Hi,

I am running a simple quantile (median) regression that has a dummy variable on the RHS. The y-variable is continuous. I am using qreg with the wls( ) option. Unfortunately, I cannot share my output nor data but the problem can be reproduced running:

use http://www.stata-press.com/data/r15/auto, clear
qreg price weight length foreign, wls(1)
qreg price weight length foreign, wls(2)
qreg price weight length foreign, wls(3)
....
qreg price weight length foreign, wls(7)
qreg price weight length foreign, wls(8)

where foreign is a dummy variable.


When the number of wls iterations is 1 to 7 I get the message that alternate solutions exist. If I have more that 7 wls iterations I no longer get the message. My questions are:

1- Why I no longer get the message for wls>7? I understand that changing wls() changes the set of initial conditions for the linear programing algorithm but I would still think that the problem has multiple solutions.

2- Does it matter after which iteration of the linear programming algorithm the mesage of "alternate solutions exists" appears? Is that informative of anything?

3- Regardless of the answer to 1- and 2-, I need to retrieve the alternate solutions when wls =1,2,3,..,7. How can I do that? I've been reading quite a bit about this issue. Some sugegsted a small perturbation on the quantile but it is unclear (to me) what that perturbation should be. I would want a perturbation such that all coefficients are the same except for the dummy variable one which has multiple solutions. Is there a more straightforward way to retrieve these alternate solutions than a perturbation?

4- I was surprised to see that when changing the number of wls iterations from 1,...,100 the reported solution is always the same for the dummy variable. In a few different exercises I tried with three other datasets I got that as I was changing the number of wls iterations the dummy variable would oscilate between two possible solutions all else equal. I thought this was expected behavior and it seems it may not be?


Thank you so much for your help. I am happy to clarify anything needed.

Daniela