The joint aggregation of beliefs and degrees of belief

Abstract

The article proceeds upon the assumption that the beliefs and degrees of belief of rational agents satisfy a number of constraints, including: (1) consistency and deductive closure for belief sets, (2) conformity to the axioms of probability for degrees of belief, and (3) the Lockean Thesis concerning the relationship between belief and degree of belief. Assuming that the beliefs and degrees of belief of both individuals and collectives satisfy the preceding three constraints, I discuss what further constraints may be imposed on the aggregation of beliefs and degrees of belief. Some possibility and impossibility results are presented. The possibility results suggest that the three proposed rationality constraints are compatible with reasonable aggregation procedures for belief and degree of belief.

Appendices

Appendix A: Lin and Kelly’s Symmetric Camera Shutter Rule

Motivated by the fact that there is no consistent context independent version of the Lockean Thesis with a threshold r < 1, Lin and Kelly (2012) proposed what they call the symmetric camera shutter rule. Like the Lockean Thesis the rule of Lin and Kelly includes a parameter r (0 < r < 1) that reflects an agent’s degree of credulity. Given a particular value r, the symmetric camera shutter rule demands the following relation between belief an degree of belief, where σ(w_i) = p(w_i)/max_j p(w_j):

$$ b\left( {w_{i} } \right) = 0\;if\;and\;only\;if\;\sigma\left( {w_{i} } \right) \le 1 - r. $$

In line with this condition, we can say that 〈b, p〉 is shutter fit at r (abbreviated SF_r(b, p)) just in case for all i: b(w_i) = 0 if and only if σ(w_i) ≤ 1  − r.

It is straightforward to define analogues of Lockean Credulity and PLC that apply to the symmetric camera shutter rule:

Definition

〈b, p〉 is credulous shutter fit (abbreviated CSF(b, p)) if and only if (1) there exists an r (0 < r < 1): 〈b, p〉 is shutter fit at r, and (2) for all b′: if φ_b′ ⊂ φ_b, then for all s (0 < s < 1): 〈b′, p〉 is not shutter fit at s.

Preservation of Shutter Fit Credulity (PSFC):

For all B, P: if for all i: CSF(b_i, p_i), then CSF(b_B, p_P).

It is, then, straightforward to demonstrate the analogue of Theorem 3 (and similarly for Theorems 5, 6, and 7):

Theorem 10

For all F: if {F = F_bel, U_b, A_b, N_b} is consistent, then {F = F_bel, U_p, A_p, N_p, PSFC} is consistent.²¹

Beyond the preceding, it turns out that the symmetric camera shutter rule is ‘more stable’ than the Lockean Thesis, as demonstrated by the behavior of the following analogues of the instances of L_r:

Preservation of Shutter Fit at Level r (SF_r):

For all B and P: if for all i: 〈b_i, p_i〉 is shutter fit at r, then 〈b_B, p_P〉 is shutter fit at r.

Unlike instances of L_r (for all r in (0.5, 1)), instances of SF_r are consistent with {U_b, A_b, N_b, U_p, A_p, N_p}, and more generally:

Theorem 11

For all F, r in (0, 1): if {F = F_bel, U_b, A_b, N_b} is consistent, then {F = F_bel, U_p, A_p, N_p, SF_r} is consistent.

However, both PSFC, along with all instances of SF_r, are still incompatible with LW (in the presence of other plausible principles):

Theorem 12

{U_b, A_b, N_b, U_p, A_b, N_b, LW, PSFC} is inconsistent.²²

Theorem 13

For all r in (0, 1): {U_b, A_b, N_b, U_p, LW, SF_r} is inconsistent.

