# Brownion motion problems

1. Jun 16, 2012

### boliobolo

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

2. Jun 16, 2012

### mathman

It dosn't look right. There is no x in the answer.

3. Jun 17, 2012

### boliobolo

it is the answer. with no x

4. Jun 17, 2012

### mathman

I am not familiar with your notation. What is the definition of B(u)?

5. Jun 19, 2012

### tait

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.

6. Jun 20, 2012

### mathman

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.

Last edited: Jun 20, 2012