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I am kind of new to this matrix notation and properties so I would like to see the algebraic part of the solution it helps to understand so I appreciate your understanding.

My question is basically:

here from $y=X\beta+u$

we have that expression from GLS estimation and Central Limit Theorem.

$ \sqrt{n}\cdot(\hat{\beta} -\beta) = \left(\dfrac{(X^T \Omega^{-1}X)}{n} \right)^{-1} \cdot \dfrac{1}{\sqrt{n}} X^T \Omega^{-1} u$

We have that:

$\sqrt{n}\cdot(\hat{\beta} -\beta) \xrightarrow[]{d} N\left(0,\sigma^2 \left(\dfrac{(X^T \Omega^{-1}X)}{n} \right)^{-1}\right) $

do we have to assume $\lim$ of $ \left(\dfrac{(X^T \Omega^{-1}X)}{n} \right)^{-1}$ should be equal to some Positive definite, finite matrix like $Q$ as in the case of OLS like $\left(\dfrac{(X^T X)}{n} \right)^{-1} $ and how we can really reach that result ? or is there something different I should learn ? Thanks.

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    $\begingroup$ Take a look at this note. $\endgroup$ Commented Jan 6, 2023 at 19:37
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    $\begingroup$ Yes, we have to assume that it is finite and nonsingular (if it's nonsingular, it's positive definite by construction.) What is your concern - that you don't understand how you get to the limiting distribution of $\sqrt{n}\cdot (\hat{\beta}-\beta)$, or... $\endgroup$
    – jbowman
    Commented Jan 6, 2023 at 21:20
  • $\begingroup$ Yes, I was just confused about how to deal with the Omega term but the notes above are smooth and useful. thanks, both of you. $\endgroup$
    – Tatanik501
    Commented Jan 7, 2023 at 6:48

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