SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Finally, we have a contradiction. So the second condition is also necessary.
Proposition D.6. Suppose  $X' \subseteq X$  is a valid subset with  $|X'| \geq 2$  and every pair of nodes in  $X'$  has only one non-dominated path. And the latent DAG only has one source node. Then the followings are true:
(1) there exists a hidden node  $H^{\prime}$  with  $X^{\prime}\subseteq \mathrm{ch}_{X}(H^{\prime})$
(2) for any node  $X_{i} \in X'$ , one can find at least another node  $X_{j} \in X'$  such that  $X_{i} \leftarrow H' \rightarrow X_{j}$  is a non- $(X_{i}, X_{j})$ -dominated path.
Proof. We'll show this by induction.
Base step: This is true because of Lemma D.2.
Inductive Hypothesis: Suppose the statement is true for all  $X'$  with  $|X'| = k$ .
Inductive Step: We will show the statement is true for  $|X'| = k + 1$ . Without loss of generality, let's suppose  $X' = \{X_1, \ldots, X_{k+1}\}$ . Consider the subset  $X'' = \{X_1, \ldots, X_k\}$ . By the inductive hypothesis, there exists a hidden node  $H'$  with  $X' \subseteq \mathrm{ch}_X(H')$  and again without loss of generality, two nodes  $X_1$  and  $X_2$ , such that  $X_1 \gets H' \to X_2$  is a non- $(X_1, X_2)$ -dominated path.
Now consider another subset  $X_{1,k+1} = \{X_1, X_{k+1}\}$ . By our assumption and Lemma D.2, there exists a hidden node  $H_{1,k+1}$  such that  $X_1 \gets H_{1,k+1} \to X_{k+1}$  is a non- $(X_1, X_{k+1})$ -dominated path. We can define  $H_{2,k+1}$  in a similar way.
If  $H_{1,k + 1} = H'$ , then we are done.
Let's now consider the cases when  $H_{1,k+1} \neq H'$ . Because there is only one single source node, there must exist at least one common ancestor of  $H_{1,k+1}$  and  $H'$ :  $H_*$ . Now let's consider three cases:
(1)  $H_{1,k+1}$  is an ancestor of  $H'$ . Then there is a directed path  $P$  from  $H_{1,k+1}$  to  $H'$ . Now we have an active path between  $X_2$  and  $X_{k+1}$  via  $P$ . Because this path is  $(X_2, X_{k+1})$ -dominated, we have a few cases.
(a) There exists a common parent of  $X_{2}$ ,  $X_{k + 1}$  that is in  $\overline{\mathrm{an}}(H_{1,k + 1})$ , which means that  $H_{2,k + 1}$  is an ancestor of  $H_{1,k + 1}$  by Lemma D.4 (case (c) covers when  $H_{2,k + 1} = H_{1,k + 1}$ ). By Lemma D.3, this is not possible because we have a path between  $X_{1}$  and  $X_{2}$  via the directed path from  $H_{2,k + 1}$  to  $H_{1,k + 1}$  but  $H'$  is a descendant of  $H_{1,k + 1}$ .
(b)  $H_{2,k+1}$  is on  $P$  but it is not the same as  $H_{1,k+1}$  nor  $H'$ . By the same argument as the case (a), this is also not possible because of Lemma D.3.
(c)  $H_{2,k+1} = H_{1,k+1}$ . This means that  $H_{1,k+1}$  is a common parent of  $X_1, X_2$ . So it cannot be an ancestor of  $H'$ . This is a contradiction.
(d) When  $H_{2,k+1} = H'$ , then  $H'$  is the hidden node we are looking for because  $X' \subseteq \operatorname{ch}_X(H')$  and for any node  $X_i \in X'$ , there exists another node  $X_j \in X'$  such that  $X_i \gets H' \to X_j$  is a non- $(X_i, X_j)$ -dominated path (for  $X_{k+1}$ , one can choose  $X_2$  such that  $X_2 \gets H' \to X_{k+1}$  is a non-dominated path).
(2)  $H'$  is an ancestor of  $H_{1,k+1}$ . Then there is a directed path  $P$  from  $H'$  to  $H_{1,k+1}$ . Let  $X_q$  be an arbitrary node in  $X''$  that is not  $X_1$ . Then we can define hidden node  $H_{q,k+1}$  such that  $X_q \gets H_{q,k+1} \to X_{k+1}$  is a non- $(X_q, X_{k+1})$ -dominated path. Now we have an active path between  $X_q$  and  $X_{k+1}$  via  $P$ . Because this path is  $(X_q, X_{k+1})$ -dominated, we have a few cases.
