Distribution of sample variance from normal distribution

Question

Assuming $N$ samples $\{x_1,...,x_N\}$ are taken from a normal distribution with mean $\mu$ and variance $\sigma^2$, then the variance can be estimated using \begin{equation} s_1^2=\frac{1}{N-1}\sum_{i=1}^N(x_i-\overline{x})^2. \end{equation} According to wikipedia $(N-1)s_1^2/\sigma^2$ has a chi-squared distribution with $(N-1)$ degrees of freedom.

In my case I know that $\mu=0$ and want to consider \begin{equation} s_2^2=\frac{1}{N}\sum_{i=1}^Nx_i^2 \end{equation} instead. Can $s_2^2$ also be linked to some known distribution?

Answer 1

$\displaystyle \sum_{i=1}^N \left(\frac{x_i-\mu}{\sigma}\right)^2$ has a $\chi_N^2$ distribution (chi-squared with $N$ degrees of freedom) as as the sum of $N$ independent standard normal random variables.

So if $\mu=0$ and $\displaystyle s_2^2=\frac{1}{N}\sum_{i=1}^Nx_i^2$ then $N s_2^2 / \sigma^2$ also has a $\chi_N^2$ distribution.