SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} \mathbb {E} _ {\mathbb {P}} \big ((U _ {i} + U _ {S} C ^ {\top}) ^ {m} \big) \\ = \mathbb {E} _ {\mathbb {P}} (U _ {i} ^ {m}) + \sum_ {l = 0} ^ {m - 1} \binom {m} {l} \mathbb {E} _ {\mathbb {P}} \big (U _ {i} ^ {l} (U _ {S} C ^ {\top}) ^ {m - l} \big) \\ = \mathbb {E} _ {\mathbb {P}} (U _ {i} ^ {m}) + \sum_ {l = 0} ^ {m - 1} \binom {m} {l} \mathbb {E} _ {\mathbb {P}} \Big (\mathbb {E} _ {\mathbb {P}} (U _ {i} ^ {l} | U _ {S}) \cdot (U _ {S} C ^ {\top}) ^ {m - l} \Big) \quad (l a w o f t o t a l e x p e c t a t i o n) \\ = \mathbb {E} _ {\mathbb {P}} (U _ {i} ^ {m}) + \sum_ {l = 0} ^ {m - 1} \binom {m} {l} \mathbb {E} _ {\mathbb {P}} \Big (\mathbb {E} _ {\mathbb {P}} (U _ {i} ^ {l} | U _ {\mathrm {p a} _ {\mathcal {G}} (i)}) \cdot (U _ {S} C ^ {\top}) ^ {m - l} \Big) \quad (\text {s i n c e} U _ {i} \perp U _ {S \backslash \mathrm {p a} _ {\mathcal {G}} (i)} \mid U _ {\mathrm {p a} _ {\mathcal {G}} (i)}) \\ \neq \mathbb {E} _ {\mathbb {P} ^ {I}} (U _ {i} ^ {m}) + \sum_ {l = 0} ^ {m - 1} \binom {m} {l} \mathbb {E} _ {\mathbb {P}} \Big (\mathbb {E} _ {\mathbb {P} ^ {I}} (U _ {i} ^ {l} | U _ {\mathrm {p a} _ {\mathcal {G}} (i)}) \cdot (U _ {S} C ^ {\top}) ^ {m - l} \Big) \\ = \mathbb {E} _ {\mathbb {P} ^ {I}} (U _ {i} ^ {m}) + \sum_ {l = 0} ^ {m - 1} \binom {m} {l} \mathbb {E} _ {\mathbb {P} ^ {I}} \Big (\mathbb {E} _ {\mathbb {P} ^ {I}} (U _ {i} ^ {l} | U _ {\mathrm {p a} _ {\mathcal {G}} (i)}) \cdot (U _ {S} C ^ {\top}) ^ {m - l} \Big) = \mathbb {E} _ {\mathbb {P} ^ {I}} \big ((U _ {i} + U _ {S} C ^ {\top}) ^ {m} \big), \\ \end{array}
$$
where the inequality is because of  $\mathbb{E}_{\mathbb{P}}(U_i^m)\neq \mathbb{E}_{\mathbb{P}^I}(U_i^m)$  and  $\mathbb{E}_{\mathbb{P}}(U_i^l |U_{\mathrm{pa}_G(i)}) = \mathbb{E}_{\mathbb{P}^I}(U_i^l |U_{\mathrm{pa}_G(i)})$  for any  $l < m$ . Therefore  $\mathbb{P}(U_i + U_SC^\top)\neq \mathbb{P}^I (U_i + U_SC^\top)$ .
If  $j \neq i$  in Assumption 2, then  $\mathbb{E}_{\mathbb{P}}(U_j) \neq \mathbb{E}_{\mathbb{P}^I}(U_j)$  implies  $\mathbb{E}_{\mathbb{P}}(U_j + U_SC^\top) \neq \mathbb{E}_{\mathbb{P}^I}(U_j + U_SC^\top)$ , which proves that  $\mathbb{P}(U_j + U_SC^\top) \neq \mathbb{P}^I(U_j + U_SC^\top)$ .
