I'm trying to calculate the expected value of this stochastic process that has the Wiener process. $E(U(t)) = e^{9t/8}$ is the answer.
$ E(U(t)) = E(e^{t+W(t)/2})$ where $W(t)$ is the Wiener process. So far I have: $$ E(U(t)) = E(e^{t+W(t)/2})$$ $$ = e^t E(e^{W(t)/2}) $$
I think I can use the integral of $W(t) = \int_{-\infty}^\infty \frac{e^{-t^2/2}}{\sqrt{(2\pi)}} dt$ but what's throwing me is that it's divided by $2$. Any help is appreciated.
Ian above is right. MGF is the way to go. Look up MGF, $E(e^{xt})$, for Normal and set $t=1$. You will see when $x$ is Normal with mean $\mu$ and variance $\sigma^2$, the expectation of $e^x$ is $e^{\mu+\sigma^2/2}$. For your process, $\mu$ is $t$ and Var($x$) is $t/4$.