Bridging Ranking Theory and the Stability Theory of Belief

Abstract

In this paper we compare Leitgeb’s stability theory of belief (Annals of Pure and Applied Logic, 164:1338-1389, 2013; The Philosophical Review, 123:131-171, [2014]) and Spohn’s ranking-theoretic account of belief (Spohn, 1988, 2012). We discuss the two theories as solutions to the lottery paradox. To compare the two theories, we introduce a novel translation between ranking (mass) functions and probability (mass) functions. We draw some crucial consequences from this translation, in particular a new probabilistic belief notion. Based on this, we explore the logical relation between the two belief theories, showing that models of Leitgeb’s theory correspond to certain models of Spohn’s theory. The reverse is not true (or holds only under special constraints on the parameter of the translation). Finally, we discuss how these results raise new questions in belief theory. In particular, we raise the question whether stability (a key ingredient of Leitgeb’s theory) is rightly thought of as a property pertaining to belief (rather than to knowledge).

Appendix A: Proofs

1.1 A.1 Prerequisites

Lemma 5 (cf. lemma 2)

Let P be a probability over a finite algebra $\mathcal {A}=\wp ({\Omega })$ . If P(A)≠1, then the following are equivalent:

• 1. $A \in \mathcal {A}$ is P-stable,

• 2. $\forall w \in A, P(w) > P(\overline {A})$,

• 3. $\min _{A} P(w) > P(\overline {A})$,

• 4. For all non-empty $D \subseteq A$ and all $E \subseteq \overline {A}$ , P(D)>P(E).

If P(A)=1 these equivalences hold, with > replaced by ≥.

Proof

First assume P(A)≠1.

$(1 \Rightarrow 2)$ :

Let $A \in \mathcal {A}$ be P-stable. If A = ∅ then (2) is vacuously satisfied. Suppose A≠∅. Let w∈A and $B = \{w\} \cup \overline {A}$, so that $B \cap A= \{w\} \neq \emptyset$. P(B)=0 is impossible, since then P(w)=0 and $P(\overline {A})= 0$, contradicting P(A)≠1. Therefore P(B)>0. Thus the stability condition applies and $P(A|B)> \frac {1}{2}$. Therefore the following are equivalent:

$$ P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{P(w)}{P(w) + P(\overline{A})} > \frac{1}{2} $$

(A.1)

$$ P(w) > \frac{1}{2}(P(w) + P(\overline{A})) $$

(A.2)

$$ \frac{1}{2}P(w) > \frac{1}{2}P(\overline{A}) $$

(A.3)

$$ P(w) > P(\overline{A}) $$

(A.4)

$(2 \Rightarrow 3)$ :

If Eq. A.4 holds for all w∈A then it holds for w∈A s.t. P(w) is minimal.

$(3 \Rightarrow 4)$ :

Let $p_{A} = \min _{A} P(w) > P(\overline {A})$. Consider $\emptyset \neq D \subseteq A$ and $E \subseteq \overline {A}$. By sub-addivity and minimality:

$$ P(D) \geq p_{A} > P(\overline{A}) \geq P(E) $$

(A.5)

$(4 \Rightarrow 1)$ :

If A = ∅, then A is vacuously P-stable. Let us therefore consider A≠∅. Let $B \in \mathcal {A}$, such that $B \cap A \neq \emptyset$ and P(B)>0. Define $D = B \cap A$ and E = B D. We have $\emptyset \neq D \subseteq A$ and $E \subseteq \overline {A}$. So that (4) applies and we have the following equivalent expressions

$$ P(D) > P(E) $$

(A.6)

$$ \frac{1}{2} P(D) > \frac{1}{2} P(E) $$

(A.7)

$$ P(D) - \frac{1}{2} P(D) > \frac{1}{2} P(E) $$

(A.8)

$$ P(D) > \frac{1}{2} (P(E) +P(D)) $$

(A.9)

$$ \frac{P(D)}{P(D) + P(E)} > \frac{1}{2} $$

(A.10)

$$ P(A|B) > \frac{1}{2} $$

(A.11)

If P(A)=1, then in $(1 \Rightarrow 2)$ P(B)=0 is possible, yielding $P(w) \geq P(\overline {A})=0$ for w∈A. This weak inequality transfers to (3) and (4). $(4 \Rightarrow 1)$ since P(D)≥P(E)=0 (4). Thus $P(A|B) = \frac {P(D)}{P(D) + P(E)} =1> \frac {1}{2}$. □

Proof Proof 2 of Theorem 1

Assume $\mathcal {A}=\wp ({\Omega })$ (finite), B a belief operator (with core C) and P a probability function over $\mathcal {A}$.

$(1) \Rightarrow (2)$ :

Suppose B,P satisfy the Lockean thesis, for t = P(C).

• i To derive a contradiction, suppose that C is not P-stable. Therefore there is v∈C such that $P(v) \leq P(\overline {C}) = 1- P(C)$ (Lemma 5.2 or even $P(v) < P(\overline {C})=0$ if P(C)=1 creating an immediate contradiction). Consider $B=\overline {C} \cup (C \setminus \{v\})$ which is not a superset of C. Therefore it is not believed. Yet, P(B)=1−P(v)≥P(C) = t, contradicting the Lockean thesis.

