<blockquote>
<p>If I include <span class="math-container">$z_1$</span> in the model, like this:
<span class="math-container">$$
&gt; y = \beta_0 + \beta_1 x + \beta_2 z_1 + e,
&gt; $$</span>
Does that mean that <span class="math-container">$\beta_1$</span> is predominantly capturing the effect of <span class="math-container">$z_2$</span>?</p>
</blockquote>
<p>Yes. This can be seen using the <a href="https://en.wikipedia.org/wiki/Frisch%E2%80%93Waugh%E2%80%93Lovell_theorem" rel="noreferrer">Frish-Waugh-Lovell theorem</a>:</p>
<p>If you regress:
<span class="math-container">$$
y = \beta_0 + \beta_1 x + \beta_2 z_1 + e,
$$</span>
then <span class="math-container">$\beta_1$</span> will be the same as the corresponding coefficient of a modified regression:
<span class="math-container">$$
\hat y = \gamma_0 + \beta_1 \hat x + \hat e \tag{1}
$$</span>
where <span class="math-container">$\hat x$</span> is the residual from regressing <span class="math-container">$x$</span> on <span class="math-container">$z_1$</span> and the same for <span class="math-container">$\hat y$</span>. Now if we regress <span class="math-container">$x$</span> on <span class="math-container">$z_1$</span> then the residual is equal to:
<span class="math-container">$$
M_{z_1} x,
$$</span>
where <span class="math-container">$M_{z_1} = 1 - z_1(z_1'z_1)^{-1}z_1'$</span> is the annihilator matrix.  If <span class="math-container">$x = z_1 + z_2$</span> then:
<span class="math-container">$$
M_{z_1} x = (1 - z_1(z_1'z_1)z_1')(z_1 + z_2) = M_{z_1}z_2
$$</span>
As such, substituting in <span class="math-container">$(1$</span>), we have:
<span class="math-container">$$
\hat y = \gamma_0 + \beta_1 \hat z_2 + \hat e,
$$</span>
where <span class="math-container">$\hat z_2$</span> is now the residual from regressing <span class="math-container">$z_2$</span> on <span class="math-container">$z_1$</span>. Using the Frish-Waugh-Lovell theorem, in the reverse direction, this gives that <span class="math-container">$\beta_1$</span> is also equal to the coefficient in the following regression:
<span class="math-container">$$
y = \delta_0 + \beta_1 z_2 + \delta_2 z_1 + \varepsilon.
$$</span>
In other words, <span class="math-container">$\beta_1$</span> will also be equal to the coefficient for <span class="math-container">$z_2$</span> for a regression of <span class="math-container">$y$</span> on both <span class="math-container">$z_2$</span> and <span class="math-container">$z_1$</span>. Notice, however, that in general <span class="math-container">$\beta_2 \ne \delta_2$</span> (so the coefficients for <span class="math-container">$z_1$</span> will not be equal in the two regressions).</p>
<p>Another way to see this is by immediately substituting <span class="math-container">$x = z_1 + z_2$</span> into the regression <span class="math-container">$(1)$</span> then:
<span class="math-container">$$
\begin{align*}
y &amp;= \beta_0 + \beta_1(z_1 + z_2) + \beta_2 z_1 + e,\\
&amp;= \beta_0 + \beta_1 z_2 + (\beta_1 + \beta_2) z_1 + e.
\end{align*}
$$</span>
So the coefficient on <span class="math-container">$z_2$</span> in the new regression is identical to the coefficient on <span class="math-container">$x$</span> in the original one <span class="math-container">$(\beta_1)$</span> , while the coefficient on <span class="math-container">$z_1$</span> in the new regression is the sum of the coefficients on <span class="math-container">$x$</span> and <span class="math-container">$z_1$</span> in the original regression <span class="math-container">$(\beta_1 + \beta_2)$</span>.</p>
<blockquote>
<p>And a follow up question is: if <span class="math-container">$z_1$</span> is correlated with the error term, will including it in the model bias the estimate of <span class="math-container">$\beta_1$</span>?</p>
</blockquote>
<p>The opposite is true. Including <span class="math-container">$z_1$</span> into the regression will make the estimate of <span class="math-container">$\beta_1$</span> unbiased. Consider the following data generating process:
<span class="math-container">$$
y = \beta_0 + \beta_1 x + e, \tag{2}
$$</span>
and assume that <span class="math-container">$e$</span> is correlated with <span class="math-container">$z_1$</span>. Then we can write:
<span class="math-container">$$
e = \gamma z_1 + \varepsilon.
$$</span>
where <span class="math-container">$\varepsilon$</span> is now uncorrelated with <span class="math-container">$z_1$</span>. (and where <span class="math-container">$\gamma = \mathbb{E}(e z_1)/\mathbb{E}((z_1)^2) \ne 0$</span>.  Assume for simplicity that <span class="math-container">$e$</span> is uncorrelated with <span class="math-container">$z_2$</span>.</p>
<p>Then the estimate of <span class="math-container">$\beta_1$</span> will be biased as the orthogonality condition <span class="math-container">$\mathbb{E}(e x) = 0$</span> is not satisfied. Indeed:
<span class="math-container">$$
\begin{align*}
\mathbb{E}(ex) &amp;= \mathbb{E}(e z_2) + \mathbb{E}(\gamma z_1 z_1) + \mathbb{E}(\varepsilon z_1),\\
&amp;= \mathbb{E}(\gamma (z_1)^2) \ne 0
\end{align*}
$$</span>
If we include <span class="math-container">$z_1$</span> into the regression. Then substituting <span class="math-container">$e = \gamma z_1 + \varepsilon$</span>, into <span class="math-container">$(2)$</span> we can write:
<span class="math-container">$$
y = \beta_0 + \beta_1 x + \gamma z_1 + \varepsilon.
$$</span>
And <span class="math-container">$\mathbb{E}(\varepsilon) = \mathbb{E}(\varepsilon x) = \mathbb{E}(\varepsilon z_1) = 0$</span>. So by including <span class="math-container">$z_1$</span> into the regression, we can guarantee the residual <span class="math-container">$\varepsilon$</span> to be uncorrelated with all covariates. This means that <span class="math-container">$\beta_1$</span> is identified and its estimate will be unbiased.</p>

