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Math Help - conditional expectations

  1. #1
    Junior Member
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    conditional expectations

    Hi, can someone please explain to me how they got the answer to this

    Question: Find E(X|Y) and E[X - E(X|Y)]^2. Let
     \left(\begin{array}{cc}X\\Y\end{array}\right) ~  N( \left (\begin{array}{cc}1\\-1\end{array}\right) ,  \left (\begin{array}{cc}5 & 2 \\ 2& 1 \end{array}\right) )

    I don't understand how they got the answer..

    E(X|Y) = 1 + 2(Y + 1) = 2Y + 3
    E[X - E(X|Y)]^2 = 5 - 4 = 1

    Where did they get "1 + 2(Y + 1) " from?? and "5 - 4 " from??

    can you guys please help explain this to me?
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  2. #2
    MHF Contributor matheagle's Avatar
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    You can see that E[X - E(X|Y)]^2=1 by plugging in the 1 + 2(Y + 1)

    E[X - E(X|Y)]^2=E[X - (2Y + 3)]^2 =E[X - 2Y - 3)]^2=V[X - 2Y - 3)]+ [E[X - 2Y - 3)]]^2

    =V[X]+4V[Y] -4cov(X,Y)+ [E[X - 2Y - 3)]]^2

    =5+4-8 +(0)^2=1

    I was going to prove the first part from scratch but I knew it had to be somewhere online.
    Look at.... Conditional distributions at ....http://en.wikipedia.org/wiki/Multiva...l_distribution

    Plugging into that, you can see that 1 + 2(Y + 1) is correct
    the 1 is the mean of X and the 2 is the ratio of the covariance over the variance of Y...
    Last edited by matheagle; October 2nd 2009 at 11:57 PM.
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