How Can I Solve This Complex Problem Involving Brownian Motion Variables?

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The discussion focuses on solving the expectation E[B(u)B(u+v)B(u+v+w)B(u+v+w+x)] in the context of Brownian motion variables. The solution is derived as 3u^2 + 3uv + uw, with the absence of x explained by the properties of conditional expectations and the independence of increments in Brownian motion. Participants clarify the notation and confirm that the Brownian motion variables are normally distributed with mean 0 and variances corresponding to time intervals. The evaluation of the expectation reveals that terms with any variable to the first power yield an expectation of 0, simplifying the calculation. Understanding the notation and properties of Brownian motion is crucial for solving such problems effectively.
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can someone show me the solution for E[B(u)B(u+v)B(u+v+w)B(u+v+w+x)], for 0<u<u+v<u+v+w<u+v+w+x .the answer is 3u^2+3uv+uw
 
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It dosn't look right. There is no x in the answer.
 
it is the answer. with no x
 
I am not familiar with your notation. What is the definition of B(u)?
 
The solution is presumably judicious applications of conditional expectations and properties of Brownian motion. It's certainly worth having a go yourself and if you don't mind me asking how far have you got? As a (rather unhelpful) hint your first move is the reason x doesn't turn up in the solution.
 
Sorry I'm late - finally understood the notation. To refresh, Brownian motion variables are normally distributed with mean 0, variance ~ time, and independent increments.

For the problem stated: B(u) = U, B(u+v) = U + V, B(u+v+w) = U + V + W, and B(u+v+w+x) = U + V + W + X, where U, V, W, X are independent normally distributed random variables with mean 0 and variances u, v, w, x.

The question is then E(U(U+V)(U+V+W)(U+V+W+X)). To evaluate this, note that for any term in the expanded polynomial with any of the variables to the first power, the expectation (E) will be 0.
So we are then left with E(U4 + 3U2V2 + U2W2) = 3u2 + 3uv + uw.
 
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