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    help with chi square freedom sums




    How do i show that the a [X1 has a chi square distribution with n degrees of freedom] + [X2 has a chi square distribution with m degrees of freedom] is a [X1+X2 has a chi square distribution with n+m degrees of freedom]?

    How can i use moment generating functions to do this?

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    Re: help with chi square freedom sums

    The MGF of the sum of independent random variables is the product of the MGFs of the random variables. You should be able to show this pretty easily using the definition of independence and the definition of the MGF.

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    Re: help with chi square freedom sums

    the mgf for a chi square with n freedoms is
    (1-2t)^(-n/2)
    the mgf for a chi square with m freedoms is
    (1-2t)^(-m/2)

    The MGF of the sum of independent random variables is the product of the MGFs of the random variables. So the product is
    (1-2t)^(-(n+m)/2)
    which is a chi squared distribution with n+m degrees of freedom

    so in conclusion [X1 has a chi square distribution with n degrees of freedom] + [X2 has a chi square distribution with m degrees of freedom] is a [X1+X2 has a chi square distribution with n+m degrees of freedom]



    Is this right?

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    Re: help with chi square freedom sums

    Looks fine to me. But it'll only pass as a proof for a class if you've proved or at least have been shown that fact about MGFs that I mentioned before. Otherwise you'll have to derive that result yourself.

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    Re: help with chi square freedom sums

    Quote Originally Posted by Dason View Post
    Looks fine to me. But it'll only pass as a proof for a class if you've proved or at least have been shown that fact about MGFs that I mentioned before. Otherwise you'll have to derive that result yourself.
    would the proof of that be:
    m[X1+X2](t) = E(e^(t(X1+X2)))
    =E(e^(tX1)e^(tX2))
    =E(e^(tX1))E(e^(tX2)) (since they are independent)
    =m[X1](s)m[X2](s)

    is this fine?

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    Re: help with chi square freedom sums


    Looks good to me.

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