# Asymptotic distribution of MLE of $\sigma$ for $N(0,\sigma^2)$

I know that given $$X_1,...,X_n \sim N(0,\sigma^2)$$, the MLE for $$\sigma$$ is $$\hat{\sigma} = \sqrt{\frac{1}{n}\sum_{i=1}^n (X_i - \bar{X})^2}$$. I want to find the asymptotic distribution for $$\hat{\sigma}$$. If I can find the asymptotic distribution $$\frac{1}{n}\sum_{i=1}^n (X_i - \bar{X})^2$$ then I could probably use the delta method to get what I am looking for. The problem is that I don't see how to apply the CLT to $$\frac{1}{n}\sum_{i=1}^n (X_i - \bar{X})^2$$ . Any suggestions?

• Why are you messing with the sample mean $\bar{X}$ when you know the mean is zero? Commented Apr 21 at 2:19

Your choice of MLE suggests you didn't actually write the likelihood for the specific model in question, which is $$\mathcal L (\sigma \mid x_1, \ldots, x_n) \propto \sigma^{-n} \exp \left( -\frac{\sum_{i=1}^n x_i^2}{2\sigma^2} \right),$$ hence the log-likelihood is $$\ell (\sigma \mid x_1, \ldots, x_n) \propto -n \log \sigma - \frac{\sum_{i=1}^n x_i^2}{2\sigma^2}.$$ Maximizing $$\ell$$ for a fixed sample yields a different MLE than what you wrote.
That said, you should know that $$(X_i/\sigma)^2 \sim \chi^2(1)$$, so $$\frac{1}{\sigma^2} \sum_{i=1}^n X_i^2 \sim \chi^2(n).$$ This is the key result you need to compute the distribution of $$\hat \sigma$$.