# Find expectation of brownian motion

How would I do the following question. I know how to do it with two variables (just B(U) and B(U + V) but I do not know how to figure this out with 4 (or even 3) terms) Thanks for the help.

E [B(U)B(U+V)B(U+V+W)B(U+V+W+x)] where U + V + W > U + V > U and x > 0.

Thanks for the help. By the way the answer is 3U^2 + 3UV + UW.

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Here is a sketch of a direct proof. Define four independent random variables $X(U):=B(U)$, $X(V):=B(U+V)-B(U)$, $X(W):=B(U+V+W)-B(U+V)$, and $X(x)=B(U+V+W+x)-B(U+V+W)$. Now multiply out $$X(U)(X(U)+X(V))(X(U)+X(V)+X(W))(X(U)+X(V)+X(W)+X(x))$$ and calculate the expectation of the result.