Expected values $E[W(1)W^2(2)]$ and $E[W^3(1)]$ for Wiener process

Question

I need to compute $E(W(1)W^2(2))$

$$W(1)W^2(2) = W(1)(W(2)-W(1)+W(1))^2 $$ $$= W(1) \left[(W(2)-W(1))^2 + 2(W(2)-W(1))W(1) + W^2(1) \right] $$ $$ = W(1)(W(2)-W(1))^2 + 2(W(2)-W(1))W^2(1) + W^3(1) $$ Therefore due to independence because $W(1)=W(1)-W(0)$ $$E(W(1)W^2(2)) = E[W(1)(W(2)-W(1))^2] + 2E[(W(2)-W(1))W^2(1)] + E[W^3(1)] $$ $$ E[W(1)]\cdot E[(W(2)-W(1))^2] + 2E[W(2)-W(1)]\cdot E[W^2(1)] + E[W^3(1)] $$ First addend is equal to 0 because $E[W(1)]=E[W(1)-W(0)]=0$ second one is also equal to 0 but i have no idea what to do with the third one. Should I also try to find increments. What did I forget?

Answer 1

$E\left[ W^3(1) \right] = 0$, since $W(1) \sim \mathcal{N}(0,1)$ by definition of Wiener process, and if $\xi \sim \mathcal{N}(0,1)$ then $$E \xi^k = \int\limits_{-\infty}^{+\infty} x^k \frac{1}{\sqrt{2\pi}} e^{-x^2/2}= 0,$$ if $k$ is odd, since we integrate an odd function over the whole line in this case.

Answer 2

To show $$\mathbb E\big[W_tW_T^2\big]=0$$ for $t\le T$ we can also use Ito calculus to get \begin{align} W^2_T&=2\int_0^TW_s\,dW_s+T\,,\\ W_tW^2_T&=2W_t\int_0^TW_s\,dW_s+W_tT\,. \end{align} Now use the the martingale property to conclude that \begin{align} \mathbb E\Big[W_t\int_0^TW_s\,dW_s\Big]=\mathbb E\Big[W_t\int_0^tW_s\,dW_s\Big]\,. \end{align} Finally, by the integration-by-parts formula, \begin{align} W_t\int_0^tW_s\,dW_s=\int_0^tW_s^2\,dW_s+\int_0^t\int_0^uW_s\,dW_s\,dW_u+\int_0^tW_s\,ds \end{align} of which all terms have expectation zero.