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.
 A: In order to compute the other expectations you need to insert some more terms so that you can use once more the independence of the increments, for instance in the second term you have to write
$$W(u+v+w) = (W(u+v+w)-W_u) + W_u$$
which will allow you to use that $W_u$ and $W(u+v+w)-W(u)$ are independent.
You can avoid these (somewhat messy) calculations by using the fact that $(W_t)_{t \geq 0}$ and $(W_t^2-t)_{t \geq 0}$ are martingales. By the tower property of conditional expectation, we have
$$\mathbb{E}(W(u) W(u+v) W(u+v+w)) = \mathbb{E} \bigg[ \mathbb{E}(W(u) W(u+v) W(u+v+w) \mid \mathcal{F}_{u+v}) \bigg]$$
where $(\mathcal{F}_{t})_{t \geq 0}$ is the canonical filtration of $(W_t)_{t \geq 0}$. As $W(u+v)$ and $W(u)$ are $\mathcal{F}_{u+v}$-measurable this gives
\begin{align*} \mathbb{E}(W(u) W(u+v) W(u+v+w)) &= \mathbb{E}\bigg[ W(u) W(u+v) \underbrace{\mathbb{E}(W(u+v+w) \mid \mathcal{F}_{u+v})}_{\stackrel{(\ast)}{=}W(u+v)} \bigg] \\ &= \mathbb{E}(W(u) W(u+v)^2)\end{align*}
where we have used in $(\ast)$ the fact that $(W_t)_{t \geq 0}$ is a martingale. Using once more the tower property we get
\begin{align*} \mathbb{E}(W(u) W(u+v) W(u+v+w)) &= \mathbb{E} \bigg[ \mathbb{E}(W(u) W(u+v)^2 \mid \mathcal{F}_u) \bigg] \\ &= \mathbb{E} \bigg[ W(u) \underbrace{\mathbb{E}(W(u+v)^2 \mid \mathcal{F}_u)}_{\stackrel{(\ast)}{=} (W_u^2-u)+(u+v)} \bigg] \\ &= \mathbb{E}(W_u^3)- v \mathbb{E}(W_u) \end{align*}
where we used in $(\ast)$ that $(W_t^2-t)_{t \geq 0}$ is a martingale. Hence, as $W_u \sim N(0,u)$,
$$\mathbb{E}(W(u) W(u+v) W(u+v+w)) = 0.$$
Remark: It is not at all surprising that the expectation equals zero. Why? Because of the symmetry of Brownian motion. It is well known that $B_t := -W_t$ is also a Brownian motion, and in particular, $(W_t)_{t \geq 0}$ equals $(B_t)_{t \geq 0}$ in distribution. This implies
$$\mathbb{E}(W(u)W(u+v)W(u+v+w)) = \mathbb{E}(B(u)B(u+v)B(u+v+w)).$$
By the very definition of $(B_t)_{t \geq 0}$, the right-hand side equals
$$- \mathbb{E}(W(u)W(u+v)W(u+v+w)),$$
and therefore we get
$$\mathbb{E}(W(u)W(u+v)W(u+v+w)) = - \mathbb{E}(W(u)W(u+v)W(u+v+w))$$
which clearly implies
$$\mathbb{E}(W(u)W(u+v)W(u+v+w)) =0.$$ 
A: 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?
