# A Good Instrument is a Bad Control

Here’s a puzzle for you.
What will happen if we regress some outcome of interest on *both* an endogenous regressor *and* a valid instrument for that regressor?
I hadn’t thought about this question until 2018, when one of my undergraduate students asked it during class.
If memory serves, my off-the-cuff answer left much to be desired.^{1}
Five years later I’m finally ready to give a fully satisfactory answer; better late than never I suppose!

# The Model

We’ll start by being a bit more precise about the setup.
Suppose that \(Y\) is related to \(X\) according to the following **linear causal model** \[
Y \leftarrow \alpha + \beta X + U
\]
where \(\beta\) is the causal effect of interest and \(U\) represents unobserved causes of \(Y\) that may be related to \(X\).
Now, for *any* observed random variable \(Z\), we can define
\[
V \equiv X - (\pi_0 + \pi_1 Z), \quad \pi_0 \equiv \mathbb{E}[X] - \pi_1 \mathbb{E}[Z], \quad \pi_1 \equiv \frac{\text{Cov}(X,Z)}{\text{Var}(Z)}.
\]
This is the **population linear regression** of \(X\) on \(Z\).
By construction it satisfies \(\mathbb{E}[V] = \text{Cov}(Z,V) = 0\).^{2}
Thus we can write,
\[
X = \pi_0 + \pi_1 Z + V, \quad \mathbb{E}[V] = \text{Cov}(Z,V) = 0
\]
for *any* random variables \(X\) and \(Z\), simply by constructing \(V\) as described above.
If \(\pi_1 \neq 0\), we say that \(Z\) is **relevant**.
If \(\text{Cov}(Z,U) = 0\), we say that \(Z\) is **exogenous**.
If \(Z\) is both relevant and exogenous, we say that it is a **valid instrument** for \(X\).

As we’ve defined it above, \(V\) is simply a regression residual.
But if \(Z\) is a valid instrument, it turns out that we can think of \(V\) as the “endogenous part” of \(X\).
To see why, expand \(\text{Cov}(X,U)\) as follows:
\[
\text{Cov}(X,U) = \text{Cov}(\pi_0 + \pi_1 Z + V, \,U) = \pi_1 \text{Cov}(Z,U) + \text{Cov}(U,V) = \text{Cov}(U,V)
\]
since we have assumed that \(\text{Cov}(Z,U) = 0\).
In words, the endogeneity of \(X\) is *precisely the same thing* as the covariance between \(U\) and \(V\).

Here’s a helpful way of thinking about this.
If \(Z\) is exogenous then our regression of \(X\) on \(Z\) *partitions* the overall variation in \(X\) into two components: the “good” (exogenous) variation \(\pi_1 Z\) is uncorrelated with \(U\), while the “bad” (endogenous) variation \(V\) is correlated with \(U\).
The logic of two-stage least squares is that regressing \(Y\) on the “good” variation, \(\pi_1 Z\) allows us to recover \(\beta\), the causal effect of interest.^{3}

# A Simulation Example

Using the model and derivations from above, let’s run a little simulation.
To simulate a valid instrument \(Z\) and an endogenous regressor \(X\) we can proceed as follows.
First generate independent standard normal draws \(\{Z_i\}_{i=1}^n\).
Next independently generate pairs of correlated standard normal draws \(\{(U_i, V_i)\}_{i=1}^n\) with \(\text{Corr}(U_i, V_i) = \rho\).
Finally, set
\[
X_i = \pi_0 + \pi_1 Z_i + V_i \quad \text{and} \quad
Y_i = \alpha + \beta X_i + U_i
\]
for each value of \(i\) between \(1\) and \(n\).^{4}
The following chunk of R code runs this simulation with \(n = 5000\), \(\rho = 0.5\), \(\pi_0 = 0.5\), \(\pi_1 = 0.8\), \(\alpha = -0.3\) and \(\beta = 1\):

```
set.seed(1234)
n <- 5000
z <- rnorm(n)
library(mvtnorm)
Rho <- matrix(c(1, 0.5,
0.5, 1), 2, 2, byrow = TRUE)
errors <- rmvnorm(n, sigma = Rho)
u <- errors[, 1]
v <- errors[, 2]
x <- 0.5 + 0.8 * z + v
y <- -0.3 + x + u
```