CC BY-SA 4.0
2021-06-29T12:00:07.813Z
Regarding *Another way to see this...*, if I simulate the data according to $y=\beta_0+\beta_1x+e$ with $x=z_1+z_2$ and then regress $y$ on $x$ and $z_1$, the estimated coefficient on $z_1$ is around zero and insignificant -- and far from $\beta_1+\beta_2=\beta_1+0=\beta_1$. Is that in line with what you are saying?
Richard Hardy
CC BY-SA 4.0
2021-06-29T12:53:02.833Z
@Richard Hardy What I am saying is that if you first regress $y$ on $x$ and $z_1$ and then $y$ on $z_2$ and $z_1$, then the coefficient on $z_1$ for the second regression is the sum of the coefficients for $x$ and $z_1$ of the first regression. In addition, the coefficient on $x$ in the first regression should be the same as the coefficient on $z_2$ in the second regression.  In your case the coefficient on $z_1$ is zero as it is zero in your data generating process.
tdm
CC BY-SA 4.0
2021-06-29T13:52:34.257Z
OK, thanks! I think my case is the same as OP's case, as I am following the setup in there. But even if that is not the case, you explanation is helpful.
Richard Hardy
CC BY-SA 4.0
2021-06-29T07:08:15.650Z
2021-06-29T07:08:15.650Z
tdm
<p>I am interested in the effect of an independent variable <span class="math-container">$x$</span> on a dependent variable <span class="math-container">$y$</span>, like so</p>
<p><span class="math-container">$$ y = \beta_0 + \beta_1 x + e $$</span></p>
<p>where <span class="math-container">$e$</span> is the error term. Now <span class="math-container">$x$</span> includes two effects <span class="math-container">$z_1$</span> and <span class="math-container">$z_2$</span>. For simplicity, let say <span class="math-container">$x = z_1 + z_2$</span>. If I include <span class="math-container">$z_1$</span> in the model, like this</p>
<p><span class="math-container">$$ y = \beta_0 + \beta_1 x + \beta_2 z_1 + e$$</span></p>
<p>Does that mean that <span class="math-container">$\beta_1$</span> is predominantly capturing the effect of <span class="math-container">$z_2$</span>?</p>
<p>And a follow up question is: if <span class="math-container">$z_1$</span> is correlated with the error term, will including it in the model bias the estimate of <span class="math-container">$\beta_1$</span>?</p>

<econometrics><applied-econometrics>
Including an endogenous covariate in a regression model as a control to estimate the effect of another variable of interest
Stein Monteiro
# Including an endogenous covariate in a regression model as a control to estimate the effect of another variable of interest

(Asked by: username_0 on 2021-06-28)

I am interested in the effect of an independent variable $x$ on a dependent variable $y$, like so

$$ y = \beta\_0 + \beta\_1 x + e $$

where $e$ is the error term. Now $x$ includes two effects $z\_1$ and $z\_2$. For simplicity, let say $x = z\_1 + z\_2$. If I include $z\_1$ in the model, like this

$$ y = \beta\_0 + \beta\_1 x + \beta\_2 z\_1 + e$$

Does that mean that $\beta\_1$ is predominantly capturing the effect of $z\_2$?

And a follow up question is: if $z\_1$ is correlated with the error term, will including it in the model bias the estimate of $\beta\_1$?

**Tags:** econometrics, applied-econometrics

---

## 1 answers

### Answer by username_1 on 2021-06-29. Score: 8 | Accepted answer

> If I include $z\_1$ in the model, like this:
> $$
> > y = \beta\_0 + \beta\_1 x + \beta\_2 z\_1 + e,
> > $$
> Does that mean that $\beta\_1$ is predominantly capturing the effect of $z\_2$?

