Why Geometric Brownian Motion Has Lower Log-Return Drift
Summary
The document explains why the drift of log returns under geometric Brownian motion is lower than the drift of percentage returns. Starting from the stochastic price process, it uses a small-change approximation for the logarithm of one plus a return. The quadratic term in that expansion contributes a volatility adjustment, producing log-return drift equal to the price process drift minus half the variance rate.
The answer gives an intuitive discrete-step argument: the random return includes a shock proportional to the square root of the time step, whose squared term is proportional to the time step. Replacing the squared standardized shock by its expected value yields the adjustment. This connects the intuition to Itô’s lemma. The derivation is an approximation for small time intervals and neglects higher-order terms; the document raises, but does not answer, whether the same equations can be used directly for finite discretization intervals.
Key ideas
- Geometric Brownian motion models percentage returns with drift and a volatility-scaled random shock.
- Expanding the logarithm of one plus a small return introduces a negative half-squared-return term.
- The expected squared standardized shock creates the volatility adjustment to log-return drift.
- The resulting log-return drift is the price drift minus half the variance rate.
- The discrete argument is approximate and relies on small time steps and neglected higher-order terms.
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# Geometric Brownian Motion: percentage returns vs log-returns
# 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*}
Thank you in advance.
## Answer by Gordon (score 8, accepted)
https://quant.stackexchange.com/a/32254
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*}Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.