Estimate variance, how to find expected value of $x^2 [n]$

Question

We have data $x_0, x_1, \ldots, x_{N-1}$ where the $x_n$'s are independent and identically distributed as ${\rm Normal}(0,\sigma^2)$. The estimate of $\sigma^2$ is

$$\hat \sigma^2 = \frac{1}{N} \sum_{n=0}^{N-1} x_n^2 $$

To find its expected value, I don't understand how mean of $x_n^2$ comes. moreover, how to find variance of above equation, what is central moment?

Answer 1

If I understood correctly you need to find the mean and the variance of $\hat \sigma^2$?
Not dealing with the fact why did you use this estimate, one can do the following: $$\mathbb{E}\left[\hat \sigma^2\right]=\mathbb{E}\left[\frac{1}{N} \sum_{n=0}^{N-1} x_n^2\right]=\frac{1}{N} \sum_{n=0}^{N-1}\mathbb{E}\left[ x_n^2\right]=\mu_2'.$$ where $\mu_2'$ is the second raw moment.
And $$ \begin{eqnarray} \operatorname{Var}\left[\hat \sigma^2\right]&=&\mathbb{E}\left[\left(\hat \sigma^2-\mathbb{E}\left[\hat \sigma^2\right]\right)^2\right]=\mathbb{E}\left[\left(\hat \sigma^2\right)^2\right]-\left(\mathbb{E}\left[\hat \sigma^2\right]\right)^2 =\mathbb{E}\left[\left(\frac{1}{N} \sum_{n=0}^{N-1} x_n^2\right)^2\right]-\left(\mu_2'\right)^2=\\ &=& \frac{1}{N^2}\mathbb{E}\left[\left( \sum_{n=0}^{N-1} x_n^2\right)\left( \sum_{n'=0}^{N-1} x_{n'}^2\right)\right]-\left(\mu_2'\right)^2=\\&=&\!\!\frac{1}{N^2}\!\mathbb{E}\!\left[\!\sum_{n=0}^{N-1}\!x_n^4\!+ \!\sum_{n=0}^{N-1}\! \sum_{\substack{n'=0 \\ n'\neq {n}}}^{N-1} x_n^2x_n'^2\!\right]\!-\!\left(\!\mu_2'\!\right)^2\!=\!\!\frac{1}{N^2}\!\left(\!\sum_{n=0}^{N-1}\!\mathbb{E}\!\left[\!x_n^4\!\right]\!+\!\sum_{n=0}^{N-1}\! \sum_{\substack{n'=0 \\ n'\neq {n}}}^{N-1}\!\!\mathbb{E}\!\!\left[x_n^2x_n'^2\!\!\right]\!\!\right)\!-\!\!\left(\!\mu_2'\!\right)^2\!=\\&=&\frac{1}{N^2}\left(\mu_4'+N(N-1)\left(\mu_2'\right)^2\right)-\left(\mu_2'\right)^2=\frac{1}{N}\left(\mu_4'-\left(\mu_2'\right)^2\right) \end{eqnarray} $$ where $\mu_4'$ is the fourth raw moment.