SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Next, suppose there is an open path from  $U_{PA}$  to  $Y$  when conditioning on  $X$ ,  $Z_{AS}$ , labeled  $U_{PA,i} \rightarrow Z_s \xrightarrow{\rightarrow} Z_{s'} \rightarrow \dots \rightarrow Y$ . By definition of the anchor set,  $Z_{AS}$  must contain the first variable on this path,  $Z_s$ , and  $Z_s$  is different from  $X, Y$ . Consider first the case with the arrow from  $Z_s$  outgoing,  $U_{PA,i} \rightarrow Z_s \rightarrow \dots \rightarrow Y$ . When conditioning on  $Z_{AS}$ , this path is closed since  $Z_s \in Z_{AS}$ , yielding a contradiction. Consider then the second case with the arrow incoming into  $Z_s$ ,  $U_{PA,i} \rightarrow Z_s \leftarrow Z_{s'} \rightarrow \dots \rightarrow Y$ . Since  $Z_{s'}$  points to  $Z_s$ ,  $Z_{s'}$  differs from  $X, Y$ . Furthermore, by anchor set exogenous ancestral closure, the exogenous variable of  $Z_{s'}$ , labeled  $U_{s'}$ , must also be in  $U_{PA}$ . Hence,  $Z_{AS}$  contains  $Z_{s'}$ , and  $Z_{s'}$  cannot be a collider on this path, so conditioning on  $Z_{AS}$  blocks the path, again yielding a contradiction. We conclude that no open path between  $U_{PA}$  and  $Y$  exists when conditioning on  $Z_{AS}, X$ . Therefore, it holds that
$$
P (y \mid x, u _ {P A}, z _ {A S}) = P (y \mid x, z _ {A S}). \tag {56}
$$
Finally, by plugging in Eqs. 55-56 into Eq. 54 we obtain that
$$
P (y \mid x ^ {U _ {P A}}) = \sum_ {u _ {P A}, z _ {A S}} P (u _ {P A}) P (y \mid x, z _ {A S}) \mathbb {1} \left(Z _ {A S} \left(u _ {P A}\right) = z _ {A S}\right) \tag {57}
$$
Figure 7: Markovian causal diagram used in Ex. 4 with explicitly drawn latent variables  $U_{1}, U_{2}$ .
$$
= \sum_ {z _ {A S}} P (y \mid x, z _ {A S}) \underbrace {\sum_ {u _ {P A}} P \left(u _ {P A}\right) \mathbb {1} \left(Z _ {A S} \left(u _ {P A}\right) = z _ {A S}\right)} _ {P \left(z _ {A S}\right) \text {b y d e f i n i t i o n}} \tag {58}
$$
$$
= \sum_ {z _ {A S}} P (y \mid x, z _ {A S}) P (z _ {A S}), \tag {59}
$$
therefore witnessing identifiability of  $P(y \mid x^{U_{PA}})$  and completing the proof.
B Examples
B.1 Markovian Decomposition Example
Example 4 (Latent variable attribution in a Markovian model). Consider the following SCM  $\mathcal{M}^*$ :
$$
\mathcal {M} ^ {*}: \left\{ \begin{array}{l l} Z _ {1} \leftarrow B (0. 5) & (6 0) \\ Z _ {2} \leftarrow B \left(0. 4 + 0. 2 Z _ {1}\right) & (6 1) \\ X \leftarrow B \left(0. 3 + 0. 2 Z _ {1} + 0. 2 Z _ {2}\right) & (6 2) \\ Y \leftarrow X + Z _ {1} + Z _ {2}, & (6 3) \end{array} \right.
