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Let be $X$ a population with normal distribution. Using likelihood function I get the below expression $\hat{\sigma_X}^2 = \sum_{i=1}^{n}{\dfrac{(X_i-\mu)^2}{n}}$ for variance.

I want prove that expression is consistent, i.e. $E[\hat{\sigma_X}^2]=\sigma_X^2$.

I begin ...


$\dfrac{1}{n}(E[(X_1-\mu)^2]+E[(X_2-\mu)^2]\cdots E[(X_n-\mu)^2])$

I don't know what else to do.

pdta: $\mu=\dfrac{1}{n}\sum_{i=0}^{n}{X_i}$

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Do you intend the parameter $\mu$ to be a fixed constant, of a random variable, function of $X_i$? – Sasha Aug 31 '12 at 23:31
yes, $\mu=\dfrac{1}{n}\sum_{i=0}^{n}{X_i}$ – juaninf Aug 31 '12 at 23:58
In that case, it is a well known result that $\mathsf{E}\left(\hat{\sigma}_X^2\right) = \frac{n-1}{n} \sigma_X^2$. – Sasha Sep 1 '12 at 0:02
but, in mi lecture would have to be $E[\hat{\sigma_X}^2] = \sigma_X^2$ and not $\mathsf{E}\left(\hat{\sigma}_X^2\right) = \frac{n-1}{n} \sigma_X^2$...(for the consistent)... then this estimative is not consistent? – juaninf Sep 1 '12 at 0:08
The proper term is $\hat{\sigma}_X^2$ is a biased estimator. Consistency has to do with large $n$ limit. $\hat{\sigma}_X^2$ is consistent, since the expectation approaches the population value for large $n$. – Sasha Sep 1 '12 at 0:10

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