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Suppose I want to estimate

$$Y = \beta_1 + \beta_2 X + \varepsilon$$

Now I know that $X$ and $Y$ are also reversely related

$$X = \gamma_1 + \gamma_2 Y + \xi$$

such that $Cov(\varepsilon,X) \neq 0$.

$Cov(\varepsilon,\xi) = 0$ by assumption.

$Cov(\varepsilon,X) \neq 0$ is shown as follows:

$$ \begin{align} Cov(\varepsilon, X) &= Cov(\varepsilon, \gamma_1 + \gamma_2 Y + \xi) \\ &= Cov(\varepsilon, \gamma_1 ) + Cov(\varepsilon, \gamma_2 Y) + Cov(\varepsilon, \xi) \\ &= \gamma_2 Cov(\varepsilon, Y) \\ &= \gamma_2 Cov(\varepsilon, \beta_1 + \beta_2 X + \varepsilon) \\ &= \gamma_2 Cov(\varepsilon, \beta_1) + \gamma_2 Cov(\varepsilon, \beta_2 X) + \gamma_2 Cov(\varepsilon, \varepsilon) \\ \end{align} $$

where $Cov(\varepsilon, \varepsilon) = Var(\varepsilon) > 0$. Hence $Cov(\varepsilon,X) \neq 0$.

The Control function approach tries to solve endogeneity by dividing $\varepsilon$ into an endogenous part $\beta_3\nu$ and the exogenous part $e$, where $\nu$ comes from the equation

$$X = \delta_1 + \delta_2 Z + \nu$$

with $Z$ a relevant instrumental variable, and $Cov(Z,\varepsilon) = 0$

Our equation for $Y$ then becomes:

$$Y = \beta_1 + \beta_2 X + \beta_3\nu + e$$

In words I understand that, conditional on $\nu$, $\varepsilon$ is independent of $X$.

I fail to see, however, how this ensures that $Cov(e, X) = 0$.

I would like to derive this in much the same way as shown above for $Cov(\varepsilon, X) \neq 0$.

Any help much appreciated.

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Recall that when we consider linear projections of the form $$W = \beta X + \epsilon,$$ by construction $Cov(X,\epsilon) = 0$. This, combined with the intuition that we split $\epsilon$ into an exogenous and endogenous part (as you mention), yields the result.

In particular, given $$X = \delta_1+\delta_2 Z + \nu,$$ we know that $Cov(Z,\nu) =0$. Furthermore, if we write out the linear projection of $\epsilon$ on $\nu$, we have $$\epsilon = \beta_3 \nu + e,$$ so that by construction, $Cov(\nu, e) = 0$.

By the exogeneity assumption, we know that $Cov(Z,\epsilon) = 0$, and so we can derive that $$Cov(Z,e) = Cov(Z, \epsilon - \beta_3\nu) = Cov(Z,\epsilon) - \beta_3 Cov(z,\nu) = 0.$$

This should make intuitive sense (can you put this in words?). So then, to show $Cov(e,X) = 0$, we use the fact that $X = \delta_1+\delta_2 Z + \nu$ and that we showed $Cov(Z,e) = Cov(\nu,e) = 0$ to then get that \begin{align} Cov(X,e) & = Cov(\delta_1+\delta_2 Z + \nu,e) \\ & = Cov(\delta_1,e) + \delta_2Cov(Z,e) + Cov(\nu,e) \\ & = 0, \end{align} as required.

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    $\begingroup$ Brilliant, thanks a lot! I somehow could not piece it together last night. Not sure what you’re hinting at with your question about $Cov(Z, e) = 0$ making intuitive sense, though. $\endgroup$ – o_v Mar 16 at 4:11
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    $\begingroup$ Yep! I just meant can you reason with yourself why you'd expect that to hold. Always good to have the intuition even if the math shows you the result :) $\endgroup$ – doubled Mar 16 at 4:17
  • $\begingroup$ So, I wanted to take it a step further and verified that $Cov(X, \varepsilon) = \beta_3$, thus capturing the endogenous variation in $Y = \beta_1 + \beta_2 X + \varepsilon$, where $\varepsilon = \beta_3 \nu + e$. In brief $Cov(X,\varepsilon) = Cov(X, \beta_3\nu + e)$. After substituting $X = \delta_1 + \delta_2Z + \nu$ I obtain $Cov(X, \varepsilon) = \beta_3 Var(\nu) = \frac{Cov(\nu, \varepsilon)}{Var(\nu)} Var(\nu) = Cov(\nu,\varepsilon) = \beta_3$. $\endgroup$ – o_v Mar 17 at 10:47
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    $\begingroup$ @doubled; In the initial projection: "$Cov(W, \epsilon)=0$ by construction" is wrong. $\endgroup$ – markowitz Apr 10 at 13:04
  • $\begingroup$ @markowitz eeek it should be $Cov(X,\epsilon) = 0$ by construction. Thanks for catching that! $\endgroup$ – doubled Apr 10 at 15:46

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