Expectation of a function of Brownian motions

Question

I would like to how I can compute this expectation and get the answer that is given. All terms W indicate a Wiener process.

$$E_t[W_s^3]=E_t[(W_t+(W_s-W_t))^3]=W_t^3+3W_t(s-t)$$

Answer 1

The first step is clear from $W_s = W_t + (W_s-W_t)$.

The second step comes from expanding the cubic expression and noting that $W_t$ is independent of $W_s-W_t$ (independent increments), as well as the fact that $W_s-W_t \sim N(0,(s-t))$.

\begin{align} (W_t+(W_s-W_t))^3 &= W_t^3 + 3 W_t^2 (W_s-W_t) + 3 W_t (W_s-W_t)^2 + (W_s-W_t)^3\\ E_t[(W_t + (W_s-W_t))^3] &= E_t[W_t^3] + 3 E_t[W_t^2] E_t[W_s-W_t] + 3 E_t[W_t] E_t[(W_s-W_t)^2] + E_t[(W_s-W_t)^3] \\ &= W_t^3 + 3 W_t^2 \cdot 0 + 3 W_t (s-t) + 0 \\ &= W_t^3 + 3 W_t (s-t). \end{align}