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Thread: Conditional Variance

  1. #1
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    Conditional Variance

    Hi,

    In this proof http://onlinecourses.science.psu.edu/stat414/node/118 I don't understand why the VAR(Y|X) (VAR(Y|X=x) is constant and can be moved outside the integral, isn't it being integrated over x?

    Thanks!
    Last edited by reflex; Jan 14th 2011 at 01:45 PM.
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  2. #2
    Guy
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    It is assumed that $\displaystyle \sigma_{Y|X}$ is constant, i.e. it does not depend on $\displaystyle x$. See (3) at the beginning of the notes.

    Now, (3) follows immediately if $\displaystyle (X, Y)$ is bivariate normal i.e. you can prove (3). If, on the other hand, we aren't assuming a bivariate normal, then the proof is fine and (3) is the key assumption that lets you pull it out of the integral.
    Last edited by Guy; Jan 14th 2011 at 04:30 PM.
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  3. #3
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    Thanks, I see (the notation confused me).

    How generally would I go about proving (3) knowing it's bivariate normal?
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  4. #4
    Guy
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    Quote Originally Posted by reflex View Post
    Thanks, I see (the notation confused me).

    How generally would I go about proving (3) knowing it's bivariate normal?
    I actually happen to have a proof sitting on my computer already; it derives the conditional distribution of Y|X:



    Some explanation may be in order if you haven't seen something like this before. Essentially the idea is to show that the conditional pdf is proportional (for fixed x, letting y vary) to the pdf of something you know. Then, because they both have to integrate to 1, you conclude that they are equal. (3) follows from this because $\displaystyle \sigma^2 _Y (1 - \rho^2)$ doesn't depend on $\displaystyle X$.
    Last edited by Guy; Jan 15th 2011 at 06:18 AM.
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  5. #5
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    Wow, thanks! I wasn't expecting the whole proof!

    Thanks again!
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