Expectation of Brownian motion $W^k, k = 2,4$

Question

G'day, I've just computed the expectation of $W^k$,

$W^k_{t} = \frac{k(k-1)}{2} \int_0^{t} E[W^{k-2}_{s}]ds $

However I am running into troubles when I calculate for the even moments. We know for the uneven moments the expectation is zero.

k = 2

$E[W^k_{t}] = \frac{2(2-1)}{2} \int_0^{t} E[W^{0}_{s}]ds = 1 \int_0^{t} E[W^{k-2}_{s}]ds = t $

k = 4

$E[W^k_{t}] = \frac{4(4-1)}{2} \int_0^{t} E[W^{4-2}_{s}]ds = 6 \int_0^{t} E[W^{2}_{s}]ds = 3t^2 $

Can someone explain how we get t as solution for the case k = 2 and 3$t^{2}$ for the case of k = 4?

Answer 1

I think you mean $\mathbb{E}[W^k_t]$ on the left-hand side of your formulas. Anyway, for $k=2$ you use the fact that $W^0_t = 1$, while for $k=4$ you have to use $\mathbb{E}[W^2_s] = s$.