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Agreeing to Disagree with Conditional Probability Systems

Elias Tsakas


In this note, we extend Aumann’s agreement theorem to a framework where beliefs are modelled by conditional probability systems à la Battigalli, P., and M. Siniscalchi. 1999. “Hierarchies of Conditional Beliefs and Interactive Epistemology in Dynamic Games.” Journal of Economic Theory 88: 188–230. We prove two independent generalizations of the agreement theorem, one where the agents share some common conditioning event, and one where they may not.

A Proofs

Proof of Proposition 1.

First of all observe that B covers X, since PB is a partition of X. Then, consider an arbitrary collection (αP)PP of positive reals such that PPαP=1. For each AA define


Verifying that p is a probability measure in Δ(X,A) is trivial. Moreover, notice that for every AA and every BB with p(B)>0, it is the case that


with eq. (10) following from (C3) since either ABBP or BP=.

Proof of Theorem 1.

Define TiB:={tiTi:ti(ω)=tifor some ωCK([qa]B[qb]B)supp(p)[B]}, which is by hypothesis nonempty. Since the event [qi]B is measurable with respect to the partition {[ti]|tiTi}, either [ti][qi]B or [ti][qi]B=. Now, for an arbitrary tiTiB, it is necessarily the case that [ti]CK([qi]B)[ti]Ki([qi]B)[ti][qi]B. Hence, for each tiTiB,


The second equality follows from p([B][ti])>0, which is by definition true for all tiTiB. Now, define the set M:=B×TaB×TbB. Hence,


Finally, since p(E|M) does not depend on iI, it is the case that qa=qb.

Proof of Theorem 2.

Step 1. Define Ti:={tiTi:ti(ω)=tifor some ωCK([qa][qb])supp(p)[B]}, which is by hypothesis nonempty. Then, define the set M:=B×Ta×Tb. Then, following Tsakas and Voorneveld (2011), Proof of Lemma, we define Qi as the coarsest partition of B that generates Bi, i.e., formally for every θB, the element of Qi that contains θ is defined by Qi(θ):=BBi:θBB. Thus, we obtain


with eq. (11) following from the fact that Bi is balanced. Before moving forward, let us point out that p(Q×Ta×Tb) or p(B×Ta×Tb) might in principle be equal to 0 for some QQi or BBi respectively. In such cases, the conditional probabilities given these null events take arbitrary values.

Step 2. Now, similarly to the proof of Theorem 1, it is the case that [ti][qi] for all tiTi, thus implying πti(E|B)=qi for all BBi and all tiTi.

Step 3. Now take an arbitrary BBi such that p(E|B×Ta×Tb)>0. Then, there exists some tiTi such that p([B][ti])>0, and therefore since p is a common prior, we obtain p(E|B×Ta×Tb)=qi for all BBi with p(E|B×Ta×Tb)>0. Hence, by eq. (12) it follows that


with eq. (13) following again from the fact that Bi is balanced. Indeed, the conditional probability can be seen as an indicator function, with p(B×Ta×Tb|Q×Ta×Tb)=1 if QB and p(B×Ta×Tb|Q×Ta×Tb)=0 otherwise. Finally, notice that p(E|M) does not depend on iI, thus completing the proof.

Proof of Corollary 1.

Recall from Remark 1 that every treelike Bi is also balanced. Moreover, by (T2) it follows that Ba=Bb=Θ for all iI. Hence, CK([qa][qb])supp(p) directly implies CK([qa][qb])supp(p)[B]. Thus, all the conditions of Theorem 2 are satisfied, and therefore qa=qb.


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I am indebted to three anonymous referees and the associate editor for their valuable comments. This note supersedes two previous papers by the same author, titled “Strong belief and agreeing to disagree” and “Hierarchies of conditional beliefs derived from commonly known priors” respectively.

Published Online: 2018-05-04

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