Probabilistic Opinion Pooling with Imprecise Probabilities

Abstract

The question of how the probabilistic opinions of different individuals should be aggregated to form a group opinion is controversial. But one assumption seems to be pretty much common ground: for a group of Bayesians, the representation of group opinion should itself be a unique probability distribution (Madansky [44]; Lehrer and Wagner [34]; McConway Journal of the American Statistical Association, 76(374), 410–414, [45]; Bordley Management Science, 28(10), 1137–1148, [5]; Genest et al. The Annals of Statistics, 487–501, [21]; Genest and Zidek Statistical Science, 114–135, [23]; Mongin Journal of Economic Theory, 66(2), 313–351, [46]; Clemen and Winkler Risk Analysis, 19(2), 187–203, [7]; Dietrich and List [14]; Herzberg Theory and Decision, 1–19, [28]). We argue that this assumption is not always in order. We show how to extend the canonical mathematical framework for pooling to cover pooling with imprecise probabilities (IP) by employing set-valued pooling functions and generalizing common pooling axioms accordingly. As a proof of concept, we then show that one IP construction satisfies a number of central pooling axioms that are not jointly satisfied by any of the standard pooling recipes on pain of triviality. Following Levi (Synthese, 62(1), 3–11, [39]), we also argue that IP models admit of a much better philosophical motivation as a model of rational consensus.

Appendix: Proofs

1.1 Proof of Proposition 1

Proof

We carry out McConway’s proof with minimal adjustments made for our framework (pp. 411–412 [45], Theorem 3.1).

WSFP ⇒ MP. Assume that $\mathcal {F}$ has the WSFP, i.e., there is a function $\mathcal {G}{\kern -1.5pt}:{} \mathcal {A}{\kern -.3pt}\times {\kern -1.5pt} [0,1]^{n} {\kern -1.5pt}\to {}\mathcal {P}([0,1])$ suchthat $\mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})(A) = \mathcal {G}(A, \boldsymbol {p}_{1}(A), \ldots , \boldsymbol {p}_{n}(A))$. By WSFP, we have $\mathcal {F}([\boldsymbol {p}_{1} {\upharpoonright }_{\mathcal {A}^{\prime }}], \ldots , [\boldsymbol {p}_{n}\upharpoonright _{\mathcal {A}^{\prime }}])(A) = \mathcal {G}(A, [\boldsymbol {p}_{1}\upharpoonright _{\mathcal {A}^{\prime }}](A), \ldots , [\boldsymbol {p}_{n}\upharpoonright _{\mathcal {A}^{\prime }}](A))$. Since $\mathcal {G}$ isafunctionand $\boldsymbol {p}_{i}(A) = [\boldsymbol {p}_{i} \upharpoonright _{\mathcal {A}^{\prime }}](A)$ forany $A \in \mathcal {A}^{\prime }$ (all such $A \in \mathcal {A}^{\prime }$ are also in $\mathcal {A}$), it follows that $\mathcal {G}(A, [\boldsymbol {p}_{1}\upharpoonright _{\mathcal {A}^{\prime }}](A), \ldots , [\boldsymbol {p}_{n}\upharpoonright _{\mathcal {A}^{\prime }}](A)) = \mathcal {G}(A, \boldsymbol {p}_{1}(A), \ldots , \boldsymbol {p}_{n}(A)) = \mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})(A)$.Hence,$\mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})(A) = \mathcal {F}([\boldsymbol {p}_{1}\upharpoonright _{\mathcal {A}^{\prime }}], \ldots , [\boldsymbol {p}_{n}\upharpoonright _{\mathcal {A}^{\prime }}])(A)$.

MP ⇒ WSFP. Assume that $\mathcal {F}$ has the MP. Let $A \in \mathcal {A}$. We want to show that $\mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})(A)$ depends only on A and p _i (A),i=1,…,n.

First, if A = ∅ or A=Ω, then, since the range of $\mathcal {F}$ is $\mathcal {P}(\mathbb {P}), \mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})(A)$ depends only on A and p _i (A),i=1,..,n, for any profile because, setting $\mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})(A) \,=\, \mathcal {G}(A, \boldsymbol {p}_{1}(A), \ldots , \boldsymbol {p}_{n}(A))$ and $\mathcal {F}(\boldsymbol {p}_{1}^{\prime }, \ldots , \boldsymbol {p}_{n}^{\prime })(A) \,=\, \mathcal {G}(A, \boldsymbol {p}_{1}^{\prime }$ $(A), \ldots , \boldsymbol {p}_{n}^{\prime }(A))$, it follows that $\mathcal {G}(A, \boldsymbol {p}_{1}(A), \ldots , \boldsymbol {p}_{n}(A)) = \mathcal {G}(A, \boldsymbol {p}_{1}^{\prime }(A),\ldots , \boldsymbol {p}_{n}^{\prime } (A))$.

