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

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.

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.
$$dS(t) = \mu S(t) dt$$,
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.
• 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 $e^{\mu t}$. – eSurfsnake Nov 17 '18 at 3:03