• ii To derive a contradiction, suppose P(C)=1, and that there is $D \subsetneq C$ with P(D)=1. Then the Lockean thesis implies B(D), therefore C is not the core.

$(2) \Rightarrow (1)$.:

Suppose (i) C is P-stable and (ii) if P(C)=1 then C is the smallest probability-1 set. Then $t:=P(C) = P(C|{\Omega })>\frac {1}{2}$ (since C≠∅). Let us show that (i,ii) imply the Lockean thesis. $C \subseteq A$ implies P(A)≥t (sub-additivity). Therefore B(A) implies P(A)≥t. Let us show the reverse.

Suppose there is B such that ¬B(B) and P(B)≥t (contradicting the Lockean thesis). Then $C \nsubseteq B$. There are three cases: (1) $B \subsetneq C$, (2) $B \cap C = \emptyset$ or (3) $\emptyset \neq B \cap C \notin \{C,B\}$. All lead to a contradiction. If (1) $B \subsetneq C$ then, by sub-additivity P(B)≤P(C) = t. P(B)<P(C) would contradict P(B)≥t. P(B) = P(C) would contradict ¬B(B). (2) is excluded because if $P(B)\geq t > \frac {1}{2}$, P(C) = t and $B \cap C = \emptyset$, then $P(B \cup C) \geq 2t >1$. Consider (3). Let $D= C \cap B \subseteq C$ and $E = B \setminus D \subseteq \overline {C}$. By construction and hypothesis P(B) = P(D) + P(E)≥t = P(C). Define $B^{\prime } = D \cup \overline {C}$. $B^{\prime }\supseteq B$, therefore $P(B^{\prime }) \geq P(B)$. Take a superset $D^{\prime }$ of D, such that $D^{\prime } = C \setminus \{w\}$ with w∈C (which exists because $D=B \cap C \neq C$ by (3). Define $B^{\prime \prime } = D^{\prime } \cup \overline {C}$, therefore $B^{\prime } \subseteq B^{\prime \prime }$. Then $P(C) =t\leq P(B) \leq P(B^{\prime }) \leq P(B^{\prime \prime }) = 1 - P(w)$. Which implies $P(w) \leq P(\overline {C})$. This contradicts (i) P-stability of C (Lemma 5.2) if P(C)≠1. If P(C)=1, then $P(w) = P(\overline {C})=0$ could be possible, but then C is not the smallest probability 1 set, contradicting (ii).

□

Proof Proof 3 of Lemma 3

Let a∈(0,1).

• 1. Let κ(w) be a ranking mass. Then $P^{a}_{\kappa }(w)$ is a probability mass: Its sum is 1 by the normalisation constant, and all $P^{a}_{\kappa }(w) \geq 0$. Since 0≤a ^κ(w) for a∈(0,1).

• 2. Let P(w) be a probability mass. Then ${\kappa ^{a}_{P}}(w)$ is a ranking mass: Let $p=\max _{v \in {\Omega }} P(v)$ (which exists, since Ω is finite). Set r _w = P(w)/p for w∈Ω. Then r _w ∈[0,1]. Thus $\ln r_{w} \in [-\infty , 0]$. Yet, for a∈(0,1), we have $\ln a \in (-\infty , 0)$. Therefore ${\kappa ^{a}_{P}}(w) = \log _{a} r_{w} = \frac {\ln r_{w}}{\ln a} \geq 0$. ${\kappa ^{a}_{P}}$ has a zero for w∈Ω such that P(w) = p. Since then r _w =1 and $\ln r_{w} = \ln 1 =0$.

• 3. Because Ψ _a Φ _a (κ) = κ and Φ _a Ψ _a (P) = P.

  $$ \kappa^{a}_{P^{a}_{\kappa}}(w) = \log_{a} \frac{P^{a}_{\kappa}(w)}{\max_{\Omega} P^{a}_{\kappa}(v)} = \log_{a} \frac{a^{\kappa(w)}}{\max_{\Omega} a^{\kappa(v)}} = \log_{a} \frac{a^{\kappa(w)}}{a^{0}} = \kappa(w) $$

  (A.12)
  $$P^{a}_{{\kappa^{a}_{P}}}(w) = \frac{a^{{\kappa^{a}_{P}}(w)}}{Z} =\frac{a^{\log_{a} \frac{P(w)}{\max_{\Omega} P(v)}}}{Z} =\frac{ P(w)}{\max_{\Omega} P(v) {\sum}_{u \in {\Omega}} \frac{P(u)}{\max_{\Omega} P(v)}} = P(w) $$

  (A.13)

• 4. For a∈(0,1), a ^x and $\log _{a} y$ are strictly decreasing.

□

1.2 A.2 Belief Models

Proof Proof 4 of Theorem 2

Let a∈(0,1).

