# Why is the functional form of the 1st stage in 2SLS not important?

In a presentation today the speaker made the above claim. He said that even if the first stage is mis-specified, the coefficient estimates of the second stage will still be valid. As a lowly graduate student I could not ask for an explanation, so now I begged for your help!

• It is my understanding that the only thing you care about is that your $\hat{x}$, i.e. the predicted value of the first stage, is uncorrelated with the error term of the second stage. Your first stage coefficients might be biased or yield predictions outside the unit intervall etc, but this won't induce correlation between the predicted values of your endogenous variable and the error term of the second stage. I have never seen a proof of this though, but I have seen explanations along this line from e.g. Imbens. – coffeinjunky Jun 27 '14 at 9:37
• If your x is a dummy, then I'd agree. If your x is continuous, I'd be skeptical (though I haven't seen a proof). Generally when people talk about unbiasedness, their starting point is assuming that the linear model is valid. I mean, generally they are looking to get that $E[\hat\beta] = \beta$ from $y = X'\beta$. But if $y = X'\beta$ is a garbage model then $\beta$ doesn't answer the question that you're presuming that it does. (I'm only talking about functional form, not distributional form) – generic_user Sep 12 '14 at 6:49

• So if the true DGP has $x$ and $x^2$, and we only include $x$. Does this count as poor functional form in your answer? To me, this count as both poor functional form and omitted variable bias. – Heisenberg Nov 26 '14 at 20:12