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Math Help - help with chi square freedom sums

  1. #1
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    Angry 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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  2. #2
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    Quote Originally Posted by Sneaky View Post
    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?
    Are X1 and X2 independent. If so, then use the well known theorem for the moment generating function of the sum.
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  3. #3
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    yes they are independent
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  4. #4
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    Quote Originally Posted by Sneaky View Post
    yes they are independent
    Then apply the appropriate theorem.
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  5. #5
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    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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  6. #6
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    Quote Originally Posted by Sneaky View Post
    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?
    Yes.
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  7. #7
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    Also asked and answered at another website. Thread closed.
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  8. #8
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    Quote Originally Posted by Sneaky View Post
    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?

    I know the thread is cloes and why, but I like the following proof which uses a completely different method (and I know that some will argue that its not a proof, but that is another discussion):

    Since A RV which is the sum of the squares of n independent standard normals has a chi-squared distribution with n degrees of freedom, the sum of two such independednt RV with n and m degrees of freedom respectively has a chi-squared distribution with n+m degrees of freedom since it is the sum of the squares of n+m independent standard normals..

    Since there is no way of telling how our chi-square RV were produced the result holds in general.

    CB
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