$$
and the causal diagram in Fig. 7. We wish to decompose the quantity  $\text{Exp-SE}_x(y)$  into the variations attributed to the latent variables  $U_1, U_2$ . Following the decomposition from Thm. 1 we can write
$$
\begin{array}{l} E x p - S E _ {x} (y \mid x _ {1}) = \underbrace {\mathbb {E} (y \mid x _ {1}) - \mathbb {E} (y \mid x _ {1} ^ {U _ {1}})} _ {U _ {1} c o n t r i b u t i o n} \tag {64} \\ + \underbrace {\mathbb {E} (y \mid x _ {1} ^ {U _ {1}}) - \mathbb {E} (y \mid x _ {1} ^ {U _ {1} , U _ {2}})} _ {U _ {2} c o n t r i b u t i o n}. \\ \end{array}
$$
We now need to compute the terms appearing in Eq. 64. In particular, we know that
$$
\begin{array}{l} \mathbb {E} \left(y \mid x _ {1} ^ {U _ {1}, U _ {2}}\right) = \mathbb {E} \left(y \mid d o \left(x _ {1}\right)\right) (65) \\ = 1 + \mathbb {E} \left(Z _ {1} \mid d o \left(x _ {1}\right)\right) + \mathbb {E} \left(Z _ {2} \mid d o \left(x _ {1}\right)\right) (66) \\ = 1 + \mathbb {E} \left(Z _ {1}\right) + \mathbb {E} \left(Z _ {2}\right) = 1 + 0. 5 + 0. 5 = 2. (67) \\ \end{array}
$$
Similarly, we can also compute
$$
\mathbb {E} (y \mid x _ {1}) = 1 + P \left(Z _ {1} = 1 \mid x _ {1}\right) + P \left(Z _ {2} = 1 \mid x _ {1}\right), \tag {68}
$$
where  $P(Z_{1} = 1 \mid x_{1})$  can be expanded as
$$
\begin{array}{l} P \left(Z _ {1} = 1 \mid x _ {1}\right) = \frac {P \left(Z _ {1} = 1 , X = 1\right)}{P (X = 1)} (69) \\ = \frac {P \left(Z _ {1} = 1 , X = 1 , Z _ {2} = 1\right) + P \left(Z _ {1} = 1 , X = 1 , Z _ {2} = 0\right)}{P (X = 1)} (70) \\ = \frac {0 . 5 * 0 . 6 * 0 . 7 + 0 . 5 * 0 . 4 * 0 . 5}{0 . 5} = 0. 6 2. (71) \\ \end{array}
$$
The value of  $P(Z_{2} = 1 \mid x_{1})$  is computed analogously and also equals 0.62, implying that  $\mathbb{E}(y \mid x_{1}) = 1 + 0.62 + 0.62 = 2.24$ . Finally, we want to compute  $\mathbb{E}(y \mid x_{1}^{U_{1}})$ , which equals
$$
\mathbb {E} \left(y \mid x _ {1} ^ {U _ {1}}\right) = 1 + P \left(Z _ {1} = 1 \mid x _ {1} ^ {U _ {1}}\right) + P \left(Z _ {2} = 1 \mid x _ {1} ^ {U _ {1}}\right). \tag {72}
$$
(a)  $\operatorname{Exp} - \operatorname{SE}_x^{\emptyset, U_1}(y)$ .
(b)  $\operatorname{Exp - SE}_x^{U_1,\{U_1,U_2\}}(y)$ .
Figure 8: Graphical representation of Exp-SE effect decomposition in Ex. 4.
By definition,  $P(Z_{1} = 1 \mid x_{1}^{U_{1}}) = P(Z_{1} = 1) = 0.5$ . For  $P(Z_{2} = 1 \mid x_{1}^{U_{1}})$  we write
$$
\begin{array}{l} P \left(Z _ {2} = 1 \mid x _ {1} ^ {U _ {1}}\right) = \sum_ {z _ {1}} P \left(Z _ {2} = 1 \mid x _ {1}, z _ {1}\right) P \left(z _ {1}\right) (73) \\ = \frac {1}{2} \left[ \frac {P \left(Z _ {2} = 1 , X = 1 , Z _ {1} = 1\right)}{P \left(X = 1 , Z _ {1} = 1\right)} + \frac {P \left(Z _ {2} = 1 , X = 1 , Z _ {1} = 0\right)}{P \left(X = 1 , Z _ {1} = 0\right)} \right] (74) \\ = \frac {1}{2} \left[ \frac {0 . 2 1}{0 . 3 1} + \frac {0 . 2 1}{0 . 3 1} \right] \approx 0. 6 8, (75) \\ \end{array}
$$
implying that  $\mathbb{E}(y\mid x_1^{U_1}) = 2.18$  . Putting everything together, we found that
$$
\underbrace {E x p - S E _ {x} (y \mid x _ {1})} _ {= 0. 2 4} = \underbrace {E x p - S E _ {x} ^ {\emptyset , U _ {1}} (y \mid x _ {1})} _ {= 0. 0 6 f r o m U _ {1}} + \underbrace {E x p - S E _ {x} ^ {U _ {1} , \{U _ {1} , U _ {2} \}} (y \mid x _ {1})} _ {= 0. 1 8 f r o m U _ {2}}. \tag {76}
$$
The terms appearing on the r.h.s. of Eq. 76 are shown as graphical contrasts in Fig. 8. On the left side of Fig. 8a,  $U_{1}, U_{2}$  are responding to the conditioning  $X = x$ , compared against the right side where only  $U_{2}$  is responding to the conditioning  $X = x$ . In the second term, in Fig. 8b, on the left only  $U_{2}$  responds to  $X = x$ , compared against the right side in which neither  $U_{1}$  nor  $U_{2}$  respond to  $X = x$  conditioning.
