Aggregating individual credences into collective binary beliefs: an impossibility result

Abstract

This paper addresses how multiple individual credences on logically related issues should be aggregated into collective binary beliefs. We call this binarizing belief aggregation. It is vulnerable to dilemmas such as the discursive dilemma or the lottery paradox: proposition-wise independent aggregation can generate inconsistent or not deductively closed collective judgments. Addressing this challenge using the familiar axiomatic approach, we introduce general conditions on a binarizing belief aggregation rule, including rationality conditions on individual inputs and collective outputs, and determine which rules (if any) satisfy different combinations of these conditions. Furthermore, we analyze similarities and differences between our proofs and other related proofs in the literature and conclude that the problem of binarizing belief aggregation is a free-standing aggregation problem not reducible to judgment aggregation or probabilistic opinion pooling.

Appendix — Proofs

Lemma 1

Let W be a non-empty set, ${\mathcal {A}} \subseteq {\mathcal {P}}(W)$ be an agenda, and $Bel:{\mathcal {A}} \rightarrow \{0,1\}$ be a binary belief function. Then, (i) Bel is complete and consistent iff Bel is extendable to a finitely additive 0/1-valued probability function on the algebra $\mathfrak {a}({\mathcal {A}})$ generated by ${\mathcal {A}}$. (ii) Bel is complete and $\sigma$-consistent iff Bel is extendable to a $\sigma$-additive 0/1-valued probability function on the $\sigma$-algebra $\sigma ({\mathcal {A}})$ generated by ${\mathcal {A}}$.

Proof

Proof of part (i)

($\rightarrow$) It is well-known that any $A \in \mathfrak {a}({\mathcal {A}})$ has the form of $\bigcup _{i=1}^{n}\bigcap _{j_i=1}^{m_i}E_{ij_i}^{\circ }$ where $E_{ij_i}^{\circ }$ is $E_{ij_i}$($\in {\mathcal {A}}$) or $E_{ij_i}^c$. Since the agenda ${\mathcal {A}}$ is closed under complement, we let $A=\bigcup _{i=1}^{n}\bigcap _{j_i=1}^{m_i} E_{ij_i}$. Define $Bel^{*}:\mathfrak {a}({\mathcal {A}}) \rightarrow \{0,1\}$ as

$$\begin{aligned} Bel^{*}\left( \bigcup _{i=1}^{n}\bigcap _{j_i=1}^{m_i}E_{ij_i}\right) :=\max _{i=1,...,n}\min _{j_i=1,..., m_i} Bel(E_{ij_i}) \end{aligned}$$

• (a) $Bel^*$ is well defined: Assume $\bigcup _{i=1}^{n}\bigcap _{j_i=1}^{m_i}E_{ij_i} = \bigcup _{k=1}^{l}\bigcap _{q_k=1}^{r_k}F_{kq_k}$. Then $(\bigcup _{i=1}^{n}\bigcap _{j_i=1}^{m_i}$ $E_{ij_i})^c \cap (\bigcup _{k=1}^{l}\bigcap _{q_k=1}^{r_k}F_{kq_k}) = \emptyset$. Since $(\bigcup _{i=1}^{n}\bigcap _{j_i=1}^{m_i}E_{ij_i})^c = \bigcup _{(j_i)_i \in J} \bigcap _{i=1}^{n}E_{ij_i}^c$ where $J=\{1,..., m_1 \} \times ... \times \{1,..., m_n \}$, we have

  $$\begin{aligned} \bigcup _{(j_i)_i \in J} \bigcup _{k=1}^{l} \bigcap _{i=1}^{n} \bigcap _{q_k=1}^{r_k} (E_{ij_i}^c \cap F_{kq_k} ) = \emptyset \end{aligned}$$

  (1)

