# Omitting a variable in IV estimation

I have an instrumental variable (IV) estimation where I use Z as an instrument for treatment D, to estimate a treatment effect of D on Y. After certain discussion, I find that there might be another variable, X, which is correlated both with Y and with Z.

I then decide to include X into my IV model, to control for what seems to be a violation of exclusion assumption (i.e. Z doesn't affect Y solely through X).

After re-doing the first stage of 2SLS, I find that Z coefficient is not statistically significant, as standard errors are high. I deduced that this might be due to high collinearity of X with Z, which is is 0.75. Still, however, X is significant in the first stage regression.

However, consequently using Z to instrument D, and including X as a control variable yields absolutely no statistical significance.

I therefore concluded that Z is a weak instrument if we do not control for X, and after we control for X, Z is still weak because there is no statistical significance in first stage. I am relatively new to IV estimation, and I wanted to get an opinion on whether my logic is flawed and what the next steps could entail:

• Is omitting X equivalent an example of OVB and does its inclusion remove it?
• What is the significance of high collinearity between instrument Z and control variable X?
• Does the insignificant first stage (with inclusion of X) mean that we cannot use Z as instrument?

The first thing to note is that whether you should consider $X$ or not cannot be answered without structural knowledge. For instance, consider the following model:

In this case X is correlated with Z and Y and Z is a valid instrument. However if you control for X, then Z is not a valid instrument anymore. Now consider the following variation:

Here we have the opposite case. By itself, Z is not a valid instrument, but if you condition on X, then Z is a valid instrument. Thus, just the fact that X is correlated with Z and Y tells you nothing about whether you should condition on X in your analysis. If you are not familiar with DAGS, you can interpret the directed arrows as direct causes and the bidirected arcs as dependence (such as correlation) of (structural) error terms.

• There's no way to say what the inclusion/omission of X, either in the first stage, in the second stage, or both, will do to your estimates (increase or reduce bias) without further structural knowledge about the problem.

However, to complement the answer, imagine you are in situation (2), where $Z$ is an instrument only conditional on $X$. What would happen if you omit $X$? Assume linearity and that the variables were standardized to simplify computation. The DAG corresponds to the following structural equations:

$$Y = \alpha D + \gamma X + U_y\\ D = \beta Z + U_d\\ X = U_x\\ Z = U_Z$$

Where the disturbances are uncorrelated except for $cov(U_y, U_d) = C_{yd}$ and $cov(U_x, U_z) = C_{xz}$. Our target parameter is $\alpha$ but the IV estimand without conditioning on $X$ gives you:

$$\frac{cov(Y,Z)}{cov(D, Z)} = \frac{\alpha\beta + \gamma C_{xz}}{\beta} = \alpha + \underbrace{\frac{\gamma C_{xz}}{\beta}}_{\text{bias}}$$

Thus, you can see the estimate is biased (this is the asymptotic bias). The bias is different from regular OVB however. Although the terms $C_{xz}$ and $\gamma$ of the bias are similar to the traditional OVB, note the strength of the instrument, $\beta$, also determines the bias due to the omission of $X$.

• Regarding the meaning of the high collinearity of $X$ and $Z$, the answer is the same as above. Without further knowledge, it just means they are correlated. It could be because they share a common cause, because Z causes X, because X causes Z etc.
• Being a weak instrument is a different problem altogether. And yes, you should be concerned about it. If your instruments are weak, you can have large finite sample biases even if they are valid instruments.

This other answer here might interest you.

• thanks a lot for the illustrative answer! Just to clarify from your DAG, what are the meanings of dashed and solid lines: correlation and causality? – Badalyan Apr 7 '18 at 7:26
• @Badalyan yes, $X \rightarrow Y$ means X causes Y and $X \leftrightarrow Y$ means the disturbances of $X$ and $Y$ are dependent (for instance, correlated). – Carlos Cinelli Apr 7 '18 at 18:13
• @Badalyan I complemented the answer, and included the derivation of the bias as well in the case where you should include $X$. – Carlos Cinelli Apr 7 '18 at 18:14
• Thanks! Would you be able to expand a little on why Z is not a valid instrument anymore if we control for non-causal X (1st model you explain)? Trying to get some intuition – Badalyan Apr 10 '18 at 15:30
• Hi, @Badalyan in the first graph conditioning on $X$ opens the collider path $Z \leftrightarrow X \leftrightarrow Y$ and induces an association between $Z$ and $Y$ not "via" $D$. Colliding paths are opened when conditioned upon, this other answer might help as well: stats.stackexchange.com/questions/330943/… – Carlos Cinelli Apr 10 '18 at 21:49