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.
