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I need to know how to find robust S.E. for the CF approach to endogeneity.

Consider the model: $$y_i=X_i\beta_1 + W_i\beta_2+\epsilon_i$$

Assume: $$E[X_i\epsilon_i]=0$$ $$E[W_i\epsilon_i] \neq 0$$

Thus, $W_i$ is endogenous.

Now, let: $$E[Z_iW_i] \neq 0$$ $$E[Z_i\epsilon_i]=0$$

The control function approach:

  • $W_i = \gamma_1 X_i + \gamma_2 Z_i + \phi_i $

  • $\epsilon_i = \alpha \phi_i + \chi$

now we replace $\epsilon_i$ in our original equation:

$$y_i=X_i\beta_1 + W_i\beta_2+ \phi_i \alpha + \chi$$

where now we have $E[W_i \chi]=0$

Intuition: I think I used the series of linear projections to 'control' for the endogenous portion of $W_i$.

EDIT: I originally typed this incorrectly. I have changed the relevant orthogonality condition. Here is the intuition behind the (correct) orthogonality condition $E[W_i \chi]=0$:

Since $W_i$ is a linear function of $X_i , Z_i$, and both are themselves orthogonal to $\chi$, we obtain the given orthogonality condition.

Okay - the question. I think that $\hat \beta_{2.CF} \equiv \hat \beta_{2.OLS} \equiv \frac{cov(W_i,Y_i)}{Var(W_i)}$

If this is correct, do I just use the R.S.E. form that I use in OLS if I want heteroskedasticity robust S.E. when using the C.F. approach?

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  • $\begingroup$ Since $\phi_i$ is part of $W_i$ and also of $\epsilon_i$, how do you obtain the orthogonality between $W_i$ and $\epsilon_i$? $\endgroup$ – Alecos Papadopoulos May 20 '16 at 19:04
  • $\begingroup$ by construction, $E[\phi_i \chi_i]=0$,$E[X_i \chi_i]=0$ and the endogeneity is fully reflected in $\alpha$. I wrote this out following the example provided by Woolridge's lecture slides. I think it is correct but I could have mixed some things around? $\endgroup$ – 123 May 20 '16 at 20:16
  • $\begingroup$ My question was about $$E(W_i\epsilon_i) = E([\gamma_1 X_i + \gamma_2 Z_i + \phi_i][ \alpha \phi_i + \chi]) = ... + \alpha E\phi^2_i$$ If $\phi$ is random then $E\phi^2_i$ cannot be zero, and so $E(W_i\epsilon_i) \neq 0$. Please clarify, or better, add information in the post, not in the comments. $\endgroup$ – Alecos Papadopoulos May 20 '16 at 20:52
  • $\begingroup$ @AlecosPapadopoulos - Oh man - I see what happened. Sorry for the confusion. Thanks for catching the mistake. $\endgroup$ – 123 May 20 '16 at 23:46
  • $\begingroup$ 123, the CF method is a specific case of when we have generated regressors. Check the appendix to chapter 6, pages 157-160, for the asymptotic covariance matrix of $N^{1/2}(\hat\beta_{CF}-\beta)$. The expression is truly enormous. ;) $\endgroup$ – An old man in the sea. May 21 '16 at 17:43
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123, the CF method is a specific case of when we have generated regressors. Check the appendix to chapter 6, pages 157-160, for the asymptotic covariance matrix of $N^{1/2}(\hat\beta_{CF}-\beta)$. The expression is truly enormous. ;)

Instead I will just cite wooldridge on page 160, 2nd paragraph:

«If $\mathbf{G}\neq \mathbf{0}$, then (...). Neither the usual 2SLS variance matrix estimator nor the heteroskedasticity-robust form is valid in this case.»The $\mathbf{G}\neq \mathbf{0}$ is the case when we have generated regressors.

Good luck ;)

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