<p>All is too strong, but probably some. This problem is called "smearing". Take a look at the proof in Greene's <a href="http://people.stern.nyu.edu/wgreene/Econometrics/Econometrics-I-13.pptx" rel="nofollow noreferrer">lecture notes</a> on slide 5.</p>

<p>Emily Oster has a <a href="http://brown.edu/research/projects/oster/sites/brown.edu.research.projects.oster/files/uploads/Unobservable_Selection_and_Coefficient_Stability.pdf" rel="nofollow noreferrer">nice working paper</a> (and Stata command <code>psacalc</code>) that can help bound the bias.</p>

CC BY-SA 3.0
2015-01-28T07:28:46.153Z
2018-02-28T21:03:24.927Z
2018-02-28T21:03:24.927Z
dimitriy
<p>This is an example of what statistician <a href="http://www.stat.columbia.edu/~gelman/" rel="nofollow">Andrew Gelman</a> 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. </p>

<blockquote>
  <p>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).</p>
  
  <p>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!...</p>
  
  <p><strong>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.</strong> 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.</p>
</blockquote>

<p><a href="http://andrewgelman.com/2005/12/27/daughters_liber/" rel="nofollow">Does having sons make you more conservative? Maybe, maybe not. A problem with controlling for an intermediate outcome</a> </p>

<p>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, <a href="http://en.wikipedia.org/wiki/Instrumental_variable#Definitions" rel="nofollow">has to be independent of of the error term in the second stage and be independent of everything else you control for directly</a>. 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. </p>

CC BY-SA 3.0
2015-01-30T17:09:13.967Z
2015-01-31T06:11:07.757Z
2015-01-31T06:11:07.757Z
BKay
<p>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.  </p>

<p>IV estimation does not solve the issue of bias, it only provides consistency for the estimator. But <em>nothing</em> 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, <a href="https://stats.stackexchange.com/questions">Cross Validated</a>, 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.</p>

CC BY-SA 3.0
2015-01-28T20:40:49.817Z
So the correct approach is to indeed find instruments for all the endogenous variables, right?
Heisenberg
CC BY-SA 3.0
2015-01-28T20:56:46.357Z
Yes, this is the way.
Alecos Papadopoulos
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2015-01-28T20:33:18.627Z
2015-01-29T04:50:29.917Z
2017-04-13T12:44:19.590Z
Alecos Papadopoulos
<p>As other posts have pointed out, endogenous regressors may contaminate all parameter estimates in regression when regressors are correlated.</p>

<p>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.</p>

<p>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.</p>

<p>Consider the following model (analogous to @jmbejara's notation)</p>

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

<p>$Z$ 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 $(k,l)$.</p>

<p><strong>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 $X_2$</strong>, i.e.,</p>

<p>\begin{equation}
 \frac{1}{n} x_1^{(k)\prime}Q_{X_2}z^{(l)} \overset{p}{\rightarrow}0
\end{equation}
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):</p>

<p>\begin{align*}
 \hat{\beta}_1 &amp;= (X_1'Q_{X_2}X_1)^{-1}X_1'Q_{X_2}y \\
 &amp;= \beta_1 +  (X_1'Q_{X_2}X_1)^{-1}X_1'\underbrace{Q_{X_2}X_2}_{\overset{p}{\rightarrow}0}\beta_2\\
 &amp;+ (X_1'Q_{X_2}X_1)^{-1}\underbrace{X_1'Q_{X_2}Z}_{\overset{p}{\rightarrow}0}\gamma  \\
 &amp;+ (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. </p>

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2018-03-17T10:45:30.530Z
2018-03-17T13:10:42.927Z
2018-03-17T13:10:42.927Z
Murphy
<h1>"But if any of these control variables are endogenous to some omitted variable, doesn't this contaminate the unbiasedness of ALL the independent variables?"</h1>

<p>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 $Cov(X_1^*,Z)=0$. </p>

<h1>"What can we do?"</h1>

<p>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.</p>

CC BY-SA 3.0
2015-01-28T14:30:02.010Z
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
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2015-01-28T14:38:58.320Z
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
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2015-01-28T17:52:13.920Z
@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
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2015-01-28T17:59:18.110Z
@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
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2015-01-28T19:03:48.493Z
@jmbejara I posted 1) as a [separate question](http://economics.stackexchange.com/questions/3203/alternative-way-of-deriving-ols-coefficients). 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
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2015-01-29T02:12:50.393Z
You're missing a subscript on the first $\beta$ in the DGP.
dimitriy
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2015-01-28T07:26:38.027Z
2015-01-29T02:44:44.313Z
2015-01-29T02:44:44.313Z
jmbejara
<p>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.</p>

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

<p>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.</p>

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

<p>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.</p>

CC BY-SA 3.0
2018-04-18T00:34:43.593Z
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
<econometrics><political-economy><identification><regression>
What happens if the "control variables" are also endogenous?
Heisenberg
# What happens if the "control variables" are also endogenous?

(Asked by: username_0 on 2015-01-28)

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:
\begin{equation}
Violence = Inequality + Growth + Development + \epsilon
\end{equation}
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.

**Tags:** econometrics, political-economy, identification, regression
> **Comments:**
> **username_1**: 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.  

---

## 5 answers

### Answer by username_6 on 2015-01-28. Score: 10 | Accepted answer

# "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.

> **Comments:**
> **username_7**: 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.  
> **username_0**: 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).  
> **username_6**: @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.  
> **username_6**: @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.  
> **username_0**: @jmbejara I posted 1) as a [separate question](http://economics.stackexchange.com/questions/3203/alternative-way-of-deriving-ols-coefficients). 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.  
> **username_2**: You're missing a subscript on the first $\beta$ in the DGP.  

---

### Answer by username_2 on 2015-01-28. Score: 6

All is too strong, but probably some. This problem is called "smearing". Take a look at the proof in Greene's [lecture notes](http://people.stern.nyu.edu/wgreene/Econometrics/Econometrics-I-13.pptx) on slide 5.

Emily Oster has a [nice working paper](http://brown.edu/research/projects/oster/sites/brown.edu.research.projects.oster/files/uploads/Unobservable_Selection_and_Coefficient_Stability.pdf) (and Stata command `psacalc`) that can help bound the bias.


---

### Answer by username_3 on 2015-01-30. Score: 5

This is an example of what statistician [Andrew Gelman](http://www.stat.columbia.edu/~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.

[Does having sons make you more conservative? Maybe, maybe not. A problem with controlling for an intermediate outcome](http://andrewgelman.com/2005/12/27/daughters_liber/)

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](http://en.wikipedia.org/wiki/Instrumental_variable#Definitions). 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.


---

### Answer by username_4 on 2015-01-28. Score: 5

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](https://stats.stackexchange.com/questions), 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.

> **Comments:**
> **username_0**: So the correct approach is to indeed find instruments for all the endogenous variables, right?  
> **username_4**: Yes, this is the way.  

---

### Answer by username_5 on 2018-03-17. Score: 1

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.,

\begin{equation}
\frac{1}{n} x\_1^{(k)\prime}Q\_{X\_2}z^{(l)} \overset{p}{\rightarrow}0
\end{equation}
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.


---

