"One should never bother with WLS. Using OLS with robust standard errors gives correct inference, at least asymptotically." True, false, or a bit of both? Explain carefully what the quote means and evaluate it critically

What will be an ideal response?

Answer: WLS is a special case of the GLS estimator. Furthermore, OLS is a special case of the WLS estimator. Both will produce different estimates of the intercept and the coefficients of the other regressors, and different estimates of their standard errors. WLS has the advantage over OLS, that it is (asymptotically) more efficient than OLS. However, the efficiency result depends on knowing the conditional variance function. When this is the case, the parameters can be estimated and the weights can be specified. Unfortunately in practice, as Stock and Watson put it, "the functional form of the conditional variance function is rarely known." Using an incorrect functional form for the estimation of the parameters results in incorrect statistical inference. The bottom line is that WLS should be used in those rare instances where the functional form is known, but not otherwise. Estimation of the parameters using OLS with heteroskedasticity-robust standard errors, on the other hand, leads to asymptotically valid inferences even for the case where the functional form of the heteroskedasticity is not known. It therefore seems that for real world applications the above statement is true.