SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Fig. 4 shows the full results. Given  $\tau$ , the found alteration can satisfy the probability requirement in most cases. Exceptions are settings where  $\tau = 0.9$ . Alterations with such a high success probability rarely exist, so the method can fail. In our implementation, when the recommendation is not attainable, the constraint is dropped and the method will only optimize the mutual information. Therefore the corresponding number of success rounds, as well as the success probability, is low, but the mutual information is high. The third column shows a trade-off between avoiding the undesired future and learning the SRM: smaller  $\tau$  is likely to give higher mutual information. An interesting phenomenon is that the true success probability often fluctuates around  $\tau$  instead of achieving higher values. An explanation is that if an alteration has a high success probability, then the effect of the alteration is less uncertain, so less new information can be revealed and the method will instead choose others with lower success probabilities. Finally, the last column shows that the true probabilities are bounded by the PAC-style bound with a relatively small gap, so it can be informative for guiding real decisions.
Realizing that correlation is inadequate and that causation is not always suitable, we advocate relying on the influence relation for decision-making. In this paper, we propose the first rehearsal learning framework that tackles the AUF (Avoiding Undesired Future) problem with influence modeling. The key to the proposed framework is SRM, which considers the interrelations of variables and the dynamic time-dependent nature of decision environments. By unifying rehearsals on SRMs and Bayesian inference, the framework recommends decisions that can persuasively avoid the undesired future. We also provide a PAC-style bound to quantify the associated risk of recommended decisions and further show the hardness of a basic linear instance of the framework, revealing that future work should focus on developing approximate solutions rather than exact ones.
This research was supported by National Key R&D Program of China (2022ZD0114800) and NSFC (61921006). The authors thank the reviewers for their valuable comments.
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A Rehearsal Graphs: Characterization and Learning
In this section, we present full motivations, basic properties, and preliminary learning strategies for rehearsal graphs.
The ability to describe and identify causal relations is at the heart of both human and artificial intelligence. The work of Pearl [12] on causality lays the foundation for accurately describing causal relations and making a sound inference. Despite its elegance and expressiveness in abstracting causality from human cognition, one may encounter difficulties when applying Pearl's causal framework to real-world observations. We identify three kinds of misspecifications that could make causal modeling fail: a) model misspecification b) data-collection misspecification c) adversary misspecification.
Model misspecification means that the variables used in the causal modeling process are inaccurate so there are indeterminant causal relations among the variables. For example, demand  $(D)$  and price  $(P)$  are two variables that are causally related. But there is not a determinant causal direction between  $D$  and  $P$  since intervening on  $D$  causes  $P$  to change but meanwhile intervening on  $P$  causes  $D$  to change. A workaround is modeling with finer granularity by adding a time index to  $D$  and  $P$ . Then we have  $D_{t}$  causes  $P_{t + 1}$  and  $P_{t}$  causes  $D_{t + 1}$ .
Data-collection misspecification means that the data collected are not perfectly aligned with the specified variable. For example, we may have variables  $D_{1}, D_{2}, P_{1}, P_{2}$  in the causal model that denote two measures  $D$  and  $P$  at different time. But the data collected may not rigorously follow the order since there can be time delays and mismatches of the modeling time intervals and real ones. This misspecification results in a misalignment between the model and data, hence no sound causal conclusions can be drawn from the data.
Adversary misspecification refers to ignoring the existence of adversaries in causal modeling. For example, let  $P$  and  $Q$  denote the prices of a product in two stores. If  $P$  is decreased then  $Q$  will also be decreased, otherwise, the second store will lose some consumers. From a causal view, it seems that  $P$  causes  $Q$ . But obviously,  $Q$  also causes  $P$  by symmetry. So we have a bi-directional causal relation, which is cannot be described in causal modeling. The reason for this phenomenon is that each store is an adversary that reacts to interventions from other agents.
The preceding three misspecifications are not necessarily disjoint in real problems and are hard to identify in advance. The consequence is that one cannot identify perfect causal knowledge even if with an infinite amount of data since the modeling and the data do not exhibit consistent determinant causal relations. In order to model both the non-determinant and determinant causal relations in data, we introduce a new probabilistic graphical model: rehearsal graphs.
A.1Rehearsal Graphs as a Probabilistic Graphical Model
Definition 8 (Mixed graph). Let  $G = (\mathbf{V}, \mathbf{E})$  be a graph, where  $\mathbf{V}$  denotes the vertices and  $\mathbf{E}$  the edges.  $G$  is a mixed graph if for any distinct vertices  $u, v \in \mathbf{V}$ , there is at most one edge connecting them, and the edge can be either directional or bi-directional, i.e.,  $u \rightarrow v$ ,  $u \leftarrow v$ , or  $u \leftrightarrow v$ .