B.2 Non-topological Counterexample
Example 5 (Non-identification of latent spurious decomposition). Consider two  $SCMs\mathcal{M}_1,\mathcal{M}_2$  Both SCMs have the same set of assignment equations  $\mathcal{F}$  given by
$$
\mathcal {F} := \left\{ \begin{array}{l} Z _ {1} \leftarrow U _ {1} \\ Z _ {2} \leftarrow \left\{ \begin{array}{l l} Z _ {1} & i f U _ {2} = 1 \\ 1 - Z _ {1} & i f U _ {2} = 2 \\ 1 & i f U _ {2} = 3 \\ 0 & i f U _ {2} = 4 \end{array} \right. \\ X \leftarrow \left(Z _ {1} \wedge U _ {X 1}\right) \vee \left(Z _ {2} \wedge U _ {X 2}\right) \vee U _ {X} \\ Y \leftarrow X + Z _ {1} + Z _ {2}, \end{array} \right. \tag {78}
$$
and the causal diagram given in Fig. 7. The two SCMs differ in the distribution over the latent variables. In particular, for  $\mathcal{M}_1$  we have
$$
P ^ {\mathcal {M} _ {1}} (U): \left\{ \begin{array}{c} U _ {1}, U _ {X 1}, U _ {X 2}, U _ {X} \sim \text {B e r n o u l l i} (0. 5) \\ U _ {2} \sim \operatorname {M u l t i n o m} \left(4, 1, \left(0, \frac {1}{4}, \frac {1}{2}, \frac {1}{4}\right)\right), \end{array} \right. \tag {81}
$$
and for  $\mathcal{M}_2$
$$
P ^ {\mathcal {M} _ {2}} (U): \left\{ \begin{array}{c} U _ {1}, U _ {X 1}, U _ {X 2}, U _ {X} \sim \text {B e r n o u l l i} (0. 5) \\ U _ {2} \sim \operatorname {M u l t i n o m} \left(4, 1, \left(\frac {1}{4}, \frac {1}{2}, \frac {1}{4}, 0\right)\right). \end{array} \right. \tag {83}
$$
That is, the only difference between  $P^{\mathcal{M}_1}(U)$  and  $P^{\mathcal{M}_2}(U)$  is in how  $U_2$  attains its value. In fact, one can check that the observational distributions  $P^{\mathcal{M}_1}(V)$  and  $P^{\mathcal{M}_2}(V)$  are the same. However,
Figure 9: Causal diagram appearing in Exs. 6-7.
when computing  $\mathbb{E}^{\mathcal{M}}(y\mid x_0^{U_2})$  we have that
$$
\mathbb {E} ^ {\mathcal {M} _ {1}} \left(y \mid x _ {0} ^ {U _ {2}}\right) = 1 \tag {85}
$$
$$
\mathbb {E} ^ {\mathcal {M} _ {2}} \left(y \mid x _ {0} ^ {U _ {2}}\right) = 0. 9 3, \tag {86}
$$
showing that the quantity  $\mathbb{E}^{\mathcal{M}}(y\mid x_0^{U_2})$  is non-identifiable.
