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We run OLS and found the Homoscedasticity is violated and Hence, we go for GLS.

But from variance-covariance of OLS's error - we have already found the Ω.

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Now, if we want to estimate β coefficients in GLS, we can directly substitute above Ω in the equation:

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I am unable to understand - If we already have Ω (from variance-covariance of OLS), then Why do we need to estimate Ω (by running another regression as we do in WLS). Shouldn't we directly substitute it?

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    $\begingroup$ Just to make sure we're clear: $\Omega_{OLS}$ is the variance-covariance matrix of the data (so is of size N), not the regression coefficients, whose variance-covariance matrix is given by $\sigma^2(\mathbf{X}^\top\mathbf{X})^{-1}$ in standard OLS (of size P). $\endgroup$
    – John Madden
    Commented Sep 14, 2022 at 12:54

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But from variance-covariance of OLS's error - we have already found the Ω.

This is dubious. The postulated model is

$$ \left\{\begin{align}\mathbf y &=\mathbf X\mathbf b +\boldsymbol\varepsilon,\\\mathbb E[\boldsymbol\varepsilon|\mathbf X]&= \mathbf 0,\\\mathbb E\left[\boldsymbol\varepsilon\boldsymbol\varepsilon ^\mathsf T |\mathbf X\right]&= \sigma^2\mathbf \Omega\end{align}\right\}.\tag{GLRM} $$

One cannot know, in general, beforehand, what the structure of $\bf \Omega$ be. If it can be known, then very well. Otherwise, one must need additional assumptions and structures for estimating the unknown parameters.

If we already have Ω, then Why do we need to estimate Ω

We don't. It is hardly the case that $\mathbf\Omega$ is known; once in a blue moon do we have the complete knowledge of $\mathbf\Omega.$ In case, it is known, symmetric, positive definite, then GLS can be employed to reach the Aitken estimator

$$\hat{\boldsymbol\beta} =\left(\mathbf X^\mathsf T\mathbf \Omega^{-1}\mathbf X\right)^{-1}\mathbf X^\mathsf T\mathbf \Omega^{-1}\mathbf y.\tag 1 $$

(by running another regression as we do in WLS).

As I said, if $\bf\Omega$ is not known, we cannot apply $(1): $ GLS is not feasible. The unknown parameters that it contains need to be estimated: since there are $n(n+1) /2$ unknown parameters and only $n$ observations, additional structure on the model has to be taken into consideration.

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  • $\begingroup$ But we have Ω from variance-covariance of OLS (see equation 1 in the question) $\endgroup$
    – pqrz
    Commented Sep 14, 2022 at 10:51
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    $\begingroup$ What is $\varepsilon_{\text{OLS}}$ in the expression? Is it $\mathbf e$ in $\mathbf y =\mathbf X\mathbf b +\mathbf e? $ $\endgroup$
    – User1865345
    Commented Sep 14, 2022 at 11:08
  • $\begingroup$ Yes, you are right .... ε = Residuals of OLS $\endgroup$
    – pqrz
    Commented Sep 14, 2022 at 12:00
  • $\begingroup$ I am assuming you are talking about $\mathbb E\left[\boldsymbol\varepsilon\boldsymbol\varepsilon ^\mathsf T \right]$ and not $\mathbb E\left[\hat{\boldsymbol\varepsilon}\hat{\boldsymbol\varepsilon} ^\mathsf T \right].$ $\endgroup$
    – User1865345
    Commented Sep 14, 2022 at 12:07
  • $\begingroup$ Yes, the predicted ε .... (since we can never know the true ε) $\endgroup$
    – pqrz
    Commented Sep 14, 2022 at 12:10

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