Suppose (2) holds true. If  $j = i$  in Assumption 2, then  $\mathbb{E}_{\mathbb{P}}(U_i) \neq \mathbb{E}_{\mathbb{P}^I}(U_i)$  implies  $\mathbb{E}_{\mathbb{P}}(U_i + U_S C^\top) \neq \mathbb{E}_{\mathbb{P}^I}(U_i + U_S C^\top)$ , which proves that  $\mathbb{P}(U_i + U_S C^\top) \neq \mathbb{P}^I(U_i + U_S C^\top)$ .
If  $j \neq i$  in Assumption 2, then for  $C \in \mathbb{R}^{|S|}$ , if the coordinate for  $U_i$  is not  $c_j$ , then  $\mathbb{E}_{\mathbb{P}}(U_i + U_SC^\top) = \mathbb{E}_{\mathbb{P}}(U_i) + \mathbb{E}_{\mathbb{P}}(U_S)C^\top \neq \mathbb{E}_{\mathbb{P}^I}(U_i) + \mathbb{E}_{\mathbb{P}^I}(U_S)C^\top = \mathbb{E}_{\mathbb{P}^I}(U_i + U_SC^\top)$ , since  $\mathbb{E}_{\mathbb{P}}(U_{S \setminus \{i\}}) = \mathbb{E}_{\mathbb{P}^I}(U_{S \setminus \{i\}})$ . If the coordinate for  $U_i$  in  $C$  is  $c_j$ , denote  $U_SC^\top = U_{S \setminus \{i\}}C_{-j}^\top + c_jU_i$ , and then similar to above we obtain
$$
\begin{array}{l} \mathbb {E} _ {\mathbb {P}} \bigl ((U _ {j} + U _ {S} C ^ {\top}) ^ {m} \bigr) \\ = \mathbb {E} _ {\mathbb {P}} \left(\left(U _ {j} + c _ {j} U _ {i} + U _ {S \backslash \{i \}} C _ {- j} ^ {\top}\right) ^ {m}\right) \\ = \mathbb {E} _ {\mathbb {P}} \big ((U _ {i} + c _ {j} U _ {i}) ^ {m} \big) + \sum_ {l = 0} ^ {m - 1} \binom {m} {l} \mathbb {E} _ {\mathbb {P}} \big ((U _ {i} + c _ {j} U _ {i}) ^ {l} (U _ {S \backslash \{i \}} C _ {- j} ^ {\top}) ^ {m - l} \big) \\ = \mathbb{E}_{\mathbb{P}}\bigl((U_{i} + c_{j}U_{i})^{m}\bigr) + \sum_{l = 0}^{m - 1}\binom {m}{l}\mathbb{E}_{\mathbb{P}}\Bigl( \mathbb{E}_{\mathbb{P}}\bigl((U_{i} + c_{j}U_{i})^{l}|U_{S\setminus \{i\}}\bigr)\cdot (U_{S\setminus \{i\}}C_{-j}^{\top})^{m - l}\Bigr) \\ \neq \mathbb{E}_{\mathbb{P}^{I}}\bigl((U_{i} + c_{j}U_{i})^{m}\bigr) + \sum_{l = 0}^{m - 1}\binom {m}{l}\mathbb{E}_{\mathbb{P}}\Bigl( \mathbb{E}_{\mathbb{P}^{I}}\bigl((U_{i} + c_{j}U_{i})^{l}|U_{S\setminus \{i\}}\bigr)\cdot (U_{S\setminus \{i\}}C_{-j}^{\top})^{m - l}\Bigr) \\ = \mathbb {E} _ {\mathbb {P} ^ {I}} \left(\left(U _ {i} + c _ {j} U _ {i}\right) ^ {m}\right) + \sum_ {l = 0} ^ {m - 1} \binom {m} {l} \mathbb {E} _ {\mathbb {P} ^ {I}} \left(\mathbb {E} _ {\mathbb {P} ^ {I}} \left(\left(U _ {i} + c _ {j} U _ {i}\right) ^ {l} \mid U _ {S \backslash \{i \}}\right) \cdot \left(U _ {S \backslash \{i \}} C _ {- j} ^ {\top}\right) ^ {m - l}\right) \\ = \mathbb {E} _ {\mathbb {P} ^ {I}} \left(\left(U _ {j} + U _ {S} C ^ {\top}\right) ^ {m}\right). \\ \end{array}
$$
Thus  $\mathbb{P}(U_j + U_SC^\top) \neq \mathbb{P}^I(U_j + U_SC^\top)$ , which completes the proof.
