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Thread: Can someone please troubleshoot this method for me?

  1. #1
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    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

    $\displaystyle X(t)=\int_{0}^{t}sign(B(s))dB(s)$ 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
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  2. #2
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    First you need to observe that regardless of the realization from time 0 to t
    $\displaystyle X(t) = |B(t)|$

    And
    $\displaystyle E[X(t)B(t)^2]=E[E[X(t)B(t)^2|B(t)]]$
    $\displaystyle =P(B(t)\geq 0)E[X(t)B(t)^2|B(t)\geq 0]+P(B(t) < 0)E[X(t)B(t)^2|B(t) < 0]$
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  3. #3
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    Quote Originally Posted by cl85 View Post
    First you need to observe that regardless of the realization from time 0 to t
    $\displaystyle X(t) = |B(t)|$

    And
    $\displaystyle E[X(t)B(t)^2]=E[E[X(t)B(t)^2|B(t)]]$
    $\displaystyle =P(B(t)\geq 0)E[X(t)B(t)^2|B(t)\geq 0]+P(B(t) < 0)E[X(t)B(t)^2|B(t) < 0]$
    $\displaystyle =\frac{1}{2}E[B(t)^3]+\frac{1}{2}E[-B(t)^3]$

    $\displaystyle =0+0$

    (Brownian motion is normally distributed with mean 0)

    The correct answer is supposed to be "not zero"
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  4. #4
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    $\displaystyle E[X(t)B(t)^2|B(t)\geq 0] \neq E[X(t)B(t)^2]$

    Also, both $\displaystyle E[X(t)B(t)^2|B(t)\geq 0]$ and $\displaystyle E[X(t)B(t)^2|B(t) < 0]$ are positive.
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  5. #5
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    Quote Originally Posted by cl85 View Post
    $\displaystyle E[X(t)B(t)^2|B(t)\geq 0] \neq E[X(t)B(t)^2]$
    But it DOES equal $\displaystyle E[B(t)^3]$ doesn't it? Because if B(t) is greater than or equal to zero, then $\displaystyle X(t)=|B(t)|=B(t)$

    Where am I going wrong here?
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