SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
q _ {b} (b \mid c) := p _ {y} (\tilde {g} (b) \mid \tilde {h} (c)) \left| \frac {\mathrm {d} \tilde {g}}{\mathrm {d} b} (b) \right|. \tag {C.6}
$$
Since  $p_z, p_x$ , and  $p_y$  are valid densities (non-negative and integrating to one), so are  $q_c, q_a$ , and  $q_b$ . Substitution into (C.3) yields for all  $(a, b, c)$ ,
$$
q (a, b, c) = q _ {c} (c) q _ {a} (a \mid c) q _ {b} (b \mid c), \tag {C.7}
$$
which implies that  $A \perp B \mid C$ , concluding the proof.
C.2 Proof of Thm. 3.2
Theorem 3.2 (Bivariate identifiability up to  $\sim_{\mathrm{CRL}}$  from one perfect stochastic intervention per node). Suppose that we have access to multiple environments  $\{P_X^e\}_{e\in \varepsilon}$  generated as described in § 2 under Asms. 2.2, 2.5, 2.8 and 2.9 with  $n = 2$ . Let  $(h,G')$  be any candidate solution such that the inferred latent distributions  $Q_Z^e = h_*(P_X^e)$  of  $Z = h(X)$  and the inferred mixing function  $h^{-1}$  satisfy the above assumptions w.r.t. the candidate causal graph  $G'$ . Assume additionally that
$$
\mathbb {E} _ {\boldsymbol {v} \sim P _ {V} ^ {\epsilon_ {0}}} \left[ \varphi \left(\frac {\tilde {p} _ {2} \left(v _ {2}\right)}{p _ {2} \left(v _ {2} \mid v _ {1}\right)}\right) \right] \neq \mathbb {E} _ {\boldsymbol {v} \sim P _ {V} ^ {\epsilon_ {1}}} \left[ \varphi \left(\frac {\tilde {p} _ {2} \left(v _ {2}\right)}{p _ {2} \left(v _ {2} \mid v _ {1}\right)}\right) \right] \tag {3.2}
$$
Proof. From the assumption of a shared mixing  $f$  and shared encoder  $h$  across all environments, we have that
$$
\boldsymbol {Z} = h (\boldsymbol {X}) = h (f (\boldsymbol {V})) = h \circ f (\boldsymbol {V}). \tag {C.8}
$$
Let  $\psi = f^{-1} \circ h^{-1}: \mathbb{R}^n \to \mathbb{R}^n$  such that
$$
\boldsymbol {V} = \psi (\boldsymbol {Z}).
$$
By Asm. 2.5,  $f$ ,  $h$ , and thus also  $h \circ f$  are diffeomorphisms. Hence,  $\psi$  is well-defined and also diffeomorphic.
By the change of variable formula, for all  $e$  and all  $z$  the density  $q^{e}(z)$  is given by
$$
q ^ {e} (\boldsymbol {z}) = p ^ {e} (\psi (\boldsymbol {z})) | \det  \boldsymbol {J} _ {\psi} (\boldsymbol {z}) | \tag {C.9}
$$
where  $(J_{\psi}(z))_{ij} = \frac{\partial\psi_i}{\partial z_j} (z)$  denotes the Jacobian of  $\psi$
We now consider two separate cases, depending on whether the intervention targets in  $q^{e_i}$  for  $e_i \in \{e_1, e_2\}$  match those in  $p^{e_i}$  (Case 1) or not (Case 2).
Case 1: Aligned Intervention Targets. According to Asm. 2.8 and (A2), (C.9) applied to the known observational environment  $e_0$  and the interventional environments  $e_1, e_2$  leads to the system of equations:
$$
q _ {1} \left(z _ {1}\right) q _ {2} \left(z _ {2} \mid z _ {\mathrm {p a} \left(2; G ^ {\prime}\right)}\right) = p _ {1} \left(\psi_ {1} (\boldsymbol {z})\right) p _ {2} \left(\psi_ {2} (\boldsymbol {z}) \mid \psi_ {\mathrm {p a} (2)} (\boldsymbol {z})\right) | \det  J _ {\psi} (\boldsymbol {z}) | \tag {C.10}
$$
$$
\tilde {q} _ {1} \left(z _ {1}\right) q _ {2} \left(z _ {2} \mid z _ {\mathrm {p a} \left(2; G ^ {\prime}\right)}\right) = \tilde {p} _ {1} \left(\psi_ {1} (\boldsymbol {z})\right) p _ {2} \left(\psi_ {2} (\boldsymbol {z}) \mid \psi_ {\mathrm {p a} (2)} (\boldsymbol {z})\right) | \det  J _ {\psi} (\boldsymbol {z}) | \tag {C.11}
$$
$$
q _ {1} \left(z _ {1}\right) \tilde {q} _ {2} \left(z _ {2}\right) = p _ {1} (\psi_ {1} (z)) \tilde {p} _ {2} (\psi_ {2} (z)) | \det  J _ {\psi} (z) | \tag {C.12}
$$
where  $z_{\mathrm{pa}(2;G')}^{\prime}$  denotes the parents of  $z_{2}$  in  $G'$ , and  $\mathrm{pa}(2)$  denotes the parents of  $V_{2}$  in  $G$ .
