Does GBM stock price model have E[S(t)] unaffected by volatility?

Question

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.

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.