SlowGuess/NeurIPS_2023_OCR huggingface dataset:
$$
\begin{array}{l} p _ {[ n - 1 ]} \left(\mathbf {z} _ {[ n - 1 ]}\right) = q _ {[ n - 1 ]} \left(\varphi_ {[ n - 1 ]} \left(\mathbf {z} _ {[ n - 1 ]}\right)\right) \prod_ {i = 1} ^ {n - 1} \left| \frac {\partial \varphi_ {i}}{\partial z _ {i}} \left(z _ {[ i ]}\right) \right| \\ = q _ {[ n - 1 ]} \left(\boldsymbol {\varphi} _ {[ n - 1 ]} \left(\mathbf {z} _ {[ n - 1 ]}\right)\right) | \det  D \boldsymbol {\varphi} _ {[ n - 1 ]} \left(\mathbf {z} _ {[ n - 1 ]}\right) |, \\ \end{array}
$$
which is the density equation for push-forward measures that we want to prove.
Back to the proof of (ii). For all  $i \in [d]$ ,  $(\varphi_1, \ldots, \varphi_{i-1})$  is a diffeomorphism because  $|\operatorname{det} D\varphi_{[i-1]}(\mathbf{z}_{[i-1]})| = \prod_{j=1}^{i-1} \left|\frac{\partial \varphi_j}{\partial z_j}(\mathbf{z})\right| \neq 0$ .
We will prove by induction on the reverse order of  $[d]$  that the i-th row off-diagonal entries of  $D\varphi(\mathbf{z})$  are zero for all  $\mathbf{z} \in \mathbb{R}^d$ .
In the interventional regime  $d$ , by the assumption on the indices of  $V(G) = [d]$  in the proof  $(i)$ , the node  $d$  is not a parent of any node in  $[d - 1]$ . Thus the perfect stochastic intervention on  $z_{d}$  leads to the density  $p^d$  and  $q^d$  factorized as follows:
$$
p _ {[ d - 1 ]} \left(\mathbf {z} _ {[ d - 1 ]}\right) \tilde {p} _ {d} \left(z _ {d}\right) = q _ {[ d - 1 ]} \left(\varphi_ {[ d - 1 ]} \left(\mathbf {z} _ {[ d - 1 ]}\right)\right) \tilde {q} _ {d} \left(\varphi_ {d} (\mathbf {z})\right) \left| \det  D \varphi_ {[ d - 1 ]} \left(\mathbf {z} _ {[ d - 1 ]}\right) \right| \left| \frac {\partial \varphi_ {d}}{\partial z _ {d}} (\mathbf {z}) \right|.
$$
Since  $D\varphi$  is lower triangular everywhere, cancel the terms of coordinate  $[d - 1]$  on both sides by Lemma B.4,
$$
\tilde {p} _ {d} \left(z _ {d}\right) = \tilde {q} _ {d} \left(\varphi_ {d} \left(\mathbf {z} _ {[ d - 1 ]}, z _ {d}\right)\right) \left| \frac {\partial \varphi_ {d}}{\partial z _ {d}} \left(\mathbf {z} _ {[ d - 1 ]}, z _ {d}\right) \right|, \tag {24}
$$
$$
\forall \mathbf {z} _ {[ d - 1 ]} \in \mathbb {R} ^ {d - 1}, \quad \widetilde {\mathbb {Q}} _ {d} = \varphi_ {d} \left(\mathbf {z} _ {[ d - 1 ]}, \cdot\right) _ {*} \widetilde {\mathbb {P}} _ {d}. \tag {25}
$$
By Lemma B.1,  $\varphi_d$  is constant in the first  $d - 1$  variables.