  Now suppose $Bel^*(\bigcup _{i=1}^{n}\bigcap _{j_i=1}^{m_i}E_{ij_i})=0$ and $Bel^*(\bigcup _{k=1}^{l}\bigcap _{q_k=1}^{r_k}F_{kq_k})=1$. Then for all $i \in \{1,..., n\}$ we can pick a $j_i \in \{1,..., m_i \}$ such that $Bel(E_{ij_i}^c)=1$. Note that this tuple $(j_1,...,j_n)$ is contained in J defined above. Moreover, there exists $k \in \{1,..., l\}$ such that for all $q_k \in \{1,..., r_k\}$, $Bel(F_{kq_k})=1$. Since Bel is consistent, $\bigcap _{i=1}^{n} \bigcap _{q_k=1}^{r_k} ( E_{ij_i}^c \cap F_{kq_k} ) \ne \emptyset$, which contradicts Equation (1). Similarly, we can prove that it is not the case that $Bel^*(\bigcup _{i=1}^{n}\bigcap _{j_i=1}^{m_i}E_{ij_i})=1$ and $Bel^*(\bigcup _{k=1}^{l}\bigcap _{q_k=1}^{r_k}F_{kq_k})=0$. Therefore, $Bel^*(\bigcup _{i=1}^{n}\bigcap _{j_i=1}^{m_i}E_{ij_i})= Bel^*(\bigcup _{k=1}^{l}\bigcap _{q_k=1}^{r_k}F_{kq_k})$.

• (b) By construction, $Bel^*(A)=Bel(A)$ for all $A \in {\mathcal {A}}$.

• (c) $Bel^*$ is a finitely additive probability function: $Bel^*(W) = Bel(E \cup E^c) = max \{ Bel(E)$ $, Bel(E^c)\}=1$ for some $E\in {\mathcal {A}}$($\ne \emptyset$) since Bel is complete. Let $A=\bigcup _{i=1}^{n} A_i$ where $A_i=\bigcap _{j_i=1}^{m_i}E_{ij_i}$ and $B=\bigcup _{k=1}^{l} B_k$ where $B_k=\bigcap _{q_k=1}^{r_k}E_{kq_k}$ and assume that A and B is disjoint. Then we have

  $$\begin{aligned} Bel^{*}(A \cup B)&= Bel^{*}\left( \bigcup _{i=1}^{n} \bigcap _{j_i=1}^{m_i}E_{ij_i} \cup \bigcup _{k=1}^{l} \bigcap _{q_k=1}^{r_k}F_{kq_k}\right) \\&= \max \{Bel^{*}(A_1), ..., Bel^{*}(A_n), Bel^{*}(B_1), ..., Bel^{*}(B_l) \} \\&= \max \{Bel^{*}(A), Bel^{*}(B) \} = Bel^{*}(A) + Bel^{*}(B) \end{aligned}$$

  The last equality follows from $Bel^{*}(A)=0$ or $Bel^{*}(B)=0$. This holds, for otherwise there would exist $i \in \{1,...,n\}$ such that for all $j_i \in \{1,...,m_i\}$ $Bel(E_{ij_i})=1$ and also $k \in \{1,...,l\}$ such that for all $q_k \in \{1,...,r_k\}$ $Bel(F_{kq_k})=1$, from which it follows, by consistency of Bel, that $\bigcap _{j_i=1}^{m_i}E_{ij_i} \cap \bigcap _{q_k=1}^{r_k}F_{kq_k} \ne \emptyset$, a contradiction to $A \cap B = \emptyset$.

($\leftarrow$) Let P be a finitely additive 0/1-valued probability function extending Bel. (a) P is complete and so is Bel since $P(A)=1$ or $P(A)=0$, i.e., $P(A^c)=1$. (b) Let $P(A_1),...,P(A_k)=1$. Then $P(\bigcap _{i=1}^{k} A_i) =1$ and hence $\bigcap _{i=1}^{k} A_i \ne \emptyset$. So P is consistent and so is Bel.

Proof of part (ii)

($\rightarrow$) (a) If Bel is complete and $\sigma$-consistent, then the extension $Bel^*$ on $\mathfrak {a}({\mathcal {A}})$ defined in the proof of part (i)(a) is also complete and $\sigma$-consistent: completeness of $Bel^*$ follows from the finite-additivity of $Bel^*$ in the proof of part (i)(c). For $\sigma$-consistency, assume $Bel^*(A_n)=1$ for all $n \in {\mathbb {N}}$. Let $A_n = \bigcup _{i=1}^{l(n)} \bigcap _{j=1}^{m_i} E_{ij}$. Then there exists $i(n) \in \{1,...,l(n)\}$ such that $Bel( E_{i(n)j} )=1$ for all $j \in \{1,...,m_{i(n)}\}$. Note that $A_n \supseteq \bigcap _{j=1}^{m_{i(n)}} E_{i(n)j}$ and thus $\bigcap _{n \in {\mathbb {N}}} A_n \supseteq \bigcap _{n \in {\mathbb {N}}} \bigcap _{j=1}^{m_{i(n)}} E_{i(n)j}$. Since Bel is $\sigma$-consistent, $\bigcap _{n \in {\mathbb {N}}} \bigcap _{j=1}^{m_{i(n)}} E_{i(n)j}$ is not empty and neither is $\bigcap _{n \in {\mathbb {N}} } A_n$. Thus $Bel^*$ is $\sigma$-consistent. (b) $Bel^*$ is $\sigma$-additive: since it follows from completeness and $\sigma$-consistency of $Bel^*$ that $Bel^*(\bigcup_{n \in {\mathbb {N}}}A_n)=0$ iff $Bel^*(A_n^c)=1$ for all $n\in {\mathbb {N}}$, where $A_n$s are pairwise disjoint, we obtain $Bel^*(\bigcup_{n \in {\mathbb {N}}}A_n)= \sum _{n \in {\mathbb {N}}} Bel^*(A_n)$. (c) Combining this with Carathéodory extension theorem yields the claim. ($\leftarrow$) It can be shown similarly to the proof of ($\leftarrow$) in part (i). $\square$