Note that neither side of the previous equations can be zero due to the full support assumption $^{12}$  (A1) and  $\psi$  being diffeomorphic implying the determinant is non-zero.
This can also be relaxed to fully supported on a Cartesian product of intervals.
Taking quotients of (C.11) and (C.10), yields
$$
\frac {\tilde {q} _ {1}}{q _ {1}} \left(z _ {1}\right) = \frac {\tilde {p} _ {1}}{p _ {1}} \left(\psi_ {1} (z)\right). \tag {C.13}
$$
Next, we take the partial derivative w.r.t.  $z_{2}$  on both sides and use the chain rule to obtain:
$$
0 = \left(\frac {\tilde {p} _ {1}}{p _ {1}}\right) ^ {\prime} (\psi_ {1} (z)) \frac {\partial \psi_ {1}}{\partial z _ {2}} (z). \tag {C.14}
$$
Now, by assumption (A3), the first term on the RHS of (C.14) is non-zero everywhere. Hence,
$$
\forall z: \quad \frac {\partial \psi_ {1}}{\partial z _ {2}} (z) = 0. \tag {C.15}
$$
It follows that  $\psi_{1}$  is, in fact, a scalar function, and
$$
V _ {1} = \psi_ {1} \left(Z _ {1}\right). \tag {C.16}
$$
Since  $\psi$  is a diffeomorphism,  $\psi_{1}$  must also be diffeomorphic. Hence, by the change of variable formula applied to (C.16), the marginal density  $q_{1}(z_{1})$  is given by
$$
q _ {1} \left(z _ {1}\right) = p _ {1} \left(\psi_ {1} \left(z _ {1}\right)\right) \left| \frac {\partial \psi_ {1}}{\partial z _ {1}} \left(z _ {1}\right) \right|. \tag {C.17}
$$
Further, since  $J_{\psi}$  is triangular due to (C.15), its determinant is given by
$$
\left| \det  J _ {\psi} (z) \right| = \left| \frac {\partial \psi_ {1}}{\partial z _ {1}} \left(z _ {1}\right) \frac {\partial \psi_ {2}}{\partial z _ {2}} \left(z _ {1}, z _ {2}\right) \right|. \tag {C.18}
$$
Substituting (C.17) and (C.18) into (C.12) yields (after cancellation of equal terms):
$$
\tilde {q} _ {2} \left(z _ {2}\right) = \tilde {p} _ {2} \left(\psi_ {2} \left(z _ {1}, z _ {2}\right)\right) \left| \frac {\partial \psi_ {2}}{\partial z _ {2}} \left(z _ {1}, z _ {2}\right) \right|. \tag {C.19}
$$
The expression in (C.19) implies that, for all  $z_{1}$ , the mapping  $\psi_2(z_1,\cdot):\mathbb{R}\to \mathbb{R}$  is measure preserving for the differentiable  $\tilde{q}_2$  and  $\tilde{p}_2$ . By Lemma C.1 (Lemma 2 of Brehmer et al. [18, § A.2]), it then follows that  $\psi_{2}$  must, in fact, be constant in  $z_{1}$ , that is
$$
\forall z: \quad \frac {\partial \psi_ {2}}{\partial z _ {1}} (z) = 0. \tag {C.20}
$$
Note that this last step is where the assumption of perfect interventions (Asm. 2.9) is leveraged: the conclusion would not hold for arbitrary imperfect interventions for which (3.8) would involve  $\tilde{q}_2(z_2\mid z_1)$  and  $p_2\left(\psi_2(z_1,z_2)\mid \psi_1(z_1)\right)$ .