Suppose the off-diagonal entries are zero for the  $i,i + 1,\ldots ,d$  rows of  $D\varphi (\mathbf{z})$
By the assumption on the indices of  $V(G)$  in the proof  $(i)$ , the node  $i$  is not a parent of any node in  $[i - 1]$ . Thus the perfect stochastic intervention on  $z_{i}$  leads to the density  $p^i$  and  $q^{i}$  factorized as follows:
$$
\begin{array}{l} \int_ {\mathbb {R} ^ {d - i}} p (\mathbf {z}) d z _ {i + 1} \dots d z _ {d} = \int_ {\mathbb {R} ^ {d - i}} q (\boldsymbol {\varphi} (\mathbf {z})) \prod_ {j = 1} ^ {d} \left| \frac {\partial \varphi_ {j}}{\partial z _ {j}} (\mathbf {z} _ {[ j ]}) \right| d z _ {i + 1} \dots d z _ {d} \\ = \prod_ {j = 1} ^ {i} \left| \frac {\partial \varphi_ {j}}{\partial z _ {j}} \left(\mathbf {z} _ {[ j ]}\right) \right| \int_ {\mathbb {R} ^ {d - i}} q \left(\varphi_ {[ i ]} \left(\mathbf {z} _ {[ i ]}\right), \varphi_ {i + 1} \left(z _ {i + 1}\right), \dots , \varphi_ {d} \left(z _ {d}\right)\right) \prod_ {k = i + 1} ^ {d} \left| \frac {\partial \varphi_ {k}}{\partial z _ {k}} \left(z _ {k}\right) \right| d z _ {i + 1} \dots d z _ {d}. \tag {26} \\ \end{array}
$$
By a change of variables  $\left\{ \begin{array}{c}u_{i + 1} = \varphi_{i + 1}(z_{i + 1})\\ \vdots \\ u_d = \varphi_d(z_d) \end{array} \right.$  , we get
$$
\begin{array}{l} p _ {[ i ]} (\mathbf {z} _ {[ i ]}) = \prod_ {j = 1} ^ {i} \left| \frac {\partial \varphi_ {j}}{\partial z _ {j}} \left(\mathbf {z} _ {[ j ]}\right) \right| \int_ {\mathbb {R} ^ {d - i}} q \left(\varphi_ {[ i ]} \left(\mathbf {z} _ {[ i ]}\right), u _ {i + 1}, \dots , u _ {d}\right) d u _ {i + 1} \dots d u _ {d} \\ = q _ {[ i ]} \left(\varphi_ {[ i ]} \left(\mathbf {z} _ {[ i ]}\right)\right) | \det  D \varphi_ {[ i ]} \left(z _ {[ i ]}\right) | \\ = q _ {[ i - 1 ]} \left(\varphi_ {[ i - 1 ]} \left(\mathbf {z} _ {[ i - 1 ]}\right)\right) \tilde {q} _ {i} \left(\varphi_ {i} (\mathbf {z})\right) \left| \det  D \varphi_ {[ i - 1 ]} \left(\mathbf {z} _ {[ i - 1 ]}\right) \right| \left| \frac {\partial \varphi_ {i}}{\partial z _ {i}} (\mathbf {z}) \right|. \\ \end{array}
$$
By Lemma B.4,  $p_{[i-1]}(\mathbf{z}_{[i-1]}) = q_{[i-1]}(\varphi_{[i-1]}(\mathbf{z}_{[i-1]})) \mid \det D\varphi_{[i-1]}(\mathbf{z}_{[i-1]})$ . By Lemma B.1,  $\varphi_{[i]}$  is constant in the first  $i-1$  variables.
In addition,  $D\varphi (\mathbf{z})$  is lower triangular for all  $z$ , so we have proven that  $\varphi \in S_{\mathrm{scaling}}$ .
B.3 Proof of Prop. 4.3
Proof. Without loss of generality by rearranging  $\left(\mathbb{P}^i,\tau_i\right)_{i\in [0,d - 1]}$ , suppose that the unintervened variable is the node  $d$ . Fix any  $\left(G,\mathbf{f},\left(\mathbb{P}^i,\tau_i\right)_{i\in [0,d - 1]}\right)$ , s.t.  $d - 1$  is a parent of  $d$ , and s.t. the causal mechanism of  $Z_{d}$  only has one conditional variable  $Z_{d - 1}$ , and  $Z_{d}\sim \mathcal{N}(Z_{d - 1},1)$ , namely,
$$
p _ {d} \left(z _ {d} \mid z _ {d - 1}\right) = \frac {1}{\sqrt {2 \pi}} \exp \left(- \frac {\left(z _ {d} - z _ {d - 1}\right) ^ {2}}{2}\right).
$$
We now construct  $\left(G, \mathbf{f}', \left(\mathbb{Q}^i, \tau_i\right)_{i \in \llbracket 0, d - 1\rrbracket}\right)$  such that  $\mathbf{f}_* \mathbb{P}^i = \mathbf{f}_*' \mathbb{Q}^i$  and  $\mathbf{f}'^{-1} \circ \mathbf{f} \notin S_{\text{reparam}}$ .
