Expectation of standard Brownian motion
Let Let $\{W_t\}_{t\ge 0}$ be a standard Brownian motion on some filtered probability space $(\Omega , \mathcal{F}_{t}, \{\mathcal{F}_{t}\}_{t\ge 0}, \mathbb{P}).$
How can we show that the expectation $$\mathbb{E}[W_{t}^{2k}]=\frac{(2k)!}{2^kk!}t^k,$$
where $k$ is a positive integer.
Any help?
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$\begingroup$HINT :
$$\mathbb{E}[W_{t}^{2k}]=(\sqrt{t})^{2k}E[Y^{2k}]$$
where Y is a standard normal variable, we use the fact that $W_t \sim \mathcal{N}(0,t)$
Introduce the function $f$ such as $$f(t)=E(e^{tY})$$
We have that $tY \sim \mathcal{N}(0,t^2)$ therefore $f(t)=e^{E(tY)+\frac{1}{2}var(tY)}=e^{\frac{1}{2}t^2}$
Using the Leibniz integral rule, you can also write that
$$f^{(n)}(t)=E(Y^{n}e^{tY})=\frac{d^{n}\left(e^{\frac{1}{2}t^2}\right)}{dt^n}$$
Study the case $t=0$, and conclude by induction.
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