This lemma gives a sufficient condition for Assumption 2 to hold. Since it involves only finite moments of the variables, one can easily check if this is satisfied for a given SCM associated with soft interventions. Note that Example 5 satisfies the first condition of Lemma 3 for  $m = 2$ .
Next we show that Assumption 3 is satisfied on a tree graph if Assumption 2 holds, under mild regularity conditions such as that the interventional support lies within the observational support. Recall Assumption 3.
Lemma 4. Suppose  $\mathcal{G}$  is a polytree and Assumption 2 holds for an intervention  $I$  targeting node  $i$ . Then for any edge  $i \to j \in \mathcal{G}$ , Assumption 3 holds if
$$
\mathbb {P} \left(U _ {i} = u \mid U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}\right) = 0 \quad \Rightarrow \quad \mathbb {P} ^ {I} \left(U _ {i} = u \mid U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}\right) = 0, \tag {7}
$$
for almost every  $u$  and all realizations of  $U_{\mathrm{pa}_{\mathcal{G}}(j)\backslash \{i\}}$ .
Proof. Suppose  $\mathcal{G}$  is a tree graph and Assumption 2 holds for  $I$  targeting  $i$ . For any edge  $i \to j \in \mathcal{G}$ , since there is only one undirected path between  $i$  and  $j$ , we have  $S = \mathrm{pa}_{\mathcal{G}}(j) \cap \mathrm{de}_{\mathcal{G}}(i) = \emptyset$ . Therefore we only need to show that  $U_i$  and  $U_j + c_j U_i$  are not conditionally independent given  $U_{\mathrm{pa}_{\mathcal{G}}(j) \setminus \{i\}}$  for any  $c_j \in \mathbb{R}$ .
The regularity condition in Eq. (7) ensures that
$$
\int_ {\mathbb {P} \left(U _ {i} = r \mid U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}\right) \neq 0} \mathbb {P} ^ {I} \left(U _ {i} = r \mid U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}\right) d r = 1, \tag {8}
$$
for any realization of  $U_{\mathrm{pa}_{\mathcal{G}}(j)\backslash \{i\}}$
Suppose  $U_i$  and  $U_j + c_j U_i$  are conditionally independent given  $U_{\mathrm{pa}_{\mathcal{G}}(j) \setminus \{i\}}$ . Then for any  $l \in \mathbb{R}$  and realization of  $U_{\mathrm{pa}_{\mathcal{G}}(j) \setminus \{i\}}$ ,  $U_i$  (denote the realization of  $U_i$  as  $r$ ),
$$
\begin{array}{l} \mathbb {P} \left(U _ {j} + c _ {j} U _ {i} = l \mid U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}\right) = \mathbb {P} \left(U _ {j} + c _ {j} U _ {i} = l \mid U _ {i} = r, U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}\right) \\ = \mathbb {P} (U _ {j} = l - c _ {j} r \mid U _ {i} = r, U _ {\mathrm {p a} _ {\varphi} (j) \backslash \{i \}}). \\ \end{array}
$$
Since this is true for any  $r$  with  $\mathbb{P}(U_i = r \mid U_{\mathrm{pa}_G(j) \setminus \{i\}}) \neq 0$ , by Eq. (8), we have
$$
\begin{array}{l} \mathbb {P} \left(U _ {j} + c _ {j} U _ {i} = l \mid U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}\right) \\ = \int_ {\mathbb {P} (U _ {i} = r | U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}) \neq 0} \mathbb {P} (U _ {j} + c _ {j} U _ {i} = l \mid U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}) \cdot \mathbb {P} ^ {I} (U _ {i} = r \mid U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}) d r \\ = \int_ {\mathbb {P} (U _ {i} = r | U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}) \neq 0} \mathbb {P} (U _ {j} = l - c _ {j} r \mid U _ {i} = r, U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}) \cdot \mathbb {P} ^ {I} (U _ {i} = r \mid U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}) d r. \\ \end{array}