Set  $\mathbb{Q}_i\left(Z_i\mid \mathbf{Z}_{\mathrm{pa}(i)}\right):\stackrel {(d)}{=}\mathbb{P}_i\left(Z_i\mid \mathbf{Z}_{\mathrm{pa}(i)}\right),\widetilde{\mathbb{Q}_i}\left(Z_i\right):\stackrel {(d)}{=}\widetilde{\mathbb{P}_i}\left(Z_i\mid \mathbf{Z}_{\mathrm{pa}(i)}\right)\quad \forall i\in [d - 1].$
Set  $\mathbb{Q}_d(Z_d\mid Z_{d - 1})\stackrel {(d)}{:=}\mathcal{N}(-Z_{d - 1},1),\quad \pmb {\varphi}(\mathbf{z})\quad := \quad (z_1,\dots , - z_{d - 1},z_d - 2z_{d - 1}),$  thus  $|\operatorname {Det}D\pmb {\varphi}(\mathbf{z})| = 1\quad \forall \mathbf{z}\in \mathbb{R}^{d}.$
$$
\begin{array}{l} q _ {d} \left(\varphi_ {d} (\mathbf {z}) \mid \varphi_ {d - 1} (\mathbf {z})\right) = \frac {1}{\sqrt {2 \pi}} \exp \left(- \frac {\left(\varphi_ {d} (\mathbf {z}) - \varphi_ {d - 1} (\mathbf {z})\right) ^ {2}}{2}\right) \\ = \frac {1}{\sqrt {2 \pi}} \exp \left(- \frac {\left(z _ {d} - 2 z _ {d - 1} + z _ {d - 1}\right) ^ {2}}{2}\right) \\ = p _ {d} \left(z _ {d} \mid z _ {d - 1}\right). \\ \end{array}
$$
From the above equation, we infer that for the unintervened regime,
$$
\prod_ {j = 1} ^ {d} p _ {j} \left(z _ {j} \mid \mathbf {z} _ {\mathrm {p a} (j)}\right) = \left[ \prod_ {j = 1} ^ {d} q _ {j} \left(\varphi_ {j} (\mathbf {z}) \mid \varphi_ {\mathrm {p a} (j)} (\mathbf {z})\right) \right] | \det  D \varphi (\mathbf {z}) |,
$$
and for the  $d - 1$  interventional regimes,
$$
\forall i \in [ d - 1 ], \tilde {p} _ {i} (z _ {i}) \prod_ {j \neq i} p _ {j} (z _ {j} \mid \mathbf {z} _ {\mathrm {p a} (j)}) = \tilde {q} _ {i} (\varphi_ {i} (z)) \left[ \prod_ {j \neq i} q _ {j} (\varphi_ {j} (z) \mid \varphi_ {\mathrm {p a} (j)} (z)) \right] | \det D \varphi (z) |,
$$
$$
\varphi_ {*} \mathbb {P} ^ {i} = \mathbb {Q} ^ {i} \quad \forall i \in [   [ 0, d - 1 ]   ].
$$
$$
\mathbf {f} _ {*} \mathbb {P} ^ {i} = \left(\mathbf {f} \circ \varphi^ {- 1}\right) _ {*} \mathbb {Q} ^ {i}
$$
However,  $\left(\mathbf{f} \circ \varphi^{-1}\right)^{-1} \circ \mathbf{f} = \varphi \notin S_{\text{reparam}}$ .
B.4 Proof of Thm. 4.5
Then  $\text{CauCA}$  in  $(G, \mathcal{F}, \mathcal{P}_G)$  is block-identifiable (following [54]): namely, if for all  $k \in [K]$ ,  $\mathbf{f}_* \mathbb{P}^k = \mathbf{f}_*' \mathbb{Q}^k$ , then for  $\varphi := \mathbf{f}'^{-1} \circ \mathbf{f}$ , for all  $k \in [K]$ ,
$$
[ \varphi (\mathbf {z}) ] _ {\tau_ {k}} = \varphi_ {\tau_ {k}} \left(\mathbf {z} _ {\tau_ {k}}\right). \tag {8}
$$
Proof. Fix  $k \in [K]$ . Then the equation of the  $k$ -th interventional regime is  $p^k(\mathbf{z}) = q^k(\varphi(\mathbf{z}))|\operatorname{det}\varphi(\mathbf{z})|$ .
- If  $\tau_{k}$  has no parents from  $[d] \setminus \tau_{k}$ , then in the unintervened regime,  $p_{k}^{0}$  and  $q_{k}^{0}$  have no conditional. Since interventions over  $\tau_{k}$  are perfect,  $p_{k}^{s}$  and  $q_{k}^{s}$  have no conditional for all  $s \in [1, n_{k}]$ .
