I understand that instrumental variable is used to address endogeneity bias since there could be correlation between the variable of interest and the error term.

Suppose now we want to see the estimated effect of X on Y. After some searching, I still do not know why we can't simply include the instrumental variable (Z) as another control variable in the OLS regression. Yes, this variable can increase the multicollinearity because it's correlated with X and it's not correlated with the error term. But this can only reduce precision of the estimate on X but not produce much bias.

If I consider Z as another control variable and omit it from the model of Y on X, does it increase bias? I think it does because Z and X are correlated.

What am I misunderstanding here?

  • 2
    $\begingroup$ Part of the definition of an instrumental variable $z$ is that it does not affect $y$ directly (i.e., the so-called exclusion restriction), which means it should not belong in your regression model. $\endgroup$
    – Durden
    Commented Feb 24 at 23:03

2 Answers 2


There is already a good answer, but I will try to explain the problem more explicitly with mathematics. However, to grasp the problem, you have to think of it in terms of causality, not just correlation; for this, DAGs can be quite helpful.

Here are the equations from the DAG in the previous answer:

$$ Y = \beta X + \gamma U + e $$ $$ X = \delta U + \theta Z + v$$

I've also introduced the independent error terms $e$ and $v$, which means that there is some exogenous variation in $Y$ and $X$ that is not explained by $U$ or $Z$. Note that we're also assuming that $U$ and $Z$ are independent.

If we regres $Y$ on $X$ by OLS, the theoretical analogue of the OLS-estimator is1

$$ \frac{Cov(X,Y)}{Var(X)} = \underbrace{\beta \vphantom{\frac{Var(U)}{Var(X)}}}_{\text{true parameter}} + \underbrace{\gamma\delta \frac{Var(U)}{Var(X)}}_{\text{omitted variable bias}}$$

Notice that $Z$ has nothing to do with the omitted variable bias (OVB); the only source of OVB is that $U$ causes both $X$ ($\delta$) and $Y$ ($\gamma$). To see how the OVB would be affected by adding $Z$ as a control variable we can use the Frisch–Waugh–Lovell (FWL) theorem. According to FWL, regressing $Y$ on $X$ with $Z$ as a control variable is the same as regressing $X$ on $Z$, saving the residuals and regressing $Y$ on those residuals. Let's denote the residuals $\tilde{X} = \delta U + v$.2 Now, a regression of $Y$ on $\tilde{X}$ yields

$$ \frac{Cov(\tilde{X},Y)}{Var(\tilde{X})} = \beta + \gamma\delta \frac{Var(U)}{Var(\tilde{X})}$$

Notice that the expression is essentially unchanged. $Z$ has neither removed nor introduced bias. However, because $Var(X)>Var(\tilde{X})$3, adding $Z$ as a control variable will amplify the omitted variable bias from $U$.

Here's how I would explain it in less technical terms: There is both good and bad variation in $X$. If it were possible, we would like to control for $U$ because that would remove bad variation from $X$ and reduce (get rid of) OVB. Controlling for $Z$ instead removes good variation from $X$. The situation had been different if

$$ Y = \beta X + \gamma U + \phi Z + e $$

In this case, $Z$ is not a valid instrument. Both $U$ and $Z$ are confounders and the only good (exogeneous) variation in $X$ that can be used to estimate its effect on $Y$ comes from $v$.

1 $$ \frac{Cov(X,Y)}{Var(X)} = \frac{Cov(\delta U + \theta Z + v, (\beta\delta + \gamma)U + \beta\theta Z + \beta v + e)}{Var(\delta U + \theta Z + v)} = \frac{\beta \left[\delta^2 Var(U) + \theta^2 Var(Z) + Var(v)\right] + \gamma\delta Var(U)}{\delta^2 Var(U) + \theta^2 Var(Z) + Var(v)} = \beta + \gamma\delta \frac{Var(U)}{Var(X)}$$ 2Note that I'm cheating a bit here, becasue the residuals are actually $\tilde{X} = X - \hat{\theta}Z = \left(\theta - \hat{\theta}\right)Z + \delta U + v $. I believe the result still holds (at least approximately), but it's harder to prove.

3 $Var(X)-Var(\tilde{X}) = \delta^2 Var(U) + \theta^2 Var(Z) + Var(v) - \delta^2 Var(U) - Var(v) = \theta^2 Var(Z)$

  • $\begingroup$ Thank you for the mathematics. This makes a lot of sense! I only want some confirmations. How can we know that $Var(X) > Var(\tilde{X})$? Is this because we have more variables in the regression of $X$, something related to lowering the variance (homoskedasticity)? Also, when you regressed Y on $\tilde{X}$, should $\tilde{X}$ be regressed on $Z$ instead of $U$? Am I misunderstanding? $\endgroup$
    – hiu
    Commented Mar 2 at 16:03
  • 1
    $\begingroup$ 1. Because we have removed the variation from $Z$; see footnote 3 in edit. 2. A regression of $X$ on $Z$ is what produces the residual $\tilde{X}$, which still contains variation from $U$ but no longer from $Z$. I think you might be confusing the definition of a variable (or "true model" / data generating process) and a regression $\endgroup$
    – Jonathan
    Commented Mar 4 at 9:04

The whole problem is that the effect of X on Y is confounded. If you include the instrument in a regression, you do not isolate any effect of X on Y.

Consider the following DAG.
enter image description here

This is a typical DAG for an instrumental variable. Here, we condition on the instrument and X, and there still exists a biasing path X <- U -> Y which the instrument does not block.

If I consider Z as another control variable and omit it from the model of Y on X, does it increase bias?

I'm not sure if it increases bias, but it certainly doesn't address confounding.

  • $\begingroup$ Thank you. It indeed does not address the correlation between X and the error term. However, I kept pondering about the bias if I included the IV. I think it's likely bias because Z and X are correlated. Even if X is now only the variation caused by Z, it's still correlated with Z, so the estimated coefficient on X must be biased when I omit Z. wiki does not say about this clearly for me $\endgroup$
    – hiu
    Commented Feb 25 at 5:11

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