Next, suppose that ∅≠A≠Ω. Consider the σ-algebra $\mathcal {A}^{\prime } = \{\emptyset , A, A^c, {\Omega }\}$. $\mathcal {A}$ contains A and has $\mathcal {A}^{\prime }$ as a sub-algebra. By MP, then

$$\mathcal{F}(\boldsymbol{p}_{1}, \ldots, \boldsymbol{p}_{n})(A) = \mathcal{F}([\boldsymbol{p}_{1}\upharpoonright_{\mathcal{A}^{\prime}}], \ldots, [\boldsymbol{p}_{n}\upharpoonright_{\mathcal{A}^{\prime}}])(A). $$

$\mathcal {A}^{\prime }$ is uniquely defined by A and any probability over $\mathcal {A}^{\prime }$ is uniquely determined by the probability of A under that distribution. So the righthand side of the equation above is determined by A and $\boldsymbol {p}_{i}\upharpoonright _{\mathcal {A}^{\prime }}(A) = [\boldsymbol {p}_{i}\upharpoonright _{\mathcal {A}^{\prime }}](A) = \boldsymbol {p}_{i}(A)$. □

1.2 Proof of Lemma 1

Proof

Let $Y = \{\boldsymbol {p} : \boldsymbol {p} = {\sum }_{i=1}^{n} \alpha _{i}\boldsymbol {p}_{i}$ such that α _i ≥0 for i=1,…,n and ${\sum }_{i=1}^{n} \alpha _{i} = 1\}$. We want to show the following:

$$\mathcal{F}(\boldsymbol{p}_{1}, \ldots, \boldsymbol{p}_{n}) = conv\{\boldsymbol{p}_{i} : i = 1, \ldots, n\} = Y $$

The first equality we have by definition. In order to show the second equality, we have to show that Y is the smallest convex set containing {p _i :i=1,…,n}. To show convexity, we show that for any two functions in Y, any convex combination of those functions is in Y. Suppose that p,p ^′∈Y. By assumption, $\boldsymbol {p} = {\sum }_{i = 1}^{n} \alpha _{i} \boldsymbol {p}_{i}$ and $\boldsymbol {p}^{\prime } = {\sum }_{i = 1}^{n} \beta _{i} \boldsymbol {p}_{i}$. Consider $\boldsymbol {p}^{\star } = \gamma \boldsymbol {p} + (1 - \gamma ) \boldsymbol {p}^{\prime } = \gamma ({\sum }_{i = 1}^{n} \alpha _{i} \boldsymbol {p}_{i}) + (1 - \gamma ) {\sum }_{i = 1}^{n} \beta _{i} \boldsymbol {p}_{i}$.

$$\begin{array}{@{}rcl@{}} \boldsymbol{p}^{\star} &=& \gamma {\sum}_{i = 1}^{n} \alpha_{i} \boldsymbol{p}_{i} + (1 - \gamma) {\sum}_{i = 1}^{n} \beta_{i} \boldsymbol{p}_{i}\\ &=& {\sum}_{i = 1}^{n} \gamma \alpha_{i} \boldsymbol{p}_{i} + {\sum}_{i = 1}^{n} (1 - \gamma) \beta_{i} \boldsymbol{p}_{i}\\ &=& {\sum}_{i = 1}^{n} [\gamma \alpha_{i} \boldsymbol{p}_{i} + (1 - \gamma) \beta_{i} \boldsymbol{p}_{i}]\\ &=& {\sum}_{i = 1}^{n} [\gamma \alpha_{i} + (1 - \gamma) \beta_{i}] \boldsymbol{p}_{i}\\ &=& {\sum}_{j = 1}^{n} \delta_{j} \boldsymbol{p}_{j} \end{array} $$