- If  $\tau_{k}$  has parents from  $[d] \setminus \tau_{k}$ , since in this case there are  $n_{k} + 1$  perfect interventions, enumerated by  $s \in [0, n_{k}]$ ,  $p_{k}^{s}$  and  $q_{k}^{s}$  have no conditional for all  $s \in [0, n_{k}]$ .
In both cases,  $p_k^s$  and  $q_k^s$  have no conditional for all  $s \in [[0, n_k]]$ .
We write the equality of pushforward densities just as Prop. 4.6, and subtract the  $k$ -th interventional regime by the unintervened regime:
$$
\ln p ^ {s} \left(\mathbf {z} _ {\tau_ {k}}\right) - \ln p ^ {0} \left(\mathbf {z} _ {\tau_ {k}}\right) = \ln q ^ {s} \left(\varphi_ {\tau_ {k}} (\mathbf {z})\right) - \ln q ^ {0} \left(\varphi_ {\tau_ {k}} (\mathbf {z})\right).
$$
For all  $i \in [d] \backslash \tau_k$  (nonempty by assumption), take the partial derivative of  $z_i$ :
$$
0 = \sum_ {j = 1} ^ {n _ {k}} \left[ \frac {\partial}{\partial x _ {j}} (\ln q _ {\tau_ {k}} ^ {1}) (\pmb {\varphi} _ {\tau_ {k}} (\mathbf {z} _ {\tau_ {k}})) - \frac {\partial}{\partial x _ {j}} (\ln q _ {\tau_ {k}} ^ {0}) (\pmb {\varphi} _ {\tau_ {k}} (\mathbf {z} _ {\tau_ {k}})) \right] \frac {\partial \varphi_ {\tau_ {k} , j}}{\partial z _ {i}} (\mathbf {z})
$$
By assumption, for all  $s \in [n_k]$ , there is one interventional regime in which the above equation holds. Those  $n_k$  equations form a linear system  $\mathbf{0} = \mathbf{M}_{\tau_k}(\mathbf{z}_{\tau_k}) \frac{\partial \varphi_{\tau_k}}{\partial z_i}(\mathbf{z}_{\tau_k})$ , where  $\mathbf{M}_{\tau_k}(\mathbf{z}_{\tau_k})$  is defined in the statement of theorem. Since  $\mathbf{M}_{\tau_k}(\mathbf{z})$  is invertible a.e. by assumption, the vector  $\frac{\partial \varphi_{\tau_k}}{\partial z_i}(\mathbf{z}) = \mathbf{0} \forall \mathbf{z} \in \mathbb{R}^d$  a.e., which is furthermore strictly everywhere since  $\varphi := \mathbf{f}'^{-1} \circ \mathbf{f}$  is  $\mathcal{C}^1$ . Such a result is valid for all  $i \in [d] \backslash \tau_k$ . Since  $\bigcup_{k \in I} \tau_k = [d]$ , all the non-diagonal entries of  $D\varphi(\mathbf{z})$  that are not in the blocks of  $\tau_k \times \tau_k$  are 0. We conclude that  $\varphi_{\tau_i}$  only depends on  $\mathbf{z}_{\tau_i}$ , i.e.,  $[\varphi(\mathbf{z})]_{\tau_k} = \varphi_{\tau_k}(\mathbf{z}_{\tau_k})$ .
For all  $k \in [K]$ ,  $[d] \setminus \tau_k \neq \emptyset$ . Suppose there exists  $\mathbf{z} \in \mathbb{R}^d$  such that  $\det(D\varphi_{\tau_k}(\mathbf{z})) = 0$ . Since  $[\varphi(\mathbf{z})]_{\tau_i} = \varphi_{\tau_i}(\mathbf{z}_{\tau_i}) \forall i \in [n]$ , the vector  $\frac{\partial \varphi_{\tau_k}}{\partial z_i}(\mathbf{z}) = \mathbf{0}$  for all  $i \notin \tau_k$ . Thus the rows  $\tau_k$  of  $D\varphi(\mathbf{z})$  are linearly dependent, which implies  $\det(D\varphi(\mathbf{z})) = 0$ , which contradicts with  $\varphi$  invertible. Thus  $\varphi_{\tau_k}$  is a diffeomorphism.
B.5 Proof of Prop. 4.6
Proof. Without loss of generality by rearranging  $\left(\mathbb{P}^i,\tau_i\right)_{i\in [0,d - 1]}$ , suppose that the unintervened variable is the node  $d$ . We apply the induction in the proof of Thm. 4.2 ( $i$ ). Since there are  $d - 1$  interventions, the induction stops at  $d - 1$ , and we can infer that for  $\varphi \coloneqq \mathbf{f}^{\prime -1}\circ \mathbf{f}$ , for all  $i\in [d - 1]$ ,  $\frac{\partial\varphi_i}{\partial z_j} (\mathbf{z}) = 0$  a.e.  $\forall j\neq i$ .
