Results 1 to 5 of 5

Math Help - Can someone please troubleshoot this method for me?

  1. #1
    Junior Member
    Joined
    Apr 2009
    Posts
    38

    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}<br />
-1 & ,\ B(s)<0\\<br />
1 & ,\ B(s)\geq 0 \\<br />
\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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Nov 2008
    Posts
    58
    Thanks
    1
    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]
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2009
    Posts
    38
    Quote Originally Posted by cl85 View Post
    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"
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2008
    Posts
    58
    Thanks
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2009
    Posts
    38
    Quote Originally Posted by cl85 View Post
    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?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. I can't troubleshoot this
    Posted in the Algebra Forum
    Replies: 2
    Last Post: March 18th 2010, 07:03 PM
  2. Replies: 2
    Last Post: August 17th 2008, 01:02 PM
  3. Replies: 3
    Last Post: November 3rd 2007, 02:43 PM
  4. Help me troubleshoot my answer
    Posted in the Statistics Forum
    Replies: 2
    Last Post: June 11th 2007, 08:01 AM
  5. Replies: 0
    Last Post: January 4th 2007, 02:29 PM

Search Tags


/mathhelpforum @mathhelpforum