The example illustrates that even in the Markovian case, when the variables are not considered in a topological order (in the example above, the variable  $U_{2}$  was considered without the variable  $U_{1}$  being added first), we might not be able to identify the decomposition of the spurious effects.
B.3 Semi-Markovian Decomposition Example
Example 6 (Semi-Markovian spurious decomposition). Consider the following SCM  $\mathcal{M}$ :
$$
\mathcal {F}, P (U): \left\{ \begin{array}{l} Z _ {1} \leftarrow U _ {1} \wedge U _ {1 X} \\ Z _ {2} \leftarrow U _ {2} \vee U _ {2 X} \\ X \leftarrow U _ {X} \wedge \left(U _ {1 X} \vee U _ {2 X}\right) \\ Y \leftarrow X + Z _ {1} + Z _ {2} \\ U _ {1}, U _ {2}, U _ {1 X}, U _ {2 X}, U _ {X} \stackrel {{i. i. d.}} {{\sim}} \text {B e r n o u l l i} (0. 5). \end{array} \right. \tag {87}
$$
The causal diagram  $\mathcal{G}$  associated with  $\mathcal{M}$  is given in Fig. 9. The exogenous variables that lie on top of a spurious trek are  $U_{1X}, U_{2X}$ . Therefore, following the decomposition from Thm. 3, we can attribute spurious variations to these two variables:
$$
\begin{array}{l} E x p - S E _ {x} (y \mid x _ {1}) = \underbrace {\mathbb {E} (y \mid x _ {1}) - \mathbb {E} (y \mid x _ {1} ^ {U _ {1 X}})} _ {U _ {1 X} c o n t r i b u t i o n} \tag {92} \\ + \underbrace {\mathbb {E} (y \mid x _ {1} ^ {U _ {1 X}}) - \mathbb {E} (y \mid x _ {1} ^ {U _ {1 X} , U _ {2 X}})} _ {U _ {2 X} c o n t r i b u t i o n}. \\ \end{array}
$$
We now compute the terms appearing in Eq. 92. In particular, we know that
$$
\begin{array}{l} \mathbb {E} \left(y \mid x _ {1} ^ {U _ {1 X}, U _ {2 X}}\right) = \mathbb {E} \left(y \mid d o \left(x _ {1}\right)\right) = 1 + \mathbb {E} \left(Z _ {1} \mid d o \left(x _ {1}\right)\right) + \mathbb {E} \left(Z _ {1} \mid d o \left(x _ {1}\right)\right) (93) \\ = 1 + \mathbb {E} \left(Z _ {1}\right) + \mathbb {E} \left(Z _ {2}\right) = 1 + 0. 2 5 + 0. 7 5 = 2. (94) \\ \end{array}
$$
$$
\mathbb {E} (y \mid x _ {1}) = 1 + P \left(Z _ {1} = 1 \mid x _ {1}\right) + P \left(Z _ {2} = 1 \mid x _ {1}\right), \tag {95}
$$
Now,  $P(Z_{1} = 1 \mid x_{1}) = \frac{P(Z_{1} = 1, x_{1})}{P(x_{1})}$ , and we know that  $X = 1$  if and only if  $U_{X} = 1$  and  $U_{1X} \vee U_{2X} = 1$ , which happen independently with probabilities  $\frac{1}{2}$  and  $\frac{3}{4}$ , respectively. Next,  $Z_{1} = 1, X = 1$  happens if and only if  $U_{X} = 1, U_{1X} = 1$  and  $U_{1} = 1$ , which happens with probability  $\frac{1}{8}$ . Therefore, we can compute
$$
P \left(Z _ {1} = 1 \mid x _ {1}\right) = \frac {\frac {1}{8}}{\frac {1}{2} * \frac {3}{4}} = \frac {1}{3}. \tag {96}
$$
Furthermore, we similarly compute that  $Z_{2} = 1$ ,  $X = 1$  happens if either  $U_{X} = 1$ ,  $U_{2X} = 1$  or  $U_{X} = 1$ ,  $U_{2X} = 0$ ,  $U_{2} = 1$ ,  $U_{1X} = 1$  which happens disjointly with probabilities  $\frac{1}{4}$ ,  $\frac{1}{16}$ , respectively. Therefore,
