Question about biased estimator

Question

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

I want prove that, $E[\hat{\sigma_X}^2]=\dfrac{n-1}{n}\sigma_X^2$.

I begin ...

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

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

pdta: $\mu$ is a theorical mean (not estimator of mean)

Answer 1

Hint: expand out $(X_i - \mu)^2 = X_i^2 - 2 X_i \mu + \mu^2$. Now compute $E[X_i^2]$, $E[X_i \mu]$ and $E[\mu^2]$.

EDIT: If $\mu$ is the actual mean rather than an estimator, then the statement is wrong: $E[(X_i - \mu)^2] = \sigma^2$ and $E[\hat{\sigma}_X^2] = \sigma^2$.