$\mathbb{E}[B^4(t)]$ with $B$= brownian motion Can anyone help me to find:
$\mathbb{E}[B^4(t)]$ where $B$ is a brownian motion?
I thought using this density function:
$f_{B_t}(x) = \frac{1}{\sqrt{2 \pi t}} e^{-\frac{x^2}{2t}}$,
but I don't know how to apply it.
 A: Note that
$$\mathbb{E}(B_t^4) = \frac{1}{\sqrt{2\pi t}} \int_{\mathbb{R}} x^4 \exp \left(- \frac{x^2}{2t} \right) \, dx = \frac{2}{\sqrt{2\pi t}} \int_{0}^{\infty} x^4 \exp \left(- \frac{x^2}{2t} \right) \, dx.$$
If we set $y := x^2/2t$, then
$$\begin{align*} \mathbb{E}(B_t^4) &= \frac{2}{\sqrt{2\pi t}} \int_0^{\infty} (2t y)^2 \cdot e^{-y} \frac{\sqrt{t}}{\sqrt{2y}} \, dy = \frac{4t^2}{\sqrt{\pi}} \int_0^{\infty} y^{3/2} e^{-y} \, dy \\ &= \frac{4t^2}{\sqrt{\pi}} \Gamma \left( \frac{5}{2} \right) \end{align*}$$
where $\Gamma$ denotes the Gamma function. Using
$$\Gamma \left( \frac{5}{2} \right) = \frac{3}{2} \Gamma \left( \frac{3}{2} \right) = \frac{3}{2} \frac{1}{2} \Gamma \left( \frac{1}{2} \right) = \frac{3}{2} \frac{1}{2} \sqrt{\pi}$$
we get
$$\mathbb{E}(B_t^4) = 3t^2.$$
Remark This calculation holds for any Gaussian random variable with mean $0$ and variance $t$. Since the moments of Gaussian random variables are known you do not need to perform this calculations - unless you are interested in the proof of this well-known result.
A: Another quick solution is given as follows if you know that the fourth moment of a standard normal random variable is 3.
$B_t \sim N(0, t)$, then $\frac{1}{\sqrt{t}}B_t \sim N(0,1)$, then 
$$\mathbb{E}\left[ \,\,  \left(\,\,  \frac{1}{\sqrt{t}}B_t  \,\, \right)^4   \,\, \right] = 3 \Longrightarrow \mathbb{E}\Big[  (  B_t )^4 \Big] = 3 t^2 \,\, .$$
Q.E.D.
