Consider a standard Brownian motion $\{B(t); t\ge 0\}$ (zero mean and $\sigma^{2} = 1 $) at times
$0 < u < u+v < u+v+w < u+v+w+x $, where $u, v, w, x > 0$
Evaluate the product moment:
$$E[B(u)B(u+v)B(u+v+w)B(u+v+w+x)] $$
I tried using the same trick employed when solving $E[B(t)B(s)] = \sigma^{2} min\{s,t\}$.
Namely for $0<s \le t$,
$$E[B(t)B(s)] = E[B(s)\{B(t)-B(s)+B(s)]=E[B(s)^{2}]+E[B(s)]E[B(t))-B(s)]=\sigma^{2}s$$
Using the same method on the above product moment and starting with the 4th term I got
$$E[B(u)B(u+v)B(u+v+w)B(u+v+w+x)]\\ = \\ E[B(u)B(u+v)B(u+v+w)\{B(u+v+w+x)-B(u+v+w)+B(u+v+w)\}]$$ $$\\ = \\ E[B(u)B(u+v)B(u+v+w)^{2}] + E[B(u)B(u+v)B(u+v+w)\{B(u+v+w+x)-B(u+v+w)\}]$$
I then used the same trick with the other elements eventually I managed to end up with
$$E[B(u)^{2}B(u+v+w)^{2}] + E[B(u)\{B(u+v)-B(u)\}B(u+v+w)^{2}] \\+E[B(u)B(u+v)^{2}\{B(u+v+w+x)-B(u+v+w)\}] \\+ E[B(u)^{2}\{B(u+v+w) -B(u+v)\}\{ B(u+v+w+x)-B(u+v+w)\}] \\+ E[B(u)\{B(u+v)-B(u)\}\{B(u+v+w) -B(u+v)\}\{ B(u+v+w+x)-B(u+v+w)\}]$$
But I've gotten lost and not sure if I am going the right way, And I don't know how I would evaluate elements such as $E[B(u)^2B(u+v+w)^{2}]$ or $E[B(u)B(u+v)^{2}\{B(u+v+w+x)-B(u+v+w)\}]$. I Believe the very last term is equal to zero by the definition of Brownian motion?
Any help would be appreciated.
So I managed to figure it out so I thought I would post the solution in case anyone else is having difficulty with the same question.
Firstly one needs to use the law of total Expectation i.e. $E[E[X|Y]]=E[X]$ and apply it iteratively. So starting from the top.
$$E[B(u)B(u+v)B(u+v+w)B(u+v+w+x)] = E[B(u)B(u+v)B(u+v+w)E[B(u+v+w+x)|\mathcal{F}_{u+v+w}]]$$ Where $\mathcal{F}_{u}:= \sigma (\{B(t);t \le u\}) $.
Using $E[B(u+v+w+x)|\mathcal{F}_{u+v+w}]]=B(u+v+w)$ we get,
$$E[B(u)B(u+v)B(u+v+w)E[B(u+v+w+x)|\mathcal{F}_{u+v+w}]]= E[B(u)B(u+v)B^{2}(u+v+w)]$$
Applying the law of total probability again we have
$$E[B(u)B(u+v)B^{2}(u+v+w)] = E[B(u)B(u+v)E[B^{2}(u+v+w)|\mathcal{F}_{u+v}]]$$
Where $E[B^{2}(u+v+w)|\mathcal{F}_{u+v}] = B^2(u+v)+(u+v+w)-(u+v) = B^2(u+v)+w$ , so we have,
$$E[B(u)B(u+v)E[B^{2}(u+v+w)|\mathcal{F}_{u+v}]]= E[B(u)B(u+v)[B^2(u+v)+w]]$$
Expanding the right hand side we have
$$ E[B(u)B(u+v)[B^2(u+v)+w]] = E[B(u)B^3(u+v)] + wE[B(u)B(u+v)]$$
Applying the Law of total probability again to both parts of the RHS we end up with
$$E[B(u)B^3(u+v)] + wE[B(u)B(u+v)] = E[B(u)E[B^3(u+v)| \mathcal{F}_{u}]] + wE[B(u)E[B(u+v)| \mathcal{F}_{u}]] $$
where, $E[B^3(u+v)| \mathcal{F}_{u}] = B^3(u) +3B(u)((u+v)-u))=B^3(u)+3B(u)v$ insertiing this into the first term on the RHS we get
$$E[B(u)E[B^3(u+v)| \mathcal{F}_{u}]] =E[B(u)(B^3(u)+3B(u)v)] = E[B^4(u)]+3vE[B^2(u)] = 3u^2+3uv$$
Looking at the second term on the RHS we have
$$ wE[B(u)E[B(u+v)| \mathcal{F}_{u}]]=wE[B^2(u)]=wu$$
Where we used the result $E[B(u+v)| \mathcal{F}_{u}]=B(u)$
Hence we end up with the full answer as
$$E[B(u)B(u+v)B(u+v+w)B(u+v+w+x)] = 3u^2+3uv+uw $$