Similar to Thm. 4.2 (i), since  $\frac{\partial\varphi_i}{\partial z_j}$  is continuous, it equals zero everywhere. Thus  $D\varphi (z)$  is lower triangular for all  $z\in \mathbb{R}^d$
By Lemma B.4,  $\left(\pmb{\varphi}_{[d - 1]}\right)_*\mathbb{P}_{[d - 1]}\left(\mathbf{Z}_{[d - 1]}\right) = \mathbb{Q}_{[d - 1]}\left(\mathbf{Z}_{[d - 1]}\right)$ . Namely,
$$
\prod_ {j = 1} ^ {d - 1} p _ {j} \left(z _ {j}\right) = \prod_ {j = 1} ^ {d - 1} q _ {j} \left(\varphi_ {j} (\mathbf {z})\right) | \det  D \varphi_ {[ d - 1 ]} \left(\mathbf {z} _ {[ d - 1 ]}\right) |. \tag {27}
$$
Since for all  $i \in [d - 1]$ ,  $\forall j \neq i$ ,  $\frac{\partial \varphi_i}{\partial z_j}(\mathbf{z}) = 0$ ,  $D\varphi(\mathbf{z})$  is lower triangular for all  $z \in \mathbb{R}^d$ . Thus  $|\det D\varphi(\mathbf{z})| = \prod_{j=1}^{d} |\partial_j \varphi_j(\mathbf{z})|$ . Moreover, in the unintervened dataset,
$$
\prod_ {j = 1} ^ {d} p _ {j} \left(z _ {j}\right) = \prod_ {j = 1} ^ {d} q _ {j} \left(\varphi_ {j} (\mathbf {z})\right) | \partial_ {j} \varphi_ {j} (\mathbf {z}) |. \tag {28}
$$
Divide (28) by (27),
$$
p _ {d} \left(z _ {d}\right) = q _ {d} \left(\varphi_ {d} (\mathbf {z})\right) \left| \frac {\partial \varphi_ {d}}{\partial z _ {d}} \left(\mathbf {z} _ {[ d - 1 ]}, z _ {d}\right) \right|,
$$
$$
\forall \mathbf {z} _ {[ d - 1 ]} \in \mathbb {R} ^ {d - 1}, \quad \widetilde {\mathbb {Q}} _ {d} = \varphi_ {d} (\mathbf {z} _ {[ d - 1 ]}, \cdot) _ {*} \widetilde {\mathbb {P}} _ {d}. \tag {29}
$$
By Lemma B.1,  $\varphi_d$  is constant in the first  $d - 1$  variables. We have proven that  $\varphi \in S_{\mathrm{scaling}}$ .
B.6 Proof of Prop. 4.7
Proof. Without loss of generality by rearranging  $\left(\mathbb{P}^i,\tau_i\right)_{i\in [0,d - 2]}$ , suppose that the two unintervened variables are the nodes  $d - 1$ ,  $d$ . Fix any  $\mathbf{f}$  and  $\left(\mathbb{P}_i,\widetilde{\mathbb{P}}_i\right)_{i\in [0,d - 2]}$  such that for all  $i\in [d - 2]$ ,  $\mathbb{P}_i,\widetilde{\mathbb{P}}_i$  have any distribution that is absolutely continuous and full support in  $\mathbb{R}$  with a differentiable density, and such that Asm. 4.1 is satisfied. We will prove that whether we suppose independent Gaussian distributions are in the class of latent distributions or not, CauCA in  $(G,\mathcal{F},\mathcal{P}_G)$  is not identifiable up to  $S_{\mathrm{reparam}}$ .
Case 1: Independent Gaussian distributions are in  $\mathcal{P}_G$ .