Yes. This can be seen using the [Frish-Waugh-Lovell theorem](https://en.wikipedia.org/wiki/Frisch%E2%80%93Waugh%E2%80%93Lovell_theorem):

If you regress:
$$
y = \beta\_0 + \beta\_1 x + \beta\_2 z\_1 + e,
$$
then $\beta\_1$ will be the same as the corresponding coefficient of a modified regression:
$$
\hat y = \gamma\_0 + \beta\_1 \hat x + \hat e \tag{1}
$$
where $\hat x$ is the residual from regressing $x$ on $z\_1$ and the same for $\hat y$. Now if we regress $x$ on $z\_1$ then the residual is equal to:
$$
M\_{z\_1} x,
$$
where $M\_{z\_1} = 1 - z\_1(z\_1'z\_1)^{-1}z\_1'$ is the annihilator matrix. If $x = z\_1 + z\_2$ then:
$$
M\_{z\_1} x = (1 - z\_1(z\_1'z\_1)z\_1')(z\_1 + z\_2) = M\_{z\_1}z\_2
$$
As such, substituting in $(1$), we have:
$$
\hat y = \gamma\_0 + \beta\_1 \hat z\_2 + \hat e,
$$
where $\hat z\_2$ is now the residual from regressing $z\_2$ on $z\_1$. Using the Frish-Waugh-Lovell theorem, in the reverse direction, this gives that $\beta\_1$ is also equal to the coefficient in the following regression:
$$
y = \delta\_0 + \beta\_1 z\_2 + \delta\_2 z\_1 + \varepsilon.
$$
In other words, $\beta\_1$ will also be equal to the coefficient for $z\_2$ for a regression of $y$ on both $z\_2$ and $z\_1$. Notice, however, that in general $\beta\_2 \ne \delta\_2$ (so the coefficients for $z\_1$ will not be equal in the two regressions).

Another way to see this is by immediately substituting $x = z\_1 + z\_2$ into the regression $(1)$ then:
$$
\begin{align\*}
y &= \beta\_0 + \beta\_1(z\_1 + z\_2) + \beta\_2 z\_1 + e,\\
&= \beta\_0 + \beta\_1 z\_2 + (\beta\_1 + \beta\_2) z\_1 + e.
\end{align\*}
$$
So the coefficient on $z\_2$ in the new regression is identical to the coefficient on $x$ in the original one $(\beta\_1)$ , while the coefficient on $z\_1$ in the new regression is the sum of the coefficients on $x$ and $z\_1$ in the original regression $(\beta\_1 + \beta\_2)$.

> And a follow up question is: if $z\_1$ is correlated with the error term, will including it in the model bias the estimate of $\beta\_1$?

The opposite is true. Including $z\_1$ into the regression will make the estimate of $\beta\_1$ unbiased. Consider the following data generating process:
$$
y = \beta\_0 + \beta\_1 x + e, \tag{2}
$$
and assume that $e$ is correlated with $z\_1$. Then we can write:
$$
e = \gamma z\_1 + \varepsilon.
$$
where $\varepsilon$ is now uncorrelated with $z\_1$. (and where $\gamma = \mathbb{E}(e z\_1)/\mathbb{E}((z\_1)^2) \ne 0$. Assume for simplicity that $e$ is uncorrelated with $z\_2$.

Then the estimate of $\beta\_1$ will be biased as the orthogonality condition $\mathbb{E}(e x) = 0$ is not satisfied. Indeed:
$$
\begin{align\*}
\mathbb{E}(ex) &= \mathbb{E}(e z\_2) + \mathbb{E}(\gamma z\_1 z\_1) + \mathbb{E}(\varepsilon z\_1),\\
&= \mathbb{E}(\gamma (z\_1)^2) \ne 0
\end{align\*}
$$
If we include $z\_1$ into the regression. Then substituting $e = \gamma z\_1 + \varepsilon$, into $(2)$ we can write:
$$
y = \beta\_0 + \beta\_1 x + \gamma z\_1 + \varepsilon.
$$
And $\mathbb{E}(\varepsilon) = \mathbb{E}(\varepsilon x) = \mathbb{E}(\varepsilon z\_1) = 0$. So by including $z\_1$ into the regression, we can guarantee the residual $\varepsilon$ to be uncorrelated with all covariates. This means that $\beta\_1$ is identified and its estimate will be unbiased.

> **Comments:**
> **username_2**: Regarding *Another way to see this...*, if I simulate the data according to $y=\beta_0+\beta_1x+e$ with $x=z_1+z_2$ and then regress $y$ on $x$ and $z_1$, the estimated coefficient on $z_1$ is around zero and insignificant -- and far from $\beta_1+\beta_2=\beta_1+0=\beta_1$. Is that in line with what you are saying?  
> **username_1**: @Richard Hardy What I am saying is that if you first regress $y$ on $x$ and $z_1$ and then $y$ on $z_2$ and $z_1$, then the coefficient on $z_1$ for the second regression is the sum of the coefficients for $x$ and $z_1$ of the first regression. In addition, the coefficient on $x$ in the first regression should be the same as the coefficient on $z_2$ in the second regression.  In your case the coefficient on $z_1$ is zero as it is zero in your data generating process.  
> **username_2**: OK, thanks! I think my case is the same as OP's case, as I am following the setup in there. But even if that is not the case, you explanation is helpful.  

---

