Subjective causal networks and indeterminate suppositional credences

Abstract

This paper has two main parts. In the first part, we motivate a kind of indeterminate, suppositional credences by discussing the prospect for a subjective interpretation of a causal Bayesian network (CBN), an important tool for causal reasoning in artificial intelligence. A CBN consists of a causal graph and a collection of interventional probabilities. The subjective interpretation in question would take the causal graph in a CBN to represent the causal structure that is believed by an agent, and interventional probabilities in a CBN to represent suppositional credences. We review a difficulty noted in the literature with such an interpretation, and suggest that a natural way to address the challenge is to go for a generalization of CBN that allows indeterminate credences. In the second part, we develop a decision-theoretic foundation for such indeterminate suppositional credences, by generalizing a theory of coherent choice functions to accommodate some form of act-state dependence. The upshot is a decision-theoretic framework that is not only rich enough to, so to speak, ground the probabilities in a subjectively interpreted causal network, but also interesting in its own right, in that it accommodates both act-state dependence and imprecise probabilities.

Appendix

In this appendix, we prove Propositions 1–5. A proof of Proposition 1 and Proposition 2 was given in Zhang and Spirtes (2011). We include a proof here for the sake of Propositions 3 and 4.

Proof of Proposition 1

Suppose for the sake of contradiction that there exist $X, Y\in {\mathbf {V}}$ such that there is no arrow from X to Y in ${\mathcal {G}}$ (i.e., $X\notin \mathbf {Pa}_{\mathcal {G}}(Y)$), but there exist two distinct possible values, $x_1\ne x_2$, for X and a value configuration ${\mathbf {z}}$ for ${\mathbf {Z}} = {\mathbf {V}} - \{X, Y\}$ such that

$$\begin{aligned} P^{do(X=x_1\wedge {\mathbf {Z}}={\mathbf {z}})}(Y) \ne P^{do(X=x_2\wedge {\mathbf {Z}}={\mathbf {z}})}(Y). \end{aligned}$$

Since $X\notin \mathbf {Pa}_{\mathcal {G}}(Y)$, $\mathbf {Pa}_{\mathcal {G}}(Y)\subseteq {\mathbf {Z}}$. Then, since according to the intervention principle that defines the CBN, we have

$$\begin{aligned} P^{do(X=x_1\wedge {\mathbf {Z}}={\mathbf {z}})}(Y) = P^{do(\emptyset )}(Y | \mathbf {Pa}_{\mathcal {G}}(Y) = {\mathbf {s}}) = P^{do(X=x_2\wedge {\mathbf {Z}}={\mathbf {z}})}(Y), \end{aligned}$$

where ${\mathbf {s}}$ is the value configuration of $\mathbf {Pa}_{\mathcal {G}}(Y)$ that is consistent with ${\mathbf {Z}}={\mathbf {z}}$. (Recall that $P^{do(\emptyset )}$ is assumed to be positive, so $P^{do(\emptyset )}(Y | \mathbf {Pa}_{\mathcal {G}}(Y) = {\mathbf {s}})$ is defined.) Contradiction. $\square$

Proof of Proposition 2¹⁵

Proof

Since $({\mathcal {G}}, \{P^{do({\mathbf {S}}={\mathbf {s}})}\})$ is a CBN over ${\mathbf {V}}$, by Proposition 1, it contains ${\mathcal {G}}_I$ as a subgraph. If ${\mathcal {G}}={\mathcal {G}}_I$, the desired conclusion follows. Otherwise, ${\mathcal {G}}$ contains arrows that do not satisfy the condition in Proposition 1. Since ${\mathcal {G}}$ is finite, the number of such arrows is finite. Let these arrows be ordered as $E_1, ..., E_m$, and let ${\mathcal {G}}_j$ ($1\le j\le m$) denote the graph resulting from taking away the set of arrows $\{E_1, ..., E_j\}$ from ${\mathcal {G}}$. Obviously ${\mathcal {G}}_m = {\mathcal {G}}_I$. Let ${\mathcal {G}}_0 = {\mathcal {G}}$. We show by induction that for all $0\le j\le m$, $({\mathcal {G}}_j, \{P^{do({\mathbf {S}}={\mathbf {s}})}\})$ is a CBN over ${\mathbf {V}}$.

