For a question like this, would I use WLS or Feasible GLS? I am leaning towards WLS but my lecture slides tell me GLS is used when there is more than one variable that drives the residual variance.
$\begingroup$
$\endgroup$
2
-
$\begingroup$ Hi @theshadowers. Potentially WLS when the variance of the error is known? (i.e multiplicative) $\endgroup$– EB3112May 30, 2022 at 20:31
-
1$\begingroup$ WLS = GLS in that case (with $w_i$ observed). The method mentioned in your lecture slides should be Feasible GLS, not GLS, I guess. FGLS = (WLS using estimated weights); GLS = (WLS using observed weights). It's a matter of the definitions of WLS, GLS, and FGLS. $\endgroup$– chan1142May 31, 2022 at 5:57
Add a comment
|