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I want to find an expression for the asymptotic variance of the OLS estimator given that the errors are heteroskedastic.

I have understood the derivation using CLT for the homoskedastic case. I.e.,

$\sqrt N (\hat{\beta} - \beta) \stackrel{a}{\sim}N(0,Q_{xx}^{-1}BQ_{xx}^{-1}) $

Where $Q_{xx} = E(x'x)$ and $B = E(u^2 x'x)$.

So under homoskedasticity I can use the law of iterated expectation to simplify $B$ as $E(E(u^2|x)x'x) = \sigma^2E(x'x)$. Where $E(u^2|x) = \sigma^2$.

So the asymptotic variance becomes $Avar(\hat{\beta}) = Q_{xx}^{-1}\sigma^2$

Now consider the case of heteroskedastic errors. E.g., $E(u^2|x) = f(x) = \theta x^2$.

Under this specification, the previous asymptotic distribution still holds:

$\sqrt N (\hat{\beta} - \beta) \stackrel{a}{\sim}N(0,Q_{xx}^{-1}BQ_{xx}^{-1}) $

But what I am having trouble with is how to simplify $B$. I see no reason why I can't use LIE again to express $B$ the same way as before $E(E(u^2|x)x'x) = E(u^2|x)E(x'x) = \theta x^2 E(x'x)$.

If I can make this substitution the rest of the derivation should just be some algebra. The reason I am confused is because in Wooldridge he says

A $\textbf{homoskedasticity}$ assumption simplifies the form of OLS asymptotic variance: $E(u^2x'x) = \sigma^2 E(x'x)$

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  • $\begingroup$ Hi: Under heteorscedasitcity, each "observation-row" has its own $\sigma^2_{i}$ so there no longer is a $\sigma^2$. Halbert White proved that the asymptotic variance of his expression is consistent so look at his econometrica paper. The title is something like "consistent heteroscedastic autocovariance estimation." or just google for "halbert white covariance matrix". if you can't find it, let me know. $\endgroup$ – mark leeds Oct 31 '18 at 22:54
  • $\begingroup$ It's here for free. pdfs.semanticscholar.org/f9cc/… $\endgroup$ – mark leeds Oct 31 '18 at 22:58
  • $\begingroup$ Thanks for the resource, I'll check it out and see if it clarifies things. $\endgroup$ – BenBernke Nov 1 '18 at 18:31
  • $\begingroup$ Hi: it's one of his seminal papers so not an easy read but I think the idea is what's important. IIRC, the one residual is used as the estimator. $\endgroup$ – mark leeds Nov 1 '18 at 21:38

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