# Can someone please troubleshoot this method for me?

• May 26th 2009, 08:50 PM
HD09
Can someone please troubleshoot this method for me?
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

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

$sign(B(s))=\begin{array}{clcr}
-1 & ,\ B(s)<0\\
1 & ,\ B(s)\geq 0 \\
\end{array}$

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

My attempt:

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

But this is wrong... Why?

Thanks for any help
• May 29th 2009, 08:45 AM
cl85
First you need to observe that regardless of the realization from time 0 to t
$X(t) = |B(t)|$

And
$E[X(t)B(t)^2]=E[E[X(t)B(t)^2|B(t)]]$
$=P(B(t)\geq 0)E[X(t)B(t)^2|B(t)\geq 0]+P(B(t) < 0)E[X(t)B(t)^2|B(t) < 0]$
• May 29th 2009, 04:07 PM
HD09
Quote:

Originally Posted by cl85
First you need to observe that regardless of the realization from time 0 to t
$X(t) = |B(t)|$

And
$E[X(t)B(t)^2]=E[E[X(t)B(t)^2|B(t)]]$
$=P(B(t)\geq 0)E[X(t)B(t)^2|B(t)\geq 0]+P(B(t) < 0)E[X(t)B(t)^2|B(t) < 0]$

$=\frac{1}{2}E[B(t)^3]+\frac{1}{2}E[-B(t)^3]$

$=0+0$

(Brownian motion is normally distributed with mean 0)

The correct answer is supposed to be "not zero"
• May 29th 2009, 07:03 PM
cl85
$E[X(t)B(t)^2|B(t)\geq 0] \neq E[X(t)B(t)^2]$

Also, both $E[X(t)B(t)^2|B(t)\geq 0]$ and $E[X(t)B(t)^2|B(t) < 0]$ are positive.
• May 29th 2009, 08:44 PM
HD09
Quote:

Originally Posted by cl85
$E[X(t)B(t)^2|B(t)\geq 0] \neq E[X(t)B(t)^2]$

But it DOES equal $E[B(t)^3]$ doesn't it? Because if B(t) is greater than or equal to zero, then $X(t)=|B(t)|=B(t)$

Where am I going wrong here?