Hence, we have shown that  $\psi$  is an element-wise function:
$$
\boldsymbol {V} = \left(V _ {1}, V _ {2}\right) = \psi (\boldsymbol {Z}) = \left(\psi_ {1} \left(Z _ {1}\right), \psi_ {2} \left(Z _ {2}\right)\right). \tag {C.21}
$$
Finally, since  $\psi$  is a diffeomorphism, (C.21) implies that
$$
V _ {1} \perp V _ {2} \Longleftrightarrow Z _ {1} \perp Z _ {2}. \tag {C.22}
$$
It then follows from the faithfulness assumption (Asm. 2.2) that we also must have  $G = G'$ .
This concludes the proof of Case 1, as we have shown that  $(h,G^{\prime})\sim_{\mathrm{cRL}}(f^{-1},G)$  with  $G^{\prime} = \pi (G) = G$  ( $\pi$  being the identity permutation) and  $h\circ f = \psi^{-1}\eqqcolon \phi$  an element-wise diffeomorphism.
Case 2: Misaligned Intervention Targets. Writing down the system of equations similar to (C.10)-(C.12), but for the case with misaligned intervention targets across  $p$  and  $q$  yields:
$$
q _ {1} \left(z _ {1}\right) q _ {2} \left(z _ {2} \mid z _ {\mathrm {p a} (2; G ^ {\prime})}\right) = p _ {1} (\psi_ {1} (z)) p _ {2} (\psi_ {2} (z) \mid \psi_ {\mathrm {p a} (2)} (z)) | \det  J _ {\psi} (z) | \tag {C.23}
$$
$$
\tilde {q} _ {1} \left(z _ {1}\right) q _ {2} \left(z _ {2} \mid z _ {\mathrm {p a} \left(2; G ^ {\prime}\right)}\right) = p _ {1} (\psi_ {1} (z)) \tilde {p} _ {2} (\psi_ {2} (z)) | \det  J _ {\psi} (z) | \tag {C.24}
$$
$$
q _ {1} \left(z _ {1}\right) \tilde {q} _ {2} \left(z _ {2}\right) = \tilde {p} _ {1} (\psi_ {1} (z)) p _ {2} (\psi_ {2} (z) \mid \psi_ {\mathrm {p a} (2)} (z)) | \det  J _ {\psi} (z) |. \tag {C.25}
$$
Taking ratios of  $e_1$  and  $e_2$  with  $e_0$  yields
$$
\frac {\tilde {q} _ {1}}{q _ {1}} \left(z _ {1}\right) = \frac {\tilde {p} _ {2} (\psi_ {2} (z))}{p _ {2} \left(\psi_ {2} (z) \mid \psi_ {\mathrm {p a} (2)} (z)\right)} \tag {C.26}
$$
$$
\frac {\tilde {q} _ {2} \left(z _ {2}\right)}{q _ {2} \left(z _ {2} \mid z _ {\mathrm {p a} \left(2 ; G ^ {\prime}\right)}\right)} = \frac {\tilde {p} _ {1}}{p _ {1}} (\psi_ {1} (z)). \tag {C.27}
$$
We separate the remainder of the proof of Case 2 into different subcases depending on  $G$  and  $G'$ : as we will see, we can use a similar reasoning as in Case 1, except for the case where both  $G$  and  $G'$  are missing no edge.
Case 2a:  $V_{1} \nrightarrow V_{2}$  in  $G$ , that is  $\mathrm{pa}(2) = \emptyset$ . Then we can proceed similarly to Case 1. First, we take the partial derivative of (C.26) w.r.t.  $z_{2}$  to arrive at:
$$
0 = \left(\frac {\tilde {p} _ {2}}{p _ {2}}\right) ^ {\prime} (\psi_ {2} (z)) \frac {\partial \psi_ {2}}{\partial z _ {2}} (z). \tag {C.28}
$$
Using (A3), this implies that  $\psi_{2}$  does not depend on  $Z_{2}$ , that is,  $V_{2} = \psi_{2}(Z_{1})$ .
