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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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    What do you know about the distribution of $W_t$? – 2017-02-09
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    @saz $W_t$ is a standard Brownian motion(i.e. it satisfies four properties). – 2017-02-09
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    I know that it is a Brownian motion; I don't want you to repeat the definition. What do you know about the distribution of $W_t$? – 2017-02-09
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    So have a look at the definition and think about it. (Honestly, replying after 1 minute that you don't know means simply that you haven't even properly thought about it.) – 2017-02-09
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    I already gave you a hint. Go back to the definition and look what it is telling you about the distribution of $W_t$. That's the very first step to solve this exercise. – 2017-02-09

1 Answers 1

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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.