# Heteroskedasticity-consistent standard errors

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 $$\hat{\sigma}^2 = \frac{1}{N}\sum_{i=1}^{n}\hat{u}_i^2$$?

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.

• Where did I claim that sample size is one? The estimator indicates that there are n observations available. – Snoopy Oct 30 '18 at 12:44
• See update. $\hat u_i$ is predominantly based on one observation only. – StasK Oct 30 '18 at 12:55
• 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. – Snoopy Oct 30 '18 at 13:10
• 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. – StasK Oct 30 '18 at 13:42
• That is already in the Wikipedia article. I do not see the point. – Snoopy Oct 30 '18 at 14:01