where δ _j = γ α _j +(1−γ)β _j . δ _j ≥0 for j=1,…,n because every term is nonnegative. ${\sum }_{j = 1}^{n} \delta _{j} = {\sum }_{i = 1}^{n} [\gamma \alpha _{i} + (1 - \gamma ) \beta _{i}] = {\sum }_{i = 1}^{n} \gamma \alpha _{i} + {\sum }_{i = 1}^{n} (1 - \gamma ) \beta _{i} = \gamma {\sum }_{i = 1}^{n} \alpha _{i} + (1 - \gamma ) {\sum }_{i = 1}^{n} \beta _{i} = \gamma (1) + (1-\gamma )1 = 1$. Hence, p ^⋆∈Y, so Y is convex. If Y were not the smallest such set, then there would be some convex $Z \subsetneq Y$ such that {p _i :i=1,…,n}⊆Z. But for any p∈Y,p is a convex combination of the elements in {p _i :i=1,…,n}. Since Z is convex and contains the p _i , it follows that p∈Z, which is a contradiction. □

1.3 Proof of Proposition 2

Proof

Since $\mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})(A) = \{\boldsymbol {p}(A): \boldsymbol {p} \in \mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})\}$, we let $\mathcal {G}$ of the SSFP be the convex hull operation applied to {p _i (A):i=1,…,n}. It is clear that $\mathcal {G}$ depends just on the individual probabilities for A. We need to show that

$$\{\boldsymbol{p}(A): \boldsymbol{p} \in \mathcal{F}(\boldsymbol{p}_{1}, \ldots, \boldsymbol{p}_{n})\} = conv\{\boldsymbol{p}_{i}(A): i = 1, \ldots, n\}. $$

Trivially, the lefthand side includes {p _i (A):i=1,…,n}. Suppose $\boldsymbol {p}(A), \boldsymbol {p}^{\prime }(A) \in \{\boldsymbol {p}(A): \boldsymbol {p} \in \mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})\}$. Since $\mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$ is convex, it follows immediately that any convex combination of p(A),p ^′(A) is in $\{\boldsymbol {p}(A): \boldsymbol {p} \in \mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})\}$. Finally, suppose that there is some convex $Z \subsetneq \{\boldsymbol {p}(A): \boldsymbol {p} \in \mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})\}$ which contains {p _i (A):i=1,…,n}. But for any $\boldsymbol {p}(A) \in \{\boldsymbol {p}(A): \boldsymbol {p} \in \mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})\}$, p(A) is a convex combination of the p _i (A) since every $\boldsymbol {p} \in \mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$ is such a convex combination of the p _i (Lemma 1). Hence, p(A)∈Z, contrary to our supposition. So, the equality holds and the SWFP is satisfied.

But since SWFP clearly implies WSFP, WSFP is satisfied, too. By Proposition 1, it follows immediately that $\mathcal {F}$ has the MP.

Because $\mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$ is a set of probability functions, $\mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})(\emptyset ) = \{0\}$. Let p _i (A)=0, i=1,…,n. Since there is a function, $\mathcal {G}$, such that $\mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})(A) = \mathcal {G}(\boldsymbol {p}_{1}(A), \ldots , \boldsymbol {p}_{n}(A))$, we have it that

$$\begin{array}{@{}rcl@{}} \{0\} &=& \mathcal{F}(\boldsymbol{p}_{1}, \ldots, \boldsymbol{p}_{n})(\emptyset)\\ &=& \mathcal{G}(\boldsymbol{p}_{1}(\emptyset), \ldots, \boldsymbol{p}_{n}(\emptyset))\\ &=& \mathcal{G}(\boldsymbol{p}_{1}(A), \ldots, \boldsymbol{p}_{n}(A))\\ &=& \mathcal{F}(\boldsymbol{p}_{1}, \ldots, \boldsymbol{p}_{n})(A) \end{array} $$

So, ZPP follows from SWFP.

For any profile $(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n}) \in \mathbb {P}^{n}$, if all p _i are identical, then the convex hull is just {p _i }. So $\mathcal {F}$ satisfies unanimity preservation. □

1.4 Proof of Lemma 2

We generalize a proof of a result due originally to Girón and Rios and Levi [25, 37] for updating on an event to updating on a common likelihood function.