Definition 9 (Bi-directional clique). A bi-directional clique  $C = (\mathbf{V}^c, \mathbf{E}^c)$  of a mixed graph  $G = (\mathbf{V}, \mathbf{E})$  is a complete induced subgraph such that any edge  $e \in \mathbf{E}^c$  is bi-directional. A single vertex also constitutes a valid bi-directional clique.  $C$  is maximal if adding any other vertex does not induce a bi-directional clique.
Definition 10 (Rehearsal graph). Let  $G = (\mathbf{V}, \mathbf{E})$  be a mixed graph. Let  $\{C_i\}_{i=1}^l$  denote all maximal bi-directional cliques of  $G$ , where  $C_i = (\mathbf{V}_i^c, \mathbf{E}_i^c)$ .  $G$  is a rehearsal graph if and only if:
1.  $\mathbf{V}_i^c\cap \mathbf{V}_j^c = \emptyset$  for any  $i\neq j$
2.  $\forall i\in [l],u\in \mathbf{V}\setminus \mathbf{V}_i^c$  , if there is any edge pointing from  $u$  to  $\mathbf{V}_i^c$  , then  $\forall v\in \mathbf{V}_i^c,u\to v$
3. The directional edges permit a topological ordering for  $\{C_i\}_{i = 1}^l$
Definition 11 (Factorization). Let  $G = (\mathbf{V}, \mathbf{E})$  be a rehearsal graph and  $C_i = (\mathbf{V}_i^c, \mathbf{E}_i^c), i \in [k]$  be maximal bi-directional cliques of  $G$ . We say that a joint probability  $P$  over  $\mathbf{V}$  factorizes according to  $G$  if  $P$  can be expressed as a product
$$
P (\mathbf {V}) = \prod_ {i = 1} ^ {l} P \left(\mathbf {V} _ {i} ^ {c} \mid \mathbf {P A} _ {i}\right), \tag {10}
$$
where  $\mathbf{PA}_i = \{u\mid \exists v\in \mathbf{V}_i^c,u\to v$  in  $G\}$
A.2  $c$ -Separation
As a probabilistic graphical model, like Bayesian networks or Markov networks, rehearsal graphs also have a graphical characterization of independence relations. Mimicking the  $d$ -separation for Bayesian networks, we present  $c$ -separation for rehearsal graphs. Here the "  $c$  " means "cliques".
Definition 12 (c-separation). A path  $p$  in a rehearsal graph  $G$  is said to be  $c$ -separated (or blocked) by a set of vertices  $Z$  if and only if
1.  $p$  contains a chain  $i \to m_1 \leftrightarrow m_2 \leftrightarrow \dots \leftrightarrow m_l \to j$  or a fork  $i \leftarrow m_1 \leftrightarrow m_2 \leftrightarrow \dots \leftrightarrow m_l \to j$  such that the bi-directionally connected vertices  $\{m_i\}_{i=1}^l \subseteq Z$ , or
2.  $p$  contains an inverted fork (or collider)  $i \to m_1 \leftrightarrow m_2 \leftrightarrow \dots \leftrightarrow m_l \gets j$  such that no vertices in  $\{m_i\}_{i=1}^l$  are in  $Z$  and such that no descendants of  $\{m_i\}_{i=1}^l$  are in  $Z$ .
A set  $Z$  is said to  $c$ -separate  $X$  from  $Y$  if and only if  $Z$  blocks every path from a vertex in  $X$  to a vertex in  $Z$ . We write  $X \perp_{G} Y \mid Z$  when  $Z$ $c$ -separates  $X$  from  $Y$  in  $G$ .
Lemma 13. Let  $G = (\mathbf{V}, \mathbf{E})$  be a rehearsal graph,  $x, y \in \mathbf{V}$  be two distinct vertices, and  $Z$  a set of vertices. Let  $C(v)$  denote the vertices of the maximal bi-directional clique to which  $v$  belongs. If  $x \perp_{G} y \mid Z$ , then  $C(x) \perp_{G} C(y) \mid Z$ .
Proof. We prove this by contradiction. Assume that  $C(x) \perp_{G} C(y) \mid Z$  does not hold given  $x \perp_{G} y \mid Z$ , meaning that there exists a path  $p$  unblocked by  $Z$  between some  $u \in C(x)$  and some  $v \in C(y)$ . Then there is an unblocked path  $x \leftrightarrow p \leftrightarrow y$  between  $x$  and  $y$ , which is a contradiction since  $Z$  is supposed to block every path between  $x$  and  $y$ .
