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st: lags of the x's in xtabond
You have a good point, that's probably what's going on.
Date: Tue, 4 May 2004 12:32:06 +0200
From: Giovanni Bruno <email@example.com>
Subject: Re: st: lags of the x's in xtabond
It seems to me that
1. xtabond y , lags(1) pre(x, lag(1,1) )
2. xtabond y , lags(1) pre(x lag_x, lag(0,1) )
actually provide different estimators for the same model:
y(it) = y(i t-1) + x(it) + x(i t-1) + e(it).
While method 1 includes L1.x (but not further lags) as instruments, method 2
includes both L1.x and L1.lag_x, which is L2.x, as instruments. This also
explains why method 2. is equivalent to
3. xtabond y , lags(1) pre(x, lag(1,2) )
which still considers the same model but includes L1.x and L2.x as
instruments, exactly as method 2.
Dept. of Agricultural and Applied Economics
University of Wisconsin - Madison
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Madison, WI 53706
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