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Math Help - Relation between degress-of-freedom for a sum of chi-square random variables

  1. #1
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    Relation between degrees-of-freedom for a sum of chi-square random variables

    Hello, everyone.

    Sorry, my LaTeX is working "part-time"

    Let

    X1 ~ U(0,1)
    X2 ~ U(0,1)

    X1 and X2 are independent.

    Now, consider

    x1 = 0.05
    x2 = 0.05

    Let:

    Z = \left [\chi^2_3 \right ]^{-1}\left ( 1-x_1 \right )
    Y = \left [\chi^2_1 \right ]^{-1}\left ( 1-x_2 \right )

    W = Z+Y

    where

    W ~ \chi^2_4

    In this case,

    W = 7.814+3.841 = 11.656

    Prob [ W> \chi^2_4 ] = 0.00209975 for an alpha level = 5%.


    So, my question is:

    Is there a way to create

    Z = \left [\chi^2_n \right ]^{-1}\left ( 1-x_1 \right )
    Y = \left [\chi^2_m \right ]^{-1}\left ( 1-x_2 \right )

    where n and m are large numbers (say, n>100000) - so that

    W ~ \chi^2_{m+n} and


    Prob [ W> \chi^2{m+n} ] is still equal (or approximately equal ) to 0.00209975 for an alpha level = 5%?
    Last edited by DerWundermann; July 24th 2013 at 09:25 AM.
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  2. #2
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    Re: Relation between degress-of-freedom for a sum of chi-square random variables

    Hey DerWundermann.

    Do you need to use simulation or an analytic approach?

    For an analytic approach, you will have to use theorems about product distributions. The distribution of 1 - X where X = U(0,1) still has the same distribution.

    The typical way of finding product distributions is to use either theorems regarding general product distributions or to use the characteristic transform which takes a distribution to Fourier space and back again, much in the same way that you take differential equations to fourier or laplace space and back to function space.

    Have you come across these techniques before?
    Thanks from DerWundermann
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  3. #3
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    Re: Relation between degress-of-freedom for a sum of chi-square random variables

    Many thanks, Chiro! No, I haven't. I am still dealing with basic issues in Statistics.
    But I need something less complex. So, if there is no simple solution, it is already solved.
    Thank you again.
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