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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?

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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 $0<Var((X_i-\mu)^2) = \varsigma^2 <\infty$, 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 $0<Var((X_i-\mu)^2) = \varsigma^2 <\infty$ 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*}

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