Assume $X$ is a random variable from a population with normal distribution. Using the likelihood function I get the expression below: $\hat{\sigma}_X^2 = \sum_{i=1}^n \frac{(X_i-\mu)^2} n$ for variance.
I want prove that, $\operatorname{E}[\hat{\sigma}_X^2]=\frac{n-1} n \sigma_X^2$.
I begin ...
$\operatorname{E}\left[\hat{\sigma}_X^2\right]=\dfrac 1 n \operatorname{E} \left[\sum_{i=1}^n (X_i-\mu)^2\right]$
$\frac 1 n \left( \operatorname{E}\left[(X_1-\mu)^2\right] + \operatorname{E}\left[(X_2-\mu)^2\right] + \cdots + \operatorname{E} \left[ \left( X_n-\mu \right)^2 \right]\right)$
pdta: $\mu$ is a theorical mean (not estimator of mean)