Consider a model $Y=X\beta+e$ and define the OLS error variance estimator as $\hat{\sigma^2}=1/n\sum_{i=1}^n \hat{e}_i^2$, assuming that $ E[e_i^2]=\sigma^2 $ More precisely I want to derive the asymptotic distribution of $\sqrt{n}(\hat{\sigma^2}-\sigma^2)$.

I can show that $E[\hat{\sigma^2}]=(n-k/n) \sigma^2$, and also $\hat{\sigma^2}\xrightarrow{p}\sigma^2$ . Maybe I missed something but it seems quite weird to derive the variance of $\sigma^2)$, as I did not see any one do so. Please help me if there is any other way to derive the asymptotic distribution.

  • $\begingroup$ What you define as OLS variance is not what textbooks and academic paper define as the OLS variance estimator. The OLS estimator is unbiased, while yours is biased. Yours is the maximum likelihood variance under a Gaussian likelihood. I suggest you edit your post to avoid conflict with the standard terminology. Also, in the OLS setting, $E[e_i^2]=\sigma^2$ holds by definition, so you do not need to assume that. Also, if I remember correctly, the quantity you are after will depend on the distributional assumption for $e$. What distribution do you assume $e$ to follow? $\endgroup$ Dec 3, 2023 at 7:28
  • $\begingroup$ Finally, Cross Validated Stack Exchange might be a better fit for your question. It has actually been asked there: see this and this. And here is a somewhat related thread. $\endgroup$ Dec 3, 2023 at 7:46
  • $\begingroup$ Thanks! It's really helpful! $\endgroup$
    – Paul Huang
    Dec 3, 2023 at 8:40


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