$$
Note that  $\mathbb{P}(U_j\mid U_i,U_{\mathrm{pa}_{\mathcal{G}}(j)\backslash \{i\}}) = \mathbb{P}^I (U_j\mid U_i,U_{\mathrm{pa}_{\mathcal{G}}(j)\backslash \{i\}})$  since  $I$  targets  $i$ , and we therefore have
$$
\begin{array}{l} \mathbb {P} \left(U _ {j} + c _ {j} U _ {i} = l \mid U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}\right) \\ = \int_ {\mathbb {P} (U _ {i} = r | U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}) \neq 0} \mathbb {P} ^ {I} (U _ {j} = l - c _ {j} r \mid U _ {i} = r, U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}) \cdot \mathbb {P} ^ {I} (U _ {i} = r \mid U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}) d r \\ = \int_ {\mathbb {P} (U _ {i} = r | U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}) \neq 0} \mathbb {P} ^ {I} (U _ {j} = l - c _ {j} r, U _ {i} = r \mid U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}) d r \\ = \int_ {\mathbb {P} ^ {I} (U _ {i} = r | U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}) \neq 0} \mathbb {P} ^ {I} (U _ {j} = l - c _ {j} r, U _ {i} = r \mid U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}) d r \\ = \mathbb {P} ^ {I} (U _ {j} + c _ {j} U _ {i} = l \mid U _ {\mathrm {p a} _ {\mathcal {G}} (j) \backslash \{i \}}), \\ \end{array}
$$
where the second-to-last equality uses the regularity condition in Eq. (7).
For simplicity, we assume  $U$  is continuous and treat  $\mathbb{P}$  as the density. For discrete  $U$ , the proofs extend by replacing  $\int$  with  $\sum$ .
Since  $\mathrm{pa}_{\mathcal{G}}(j) \cap \mathrm{de}_{\mathcal{G}}(i) = \emptyset$ , it holds that  $\mathbb{P}(U_{\mathrm{pa}_{\mathcal{G}}(j) \backslash \{i\}}) = \mathbb{P}^I(U_{\mathrm{pa}_{\mathcal{G}}(j) \backslash \{i\}})$ , and thus  $\mathbb{P}(U_j + c_j U_i) = \mathbb{P}^I(U_j + c_j U_i)$ , which is a contradiction to linear interventional faithfulness of  $I$ . Therefore, we must have that  $U_i$  and  $U_j + c_j U_i$  are not conditionally independent given  $U_{\mathrm{pa}_{\mathcal{G}}(j) \backslash \{i\}}$ , which completes the proof.
Essentially, Assumption 2 guarantees influentiality and Assumption 3 guarantees adjacency faithfulness. These assumptions differ from existing faithfulness conditions (c.f., [57, 58, 67]) due to the fact that we can only observe a linear mixing of the causal variables.
B.2 Summary of representations
In the remainder of this appendix, we will develop a series of representations which are increasingly related to the underlying representation  $U$ . These representations are summarized in Table 1.
Symbol	Definition	Section
U	Section 2
X	X = UΛ + b, Λ ∈ Rp×p, b ∈ Rp	Section 2
ˆU	ˆU = UˆΓ + ā, ˆΓ = ΛΠ, ā = bΠ for Π ∈ Rp×p	Appendix B.3.1
ˆU	ˆU = UˆΓ + ā, ˆΓ = ˆΓˆR, ā = āˆR forˆR ∈ Rp×pupp. tri.	Appendix B.3.2
ˆU	ˆU = UˆΓ + ā, ˆΓ = ˆΓˆR, ā = āˆR forˆR ∈ Rp×pupp. tri.	Appendix B.4
Table 1: Representations of  $U$  that are used in this appendix. Note that, under Assumption 1, we can assume  $X = U\Lambda + b$  without loss of generality, by Lemma 1 and Remark 1.
In the main text (Section 4.2), we laid out an illustrative procedure to identify the transitive closure of  $\mathcal{G}$  when we consider a simpler setting with  $K = p$ . This process relies on iteratively finding source nodes of  $\mathcal{G}$ . In the generalized setting with  $K \geq p$ , the proof works in the reversed way, where we iteratively identify the sink nodes<sup>10</sup> of  $\mathcal{G}$ .