Next, we again write  $q(z_{1})$  in terms of  $p_2(\psi_2(z_1))$  using the univariate change of variable formula, substitute into (C.25), cancel the corresponding terms, and arrive at:
$$
\tilde {q} _ {2} \left(z _ {2}\right) = \tilde {p} _ {1} \left(\psi_ {1} \left(z _ {1}, z _ {2}\right)\right) \left| \frac {\partial \psi_ {1}}{\partial z _ {2}} \left(z _ {1}, z _ {2}\right) \right| \tag {C.29}
$$
Lemma C.1 applied to  $\psi_{1}(z_{1},\cdot)$  which preserves  $\tilde{q}_2$  and  $\tilde{p}_1$  for all  $z_{1}$  shows that  $\psi_{1}$  is constant in  $Z_{1}$ , that is
$$
\boldsymbol {V} = \left(V _ {1}, V _ {2}\right) = \psi (\boldsymbol {Z}) = \left(\psi_ {1} \left(Z _ {2}\right), \psi_ {2} \left(Z _ {1}\right)\right). \tag {C.30}
$$
Since  $V_{1} \perp V_{2}$  by the assumption of Case 2a, it follows from the invertible element-wise reparametrisation above that  $Z_{1} \perp Z_{2}$  and hence, by faithfulness,  $Z_{1} \nrightarrow Z_{2}$  in  $G'$ .
Finally, note that there is no partial order on the empty graph and so  $G' = \pi(G) = G$  and  $Z = P_{\pi^{-1}} \cdot \psi^{-1}(V)$  where  $\pi$  is the nontrivial permutation of  $\{1,2\}$ .
Case 2b:  $V_{1} \to V_{2}$  in  $G$ , that is  $\mathrm{pa}(2) = \{1\}$ . If  $G' \neq G$ , that is  $Z_{1} \nrightarrow Z_{2}$  in  $G'$ , then the same argument as in Case 2a, this time starting by taking the partial derivative of (C.27) w.r.t.  $z_{1}$ , can be used to reach the same conclusion in (C.30). However, this contradicts faithfulness since  $V_{1} \not\sqsubseteq V_{2}$  in  $G$ .
Hence, we must have  $G' = G$ , and the following two equations must hold for all  $z$ :
$$
\frac {\tilde {q} _ {1} \left(z _ {1}\right)}{q _ {1} \left(z _ {1}\right)} = \frac {\tilde {p} _ {2} \left(\psi_ {2} (\boldsymbol {z})\right)}{p _ {2} \left(\psi_ {2} (\boldsymbol {z}) \mid \psi_ {1} (\boldsymbol {z})\right)} \tag {C.31}
$$
$$
\frac {\tilde {q} _ {2} \left(z _ {2}\right)}{q _ {2} \left(z _ {2} \mid z _ {1}\right)} = \frac {\tilde {p} _ {1} \left(\psi_ {1} (\boldsymbol {z})\right)}{p _ {1} \left(\psi_ {1} (\boldsymbol {z})\right)} \tag {C.32}
$$
The remainder of the proof consists of exploring the implications of (C.32) and (C.31), ultimately resulting in a violation of the genericity condition (A4).
To ease notation, define the following auxiliary functions:
$$
a \left(z _ {1}\right) := \frac {\tilde {q} _ {1} \left(z _ {1}\right)}{q _ {1} \left(z _ {1}\right)}, \tag {C.33}
$$
$$
b (\boldsymbol {v}) := \frac {\tilde {p} _ {2} \left(v _ {2}\right)}{p _ {2} \left(v _ {2} \mid v _ {1}\right)}, \tag {C.34}
$$
$$
c (\boldsymbol {z}) := \frac {\tilde {q} _ {2} \left(z _ {2}\right)}{q _ {2 | 1} \left(z _ {2} \mid z _ {1}\right)}, \tag {C.35}
$$
$$
d \left(v _ {1}\right) := \frac {\tilde {p} _ {1} \left(v _ {1}\right)}{p _ {1} \left(v _ {1}\right)}. \tag {C.36}
$$
With this, (C.31) and (C.32) take the following form:
$$
a \left(z _ {1}\right) = b (\psi (z)). \tag {C.37}
$$
$$
c (\boldsymbol {z}) = d \left(\psi_ {1} (\boldsymbol {z})\right), \tag {C.38}
$$
Next, define the following maps:
$$
\kappa : \boldsymbol {z} \mapsto \left[ \begin{array}{l} a \left(z _ {1}\right) \\ c (\boldsymbol {z}) \end{array} \right] \tag {C.39}
$$
$$
\rho : \boldsymbol {v} \mapsto \left[ \begin{array}{l} b (\boldsymbol {v}) \\ d (v _ {1}) \end{array} \right] \tag {C.40}
$$
Then, (C.37) and (C.38) together imply that
$$
\kappa = \rho \circ \psi . \tag {C.41}
$$