(a) There exists a common parent of  $X_{q}$ ,  $X_{k + 1}$  that is in  $\overline{\mathrm{an}}(H')$ , which means that  $H_{q,k + 1}$  is an ancestor of  $H'$  by Lemma D.4 (case (c) covers when  $H_{q,k + 1} = H'$ ). By Lemma D.3, this is not possible because we have a path between  $X_{1}$  and  $X_{k + 1}$  via the directed path from  $H_{q,k + 1}$  to  $H'$  but  $H_{1,k + 1}$  is a descendant of  $H'$ .
(b)  $H_{q,k+1}$  is on  $P$  but it is not the same as  $H_{1,k+1}$  nor  $H'$ . By the same argument as the case (a), this is also not possible because of Lemma D.3.
(c)  $H_{q,k+1} = H'$ . This means  $H'$  is a common parent of  $X_1, X_{k+1}$ . So it cannot be an ancestor of  $H_{1,k+1}$ . This is a contradiction.
(d)  $H_{q,k+1} = H_{1,k+1}$ . This is possible. But because  $X_q$  is chosen arbitrarily. We have  $H_{q,k+1} = H_{1,k+1}$  for  $q = 2, \ldots, k$ . Therefore  $X' \subseteq \operatorname{ch}_X(H_{1,k+1})$ . And for any node  $X_i \neq X_{k+1}$ , we have that  $X_i \gets H_{1,k+1} \to X_{k+1}$  is a non-dominated path.
(3) There exist a common ancestor  $H_*$  that is neither  $H_{1,k+1}$  nor  $H'$ ,  $H'$  is not an ancestor of  $H_{1,k+1}$  and  $H_{1,k+1}$  is not an ancestor of  $H'$ .  $H_*$  is a common ancestor of  $X_2$  and  $X_{k+1}$ . Therefore, there is a common ancestor path between  $X_2$  and  $X_{k+1}$ . But this path is a  $(X_2, X_{k+1})$ -dominated path.
There are two cases: (a)  $H_{2,k+1} \in \overline{\mathrm{an}}(H_{1,k+1})$  when  $H_{2,k+1}$  is on the directed path from  $H_*$  to  $H_{1,k+1}$  or there is another common parent of  $X_2$ ,  $X_{k+1}$  that is in  $\overline{\mathrm{an}}(H_*)$  (Lemma D.4), (b)  $H_{2,k+1} \in \overline{\mathrm{an}}(H')$  when  $H_{2,k+1}$  is on the directed path from  $H_*$  to  $H'$  or there is another common parent of  $X_2$ ,  $X_{k+1}$  that is in  $\overline{\mathrm{an}}(H_*)$  (Lemma D.4). For case (a), we have a  $(X_1, X_2)$ -dominated path caused by the path from  $H_{2,k+1}$  to  $H_{1,k+1}$ . In this case, because  $X_1 \gets H' \to X_2$  is a non- $(X_1, X_2)$ -dominated path,  $H'$  must be an ancestor for  $H_{1,k+1}$ , which is a contradiction. By a similar argument, case (b) is also not possible.
Figure 10: In this example, the non-imaginary subset  $\{X_5,X_6,X_7\}$  is fractured
(b)
Figure 11: An example where two maximal measurement models share the same UDG
D.3 Fractured subset condition
A necessary condition for a non-replaceable imaginary subset is that the subset is fractured. The definition is restated below. Intuitively, a fractured subset provides redundant information as every connection between nodes in the fractured subset can be explained by other maximal valid subsets.
Definition 4.5 (Fractured subset). Given a collection of maximal valid subsets  $S$ , a clique  $C$  is called shattered by  $S$  if there exists a subset  $S' \subseteq S$  such that  $\cup S' = C$ . A collection of maximal valid subsets  $\{S_i\}_{i=1}^k$  is complete if for every intervention target  $I \in \mathcal{I}$ , all shattered cliques in  $D(P^{(I)})$  form an edge cover of  $D(P^I)$ . A maximal valid subset  $X' \subseteq X$  is fractured if there exists a complete collection  $\{S_i\}_{i=1}^k$  such that  $S_i \nsubseteq X'$  for all  $S_i$ .
