# Geometric Brownian Motion: percentage returns vs log-returns

In classical calculus, we know that the limit of percentage return (ie $dS/S$) equals that of the log return (ie. $dln(S)$ ).

With uncertainty, we rely on Ito Lemma to draw a relationship between the two:

\begin{equation*} dS = \mu S dt + \sigma Sdz \end{equation*}

and

\begin{equation*} dln(S) = (\mu - \sigma^2/2) dt + \sigma dz \end{equation*}

I understand the mathematics behind but I would like to know more about the intuition, mainly

with uncertainties, when we "switch" from percentage return to log return, why do we have a smaller drift $(\mu - \sigma^2/2)$? Is there any intuition or financial sense behind?

Moreover, when we discretize the process, can we draw the same relationship and say something like \begin{equation*} \Delta S = \mu S \Delta t + \sigma S \Delta z \end{equation*}

and \begin{equation*} \Delta ln(S) = (\mu - \sigma^2/2) \Delta t + \sigma \Delta z \end{equation*}

The percentage return over the infinitesimal interval $[t, t+dt]$ is given by \begin{align*} \frac{S_{t+dt} - S_t}{S_t} \approx \mu dt + \sigma \sqrt{dt} \xi, \end{align*} where $\xi$ is a standard normal random variable. On the log-return, note that, for $x$ sufficiently small, \begin{align*} \ln (1+x) \approx x -\frac{x^2}{2}, \end{align*} then, by ignoring the higher order terms (relative to $dt$), \begin{align*} \ln \frac{S_{t+dt}}{S_t} &= \ln \left(1+ \frac{S_{t+dt} - S_t}{S_t} \right)\\ &\approx \frac{S_{t+dt} - S_t}{S_t} -\frac{1}{2} \left( \frac{S_{t+dt} - S_t}{S_t}\right)^2\\ &\approx \mu dt + \sigma \sqrt{dt} \xi -\frac{1}{2} \left(\mu dt + \sigma \sqrt{dt} \xi\right)^2\\ &\approx \mu dt + \sigma \sqrt{dt} \xi -\frac{1}{2}\sigma^2\xi^2 dt\\ &\approx \left(\mu - \sigma^2/2 \right)dt + \sigma \sqrt{dt} \xi. \end{align*} Here, we assume that \begin{align*} \xi^2 \approx E(\xi^2) = 1. \end{align*}
• @Gordon Thank you for your great explanation. I just wanna ask why can we safely assume that $\xi^2 \approx E(\xi^2) = 1$? – Jason chiu Feb 7 '17 at 15:35
• And can I intuitively understand the fact that log return has a smaller drift by the curvature of $\exp(x)$ :because $\exp(x)$ is increasing exponentially, so when we transform a normal random variable in such a way, the original normal distribution is skewed by $\exp(x)$ and hence shifting the mean to the right. The degree of Skewness will depend on the curvature of $\exp(x)$ of the region, which is in turn affected by the variance of the normal random variable. When the variance is zero, the transformation is linear and hence there is no shifting of mean. – Jason chiu Feb 7 '17 at 15:47