulations, $$K$$ bounces around $$3$$ (for small $$\delta$$). Note that $$W_{t+s}-W_s$$ is independent of $$s$$. Any help would be much appreciated.

• If you use logarithmic returns instead of relative returns, the kurtosis is exactly 3. Apr 8, 2022 at 14:59
• i.e. $\text{kurt}(\ln(S_{t+\delta}/S_t))=3$, since returns are lognormal? Apr 8, 2022 at 15:23
• Yes. Prices are lognormal, relative returns are (shifted) lognormal, and log returns are of course normal; hence kurtosis equal to 3 Apr 8, 2022 at 16:05
• @Kermittfrog Thank you very much :D Apr 8, 2022 at 16:27