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Math Help - Cov(X,YZ)

  1. #1
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    Cov(X,YZ)

    Hi there

    i posted this in the advanced stats page, but now seems i really have to figure this out asap, so ive posed it here.

    I was wondering if someone can explain to me why the expansion of Cov(X,YZ) is

    <(X-<X>)(Y-<Y>)(Z-<Z>)> + <Y>Cov(X,Z) + <Z>Cov(X,Y)

    i think it has something to do with the identities. Ive tried to mimc the expansion of Cov(X+Y,Z) as this should be similar, but i dont get the answer

    anyhelp would be much appreciated
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by chogo View Post
    Hi there

    i posted this in the advanced stats page, but now seems i really have to figure this out asap, so ive posed it here.

    I was wondering if someone can explain to me why the expansion of Cov(X,YZ) is

    <(X-<X>)(Y-<Y>)(Z-<Z>)> + <Y>Cov(X,Z) + <Z>Cov(X,Y)

    i think it has something to do with the identities. Ive tried to mimc the expansion of Cov(X+Y,Z) as this should be similar, but i dont get the answer

    anyhelp would be much appreciated
    This is a bit fiddly to type without error but expand the covariances as:

    Cov(A,B)=<(A-<A>)(B-<B>)> in both expressions and their equality should fall
    out.

    RonL
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  3. #3
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    appologies but i dont quite understant what you mean. are u suggesting i work backwards from the expanded form?

    Thank you very much for your assitance it truly appreciated
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by chogo View Post
    appologies but i dont quite understant what you mean. are u suggesting i work backwards from the expanded form?

    Thank you very much for your assitance it truly appreciated
    Not quite I'm suggesting you expand both of the forms to get rid of the Cov's
    and reduce them to expectations X, Y, Z and their products.

    RonL
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  5. #5
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    Ahh ok done it, many thanks
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