The base case is trivial. Suppose $({\mathcal {G}}_k, \{P^{do({\mathbf {S}}={\mathbf {s}})}\})$ is a CBN over ${\mathbf {V}}$, we now argue that $({\mathcal {G}}_{k+1}, \{P^{do({\mathbf {S}}={\mathbf {s}})}\})$ is also a CBN over ${\mathbf {V}}$. Note that the only difference between ${\mathcal {G}}_{k}$ and ${\mathcal {G}}_{k+1}$ is that the former, but not the latter, contains the arrow $E_{k+1}$. Let $X\rightarrow Y$ denote the arrow $E_{k+1}$. To show that $({\mathcal {G}}_{k+1}, \{P^{do({\mathbf {S}}={\mathbf {s}})}\})$ is a CBN, it suffices to show that for every $V\in {\mathbf {V}}$, $P^{do(\emptyset )}(V|\mathbf {Pa}_{{\mathcal {G}}_{k+1}}(V)) = P^{do(\emptyset )}(V|\mathbf {Pa}_{{\mathcal {G}}_k}(V))$, because $({\mathcal {G}}_k, \{P^{do({\mathbf {S}}={\mathbf {s}})}\})$ is a CBN by the inductive hypothesis. This is obviously the case for every $V\ne Y$, which has the exact same parents in both ${\mathcal {G}}_{k+1}$ and ${\mathcal {G}}_k$. Regarding Y, $\mathbf {Pa}_{{\mathcal {G}}_k}(Y) = \mathbf {Pa}_{{\mathcal {G}}_{k+1}}(Y)\cup \{X\}$. Since the arrow $X\rightarrow Y$ does not satisfy the condition in Proposition 1, it follows that for every value configuration ${\mathbf {s}}$ for ${\mathbf {S}}=\mathbf {Pa}_{{\mathcal {G}}_{k+1}}(Y)$, value configuration ${\mathbf {t}}$ for ${\mathbf {T}} = {\mathbf {V}}-\{X, Y\} - \mathbf {Pa}_{{\mathcal {G}}_{k+1}}(Y)$, and distinct values $x_1\ne x_2$ for X, we have

$$\begin{aligned} P^{do(X=x_1\wedge {\mathbf {S}}={\mathbf {s}}\wedge {\mathbf {T}}={\mathbf {t}})}(Y) = P^{do(X=x_2\wedge {\mathbf {S}}={\mathbf {s}}\wedge {\mathbf {T}}={\mathbf {t}})}(Y). \end{aligned}$$

On the other hand, since $({\mathcal {G}}_k, \{P^{do({\mathbf {S}}={\mathbf {s}})}\})$ is a CBN, for both $i=1, 2$ we have

$$\begin{aligned} P^{do(X=x_i\wedge {\mathbf {S}}={\mathbf {s}}\wedge {\mathbf {T}}={\mathbf {t}})}(Y) = P^{do(\emptyset )}(Y | X=x_i\wedge {\mathbf {S}}={\mathbf {s}}). \end{aligned}$$

because $\mathbf {Pa}_{{\mathcal {G}}_k}(Y) = {\mathbf {S}}\cup \{X\}$. Hence, for any distinct values $x_1, x_2$ for X and any value configuration ${\mathbf {s}}$ for $\mathbf {Pa}_{{\mathcal {G}}_{k+1}}(Y)$,

$$\begin{aligned} P^{do(\emptyset )}(Y | X=x_1\wedge \mathbf {Pa}_{{\mathcal {G}}_{k+1}}(Y)={\mathbf {s}}) = P^{do(\emptyset )}(Y | X=x_2\wedge \mathbf {Pa}_{{\mathcal {G}}_{k+1}}(Y)={\mathbf {s}}). \end{aligned}$$

Thus, $P^{do(\emptyset )}(Y|\mathbf {Pa}_{{\mathcal {G}}_k}(Y)) = P^{do(\emptyset )}(Y|\mathbf {Pa}_{{\mathcal {G}}_{k+1}}(Y)\cup \{X\}) = P^{do(\emptyset )}(Y|\mathbf {Pa}_{{\mathcal {G}}_{k+1}}(Y))$. Therefore, for every $V\in {\mathbf {V}}$, $P^{do(\emptyset )}(V|\mathbf {Pa}_{{\mathcal {G}}_{k+1}}(V)) = P^{do(\emptyset )}(V|\mathbf {Pa}_{{\mathcal {G}}_k}(V))$, which implies that $({\mathcal {G}}_{k+1}, \{P^{do({\mathbf {S}}={\mathbf {s}})}\})$ is also a CBN.

