Ito integral for simple stochastic process

Question

I need for $l=1,2......$ prove that $E[W^{2l} (t)]=$ $\frac{(2l)!}{2^l l!}$ and $E[W^{2l+1} (t)]=0$

I know that Ito integral for simple stochastic process satisfies $E[I^2 (t)]=E\int_0^t\Delta^2(u)du$ - Ito Isometry.

I am trying to find the ways of solving the task from Stochastic calculus, but it seems to be very difficult to start with.

I am really appreciate any hints and thoughts about solution.

Answer 1

I assume you want to prove it using Ito calculus, otherwise it is just moments of a gaussian random variable.

By Ito: $$W^{n}_t=\int_{0}^t n W^{n-1}_s dW_s+\frac{n(n-1)}{2}\int_{0}^tW^{n-2}_sds $$ using induction hypothesis (to get the local martingale part being a true martingale), I have: $$\mathbb{E}[W^{n}_t]=\frac{n(n-1)}{2}\int_{0}^t\mathbb{E}[W^{n-2}_s]ds$$

so if I denote $f_n(t)=\mathbb{E}[W^{n}_t]$

I have $$f'_n(t)=\frac{n(n-1)}{2}f_{n-2}(t)$$ so by induction hypothesis (the even case being the hard one) $$f'_{2l}(t)=\frac{2l(2l-1)}{2} \frac{(2l-2)!}{2^{l-1} (l-1)!} t^{l-1} = \frac{(2l)!}{2^l l!}~l t^{l-1}$$ and by integrating between $0$ and $t$ with $f_n(0)=0$, I have the wanted result.