In Section B.3.1, we introduce the concept of a topological representation: a representation  $\tilde{U}$  of the data for which marginal distributions change in a way consistent with an assignment  $\rho_1,\ldots ,\rho_p$  of intervention targets. In Lemma 5, we show that under Assumptions 1 and 2, a topological representation is guaranteed to exist. In Lemma 6, we show that any topological representation is also topologically consistent in a natural way with the underlying representation  $U$ .
In Section B.3.2, we consider transforming a topological representation  $\tilde{U}$  into a different topological representation  $\hat{U}$ . For any such representation  $\hat{U} = \tilde{U}\hat{R}'$ , we define an associated graph  $\hat{\mathcal{G}}^{\hat{R}'}$ . In Lemma 7, we show that picking  $\hat{R}$  so that  $\hat{\mathcal{G}}^{\hat{R}'}$  has the fewest edges will yield that  $\hat{\mathcal{G}}^{\hat{R}} = \mathcal{T}\mathcal{S}(\mathcal{G}_{\tau})$ .
Together, these results are used to prove Theorem 1: that we can identify  $\mathcal{G}$  up to transitive closure.
B.3.1 Topologically ordered representations
We begin by introducing the concept of a topological representation.
Definition 2. Suppose  $X = U\Lambda + b$ . Let  $\Pi \in \mathbb{R}^{p\times p}$  be a non-singular matrix, let  $\tilde{U} = X\Pi$ , and let  $\rho_1,\ldots ,\rho_p\in [K]$ . We call  $\tilde{U}$  a topological representation of  $X$  with intervention targets  $\rho_{1},\dots ,\rho_{p}$  if the following two conditions are satisfied for all  $j\in [p]$ :
Condition 1)  $\mathbb{P}(\tilde{U}_j)\neq \mathbb{P}^{I_{\rho_j}}(\tilde{U}_j)$
(Condition 2)  $\mathbb{P}(\tilde{U}_{1:j-1}C^{\top}) = \mathbb{P}^{I_{\rho_j}}(\tilde{U}_{1:j-1}C^{\top})$  and for any  $C \in \mathbb{R}^{j-1}$ .
Here,  $\mathbb{P}(\tilde{U}),\mathbb{P}^{I_k}(\tilde{U})$  are the induced distributions for  $\tilde{U}$  when  $X\sim \mathbb{P}_X$  and  $X\sim \mathbb{P}_X^{I_k}$ , respectively.
10 A sink node is a node without children
The next result shows that a topological representation always exists. In particular, we show that a topological representation can be recovered simply by re-ordering the nodes of  $\mathcal{G}$ .
Lemma 5. Suppose that Assumption 2 hold. Then, there exists a topological representation of  $X$ .
Proof. Assume without loss of generality that  $\mathcal{G}$  has topological order  $\tau = (1,2,\dots,p)$ , i.e.,  $i\rightarrow j\in \mathcal{G}$  only if  $i < j$ . Let  $\Pi = \Lambda^{-1}$ , then  $\tilde{U} = U + \tilde{c}$  for constant vector  $\tilde{c} = b\Pi$ . Set  $\rho_1,\ldots ,\rho_p$  to be such that  $T(I_{\rho_j}) = j$  for  $j\in [p]$ . Let  $j\in [p]$  and  $C\in \mathbb{R}^{j - 1}$ .
Condition 1. Since  $I_{\rho_j}$  targets  $U_{j}$ , by Assumption 2, we have  $\mathbb{P}(U_j)\neq \mathbb{P}^{I_{\rho_j}}(U_j)$ , and thus  $\mathbb{P}(\tilde{U}_j)\neq \mathbb{P}^{I_{\rho_j}}(\tilde{U}_j)$ .
Condition 2. Since  $I_{\rho_j}$  targets  $U_{j}$  and  $U_{1:j - 1} \subset U_{[p] \setminus \overline{\operatorname{de}} g(j)}$ , we have  $\mathbb{P}(U_{1:j - 1}C^{\top}) \neq \mathbb{P}^{I_{\rho_j}}(U_{1:j - 1}C^{\top})$ , and thus  $\mathbb{P}(\tilde{U}_{1:j - 1}C^{\top}) \neq \mathbb{P}^{I_{\rho_j}}(\tilde{U}_{1:j - 1}C^{\top})$ .