The aforementioned necessity is shown by the following:
Proof. By Lemma D.16, we know that  $\{\mathrm{ch}_X(H_i)\}_{i=1}^m$  is a complete cover. And because  $X'$  is an imaginary subset,  $X' \not\subsetneq \mathrm{ch}_X(H_i)$  for any  $H_i$ .
One might suggest getting rid of all fractured subsets. However, even a non-imaginary subset can be a fractured subset (Example 9).
Example 9 (Non-imaginary set can be fractured). Consider Figure 10. If the intervention target  $I$  is  $\{H_1\}$ ,  $\{H_2\}$ ,  $\emptyset$ , then we have two maximal cliques:  $\{X_1, X_3, X_4, X_5, X_6, X_7\}$ ,  $\{X_2, X_3, X_4, X_5, X_6, X_7\}$ . If the intervention target  $I$  is  $\{H_3\}$ , then there are three maximal cliques:  $\{X_3, X_5, X_7\}$ ,  $\{X_2, X_4, X_5, X_6, X_7\}$ ,  $\{X_1, X_4, X_5, X_6, X_7\}$ . If the intervention target  $I$  is  $\{H_4\}$ , then there are three maximal cliques:  $\{X_4, X_6, X_7\}$ ,  $\{X_2, X_3, X_5, X_6, X_7\}$ ,  $\{X_1, X_3, X_5, X_6, X_7\}$ . When the intervention target is  $\{H_5\}$ , there are three maximal cliques:  $\{X_1, X_5, X_6\}$ ,  $\{X_2, X_5, X_6\}$ ,  $\{X_3, X_4, X_5, X_6, X_7\}$ . By Proposition 4.1, all these five valid subsets are maximal:  $\{X_1, X_5, X_6\}$ ,  $\{X_2, X_5, X_6\}$ ,  $\{X_3, X_5, X_7\}$ ,  $\{X_4, X_6, X_7\}$  and  $\{X_5, X_6, X_7\}$ . But  $\{X_5, X_6, X_7\}$  is fractured. In fact,  $\{X_1, X_5, X_6\}$ ,  $\{X_2, X_5, X_6\}$ ,  $\{X_3, X_5, X_7\}$ ,  $\{X_4, X_6, X_7\}$  constitutes a complete collection of maximal valid subsets.
Even under the maximality condition, it is still possible that two DAGs can share the same UDG (Example 10). The no fractured subset condition captures such ambiguities but is overkill.
Example 10. Consider Figure 11. For both  $G_{(a)}$  and  $G_{(b)}$ , under any intervention target, there are four maximal cliques:  $\{X_1, X_4, X_5\}$ ,  $\{X_2, X_5, X_6\}$ ,  $\{X_3, X_4, X_6\}$ ,  $\{X_4, X_5, X_6\}$ . And one can check that both  $G_{(a)}$  and  $G_{(b)}$  satisfy the maximality condition.
D.4 Pure child assumption
We can also use the pure child assumption (Assumption 3) to get identifiability. The pure child assumption has been made in many existing works [6, 22, 66] with additional parametric assumptions to get identifiability.
Assumption 3 (Pure child). For every  $H_{i}\in H$ , there exists at least one  $X_{i}\in X$  with  $\mathrm{pa}(X_i) = \{H_i\}$ , i.e.  $X_{i}$  only has one parent and that parent is  $H_{i}$ .
Note that under Assumption 3, there still could be imaginary subsets.
Example 11. Figure 12 satisfies Assumption 3 but there is an imaginary subset  $\{X_5, X_6\}$ .
Proof. By Assumption 1(a) and Assumption 1(b), two observed variables  $X_{i}$  and  $X_{j}$  are dependent if and only if  $X_{i}$  and  $X_{j}$  are  $d$ -connected. Therefore,  $D(G^{(I)})$  is the same as  $D(P^{(I)})$  under any intervention target  $I$ .
By Assumption 3, let's associate each hidden node  $H_{i}$  with one pure child  $X_{i}$ .
First of all, there does not exist an imaginary subset with two pure children  $X_{i}$  and  $X_{j}$  of different hidden parents. If the opposite is true, then  $\{X_{i}, X_{j}\}$  must be valid which is impossible by Lemma D.14. Therefore, every complete collection must have at least  $m$  maximal subsets.