Proof

We want to show that $\mathcal {F}^{\lambda }(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$ is convex. That is, given any two members, $\boldsymbol {p}^{\lambda }, \boldsymbol {p}^{\prime \lambda } \in \mathcal {F}^{\lambda }(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$ and α∈[0,1], p ^⋆ = α p ^λ+(1−α)p ^′λ is in $\mathcal {F}^{\lambda }(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$. If there is a convex combination of p and p ^′, p _⋆, such that p ^⋆λ = p ^⋆, then the convexity of $\mathcal {F}^{\lambda }(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$ is established as a consequence of the convexity of $\mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$. Where $\boldsymbol {p}^{\star \lambda }(\cdot ) = \frac {\boldsymbol {p}_{\star }(\cdot )\lambda (\cdot )}{{\sum }_{\omega ^{\prime } \in {\Omega }}\boldsymbol {p}_{\star }(\omega ^{\prime })\lambda (\omega ^{\prime })} = \frac {\beta \boldsymbol {p}(\cdot )\lambda (\cdot )+(1-\beta ) \boldsymbol {p}^{\prime }(\cdot )\lambda (\cdot )}{\beta {\sum }_{\omega ^{\prime } \in {\Omega }}\boldsymbol {p}(\omega ^{\prime })\lambda (\omega ^{\prime })+(1-\beta ) {\sum }_{\omega ^{\prime } \in {\Omega }}\boldsymbol {p}^{\prime }(\omega ^{\prime })\lambda (\omega ^{\prime })}$, for any α we want to find some β such that

$$\boldsymbol{p}^{\star}(\cdot) = \alpha \boldsymbol{p}^{\lambda}(\cdot) + (1 - \alpha) \boldsymbol{p}^{\prime \lambda}(\cdot) = \frac{\beta \boldsymbol{p}(\cdot)\lambda(\cdot)+(1-\beta) \boldsymbol{p}^{\prime}(\cdot)\lambda(\cdot)}{\beta {\sum}_{\omega^{\prime} \in {\Omega}}\boldsymbol{p}(\omega^{\prime})\lambda(\omega^{\prime})+(1-\beta) {\sum}_{\omega^{\prime} \in {\Omega}}\boldsymbol{p}^{\prime}(\omega^{\prime})\lambda(\omega^{\prime})} = \boldsymbol{p}_{\star}^{\lambda}(\cdot). $$

For $\beta =\frac {\alpha {\sum }_{\omega ^{*} \in {\Omega }}\boldsymbol {p}^{\prime }(\omega ^{*})\lambda (\omega ^{*})}{\alpha {\sum }_{\omega ^{*} \in {\Omega }} \boldsymbol {p}^{\prime }(\omega ^{*})\lambda (\omega ^{*})+(1-\alpha ) {\sum }_{\omega ^{*} \in {\Omega }}\boldsymbol {p}(\omega ^{*})\lambda (\omega ^{*})}$, the equality is verifiable with some tedious algebra. □

1.5 Proof of Proposition 3

Proof

We must show that convex IP pooling functions are externally Bayesian, i.e., $\mathcal {F}(\boldsymbol {p}_{1}^{\lambda }, \ldots , \boldsymbol {p}_{n}^{\lambda }) = \mathcal {F}^{\lambda }(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$ (provided the relevant profiles are in the domain of $\mathcal {F}$).

$\mathcal {F}(\boldsymbol {p}_{1}^{\lambda }, \ldots , \boldsymbol {p}_{n}^{\lambda }) \subseteq \mathcal {F}^{\lambda }(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$. Trivially, for each i=1,…,n, $\boldsymbol {p}_{i}^{\lambda } \in \mathcal {F}^{\lambda }(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$. By Lemma 2, $\mathcal {F}^{\lambda }(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$ is convex. It follows that $conv\{\boldsymbol {p}_{i}^{\lambda } : i =1, \ldots , n\} = \mathcal {F}(\boldsymbol {p}_{1}^{\lambda }, \ldots , \boldsymbol {p}_{n}^{\lambda }) \subseteq \mathcal {F}^{\lambda }(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$.

