In a previous post I have asked about the derivation of $Cov(e, X) = 0$, when using the control function approach.

A follow-up question on this, or, actually, the real question, is: how does the control function approach ensure that the endogeneity is captured by the parameter on the estimated residuals included in the main model. Let me elaborate.

The problem is as follows. I want to estimate

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

Now I know that $X$ and $Y$ are also reversely related, such that $Cov(\varepsilon,X) \neq 0$.

The Control function approach tries to solve endogeneity by dividing $\varepsilon$ into an endogenous part and the exogenous part.

What we need is a relevant instrument $Z$, such that

$$X = \delta_1 + \delta_2 Z + \nu,\tag{ins}\label{ins}$$

where $Cov(\nu, Z) = 0$, to obtain an estimation of the endogenous part of $\varepsilon$.


$$\varepsilon = \beta_3 \nu + e,$$

where $\beta_3 = \dfrac{Cov(\nu,\varepsilon)}{Var(\nu)}$ is the population regression coefficient, we can use the estimated residuals $\nu$ from \eqref{ins} and plug it into our main model:

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

We can then derive $E[\varepsilon | X]$ as follows:

$$ \begin{aligned} E[\varepsilon | X] &= Cov(\beta_3 \nu + e, X) \\ & = Cov(\beta_3 \nu, X) + Cov(e, X) \end{aligned} $$

From the answer on the previous post, we know that $Cov(e, X) = 0$, and so

$$ \begin{aligned} Cov(\beta_3 \nu, X) + Cov(e, X) &= Cov(\beta_3 \nu, X) \\ & = Cov(\beta_3 \nu, \gamma_1 + \gamma_2 Z + \nu) \\ & = Cov(\beta_3 \nu, \gamma_1) + Cov(\beta_3 \nu, \gamma_2 Z) + Cov(\beta_3 \nu, \nu) \\ & = \beta_3 Var(\nu) \\ & = \dfrac{Cov(\nu, \varepsilon)}{Var(\nu)} Var(\nu) \\ & = Cov(\nu, \varepsilon) \\ & = \beta_3. \end{aligned} $$

Now, my question is about the very last step. On one hand, intuitively, this has to hold. On the other hand $\beta_3$ cannot simultaneously be $Cov(\nu, \varepsilon)$ and $\dfrac{Cov(\nu,\varepsilon)}{Var(\nu)}$, unless $Var(\nu) = 1$.

Could anyone confirm that this makes sense, or otherwise enlighten me how $Cov(\varepsilon, X) = \beta_3$ holds?


1 Answer 1


As it turns out, this derivation needs one additional assumption. This is pointed out by Amil Petrin and Kyoo il Kim in this NBER working paper (footnote on page 5).

The assumption being $E[\varepsilon | \nu] = \beta_3 \nu$, i.e. $Cov(\nu,\varepsilon) = \beta_3 \nu$ concludes the above derivation.


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