0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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 $-1\le R_A \le\infty$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.