I can't see where I'm going wrong, but I don't get the correct answer.

B(t) is a brownian motion started at 0

$\displaystyle X(t)=\int_{0}^{t}sign(B(s))dB(s)$ where

$\displaystyle sign(B(s))=\begin{array}{clcr}

-1 & ,\ B(s)<0\\

1 & ,\ B(s)\geq 0 \\

\end{array}$

I want to find $\displaystyle E[X(t)B(t)^2]$

My attempt:

$\displaystyle E[X(t)B(t)^2]=E[E[X(t)B(t)^2|B(s)\geq 0]]=E[E[\int_{0}^{t}1dB(s)*B(t)^2]]$$\displaystyle =E[E[B(t)^3]]=E[0]=0$

But this is wrong... Why?

Thanks for any help