By the famous result in linear ICA, if  $\mathbb{P}_{d - 1}$ ,  $\mathbb{P}_d$  form an isotropic Gaussian vector  $\mathcal{N}(\mathbf{0},\Sigma)$ , i.e.,  $\boldsymbol{\Sigma}$  is diagonal with the same variances on each dimension, then any rotation form a spurious solution. Namely, let  $\varphi (\mathbf{z}) = (z_{1},\dots ,z_{d - 2},z_{d - 1}\cos (\theta) - z_{d}\sin (\theta),z_{d}\cos (\theta) + z_{d - 1}\sin (\theta))$ ,  $\theta \neq k\pi$ , then
$$
\forall i \in [   [ 0, d - 2 ]   ] \quad \varphi_ {*} \mathbb {P} ^ {i} = \mathbb {P} ^ {i}
$$
$$
\mathbf {f} _ {*} \mathbb {P} ^ {i} = \left(\mathbf {f} \circ \boldsymbol {\varphi} ^ {- 1}\right) _ {*} \mathbb {P} ^ {i}
$$
However,  $\left(\mathbf{f} \circ \varphi^{-1}\right)^{-1} \circ \mathbf{f} = \varphi \notin S_{\mathrm{reparam}}$ .
Case 2: Independent Gaussian distributions are not in  $\mathcal{P}_G$ .
Suppose that for all  $i\in \{d - 1,d\}$ $\mathbb{P}_i$  has the same density  $p_a$  ..
$$
p _ {a} (z) = \left\{ \begin{array}{l l} \exp (- a z ^ {2}) & z <   0 \\ 1 & 0 \leq z \leq 1 - \sqrt {\frac {\pi}{a}} \\ \exp \left(- a \left(z - \left(1 - \sqrt {\frac {\pi}{a}}\right)\right) ^ {2}\right) & z > 1 - \sqrt {\frac {\pi}{a}} \end{array} \right.
$$
where  $\sqrt{\frac{\pi}{a}} < 1$ . One can verify that  $p_a$  is a smooth p.d.f.
We construct a measure-preserving automorphism inspired by [19].
$$
\boldsymbol {\varphi} (\mathbf {Z}) = \left\{ \begin{array}{l l} \mathbf {Z} & | | \mathbf {Z} _ {[ [ d - 1, d ]} | | \geq R \\ \left( \begin{array}{c} \mathbf {Z} _ {[ d - 2 ]} \\ \cos (\alpha (| | \mathbf {Z} _ {[ [ d - 1, d ]} - \mathbf {C} | | - R)) Z _ {d - 1} \\ - \sin (\alpha (| | \mathbf {Z} _ {[ [ d - 1, d ]} - \mathbf {C} | | - R)) Z _ {d} \\ \cos (\alpha (| | \mathbf {Z} _ {[ [ d - 1, d ]} - \mathbf {C} | | - R)) Z _ {d} \\ + \sin (\alpha (| | \mathbf {Z} _ {[ [ d - 1, d ]} - \mathbf {C} | | - R)) Z _ {d - 1} \end{array} \right) & | | \mathbf {Z} _ {[ [ d - 1, d ]} | | <   R \end{array} \right.
$$
where  $\alpha \neq 0, \mathbf{C} = \left(\frac{1}{2}\left(1 - \sqrt{\frac{\pi}{a}}\right), \frac{1}{2}\left(1 - \sqrt{\frac{\pi}{a}}\right)\right), R \in \left(0, \frac{1}{2}\left(1 - \sqrt{\frac{\pi}{a}}\right)\right]$ .
Now let us prove that  $\varphi$  preserves  $\mathbb{P}_{[[d-1,d]]}$ . By shifting the center of  $p_a$  to the origin, we only need to prove that the shifted  $\varphi_{[[d-1,d]]}$  preserves the uniform distribution over  $[-R,R]^2$ . One can verify that  $\varphi_{[[d-1,d]]}$  is a diffeomorphism over the 2-dimensional open disk  $D^2(\mathbf{0},R) \setminus \{\mathbf{0}\} \to D^2(\mathbf{0},R) \setminus \{\mathbf{0}\}$  and  $|\det(D\varphi_{[[d-1,d]]}(\mathbf{z}))| = 1$ . Thus  $p_a(\mathbf{z}) = p_a(\varphi_{[[d-1,d]]}(\mathbf{z}))|\det(D\varphi_{[[d-1,d]]}(\mathbf{z}))| \forall \mathbf{z} \in D^2(\mathbf{0},R) \setminus \{\mathbf{0}\}$ . Since  $\varphi = Id$  outside of the disk, this change of variables formula holds almost everywhere in  $\mathbb{R}^2$ , thus  $\mathbb{P}_{[[d-1,d]]} = (\varphi_{[[d-1,d]]})_*\mathbb{P}_{[[d-1,d]]}$ , namely,  $\varphi_{[[d-1,d]]}$  preserves  $\mathbb{P}_{[[d-1,d]]}$  on  $\mathbb{R}^2$ .