Now, we show that any topological representation is also consistent with the underlying representation  $U$  up to some linear transformation which respects the topological ordering.
Lemma 6. Suppose that Assumptions 2 hold. Let  $\tilde{U} = X\Pi$  be a topological representation of  $X$  and denote  $\tilde{\Gamma} = \Lambda \Pi$ . Then, there exists a topological ordering  $\tau$  of  $\mathcal{G}$  such that for any  $j \in [p]$ , we have that
$$
i <   j \quad \Longrightarrow \quad \tilde {\Gamma} _ {\tau (j), i} = 0 \quad \text {a n d} \quad \tilde {\Gamma} _ {\tau (j), j} \neq 0. \tag {9}
$$
Proof. We prove by induction. Let  $\tilde{c} = b\Pi$ . Note that  $\tilde{U} = U\tilde{\Gamma} + \tilde{c}$ .
Consider  $I_{\rho_p}$ . Let  $T(I_{\rho_p}) = i$ . We will show that  $i$  must be a sink node in  $\mathcal{G}$ .
Suppose  $i$  is not a sink node, and let  $j \in \mathrm{mch}_{\mathcal{G}}(i)$ . Since  $\tilde{\Gamma}$  is nonsingular,  $\mathrm{rank}(\mathrm{span}(\tilde{\Gamma}_{:,1}, \ldots, \tilde{\Gamma}_{:,p-1})) = p-1$ . Therefore, we must have  $\mathrm{span}(e_i, e_j) \cap \mathrm{span}(\tilde{\Gamma}_{:,1}, \ldots, \tilde{\Gamma}_{:,p-1}) \neq \{0\}$ . Thus,
$$
\tilde {\Gamma} \gamma^ {\top} = a e _ {i} ^ {\top} + b e _ {j} ^ {\top} \quad \text {f o r s o m e} a, b \in \mathbb {R}, \gamma \in \mathbb {R} ^ {p} \text {s u c h t h a t} a ^ {2} + b ^ {2} \neq 0, \gamma_ {p} = 0
$$
By Condition 2, we have that  $\mathbb{P}(\tilde{U}\gamma^{\top}) = \mathbb{P}_{I_{\rho_p}}(\tilde{U}\gamma^{\top})$ . Since  $\tilde{c}\gamma^{\top}$  is a constant, this implies that  $\mathbb{P}(aU_i + bU_j) = \mathbb{P}^{I_{\rho_p}}(aU_i + bU_j)$ . However, this contradicts Assumption 2. Thus,  $i$  must be a sink node, which we denote by  $\tau(p)$ .
We also have that  $\tilde{\Gamma}_{\tau(p),i} = 0$  for any  $i < p$ . Otherwise suppose  $\tilde{\Gamma}_{\tau(p),i} \neq 0$ , then  $\tilde{U}_i = U\tilde{\Gamma}_{:,i} + h_i$  can be written as  $\tilde{\Gamma}_{\tau(p),i} \cdot U_{\tau(p)} + U_SC^\top + \tilde{c}_i$  with  $S = [p] \setminus (\{\tau(p)\}) = [p] \setminus \overline{\mathrm{de}}_G(\tau(p))$ . By Assumption 2, we have  $\mathbb{P}(\tilde{U}_i) \neq \mathbb{P}^{I_{\rho_p}}(\tilde{U}_i)$ , a contradiction to Condition 2.
Induction step.
Suppose that we have proven the statement for  $q \leq p$ . Denote the intervention targets of  $I_{\rho_q}, \ldots, I_{\rho_p}$  as  $\tau(q), \ldots, \tau(p)$ , respectively. Let  $K = [p] \setminus \{\tau(q), \ldots, \tau(p)\}$ .
Consider  $I_{\rho_{q-1}}$  with  $T(I_{\rho_{q-1}}) = i$ . Let  $\mathcal{G}_q$  denote the graph  $\mathcal{G}$  after removing the nodes  $\tau(q), \ldots, \tau(p)$ . We will show that  $i$  must be a sink node in  $\mathcal{G}_q$ .