$\mathcal {F}^{\lambda }(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n}) \subseteq \mathcal {F}(\boldsymbol {p}_{1}^{\lambda }, \ldots , \boldsymbol {p}_{n}^{\lambda })$. By Lemma 1, any $\boldsymbol {p} \in \mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$ can be expressed as the convex combination of the n extreme points generating $\mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$, i.e., $\boldsymbol {p} = {\sum }_{i=1}^{n} \alpha _{i} \boldsymbol {p}_{i}$ where α _i ≥0 for i=1,…,n and ${\sum }_{i=1}^{n}\alpha _{i}=1$. By definition,

$$\mathcal{F}^{\lambda}(\boldsymbol{p}_{1}, \ldots, \boldsymbol{p}_{n}) = \left\{\boldsymbol{p}^{\lambda}: \boldsymbol{p} \in \mathcal{F}(\boldsymbol{p}_{1}, \ldots, \boldsymbol{p}_{n})\ \text{and}\ \boldsymbol{p}^{\lambda}(\cdot) = \frac{\boldsymbol{p}(\cdot)\lambda(\cdot)}{{\sum}_{\omega^{\prime} \in {\Omega}} \boldsymbol{p}(\omega^{\prime})\lambda(\omega^{\prime})}\right\} $$

We show that any member of $\mathcal {F}^{\lambda }(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$ is identical to some member of $\mathcal {F}(\boldsymbol {p}_{1}^{\lambda }, \ldots , \boldsymbol {p}_{n}^{\lambda })$.

$$\begin{array}{llll} \boldsymbol{p}^{\lambda}(\omega) &=& \frac{\boldsymbol{p}(\omega)\lambda(\omega)}{{\sum}_{\omega^{\prime} \in {\Omega}} \boldsymbol{p}(\omega^{\prime})\lambda(\omega^{\prime})} & [\text{Definition}]\\ &=& \frac{{\sum}_{i=1}^{n} \alpha_{i} \boldsymbol{p}_{i}(\omega)\lambda(\omega)}{{\sum}_{\omega^{\prime} \in {\Omega}} {\sum}_{i=1}^{n} \alpha_{i} \boldsymbol{p}_{i}(\omega^{\prime})\lambda(\omega^{\prime})} & [\text{Lemma~1}]\\ &=& \frac{{\sum}_{i=1}^{n} \alpha_{i} \boldsymbol{p}_{i}^{\lambda}(\omega) \cdot {\sum}_{\omega^{\prime} \in {\Omega}} \boldsymbol{p}_{i}(\omega^{\prime})\lambda(\omega^{\prime})}{{\sum}_{\omega^{\prime} \in {\Omega}} {\sum}_{i=1}^{n} \alpha_{i} \boldsymbol{p}_{i}(\omega^{\prime})\lambda(\omega^{\prime})} & [\boldsymbol{p}_{i}(\omega)\lambda(\omega) = \boldsymbol{p}_{i}(\omega)^{\lambda} \cdot {\sum}_{\omega^{\prime} \in {\Omega}} \boldsymbol{p}_{i}(\omega^{\prime})\lambda(\omega^{\prime})]\\ &=& {\sum}_{j=1}^{n} \beta_{j} {\boldsymbol{p}}_{j}^{\lambda}(\omega) \in \mathcal{F}(\boldsymbol{p}_{1}^{\lambda}, \ldots, \boldsymbol{p}_{n}^{\lambda}) & [\text{Algebra}] \end{array} $$

where $\beta _{j} = \frac {\alpha _{j} \cdot {\sum }_{\omega ^{\prime } \in {\Omega }} \boldsymbol {p}_{j}(\omega ^{\prime }) \lambda (\omega ^{\prime })}{{\sum }_{\omega ^{\prime } \in {\Omega }} {\sum }_{i=1}^{n}\alpha _{i} \boldsymbol {p}_{i}(\omega ^{\prime }) \lambda (\omega ^{\prime })}$ with β _j ≥0 for all j=1,…,n and ${\sum }_{j=1}^{n} \beta _{j} = 1$. □

