Expected value of the product $E[W(u)W(u+v)W(u+v+w)]$ for Brownian motion I am trying to understand the Wiener process, and I am not sure if my assumptions make sense.

Let $W(t)$ be a Wiener process ($W(0)=0$ and for $t \le s, (W(s)-W(t))$ is a normal distribution with mean $0$ and variance $(s-t)$). For $u,v,w>0$, I am trying to find the value of $E[W(u)W(u+v)W(u+v+w)]$.

My approach involves rearraging this product in a way that I can exploit the property of independence of two disjoint sets:
$$E[W(u)\cdot (W(u+v)-W(u))\cdot (W(u+v+w)-W(u+v))+W(u)^2 \cdot W(u+v+w)-W(u)^2 \cdot W(u+v)+W(u+v)^2 \cdot W(u)]$$
Which can be rewritten as:
$$E[W(u) \cdot (W(u+v)-W(u))\cdot (W(u+v+w)-W(u+v))]+E[W(u)^2\cdot W(u+v+w)]-E[W(u)^2 \cdot W(u+v)]+E[W(u+v)^2 \cdot W(u)]$$
Apparently, the first expected value is $0$, but I am not sure what to do with the other terms.

If we set $N_u=W(u)$, $N_v=W(u+v)-W(u)$ and $N_w=W(u+v+w)-W(u+v)$ then these random variables are independent, have normal distribution, have mean $0$ and satisfy $\mathbb EN_x^2=x$ for $x=u,v,w$
Further:


*

*$W(u)=N_u$

*$W(u+v)=N_u+N_v$

*$W(u+v+w)=N_u+N_v+N_w$
So to be found is: $$\mathbb EN_u(N_u+N_v)(N_u+N_v+N_w)$$
Can you finish this?
