Expected value of geometric Brownian motion

Question

So everyone knows that the expected value of GBM is given by:

$X_0 \exp(\mu t)$

My question is that what does this say about such stochastic processes? Since $X_0$ and $\mu$ are within "my control" to choose, could I not always just manipulate this expression to get the expected value I want? Therefore, are these processes "really" stochastic?

Answer 1

If $\{B_t:t\in\mathbb R_+\}$ is a standard Brownian motion, then $$X_t := X_0\exp\left(\left(\mu-\frac12\sigma^2\right)t+\sigma B_t\right) $$ (where $X_0$, $\mu$, and $\sigma$ are constants) defines a geometric Brownian motion. It is clear that if $X_0=0$ then $X_t$ is identically zero, and similarly if $\sigma=0$ then $X_t = X_0 e^{\mu t}$ with probability one. Assuming that $X_0\ne 0$ and $\sigma>0$, though, $X_t$ has log-normal distribution with \begin{align} \mathbb E[X_t] &= X_0 e^{\mu t},\\ \operatorname{Var}(X_t) &= X_0^2 e^{2\mu t}\left(e^{\sigma^2 t} -1\right). \end{align} So $X=\{X_t:t\in\mathbb R_+\}$ is a collection of random variables indexed by a totally ordered set, and indeed is a stochastic process. The limiting behavior of $X_t$ is determined by the value of the parameters, however. From the law of the iterated logarithm $$\mathbb P\left(\limsup_{n\to\infty}\frac{B_t}{\sqrt{2t\log\log t}}=1\right)=1, $$ and so almost sure convergence of $X_t$ is determined by the $\left(\mu-\frac12\sigma^2\right)t$ term:

• If $\left(\mu-\frac12\sigma^2\right)>0$, then $\lim_{t\to\infty} \left(\mu-\frac12\sigma^2\right)t=\infty$ and so $X_t\stackrel{\mathrm{a.s.}}\longrightarrow\infty$.
• If $\left(\mu-\frac12\sigma^2\right)<0$, then $\lim_{t\to\infty} \left(\mu-\frac12\sigma^2\right)t=0$ and so $X_t\stackrel{\mathrm{a.s.}}\longrightarrow0$.
• If $\left(\mu-\frac12\sigma^2\right)=0$, then $X_t$ does not converge almost surely (since $B_t$ does not). Also, this condition implies that $X_t$ is a martingale (verifying this is a good exercise).