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Math Help - Finding the correlation coefficient

  1. #1
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    Finding the correlation coefficient

    Hello

    I've been working on this problem all day but, honestly I don't know what else to try.

    Let Z~N(0,1) and Y= a+bZ+cZ^2

    find \rho(Y,Z) (Correlation coefficient



    I think I need to find the the joint PDF of X and Y then calculate the Cov(XY), maybe this will let me cancel what's left of the denominator and obtain a number. I'm quite lost.


    *****EDIT******


    I should probably mentioned that i tried the linear properties of expectation and arrived to the conclucion that it was not correlated but i think my application is wrong.
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  2. #2
    MHF Contributor matheagle's Avatar
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    You will need the covariance divided by the st. deviations.

    The covariance is E(YZ)-E(Y)E(Z)=E(YZ) since E(Z)=0.

    Now E(YZ)= E((a+bZ+cZ^2)Z) =aE(Z)+bE(Z^2)+cE(Z^3)=0+b+0=b
    since Z is symmetric and the odd moments should be zero.

    Next \sigma_Z=1 and \sigma_Y=\sqrt{Var(a+bZ+cZ^2)}

    Then Var(a+bZ+cZ^2)= Var(bZ+cZ^2)= b^2Var(Z)+c^2Var(Z^2)+2bcCov(Z,Z^2)

    Where you will need Cov(Z,Z^2)=E(Z^3)-E(Z)E(Z^2)=0-(0)(1)=0.

    That's most of this problem. You should not even consider integrating it, whatsoever.

    Fine, no one's going to finish this... Var(Z^2)=E(Z^4)-(E(Z^2))^2.

    I was too lazy to compute that fourth moment, so I looked it up at http://en.wikipedia.org/wiki/Normal_distribution
    It's 3, so Var(Z^2)=E(Z^4)-(E(Z^2))^2=3-1=2.

    This seems to be (or not 2b, that is your question) {b\over \sqrt{b^2+2c^2}}.
    Last edited by matheagle; May 17th 2009 at 02:23 PM.
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  3. #3
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    Thanks very much for your answer. I see it clearly now. This should help me inn next week's final
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  4. #4
    MHF Contributor matheagle's Avatar
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    that was fun, any more of these?
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