# st: RE: xtabond machinations

 From "Salvati, Jean" <[email protected]> To <[email protected]> Subject st: RE: xtabond machinations Date Wed, 8 Dec 2004 16:38:07 -0500

```> 1) I thought that <xtabond> would run the following 4
> regressions in the first stage:
>
> y2-y1 = a * y1 + b * x1 --> predict y2-y1
> x2-x1 = a * y1 + b * x1 --> predict x2-x1
>
> y3 - y2 = a * y1 + b * y2 + c * x1 + d * x2 --> predict y3 - y2
> x3 - x2 = a * y1 + b * y2 + c * x1 + d * x2 --> predict x3 - x2
>
> Then I thought it would use those predictions in the second
> stage to estimate the first eqn above. When carry out these
> steps manually, however, I don't get the same results as when
> I just use <xtabond>. The code for using <xtabond> would look
> like this:
>
> ge x_lag = L.x
> xtabond y, pre(x_lag, endog)

This is not at all what xtabond does.

> Which part of what <xtabond> am I misunderstanding?

xtabond is a GMM estimator with "dynamic" instruments. The instruments
are "dynamic" in the sense that there are more instruments available at
time t than at time t-1.

The instruments for all time periods are used to establish
moment/orthogonality conditions. These moment conditions are stacked in
a vector. The square norm of this vector is minimized (you want the left
hand side of the moment conditions to be as close to 0 as possible). The
first-order conditions of the minimization problem form a system that
can be solved for the model's coefficients (assuming that you have at
least as many non-colinear instruments as parameters to estimate).

> 2) If x(t) is predetermined instead of endog in my model,
> there is no need to instrument for x(t-1) - x(t-2). Will
> Stata know this?

It's your responsibility to tell xtabond whether x is predetermined,
endogenous, or exogenous. How to do this is explained in the doc.

Exogenous regressors can (and should) be used as instruments (for
themselves). xtabond automatically does that.