1.6 Proof of Proposition 4

Proof

We provide a very simple type of counterexample to individualwise Bayesianity, though counterexamples are plentiful. Consider the profile (p ₁,p ₂) for n=2 agents such that p ₁ = p ₂. Individualwise Bayesianity requires that $\mathcal {F}(\boldsymbol {p}_{1}, \boldsymbol {p}_{2}^{\lambda }) = \mathcal {F}^{\lambda }(\boldsymbol {p}_{1}, \boldsymbol {p}_{2})$ (provided both (p ₁,p ₂) and $(\boldsymbol {p}_{1}, \boldsymbol {p}_2^{\lambda })$ are in the domain of $\mathcal {F}$). By Proposition 3 (external Bayesianity), it follows that $\mathcal {F}^{\lambda }(\boldsymbol {p}_{1}, \boldsymbol {p}_{2}) = \mathcal {F}(\boldsymbol {p}_{1}^{\lambda }, \boldsymbol {p}_{2}^{\lambda })$. But since p ₁ = p ₂, it follows that $\boldsymbol {p}_{1}^{\lambda } = \boldsymbol {p}_{2}^{\lambda }$. By unanimity (Proposition 2), then, we have $\mathcal {F}(\boldsymbol {p}_{1}^{\lambda }, \boldsymbol {p}_{2}^{\lambda }) = \{\boldsymbol {p_{i}}^{\lambda }\},$ where $\boldsymbol {p}_{i}^{\lambda } = \boldsymbol {p}_{1}^{\lambda } = \boldsymbol {p}_{2}^{\lambda }$. However, in general $\boldsymbol {p}_{i} \neq \boldsymbol {p}_{i}^{\lambda }$ and so $\mathcal {F}(\boldsymbol {p}_{1}, \boldsymbol {p}_{2}^{\lambda })$ is not a singleton. It follows that, in general, $\mathcal {F}(\boldsymbol {p}_{1}, \boldsymbol {p}_{2}^{\lambda }) \neq \mathcal {F}^{\lambda }(\boldsymbol {p}_{1}, \boldsymbol {p}_{2})$. □

1.7 Proof of Proposition 5

Proof

Suppose that p _i (A|B) = p _i (A) for i=1,…,n. We want to show that $\mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})(A) = \mathcal {F}^{B}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})(A)$. Consider $\boldsymbol {p}^{\star }(A) \in \mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})(A)$ and $\boldsymbol {p}_{\star }(A) \in \mathcal {F}^{B}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})(A)$. By Lemma 1, $\boldsymbol {p}^{\star }(A) = {\sum }_{i = 1}^{n} \alpha _{i} \boldsymbol {p}_{i}(A)$, for appropriate α _i . By Proposition 3 (external Bayesianity), $\mathcal {F}^{B}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})(A) = \mathcal {F}(\boldsymbol {p}_{1}^{B}, \ldots , \boldsymbol {p}_{n}^{B})(A)$ (Proposition 3 holds for standard conditionalization since standard conditionalization is a special case of updating on a likelihood function, as noted in the body of the paper). So, we have $\boldsymbol {p}_{\star }(A) = {\sum }_{i = 1}^{n} \beta _{i} \boldsymbol {p}_{i}^{B}(A)$, for appropriate β _i , again by Lemma 1. By hypothesis $\boldsymbol {p}_{i}^{B}(A) = \boldsymbol {p}_{i}(A)$ for i=1,…,n. Hence, $\boldsymbol {p}_{\star }(A) = {\sum }_{i = 1}^{n} \beta _{i} \boldsymbol {p}_{i}^{B}(A) = {\sum }_{i=1}^{n} \beta _{i} \boldsymbol {p}_{i}(A)$. Letting α _i = β _i , it follows that $\mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})(A) = \mathcal {F}^{B}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})(A)$. □

1.8 Proof of Proposition 6

Proof

We show first that $\mathcal {M}_{n}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n}) \subseteq conv(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$ for all $(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n}) \in \mathbb {P}^{n}$. Since there are at least three disjoint events, $A_{1}, A_2, A_3 \in \mathcal {A}$, following Lehrer and Wagner [34, Theorems 6.4, 6.7] and McConway [45, Theorem 3.3], we can exploit techniques and results for functional equations. For any numbers a _i ,b _i ∈[0,1] with a _i + b _i ∈[0,1], define a sequence of probability measures, p _i , i=1,…,n by setting

$$\begin{array}{@{}rcl@{}} \boldsymbol{p}_{i}(A_{1}) &=& a_{i}\\ \boldsymbol{p}_{i}(A_2) &=& b_{i}\\ \boldsymbol{p}_{i}(A_3) &=& 1-a_{i}-b_{i} \end{array} $$