Therefore, $({\mathcal {G}}_j, \{P^{do({\mathbf {S}}={\mathbf {s}})}\})$ is a CBN over ${\mathbf {V}}$ for all $0\le j\le m$. The desired conclusion follows since ${\mathcal {G}}_m = {\mathcal {G}}_I$. $\square$

Proof of Proposition 3

By the definition of a CCN (Definition 2), for every $\{P^{do({\mathbf {S}}={\mathbf {s}})}\}\in \mathbf {CK}$, $({\mathcal {G}}, \{P^{do({\mathbf {S}}={\mathbf {s}})}\})$ is a CBN. Then the argument for Proposition 1 can be used here to establish the desired conclusion.

Proof of Proposition 4

Again, for every $\{P^{do({\mathbf {S}}={\mathbf {s}})}\}\in \mathbf {CK}$, $({\mathcal {G}}, \{P^{do({\mathbf {S}}={\mathbf {s}})}\})$ is a CBN. Therefore, if an arrow in ${\mathcal {G}}$ does not satisfy the condition in Proposition 3, then the argument for Proposition 2 can be applied for every $\{P^{do({\mathbf {S}}={\mathbf {s}})}\}\in \mathbf {CK}$, establishing that the DAG resulting from taking away the arrow from ${\mathcal {G}}$ still constitutes a CBN with every $\{P^{do({\mathbf {S}}={\mathbf {s}})}\}\in \mathbf {CK}$, and so still constitutes a CCN with $\mathbf {CK}$. $\square$

Proof of Proposition 5

The proof is a generalization of a simplified version of SSK’s argument for their uniqueness result (Seidenfeld et al. 2010, pp. 161–162). Consider an arbitrary vector $\langle P^1, \dots , P^m\rangle$ of distributions over $\Omega$. Each $P^j$ can itself be written as a probability vector with n coordinates. Let $p_*$ be the smallest positive coordinate in the m probability vectors, and let ${\underline{p}}=\frac{1}{2}p_*$.¹⁶ Let a denote the vNM lottery ${\underline{p}}B\oplus (1-{\underline{p}})W$. For each $k = 1,\dots , n$ and $j = 1, \dots , m$, define option $o_k^j$ (i.e., option $o_k$ with index j) as follows, where $p_k^j$ denotes the kth coordinate of the probability vector $P^j$, namely, $P^j(w_k)$—the probability mass $P^j$ assigns to state $w_k$:

$$\begin{aligned} o_k^j(\omega _i) = \left\{ \begin{array}{ll} B &{} \text {if } i = k \text { and } p^j_k = 0, \\ a &{} \text {if } i \ne k \text { and } p^j_k=0, \\ \frac{{\underline{p}}}{p^j_k}B \oplus \left( 1-\frac{{\underline{p}}}{p^j_k}\right) W &{} \text {if } i = k \text { and } p^j_k> 0, \\ W &{} \text {if } i \ne k \text { and } p^j_k > 0. \end{array} \right. \end{aligned}$$

Let $O_{\langle P^1, \dots , P^m\rangle } = \{o_k^j \mid k = 1, \dots , n; j=1, \dots , m \}\cup \{{\mathbf {a}}^1\}$, where ${\mathbf {a}}^1$ denotes the constant horse lottery (with index 1) that assigns to each state the vNM lottery a. Suppose C is a coherent choice function over ${\mathcal {H}}^\dag$ that is represented by a set of vectors of probability distributions ${\mathbf {Q}}$ over $\Omega$ and a (state-independent and index-independent) utility function u over the set of vNM lotteries. Without loss of generality, suppose $u(B) = 1$ and $u(W)=0$ (recall that a strict preference of B to W is assumed), so that $u(a) = {\underline{p}}$. We now show that $\langle P^1, \dots , P^m\rangle \in {\mathbf {Q}}$ if and only if ${\mathbf {a}}^1\in C(O _{\langle P^1, \dots , P^m\rangle })$.