By Lemma D.16,  $\{\mathrm{ch}_X(H_i)\}_{i = 1}^m$  is a complete collection with  $m$  maximal subsets.
In fact,  $\{\mathrm{ch}_X(H_i)\}_{i = 1}^m$  is the only complete collection with  $m$  maximal subsets. Suppose there exists a complete collection with imaginary subsets. There are two cases:
A. None of the imaginary subsets contains a pure child. Then we still need at least  $m$  nonimaginary subsets to cover pure children.
B. At least one of the imaginary subsets contains a pure child. Suppose one such imaginary subset is  $S_{i}$  and it contains  $X_{i}$ . Then there must exist another maximal subset from the collection that contains  $X_{i}$ .
Suppose the opposite is true. Because  $S_{i}$  is an imaginary subset, there exists  $X_{k} \in S_{i}$  such that  $X_{k}$  and  $X_{i}$  do not share the same parent. And for any intervention target  $I$ , if there is any edge between  $X_{i}$  and  $X_{j}$  in  $D(G^{(I)})$ , by definition, there exists a shattered clique  $C \in D(G^{(I)})$  such that  $\{X_{i}, X_{j}\} \subseteq C$  and  $S_{i} \subseteq C$ . Therefore, there is an edge between  $X_{j}$  and  $X_{k}$  as well. Now consider graph  $G' = (V, E')$  where  $E' = E \cup \{H_{i} \to X_{k}\}$ . Under any intervention target  $I$ ,  $D(G'^{(I)})$  can only have more edges than  $D(G^{(I)})$  and if there exists a new edge, these edges must involve  $X_{k}$ . Suppose one of these newly added edges is between  $X_{k}$  and  $X_{j}$ . Then it must be that a parent of  $X_{j}$  is  $d$ -connected to  $H_{i}$ . This would mean that  $X_{i}$  is  $d$ -connected to  $X_{j}$  in  $D(G^{(I)})$ . By the previous argument,  $X_{k}$  is also  $d$ -connected to  $X_{j}$  in  $D(G^{(I)})$ . Thus,  $G'$  would violate the maximality condition by Lemma D.10 and Lemma D.17 which is a contradiction. Therefore, if there is one
Figure 12: In this example, the graph satisfies pure child but  $\{X_5, X_6\}$  is an imaginary subset.
imaginary subset in the complete collection that contains a pure child, there must be at least two imaginary subsets that contain such a pure child. This means that to cover all  $m$  pure children, we need more than  $m$  maximal subsets.
On the other hand, suppose there exists a complete collection with at least one maximal subset  $X'$  such that  $X' \subseteq \operatorname{ch}_X(H_i)$ . Then  $X'$  must not include the pure child of  $H_i$ :  $X_i$ . This is because, if  $X_i \in X'$ , then for any maximal clique  $C \in \Omega$  such that  $X' \subseteq C$ , we have that  $\{X_i\} \subseteq X' \subseteq \operatorname{ch}_X(H_i) \subseteq C$  because any node in  $C \setminus \{X_i\}$  is  $d$ -connected by  $X_i$  via  $H_i$  since  $X_i$  is a pure child. Therefore,  $X'$  is not maximal which is a contradiction. Because  $X'$  does not contain a pure child of  $H_i$ , we still need to cover all  $m$  pure children. We need more than  $m$  maximal subsets.
For algorithmic purposes, under the pure child assumption, one only needs to check if shattered maximal cliques form an edge cover.
Lemma D.7. Under Assumptions 1-3, for any intervention target  $I$ , all maximal cliques  $C \in \Omega_P^{(I)}$  that is shattered by  $\{\mathrm{ch}_X(H_i)\}_{i=1}^m$  is an edge cover of  $D(P^{(I)})$ .
Proof. By Assumption 1(a) and Assumption 1(b), two observed variables  $X_{i}$  and  $X_{j}$  are dependent if and only if  $X_{i}$  and  $X_{j}$  are  $d$ -connected. Therefore,  $D(G^{(I)})$  is the same as  $D(P^{(I)})$  under any intervention target  $I$ . By Assumption 3, let's associate each hidden node  $H_{i}$  with one pure child  $X_{i}$ .
