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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?

Directed acyclic graph with multiple instrumental and endogenous variables

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I think this is a case where you have to consider all the variables at once, instead of one-at-a-time. Consider the classic setup for instrumental variables:

enter image description here

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:

enter image description here

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 $X.$

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  • $\begingroup$ 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. $\endgroup$
    – bilgelm
    Nov 4, 2021 at 16:45
  • $\begingroup$ 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. $\endgroup$ Nov 4, 2021 at 17:14

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