In the simulation \(Z\) is a valid instrument, \(X\) is an endogenous regressor, and the true causal effect of interest equals one. Using our simulation data, let’s test out three possible estimators:

- \(\widehat{\beta}_\text{OLS}\equiv\) the slope coefficient from an OLS regression of \(Y\) on \(X\).
- \(\widehat{\beta}_\text{IV}\equiv\) slope coefficient from an IV regression of \(Y\) on \(X\) with \(Z\) as an instrument.
- \(\widehat{\beta}_{X.Z}\equiv\) the coefficient on \(X\) in an OLS regression of \(Y\) on \(X\) and \(Z\).

```
c(truth = 1,
b_OLS = cov(x, y) / var(x),
b_IV = cov(z, y) / cov(z, x),
b_x.z = unname(coef(lm(y ~ x + z))[2])) |> # unname() makes the names prettier!
round(2)
```

```
## truth b_OLS b_IV b_x.z
## 1.00 1.31 1.01 1.49
```

As expected, OLS is far from the truth while IV pretty much nails it.
Interestingly, the regression of `y`

on `x`

and `z`

gives the worst performance of all! Is this just a fluke?
Perhaps it’s an artifact of the simulation parameters I chose, or just bad luck arising from some unusual simulation draws.
To find out, we’ll need a bit more algebra.
But stay with me: the payoff is worth it, and there’s not too much extra math required!

# The General Result

## Regression of \(Y\) on \(X\) and \(Z\)

The coefficient on \(X\) in a population linear regression of \(Y\) on \(X\) and \(Z\) is given by
\[
\beta_{X.Z} = \frac{\text{Cov}(\tilde{X}, Y)}{\text{Var}(\tilde{X})}
\]
where \(\tilde{X}\) is defined as the *residual* in another population linear regresasion: the regression of \(X\) on \(Z\).^{5}
But wait a minute: we’ve *already seen* this residual!
Above we called it \(V\) and used it to write \(X = \pi_0 + \pi_1 Z + V\).
Using this equation, along with the linear causal model relating \(Y\) to \(X\) and \(U\), we can re-express \(\beta_{X.Z}\) as
\[
\begin{align*}
\beta_{X.Z} &= \frac{\text{Cov}(V, Y)}{\text{Var}(V)} = \frac{\text{Cov}(V, \alpha + \beta X + U)}{\text{Var}(V)}\\
&= \frac{\text{Cov}(U,V) + \beta\text{Cov}(V, \pi_0 + \pi_1 Z + V)}{\text{Var}(V)}\\
&= \beta + \frac{\text{Cov}(U,V)}{\text{Var}(V)}
\end{align*}
\]
since \(\text{Cov}(Z, V) = 0\) by construction.
We have some simulation data at our disposal, so let’s check this calculation.
In the simulation \(\beta = 1\) and
\[
\frac{\text{Cov}(U, V)}{\text{Var}(V)} = 0.5
\]
since \(\text{Var}(U) = \text{Var}(V) = 1\) and \(\text{Cov}(U, V) = 0.5\).
Therefore \(\beta_{X.Z} = 1.5\).
And, indeed, this is almost *exactly* the value of our estimate from our simulation above.

## Regression of \(Y\) on \(X\) Only

So far so good. Now what about the “usual” OLS estimator? A quick calculation gives \[ \beta_{\text{OLS}} = \beta + \frac{\text{Cov}(X,U)}{\text{Var}(X)} = \beta + \frac{\text{Cov}(V,U)}{\text{Var}(X)} \] using the fact that \(\text{Cov}(X,U) = \text{Cov}(U,V)\), as explained above. Again, we can check this against our simulation results. We know that \(\text{Cov}(V,U) = 0.5\) and \[ \text{Var}(X) = \text{Var}(\pi_0 + \pi_1 Z + V) = \pi_1^2 \text{Var}(Z) + \text{Var}(V) = (0.8)^2 + 1 = 41/25 \] since \(Z\) and \(V\) are uncorrelated by construction, \(\text{Var}(Z) = \text{Var}(V) = 1\) and \(\pi_1 = 0.8\) in the simulation design. Hence, \(\beta_{\text{OLS}} = 1 + 25/82 \approx 1.305\). Again, this agrees almost perfectly with our simulation.

