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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.

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

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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 $\operatorname{Var}(u_1^2)=\mathsf{E}u_1^4-\sigma^4$.

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