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Math Help - Chi-Square Variance Ratio Proof

  1. #1
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    Chi-Square Variance Ratio Proof

    How would you go about proving that

    [ ( n - 1 ) * s^2 ] / σ^2


    has a Chi-Square Distribution?
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  2. #2
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    Hello,

    Some information necessary about s^2
    Anyway, I can just guess what it is. Note that aX~N(0,a), where X~N(0,1)
    Now think about what happens if you divide a normal distribution N(0,σ) by σ !

    And finally, remember that a chi-square distribution is the sum of squares of iid random variables following a N(0,1)
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  3. #3
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    One way to get at this is to assume \sigma^2 = 1, prove it for the case n = 2, then induct by showing that

    (n - 1) S^2_n = (n - 2) S^2_{n - 1} + \left(\frac{n - 1}{n}\right)(X_n - \bar{X}_{n - 1})^2,

    which will get you where you need since you will have the sum of two independent chi-squares. Some facts that you will use are that the sum of independent chi-squares is chi-square, that the square of an N(0, 1) is \chi^2_1, and that \bar{X}_n \perp S^2_n. The definitions are:

    S^2 _n = \frac{\sum_{i = 1} ^ n (X_i - \bar{X}_{n})^2}{n - 1}<br />
    \bar{X}_n = \sum_{i = 1} ^ n \frac{X_i}{n}

    Once you have this, it's just a matter of undoing the assumption that \sigma^2 = 1, which isn't really a big deal. To be honest, if this is just for kicks, it's actually more of a pain in the head than it appears, particularly in proving the identity above.
    Last edited by mr fantastic; May 14th 2010 at 09:45 PM.
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