$$
P \left(Z _ {2} = 1 \mid x _ {1}\right) = \frac {\frac {1}{4} + \frac {1}{1 6}}{\frac {1}{2} * \frac {3}{4}} = \frac {5}{6}. \tag {97}
$$
Putting everything together we obtain that
$$
\mathbb {E} (y \mid x _ {1}) = 1 + \frac {1}{3} + \frac {5}{6} = \frac {1 3}{6}. \tag {98}
$$
Finally, we want to compute  $\mathbb{E}(y\mid x_1^{U_1X})$ , which equals
$$
\mathbb {E} \left(y \mid x _ {1} ^ {U _ {1} X}\right) = 1 + P \left(Z _ {1} = 1 \mid x _ {1} ^ {U _ {1} X}\right) + P \left(Z _ {2} = 1 \mid x _ {1} ^ {U _ {1} X}\right). \tag {99}
$$
Now, to evaluate these expressions, we distinguish two cases, namely (i)  $U_{1X} = 1$  and (ii)  $U_{1X} = 0$ . In the first case,  $P(Z_1 \mid x_1) = \frac{1}{2}$  and  $P(Z_2 = 1 \mid x_1) = \frac{3}{4}$ . In the second case,  $P(Z_1 \mid x_1) = 0$  and  $P(Z_2 = 1 \mid x_1) = 1$ . Therefore, we can compute
$$
P \left(Z _ {1} = 1 \mid x _ {1} ^ {U _ {1 X}}\right) = \frac {1}{2} P _ {U _ {1 X} = 1} \left(Z _ {1} \mid x _ {1}\right) + \frac {1}{2} P _ {U _ {1 X} = 0} \left(Z _ {1} \mid x _ {1}\right) = \frac {1}{4} \tag {100}
$$
$$
P \left(Z _ {2} = 1 \mid x _ {1} ^ {U _ {1 X}}\right) = \frac {1}{2} P _ {U _ {1 X} = 1} \left(Z _ {2} \mid x _ {1}\right) + \frac {1}{2} P _ {U _ {1 X} = 0} \left(Z _ {2} \mid x _ {1}\right) = \frac {7}{8}, \tag {101}
$$
which implies that  $\mathbb{E}(y\mid x_1^{U_{1X}}) = \frac{17}{8}$ . Finally, this implies that
$$
\underbrace {E x p - S E _ {x} \left(y \mid x _ {1}\right)} _ {= \frac {1}{6}} = \underbrace {E x p - S E _ {x} ^ {\emptyset , U _ {1 X}} \left(y \mid x _ {1}\right)} _ {= \frac {1}{2 4} f r o m U _ {1 X}} + \underbrace {E x p - S E _ {x} ^ {U _ {1 X} , \left\{U _ {1 X} , U _ {2 X} \right\}} \left(y \mid x _ {1}\right)} _ {= \frac {1}{8} f r o m U _ {2 X}}. \tag {102}
$$
The terms appearing on the r.h.s. of Eq. 102 are shown as graphical contrasts in Fig. 4. On the left side of Fig. 4a,  $U_{1X}, U_{2X}$  are responding to the conditioning  $X = x$ , compared against the right side where only  $U_{2X}$  is responding to the conditioning  $X = x$ . In the second term, in Fig. 4b, on the left only  $U_{2X}$  responds to  $X = x$ , compared against the right side in which neither  $U_{1X}$  nor  $U_{2X}$  respond to  $X = x$  conditioning.
B.4 Semi Markovian Identification Examples
Example 7 (Spurious Treks). Consider the causal diagram in Fig. 7. In the diagram, latent variables  $U_{1}, U_{2}$  both lie on top of a spurious trek because:
$$
X \leftarrow Z _ {1} \leftarrow U _ {1} \rightarrow Z _ {1} \rightarrow Y \text {i s a s p u r i o u s t r e k w i t h t o p} U _ {1}
$$
$$
X \leftarrow Z _ {2} \leftarrow U _ {2} \rightarrow Z _ {2} \rightarrow Y \text {i s a s p u r i o u s t r e k w i t h t o p} U _ {2}.
$$
There are also other spurious treks with  $U_{1}$  on top, such as  $X \gets Z_{1} \gets U_{1} \to Z_{1} \to Z_{2} \to Y$ .
