Let $(W(t))_{t \geq 0}$ be standard Brownian motion, and let $t_1, t_2, t_3 \in \mathbb{R}_{> 0}$ with $t_1 < t_2 < t_3$ be arbitrary. Compute: $$ \mathbb{E} [W(t_1) * W(t_1 + t_2) * W(t_1 + t_2 + t_3)] $$ using the properties of $W(t)$.
This question has been haunting me for days. I know you can write $$ \mathbb{E} [W(t_1) * W(t_1 + t_2)] = \mathbb{E} [W(t_1) * ( W(t_1 + t_2) - W(t_1) + W(t_1) )] = \mathbb{E} [W(t_1) * ( W(t_1 + t_2) - W(t_1) )] + \mathbb{E} [W(t_1)^2] $$ ("sneakily" using the independence of increments). Also, $$\mathbb{E} [W(t_1) * ( W(t_1 + t_2) - W(t_1) )] = \mathbb{E} [(W(t_1) - W(0)) * ( W(t_1 + t_2) - W(t_1) )] = \mathbb{E} [(W(t_1 + 0) - W(0)] * \mathbb{E} [ W(t_1 + t_2) - W(t_1) ] = 0 $$ (where the first equality follows from the fact that $W(0) = 0$, the second from the fact that Brownian motion has independent increments, the third from the $N(0, s)$ distribution of any increment $W(t + s) - W(t)$ for all $t \geq 0, s > 0$).
So, $\mathbb{E} [W(t_1) * W(t_1 + t_2)] = \mathbb{E} [W(t_1)^2]$.
So far I have reduced the problem to $$\mathbb{E} [W(t_1) * W(t_1 + t_2) * W(t_1 + t_2 + t_3)] = \mathbb{E} (W(t_1)^3)$$ but I don't know how to compute this last quantity, nor if this is the right approach, nor if it is even correct.
Am I missing something crucial?