Suppose the statement is not true. Then for some intervention target  $I \in \mathcal{I}$ , there exists one clique  $C$  in  $\Omega_G^{(I)}$  that is not shattered and that clique contains one edge that is not covered by other shattered cliques in  $\Omega_G^{(I)}$ . Suppose that edge is between  $X_1$  and  $X_2$ , then there exist  $H_1$  and  $H_2$  with  $X_1 \in \mathrm{ch}_X(H_1)$  and  $X_2 \in \mathrm{ch}_X(H_2)$  and  $H_1$  is  $d$ -connected to  $H_2$ . Let one such active path between  $H_1$  and  $H_2$  be  $P_{1,2}$ . By Lemma E.3, there exists one source node  $\mathrm{so}(P_{1,2})$  (If  $H_1 = H_2$ , then  $\mathrm{so}(P_{1,2})$  is just  $H_1$ ). Note that we can always find one node  $H_*$  in  $\overline{\mathrm{an}}(\mathrm{so}(P_{1,2}))$  that has no incoming edge. Let's denote that the pure child of  $H_*$  is  $X_*$ .
Then  $C_* = \cup_{H_i \in \overline{\mathrm{de}}(H_*)} \mathrm{ch}_X(H_i)$  must be a shattered maximal clique in  $\Omega_G^{(I)}$ . Suppose  $C_*$  is not a maximal clique. Then the maximal clique that contains  $C_*$  must contain another element  $X_3$  not in  $C_*$ . There must exist another hidden variable  $H_3$  with  $X_3 \in \mathrm{ch}_X(H_3)$  that  $H_*$  is  $d$ -connected to because  $X_3$  is connected to the pure child of  $H_*$ . But this is not possible by Lemma E.3 and the fact that  $H_*$  is a source node while  $X_3$  is not a descendant of  $H_*$ .
D.5 Examples of measurement model with no non-replaceable fractured subsets
Lemma 4.6 shows that an imaginary subset is a fractured subset. Because a replaceable imaginary subset can be found, our concern should focus solely on the non-replaceable ones (Theorem 4.3). In fact, Corollary 4.7 can be strengthened to say that as long as there are no non-replaceable fractured subsets, then the bipartite graph can be identified. In this section, we will show a class of measurement models with no non-replaceable fractured subsets (see Remark 4.1).
Let's first present a simple example where there are multiple source nodes and the pure child assumption is violated, and there is still no non-replaceable fractured subset.
Example 12. Consider Figure 13. If the intervention target  $I$  is  $\emptyset$ ,  $H_{1}$  or  $H_{3}$ , then two maximal cliques are  $\{X_{1}, X_{2}, X_{3}\}$  and  $\{X_{2}, X_{3}, X_{4}\}$ . If the intervention target is  $H_{2}$ , then there are three maximal clique:  $\{X_{1}, X_{2}\}$ ,  $\{X_{2}, X_{3}\}$  and  $\{X_{3}, X_{4}\}$ . One can easily check that there is no non-replaceable fractured subset.
Figure 13: In this example, there is no fractured set.
In fact, we can extend Figure 13 to a family of measurement models as follows:
Definition D.2 (Sparse Measurement Model). For a given measurement model  $G = (V, E)$  and  $V = X \cup H$ , if  $X_1, X_2 \in X$  is  $d$ -connected, then it must be that
1. There exists at most one hidden node  $H_{i}$  such that  $\{X_1,X_2\} \subseteq \mathrm{ch}_X(H_i)$  (Type-1 path).
2. There exists at most one active path  $P$  between  $X_{1}$  and  $X_{2}$  such that  $\mathrm{pa}_P(X_1)\neq \mathrm{pa}_P(X_2)$  (Type-2 path).
And for any tuple  $(X_{1},X_{2},X_{3})\subseteq X$  such that  $\{X_1,X_2\} \subseteq \mathrm{ch}_X(H_2)$  and  $\{X_1,X_3\} \subseteq \mathrm{ch}_X(H_3)$  with  $H_{2}\neq H_{3}$ ,  $X_{2}$  and  $X_{3}$  can only be  $d$ -connected via type-2 path  $P$  with the property that  $\mathrm{pa}_P(X_2) = H_2$  and  $\mathrm{pa}_P(X_3) = H_3$ .
Theorem D.8. Under Assumptions 1-2 and with maximal sparse measurement model, there is no imaginary subset.