Example 7 (continued -  $U_{sToT}$  construction). We continue with Ex. 6 and the causal graph in Fig. 9 and perform the steps as follows:
(i) initialize  $U_{sToT} = \emptyset$
(ii) note that  $\{X,Z_1\}$  create a maximal clique, since:
(a) they are connected with a bidirected edge and thus form a clique,
(b)  $\{X,Z_1,Z_2\}$  do not form a clique, due to the bidirected edge  $Z_{1}\leftrightarrow Z_{2}$  not being present,
(c)  $\{X,Z_1,Y\}$ ,  $\{X,Z_1,Z_2,Y\}$  do not form a clique, since  $Y$  is not incident to any bidirected edges,
(d) thus, the clique  $\{X,Z_1\}$  is also maximal.
Let the variable  $U_{1X}$  be associated with this clique, pointing to  $X$ ,  $Z_1$ , and note that  $U_{1X}$  lies on top of a spurious trek between  $X, Y$ ,
(iii) similarly,  $\{X,Z_2\}$  also create a maximal clique, associated with the variable  $U_{X2}$ , pointing to  $X,Z_2$ , that lies on top of a spurious trek between  $X,Y$ ,
(iv) the node  $Y$  also forms a maximal clique and is associated with the variable  $U_{Y}$  that does not lie on top of a spurious trek (it does not have a path to  $X$ ).
Therefore, we have constructed the set  $U_{sToT} = \{U_{1X}, U_{2X}\}$ .
(a) Causal diagram in Ex. 8.
(b) Causal diagram in Ex. 9.
Figure 10: Causal diagrams in Exs. 8-9.
Example 7 (continued - anchor set). For the set  $U_{sToT} = \{U_{1X}, U_{2X}\}$  associated with the causal diagram in Fig. 9, the anchor sets can be computed as follows:
$$
A S \left(U _ {1 X}\right) = \{X, Z _ {1} \}, \tag {103}
$$
$$
A S \left(U _ {2 X}\right) = \{X, Z _ {2} \}, \tag {104}
$$
$$
A S \left(U _ {1 X}, U _ {2 X}\right) = \{X, Z _ {1}, Z _ {2} \}. \tag {105}
$$
Example 7 (continued - anchor set exogenous ancestral closure). Consider the causal diagram in Fig. 9. With respect to the diagram, we have that
$$
\left. \operatorname {a n} ^ {e x} \left(A S \left(U _ {1 X}\right)\right) = \operatorname {a n} ^ {e x} (X, Z _ {1}) = \left\{U _ {1 X}, U _ {2 X} \right\}, \right. \tag {106}
$$
$$
\left. \operatorname {a n} ^ {e x} \left(A S \left(U _ {2 X}\right)\right) = \operatorname {a n} ^ {e x} (X, Z _ {2}) = \left\{U _ {1 X}, U _ {2 X} \right\}, \right. \tag {107}
$$
$$
\operatorname {a n} ^ {e x} \left(A S \left(\left\{U _ {1 X}, U _ {2 X} \right\}\right)\right) = \operatorname {a n} ^ {e x} (X, Z _ {1}, Z _ {2}) = \left\{U _ {1 X}, U _ {2 X} \right\}. \tag {108}
$$
Therefore,  $\{U_{1X}, U_{2X}\}$  satisfies anchor set exogenous ancestral closure, whereas  $U_{1X}$  and  $U_{2X}$  do not, since for instance  $U_{1X}$  has  $X$  in its anchor set, but  $X$  has  $U_{2X}$  as its ancestor.
