Suppose there is a common priorµand a player assigns zero posterior to some state
ω ∈Ω0. The next result proves that every common prior must assign zero probability
to the state ω. In other words, if there is a common prior, it is not possible that this prior assigns positive probability to a state while a player assigns zero posterior to this same state.
Lemma 5.1. Consider a modelM =Ω, J,{Πj, θj} j∈J
which admits a common priorµ. If
ω ∈Ω0, thenµω = 0.
Proof. Becauseω∈ Ω0, there is a playerj ∈ J such thatθjω = 0. Letπ =πj(ω) =πj(ω).
Suppose by contradiction thatµω >0. Hence,µ(π) ≥µω >0. Becauseµis a prior for
playerj,µω >0andµ(π) >0, thenθωj = µω/µ(π) >0, contradicting the assumption
θj
The task of finding a common prior onΩreduces to finding a common prior on the model restricted toΩ\Ω0. In particular, ifΩ = Ω0, there is no common prior.
Lemma 5.2. Consider a modelM =Ω, J,{Πj, θj} j∈J
which admits a common priorµ, and suppose there exists a stateω0 ∈Ωsuch thatµω0 = 0. Fixω ∈Ω. If there is a positive pathp fromωtoω0, thenµω = 0.
Proof. Letp = {hj1;ω, ω1i, hj2;ω1, ω2i, · · ·,hjt;ωt−1, ω0i} be a positive path. Suppose,
by way of contradiction, that µωt−1 6= 0. Then, µ(π
jt(ω
0)) ≥ µωt−1 > 0, so θω0 =
µω0/µ(π
jt(ω
0)) = 0/µ(πjt(ω0)) = 0. Butpis a positive path, soθωjt0 6= 0. Hence,µωt−1 = 0. By induction alongp,µω = 0.
ChooseωP ∈ΩP. Asµω0 = 0for allω0 ∈Ω0, and there exists a path fromωP toω0, therefore by Lemma 5.2,µωP = 0.
Corollary 5.1. Consider a modelM =
Ω, J,{Πj, θj}j∈Jwhich admits a common priorµ. Ifω ∈ΩP, thenµω = 0.
Fix a model M = Ω, J,{Πj, θj} j∈J
such thatθj
ω > 0for allj ∈ J, ω ∈ Ω. In this
case, the cycles approach established that there is a common prior with full support if and only if CI = ∅; that is, when there is no inconsistent cycle. When the cycle equation of a cycle c ∈ C does not hold, the only hope to find a common prior is to assign probability zero to all states along c. If θj
ω > 0 for allj ∈ J and ω ∈ Ω, it is
possible to use this trick of assigning zero probability to states along inconsistent cycles so long as there is at least one state left in the end to receive probabilistic mass.
Lemma 5.3. Consider a modelM =Ω, J,{Πj, θj}j∈Jwhich admits a common priorµ. If
ω ∈ΩI, thenµω = 0.
Proof. LetcI ∈ CI. There are two potential cases: (i)µassigns positive probability to
all states incIor;(ii)µassigns zero probability to at least one state incI.
Case (i): This case is impossible. Consider the submodel consisting only of those states incI. Then Rodrigues-Neto [2009] tells us that there is no possible common prior
which places positive mass on every state.
Case (ii): Asω ∈ΩI, there exists a cyclecI ∈ CI such thatω is a node in this cycle.
positive path, then, by Lemma 5.2, µω = 0. Hence,µω = 0. Ifpis not a positive path,
construct instead the pathpefromωtoω
0
along the edges of−cI. AscI is inconsistent,
and aspis not a positive path, thenpeis a positive path and so by Lemma 5.2, µω = 0.
Therefore if anyω0 ∈cI hasµω0 = 0, then allω ∈cI haveµω = 0.
There are only three reasons for a common prior to have to assign zero probability to a state ω ∈ Ω: some player assigns zero posterior to ω; there is a positive path fromωto a state with zero prior; and stateωbelongs to an inconsistent cycle. Because
ΩN = ΩI∪ΩP∪Ω0, states inΩN must have zero prior. IfΩN = Ω(equivalently,ΩY =∅),
then it is not possible to find any common prior, not even by accepting priors without full support. IfΩY 6= ∅, it is always possible to find a common prior with support in
ΩY.
Remark 5.1. It may seem as though we would also need to consider states ω which have positive paths to someωI ∈ΩI. Fortunately, these are covered by the previous three conditions.
Let ω /∈ Ω0 have a positive pathptoωI ∈ ΩI. Then eitherpcontains states inΩ0, or it does
not. If it does, then simply truncate the path at the first state inΩ0, and thenω ∈ΩP. Ifpdoes
not contain any points inΩ0, then by appendingpto any inconsistent cycle which containsωI
we can construct a new inconsistent cycle which containsω. So in either case,ω∈ΩN.