Lemma 14. Let  $G = (\mathbf{V}, \mathbf{E})$  be a rehearsal graph,  $C_a, C_b \subseteq V$  be a set of vertices of two distinct maximal bi-directional cliques, and  $Z$  a set of vertices. If  $C_a \perp_{G} C_b \mid Z$ , then for all probability functions  $P$  that factorizes according to  $G, C_a \perp_{P} C_b \mid Z$
Proof. We construct a DAG  $\tilde{G} = (\tilde{V},\tilde{E})$  from  $G$  with an one-to-one mapping  $f$ : every maximal cliques  $C_i$  of  $G$  is represented with a vertex  $\tilde{v}_i\in \tilde{G}$ ,  $\tilde{v}_i = (v_i^1,\dots,v_i^{|C_i|})$ , and the edge directions between  $C_i$ s are kept. For any path  $p$ $\tilde{v}_{\alpha_1} - \tilde{v}_{\alpha_2} - \dots -\tilde{v}_{\alpha_l}$  (here we omit the edge directions) in  $\tilde{G}$ , where  $\{\alpha_{i}\}_{i}$  is an index set, there is a counterpart  $q$  in the original  $G$  that reads  $v_{\alpha_1}^1\leftrightarrow v_{\alpha_1}^2\leftrightarrow \dots \leftrightarrow v_{\alpha_1}^{|C_{\alpha_1}|} - v_{\alpha_2}^1\leftrightarrow \dots \leftrightarrow v_{\alpha_2}^{|C_{\alpha_2}|} - \dots -v_{\alpha_l}^1\leftrightarrow v_{\alpha_l}^2\leftrightarrow \dots \leftrightarrow v_{\alpha_l}^{|C_{\alpha_l}|}$ , where  $v_{\alpha_k}^j$  is the  $j$ -th vertex in  $C_{\alpha_k}$ . For any  $Z\subseteq V$ , we map it to a set of vertices in  $\tilde{G}$  with  $h(Z) = \{\tilde{v}\in \tilde{V}\mid f^{-1}(\tilde{v})\subseteq Z\}$ . From the definition of  $c$ -separation for rehearsal graphs and  $d$ -separation for Bayesian networks, we see that if  $q$  is  $c$ -separated by  $Z$  in  $G$ , then  $p$  is  $d$ -separated by  $h(Z)$  in  $\tilde{G}$ . Therefore, if  $C_a\perp_L C_b\mid Z$ , then every path between  $\tilde{v}_a$  and  $\tilde{v}_b$  is  $d$ -separated by  $h(Z)$ . Due to the Markov property of Bayesian networks, for all  $P$  that factorizes according to  $\tilde{G}$  (which also factorizes according to  $G$  by definition), we have  $\tilde{v}_a\perp_P\tilde{v}_b\mid h(Z)$ , i.e.,  $C_a\perp_P C_b\mid f^{-1}(h(Z))$ . Since  $f^{-1}(h(Z))\subseteq Z$ , we have  $C_a\perp_P C_b\mid Z$ .
Theorem 15. Let  $G = (\mathbf{V}, \mathbf{E})$  be a rehearsal graph. For all probability functions  $P$  that factorizes according to  $G$  and any three disjoint subsets of vertices  $X, Y, Z \subseteq \mathbf{V}$ , if  $X \perp_{G} Y \mid Z$ , then  $X \perp_{P} Y \mid Z$ .
$$
\begin{array}{l} X \perp_ {G} Y \mid Z \Rightarrow \forall x \in X, y \in Y, x \perp_ {G} y \mid Z (11) \\ \xrightarrow {\text {L e m .} 1 3} \quad \forall x \in X, y \in Y, C (x) \perp_ {G} C (y) \mid Z (12) \\ \xrightarrow {\text {L e m .} 1 4} \quad \forall x \in X, y \in Y, C (x) \perp_ {P} C (y) \mid Z (13) \\ \Rightarrow \quad \forall x \in X, y \in Y, x \perp_ {P} y \mid Z (14) \\ \Rightarrow X \perp_ {P} Y \mid Z. (15) \\ \end{array}
$$
(a) A rehearsal graph  $G$ .