We now consider an example of effect identification based on Thm. 4:
Example 8 (Thm. 4 Application). Consider the causal diagram in Fig. 10a. Consider the query  $\mathbb{E}(y \mid x_1^{U_{12}})$  associated with a partially abducted submodel in which the noise variable  $U_{12}$  determining the values of  $Z_1, Z_2$  is not updated according to evidence. Based on Thm. 4, we verify that
(i)  $X,Y$  are not in the anchor set  $AS(U_{12}) = \{Z_1,Z_2\}$
(ii)  $\mathrm{an}^{ex}(AS(U_{12})) = \mathrm{an}^{ex}(Z_1,Z_2) = U_{12}$  meaning that  $U_{12}$  satisfies anchor set exogenous ancestral closure (ASEAC).
Therefore, the query  $\mathbb{E}(y\mid x_1^{U_{12}})$  is identifiable from observational data. To witness, we expand the query as:
$$
\begin{array}{l} \mathbb {E} \left(y \mid x _ {1} ^ {U _ {1 2}}\right) = \sum_ {u _ {1 2}} P \left(u _ {1 2}\right) \mathbb {E} \left(y \mid x _ {1}, u _ {1 2}\right) (109) \\ = \sum_ {u _ {1 2}, z _ {1}, z _ {2}} P (u _ {1 2}) \mathbb {E} (y \mid x _ {1}, u _ {1 2}, z _ {1}, z _ {2}) \mathbb {1} \left(Z _ {1} (u _ {1 2}) = z _ {1}, Z _ {2} (u _ {1 2}) = z _ {2}\right) (110) \\ = \sum_ {u _ {1 2}, z _ {1}, z _ {2}} P (u _ {1 2}) \mathbb {E} (y \mid x _ {1}, z _ {1}, z _ {2}) \mathbb {1} \left(Z _ {1} \left(u _ {1 2}\right) = z _ {1}, Z _ {2} \left(u _ {1 2}\right) = z _ {2}\right) (111) \\ = \sum_ {z _ {1}, z _ {2}} \mathbb {E} (y \mid x _ {1}, z _ {1}, z _ {2}) \sum_ {u _ {1 2}} P \left(u _ {1 2}\right) \mathbb {1} \left(Z _ {1} \left(u _ {1 2}\right) = z _ {1}, Z _ {2} \left(u _ {1 2}\right) = z _ {2}\right) (112) \\ = \sum_ {z _ {1}, z _ {2}} \mathbb {E} (y \mid x _ {1}, z _ {1}, z _ {2}) P (z _ {1}, z _ {2}) (113) \\ = \sum_ {z _ {1}, z _ {2}, z _ {3}} \mathbb {E} (y \mid x _ {1}, z _ {1}, z _ {2}, z _ {3}) P \left(z _ {1}, z _ {2}\right) P \left(z _ {3} \mid x _ {1}, z _ {1}, z _ {2}\right), (114) \\ \end{array}
$$
providing an identification expression from observational data.
C Discussion of Thm. 4
Thm. 4 introduces a sufficient condition for the identification of quantities under partial abduction (Def. 2). Here, we discuss why some of the conditions in the theorem are necessary, and provide an example that further elucidates the theorem's scope.
Example 9 (Non-Identification in Semi-Markovian Models). Consider the causal diagram in Fig. 10b and consider two SCMs  $\mathcal{M}_1, \mathcal{M}_2$  constructed as follows. Both SCMs have the same set of assignment equations  $\mathcal{F}$ , given by
$$
\mathcal {F} := \left\{ \begin{array}{l l} Z _ {1} \leftarrow \left\{ \begin{array}{l l} 1 & \text {i f} U _ {1 X} > 4 \\ 0 & \text {i f} U _ {1 X} \leq 4 \end{array} \right. \\ Z _ {2} \leftarrow U _ {2} \\ X \leftarrow \left\{ \begin{array}{l l} 1 & \text {i f} U _ {1 X} \in \{1, 5 \} \\ Z _ {2} & \text {i f} U _ {1 X} \in \{2, 6 \} \\ 1 - Z _ {2} & \text {i f} U _ {1 X} \in \{3, 7 \} \\ 0 & \text {i f} U _ {1 X} \in \{4, 8 \} \end{array} \right. \\ Y \leftarrow Z _ {1} \vee Z _ {2}. \end{array} \right. \tag {118}
