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Math Help - Proving question

  1. #1
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    Proving question

    Hi,

    given that: S^2=\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\overline X)^2.

    I want to show that E[S^2]=\sigma^2

    However, when i reached: \frac{n}{n-1}(\frac{1}{n}\sum_{i=1}^{n}E[X_i^2]-E[\overline X^2]), i was puzzled on finding E[\overline X^2]?

    Since E[\overline X^2]=Var(\overline X) + [E(\overline X)]^2 which variance should I use, s.r.s or i.i.d?

    or maybe is there any better way to prove it?
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  2. #2
    Super Member Random Variable's Avatar
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     \frac{(n-1)S^{2}}{\sigma^{2}} is  \chi^{2}_{n-1} (chi-square distriubtion with (n-1) degrees of freedom)

    If you want to try and prove that statement, use the fact that  S^{2} and \bar{X} are independent.

    So  E(S^{2}) = E\Big(\frac{\sigma^{2}}{n-1} \ \frac{(n-1)S^{2}}{\sigma^{2}} \Big) = \frac{\sigma^{2}}{n-1} E\Big(\frac{(n-1)S^{2}}{\sigma^{2}} \Big) = \frac{\sigma^{2}}{n-1} (n-1) = \sigma^{2}

    since the expected value of a chi-square distribution with k degrees of freedom is k
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  3. #3
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    Hi tks for the reply.

    In addition to this, I've read book stating that the best estimator is (1-\frac{1}{N})s^2 as compared to this answer showing that E[s^2]=\sigma ^2.

    Any explaination for this?
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  4. #4
    Super Member Random Variable's Avatar
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    Here's a much better proof that doesn't involve knowing anything about the chi-square distribution:

     \sum^{n}_{i=1} (X_{i} -\overline{X})^{2} = \sum^{n}_{i=1} X_{i}^{2} -n\overline{X}^{2}

    so  E\Big(\sum^{n}_{i=1} (X_{i} -\overline{X})^{2}\Big) = E\Big(\sum^{n}_{i=1} X_{i}^{2}\Big) - E(n\overline{X}^{2}) = \sum^{n}_{i=1} E(X_{i}^{2}) - nE(\overline{X}^{2})

     = \sum^{n}_{i=1}(\sigma^{2}+ \mu^{2})-n \Big(\frac{\sigma^{2}}{n}+ \mu^{2}\Big) = n(\sigma^{2}+ \mu^{2})- \sigma^{2}- n\mu^{2} = (n-1)\sigma^{2}

    then  E(S^{2}) = E\Big(\frac{1}{n-1} \sum^{n}_{i=1} (X_{i} -\overline{X})^{2}\Big) = \frac{1}{n-1} E\Big(\sum^{n}_{i=1} (X_{i} -\overline{X})^{2}\Big) = \frac{1}{n-1}(n-1)\sigma^{2} = \sigma^{2}
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