# Asymptotic distribution of the sample variance

Consider the linear model $$y_i = \beta x_i +u_i$$ for $$i=1,...,n$$ where $$(x_i,y_i)$$ are i.i.d. and $$E(u_i\mid x_i)=0$$ while $$E(x_i^4)<\infty$$ and $$E(u_i^4)<\infty$$. Let $$n$$ be large.

Derive the asymptotic distribution for the sample variance $$s^2$$.

I have managed to show that $$E(s^2)=0$$

I have also shown that you can write the sample variance as $$s^2 = \frac{n}{n-1}[\frac{u'u}{n}-(\frac{u'X}{n})(\frac{X'X}{n})^{-1}(\frac{X'u}{n})]$$

However, I'm not sure how I would find the variance of $$\sqrt{n}(s^2-\sigma^2)$$, which I need to find the asymptotic distribution according to the CLT.

• How do you define $s^2$? Nov 6, 2021 at 19:17

You've shown that $$\sqrt{n}(\tilde{s}^2-\sigma^2)=\frac{1}{\sqrt{n}}\sum_{i=1}^n [u_i^2-\mathsf{E}u_1^2]+R_n,$$ where $$\tilde{s}^2=\boldsymbol{u}^{\top}\boldsymbol{u}/n$$ and $$R_n$$ is the remainder term. If $$R_n=o_p(1)$$, then the asymptotic distribution of $$\sqrt{n}(\tilde{s}^2-\sigma^2)$$, and, hence, of $$\sqrt{n}(s^2-\sigma^2)$$ is $$\mathcal{N}(0,\operatorname{Var}(u_1^2))$$, where $$\operatorname{Var}(u_1^2)=\mathsf{E}u_1^4-\sigma^4$$.