# What happens if the “control variables” are also endogenous?

I work in Political Economy, and a lot of the models include "innocent" control variables such as population, inequality, colonial legacy, etc. so that the author can claim unbiasedness on their independent variable of interest.

But if any of these control variables are endogenous to some omitted variable, doesn't this contaminate the unbiasedness of ALL the independent variables?

If that's true, then what can we do? Leave those control variables out and they lead to omitted variable bias themselves. Include those in and they will contaminate everything in the model.

Example: A researcher wants to know if inequality leads to violence, and he controls for a few things: $$Violence = Inequality + Growth + Development + \epsilon$$ Seeing that Inequality is likely to be endogenous (because of the omitted variable Level of altruism), he will try to find a instrumental variable for Inequality. But aren't Growth and Development likely to be endogenous (i.e. correlated with Level of altruism) too?

This example may look silly, but my point is in Political Economy / Development work, there are so many factors at play (yet omitted) that I'm afraid many variables included on the LHS are endogenous. Yet often, the researcher only looks for an instrument for his pet independent variable only.

• Yet another thing to consider is the so-called "bad control" issue - a situation when the control is an outcome variable itself. I would suggest you to read Section 3.2.3 in Angrist and Pischke's celebrated "Mostly Harmless Econometrics" to get a grasp of this topic and why it matters if you want to have a better understanding of your question. – MauOlivares Apr 18 '18 at 0:34

# "But if any of these control variables are endogenous to some omitted variable, doesn't this contaminate the unbiasedness of ALL the independent variables?"

I don't want to emphasize this too much, but it's worth mentioning that this is not true in general. The following derivation will hopefully provide some understanding of the "contamination" you mention. As a simple counterexample, suppose that the data generating process is given by $$Y = X_1 \beta_1 + X_2 \beta_2 + Z \gamma + \varepsilon,$$ where $Z$ is unobserved. Let $Cov(X_1,Z) = 0$, $Cov(X_2, Z) \neq 0$, and $Cov(X_1,X_2) = 0$. Then, it is clear that $X_2$ is "endogenous." But notice that because $Cov(X_1,Z) = 0$, our estimate of $\beta_1$ will still be ok: $$\text{plim}\, \hat \beta_{1} = \beta_1 + \gamma \frac{Cov(X_1^*, Z)}{Var(X_1^*)} = \beta_1,$$ where $X_1^* = M_2 X_1$ and $M_2 = [I - X_2(X_2'X_2)^{-1}X_2']$. Because $Cov(X_1,X_2) = 0$, $X_1^* = X_1$. So $Cov(X_1^*,Z)=0$.

# "What can we do?"

One of the mains challenges of doing good econometrics is thinking of potential identification strategies. In the type of situation you describe, there is probably nothing you can do but to try to approach the problem a different way.

• While you're technically right, I would not emphasize this point. I'd rather say that in general, we cannot rule out biasedness of any of the variables, instead of saying in some scenarios its ok, well, because we usually don't know the DGP. – FooBar Jan 28 '15 at 14:30
• 1) Could you point me to a reference where the $\hat\beta$ is derived this way? I wasn't taught this in my econometrics. 2) Where do you use $Cov(X_1, Z)=0$ in the proof? It seems like $Cov(X_1, X-2)=0$ is sufficient. 3) I agree with @FooBar that $Cov(X_1, X_2)=0$ are the exception, not the norm. Indeed, if $Cov(X_1, X_2)=0$ we wouldn't bother to control for $X_2$ in the first place (except to increase precision). – Heisenberg Jan 28 '15 at 14:38
• @FooBar, I agree. I've updated the post to emphasize that this is a special case. As far as the point about not knowing the DGP, that is true. But that's not the point. Any analysis has to make assumptions about the DGP and the quality of the analysis depends on the quality of the assumptions. The derivation I gave just serves to illustrate an example of the assumptions (albeit, very strong assumptions) that could get you where you'd want to go. – jmbejara Jan 28 '15 at 17:52
• @Heisenberg: 1) Could you open a new question in main about this? If you just copy and paste the derivation and present your question, that would be best. 2) $Cov(X_1, Z) = 0$ is needed when I say that $Cov(X_1^*,Z) = 0$. 3) You're right. If we we're interested in predicting $Y$, it would be important. But, yeah, that's a good point. On the other hand, it's maybe useful to note that the size of the bias depends on how correlated you believe $X_1$ and $X_2$ to be. – jmbejara Jan 28 '15 at 17:59
• @jmbejara I posted 1) as a separate question. Please feel free to edit my question / title, since I don't know how to phrase the title intelligently and useful for Googler in this case. – Heisenberg Jan 28 '15 at 19:03

All is too strong, but probably some. This problem is called "smearing". Take a look at the proof in Greene's lecture notes on slide 5.

Emily Oster has a nice working paper (and Stata command psacalc) that can help bound the bias.

In the context of Least-squares estimation, the way we have to (attempt to) deal with possible endogeneity of regressors is through Instrumental Variables estimation. This approach does not depend on having just one endogenous regressor -you may have many. In such a case of course you need to find more instruments which make things harder -but in principle, the method will work the same way.

