<p>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. ;)</p>

<p>Instead I will just cite wooldridge on page 160, 2nd paragraph:</p>

<p>«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.</p>

<p>Good luck ;)</p>

CC BY-SA 3.0
2016-05-21T17:48:11.523Z
2016-05-21T17:48:11.523Z
An old man in the sea.
<p>I need to know how to find robust S.E. for the CF approach to endogeneity. </p>

<p>Consider the model: 
$$y_i=X_i\beta_1 + W_i\beta_2+\epsilon_i$$</p>

<p>Assume: 
$$E[X_i\epsilon_i]=0$$
$$E[W_i\epsilon_i] \neq 0$$</p>

<p>Thus, $W_i$ is endogenous. </p>

<p>Now, let: 
$$E[Z_iW_i] \neq 0$$
$$E[Z_i\epsilon_i]=0$$ </p>

<p>The control function approach: </p>

<ul>
<li><p>$W_i = \gamma_1 X_i + \gamma_2 Z_i + \phi_i  $</p></li>
<li><p>$\epsilon_i = \alpha \phi_i + \chi$ </p></li>
</ul>

<p>now we replace $\epsilon_i$ in our original equation: </p>

<p>$$y_i=X_i\beta_1 + W_i\beta_2+ \phi_i \alpha + \chi$$</p>

<p>where now we have $E[W_i \chi]=0$</p>

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

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

<p>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. </p>

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

<p>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? </p>

CC BY-SA 3.0
2016-05-20T19:04:41.823Z
Since $\phi_i$ is part of $W_i$ and also of $\epsilon_i$, how do you obtain the orthogonality between $W_i$ and $\epsilon_i$?
Alecos Papadopoulos
CC BY-SA 3.0
2016-05-20T20:16:58.580Z
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?
123
CC BY-SA 3.0
2016-05-20T20:52:46.553Z
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.
Alecos Papadopoulos
CC BY-SA 3.0
2016-05-20T23:46:20.697Z
@AlecosPapadopoulos - Oh man - I see what happened. Sorry for the confusion. Thanks for catching the mistake.
123
CC BY-SA 3.0
2016-05-21T17:43:38.267Z
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. ;)
An old man in the sea.
<econometrics>
Robust Standard Errors for Control Function Approach?
123
# Robust Standard Errors for Control Function Approach?

(Asked by: username_0 on 2016-05-20)

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?

**Tags:** econometrics

> **Comments:**
> **username_1**: Since $\phi_i$ is part of $W_i$ and also of $\epsilon_i$, how do you obtain the orthogonality between $W_i$ and $\epsilon_i$?  
> **username_0**: 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?  
> **username_1**: 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.  
> **username_0**: @AlecosPapadopoulos - Oh man - I see what happened. Sorry for the confusion. Thanks for catching the mistake.  
> **username_2**: username_0, 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. ;)  

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## 1 answers

### Answer by username_2 on 2016-05-21. Score: 2

username_0, 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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