Estimating the variance of the GLS estimator

Question

Consider the linear regression model where $y = XB + u$. Assume that $\mathrm{E}[u \mid X] = 0$. Assume that $\mathrm{V}[u \mid X] = \sigma^2I$. This is the simple linear model. Now relax the homoskedasticity assumption so that $\mathrm{V}[u \mid X] = \sigma^2 \Delta$. Assume that $\Delta$ is known but $\sigma^2$ is not known.

Question 1: Can the GLS estimator be used? The GLS estimator is given by $\hat{\beta}_{GLS} = [X^{\top}(\sigma^2\Delta)^{-1}X]^{-1}X^{\top}(\sigma^2\Delta)^{-1}y = [X^{\top}\Delta^{-1}X]^{-1}X^{\top}\Delta^{-1}y$. So the unknown $\sigma^2$ drops and $\Delta$ is known. So the GLS estimator can be used. We do not need the FGLS estimator.

Question 2: Can the variance of the GLS estimator be used? The variance of the GLS estimator is given by $\hat{\beta}_{GLS}$ given by $\mathrm{Var}[\hat{\beta}_{GLS} \mid X] = \sigma^2 (X'\Omega^{-1}X)^{-1}$. $\Delta$ is known but $\sigma^2$ is not known. So we have to estimate $\sigma^2$ first and use it to estimate $\mathrm{Var}[\hat{\beta}_{GLS} \mid X]$. $\sigma^2$ can be consistently estimated by $\frac{1}{N-K}\sum{\hat{u}_i}^2$ where $\hat{u}_i$ is the residual from the OLS estimation of $y = XB +u$.

Are these answers all correct? Especially the answer of the second question?

Answer 1

Assume that $\Delta$ is known, everything is simple.

Original Model $$Y=X\beta + u$$ can be convert to $$ Y_n=X_n\beta + \epsilon $$ where $Y_n=\Delta^{-1/2}Y$, $X_n=\Delta^{-1/2}X$, and $\epsilon = \Delta^{-1/2}u$. Obviously, $\epsilon \sim N(0, \sigma^2 I)$. Then the new model met the general linear model requirements. From here you can estimate $\beta$s and $\sigma^2$ following the linear model. $\hat \beta_{GLS}$ in your question is correct. But the estimate of $\sigma^2$ should come from the residual of $Y_n$, i.e., $Y_n-X_n\hat \beta_{GLS}$.