Recalling that by (A1) all densities are continuously differentiable, the Jacobians of  $\kappa$  and  $\rho$  are given by:
$$
\boldsymbol {J} _ {\kappa} (\boldsymbol {z}) = \left[ \begin{array}{c c} a ^ {\prime} \left(z _ {1}\right) & 0 \\ \frac {\partial c}{\partial z _ {1}} (\boldsymbol {z}) & \frac {\partial c}{\partial z _ {2}} (\boldsymbol {z}) \end{array} \right], \tag {C.42}
$$
$$
\boldsymbol {J} _ {\rho} (\boldsymbol {v}) = \left[ \begin{array}{c c} \frac {\partial b}{\partial v _ {1}} (\boldsymbol {v}) & \frac {\partial b}{\partial v _ {2}} (\boldsymbol {v}) \\ d ^ {\prime} (v _ {1}) & 0 \end{array} \right], \tag {C.43}
$$
and the corresponding determinants are given by
$$
\left| \det  J _ {\kappa} (\boldsymbol {z}) \right| = \left| a ^ {\prime} \left(z _ {1}\right) \frac {\partial c}{\partial z _ {2}} (\boldsymbol {z}) \right| \neq 0 \tag {C.44}
$$
$$
\left| \det  \boldsymbol {J} _ {\rho} (\boldsymbol {v}) \right| = \left| d ^ {\prime} \left(v _ {1}\right) \frac {\partial b}{\partial v _ {2}} (\boldsymbol {v}) \right| \neq 0 \tag {C.45}
$$
where the inequalities for all  $z$  follow since, by assumption (A3), the derivatives of ratios of intervened and original mechanisms are non-vanishing everywhere:
$$
a ^ {\prime} \left(z _ {1}\right) \neq 0 \neq \frac {\partial c}{\partial z _ {2}} (z) \quad \text {a n d} \quad d ^ {\prime} \left(v _ {1}\right) \neq 0 \neq \frac {\partial b}{\partial v _ {2}} (v), \tag {C.46}
$$
This implies that the following families of maps are continuously differentiable, monotonic, and invertible,
$$
a: z _ {1} \mapsto a \left(z _ {1}\right), \tag {C.47}
$$
$$
b _ {v _ {1}}: v _ {2} \mapsto b \left(v _ {1}, v _ {2}\right), \tag {C.48}
$$
$$
c _ {z _ {1}}: z _ {2} \mapsto c \left(z _ {1}, z _ {2}\right), \tag {C.49}
$$
$$
d: v _ {1} \mapsto d \left(v _ {1}\right), \tag {C.50}
$$
with continuously differentiable inverses
$$
a ^ {- 1}: w _ {1} \mapsto a ^ {- 1} \left(w _ {1}\right), \tag {C.51}
$$
$$
b _ {v _ {1}} ^ {- 1}: w _ {1} \mapsto b _ {v _ {1}} ^ {- 1} \left(w _ {1}\right), \tag {C.52}
$$
$$
c _ {z _ {1}} ^ {- 1}: w _ {2} \mapsto c _ {z _ {1}} ^ {- 1} (w _ {2}), \tag {C.53}
$$
$$
d ^ {- 1}: w _ {2} \mapsto d ^ {- 1} (w _ {2}). \tag {C.54}
$$
This implies that  $\rho$  and  $\kappa$  are valid diffeomorphisms onto their image and their inverses are given by:
$$
\kappa^ {- 1}: \boldsymbol {w} \mapsto \left[ \begin{array}{c} a ^ {- 1} \left(w _ {1}\right) \\ c _ {a ^ {- 1} \left(w _ {1}\right)} ^ {- 1} \left(w _ {2}\right) \end{array} \right], \tag {C.55}
$$
$$
\rho^ {- 1}: \boldsymbol {w} \mapsto \left[ \begin{array}{c} d ^ {- 1} \left(w _ {2}\right) \\ b _ {d ^ {- 1} \left(w _ {2}\right)} ^ {- 1} \left(w _ {1}\right) \end{array} \right]. \tag {C.56}
$$
Since  $V = \psi (Z)$  , by (C.41) we have
$$
\boldsymbol {W} := \rho (\boldsymbol {V}) = \rho \circ \psi (\boldsymbol {Z}) = \kappa (\boldsymbol {Z}). \tag {C.57}
$$
Denote the distributions of  $\mathbf{W}$  by  $R_{\mathbf{W}}$  and its density by  $r(\mathbf{w})$ . Since for all  $e$ , we have
$$
P _ {\boldsymbol {V}} ^ {e} = \psi_ {*} \left(Q _ {\boldsymbol {Z}} ^ {e}\right) \tag {C.58}
$$
it follows from (C.57) that
$$
R _ {\boldsymbol {W}} ^ {e} = \rho_ {*} \left(P _ {\boldsymbol {V}} ^ {e}\right) = \kappa_ {*} \left(Q _ {\boldsymbol {Z}} ^ {e}\right). \tag {C.59}
$$
This provides two different ways of applying the change of variable formula to compute  $r(\boldsymbol{w})$ .
