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My book introduces $\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n \hat{e}_i^2$, which makes sense to me since to my understanding $\hat{\sigma}^2$ is the error variance estimator. However, I have also seen $\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \overline{X})^2$ such as in this problem (Asymptotic distribution of $\sqrt{n}\left(\hat{\sigma_{1}^{2}}-\sigma^2\right)$) that I was working on initially and $\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (Y_i - \overline{Y})^2$ ( https://math.stackexchange.com/questions/688035/properties-of-hat-sigma2-bias-and-variance). Are these all referring to the same thing? If so, can anyone point me in the direction of proving their equality?

Thank you all very much.

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No, assuming typical regression notation is used, $\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n \hat{e}_i^2$ is an estimate for residual variance (since $\bar e = 0$), $\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \overline{X})^2$ is an estimate for feature/independent variable variance, and $\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (Y_i - \overline{Y})^2$ is an estimate for the target variable's variance.

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    $\begingroup$ Ok, thank you! So if a problem asks me to find the asymptotic distribution of $\sqrt{n}(\hat{\sigma}^2 - \sigma^2)$ in a regression context, do I assume it to be the residual variance? $\endgroup$
    – stressed
    Mar 28, 2021 at 21:20
  • $\begingroup$ I'd assume so, yes. That's the most likely one. $\endgroup$
    – gunes
    Mar 28, 2021 at 21:37

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