(d)  $G_3^{\prime}$
(b)  $G_1^\prime$
(c)  $G_2^\prime$
(f)  $G_5^{\prime}$
(e)  $G_4^{\prime}$
(g)  $G_6^{\prime}$
Figure 5: An example rehearsal graph and its Bayesian network counterparts that encode the same conditional independence relations. We use plate notation to simplify the edges. For example, in (a), vertex A having an edge pointing to the rectangle containing three vertices, implies that  $\mathrm{A}\rightarrow \mathrm{E}$ $\mathrm{A}\to \mathrm{F}$  and  $\mathrm{A}\to \mathrm{G}$ .
Theorem 16. Let  $G = (\mathbf{V}, \mathbf{E})$  be a rehearsal graph. Let  $G' = (\mathbf{V}, \mathbf{E}')$  be any DAG that has identical skeleton with  $G$  (see Fig. 5 for an example). If all directional edges in  $G$  appear in  $G'$ , then for any disjoint set of vertices  $X, Y$  and  $Z$ , we have  $X \perp_{G} Y \mid Z$  if and only if  $X \perp_{G'} Y \mid Z$ , where  $\perp_{G}$  means c-separation for rehearsal graphs and  $\perp_{G'}$  d-separation for DAGs.
Proof. We first prove  $X \perp_{G} Y \mid Z \Rightarrow X \perp_{G'} Y \mid Z$ .
Every path  $p$  in  $G$  has a unique counterpart  $p'$  in  $G'$  that shares the same vertex sequence.  $p$  and  $p'$  only differ in some edges: a bi-directional edge  $u \leftrightarrow v$  in  $G$  is a directed one  $u \rightarrow v$  or  $u \leftarrow v$  in  $G'$ . Obviously, the mapping between  $p$  and  $p'$  is one-to-one. We only need to show that for any path  $p'$  in  $G'$ , if its counterpart  $p$  in  $G$  is  $c$ -separated by a set  $Z$ , then  $p'$  is  $d$ -separated in  $G$ .
We prove it by contradiction. Assume that  $p'$  is not  $d$ -separated by  $Z$ , then according to the definition of  $d$ -separation, we have that (a) for every chain  $i \to m \to j$  and every fork  $i \leftrightarrow m \leftrightarrow j$  in  $p'$ ,  $m \notin Z$ , and that (b) for every inverted fork  $i \to m \leftrightarrow j$  in  $p'$ ,  $(\{m\} \cup des(m)) \cap Z \neq \emptyset$ . For every chain  $i \to m_1 \leftrightarrow m_2 \leftrightarrow \dots \leftrightarrow m_l \to j$  and every fork  $i \leftrightarrow m_1 \leftrightarrow m_2 \leftrightarrow \dots \leftrightarrow m_l \to j$  in  $p$ , we have  $\{m_i\}_{i=1}^l \nsubseteq Z$  since otherwise either  $* \leftrightarrow m_l \to j$  in  $p'$  and  $m_l \in Z$ , or  $* \to m_l \to j$  in  $p'$  and  $m_l \in Z$ , which violate requirement (a). For every inverted fork  $i \to m_1 \leftrightarrow m_2 \leftrightarrow \dots \leftrightarrow m_l \leftrightarrow j$  in  $p$ ,  $(\{m_i\}_{i=1}^l \cup des(\{m_i\}_{i=1}^l)) \cap Z \neq \emptyset$  since otherwise there exists an inverted fork  $* \to m_t \leftrightarrow *, t \in [l]$  such that  $(\{m_t\} \cup des(m_t)) \cap Z = \emptyset$ , which violates requirement (b). Combining the above results, we have that  $p$  is not  $c$ -separated by  $Z$ , which is a contradiction. Therefore,  $p'$  is  $d$ -separated by  $Z$ , which concludes that  $X \perp_{G} Y | Z \Rightarrow X \perp_{G'} Y | Z$ .
We next prove  $X \perp_{G'} Y \mid Z \Rightarrow X \perp_{G} Y \mid Z$ .
