If a stock price is modelled with a geometric brownian motion process with this definition:
$GBM(t)=s_0 e^{X(t)}$
where $X(t)$ is a brownian process $N(\mu – \sigma^2/2,\sigma)$, then doesn't this mean that when $t$ tends to infinite and $\mu=0$, GBM(t) tends to zero?
I am confused since $\mu=0$ is supposed to mean "no drift", but the price will always drift to zero.
This is a follow up question about different ways to model a GBM here.
Thanks
Best Answer
Yes, the set $\{\omega\mid GBM(t)(\omega)\to 0\}$ has probability 1.
This is not a contradiction to "there is no drift", because at every finite time you have a (small) nonzero probability of the stock price being sky-rocketed to compensate for much larger probability that the stock is almost worthless. Note that "there is no drift" means what happens to expectation, but you can't push the limit inside the integral because the hypotheses of dominated convergence theorem fails.
A discrete-time martingale should help you understand this. Suppose the stock price either become worthless or double on each time step, both with probability $\frac12$. So it is a martingale (therefore has no drift), but the price is almost surely 0 in the long run.