Proof

For all r in (0, 1), we show that {U_b, A_b, N_b, U_p, LW, SF_r} is inconsistent. Consider an arbitrary r in (0, 1). Take the least n such that ((1 − (³√(n)/n))/(n − 1))/(³√(n)/n) ≤ 1 − r. Let |Π| = n. Let B = 〈b₁, …, b_n〉, where, for all i, b_i has the value 1 in the ith position, and the value 0 in all other positions. Let P = 〈p₁, …, p_n〉, where p₁ has the value v in the first position, and the value (1 − v)/(n − 1) in all other positions, where v = ³√(n)/n. For all i > 1: let p_i have the value 1 in the ith position and the value 0 in all other positions. Notice that A_b and N_b imply that b_B is a series of 1 s, and so b_B(w₁) = 1. Next notice that for all i: 〈b_i, p_i〉 is shutter-fit at r. (In particular, for all i > 1: σ_p1(w_i) = (1 − (³√(n)/n))/(n − 1))/(³√(n)/n) ≤ 1 − r.) So SF_r implies that 〈b_B, p_B〉 is shutter-fit at r. For reasons that follow, this implies that c₁ > 1/n. Assume c₁ ≤ 1/n, then p_P(w₁) ≤ (³√(n)/n)(1/n), and there exists an i such that p_P(w_i) ≥ (1 − (³√(n)/n))/(n − 1))(1/n) + 1/n, so that σ_P(w₁) ≤ (³√(n)/n)(1/n)/((1 − (³√(n)/n))/(n − 1))(1/n) + 1/n). But for all n > 1: (³√(n)/n)(1/n)/((1 − (³√(n)/n))/(n − 1))(1/n) + 1/n) ≤ ((1 − (³√(n)/n))/(n − 1))/(³√(n)/n). Assume not. Then for some n > 1: ³√(n²)((n −2)/n) + ³√(n)((n + 1)/n) − n > 0, which is absurd. So σ_P(w₁) ≤ 1 − r. The latter implies that b_B(w₁) = 0, which contradicts b_B(w₁) = 1. So c₁ > 1/n. By similar reasoning concerning permutations of B and P, it follows that for all i: c_i > 1/n, which contradicts LW.□

PSFC, along with all instances of SF_r, are also incompatible with CL and IP (in the presence of reasonable principles):

Theorem 14

{U_b, A_b, N_b, Z_b, U_p, PSFC, CL} is inconsistent.

Theorem 15

For all r in (0, 1): {U_b, A_b, N_b, Z_b, U_p, SF_r, CL} is inconsistent.

Theorem 16

{U_b, A_b, N_b, Z_b, U_p, A_p, N_p, PSFC, IP} is inconsistent.

Theorem 17

For all r in (0, 1): {U_b, A_b, N_b, Z_b, U_p, A_p, N_p, SF_r, IP} is inconsistent.

The proofs of Theorems 8 and 9 are annotated to indicate the modifications required to prove Theorems 14, 15, 16, and 17.

Appendix B: Proofs

Fact 2.

If U_b, A_b, and S, then for all B, φ: b_B(φ) = 1 if and only if for all b_i in B: b_i(φ) = 1.

Proof

The right to left direction of the consequent follows from the consistency requirement on b_B. To establish the left to right direction, it is sufficient to show that for all n, m: if 0 ≤ m < n, then there exists some sets of belief functions B, such that $n_{1j}$ + … + $n_{nj}$ = n − m, and $n_{j}$ = 1. If there are such sets of belief functions, then we have, for all m, such that 0 ≤ m < n, a proposition ¬w_j, that is believed by m of n agents, but not by the collective. It is straightforward to show that all of the needed B exist. By U_b, we have for all n and m, such that 0 ≤ m < n, some B such that the columns of B are the set of permutations of m 0 s and n − m 1 s. For each such B, $n_{i}$ = 1, for all $n_{i}$ in b_B, given A_b and N_b. The preceding holds, since, for each such B, each $n_{i}$ must take the same value, given A_b and N_b, which must be 1, by the consistency requirement on b_B.□

Theorem 1

For all r: if 0.5 < r < 1, then {U_b, A _b, N_b, U_p, A_p, N_p, L_r} is inconsistent.

