Finding the expectation of a Wiener Process

Question

My question is on how to find $\mathbb E[W_t^n]$ where $n= 0,1,2,...$ and $W_t$ is a standard normal Wiener process. Would we need to use a moment generating function. Thanks.

Answer 1

Since $W_t\sim N(0,t)$ at a fixed time we can find it's moments by simply doing an integral

$$E[W_t^n]=\int_{-\infty}^{\infty}x^nf_{W_t}(x;t)dx=\frac{1}{\sqrt{2\pi t}} \int_{-\infty}^{\infty}x^n e^{-x^2/2t}dx$$ which can be shown to be zero for $n$ odd, and whenever $n=2k$ is even:

$$E[W_t^{2k}]=\frac{(2t)^{k}}{\sqrt{\pi}}\frac{d^k}{da^k}\Big(\int_{-\infty}^{\infty}e^{-ax^2}dx\Big)\Bigg|_{a=1}=(2t)^k\frac{(2k)!}{4^kk!}$$

One can obtain the same result by taking derivatives of the generating function, which are easily found since it's Taylor expansion is well known:

$$E[W_t^{2n}]=\frac{d^n}{d\lambda^n}E[e^{\lambda W_t}]\Big|_{\lambda=0}=\frac{d^n}{d\lambda^n}e^{t\lambda^2/2}\Big|_{\lambda=0}=t^n\frac{(2n)!}{n!2^n}$$