Lemma 2

In binarizing belief aggregation and judgment aggregation, the following holds:

• (1) Given CCS, CP implies ZP.

• (2) Given CCP, ZP implies CP.

• (3) Let $W, \emptyset \in {\mathcal {A}}$. Then, $F(\vec {P})(W)=1$ by CDC or CP, and $F(\vec {P})(\emptyset )=0$ by CCS or ZP.

• (4) Let $\emptyset \in {\mathcal {A}}$ and F satisfy CDC and CP. Then, F satisfies ZP iff F satisfies CCS.

Proof

• (1) Assume $\vec {P}(A)=\vec {0}$, which is equivalent to $\vec {P}(A^c)=\vec {1}$ since $\vec {P}$ is a profile of probabilistic beliefs. By CP, we have $F(\vec {P})(A^c)=1$, from which it follows that $F(\vec {P})(A)=0$ by CCS.

• (2) Assume $\vec {P}(A)=\vec {1}$, which is equivalent to $\vec {P}(A^c)=\vec {0}$ since $\vec {P}$ is a profile of probabilistic beliefs. By ZP, we have $F(\vec {P})(A^c)=0$, from which it follows that $F(\vec {P})(A)=1$ by CCP.

• (3) As noted already, CDC and CCS imply $F(\vec {P})(W)=1$ and $F(\vec {P})(\emptyset )=0$, respectively. The rest parts hold since $\vec {P}(W)=\vec {1}$ and $\vec {P}(\emptyset )=\vec {0}$.

• (4) ($\leftarrow$) ZP follows from CCS and CP by part (1). ($\rightarrow$) When $\emptyset \in {\mathcal {A}}$, from ZP it follows that $F(\vec {P})(\emptyset )=0$ by part (3), which is equivalent to CCS when we have CDC.

$\square$

Lemma 3

(Contagion Lemma) Let ${\mathcal {A}}$ be a non-trivial algebra and F be a BA with UD. If F satisfies CDC, ZP, CP, and IND, then it satisfies SYS.

Proof

By IND, we can let $F(\vec {P})(A)=G_A(\vec {P}(A))$ for all $\vec {P}$. We need to show that $G_A=G_{B}$ for all $A, B \in {\mathcal {A}}$.

(Case 1) $\emptyset \ne A \subseteq B \ne W$

By UD, $\vec {P}$ given as in Fig. 4 can be an argument of F: since there exist possible worlds in A and $B^c$, which are represented by dots in Fig. 4, due to UD, we can assign the probabilities $\vec {a}$ and $\vec {1}-\vec {a}$ to A and $B^c$, respectively. Then B has the probabilities $\vec {a}$. This gives the following:

• (i) If $G_A(\vec {a})= F(\vec {P})(A)=1$, then $F(\vec {P})(B)=G_B(\vec {a})=1$ by CDC (closure under superset).

• (ii) Since $F(\vec {P})( A \cup B^c )=1$ by CP, if $G_B(\vec {a})= F(\vec {P})(B)=1$, then $F(\vec {P})(A)=G_A(\vec {a})=1$ by CDC (closure under intersection) since $(A \cup B^c) \cap B = A$.

