# asymptotic normality of Generalized Least Square

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.

• Take a look at this note. Commented Jan 6, 2023 at 19:37
• 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... Commented Jan 6, 2023 at 21:20
• 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. Commented Jan 7, 2023 at 6:48