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RE: st: Does the xtabond command have a mistake???
Thanks for the clarification. I'll keep a look out for the new update, but meanwhile probably my best bet is to "manually" create lags a la
gen lag_x = L.x and then include that in the predetermined vars expression:
pre(x lag_x , lag(0,#) The reason I have to do it that way is because stata won't accept
pre(x L.x , lag(0,#) I have been wondering why it refuses L.x. Is it an easy fix to reprogramme xtabond so that entering time series operators in the pre( ) won't cause an error message?
Date: Thu, 06 May 2004 17:53:55 -0500
From: "David M. Drukker, StataCorp" <email@example.com>
Subject: RE: st: Does the xtabond command have a mistake???
Tewodaj Mogues <firstname.lastname@example.org> asked about a problem with
-xtabond- that occurs with lags of predermined variables. We have
researched the issue and concluded that he is correct. -xtabond- is using
levels lagged one or more periods as instruments for predetermined
variables, regardless of the number of lags of the predermined variables
included in the model. -xtabond- should only be using levels lagged p+1 or
more times, where p is the number of lags of the predetermined variables in
This problem implies that the current -xtabond- produces inconsistent
estimates of the parameters in models with lags of the predetermined
variables. For instance, estimates from the model
. xtabond y , pre(x , lag(1, .))
are not consistent.
This problem does not arise when there are no lags of the predetermined
variables. For instance, the current -xtabond- produces consistent
estimates from the model
. xtabond y , pre(x)
-xtabond- also produces consistent estimates for the models
. xtabond y , lags(2)
. xtabond y l(0/1).x1 , pre(x2)
. xtabond y l(0/1).x1 , pre(x2, lag(0, 1))
The problem will be fixed in an upcoming ado update that we expect to have
in distribution early next week.
Dept. of Agricultural and Applied Economics
University of Wisconsin - Madison
427 Lorch St. #317, Taylor Hall
Madison, WI 53706
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