Moreover, since  $\varphi_{[d - 2]}$  is identity,  $\widetilde{\mathbb{P}}_i = (\varphi_i)_*\widetilde{\mathbb{P}}_i$  and  $\mathbb{P}_i = (\varphi_i)_*\mathbb{P}_i$  for all  $i\in [d - 2]$ . Thus
$$
\forall i \in [   [ 0, d - 2 ]   ] \quad \varphi_ {*} \mathbb {P} ^ {i} = \mathbb {P} ^ {i},
$$
$$
\mathbf {f} _ {*} \mathbb {P} ^ {i} = \left(\mathbf {f} \circ \varphi^ {- 1}\right) _ {*} \mathbb {P} ^ {i}.
$$
B.7 Further constraint on the indeterminacy set
Corollary B.5. Based on the assumption of (2) of Thm. 4.2, if in every dataset we are given the set of possible intervention mechanisms:  $\mathcal{M} = (\mathcal{M}_i)_{i\in [d]}$ , then CauCA in  $(G,\mathcal{F},\mathcal{P}_G)$  is identifiable up to  $\mathcal{S}_{\mathcal{M}}\coloneqq \{\varphi :\mathbb{R}^n\to \mathbb{R}^n |\varphi \in \mathcal{S}_{\mathrm{scaling}},\forall i\in [d],\varphi_i\in \mathcal{S}_{\mathcal{M}_i}\}$  where  $\mathcal{S}_{\mathcal{M}_i}\coloneqq \{\bar{F}_{\mathbb{M}_i'}^{-1}\circ F_{\mathbb{M}_i}|\mathbb{M}_i,\mathbb{M}_i'\in \mathcal{M}_i\}$
In particular, if  $\mathcal{M}_i$  is singleton for all  $i$ , then  $\text{CauCA in } (G, \mathcal{F}, \mathcal{P}_G)$  is identifiable up to  $\mathcal{S}_{\text{reflexion}} := \{\mathbf{h} : \mathbb{R}^d \to \mathbb{R}^d | \forall i \in [d], h_i = Id \text{ or } -Id\}$ .
Proof. Based on the conclusion of (2) of Thm. 4.2, for any  $i \in [d]$ , choose any  $\mathbb{M}_i$ ,  $\mathbb{M}_i'$  in  $\mathcal{M}_i$ ,
$$
\left(\varphi_ {i}\right) _ {*} \mathbb {M} _ {i} = \mathbb {M} _ {i} ^ {\prime}.
$$
By Lemma B.2, the only possible  $\varphi_{i}$  are  $F_{\mathbb{M}_i'}^{-1} \circ F_{\mathbb{M}_i}$  and  $\bar{F}_{\mathbb{M}_i'}^{-1} \circ F_{\mathbb{M}_i}$ . Thus  $\varphi_{i} \in S_{\mathcal{M}_{i}}$ . In particular if  $\mathcal{M}_i$  is a singleton  $\{\mathbb{M}_i\}$ , then  $F_{\mathbb{M}_i}^{-1} \circ F_{\mathbb{M}_i} = \mathrm{Id}$ , and  $\bar{F}_{\mathbb{M}_i}^{-1} \circ F_{\mathbb{M}_i} = -\mathrm{Id}$ .
B.8 Proof of Corollary 4.8
Proof. This corollary is a special case of Thm. 4.5 when  $G$  has no arrow. Since  $G$  has no arrow, all the blocks of interventions are in the first case of block-interventional discrepancy in Thm. 4.5: for all  $k \in [K]$ ,  $|\tau_k| = n_k$ . To prove all the blocks satisfy the block-interventional discrepancy, it suffices to prove that the linearly independent vectors in the statement form the matrix  $\mathbf{M}_{\tau_k}$ . To see this, it suffices to notice that for all  $s \in [[0, n_k]]$ ,  $\ln q_{\tau_k}^s (\mathbf{z}_{\tau_k}) = \sum_{j \in \tau_k} \ln q_{\tau_k,j}^s (z_{\tau_k,j})$ , and therefore  $\frac{\partial}{\partial z_j} (\ln q_{\tau_k,j}^s)(\mathbf{z}_{\tau_k}) = (\ln q_{\tau_k,j}^s)'(z_{\tau_k,j})$ .
