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: $E(u^2x'x) = \sigma^2 E(x'x)$