(Only if) Suppose $\langle P^1, \dots , P^m\rangle \in {\mathbf {Q}}$. With respect to $\langle P^1, \dots , P^m\rangle$ and u, the expected utility of $\{{\mathbf {a}}^1\}$ is obviously $u(a) = {\underline{p}}$. Consider now any $o_k^j\in O_{\langle P^1, \dots , P^m\rangle }$, its expected utility with respect to $\langle P^1, \dots , P^m\rangle$ and u is given by

$$\begin{aligned} \sum _{i=1}^nP^j(w_i)u(O_k^j(w_i)) = \left\{ \begin{array}{ll} u(a) = {\underline{p}} &{} \text {if } p^j_k=0, \\ p^j_ku(O_k^j(w_k)) + (1-p^j_k)u(W) = {\underline{p}} &{} \text {if } p^j_k>0. \end{array}\right. \end{aligned}$$

Thus, every option in $O_{\langle P^1, \dots , P^m\rangle }$ has the same expected utility with respect to $\langle P^1, \dots , P^m\rangle$ and u. Since C is represented by ${\mathbf {Q}}$ and u, and $\langle P^1, ..., P^m\rangle \in {\mathbf {Q}}$, $C(O _{\langle P^1, \dots , P^m\rangle }) = O _{\langle P^1, \dots , P^m\rangle }$, and so ${\mathbf {a}}^1\in C(O _{\langle P^1, \dots , P^m\rangle })$.

(If) Suppose ${\mathbf {a}}^1\in C(O _{\langle P^1, \dots , P^m\rangle })$. Consider any vector $\langle Q^1, \dots , Q^m\rangle$ that is different from $\langle P^1, \dots , P^m\rangle$. Then there exists $1\le j\le m$ such that $P^j\ne Q^j$, which implies that there exists $1\le k\le n$ such that $P^j(w_k)=p^j_k < q^j_k=Q^j(w_k)$. Note also that by definition, ${\underline{p}}<1$. Then the expected utility of option $o_k^j$ with respect to $\langle Q^1, \dots , Q^m\rangle$ and u is given by

$$\begin{aligned} \sum _{i=1}^nQ^j(w_i)u(O_k^j(w_i)) = \left\{ \begin{array}{ll} q^j_ku(B) + (1-q^j_k)u(a)> {\underline{p}} &{} \text {if } p^j_k=0, \\ q^j_ku(O_k^j(w_k)) + (1-q^j_k)u(W) =\frac{q^j_k}{p^j_k}{\underline{p}}> {\underline{p}} &{} \text {if } p^j_k>0. \end{array}\right. \end{aligned}$$

whereas the expected utility of ${\mathbf {a}}^1$ remains ${\underline{p}}$ with respect to $\langle Q^1, \dots , Q^m\rangle$ and u. Therefore, for every $\langle Q^1, \dots , Q^m\rangle \ne \langle P^1, \dots , P^m\rangle$, ${\mathbf {a}}^1$ is not acceptable in $O _{\langle P^1, \dots , P^m\rangle }$ with respect to $\langle Q^1, \dots , Q^m\rangle$ and u. But by supposition, $\{{\mathbf {a}}^1\}\in C(O _{\langle P^1, \dots , P^m\rangle })$, which means that $\{{\mathbf {a}}^1\}$ is acceptable with respect to ${\mathbf {Q}}$ and u. It follows that $\langle P^1, \dots , P^m\rangle \in {\mathbf {Q}}$.

What we have shown is that whether an arbitrary vector of probability distributions is in the set that represents a given coherent choice function over ${\mathcal {H}}^\dag$ is fully determined by the choice function. Therefore, the set of vectors of probability distributions that represents a given coherence choice function is unique. $\square$