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Math Help - Vector of independent normal RV's - mgfs

  1. #1
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    Vector of independent normal RV's - mgfs

    Let X = (X_1,...,X_n)^T be a column vector of independent N(0,1) distributed random variables. Let A be an n \times n orthogonal matrix.

    Prove that  ||X||^2 = X_1^2 + ... + X_n^2 \sim \mathcal{X}_n^2 using moment generating functions.
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  2. #2
    MHF Contributor matheagle's Avatar
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    And where does this A come in?
    Do you want AX?

      X_1^2 + ... + X_n^2 \sim \chi_n^2 via moment generating functions.

    Each   X_i^2 \sim \chi_1^2 and the sum is \chi_n^2
    Last edited by matheagle; August 29th 2009 at 11:10 PM.
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  3. #3
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    Quote Originally Posted by matheagle View Post
    It's been proved many times on this site that

      X_1^2 + ... + X_n^2 \sim \chi_n^2 via moment generating functions.

    Each   X_i^2 \sim \chi_1^2 and the sum is \chi_n^2
    I cannot seem to find such a post. Could you please recommend one?
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  4. #4
    MHF Contributor matheagle's Avatar
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    Let X\sim\Gamma(\alpha_1,\beta) and Y\sim\Gamma(\alpha_2,\beta)

    If they are independent, then the MGF of the sum is the product of the individual MGFs.

    So M_{X+Y}(t)=M_X(t)M_Y(t)= (1-\beta t)^{-\alpha_1}(1-\beta t)^{-\alpha_2}=(1-\beta t)^{-(\alpha_1+\alpha_2)}

    Thus X+Y\sim\Gamma(\alpha_1+\alpha_2,\beta)

    NOW, let \beta=2 which shows that
    the sum of any two independent \chi^2s
    is a \chi^2, you just add the degrees of freedom.
    This can easily be extended to n random variables.
    Last edited by matheagle; August 29th 2009 at 11:16 PM.
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