What is the expectation
$$\mathbb E[\exp(W_t)],$$ $$\mathbb E[\exp(W_t^4)]$$
where $W(t)$ is a standard Brownian and $W_t∼N(0,t)$? I know that the standard Brownian motion has mean of 0 and variance having the value of $t$. Also, I understand that the first one is some form of geometric brownian motion. How do I proceed from here on? I can try and do the second one myself, if somehow the first one is explained by someone
$W_t \sim N(0,t)$. So $Ee^{W_t}=\frac 1 {\sqrt {2\pi} t} \int_{\mathbb R} e^{x} e^{-\frac {x^{2}} {2t}}dx$. Now $\int_{\mathbb R} e^{x} e^{-\frac {x^{2}} {2t}}dx=e^{t/2}\int_{\mathbb R} e^{-\frac {(x-t)^{2}} {2t}}dx=\sqrt {2\pi} t e^{t/2}$.
You can also use the general formula for the genearting function of normal distribution (*) which immediately gives $Ee^{W_t}=e^{t/2}$
(*): ( $Ee^{cX}=e^{c\mu} e^{c^{2}\sigma^{2}/2}$ where $\mu$ is the mean and $\sigma^{2}$ is the variance of $X$).
Also $Ee^{W_t^{4}} =\infty$.