## Comparing the Results

To summarize, we have shown that
\[
\beta_{X.Z} = \beta + \frac{\text{Cov}(U,V)}{\text{Var}(V)}, \quad \text{while} \quad
\beta_{\text{OLS}} = \beta + \frac{\text{Cov}(U,V)}{\text{Var}(X)}.
\]
There is only one difference between these two expressions: \(\beta_{X.Z}\) has \(\text{Var}(V)\) where \(\beta_{\text{OLS}}\) has \(\text{Var}(X)\).
Returning to our expression for \(\text{Var}(X)\) from above,
\[
\text{Var}(X) = \pi_1^2 \text{Var}(Z) + \text{Var}(V) > \text{Var}(V)
\]
as long as \(\pi_1 \neq 0\) and \(\text{Var}(Z) \neq 0\).
In other words, there is always *more* variation in \(X\) than there is in \(V\), since \(V\) is the “leftover” part of \(X\) after regressing on \(Z\).
Because the variances of \(X\) and \(V\) appear in the denominators of our expressions from above, it follows that
\[
\left| \text{Cov}(U,V)/\text{Var}(V)\right| > \left| \text{Cov}(U,V)/\text{Var}(X)\right|.
\]
In other words, \(\beta_{X.Z}\) is **always farther from the truth** than \(\beta_{OLS}\), exactly as we found in our simulation.

# Some Intuition

In our simulation, \(\widehat{\beta}_{X.Z}\) gave a *worse* estimate of \(\beta\) than \(\widehat{\beta}_{X.Z}\).
The derivations from above show that this wasn’t a fluke: adding a valid instrument \(Z\) as an additional control regressor only makes the bias in our estimated causal effect *worse* than it was to begin with.
This holds for any valid instrument and any endogenous regressor in a linear causal model.
I hope you found the derivations from above convincing.
But even so, you may be wondering if there’s an intuitive explanation for this phenomenon.
I am please to inform you that the answer is yes!

In an earlier post I described the **control function** approach to instrumental variables regression.
That post showed that the coefficient on \(X\) in a regression of \(Y\) on \(X\) and \(V\) gives the *correct* causal effect.
We don’t know \(V\), but we can estimate it by regressing \(X\) on \(Z\) and saving the residuals.
The logic of multiple regression shows that including \(V\) as a control regressor “soaks up” the portion of \(X\) that is explained by \(V\).
Because \(V\) represents the “bad” (endogenous) variation in \(X\), this solves our endogeneity problem.
In effect, \(V\) captures the unobserved “omitted variables” that play havoc with a naive regression of \(Y\) on \(X\).

Now, contrast this with a regression of \(Y\) on \(X\) and \(Z\).
In this case, we soak up the variation in \(X\) that is explained by \(Z\).
But \(Z\) represents the **good** (exogenous) variation in \(X\)!
Soaking up this variation leaves only the bad variation behind, making our endogeneity problem worse than it was to begin with.
In this example, \(Z\) is what is known as a bad control, a control regressor that makes things worse rather than better.
A common piece of advice for avoiding bad controls is to only include control regressors that are correlated with \(X\) and \(Y\) but are *not themselves* caused by \(Z\).
The example in this post shows that this advice **wrong**.
Here \(Z\) is not caused by \(X\), and is correlated with both \(X\) and \(Y\).
Nevertheless, it is a bad control
In short, a valid instrument provides a powerful way to carry out causal inference from observational data, but only if you use it in the right way.
A good instrument is a bad control!

I seem to recall saying something like “this won’t in general give us the causal effect we’re interested in, but I don’t think it’s possible to say anything more without extra assumptions.” Fortunately my lackluster response didn’t derail the student who asked the question: he’s currently pursuing a PhD in Economics at UChicago!↩︎

Check if you don’t believe me: substitute the expressions for \(\pi_0\) and \(\pi_1\), take expectations / covariances, and simplify.↩︎

See this blog post for more discussion.↩︎

We don’t necessarily need \(Z_i\) to be normally distributed, as long as it’s independent of \((U_i, V_i)\), so you could use e.g. uniform draws if you prefer. Generating \((U_i, V_i)\) from a bivariate normal distribution isn’t necessary either, but it’s a simple way of controlling the endogeneity in \(X\).↩︎

This is a special case of the so-called FWL Theorem, although I’d argue that we should call it “Yule’s Rule” since George Udny Yule was arguably the first person to popularize it, decades before F, W, or L.↩︎