Proof

Notice that for all r: if 0.5 < r < 1, then there exists n and ε: 0 < ε < 1/n and r = ((n − 1)/n) + ε. In light of the preceding, consider the instances of the following schema, for all n and ε: 0 < ε < 1/n. Let Π = {w₁, …, w_n}. Let B_n = 〈b₁, …, b_n〉, where b_i has the value 0 in the ith position, and the value 1 in all other positions. Let P_n = 〈p₁, …, p_n〉, where p_i has the value (1/n) − ε in the ith position, and the value (1/n) + (ε/(n − 1)) in all other positions. In this case, A_b and N_b imply that b_Bn has the value 1 in every position, and A_p and N_p imply p_Pn has the value 1/n in every position. Notice that for all i: 〈b_i, p_i〉 is Lockean at ((n − 1)/n) + ε, but 〈b_Bn, p_Pn〉 is not Lockean at ((n − 1)/n) + ε.□

Theorem 3

For all F: if {F = F_bel, U_b, A_b, N_b} is consistent, then {F = F_bel, U_p, A_p, N_p, PLC} is consistent.

Proof

To demonstrate the result, we show how to define a function F_prob that satisfies U_p, A_p, N_p, and PLC, given any aggregation function F_bel that satisfies A_b and N_b. Given any P, we define B* = 〈b*₁,…, b*_n〉 to be the (unique) set of belief functions, such that for all i: CL(b*_i, p_i) [or such that CSF(b*_i, p_i) or SF_r(b*_i, p_i) for the proofs of Theorems 12 and 13, respectively]. We then define F_prob by $r_{j}$ = 1/($n^{*}_{1}$ +···+ $n^{*}_{k}$), if $n^{*}_{j}$ = 1, and $r_{j}$ = 0, if $n^{*}_{j}$ = 0 (where $n^{*}_{1}$ through $n^{*}_{k}$ are determined via F_bel, given U_b). [Notice that, in the preceding step, we require that |{ $n^{*}_{j}$ | $n^{*}_{j}$ = 1}| is finite, which holds if Π is finite, but also assuming Z_b, since all p-stable sets are finite (Leitgeb 2013, p. 1366).] It follows immediately from the definition of F_prob that the combination of F_bel and F_prob satisfy PLC [or PSFC or SF_r, respectively]. The inputs to F_prob are not restricted; so U_p is also satisfied. A_p requires that for all P and g: if g: {1, …, n} → {1, …, n} is a permutation, and P′ = 〈p_g(1), …, p_g(n)〉, then p_P′ = p_P. We assume for arbitrary P and g, that g: {1, …, n} → {1, …, n} is a permutation, and P′ = 〈p_g(1), …, p_g(n)〉. We show that p_P′ = p_P. To begin with, notice that P determines B*, and P′ determines B′*, so that b′*_i = b*_g(i). But A_b, so B′* = B*, and thus p_P′ = p_P, given the definition of F_prob. So A_p is satisfied. N_p requires that for all P: if f: {1, …, k} → {1, …, k} is a permutation, P′ = 〈p′₁, …, p′_n〉, and for all i: p′_i = 〈$r_{if\left( 1 \right)}$, …, $r_{if\left( k \right)}$〉, then p_P′ = 〈$r_{f\left( 1 \right)}$, …, $r_{f\left( k \right)}$〉. We assume for arbitrary P and f, that f: {1, …, k} → {1, …, k} is a permutation, P′ = 〈p′₁, …, p′_n〉, and for all i: p′_i = 〈$r_{if\left( 1 \right)}$, …, $r_{if\left( k \right)}$〉. We show that p_P′ = 〈$r_{f\left( 1 \right)}$, …, $r_{f\left( k \right)}$〉. To begin with, notice that P determines B*, and P′ determines B′*, so that b′*_i = b*_g(i). But A_b, so B′* = B*, and thus p_P′ = p_P, given the definition of F_prob. So A_p is satisfied. N_p requires that for all P: if f: {1, …, k} → {1, …, k} is a permutation, P′ = 〈p′₁, …, p′_n〉, and for all i: p′_i = 〈$r_{if\left( 1 \right)}$, …, $r_{if\left( k \right)}$〉, then p_P′ = 〈$r_{f\left( 1 \right)}$, …, $r_{f\left( k \right)}$〉. We assume for arbitrary P and f, that f: {1, …, k} → {1, …, k} is a permutation, P′ = 〈p′₁, …, p′_n〉, and for all i: p′_i = 〈$r_{if\left( 1 \right)}$, …, $r_{if\left( k \right)}$〉. We show that p_P′ = 〈$r_{f\left( 1 \right)}$, …, $r_{f\left( k \right)}$〉. To begin with, notice that P determines B*, and P′ determines B′*, such that b*_i = 〈$n_{i1}$, …, $n_{ik}$〉 and b′*_i = 〈$n_{if\left( 1 \right)}$, …, $n_{if\left( k \right)}$〉. But N_b, so p_B′* = 〈$n_{f\left( 1 \right)}$, …, $n_{f\left( k \right)}$〉. So p_P′ = 〈$r_{f\left( 1 \right)}$, …, $r_{f\left( k \right)}$〉, given F_prob. So N_p is satisfied.□

