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Math Help - showing that a quantile relation holds

  1. #1
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    showing that a quantile relation holds

    Will somebody please help answer this question?

    Use the result of F(3,4) = 1/F(4/3) about F distribution to show that the quantile relation f'subscript'(0.95,3,4) = 1/f'subscript'(0.05,4,3).
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  2. #2
    MHF Contributor matheagle's Avatar
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    Using upper percentile points....

    Which means that a=P(F_{3,4}>F_{3,4,a})

    we have with a=.95............

    .95=P(F_{3,4}>F_{3,4,.95})

    (Note that F_{3,4} is a rv and F_{3,4,a} is a percentile point.)

    Taking the reciprocal inside the probability we have

    .95=P(F_{3,4}>F_{3,4,.95})

    =P(F^{-1}_{3,4}<F^{-1}_{3,4,.95})

    =P(F_{4,3}<F^{-1}_{3,4,.95})

    (Which follows by the definition of an F, it's a ratio of chi-squares divided by their dfs.)

    And note that

    P(F_{4,3}<F_{4,3,.05})=.95 also.

    Thus by the continuity of the densities, these percentile points must be equal....

    Hence F^{-1}_{3,4,.95} must equal F_{4,3,.05}

    and this is true for any probability, not just .05.
    Last edited by matheagle; February 20th 2010 at 12:24 AM.
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