Proposition 5.1. LetM =Ω, J,{Πj, θj} j∈J
. Consider the decomposition of the state space,
Ω = ΩY ∪ΩN, as described above. ModelM admits a common prior if and only ifΩY 6=∅, in
which caseM admits a common prior with supportΩY. Moreover,ΩY is the maximal support
for a common prior. In the particular case where ΩY = Ω, modelM admits a common prior
with full support.
Proof. Suppose that there is a common priorµ. Asµis a measure on a finite space, it has non-empty supportS. Asω ∈Shasµω >0, therefore by Lemmas 5.1, 5.2, and 5.3,
ω /∈Ω0∪ΩP ∪ΩI. Therefore,ω∈ΩY, soΩY is non-empty.
Conversely, suppose thatΩY 6= ∅. Consider the restriction of ModelM onto ΩY.
That is, consider the submodel MY =
ΩY, J, Πj,Y, θj,Y j∈J
, where each partition
Πj,Y is the restriction of partitionΠj toΩ
Y, and each vectorθj,Y is the restriction ofθj
to the coordinates that correspond to the states in ΩY. Because every stateω of MY
belongs toΩY, thenω /∈ΩN, soθjω >0and every cycle is consistent. Therefore, by the
supportΩY. Asω ∈ ΩY, in particular, ω /∈ ΩP, so no cycle can be extended in such a
way as to include a state in Ω0 and retain a positive cycle equation. Therefore when
this common prior is extended toΩby assigning probability zero to all states outside
ΩY, the extension will indeed be a common prior onΩ. This extension has support
ΩY, withΩY 6=∅, and is consistent with the posteriors of all players, and therefore, a common prior ofM.
Remark 5.2. The cycles approach is a direct method to study Harsanyi’s Consistency Problem. The extension of the cycles approach to common priors without full support that is proposed here also works in the case of non-partitional models (see Fiorini and Rodrigues-Neto [2017]).
The next result provides a direct sufficient condition for common prior existence.
Proposition 5.2. Let M =
Ω, J,{Πj, θj}j∈J. If all cycle equations are satisfied, thenM
admits a common prior.
Proof. Using Proposition 5.1, it is necessary and sufficient to show that if all cycle equations hold, thenΩY 6=∅.
LetCI = ∅, and suppose, by way of contradiction, thatΩY = ∅. Then, ΩI = ∅and
ΩN = Ω, which implies that Ω = Ω0 ∪ ΩP. We want to show that there exists an
inconsistent cycle. Letfe: Ω0 →Ωsuch that for all ω0 ∈ Ω0 there exists a playerj ∈ J
for whomθj
ω0 = 0,f(ωe 0) ∈ π
j(ω
0), andθj
e
f(ω0) > 0. This is possible asω0 ∈ Ω0, and as posteriors θωj must sum to1 overπj(ω0). There may be more than oneω ∈ Ωwhich
satisfies this property; in this case pick any. There are two cases to consider:
Case (i): Suppose there exists a sequence {ωi}Ii=1 ∈ Ω0 such thatωi+1 = fe(ωi), for
alli ∈ {1,· · · , I−1}, andω1 = f(ωe I). For eachi ∈ {1,· · · , I}, letji be a player such
thatωi+1 ∈πji(ωi),θjiωi = 0, andθjiωi+1 >0. There is a cycle(hj1, ω1, ω2i, . . . ,hjI, ωI, ωI+1i) with ωI+1 = ω1 and cycle equation θjω11· · ·θ
jI ωI = θjω12· · ·θ jI ω1. But θ j1 ω1· · ·θ jI ωI = 0, while θj1 ω2· · ·θ jI
ω1 >0. Therefore, this cycle is inconsistent, and so, there exists an inconsistent cycle.
Case (ii): Suppose no cycles of the form discussed in case (i) exist. Define a function
f : Ω0 → ΩP \Ω0 wheref(ω0)is found by repeated applications of feuntil a state in
ΩP \Ω0is reached. That is,f(ω0) = (fe◦ · · · ◦f)(ωe 0), with sufficiently manyfe’s so that
process cannot form a loop, as this would be a cycle of the kind studied in case (i), and cannot otherwise go forever becauseΩis finite.