B.9 Proof of Prop. 4.9
where  $q_{\tau_k}^s$  is the intervention of the  $s$ -th interventional regime that has the target  $\tau_k$ , and  $q_{\tau_{k,j}}^s$  is the  $j$ -th marginal of it.  $z_{\tau_k,j}$  is the  $j$ -th dimension of  $\mathbf{z}_{\tau_k}$ .  $s = 0$  denotes the unintervened regime. Then  $\varphi_{\tau_k} \in S_{\text{reparam}} := \left\{\mathbf{g} : \mathbb{R}^{n_k} \to \mathbb{R}^{n_k} \mid \mathbf{g} = \mathbf{P} \circ \mathbf{h} \text{ where } \mathbf{P} \text{ is a permutation matrix and } \mathbf{h} \in S_{\text{scaling}}\right\}$ .
The proof is based on Theorem 1 of [21].
Proof. Since the intervention targets are known, without loss of generality, suppose the interventions are on the first  $n_k$  variables. By the result of Thm. 4.5 we have  $p_k^s(\mathbf{z}_{\tau_k}) =$
$q_{k}^{s}\left(\varphi_{\tau_{k}}\left(\mathbf{z}_{\tau_{k}}\right)\right) \mid \operatorname*{det} D\varphi_{\tau_{k}}\left(\mathbf{z}_{\tau_{k}}\right)\mid$  where  $p_k^s$  denotes the joint distribution in the  $s$ -th interventional regime such that the intervention target is  $\tau_{k}$ . Factorize  $p_k^s$  and  $q_{k}^{s}$  in the change of variables formula, and take the logarithm:
$$
\sum_ {l = 1} ^ {n _ {k}} \ln p _ {k, l} ^ {s} (z _ {l}) = \sum_ {l = 1} ^ {n _ {k}} \ln q _ {k, l} ^ {s} (\varphi_ {l} (\mathbf {z} _ {\tau_ {k}})) + \ln | \det  D \boldsymbol {\varphi} _ {\tau_ {k}} (\mathbf {z} _ {\tau_ {k}}) |, \tag {30}
$$
where  $p_l^s$  is the  $l$ -th marginal of the intervention of the  $s$ -th interventional regime that has the target  $\tau_k$ .
For the unintervened regime, denote the density of the  $l$ -th marginal of  $\mathbb{P}$  as  $p_l$ . By the result of Thm. 4.5 we have
$$
\sum_ {l = 1} ^ {n _ {k}} \ln p _ {l} ^ {0} \left(z _ {l}\right) = \sum_ {l = 1} ^ {n _ {k}} \ln q _ {l} ^ {0} \left(\varphi_ {l} \left(\mathbf {z} _ {\tau_ {k}}\right)\right) + \ln | \det  D \boldsymbol {\varphi} _ {\tau_ {k}} \left(\mathbf {z} _ {\tau_ {k}}\right) |. \tag {31}
$$
Subtract the equation (30) by (31):
$$
\sum_ {l = 1} ^ {n _ {k}} \left[ \ln p _ {k, l} ^ {s} \left(z _ {l}\right) - \ln p _ {l} ^ {0} \left(z _ {l}\right) \right] = \sum_ {l = 1} ^ {n _ {k}} \left[ \ln q _ {k, l} ^ {s} \left(\varphi_ {l} \left(\mathbf {z} _ {\tau_ {k}}\right)\right) - \ln q _ {l} ^ {0} \left(\varphi_ {l} \left(\mathbf {z} _ {\tau_ {k}}\right)\right) \right]. \tag {32}
$$
For any  $j\in \tau_{k} = [n_{k}]$  , take the partial derivative over  $z_{j}$
$$
\frac {p _ {k , j} ^ {s ^ {\prime}} \left(z _ {j}\right)}{p _ {k , j} ^ {s} \left(z _ {j}\right)} - \frac {p _ {j} ^ {0 ^ {\prime}} \left(z _ {j}\right)}{p _ {j} ^ {0} \left(z _ {j}\right)} = \sum_ {l = 1} ^ {n _ {k}} \left[ \frac {q _ {k , l} ^ {s ^ {\prime}} \left(\varphi_ {l} \left(\mathbf {z} _ {\tau_ {k}}\right)\right)}{q _ {k , l} ^ {s} \left(\varphi_ {l} \left(\mathbf {z} _ {\tau_ {k}}\right)\right)} - \frac {q _ {k , l} ^ {0 ^ {\prime}} \left(\varphi_ {l} \left(\mathbf {z} _ {\tau_ {k}}\right)\right)}{q _ {k , l} ^ {0} \left(\varphi_ {l} \left(\mathbf {z} _ {\tau_ {k}}\right)\right)} \right] \frac {\partial \varphi_ {l}}{\partial z _ {j}} \left(\mathbf {z} _ {\tau_ {k}}\right). \tag {33}
$$
For any  $1 \leqslant k < j$ , take the partial derivative over  $z_k$ ,