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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 ...

$E[\hat{\sigma_X}^2]=\dfrac{1}{n}E[\sum_{i=1}^{n}{(X_i-\mu)^2}]$

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

I don't know what else to do.

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

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    sorry, now understand, please will be able to development this expression $\mathsf{E}\left(\hat{\sigma}_X^2\right) = \frac{n-1}{n} \sigma_X^2$... What steps remain? ...2012-09-01

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