Brownian motion,Expectations for $\exp(B(t))$

Question

Brownian motion process $B(t)\sim N(0,t)$

$E[e^{B(t)}]=?$

Answer 1

$\newcommand{\E}{\Bbb{E}}$If $X$ is a $\textsf{Normal}(\mu,\sigma^2)$ random variable, then $\E\left[e^{X}\right] = e^{\mu + \frac{1}{2}\sigma^2}$. (See Calculating moment generating function with normal distribution for a proof.)

In your case, $\mu = 0$ and $\sigma^2=t$, so $\E\left[e^{B(t)}\right] = e^{\frac{t}{2}}$.