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See https://en.wikipedia.org/wiki/Heteroscedasticity-consistent_standard_errors. Assume the model of interest is the linear regression model. If the errors are heteroskedastic, $\hat{\sigma}^2_i = \hat{u}_i^2$ is a consistent estimator of $\sigma^2_i$. If the errors are homoskedastic, $\hat{\sigma}^2 = \frac{1}{N}\sum_{i=1}^{n}\hat{u}_i^2$ is a consistent estimator of $\sigma^2 = \sigma^2_i$. When errors are homoskedastic, why $\hat{\sigma}^2_i = \hat{u}_i^2$ is not an estimator of $\sigma^2 = \sigma^2_i$, and that instead we have to use the estimator $\hat{\sigma}^2 = \frac{1}{N}\sum_{i=1}^{n}\hat{u}_i^2$?

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You cannot have a consistent estimator based on the sample of size 1 -- that is when you have only one residual $\hat u_i$. The expected value of that residual may be converging to $\sigma^2$ in the homoskedastic case, or to $\sigma_i^2$ in the heteroskedastic case, provided the model is correctly specified, but that expected value is an extremely confusing construct -- it surely confused you on this occasion.

Update:

$$ \mathop{\mathbb P}[ \hat u_i/\sigma^2_i > a > 1] \to \mathop{\mathbb P}[ u_i/\sigma^2_i > a] = \mbox{assume normality} = \mathop{\mathbb P}[ \chi^2_1 > a] $$

does not go to zero. Note that I gave you a huge credit in treating $\hat u_i$ as a consistent estimator of the true residual $u_i$, and frankly I am not even convinced of that.

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  • $\begingroup$ Where did I claim that sample size is one? The estimator indicates that there are n observations available. $\endgroup$ – Snoopy Oct 30 '18 at 12:44
  • $\begingroup$ See update. $\hat u_i$ is predominantly based on one observation only. $\endgroup$ – StasK Oct 30 '18 at 12:55
  • $\begingroup$ I do not understand this. That residual is not calculated based on one observation only. Furthermore I no where claimed that the residual is a consistent estimator of the error. $\endgroup$ – Snoopy Oct 30 '18 at 13:10
  • $\begingroup$ You need to make your story straight. Your third sentence is, If the errors are heteroskedastic, $\hat{\sigma}^2_i = \hat{u}_i^2$ is a consistent estimator of $\sigma^2_i$. That's the claim I was responding to. $\endgroup$ – StasK Oct 30 '18 at 13:42
  • $\begingroup$ That is already in the Wikipedia article. I do not see the point. $\endgroup$ – Snoopy Oct 30 '18 at 14:01

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