$$
The two SCMs differ in the distribution over the latent variables. In particular, for  $\mathcal{M}_1$  we have
$$
P ^ {\mathcal {M} _ {1}} (U): \left\{ \begin{array}{l} U _ {2} \sim \text {B e r n o u l l i} (0. 6) \\ U _ {1 X} \sim \text {M u l t i n o m} (8, 1, (\frac {1}{8}, \frac {1}{8}, \frac {1}{8}, \frac {1}{8}, \frac {1}{8}, \frac {1}{8}, \frac {1}{8}), \end{array} \right. \tag {119}
$$
$$
P ^ {\mathcal {M} _ {2}} (U): \left\{ \begin{array}{l} U _ {2} \sim \text {B e r n o u l l i} (0. 6) \\ U _ {1 X} \sim \text {M u l t i n o m} (8, 1, \left(0, \frac {1}{4}, \frac {1}{4}, 0, 0, \frac {1}{4}, \frac {1}{4}, 0\right)), \end{array} \right. \tag {122}
$$
That is, the only difference between  $P^{\mathcal{M}_1}(U)$  and  $P^{\mathcal{M}_2}(U)$  is in how  $U_{1X}$  attains its value. Furthermore, we can verify that  $\mathcal{M}_1$  and  $\mathcal{M}_2$  generate the same observational distribution, given by following distribution table
Z1	Z2	X	P(Z1,Z2,X)
0	0	0	0.10
1	0	0	0.10
0	1	0	0.15
1	1	0	0.15
0	0	1	0.10
1	0	1	0.10
0	1	1	0.15
1	1	1	0.15
and  $Y$  given simply as a deterministic function of  $Z_{1}, Z_{2}$ . Now, suppose we are interested in computing the conditional probability of  $Y$  given  $X = x_{1}$  in a partially abducted submodel where  $U_{1X}$  does not respond to evidence. The quantity  $\mathbb{E}(y \mid x_{1}^{U_{1X}})$  can be computed as
$$
\begin{array}{l} \mathbb {E} \left(y \mid x _ {1} ^ {U _ {1 X}}\right) = \sum_ {u _ {1 x} = 1} ^ {8} P \left(u _ {1 x}\right) \mathbb {E} \left(y \mid x _ {1}, u _ {1 x}\right) (123) \\ = \sum_ {u _ {1 x} = 1} ^ {8} P \left(u _ {1 x}\right) \mathbb {E} \left(y \mid x _ {1}, u _ {1 x}, z _ {2}\right) P \left(z _ {2} \mid x _ {1}, u _ {1 x}\right) (124) \\ = \sum_ {u _ {1 x} = 1} ^ {8} P \left(u _ {1 x}\right) P \left(z _ {2} \mid x _ {1}, u _ {1 x}\right) \left[ Z _ {1} \left(u _ {1 x}\right) \vee z _ {2} \right], (125) \\ \end{array}
$$
where  $P(Z_{2} = 1 \mid x_{1}, u_{1x}) = 0.6$  for  $u_{1x} \in \{1, 4, 5, 8\}$ ,  $P(Z_{2} = 1 \mid x_{1}, u_{1x}) = 0$  for  $u_{1x} \in \{3, 7\}$ , and  $P(Z_{2} = 1 \mid x_{1}, u_{1x}) = 1$  for  $u_{1x} \in \{4, 8\}$ . Evaluating the expression in Eq. 125 for the two
SCMs yields
$$
\mathbb {E} ^ {\mathcal {M} _ {1}} \left(y \mid x _ {1} ^ {U _ {1} X}\right) = \frac {3 1}{4 0} \tag {126}
$$
$$
\mathbb {E} ^ {\mathcal {M} _ {2}} \left(y \mid x _ {1} ^ {U _ {1 X}}\right) = \frac {3}{4}, \tag {127}
$$
demonstrating that the quantity  $\mathbb{E}(y\mid x_1^{U_{1X}})$  is not identifiable for the diagram in Fig. 10b.
The above example provides some intuition about why the variable  $X$  cannot be in the anchor set of the variables  $U_{PA}$  that are not updated according to evidence. Very similarly, an analogous example demonstrating that the variable  $Y$  cannot be in the anchor set of  $U_{PA}$  can also be constructed.