Try it.  Suppose you have, like OP suggests, two simple regression models, say:  y=x1 x2 x3 , and  y=x1 x2 x3 x4 if you fit the second model and the x4 parameter estimate is 2.3, then if you fit the model y-2.3*x4=x1 x2 x3 you get the same parameter estimates for x1-x3 and you can then combine these data sets and make a comparison of, say, the two x1 parameters. ... View more

I figured it out. Thank you very much, it works as you said. ... View more

data have; input id1 @6 id2 $ @9 time yymmdd8.; format time date9.; cards; 1 A 20131128 1 B 20140214 1 C 20140530 1 C 20140622 1 D 20140831 1 D 20141220 1 A 20150217 1 A 20150302 1 C 20150410 1 D 20150425 1 E 20150609 1 F 20151025 1 C 20160612 1 D 20180101 2 A 19900515 2 B 19900813 2 E 19910522 2 A 19910524 2 F 19910919 2 G 19920101 2 A 19930321 ;run; proc sql; select *,(select count(distinct id2) from have where id1=a.id1 and time le a.time) as count1, (select count(distinct id2) from have where id1=a.id1 and time between a.time-360 and a.time) as count2 from have as a; quit; ... View more

This has been very helpful. Thank you very much for your kind and quick replies. ... View more