Hello All,
I would like to estimate a selection model where my dependent variable of interest is binary.
Is this a proper code?
proc qlim data=a;
model y1 = x1 x2 z1 z2 / discrete; where z1 and z2 are instrumental variables
model y2 = x1 x2 / select (y1 = 1) discrete;
run;
Additionally, one of the variables in my equation of interest is endogenous and I would like to add an instrument.
I would be very grateful for help with a code.
Best regards,
Aleksandra
Hi Alexandra,
With the developer's help...
You have the selection portion of the code correct. You need to add a third equation for the endogeneity. Try this....
proc qlim data=a;
model y1 = x1 x2 z1 z2 / discrete; where z1 and z2 are instrumental variables
model y2 = x1 x2 / select (y1 = 1) discrete;
model x1 = x2 z1 z2;
run;
Now, the consistency properties of this estimator, I can't comment on, but you should be able to estimate this thing.
Best-Ken
Hi Alexandra,
With the developer's help...
You have the selection portion of the code correct. You need to add a third equation for the endogeneity. Try this....
proc qlim data=a;
model y1 = x1 x2 z1 z2 / discrete; where z1 and z2 are instrumental variables
model y2 = x1 x2 / select (y1 = 1) discrete;
model x1 = x2 z1 z2;
run;
Now, the consistency properties of this estimator, I can't comment on, but you should be able to estimate this thing.
Best-Ken
Hi Ken,
Thank you very much for your help.
I have one more problem - a technical one this time. When I run a probit model with an endogenous variable and test for endogeneity, although there is no error, SAS prints only a description of variables and class level information, but no parameter estimates. Do you know how to fix it? I have no problems of this nature with other models.
Best regards,
Aleksandra
Aleksandra,
Verify for me that you are are one SAS/ETS 13.1 or higher by running
proc product_status; run;
and look at the output. If you are on an earlier edition then perhaps you want to give SAS Cloud Edition a shot here. SAS® Logon Manager
Let me know if you have any trouble. You should be able to create an account and log on. -Ken
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