P-score in the reg?

Geo. from GA asks this interesting question 'bout the propensity score: 

I was wondering whether replacing high dimensional
covariates (X) in the regression model with their propensity scores
(p(X)) was a good idea? That is, Y = a + bT + cX + e becomes Y= a + bT
+ c(p(X)) + e. The book does not really address it unless I missed it.
What are the implications? Thanks.


George: its certainly not a crazy idea. In fact, Dehejia-Wahba (1999) tried
this (Table 5, estimates labeled quadratic in score).  But its not clear
what the theoretical justification is here; once you are using regression,
why do this two-step procedure instead of just sticking the covs you've put
in the score right into the reg (since you're implicitly assuming these are
the only source of OVB)?  Also, as we know from chpt 3, regression does not
estimate the pop ATE or the effect of treatment on the treated except under
constant effects or if the score is constant. Score fiends are often after
those parameters instead of the variance-weighted avg that regression produces.
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One Comment

  1. George
    Posted April 1, 2010 at 4:56 pm | Permalink

    The reason for my question is that in my data I have many observable high dimensional covariates. I mean in the hundreds. I don’t think throwing them all into the regression function is really feasible. It will be nice to find out more about the theoretical justification of using propensity scores. I am in the computer science field (phd student) and new to the work in econometrics but your book has been extremely helpful. Thank you very much.

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