Suppose that  $i$  is not a sink node  $\mathcal{G}_q$  and let  $j$  be a maximal child of  $i$  in  $\mathcal{G}_q$ . Since  $\tilde{\Gamma}_{[p]\backslash K,[q]} = 0$ ,  $|K| = q$ , and  $\tilde{\Gamma}$  is nonsingular, we have that  $\tilde{\Gamma}_{K,[q]}$  is nonsingular. Thus, as above,
$$
\tilde {\Gamma} \gamma^ {\top} = a e _ {i (q)} ^ {\top} + b e _ {j (q)} ^ {\top} \quad \text {f o r s o m e} a, b \in \mathbb {R}, \gamma \in \mathbb {R} ^ {p} \text {s u c h t h a t} a ^ {2} + b ^ {2} \neq 0, \gamma_ {q} = \gamma_ {q + 1} = \dots = \gamma_ {p} = 0
$$
where  $e_{i(q)}, e_{j(q)}$  are indicator vectors in  $\mathbb{R}^q$  with ones at positions of  $i, j$  in  $1, \ldots, p$  after removing  $\tau(q + 1), \ldots, \tau(p)$ , respectively. Thus, by Condition 2, we have that  $\mathbb{P}(\tilde{U}\gamma^\top) = \mathbb{P}^{I_{\rho_q}}(\tilde{U}\gamma^\top)$ , which contradicts Assumption 2. Therefore  $I_{\rho_q}$  targets a sink node of  $\mathcal{G}_q$ .
To show that  $\tilde{\Gamma}_{\tau(q),i} = 0$  for any  $i < q$ , use  $\tilde{\Gamma}_{\tau(k),i} = 0$  for all  $k \geq q$  and write  $\tilde{U}_i = U\tilde{\Gamma}_{:,i} + \tilde{c}_i$  as  $\tilde{\Gamma}_{\tau(q-1),i} \cdot U_{\tau(q-1)} + U_S C^\top + \tilde{c}_i$  with  $S = [p] \setminus \{\tau(q-1), \tau(q), ..., \tau(p)\} \subset [p] \setminus \overline{\mathrm{de}}_{\mathcal{G}}(\tau(q))$ . By Assumption 2, we have  $\mathbb{P}(\tilde{U}_i) \neq \mathbb{P}^{I_{\rho_q}}(\tilde{U}_i)$  if  $\tilde{\Gamma}_{\tau(q),i} \neq 0$ , a contradiction to Condition 2.
By induction, we have thus proven that the solution to Condition 1 and Condition 2 satisfies  $i < j \Rightarrow \tilde{\Gamma}_{\tau(j),i} = 0$ . Therefore  $\tilde{\Gamma}_{\tau,:}$  is upper triangular. Since it is also non-singular, it must hold that  $\tilde{\Gamma}_{\tau(j),j} \neq 0$ . Thus Eq. (9) holds for some unknown  $\tau$ . Furthermore, the proof shows that  $I_{\rho_1}, \dots, I_{\rho_p}$  target  $U_{\tau(1)}, \dots, U_{\tau(p)}$  respectively.
B.3.2 Sparsest topological representation
In the section, we will introduce a graph associated to any topological representation. We consider picking a topological representation such that the associated graph is as sparse is possible, and we show that this choice recovers the underlying graph  $\mathcal{G}$  up to transitive closure.
We begin by establishing the following property of a topological representation  $\tilde{U}$ , which relates ancestral relationships in the underlying graph  $\mathcal{G}$  to changes in marginals of  $\tilde{U}$ .
Proposition 1. Suppose that Assumptions 2 hold. Let  $\tilde{U}$  be a topological representation with intervention targets  $\rho_1, \ldots, \rho_p$ .
