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What is the difference between $\hat{e}^2_i$ and $\sigma^2_i$?

In regression, we assume that $var(e_i)=E(e_i^2)=var(y_i)=\sigma_i^2$.

Does this imply that $\sigma_i^2$(the sample variance) is equal to $\hat{e}_i^2$?

When we calculate FGLS (feasible generalized least squres), we consider $\ln(\hat{e}_i^2)=\ln(\sigma_i^2)+v_i$, where $v_i$ is just the error term. This seems that we distinguish $\sigma_i^2$ from $\hat{e}_i^2$.

But, when we calculate the robust robust standard error, we simply repalce $\sigma_i^2$ with $\hat{e}_i^2$. Thus, theses two are considered to be equal.

I am confused with these two terms. Can anybody clarify this?

Thank you in advance.

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Certainly the true variance is not equal to the squared residual.

In the case or robust standard errors, it has been proven that if we use $\hat{e}^2_i$ we obtain a consistent estimator of the variance-covariance matrix as a whole, even though $\hat{e}^2_i$ is not equal to $\sigma^2_i$ and even though each single $\hat{e}^2_i$ does not converge in probability to $\sigma^2_i$ (it converges to the true squared error, which is a random variable).

The fundamental reason for this result is that

$$E(\hat{e}^2_i) \to_{n \to \infty} E(e^2_i) =\sigma^2_i$$

So the OP's phrase

When we calculate the robust robust standard error, we simply repalce $\sigma^2_i$ with $\hat{e}^2_i$. Thus, theses two are considered to be equal.

is simply wrong.

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