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Math Help - Another Chi square - s^2 defintion

  1. #1
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    Another Chi square - s^2 defintion

    Dear Matheagle,
    You still there? Here's another one that I've been spending too much time on. Can you please help?

    The problem reads:

    Let X_1, X_2,..., X_n be iid r.v.s from Normal (\mu, \sigma). Given the fact that the rv

    X^2_n = \sum_{i=1}^n \left(\frac{x_i-\mu}{\sigma}\right)^2

    has a chi square distribution with degrees of freedom n.

    Show that

    \frac{(n-1)s^2}{\sigma^2} = \sum_{i=1}^n \left(\frac{x_i-\mu}{\sigma}\right)^2 - \left(\frac{\overline{x}-\mu}{\sigma/\sqrt{n}}\right)^2.

    Thanks a bunch!
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  2. #2
    MHF Contributor matheagle's Avatar
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    Dear Poisson Ivy,

    Use the definition of s^2

    {(n-1)s^2\over \sigma^2} = {\sum_{i=1}^n(x_i-\bar x)^2\over \sigma^2}

    = {\sum_{i=1}^n((x_i-\mu)+(\mu-\bar x))^2\over \sigma^2}

    = {\sum_{i=1}^n (x_i-\mu)^2+2(\mu-\bar x) \sum_{i=1}^n(x_i-\mu)+n(\mu-\bar x)^2\over \sigma^2}

    = {\sum_{i=1}^n (x_i-\mu)^2+2(\mu-\bar x) n(\bar x-\mu)+n(\mu-\bar x)^2\over \sigma^2}

    = {\sum_{i=1}^n (x_i-\mu)^2-2n(\bar x-\mu)^2+n(\bar x-\mu)^2\over \sigma^2}

    = {\sum_{i=1}^n (x_i-\mu)^2-n(\bar x-\mu)^2\over \sigma^2}

    = {\sum_{i=1}^n (x_i-\mu)^2\over \sigma^2}   -{n(\bar x-\mu)^2\over \sigma^2}

    = \sum_{i=1}^n {(x_i-\mu)^2\over \sigma^2}   -{(\bar x-\mu)^2\over \sigma^2/n}

    = \sum_{i=1}^n \left(\frac{x_i-\mu}{\sigma}\right)^2 - \left(\frac{\overline{x}-\mu}{\sigma/\sqrt{n}}\right)^2.
    Last edited by matheagle; December 13th 2009 at 12:51 AM.
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