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Math Help - Problem understanding the theorem

  1. #1
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    Problem understanding the theorem

    Theorem : Is S^2 is the variance of a random sample from a infinite population with the finite variance \sigma^2 , then E(S^2) = \sigma^2
    Proof:

    E(S^2) = E \left[ \frac{1}{n-1} . \sum_{i=1}^{n} (X_i - \bar X)^2 \right]..............Understood!!!!



    E(S^2) =  \frac{1}{n-1} . E \left[\sum_{i=1}^{n} [(X_i - \mu) - (\bar X - \mu)]^2 \right].............Understood!!!!

    E(S^2) =  \frac{1}{n-1} . \left[\sum_{i=1}^{n} E[(X_i - \mu)^2] - n E[(\bar X - \mu)^2] \right]...................Understood!!!!


    since E[(X_i - \mu)^2] = \sigma^2 and E[(\bar X - \mu)^2] = \frac{\sigma^2}{n} it follows that
    <br /> <br />
E(S^2) = \frac{1}{n-1} \left[\sum_{i=1}^{n} \sigma^2 - n \frac{\sigma^2}{n}\right]...............Understood!!!!


    = \sigma^2....................Couldnt Understood!!!!

    How did he get
    \sigma^2
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  2. #2
    MHF Contributor matheagle's Avatar
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    the sum of a constant n times is n times that constant...

    E(S^2) = \frac{1}{n-1} \left[\sum_{i=1}^{n} \sigma^2 - n \frac{\sigma^2}{n}\right]

     = \frac{1}{n-1} \left[n \sigma^2 -  \sigma^2\right]

     = \frac{\sigma^2 }{n-1} \left[n-1\right]

    = \sigma^2
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  3. #3
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    thanks
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