In finance, we often assume that the log-returns $\ln(1+R(t))$ follow a normal distribution.

Since $\ln(1+R(t)) \approx R(t)$ when $R(t)$ is small, \begin{equation*} dS/S \sim \text{Normal}. \end{equation*}

However, I have seen sometimes people assuming that \begin{equation*} \Delta S/S \sim \text{Normal}, \end{equation*}

so I wonder if the result holds in general (e.g. for percentage returns over a long time period, my understanding is that percentage return will follow a Normal distribution only when its value is small, i.e. for $dS/S$ ). In particular, what conclusion can we draw about the distribution of $\Delta S/S$ if we assume that log-returns $d\ln(s)$ follow a Normal distribution?


What is the mapping between log return $r_l$ and arithmetic return $R_A$? It is $r_l=\ln(1+R_A)$ and $R_A=e^{r_l}-1$.

If $r_l$ has the normal distribution then $e^{r_l}$ has the lognormal distribution (by definition) and $e^{r_l}-1=R_A$ has the "lognormal distribution shifted to the left by 1". I don't think there is a name for this distribution, which has support on $-1\le R_A \le\infty$.

  • $\begingroup$ Thank you for your answer. Is it correct that the assumption: $\Delta S/S∼Normal$ is valid only when $\Delta t$ or $\Delta S/S$ is small enough? In other conditions, $\Delta S/S$ will just follow a shifted lognormal distribution. $\endgroup$
    – Jason chiu
    Feb 5 '17 at 10:25
  • $\begingroup$ That's right. Try it yourself: the log of 1.02 is 1.98% (very close to 2%), the log of 1.10 is 9.53% (already somewhat different from 10%), the log od 1.4 is 33.65% (way off from 40%). So if the stock is not very volatile or the time interval is short so the movement is only a few percent, the approximation is OK. $\endgroup$
    – Alex C
    Feb 5 '17 at 16:47

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