Many an author claims that, if you model stock prices through GBM, $E[S(t)]=e^{\mu t}$, and the expectation is thus not related to volatility.

I keep running around in circles on this one. First of all, it seems intuitively to have some doubt. But I can argue it either way.

One thing that may affect this is that people are sloppy, I believe, in thinking about the solution to the SDE. the solution is

$$S(t)=S(0)e^{(\mu-\frac{\sigma^2}{2})t+\sigma W_t} $$

i.e., a lognormal distribution.

Suppose you were looking at a stock that went up from \$100 to \$105 last year with 20% volatility. It seems many people believe the parameters for the lognormal are thus $\mu=.05$ and $\sigma=.20$ But, it looks to me like the actual parameter that goes in for "mu" for the lognormal is really $.05-\frac{.20^2}{2}$, and to be more accurate it is a smaller number yet since continuous compounding has an impact (i.e., even if $\sigma$ were 0, a number slightly less than $.05$ would be the right rate, $ln(1.05)$ to be exact, so that continuous compounding gives you the 5% one-year return.

So, in that way, it seems like volatility in a GBM reduces returns,since it gets subtracted off.

On the other hand, a lognormal has a mean of $e^{\omega+\frac{\sigma^2}{2}}$, so if $\omega=\mu-\frac{\sigma^2}{2}$ you can convince yourself they do, indeed, cancel out, leaving $e^{\mu t}$. But if this is corect, is it true that the expected value of a price evolving under GBM has no dependence on volatility? If nothing else this seems hard to square with cases where vol is very high, so much so that $\mu-\sigma^2/2$ could become very negative (try $\mu=.10$ and $\sigma=.7$) thus pretty much seeming to guarantee that the $lim$ $t\to \infty$of $S(t)$ goes a.s. to zero.


1 Answer 1


Expectations is what we find when we average over all values of uncertainty. If you take a normally distributed variable $X \sim \mathcal{N}(\mu, \sigma^2)$, then it would not be a surprise that the mean is just $\mathbb{E}(X) =\mu$ no matter what $\sigma^2$ is. In this case the underlying model is that $X = \mu + \sigma \eta$ with $\eta \sim \mathcal{N}(0,1)$, we just observe it noisily.

Geometric Brownian motion is equivalently just a noisy version of a first-order ODE

$dS(t) = \mu S(t) dt$,

so it seems reasonable that the expectation would be equal to this simple underlying model of stock prices. This is just $S(t) = e^{\mu t}$ (we have assumed the constant initial price to be 1), which is exactly what you can laboriously derive from the SDE.

  • $\begingroup$ I understand what you're saying in the normal case but, in the lognormal case, the mean involves the variance. Therefore, since S(t) is assumed lognormal, one might think that it's expectation would involve sigma. Any intuition for why not ? Thanks. $\endgroup$
    – mark leeds
    Nov 16, 2018 at 17:47
  • $\begingroup$ He's right...what is confusing is that GBM has what I will call the "correction term" so that it looks like the return (at least the mean) will be affected owing to the $e^{(\mu - \frac{\sigma^2}{2}})$ in the first term. The key is knowing that $E[e^{\sigma W_t}]$ - the expectation of a lognormal - is $\mu+\sigma^2/2$ Thus, that part that seems to affect return via $\sigma$ is there so, when multiplied by the "shape" of the lognormal it doesn't drive the E[S(t)] ABOVE $e^{\mu t}$. $\endgroup$
    – eSurfsnake
    Nov 17, 2018 at 3:03

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