Theorem 4

{U_b, A_b, N_b, U_p, LW, PLC} is inconsistent.

Proof

Let Π = {w₁, w₂}. Consider B₁ = 〈〈1, 0〉, 〈0, 1〉〉, and P₁ = 〈〈0.51, 0.49〉, 〈0.01, 0.99〉〉. Next consider B₂ = 〈〈1, 0〉, 〈0, 1〉〉, and P₂ = 〈〈0.99, 0.01〉, 〈0.49, 0.51〉〉. Notice that A_b and N_b imply that b_B1 = b_B2 = 〈1, 1〉. Notice that PLC (and similarly PSFC) imply that p_P1 = 〈0.5, 0.5〉, which, according LW, holds only if 0.51·c₁ + 0.01·c₂ = 0.5, and so where c₁ = 0.98 and c₂ = 0.02. But PLC (and similarly PSFC) also imply that p_P2 = 〈0.5, 0.5〉, which, according LW, holds only if 0.01·c₁ + 0.51·c₂ = 0.5, and so where c₁ = 0.02 and c₂ = 0.98, which is a contradiction.□

Theorem 5

For all F: if {F = F_bel, U_b, A_b, N_b, UN_b} is consistent, then {F = F_bel, U_p, A_p, N_p, UN_p, PLC} is consistent.

Proof

The proof proceeds as the proof of Theorem 5, save that we make the further assumption that UN_b, and show that F_prob satisfies UN_p. Assume not. Then there is a case where (i) $r_{j}$ ≠ 0 and for all i: $r_{ij}$ = 0, or a case where (ii) $r_{j}$ ≠ 1 and for all i: $r_{ij}$ = 1. In case (i), we have for all i: $n^{*}_{ij}$ = 0, given for all i: $r_{ij}$ = 0 (by the definition of B*). So $n^{*}_{j}$ = 0, given UN_b. And so $r_{j}$ = 0, by the definition of F_prob. In case (ii), we have for all i: $n^{*}_{ij}$ = 1 and for all i, k ≠ j: $n^{*}_{ik}$ = 0, given for all i: $r_{ij}$ = 1 (by the definition of B*). So $n^{*}_{j}$ = 1, and for all k ≠ j: $n^{*}_{k}$ = 0, given U_b and UN_b. So $r_{j}$ = 1, by the definition of F_prob.□

Theorem 6

For all F: if {F = F_bel, U_b, A_b, N_b, WD_b} is consistent, then {F = F_bel, U_p, A_p, N_p, WD_p, PLC} is consistent.

Proof

The proof proceeds as the proof of Theorem 5, save that we make the further assumption that WD_b, and show that F_prob satisfies WD_p. Let P, j, and k be arbitrary, with for all i: $r_{ij}$ ≥ $r_{ik}$. In that case, for all i: $n^{*}_{ij}$≥ $n^{*}_{ik}$, by the definition of F_prob. So $n^{*}_{j}$≥ $n^{*}_{k}$, given WD_b. So $r_{j}$ ≥ $r_{k}$, by the definition of F_prob.□

Theorem 8

{U_b, A_b, N_b, Z_b, U_p, PLC, CL} is inconsistent.