(Case 2) $A {\setminus } B \ne \emptyset$ and $B {\setminus } A \ne \emptyset$

Let $v \in A {\setminus } B$ and $w \in B {\setminus } A$ as in Fig. 5. We can use $C \in {\mathcal {A}}$ such that $\{v,w\}\subseteq C\ne W$ since ${\mathcal {A}}$ is a non-trivial algebra. (Take the union of two elements in ${\mathcal {A}}$ one of which includes v and one of which includes w. We can find such two elements whose union is not W because ${\mathcal {A}}$ is non-trivial: if there were no such two elements, it means that $(A-B) \cup (B-A) = W$, hence ${\mathcal {A}}=\{ \emptyset , A,B, W \}$ where $B=A^c$, which contradicts the non-triviality of ${\mathcal {A}}$.) By the result of (Case 1), we have $G_A = G_{A \cap C}=G_C= G_{C \cap B}=G_B$.

(Case 3) We can let $G_{\emptyset } = G_A$ and $G_W = G_A$, for any $A(\ne \emptyset , W) \in {\mathcal {A}}$ since $G_A(\vec {0})=0$ by ZP and $G_A(\vec {1})=1$ by CP. $\square$

Fig. 4

A dot $\bullet$ represents a possible world

Fig. 5

The triangles $\scriptstyle \Delta$ indicate all possible locations at one of which a possible world is ensured to exist so that $C \ne W$

Theorem 4

(Triviality Result) Let ${\mathcal {A}}$ be a non-trivial algebra. The only BA satisfying UD, ZP, CP, IND, CDC, and AN is the unanimity rule.

Proof

It is easily seen that the unanimity rule satisfies all mentioned properties. For the other direction, by Lemma 3 we have SYS and thus, we can let $F(\vec {P})(A) = G(\vec {P}(A))$ where $G(\vec {1})=1$ by CDC (in particular $F(\vec {P})(W)=1$) or CP. Now suppose that $G(\vec {a})=1$ for some $\vec {a}\ne \vec {1}$ and pick up any $a_i \ne 1$ in $\vec {a}$. To derive a contradiction, we take the following three steps.

[Step 1] We show the following:

(Fact 1) if $\vec {a} \le \vec {b}$ and if $G(\vec {a})=1$, then $G(\vec {b})=1$

(Fact 2) if $\vec {a}+\vec {b}-\vec {1} \ge \vec {0}$ and if $G(\vec {a})=1$ and $G(\vec {b})=1$, then $G(\vec {a}+\vec {b}-\vec {1})=1$

Since ${\mathcal {A}}$ is non-trivial, we have at least 3 non-empty elements of ${\mathcal {A}}$ that have no intersections with each other. We represent each possible world of such elements by a dot in Fig. 6. Since ${\mathcal {A}}$ is an algebra, there are A and B in ${\mathcal {A}}$ as in the left figure and A, B and $A \cap B$ as in the right figure. By UD, we can assign to A and B, respectively, the probabilities $\vec {a}$ and $\vec {b}$ such that $\vec {a} \le \vec {b}$, as in the left figure. In the right figure, by UD we can assign to A, B and $A \cap B$, respectively, the probabilities $\vec {a}$, $\vec {b}$, and $\vec {a}+\vec {b}-\vec {1}$ where $\vec {a}+\vec {b}-\vec {1} \ge \vec {0}$. The left figure gives us (Fact 1) by closure under superset from CDC and the right figure gives us (Fact 2) by closure under intersection from CDC.

[Step 2] We show that $G(\vec {a}[a_i \mapsto 0, a_l \mapsto 1 \text { for all } l \ne i])=1.$

By (Fact 1), we can substitute $a_i$ and $a_l$ with any higher value keeping the value of G as 1 and by mixed applications of (Fact 1) and (Fact 2), we can substitute $a_i$ with any lower value using the fact that for any $\vec {a} \ge (0.5,...,0.5)$,

$$\begin{aligned} \text{ if } G(\vec {a})=1 \text{ then } G(\vec {a}+ \vec {a} - 1)=1 \end{aligned}$$

(2)

as follows. By (Fact 1), we can substitute all other components $a_l$ (i.e., $l \ne i$) that are not 1 with 1 and so we have $G((1,...,1, a_i, 1,...., 1))=1$. This process enables us to focus on the i-th component of vectors when we apply (2) because $1+1-1=1$. Now employ (Fact 1) and (2). (i) If $a_i \le 0.5$, we have $G(1,...,1, 0.5, 1,...,1)=1$ by (Fact 1) and $G(1,...,1, 0, 1,...,1)=1$ by (2). (ii) Now let $a_i > 0.5$. After applying (2) to $G((1,...,1, a_i, 1,...., 1))=1$ k times, we have $G((1,...,1, a_i^{(k)}, 1,...., 1))=1$ where $a_i^{(k)}=1-2^k(1-a_i)$ and there must be k such that $a_i^{(k)} \le 0.5$. Then, we can use (i) and obtain $G(1,...,1, 0, 1,...,1)=1$.

