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Thread: Vector of independent normal RV's - mgfs

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

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

    Prove that $\displaystyle ||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?

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

    Each $\displaystyle X_i^2 \sim \chi_1^2 $ and the sum is $\displaystyle \chi_n^2 $
    Last edited by matheagle; Aug 29th 2009 at 10:10 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

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

    Each $\displaystyle X_i^2 \sim \chi_1^2 $ and the sum is $\displaystyle \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 $\displaystyle X\sim\Gamma(\alpha_1,\beta)$ and $\displaystyle Y\sim\Gamma(\alpha_2,\beta)$

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

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

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

    NOW, let $\displaystyle \beta=2$ which shows that
    the sum of any two independent $\displaystyle \chi^2$s
    is a $\displaystyle \chi^2$, you just add the degrees of freedom.
    This can easily be extended to n random variables.
    Last edited by matheagle; Aug 29th 2009 at 10:16 PM.
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