Monotonic models and cycles

Abstract

A partitional model of knowledge is monotonic if there exists a linear order on the state space such that, for every player, each element of her partition contains only a sequence of consecutive states. In monotonic models, the absence of alternating cycles is equivalent to the property that, for every pair of players, the join of their partitions contains only singletons. Under these equivalent conditions any set of posteriors for the players is consistent (i.e., there is a common prior). When checking for consistency in a monotonic model, it is not necessary to evaluate all cycle equations; if the cycle equations corresponding to cycles of length two hold, then there is a common prior.

Appendix: Proofs

1.1 Proof of Proposition 1

In a monotonic model, suppose that, for every pair of different partitions $\Pi ^{j}\ne \Pi ^{j^{\prime }}$, their join contains only singletons; $\Pi ^{j}\vee \Pi ^{j^{\prime }}=\{\{\omega \}\left| \right. \omega \in \Omega \}$. Then, there is no pair of different states belonging to the same element of $\Pi ^{j}$ and also to the same element of $\Pi ^{j^{\prime }}$. In particular, there are no two consecutive states belonging to the same element of $\Pi ^{j}$ and, at the same time, belonging to the same element of $\Pi ^{j^{\prime }}$. This means that there is no cycle of length two; i.e., $\mathfrak I ^{*}$ is empty. Hence, this monotonic model is acyclic because, by Proposition 3, the existence of any cycle would imply the existence of a cycle of length two.

Conversely, suppose that there is no alternating cycle. Let $j_{1}\ne j_{2}\in J$ be two arbitrary players. Suppose by contradiction that there are two distinct states $\omega _{1}$ and $\omega _{2}$ in the same element of the join $\Pi ^{j_{1}}\vee \Pi ^{j_{2}}$. Because the join is a refinement of $\Pi ^{j_{1}}$, then, $\omega _{1}$ and $\omega _{2}$ belong to the same element of $\Pi ^{j_{1}}$. Hence, the meet-join diagram has the $j_{1}$-edge $\left\langle j_{1};\omega _{1},\omega _{2}\right\rangle$. Similarly, $\omega _{1}$ and $\omega _{2}$ are in the same element of $\Pi ^{j_{2}}$, and the meet-join diagram has the $j_{2}$-edge $\left\langle j_{2};\omega _{2},\omega _{1}\right\rangle$. Consider the path $c=\{\left\langle j_{1};\omega _{1},\omega _{2}\right\rangle ,\left\langle j_{2};\omega _{2},\omega _{1}\right\rangle \}$. Clearly, $c$ is an alternating cycle, which contradicts the hypothesis.

If in a monotonic model all joins of two players have only singletons, there is no cycle. By Rodrigues-Neto (2009), all posteriors of are consistent; that is, there is a common prior.