Proof

Let Π = {w₁, w₂, w₃}. Consider B = 〈〈1, 0, 0〉, 〈0, 0, 1〉〉, and P = 〈〈0.9, 0.09, 0.01〉, 〈0.01, 0.09, 0.9〉〉. Notice that A_b, N_b, and Z_b imply that b_B = 〈1, 0, 1〉. Without loss of generality, assume that p_p = 〈a, b, c〉. Then PLC implies that a > b, since CL(〈1, 0, 0〉, 〈0.9, 0.09, 0.01〉) and CL(〈0, 0, 1〉, 〈0.01, 0.09, 0.99〉) [and similarly CSF(〈1, 0, 0〉, 〈0.9, 0.09, 0.01〉) and CSF(〈0, 0, 1〉, 〈0.01, 0.09, 0.99〉)²³]. Now consider P′ = 〈〈90/99, 9/99, 0〉, 〈1/10, 9/10, 0〉〉, and B′ = 〈〈1, 0, 0〉, 〈0, 1, 0〉〉. Notice that A_b, N_b, and Z_b imply that b_B′ = 〈1, 1, 0〉. Now notice that CL implies that p_p′ = 〈a/(a + b), b/(a + b), 0〉. Finally, notice that CL(〈1, 0, 0〉, 〈90/99, 9/99, 0〉) and CL(〈0, 1, 0〉, 〈1/10, 9/10, 0〉) imply that CL(〈1, 1, 0〉, 〈a/(a + b), b/(a + b), 0〉), which is absurd, since CL(〈1, 0, 0〉, 〈a/(a + b), b/(a + b), 0〉), given a > b [and similarly with CSF in place of CL].□

Theorem 9

{U_b, A_b, N_b, Z_b, U_p, A_p, N_p, PLC, IP} is inconsistent.

Proof

Let Π = {w₁, w₂, w₃, w₄}. Consider B = 〈〈0, 0, 0, 1〉, 〈1, 0, 0, 0〉〉, and P = 〈〈0.01, 0.09, 0.09, 0.81〉, 〈0.81, 0.09, 0.09, 0.01〉〉.²⁴ Notice that A_b, N_b, and Z_b imply that b_B = 〈1, 0, 0, 1〉. Now notice that A_p and N_p imply that p_P(w₁) = p_P(w₄) and p_P(w₂) = p_P(w₃), and this implies that p_P(w₁∪w₂) = p_P(w₁∪w₃) = 1/2. Now notice that CL(〈0, 0, 0, 1〉, 〈0.01, 0.09, 0.09, 0.81〉) and CL(〈〈1, 0, 0, 0〉, 〈0.81, 0.09, 0.9, 01〉) [and similarly CSF(〈0, 0, 0, 1〉, 〈0.01, 0.09, 0.09, 0.81〉) and CSF(〈1, 0, 0, 0〉, 〈0.81, 0.09, 0.09, 0.01〉)]. So PLC implies that CL(〈1, 0, 0, 1〉, P_P) [and similarly CSF(〈1, 0, 0, 1〉, P_P)], which implies that p_P(w₁) > p_P(w₂), and thus p_P(w₁) > 1/4. Finally, notice that p₁((w₁∪w₂)∩(w₁∪w₃)) = p₁(w₁∪w₂)p₁(w₁∪w₃) and p₂((w₁∪w₂)∩(w₁∪w₃)) = p₂(w₁∪w₂)p₁(w₁∪w₃). So IP implies that p_P((w₁∪w₂)∩(w₁∪w₃)) = p_P(w₁∪w₂)p_P(w₁∪w₃). But p_P((w₁∪w₂)∩(w₁∪w₃)) = p_P(w₁), and, since p_P(w₁∪w₂) = p_P(w₁∪w₃), p_P(w₁) = 1/4, which is absurd.□