IV estimation does not solve the issue of bias, it only provides consistency for the estimator. But nothing solves the issue of bias bar strict exogeneity itself (and then there are some methods of bias reduction). But if you take a look around another SE site, Cross Validated, which is about statistics, you will see that seasoned statisticians don't really give much weight to the property of unbiasedness -they focus on Mean-Square Efficiency for finite sample properties, and on consistency for large sample properties.

• So the correct approach is to indeed find instruments for all the endogenous variables, right? – Heisenberg Jan 28 '15 at 20:40
• Yes, this is the way. – Alecos Papadopoulos Jan 28 '15 at 20:56

This is an example of what statistician Andrew Gelman calls "the fallacy of controlling for an intermediate outcome". Here is his description of this fallacy popping up when researchers ask if having more daughters changes your politics. The decision to have a second child is necessarily conditional on the previous decision to have the first child, and so seems like a clear example of controlling for decision variable that was endogenous.

Several studies have been performed in the last few years looking at the economic decisions of parents of sons, as compared to parents of daughters....A common feature of all these studies is that they control for the total number of children....At first sight, controlling for the total number of children seems reasonable. There is a difficulty, however, in that the total number of kids is an intermediate outcome, and controlling for it (whether by subsetting the data based on #kids or using #kids as a control variable in a regression model) can bias the estimate of the causal effect of having a son (or daughter).

To see this, suppose (hypothetically) that politically conservative parents are more likely to want sons, and if they have two daughters, they are (hypothetically) more likely to try for a third kid. In comparison, liberals are more likely to stop at two daughters. In this case, if you look at data on families with 2 daughters, the conservatives will be underrepresented, and the data could show a correlation of daughters with political liberalism–even if having the daughters has no effect at all!...

A solution is to apply the standard conservative (in the statistical sense!) approach to causal inference, which is to regress on your treatment variable (sex of kid) but controlling only for things that happen before the kid is born. For example, one could compare parents whose first child is a girl to parents whose first child is a boy. One can also look at the second birth, comparing parents whose second child is a girl to those whose second child is a boy–controlling for the sex of the first child. And so on for third child, etc.

Regarding your comment that "Leave those control variables out and they lead to omitted variable bias themselves.", this seems to depend what sort of instrument you get. A good instrument, one that really satisfies the requirements, has to be independent of of the error term in the second stage and be independent of everything else you control for directly. That is, the instrument changes Y only through X. So a suitable instrument for inequality has to be independent of growth and development (good luck finding that!) if we believe that the violence equation is the structural equation for violence.

As other posts have pointed out, endogenous regressors may contaminate all parameter estimates in regression when regressors are correlated.

Moreover, it may seem difficult to conceive a situation where, say, $X_1$ and $X_2$ are correlated and $X_2$ is endogenous but $X_1$ is not.

However, less than that is required to guarantee consistency of $\hat{\beta}_1$ even when $X_2$ is endogenous and $X_1$ and $X_2$ are correlated.

Consider the following model (analogous to @jmbejara's notation)

\begin{equation*} y=X_1\beta_1+X_2\beta_2+Z\gamma+\varepsilon, \end{equation*}

$Z$ unobserved, with the usual exogeneity assumptions w.r.t. $\varepsilon$, i.e., $\frac{1}{n}{x_1^{(k)\prime}}\varepsilon\overset{p}{\rightarrow}0$ and $\frac{1}{n}x_2^{(k)\prime}\varepsilon\overset{p}{\rightarrow}0$ for all $k$ regressors. $X_2$ is endogenous in the sense that $\frac{1}{n} x_{1}^{(k)\prime}z^{(l)} \overset{p}{\not\rightarrow}0$ for some pair of variables $(k,l)$.

Now if $X_2$ is endogenous but $X_1$ is not in the sense that all correlation between $X_1$ and $Z$ will be gone after controlling for $X_2$, i.e.,

$$\frac{1}{n} x_1^{(k)\prime}Q_{X_2}z^{(l)} \overset{p}{\rightarrow}0$$ for all $(k,l)$, where $Q_{X_2}$ is the projection onto the null space of $X_2$ (the residual maker''), i.e. $Q_{X_2}\equiv [I_n - X_2(X_2'X_2)^{-1}X_2']$ then we are fine. The reason is seen from the following two-step estimator of $\beta_1$ (e.g. Amemiya, 1985, pp. 6-7):

\begin{align*} \hat{\beta}_1 &= (X_1'Q_{X_2}X_1)^{-1}X_1'Q_{X_2}y \\ &= \beta_1 + (X_1'Q_{X_2}X_1)^{-1}X_1'\underbrace{Q_{X_2}X_2}_{\overset{p}{\rightarrow}0}\beta_2\\ &+ (X_1'Q_{X_2}X_1)^{-1}\underbrace{X_1'Q_{X_2}Z}_{\overset{p}{\rightarrow}0}\gamma \\ &+ (X_1'Q_{X_2}X_1)^{-1}\underbrace{X_1'Q_{X_2}\varepsilon}_{\overset{p}{\rightarrow}0} \end{align*} QED. The third line here is key, and it also shows why we are safe when $X_1$ and $X_2$ are uncorrelated/orthogonal. Happy endogenous regressions.