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.
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1$\begingroup$ If you use logarithmic returns instead of relative returns, the kurtosis is exactly 3. $\endgroup$– KermittfrogApr 8, 2022 at 14:59
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$\begingroup$ i.e. $\text{kurt}(\ln(S_{t+\delta}/S_t))=3$, since returns are lognormal? $\endgroup$– UNOwenApr 8, 2022 at 15:23
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1$\begingroup$ Yes. Prices are lognormal, relative returns are (shifted) lognormal, and log returns are of course normal; hence kurtosis equal to 3 $\endgroup$– KermittfrogApr 8, 2022 at 16:05
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$\begingroup$ @Kermittfrog Thank you very much :D $\endgroup$– UNOwenApr 8, 2022 at 16:27
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