# How to establish asymptotic normality of unbiased estimator and find asympotic variance?

Given a linear regression model with deterministic regressors $$y_{i} = x'_{i}\beta + \epsilon_i, \quad \epsilon \sim (0,\sigma^2) i.i.d., \quad i = 1, ...,n, \quad with \quad \mu_4 := \quad \mathbb{E}(\epsilon{^4_1})< \infty$$ where the regressor matrix has full rank $$k$$, and $$e$$ denotes the vector OLS residuals. How do I establish the asymptotic normality of of the unbiased estimator $$\hat{\sigma}^2 = \frac{e'e}{n-k} \quad of \quad \sigma^2$$, that is, $$\sqrt{n}(\hat{\sigma}^2{-\sigma^2}) \xrightarrow{d} N(0,v)$$ and find the asympotic variance $$v$$, using the Central Limit Theorem?

Since $$k$$ is fixed and $$\dfrac{n}{n-k} \rightarrow 1$$, it's not W.L.O.G to consider the estimator, $$\hat{\sigma}^2 = \dfrac{1}{n}e^Te$$ instead. After some matrix algebra, you can write,

$$\dfrac{1}{n}e^Te = \dfrac{1}{n}\epsilon^T\epsilon - \dfrac{1}{n}\epsilon^TH\epsilon$$

where $$\epsilon$$ is the vector containing the errors, $$\epsilon_i$$ and $$H = X(X^TX)^{-1}X^T$$ is the projection matrix in the OLS estimator. The second term converges to $$0$$ in probability by Chebyshev's inequality,

$$\begin{eqnarray} P\left(\frac{1}{n}\epsilon^TH\epsilon > \delta\right) \leq \dfrac{1}{n\delta}E[\epsilon^TH\epsilon] = \dfrac{\sigma^2}{n\delta}\text{tr}(H) \rightarrow 0, \;\; \forall \delta > 0 \end{eqnarray}$$

So you only have to consider the first term,

$$\dfrac{1}{n}\epsilon^T\epsilon = \dfrac{1}{n}\sum\limits_{i=1}^n\epsilon_i^2$$

Since $$E[\epsilon_i^4] < \infty$$, you can directly apply the CLT,

$$\sqrt{n}(\hat{\sigma}^2 - \sigma^2) = \sqrt{n}\left(\dfrac{1}{n}\sum\limits_{i=1}^n\epsilon_i^2 - \sigma^2\right) \xrightarrow{D} N(0, v)$$

where $$v = E[\epsilon_i^4] - \sigma^4$$.