LetX =f(Ω0). As each stateω ∈Xalso belongs toΩP \Ω0, there is a positive path
fromωtoΩ0. For eachω∈X, arbitrarily pick any such positive path, and call the end
of this positive pathg(ω). This defines a functiong :X →Ω0. Consider the composite
functionf ◦g :X →X. AsXis finite, the functionf◦g :X →Xmust have a cycle in the usual set-theoretic sense, namely the finite sequence{ω1, f◦g(ω1)· · ·,(f◦g)k(ω1)},
whereω1 ∈Xand(f◦g)k(ω1) =ω1. Such a set-theoretic cycle off ◦g :X →X can be
identified with a cycle of the multigraphGby the following process: letω2 = (f◦g)(ω1),
· · ·,ωk= (f◦g)(ωk−1)andω1 = (f◦g)(ωk). Stateg(ωi)can be identified with a positive
pathpifromωi tog(ωi); andf(g(ωi))can be identified with the path
qi = nD ji,1;g(ωi),fe(g(ωi) E ,Dji,2;fe(g(ωi),fe2(g(ωi) E ,· · · ,Dji,m;fem−1(g(ωi)),fem(g(ωi)) Eo ,
where fem(g(ωi)) = f(g(ωi)) and ji,l is chosen so that θ
ji,l
e
fl(g(ωi)) > 0. Hence, the path
p =p1q1p2q2· · ·pkqkis a positive path fromωtoω, and so, it is a closed positive path;
the product of posteriors alongpis strictly positive. However, the opposite path,−p, contains at least one zero posterior along its path, making the product of posteriors along −p to be zero. Hence, the cycle equation of the closed path p does not hold. Therefore,pis an inconsistent cycle because every closed path that is not a cycle has a cycle equation that holds trivially (left and right-hand sides of the equation are identical products of posteriors). Thus, there exists an inconsistent cycle, as required.
For visual and intuition purposes, in all examples a version of the meet-join diagram will be used. Versions are defined in Rodrigues-Neto [2012]. A version is a simple undirected graph, with the same nodes as the meet-join diagram, but where(i)edges of the formhj, ω, ωihave been ignored,(ii)edgeseand−ehave been merged, and(iii)
edges have been removed so there is at most onej-path between any two nodes. There is typically more than one version of a given meet-join diagram, depending on which edges are removed. However, the version always preserves the number of connected components.
Consider the model withΩ ={1,2,3,4,5},J ={A, B}andΠA ={{1,2,3},{4,5}},
andΠB ={{1,4},{2}{3,5}}. The meet-join diagram of this model, and a version, are
the edges for playerB are in red.
1
2
3
4
5
(a) Full Meet-Join Diagram
1
2
3
4
5
(b) A Version
Figure 5.2
Example 5.1 illustrates the consequences of Propositions 5.1 and 5.2 and their limi- tations.
Example 5.1. Let Ω = {1,2,3,4,5,6}, J = {A, B}, ΠA = {{1,2,5},{6},{3,4}} and
ΠB = {{1,3},{2,4},{5,6}}. A visualization of these partitions is given in Figure 5.3 with
playerAin blue and dashed, and playerB in red.
1
2
3
4
5
6
Figure 5.3:Version for Example 5.1
In each model below these features coincide, but posteriors differ.
First, suppose that posteriors are: θA1 = 0.5,θ2A = 1/3,θ3A = 0.5,θA4 = 0.5, θA5 = 1/6,
θA
6 = 1, for playerA, and for playerB,θωB = 0.5, for everyω ∈Ω. There is no common prior
with full support because the following cycle equation fails:
Indeed,θA2θB4θ3Aθ1B = 1/24, butθ3BθA4θB2θ1A= 1/16. Because all states belong to an inconsistent cycle, thenΩY =∅. For instance, states5and6belongs to the following inconsistent cycle:
{hB; 6,5i,hA; 5,1i,hB; 1,3i,hA; 3,4i,hB; 4,2i,hA; 2,5i,hB; 5,6i}. By Proposition 5.1, there is no common prior, not even one without full support.
Next, consider again the same Ω,J,ΠA and ΠB, but change the posteriors, so that now
θA
5 = 0 and θB5 > 0. State ω = 5 ∈ Ω0, and from the positive path {hB; 6,5i}, state
ω = 6∈ΩP. So, any common prior that may exist must haveµ5 =µ6 = 0. A common prior
exists (and has support in{1,2,3,4}) if and only if the cycle equationθA
2θ4BθA3θB1 =θ3BθA4θB2θ1A
is satisfied. If this cycle equation is not satisfied, thenΩN = Ω, and there is no common prior,
not even one without full support.
Finally, consider again the sameΩ,J,ΠAandΠB. Suppose thatθB
5 = 0andθ2Aθ4BθA3θB1 6=
θ3BθA4θB2θ1A. In this case,ω = 5 ∈ Ω0 and ω = 6 ∈/ ΩP, as every path starting from ω = 6
contains the edge hB; 6,5i. BecauseθB
5 = 0 andθB5 +θ6B = 1, thenθ6B = 1. Also, θA6 = 1
because stateω = 6belongs to a singleton in partitionΠA. Thus,ω = 6 ∈/ Ω0. There exists
a unique common priorµ. This common prior assigns all probabilistic mass to stateω = 6;
µ6 = 1. However, the cycle equationθA2θB4θ3Aθ1B =θB3θ4Aθ2BθA1 is not satisfied. This shows that
the converse of Proposition2is not true.