First, we consider the pushforward of  $Q_Z^{e_0}$  by  $\kappa$ :
$$
\begin{array}{l} r (\boldsymbol {w}) = q \left(\kappa^ {- 1} (\boldsymbol {w})\right) | \det  J _ {\kappa^ {- 1}} (\boldsymbol {w}) | (C.60) \\ = q _ {1} \left(a ^ {- 1} \left(w _ {1}\right)\right) q _ {2} \left(c _ {a ^ {- 1} \left(w _ {1}\right)} ^ {- 1} \left(w _ {2}\right) \mid a ^ {- 1} \left(w _ {1}\right)\right) \left| \frac {\mathrm {d}}{\mathrm {d} w _ {1}} a ^ {- 1} \left(w _ {1}\right) \frac {\mathrm {d}}{\mathrm {d} w _ {2}} c _ {a ^ {- 1} \left(w _ {1}\right)} ^ {- 1} \left(w _ {2}\right) \right| (C.61) \\ \end{array}
$$
By integrating this joint density with respect to  $w_{2}$ , we obtain the following expression for the marginal  $r_1(w_1)$ :
$$
r _ {1} \left(w _ {1}\right) = \left| \frac {\mathrm {d}}{\mathrm {d} w _ {1}} a ^ {- 1} \left(w _ {1}\right) \right| q _ {1} \left(a ^ {- 1} \left(w _ {1}\right)\right) \int q _ {2} \left(c _ {a ^ {- 1} \left(w _ {1}\right)} ^ {- 1} \left(w _ {2}\right) \mid a ^ {- 1} \left(w _ {1}\right)\right) \left| \frac {\mathrm {d}}{\mathrm {d} w _ {2}} c _ {a ^ {- 1} \left(w _ {1}\right)} ^ {- 1} \left(w _ {2}\right) \right| \mathrm {d} w _ {2}. \tag {C.62}
$$
By the diffeomorphic change of variable  $z_{2} = c_{a^{-1}(w_{1})}^{-1}(w_{2})$ , this can be written as
$$
\begin{array}{l} r _ {1} \left(w _ {1}\right) = \left| \frac {d}{d w _ {1}} a ^ {- 1} \left(w _ {1}\right) \right| q _ {1} \left(a ^ {- 1} \left(w _ {1}\right)\right) \int q _ {2} \left(z _ {2} \mid a ^ {- 1} \left(w _ {1}\right)\right) d z _ {2} (C.63) \\ = \left| \frac {d}{d w _ {1}} a ^ {- 1} \left(w _ {1}\right) \right| q _ {1} \left(a ^ {- 1} \left(w _ {1}\right)\right) (C.64) \\ \end{array}
$$
Next, we carry out the same calculation for the pushforward of  $P_V^{e_0}$  by  $\rho$ :
$$
\begin{array}{l} r (\boldsymbol {w}) = p \left(\rho^ {- 1} (\boldsymbol {w})\right) | \det  \boldsymbol {J} _ {\rho^ {- 1}} (\boldsymbol {w}) | (C.65) \\ = p _ {1} \left(d ^ {- 1} \left(w _ {2}\right)\right) p _ {2} \left(b _ {d ^ {- 1} \left(w _ {2}\right)} ^ {- 1} \left(w _ {1}\right) \mid d ^ {- 1} \left(w _ {2}\right)\right) \left| \frac {\mathrm {d}}{\mathrm {d} w _ {2}} d ^ {- 1} \left(w _ {2}\right) \frac {\mathrm {d}}{\mathrm {d} w _ {1}} b _ {d ^ {- 1} \left(w _ {2}\right)} ^ {- 1} \left(w _ {1}\right) \right|, (C.66) \\ \end{array}
$$
leading to the marginal
$$