1.2 Proof of Proposition 2

If there is no cycle, the result is trivial. Consider any arbitrary cycle $p$ of arbitrary length $k\ge 2$. Without loss of generality (see footnote $5$), $p$ is an alternating cycle. Fix any arbitrary strict linear order $\succ$ on $\Omega$ that makes the model monotonic. Let $U_{0}$ be the set of edges $\left\langle j;\omega _{\alpha },\omega _{\beta }\right\rangle$ of cycle $p$ such that $\omega _{\beta }\succ \omega _{\alpha }$. Let $D_{0}$ be the set of edges $\left\langle j;\omega _{\alpha },\omega _{\beta }\right\rangle$ of cycle $p$ such that $\omega _{\alpha }\succ \omega _{\beta }$. Because cycle $p$ is a closed path, $U_{0}\ne \varnothing$ and $D_{0}\ne \varnothing$. Let $U$ be the multiset of edges obtained from the edges of $U_{0}$ in the following way: for each edge $\left\langle j;\omega _{\alpha },\omega _{\beta }\right\rangle \in U_{0}$, include in $U$ the edges $\left\langle j;\omega _{1},\omega _{2}\right\rangle , \left\langle j;\omega _{2},\omega _{3}\right\rangle , \ldots , \left\langle j;\omega _{k-1},\omega _{k}\right\rangle$, such that $\omega _{1}=\omega _{\alpha }$ , $\omega _{k}=\omega _{\beta }$, and for every $i\in \{1,\ldots ,k-1\}, \omega _{i+1}$ and $\omega _{i}$ are consecutive according to $\succ$, with $\omega _{i+1}\succ \omega _{i}$. The monotonicity of the model implies that $\omega _{1}, \omega _{2}, \ldots , \omega _{k}$ belong to the same element of player $j$’s partition. Hence, every $\left\langle j;\omega _{i},\omega _{i+1}\right\rangle$ belongs to the meet-join diagram. Construct a multiset $D$ from the set $D_{0}$ in an analogous way. It is crucial to define $U$ and $D$ as multisets (instead of sets), so they can include multiple copies of the same edge, if necessary. Because $p$ is a closed path and because we defined $U$ and $D$ as multisets, they must always have the same number of elements, $\left| U\right| =\left| D\right|$. Because $U_{0}\ne \varnothing$ and $D_{0}\ne \varnothing$, then $U\ne \varnothing$ and $D\ne \varnothing$. Because $p$ is a closed path, there is a bijection $F$ from $U$ to $D$ such that each edge $u=\left\langle j_{U};\omega _{i},\omega _{i+1}\right\rangle$ of $U$ is associated to edge $F(u)=\left\langle j_{D};\omega _{i+1},\omega _{i}\right\rangle$ of $D, j_{U}\ne j_{D}$, (edge $u$ “moves up” from $\omega _{i}$ to $\omega _{i+1}$; edge $F(u)$ “goes down” from $\omega _{i+1}$ to $\omega _{i}$),¹⁰ or else, in this process, there is a link between two consecutive states $\omega _{i}$ to $\omega _{i+1}$ such that $j_{U}=j_{D}$. In the first case, every $\{u,F(u)\}$ , for $u\in U$, is a cycle of length two, and the collection of cycle equations corresponding to the collection of cycles $\{\{u,F(u)\}\left| \right. u\in U\}$ generates the cycle equation corresponding to cycle $p$. Hence, if all cycle equations corresponding to the collection of (length two) cycles $\{\{u,F(u)\}\left| \right. u\in U\}$ are satisfied, the cycle equation corresponding to cycle $p$ must also be satisfied. In the second case, there are edges $u=\left\langle j;\omega _{i},\omega _{i+1}\right\rangle \in U$ and $\left\langle j;\omega _{i+1},\omega _{i}\right\rangle \in D$, and then, $\{u,F(u)\}$ is not a cycle, but only a closed path. Let $Z$ be the sub-collection of $\{\{u,F(u)\}\left| \right. u\in U\}$ such that each $\{u,F(u)\}$ in $Z$ is a cycle. This sub-collection has only cycles of length two (by construction) and is non-empty because $p$ is a cycle. Moreover, if all cycle equations corresponding to cycles in $Z$ are satisfied, the cycle equation corresponding to cycle $p$ must also be satisfied.¹¹ In both cases, all cycle equations hold. Because $\Omega$ is finite, then, by Rodrigues-Neto (2009), there is a common prior.

1.3 Proof of Proposition 3

This proof follows similar steps to those in the previous argument. Given any arbitrary cycle $c$, the decomposition of $c$ proposed in the Proof of Proposition 2 leads to cycles of length two belonging to the collection $\mathfrak I ^{*}$. If $\mathfrak I ^{*}$ is non-empty, it must contain a basis for the cycle space because the previous part of this proposition states that $\mathfrak I ^{*}$ spans the cycle space. Every maximal linearly independent subset of $\mathfrak I ^{*}$ is a basis for the cycle space. There is at least one maximal linearly independent subset of $\mathfrak I ^{*}$ because the collection $\mathfrak I ^{*}$ is non-empty and finite.