Again we prove by contradiction. Assume that there exists a path  $p$  from  $x \in X$  to  $y \in Y$  in  $G$  that is not  $c$ -separated by  $Z$ . Let  $p$  be  $x - m_1^1 \leftrightarrow m_1^2 \leftrightarrow \dots \leftrightarrow m_1^{l_1} - m_2^1 \leftrightarrow m_2^2 \leftrightarrow \dots \leftrightarrow m_2^{l_2} - \dots - m_t^1 \leftrightarrow m_t^2 \leftrightarrow \dots \leftrightarrow m_t^{l_t} - y$  (we omit the direction of directional edges). We construct a path  $q$  from  $x$  to  $y$  by picking one vertex from each bi-directionally connected subpath  $m_i^1 \leftrightarrow m_i^2 \leftrightarrow \dots \leftrightarrow m_i^{l_i}$  in order such that  $q$  contains only directional edges, so  $q$  is also a path in  $G'$ . For each bi-directionally connected subpath  $m_i^1 \leftrightarrow m_i^2 \leftrightarrow \dots \leftrightarrow m_i^{l_i}$  in  $p$ , if it is a part of a chain  $* \rightarrow m_i^1 \leftrightarrow m_i^2 \leftrightarrow \dots \leftrightarrow m_i^{l_i} \rightarrow *$  or a fork  $* \leftrightarrow m_i^1 \leftrightarrow m_i^2 \leftrightarrow \dots \leftrightarrow m_i^{l_i} \rightarrow *$ , then by the definition of  $c$ -separation,  $\{m_i^j\}_{j=1}^{l_i} \setminus Z \neq \emptyset$ , so we randomly select a vertex  $v_i \in \{m_i^j\}_{j=1}^{l_i} \setminus Z$  to join  $q$ . If the bi-directionally connected subpath is a part of an inverted fork  $* \rightarrow m_i^1 \leftrightarrow m_i^2 \leftrightarrow \dots \leftrightarrow m_i^{l_i} \leftarrow *$ , then by the definition of  $c$ -separation, either some  $m_i^j \in Z$  or some descendant of  $\{m_i^j\}_{j=1}^{l_i}$  is in  $Z$ . In the former case, we select  $v_i = m_i^j$  to join  $q$ , and in the latter, we randomly select any  $v_i \in \{m_i^j\}_{j=1}^{l_i}$ . By construction,  $q = x - v_1 - \dots - v_t - y$  is a path from  $X$  to  $Y$  in  $G'$  and is not  $d$ -separated, which is a contradiction. Therefore, any path  $p$  from  $X$  to  $Y$  in  $G$  is  $c$ -separated by  $Z$ , which concludes  $X \perp_{G'} Y | Z \Rightarrow X \perp_{G} Y | Z$ .
Theorem 17. Let  $G = (\mathbf{V}, \mathbf{E})$  be a rehearsal graph. For any three disjoint subsets of vertices  $X, Y, Z \subseteq V$ , if  $X \perp_{P} Y \mid Z$  for all probability functions  $P$  that factorizes according to  $G$ , then  $X \perp_{G} Y \mid Z$ .
Proof. Let  $G' = (\mathbf{V}, \mathbf{E}')$  be any DAG that has identical skeleton with  $G$  and has all directional edges in  $G$ . Let  $V_i = \{v_j\}_{j=1}^l$  denote the set of vertices in a maximal bi-directional clique in  $G$  and  $PA_i$  the parent of  $V_i$ . Then  $V_i$  is also a clique in  $G'$ . Without loss of generality, assume the topological ordering among  $V_i$  is  $v_1 \prec v_2 \prec \dots \prec v_l$ . By the chain rule, we have  $P(V_i \mid PA_i) = \prod_{k=1}^l P(v_k \mid v_{k-1}, v_{k-2}, \ldots)$ , where the LHS is the conditional probability defined in the factorization of  $G$  and the RHS is the factorization defined by  $G'$ . So if  $P$  is a probability function that factorizes according to  $G$ , then  $p$  also factorizes according to  $G'$ , and vice versa. Let  $\mathcal{P}(\cdot)$  denote the probability family that factorizes according to  $\cdot$ . We have  $\mathcal{P}(G) = \mathcal{P}(G')$ .
$$
\begin{array}{l} \forall P \in \mathcal {P} (G), X \perp_ {P} Y \mid Z \quad \xrightarrow {\mathcal {P} (G) = \mathcal {P} \left(G ^ {\prime}\right)} \quad \forall P \in \mathcal {P} \left(G ^ {\prime}\right), X \perp_ {P} Y \mid Z (16) \\ \xrightarrow {\text {c o m p l e t e n e s s o f} d \text {- s e p a r a t i o n}} X \perp_ {G ^ {\prime}} Y \mid Z (17) \\ \xrightarrow {\text {T h m .} 1 6} X \perp_ {G} Y \mid Z. (18) \\ \end{array}
$$
A.3 Structure Learning
If we drop the bi-directional edges in a rehearsal graph, it becomes a valid Bayesian network, which means that Bayesian networks are special cases of rehearsal graphs. So learning rehearsal graphs can only be harder than learning Bayesian networks. In this subsection, we present a preliminary learning method. More efficient learning methods are left for further research.