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