Then, for any  $i < j$  such that  $\tau(j) \in \mathrm{deg}_{\mathcal{G}}(\tau(i))$ , we must have
$$
\mathbb {P} (\tilde {U} _ {j}) \neq \mathbb {P} ^ {I _ {\rho_ {k}}} (\tilde {U} _ {j}) \quad f o r s o m e i \leq k <   j \quad s u c h t h a t \tau (k) \in \overline {{\mathrm {d e}}} _ {\mathcal {G}} (\tau (i)).
$$
Proof. By Lemma 6, Eq. (9),  $\tilde{U}_j$  is a linear combination of  $U_{\tau(1)}, \ldots, U_{\tau(j)}$  with nonzero coefficient of  $U_{\tau(j)}$ . Let  $k_0$  be
(Case 1) the largest such that  $i \leq k_0 < j$  where  $\tau(k_0) \in \overline{\operatorname{de}}_{\mathcal{G}}(\tau(i))$  and the coefficient of  $U_{\tau(k_0)}$  in  $\tilde{U}_j$  is nonzero,
Case 2)  $i$ , if no  $k_{0}$  satisfies Case 1.
Then let  $k = k_{0}$  if  $\tau(j) \notin \mathrm{de}_{\mathcal{G}}(\tau(k_{0}))$ ; otherwise let  $k$  be such that  $\tau(k) \in \overline{\mathrm{de}}_{\mathcal{G}}(\tau(k_{0}))$  and  $\tau(j) \in \mathrm{mch}_{\mathcal{G}}(\tau(k))$  (such  $k$  exists by considering the parent of  $\tau(j)$  on the longest directed path from  $\tau(k_{0})$  to  $\tau(j)$  in  $\mathcal{G}$ ). Figure 8 illustrates the different scenarios for  $k_{0}, k$ . Note that we always have  $\tau(k) \in \overline{\mathrm{de}}_{\mathcal{G}}(\tau(i))$ .
(A) Case 1-1
(B) Case 1-2
(C) Case 2
Figure 8: Illustration of  $k_{0}, k$ .
(Case 1): We first show that  $\tilde{U}_j$  can be written as a linear combination of  $U_{\tau(j)}, U_{\tau(k_0)}$  and  $U_S$  for  $S \subset [p] \setminus \overline{\operatorname{de}}_{\mathcal{G}}(\tau(k_0))$  with nonzero coefficient for  $U_{\tau(j)}, U_{\tau(k_0)}$ . Consider an arbitrary  $l \in [p]$ . If the coefficient for  $U_{\tau(l)}$  in  $\hat{U}_j$  is nonzero, by Eq. (9), we have  $l \leq j$ . Also since  $k_0$  is the largest, we have  $l = k_0$  or  $l = j$  or  $l < k_0$  or  $\tau(l) \notin \overline{\operatorname{de}}_{\mathcal{G}}(\tau(i))$ . If  $l < k_0$ , then by the topological order, it holds that  $\tau(l) \notin \operatorname{de}_{\mathcal{G}}(\tau(k_0))$ . If  $\tau(l) \notin \overline{\operatorname{de}}_{\mathcal{G}}(\tau(i))$ , since  $\tau(k_0) \in \overline{\operatorname{de}}_{\mathcal{G}}(\tau(i))$ , it also holds that  $\tau(l) \notin \overline{\operatorname{de}}_{\mathcal{G}}(\tau(k_0))$ . Therefore  $\tilde{U}_j$  can be written as a linear combination of  $U_{\tau(j)}, U_{\tau(k_0)}$  and  $U_S$  with nonzero coefficient for  $U_{\tau(j)}, U_{\tau(k_0)}$ . Next, we show that  $\mathbb{P}(\tilde{U}_j) \neq \mathbb{P}^{I_{\rho_k}}(\tilde{U}_j)$  by considering two subcases of Case 1.
If  $\tau(j) \notin \mathrm{de}_{\mathcal{G}}(\tau(k_0))$ , then  $k = k_0$  (illustrated in Figure 8A). Then  $S \cup \{\tau(j)\} \subset [p] \setminus \overline{\mathrm{de}}_{\mathcal{G}}(\tau(k))$ . Therefore  $\tilde{U}_j$  can be written as a linear combination of  $U_{\tau(k)}$  and  $U_{S'}$  for  $S' \subset [p] \setminus \overline{\mathrm{de}}_{\mathcal{G}}(\tau(k))$  with nonzero coefficient for  $U_{\tau(k)}$ . By Assumption 2, we have  $\mathbb{P}(\tilde{U}_j) \neq \mathbb{P}^{I_{\rho_k}}(\tilde{U}_j)$ .