# Identifiability of multivariate instrumental variable model

I'm interested in estimating the effects of $$X_1$$ and $$X_2$$ on $$Y$$ in the directed acyclic graph below. $$U_1$$ and $$U_2$$ are unobserved confounders. Based on Definition 7.4.1 on p. 248 of Causality 2nd Ed. by Pearl, $$\boldsymbol Z = \{Z_1, Z_2, Z_3\}$$ is not admissible as an instrumental variable for estimating the effects of $$X_1$$ and $$X_2$$ on $$Y$$.

This is because, for example, to estimate the effect of $$X_1$$ on $$Y$$, we need to block the path between $$\boldsymbol Z$$ and $$Y$$ by conditioning on $$X_2$$, but doing so opens the collider $$\boldsymbol Z \rightarrow X_2 \leftarrow U_2$$, violating the exclusion assumption.

Is it nevertheless appropriate to use two stage least squares (2SLS) for estimating the effects of $$X_1$$ and $$X_2$$ on $$Y$$ for this DAG? Are there any additional assumptions that could make this problem identifiable?

Here $$U_Y$$ is unmeasured. The rule for $$I$$ being an instrument is that it is $$d$$-connected to $$X$$ in the original graph above, and $$d$$-separated from $$Y$$ in this graph:
The $$d$$-separation occurs because $$X$$ is an unconditioned collider, which prevents information flow.
Now all of these considerations hold in your case, although they only hold in the aggregate. That is, the sets \begin{align*} Z&=\{Z_1,Z_2,Z_3\}\\ X&=\{X_1,X_2\} \end{align*} obey these rules, taken as a whole. The set $$Z$$ is $$d$$-connected to $$X,$$ clearly. Now what happens if you delete all the arrows from the set $$X$$ to $$Y?$$ The set $$X$$ acts as an unconditioned collider for any causal information that might want to go from $$Z$$ to $$Y.$$ Hence $$Z$$ is $$d$$-separated from $$Y$$ in the modified graph. Therefore, $$Z$$ is an instrument for $$X.$$
• Thank you for your answer. If I understand correctly, the combined effects of $X_1$ and $X_2$ on $Y$ is identifiable, but not their individual effects on $Y$. So, I can use 2SLS to estimate the 2-D vector describing the effect of $X$ on $Y$, but I cannot interpret the individual elements of this vector as the distinct effects of $X_1$ and $X_2$ on $Y$. I suppose the unidentifiability here stems from the indistinguishability of these two effects. Commented Nov 4, 2021 at 16:45
• Yes, I would agree that the combined effect is identifiable, but not the individual effect. I'm not sure I agree that the unidentifiability stems from indistinguishability, but I would say rather from the graph connectivity. For example, imagine that $U_2$ did not exist. How would that change things? Surely you could identify at least one of $X_1$ or $X_2$ separately. Commented Nov 4, 2021 at 17:14