TITLE: Expectation of geometric brownian motion
QUESTION [19 upvotes]: I was deriving the solution to the stochastic differential equation $$dX_t = \mu X_tdt + \sigma X_tdB_t$$ where $B_t$ is a brownian motion. After finding $$X_t = x_0\exp((\mu - \frac{\sigma^2}{2})t + \mu B_t)$$ I wanted to calculate the expectation of $X_t$. However I think I'm not quite getting it. I thought that I'd just calculate $$E(x_0\exp((\mu - \frac{\sigma^2}{2})t + \mu B_t) = x_0\exp((\mu - \frac{\sigma^2}{2})t)E(\exp(\mu B_t))$$ but the book I'm using gives as answer $E(X_t) = x_0\exp(\mu t)$. I found this quite surprising as I don't quite see how $\sigma$ could just disappear. After reading Wikipedia I see that the result could be either $E(X_t) = x_0\exp((\mu + \frac{\sigma^2}{2})t)$ or $E(X_t) = x_0\exp(\mu t)$, depending on whether you use the Itô interpretation or the Stratanovich interpretation.
Since the book I use only considers the Itô formulation of stochastic integration I am interested in the latter result. But how do I obtain this? Do I just fail in calculating $E(\exp(\mu B_t))$? Thanks in advance.

REPLY [9 votes]: Even though it's pretty late I think this still may be helpful for a few of you.
First of all notice as $B_t$ is a geometric Brownian motion, by definition it is normally distributed with mean $0$ and variance $t$.
I.e. $B_t$  has the moment-generating function 
\begin{equation}\label{moment}\tag{1}
\mathbb{E}[\exp(u B_t)]=\exp(\frac{1}{2}u^2t),\qquad u\in\mathbb{R}.
\end{equation}
Now we have for $X_t$ being a geometric Brownian motion
$$X_t= x_0 \exp\Big(\big(\mu - \frac{\sigma ^2}{2}\big)t+\sigma B_t\Big).$$
Therefore, applying the expectation value yields
\begin{align}
\mathbb{E}[X_t]&= \mathbb{E}\Big[x_0 \exp\Big(\big(\mu - \frac{\sigma ^2}{2}\big)t+\sigma B_t\Big)\Big]\\
&=x_0 \exp\Big(\big(\mu - \frac{\sigma ^2}{2}\big)t\Big)\mathbb{E}\Big[ \exp\big(\sigma B_t\big)\Big].
\end{align}
So, by using equality \eqref{moment} with $u=\sigma$ we get
\begin{align}
\mathbb{E}[X_t]&= x_0 \exp\Big(\big(\mu - \frac{\sigma ^2}{2}\big)t\Big)\mathbb{E}\Big[ \exp\big(\sigma B_t\big)\Big]\\
&=x_0 \exp\Big(\big(\mu - \frac{\sigma ^2}{2}\big)t\Big)\exp\big(\frac{1}{2}\sigma^2 t\big)\\
&=x_0 \exp\big(\mu t\big).
\end{align}
This gives you the desired result.