Proof. Suppose there is an imaginary subset  $X'$ . Then  $|X'| \geq 3$ . If  $X'$  has only one element, then it must be a subset of  $\mathrm{ch}_X(H')$  for some hidden node  $H'$ . If  $X'$  has two nodes, by the definition of sparse measurement, there could only be one active path between them and there exists an intervention target that can break this path. So  $X'$  would not be valid.
Next, we'll use induction to show that for any valid subset  $X'$  with  $|X'| \geq 3$ , there exists a hidden node  $H'$  such that  $X' \subseteq \operatorname{ch}_X(H')$ . In other words,  $X'$  cannot be imaginary.
Base Step: If  $|X'| = 3$ , then let  $X' = \{X_1, X_2, X_3\}$ . Suppose  $X'$  is imaginary. Then every pair of nodes in  $X'$  must have a type-1 path and a type-2 path because there exists an intervention target that can destroy the type-2 path but every pair of nodes is valid. Therefore, we have  $H_{1,2}$  as the common parent of  $X_1, X_2, H_{2,3}$  as the common parent of  $X_2, X_3$  and  $H_{1,3}$  as the common parent of  $X_1, X_3$ . Note that these three hidden nodes are distinct, otherwise,  $X'$  would not be imaginary. This is also not possible by the definition of the sparse measurement model.
Inductive Hypothesis: For any valid subset  $X'$  with  $|X'| = k \geq 3$ , there exists a hidden node  $H'$  such that  $X' \subseteq \mathrm{ch}_X(H')$ .
Inductive Step: We'll show this is true for  $|X'| = k + 1$ . Without loss of generality, suppose  $X' = \{X_1, \ldots, X_{k+1}\}$ . By the inductive hypothesis,  $X'' = \{X_1, \ldots, X_k\}$  and  $X''' = \{X_2, \ldots, X_{k+1}\}$  are both non-imaginary. In other words, there exist  $H''$  and  $H'''$  such that  $X'' \subseteq \operatorname{ch}_X(H'')$  and  $X''' \subseteq \operatorname{ch}_X(H'''')$ . Because  $|X'| \geq 4$ , both  $H''$  and  $H'''$  are common parent of  $X_2$  and  $X_3$ . By the definition of the sparse measurement model,  $H'' = H'''$ .
Theorem D.9. Under Assumptions 1-2 and with maximal sparse measurement model, there is no non-replaceable fractured subset.
Proof. By Theorem D.8, there is no imaginary subset. Therefore, if  $X'$  is a maximal valid subset, then  $X' \subseteq \mathrm{ch}_X(H_i)$  for some hidden variable  $H_i$ . By definition, if  $X'$  is non-replaceable, there must exist a hidden variable  $H_i$  such that  $X' = \mathrm{ch}_X(H_i)$ .
In addition, if  $X'$  is fractured, there exists a complete collection  $S = \{S_j\}_{j=1}^k$  such that  $S_j \nsubseteq X'$  for all  $S_j$ . Note for any  $S_j$ , there exists a hidden node  $H_j$  such that  $S_j \subseteq \operatorname{ch}_X(H_j)$  because there is no imaginary subset.
First of all,  $X'$  must have at least two elements in it. If  $|X'| = 1$ , then by the subset condition (Lemma C.1), no other hidden variable can be a parent of the single node in  $X'$  and the aforementioned complete collection does not exist.
Let  $X_{1}, X_{2}$  be two elements in  $X'$ . Because the measurement model is sparse, by Lemma E.4, there exists an intervention target  $I$  (could be the empty set) such that, the type-2 path between  $X_{1}, X_{2}$  is destroyed. Under this intervention target, there must exist a  $C \in D(G^{(I)})$  that is shattered by  $S$  and contains  $\{X_{1}, X_{2}\}$ . Therefore, there exists a subset of  $S$ :  $\{S_{j}'\}_{j=1}^{k'}$  with  $\{X_{1}, X_{2}\} \subseteq C = \cup S_{j}'$ .