Since it is the case that $\mathfrak {m}\left (\boldsymbol {p}_{1}(\cdot ), \ldots , \boldsymbol {p}_{n}(\cdot )\right ) \in \mathbb {P}$ for all $(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n}) \in \mathbb {P}^{n}$ and every $\mathfrak {m}\left (\boldsymbol {p}_{1}(\cdot ), \ldots , \boldsymbol {p}_{n}(\cdot )\right ) \in \mathcal {M}_{n}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$, we have it that $\mathfrak {m}\left (\boldsymbol {p}_{1}(A), \ldots , \boldsymbol {p}_{n}(A)\right )=\boldsymbol {p}(A),$ for some $\boldsymbol {p} \in \mathbb {P}$ and all $A \in \mathcal {A}$. Now, by the additivity of probability measures, p(A ₁∪A ₂) = p(A ₁) + p(A ₂). Hence, $\mathfrak {m}(a_{1}+b_{1}, \ldots , a_{n} + b_{n}) = \mathfrak {m}\left (a_{1}, \ldots , a_{n}\right ) + \mathfrak {m}\left (b_{1}, \ldots , b_{n}\right )$. So, $\mathfrak {m}$ satisfies Cauchy’s multivariable functional equation. For each i=1,…,n, define $\mathfrak {m}_{i}(a) = \mathfrak {m}(0, \ldots , a, \ldots , 0),$ where a occupies the i-th position of the vector (0,…,a,…,0). It is clear that $\mathfrak {m}_{i}(a + b) = \mathfrak {m}_{i}(a) + \mathfrak {m}_{i}(b)$ for all a,b∈[0,1] with a + b∈[0,1]. Because $\mathfrak {m}$ is nonnegative, so is $\mathfrak {m}_{i},\ i = 1, \ldots , n$. By Theorem 3 of [2, p. 48], it follows that there exists a nonnegative constant α _i such that $\mathfrak {m}_{i}(a) = \alpha _{i} a$ for all a∈[0,1]. By the Cauchy equation we have

$$\begin{array}{@{}rcl@{}} \mathfrak{m}(a_{1}, \ldots, a_{n}) &=& \mathfrak{m}(a_{1}, 0, \ldots, 0) + \mathfrak{m}(0, a_2, \ldots, a_{n})\\ &=& \mathfrak{m}(a_{1}, 0, \ldots, 0) + \mathfrak{m}(0, a_2, 0, \ldots, 0) + {\ldots} + \mathfrak{m}(0, \ldots, 0, a_{n}) \end{array} $$

So we have $\mathfrak {m}(a_{1}, \ldots , a_{n}) = \mathfrak {m}_{1}(a_{1}) + {\ldots } + \mathfrak {m}_{n}(a_{n}) = \alpha _{1} a_{1} + {\ldots } + \alpha _{n} a_{n}$. And since $\mathfrak {m}(1, \ldots , 1)=1$ (by consideration of the probability of Ω), it follows that ${\sum }_{i=1}^{n} \alpha _{i}=1$. Thus, $\mathfrak {m}$ is a convex combination.

Now, we want to show that $conv(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n}) \subseteq \mathcal {M}_{n}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$. Let p be an element of c o n v(p ₁,…,p _n ). It is clear that there exists an $\mathfrak {m} \in \mathfrak {M}_{n}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$ such that $\mathfrak {m}\left (\boldsymbol {p}_{1}(\cdot ), \ldots , \boldsymbol {p}_{n}(\cdot )\right ) = \boldsymbol {p}$. And since p is just a convex combination, there exist weights α ₁,…,α _n ∈[0,1] such that ${\sum }_{i=1}^{n} \alpha _{i} = 1$ and $\boldsymbol {p} = {\sum }_{i=1}^{n} \alpha _{i}\boldsymbol {p}_{i}$. But for any other profile $(\boldsymbol {q}_{1}, \ldots , \boldsymbol {q}_{n}) \in \mathbb {P}^{n}$, taking any convex combination yields a probability measure. In particular, ${\sum }_{i=1}^{n} \alpha _{i}\boldsymbol {q}_{i} \in \mathbb {P}$. It follows that $\mathfrak {m} \in \bigcap _{\vec {\boldsymbol {q}} \in \mathbb {P}^{n}} \mathfrak {M}_{n}(\vec {\boldsymbol {q}})$. So, $\boldsymbol {p} = \mathfrak {m}\left (\boldsymbol {p}_{1}(\cdot ), \ldots , \boldsymbol {p}_{n}(\cdot )\right ) \in \mathcal {M}_{n}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n}),$ as desired.

The two inclusions above show that $\mathcal {M}_{n}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n}) = conv(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$. Hence, $\mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n}) = conv(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$ is equivalent to $\mathcal {F}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n}) = \mathcal {M}_{n}(\boldsymbol {p}_{1}, \ldots , \boldsymbol {p}_{n})$. □