\begin{array}{l} r _ {1} \left(w _ {1}\right) = \int p _ {1} \left(d ^ {- 1} \left(w _ {2}\right)\right) p _ {2} \left(b _ {d ^ {- 1} \left(w _ {2}\right)} ^ {- 1} \left(w _ {1}\right) \mid d ^ {- 1} \left(w _ {2}\right)\right) \left| \frac {\mathrm {d}}{\mathrm {d} w _ {1}} b _ {d ^ {- 1} \left(w _ {2}\right)} ^ {- 1} \left(w _ {1}\right) \right| \left| \frac {\mathrm {d}}{\mathrm {d} w _ {2}} d ^ {- 1} \left(w _ {2}\right) \right| \mathrm {d} w _ {2} (C.67) \\ = \int p _ {1} \left(v _ {1}\right) p _ {2} \left(b _ {v _ {1}} ^ {- 1} \left(w _ {1}\right) \mid v _ {1}\right) \left| \frac {\mathrm {d}}{\mathrm {d} w _ {1}} b _ {v _ {1}} ^ {- 1} \left(w _ {1}\right) \right| \mathrm {d} v _ {1}, (C.68) \\ \end{array}
$$
where the second line is obtained by the diffeomorphic change of variable  $v_{1} = d^{-1}(w_{2})$ .
Equating the two expressions for  $r(w_{1})$  in  $e_0$  in (C.68) and (C.64), we obtain for all  $w_{1}$ :
$$
\left| \frac {\mathrm {d}}{\mathrm {d} w _ {1}} a ^ {- 1} \left(w _ {1}\right) \right| q _ {1} \left(a ^ {- 1} \left(w _ {1}\right)\right) = \int p _ {1} \left(v _ {1}\right) p _ {2} \left(b _ {v _ {1}} ^ {- 1} \left(w _ {1}\right) \mid v _ {1}\right) \left| \frac {\mathrm {d}}{\mathrm {d} w _ {1}} b _ {v _ {1}} ^ {- 1} \left(w _ {1}\right) \right| \mathrm {d} v _ {1}. \tag {C.69}
$$
Applying the same approach to the environment in which  $V_{1}$  is intervened upon changing  $p_1$  to  $\tilde{p}_1$  while  $Z_{2}$  is intervened upon leaving  $q_{1}$  invariant, yields for all  $w_{1}$ :
$$
\left| \frac {\mathrm {d}}{\mathrm {d} w _ {1}} a ^ {- 1} \left(w _ {1}\right) \right| q _ {1} \left(a ^ {- 1} \left(w _ {1}\right)\right) = \int \tilde {p} _ {1} \left(v _ {1}\right) p _ {2} \left(b _ {v _ {1}} ^ {- 1} \left(w _ {1}\right) \mid v _ {1}\right) \left| \frac {\mathrm {d}}{\mathrm {d} w _ {1}} b _ {v _ {1}} ^ {- 1} \left(w _ {1}\right) \right| \mathrm {d} v _ {1}. \tag {C.70}
$$
$^{13}$ Note that:  $\int q_2(z_2(w_2))\left|\frac{\mathrm{d}z_2}{\mathrm{d}w_2}\right|\mathrm{d}w_2 = \int q_2(z_2)\mathrm{d}z_2$
Finally, by equating (C.69) and (C.70), we arrive at the following expression which must hold for all  $w_{1}$ :
$$
\int p _ {1} \left(v _ {1}\right) p _ {2} \left(b _ {v _ {1}} ^ {- 1} \left(w _ {1}\right) \mid v _ {1}\right) \left| \frac {\mathrm {d}}{\mathrm {d} w _ {1}} b _ {v _ {1}} ^ {- 1} \left(w _ {1}\right) \right| \mathrm {d} v _ {1} = \int \tilde {p} _ {1} \left(v _ {1}\right) p _ {2} \left(b _ {v _ {1}} ^ {- 1} \left(w _ {1}\right) \mid v _ {1}\right) \left| \frac {\mathrm {d}}{\mathrm {d} w _ {1}} b _ {v _ {1}} ^ {- 1} \left(w _ {1}\right) \right| \mathrm {~d} v _ {1} \tag {C.71}
$$
which we can rewrite as
$$
\int \left(\tilde {p} _ {1} \left(v _ {1}\right) - p _ {1} \left(v _ {1}\right)\right) p _ {2} \left(b _ {v _ {1}} ^ {- 1} \left(w _ {1}\right) \mid v _ {1}\right) \left| \frac {\mathrm {d}}{\mathrm {d} w _ {1}} b _ {v _ {1}} ^ {- 1} \left(w _ {1}\right) \right| \mathrm {d} v _ {1} = 0. \tag {C.72}
$$