Because  $S_{j}^{\prime} \nsubseteq X^{\prime}$ , there exist two maximal subsets  $S_{1}^{\prime}$  and  $S_{2}^{\prime}$  such that  $X_{1} \in S_{1}^{\prime}$  and  $X_{2} \in S_{2}^{\prime}$ . This also means that there exist  $H_{1}$  and  $H_{2}$  with  $X_{1} \in S_{1}^{\prime} \subseteq \mathrm{ch}_{X}(H_{1})$  and  $X_{2} \in S_{1}^{\prime} \subseteq \mathrm{ch}_{X}(H_{2})$ .
and  $H_{2}$  must be two different nodes other than  $H_{i}$  by the definition of the sparse measurement model. In addition, because  $S_1' \nsubseteq X'$ , there exists  $X_{3} \in S_{1}'$  where  $X_{3} \notin X'$ .
Because  $\{X_1, X_2\} \subseteq \mathrm{ch}_X(H_i)$  and  $\{X_1, X_3\} \subseteq \mathrm{ch}_X(H_1)$  and  $X_2, X_3$  are  $d$ -connected under the intervention target  $I$ , by the definition of the sparse measurement model,  $X_2$  and  $X_3$  are  $d$ -connected via the path between  $H_i$  and  $H_1$ . This is not possible because this path creates another type-2 path between  $X_1$  and  $X_2$ .
D.6 Useful lemmas
The first two lemmas show that we only need to care about maximal cliques of undirected dependency graphs.
Lemma D.10. For two measurement model  $G_{1}$  and  $G_{2}$  and two intervention targets  $I_{1}$  and  $I_{2}$ , we have that  $\Omega_{G_1}^{(I_1)} = \Omega_{G_2}^{(I_2)}$  if and only if  $\mathcal{T}_X(G_1^{(I_1)}) = \mathcal{T}_X(G_2^{(I_2)})$ .
Proof. If  $\mathcal{T}_X(G_1^{(I_1)}) = \mathcal{T}_X(G_2^{(I_2)})$ , then  $D(G_{1}^{(I_{1})})$  is the same as  $D(G_{2}^{(I_{2})})$  by our construction of undirected dependency graphs. The obviously  $\Omega_{G_1}^{(I_1)} = \Omega_{G_2}^{(I_2)}$ . On the other hand, if  $\Omega_{G_1}^{(I_1)} = \Omega_{G_2}^{(I_2)}$ , then  $D(G_{1}^{(I_{1})})$  is the same as  $D(G_{2}^{(I_{2})})$  by Lemma D.11. By Proposition 6 of [43],  $\mathcal{T}_X(G_1^{(I_1)}) = \mathcal{T}_X(G_2^{(I_2)})$ .
Lemma D.11. If  $G_{1} = (V, E_{1})$  and  $G_{2} = (V, E_{2})$  share the same maximal cliques, then  $E_{1} = E_{2}$ .
Proof. Let's first show  $E_1 \subseteq E_2$ . Suppose  $(A \to B) \in E_2$  and  $(A \to B) \notin E_1$ . Then there exists a maximal clique that contains  $A, B$  for  $G_2$ . But there is no clique that contains  $A, B$  at the same time for  $G_1$ . Therefore  $E_1 \subseteq E_2$ . Similarly,  $E_2 \subseteq E_1$ .
Lemma D.12. Under Assumption 1(a) and (b), two distributions  $P_{1}$  and  $P_{2}$  over two measurement models  $G_{1}$  and  $G_{2}$  share the same UDG if and only if  $\mathcal{T}_X(G_1) = \mathcal{T}_X(G_2)$ .
Proof. Because of Assumption 1(a) and (b),  $D(P_{1}) = D(G_{1})$  and  $D(P_{2}) = D(G_{2})$ . By Lemma D.10 and Lemma D.11, the statement is true.
Lemma D.13. If two hidden nodes  $H_{i}$  and  $H_{j}$  are  $d$ -connected in  $G^{(I)}$  under intervention target  $I$ , then there exists a maximal clique  $C \in \Omega_G^{(I)}$  such that  $\mathrm{ch}_X(H_i) \cup \mathrm{ch}_X(H_j) \subseteq C$
Proof. Because  $H_{i}$  and  $H_{j}$  are  $d$ -connected,  $\mathrm{ch}_X(H_i)$  and  $\mathrm{ch}_X(H_j)$  form a clique. Therefore, there exists a maximal clique that is a superset of  $\mathrm{ch}_X(H_i) \cup \mathrm{ch}_X(H_j)$ .
Lemma D.14. If a set of two observed elements  $\{X_a, X_b\}$  is valid and  $X_a, X_b$  do not share the same parent, then one of  $X_a$  and  $X_b$  must have at least two parents.