# Asymptotic distribution of $\sqrt{n}\left(\hat{\sigma_{1}^{2}}-\sigma^2\right)$

I'm trying to find a confidence interval for variance $$\sigma^{2}$$ when some sample $$X_{1},...,X_{n}$$, with mean $$\mu$$ known, may have violated normality assumption. To do this I'm investigating the asymptotic distribution of $$\sqrt{n}\left(\hat{\sigma_{1}^{2}}-\sigma^{2}\right)$$, where $$\hat{\sigma_{1}^{2}} = \frac{1}{n}\sum_{i=1}^{n}(X_{i}-\mu)^{2}$$, in the hopes that it can be used as a pivot.

However, I'm struggling with deriving the asymptotic distribution. I have a hint that you're meant to utilise the fact that $$\hat{\sigma_{1}^{2}}$$ is itself a mean of a sample from the $$(X-\mu)^{2}$$ distribution.

I've tried using the delta method, which states $$\sqrt{n}(\bar{X_{n}}-\mu)\rightarrow N(0,\sigma^2)$$, but I can't seem to get anywhere.

Any ideas on deriving the asymptotic distribution?

No delta method is required to get the asymptotic distribution of $$\widehat{\sigma}^2$$.

Asymptotic Distribution of $$\widehat{\sigma}^2 = \frac{1}{n}\sum_{i=1}^n (X_i-\overline{X})^2$$:

In the following we assume an iid sample $$X_1, \dots, X_n$$ with $$0, where $$\mathbb{E}(X_i)=\mu$$. We then have by the usual decomposition:

\begin{align*} \widehat{\sigma}^2 & = \frac{1}{n}\sum_{i=1}^n (X_i-\overline{X})^2\\ & = \frac{1}{n}\sum_{i=1}^n ((X_i- \mu) + (\mu-\overline{X}))^2\\ & = \frac{1}{n}\sum_{i=1}^n (X_i- \mu)^2 + 2 \frac{1}{n} \sum_{i=1}^n(X_i- \mu)\cdot(\mu-\overline{X}) + \frac{1}{n}\sum_{i=1}^n(\overline{X}-\mu)^2\\ & = \frac{1}{n}\sum_{i=1}^n (X_i- \mu)^2 + 2 \frac{1}{n}\sum_{i=1}^n (X_i\mu - X_i\overline{X}-\mu^2+\mu\overline{X}) + (\overline{X}^2-2\overline{X}\mu +\mu^2)\\ &=\underbrace{\frac{1}{n}\sum_{i=1}^n (X_i- \mu)^2}_{A} + \underbrace{(\overline{X}-\mu)^2}_{B} \end{align*} Part A is our main term. By assumption we have $$0 and hence, by an application of the Lindeberg-Lévy central limit theorem with $$\mathbb{E}((X_i-\mu)^2)=\sigma^2$$ we derive for Part A: $$\sqrt{n}(\widehat{\sigma}^2-\sigma^2) \stackrel{d}{\to}\mathcal{N}(0,\varsigma^2).$$

Part B on the other hand is asymptotical negligible: \begin{align*} \sqrt{n}B & = \sqrt{n}(\overline{X}-\mu)^2\\ & = (\sqrt{n}(\overline{X}-\mu))(\overline{X}-\mu) \\ & = (\sqrt{n}(\overline{X}-\mu))(0 + o_p(1)) \\ & = O_p(1)(0+o_p(1)) \\ & = o_p(1). \end{align*} In words: $$(\overline{X}-\mu)$$ converges in probablilty to the constant $$0$$. At the same time $$(\sqrt{n}(\overline{X}-\mu))$$ converges in distribution to a normal distributed random variable. Hence $$(\sqrt{n}(\overline{X}-\mu))(\overline{X}-\mu)$$ converges by the Slutsky theorem in distribution to $$0$$, and since $$0$$ is a constant, $$(\sqrt{n}(\overline{X}-\mu))(\overline{X}-\mu)$$ converges additionally to $$0$$ in probability.

We can thus conclude by another application of the Slutsky theorem that: $$\sqrt{n}(\widehat{\sigma}^2 - \sigma^2) \stackrel{d}{\to} \mathcal{N}(0,\varsigma^2),$$ where $$\varsigma^2= Var((X_i-\mu)^2) = \mathbb{E}((X_i-\mu)^4) - (\mathbb{E}(X_i-\mu)^2)^2 = \mathbb{E}((X_i-\mu)^4) - \sigma^4.$$

Extra: asymptotic $$(1-\alpha)$$-confidence interval for $$\sigma^2$$

Let $$z_{\alpha}$$ be the $$\alpha$$-quantil of a standard normal distribution. An asymptotical $$1-\alpha$$-confidence interval for $$\sigma^2$$ is then given by $$[\widehat{\sigma}^2- z_{1-\alpha/2} \frac{\varsigma}{n},\widehat{\sigma}^2 + z_{1-\alpha/2}\frac{\varsigma}{n}]$$ where we obviously have to replace the unknown $$\varsigma = \sqrt{\mathbb{E}((X_i-\mu)^4) - \sigma^4}$$ by a consistent estimator for $$\varsigma$$, i.e. using $$\widehat{\varsigma} = \sqrt{\frac{1}{n}\sum_{i=1}^n (X_i - \overline{X})^4 - \widehat{\sigma}^4}.$$

The confidence follows since: \begin{align*} 1-\alpha & = P\left(-z_{1-\alpha/2}\leq \sqrt{n}\frac{\widehat{\sigma}^2-\sigma^2}{\varsigma}\leq z_{1-\alpha/2}\right)\\ & = P\left(\widehat{\sigma}^2- z_{1-\alpha/2} \frac{\varsigma}{n} \leq \sigma^2 \leq \widehat{\sigma}^2 + z_{1-\alpha/2}\frac{\varsigma}{n}\right). \end{align*}