• 1. Let (B,κ) be a Spohn belief. Since κ is a ranking function, $P={P^{a}_{K}}$ is a probability function for any a∈(0,1) (Lemma 3.1). It therefore suffices to show that [≥y] _P is the core of B for y: = a ^z/Z(a). Note that κ(w)≤z iff P(w)≥y (Lemma 3.4). Therefore for all $A \in \mathcal {A}$:

  $$\begin{array}{rcl} A \in \mathcal{B}& \text{ iff} & \kappa(\overline{A})>z\\ &&\\ & \text{ iff} & [\leq z]_{\kappa} \subseteq A\\ &&\\ & \text{ iff} & [\geq y]_{P} \subseteq A\\ \end{array} $$

  Therefore $\mathcal {B}:=\{A \in \mathcal {A}: \mathbf {B}(A)\}$ is generated by [≥y] _P (non-empty). Thus $\mathcal {B}$ is a principal filter. Yet, a principal filter is generated by a unique element (its core). Therefore [≤y] _P is the core.

• 2. By the fact that ${\Psi }_{a} = {\Phi }^{-1}_{a}$ (Lemma 3.3).

□

Proof Proof 5 of Theorem 3

Assume (B,P) is an atomic probabilistic belief model over a finite algebra $\mathcal {A} = \wp ({\Omega })$. Therefore P is a probability and B is a belief operator over $\mathcal {A}$. Additionally B has a unique core, namely C=[≥y] _P for some y∈[0,1].

• (⇒)  Suppose (B,P) is a Leitgeb model. In particular, (B,P) satisfy the Lockean thesis, for t = P(C) with C the core. Therefore, by Theorem 1 ($1 \Rightarrow 2$), (i) C is P-stable and (ii) if P(C)=1 then C is the smallest probability-1 set. Set $y= \min _{C} P(w)$. Then $y > P(\overline {C})$ by Lemma 5.2 and P-stability of C, if P(C)≠1. If P(C)=1, then C is the smallest probability 1 set and therefore $y>0 = P(\overline {C})$.

• (⇐)  Suppose that the core C of the atomic belief model (B,P) satisfies $y>P(\overline {C})$, for y the atomic threshold. It suffices to show that the Lockean thesis is satisfied for t _C = P(C). Let t _C = P(C). Then it suffices to show that q<t _C for $q = \max \{P(A): \neg \mathbf {B}(A)\}$. Because we may then always chose t with q<t<t _C such that the Lockean thesis is satisfied. By finiteness q exists. A non-believed element with maximal probability will have the form $A = \overline {C} \cup C \setminus \{w\}$, where w∈C has minimal probability in C. Yet, by the separation assumption $P(w) \geq y > P(\overline {C})$. This implies $0 > P(\overline {C}) - P(w)$ which implies $P(C) > P(\overline {C}) + P(C) - P(w)$, which implies t _C = P(C)>P(A) = q.

□

Proof Proof 6 of Corollary 1

Let a∈(0,1).

• 1. In a Leitgeb belief model, B is a belief operator. B is closed under conjunction by assumption and B satisfies (B1-3) by the Lockean thesis. Additionally, the core C is P-stable and if P(C)=1, then it is the smallest probability 1 set. Therefore $\min _{C} P(w) > P(\overline {C})$. Therefore $y=\min _{C} P(w)$ can be chosen as atomic threshold. Therefore a Leitgeb belief model is always an atomic belief model. Thus the claim follows from Theorems 2.2 and 3.

• 2. Let (B,κ) be a Spohn belief model with core C and $z = \max _{C} \kappa (w)$ and $x = \min _{\overline {C}} \kappa (w)$. Let a∈(0,1). Then $(\mathbf {B}, P^{a}_{\kappa })$ is an atomic belief model, with

  $$ y(a,z) = \frac{a^{z}}{Z(a)} $$

  (A.14)

  by Theorem 2.1. This model is a Leitgeb model iff $y(a,z)>P(\overline {C})$ (Theorem 3). This holds iff $a^{z} > Z(\overline {C}) :={\sum }_{\overline {C}} a^{\kappa (w)}$. It therefore suffices to show this later inequality. Assume

  $$ a >|\overline{C}|^{\frac{1}{z-x}} = \left( e^{\ln |\overline{C}|}\right)^{1/(z-x)} = e^{\frac{\ln |\overline{C}|}{z-x}} $$

  (A.15)

  The following conditions are equivalent:

  $$\ln a> \frac{ \ln |\overline{C}|}{(z-x)} $$

  (A.16)
  $$(z-x)\ln a> \ln |\overline{C}| $$

  (A.17)
  $$a^{z-x}> |\overline{C}| $$

  (A.18)
  $$ a^{z} > |\overline{C}| a^{x} = \sum\limits_{\overline{C}} a^{x} $$

  (A.19)

  Therefore $a^{z} > {\sum_{\overline {C}} a^{x} \geq Z(\overline {C})}$, since $x = \min _{\overline {C}} \kappa (w)$.

□