Multiplying by any continuous function  $\varphi(w_1)$ , integrating w.r.t.  $w_1$  and applying the diffeomorphic change of variable  $v_2 = b_{v_1}^{-1}(w_1)$ , this can be expressed as:
$$
\begin{array}{l} 0 = \int \varphi \left(w _ {1}\right) \int \left(\tilde {p} _ {1} \left(v _ {1}\right) - p _ {1} \left(v _ {1}\right)\right) p _ {2} \left(b _ {v _ {1}} ^ {- 1} \left(w _ {1}\right) \mid v _ {1}\right) \left| \frac {\mathrm {d}}{\mathrm {d} w _ {1}} b _ {v _ {1}} ^ {- 1} \left(w _ {1}\right) \right| \mathrm {d} v _ {1} \mathrm {d} w _ {1} (C.73) \\ = \iint \varphi \left(b _ {v _ {1}} \left(v _ {2}\right)\right) \left(\tilde {p} _ {1} \left(v _ {1}\right) - p _ {1} \left(v _ {1}\right)\right) p _ {2} \left(v _ {2} \mid v _ {1}\right) \mathrm {d} v _ {2} \mathrm {d} v _ {1} (C.74) \\ = \iint \varphi \left(\frac {\tilde {p} _ {2} (v _ {2})}{p _ {2} \left(v _ {2} \mid v _ {1}\right)}\right) \left(\tilde {p} _ {1} \left(v _ {1}\right) - p _ {1} \left(v _ {1}\right)\right) p _ {2} \left(v _ {2} \mid v _ {1}\right) d v _ {2} d v _ {1} (C.75) \\ \end{array}
$$
where we have resubstituted the expression for  $b_{v_1}(v_2)$  in the last line.
Equivalently, this can be written as: for any continuous function  $\varphi$
$$
\mathbb {E} _ {\boldsymbol {v} \sim P _ {V} ^ {e _ {0}}} \left[ \varphi \left(\frac {\tilde {p} _ {2} \left(v _ {2}\right)}{p _ {2} \left(v _ {2} \mid v _ {1}\right)}\right) \right] = \mathbb {E} _ {\boldsymbol {v} \sim P _ {V} ^ {e _ {1}}} \left[ \varphi \left(\frac {\tilde {p} _ {2} \left(v _ {2}\right)}{p _ {2} \left(v _ {2} \mid v _ {1}\right)}\right) \right]. \tag {C.76}
$$
However, the genericity condition (A4) precisely rules this out, since the above equality must be violated for at least one  $\varphi$ , concluding this last case.
To sum up, all cases either lead to a contradiction, or imply the conclusion that  $(f^{-1}, G) \sim_{\mathrm{CRL}} (h, G')$ , concluding the proof.
C.3 Proof of Thm. 3.4
Theorem 3.4 (Identifiability up to  $\sim_{\mathrm{CRL}}$  from two paired perfect stochastic interventions per node). Suppose that we have access to multiple environments  $\{P_X^e\}_{e\in \mathcal{E}}$  generated as described in § 2 under Asms. 2.2, 2.3, 2.5, 2.8 and 2.9. Let  $(h,G')$  be any candidate solution such that the inferred latent distributions  $Q_Z^e = h_*(P_X^e)$  of  $Z = h(X)$  and the inferred mixing function  $h^{-1}$  satisfy the above assumptions w.r.t. the candidate causal graph  $G'$ . Assume additionally that
Proof. First, we show that we can extract from the  $m \geq n$  available pairs of environments a suitable subset  $\mathcal{E}_n$  of exactly  $n$  pairs, containing one pair of interventional environments for each node.