[Step 3] We show, by induction, $G((0,...,0))=1$, which contradicts ZP.

We have $G((0,1,...,1))=G((1,0,1,...,1))=...=G((1,...,1,0)$ by UD, AN and (Step 2). Let $\vec {a}_k$ be a vector (0, ..., 0, 1, ..., 1) where the first k components are 0 and the others are 1. For $k=1$, we have $G(\vec {a}_1)=1$. Assume $G(\vec {a}_k)=1$. Since $G((1,...,1,0,1,...,1))=1$ where all components are 1 except for the (k+1)-th one, which is 0, we have $G(\vec {a}_{k+1})=1$ by (Fact 2). $\square$

Fig. 6

The left one is for (Fact 1) and the right one is for (Fact 2)

Theorem 5

(Oligarchy Result) Let ${\mathcal {A}}$ be a non-trivial algebra. The only BAs satisfying UD, ZP, CP, IND, and CDC are the oligarchies.

Proof

It is obvious that an oligarchy satisfies the properties. For the other direction, to construct the set M of oligarchs in Definition 11, we employ [Step 1] and [Step 2] in the proof of Theorem 4. (By UD, ZP, CP, IND and CDC, we have SYS by Lemma 3, and [Step 1] and [Step 2] follow from UD and CDC. Note that in the proof of Theorem 4, we did not use AN except in [Step 3].) Consider the set $G^{-1}(1):=\{ \vec {a} \vert \ G(\vec {a})=1 \}$ where G is a function satisfying $F(\vec {P})(A) = G(\vec {P}(A))$. We collect individuals i such that $a_i=1$ for all $\vec {a} \in G^{-1}(1)$ and define the set M of such individuals: $M:=\{ i \in N \vert \ a_i=1 \text{ for } \text{ all } \vec {a} \text{ such } \text{ that } G(\vec {a})=1\}$. We will show (i) and (ii) in the following.

• (i) M is non-empty.

  Suppose M is empty. Then $G(0,1,...,1)=G(1,0,1,...,1)=...=G(1,...,1,0)=1$ by [Step 1] and [Step 2], and we have $G(0,...,0)= 1$ using the same way of [Step 3], which contradicts ZP.

• (ii) $a_i=1$ for all $i \in M$ iff $G(\vec {a})=1$.

  ($\leftarrow$) It is obvious by the construction of M. ($\rightarrow$) Since we have (Fact 1) in [Step 1], it is enough to show that $G((\mathbbm {1}_M (i))_{i \in N})=1$ where $\mathbbm {1}_M (i) = 1$ if $i \in M$, otherwise $\mathbbm {1}_M (i) = 0$. For any $j \notin M$, there is $\vec {a}$ such that $G(\vec {a})=1$ and $a_j \ne 1$, by definition of M. By [Step 2],

  $$\begin{aligned} G(\vec {a}[a_j \mapsto 0, a_l \mapsto 1 \text{ for } \text{ all } l \ne j ])=1 \end{aligned}$$

  (3)

  Now, we proceed by induction analogously to [Step 3]. Enumerate individuals who are not in M, like $j_1, j_2,...,j_{\vert N \vert - \vert M \vert }$ and let $\vec {a}_k$ be a profile where $a_{j_1}=0,..., a_{j_k}=0$ and other components are all 1. For $k=1$, we have $G(\vec {a}_1)=1$. Assume $G(\vec {a}_k)=1$. Since by Equation (3) we have $G(1,...,1,0,1,...,1)=1$ where 0 is the $j_{k+1}$-th component, by (Fact 2) in [Step 1], we have $G(\vec {a}_{k+1})=1$. Therefore, we have $G(\vec {a}_{\vert N \vert - \vert M \vert })=G((\mathbbm {1}_M (i))_{i \in N})=1$

$\square$

Corollary 6

(Non-existence Result) Let ${\mathcal {A}}$ be a non-trivial algebra. There is no BA satisfying UD, CP, IND, CCP, and CCS.

Proof

Since CDC follows from CCP and CCS, and ZP follows from CCS and CP, the only possible BAs satisfying the above conditions would be the oligarchies by Theorem 5, which do not satisfy collective completeness. $\square$