Countable spaces and common priors

Abstract

We show that the no betting characterisation of the existence of common priors over finite type spaces extends only partially to improper priors in the countably infinite state space context: the existence of a common prior implies the absence of a bounded agreeable bet, and the absence of a common improper prior implies the existence of a bounded agreeable bet. However, a type space that lacks a common prior but has a common improper prior may or may not have a bounded agreeable bet. As a side-benefit of the proofs here, we also obtain a constructive proof of the no betting characterisation in finite spaces.

Appendix

1.1 Chains

Definition 4

A chain ⁴ of length $n\ge 0$, for a partition profile $\mathbf \Pi$, from one state to another, is defined by induction on $n$. A state $\omega _0$ is a chain of length 0 from $\omega _0$ to $\omega _0$. A chain of length $n+1$, from $\omega _0$ to $\omega$, is a sequence $c\stackrel{i}{\rightarrow }\omega$, where $c$ is a chain of length $n$ from $\omega _0$ to $\omega '$, and $\omega \in \Pi _i(\omega ')$. Thus, a chain of positive length $n$ is a sequence $c =\omega _0\stackrel{i_0}{\rightarrow }\omega _1\stackrel{i_1}{\rightarrow }\dots \stackrel{i_{n-1}}{\rightarrow }\omega _n$, such that $\omega _{s+1}\in \Pi _{i_s}(\omega _s)$ for $s=0,\dots , n-1$.

We write $\omega \rightarrow \omega '$ when there is a chain from $\omega$ to $\omega '$, in which case we say that $\omega$ and $\omega '$ are connected by a chain. The binary relation $\rightarrow$ is the transitive closure of the union of the relations $\stackrel{i}{\rightarrow }$, and it is an equivalence relation.

Definition 5

Given a chain $c =\omega _0\stackrel{i_0}{\rightarrow }\omega _1\stackrel{i_1}{\rightarrow }\dots \stackrel{i_{n-1}}{\rightarrow }\omega _n$, its reverse chain $c^{-1}$ is defined as

$$\begin{aligned} c^{-1} \mathrel {\mathop :}= \omega _n\stackrel{i_{n-1}}{\rightarrow }\omega _{n-1} \stackrel{i_{n-2}}{\rightarrow }\dots \stackrel{i_{0}}{\rightarrow }\omega _0. \end{aligned}$$

A chain $c$ is alternating if no two consecutive states $\omega _s$ and $\omega _{s+1}$ in $c$ are the same, and no two consecutive agents $i_s$ and $i_{s+1}$ in $c$ are the same.

Hellman and Samet (2012) prove that a partition profile $\mathbf{\Pi }$ is connected if and only if every two states are connected by at least one chain.

1.2 Positive, zero, and singular states

Definition 6

Given a type space $\tau$, a state $\omega \in {\Omega }$ is:

• positive if $t_i(\omega ) > 0$ for all $i \in I$;

• zero if $t_i(\omega ) = 0$ for all $i \in I$;

• singular if it is neither positive nor zero.

Based on the categorisation of states in Definition 6, define the following:

Definition 7

Given a type space $\tau$,

• A subset $S \subseteq {\Omega }$ is $i$-positive if $t_i(\omega ) > 0$ for all $\omega \in S$.

• A subset $S \subseteq {\Omega }$ is positive if it is $i$-positive for all $i$ (equivalently, if every $\omega \in S$ is a positive state). A chain $c$ satisfying the condition that every element $\omega \in c$ is a positive state is a positive chain.

• A subset $S \subseteq {\Omega }$ is $i$-non-singular if $t_i(\omega )= 0$ for every singular $\omega \in S$.

• A subset $S \subseteq {\Omega }$ is non-singularly positive if it is positive, and every maximal chain $c$ entirely contained in $S$ satisfies the property that for every $\omega \in c$ and every $i \in I, \pi _i(\omega )$ is $i$-non-singular.

A subset $S$ of ${\Omega }$ is thus non-singularly positive if it is positive, and for every $\omega \in S$, and every $i \in I$, every $\omega ' \in \Pi _i(\omega )$ satisfies the condition that either

• $\omega ' \in S$, or

• $\omega '$ is a zero state, or

• $\omega '$ is a singular state such that $t_i(\omega ') = 0$.

Note that it is immediate by definition that if ${\Omega }$ is a positive state space, then trivially the entire space ${\Omega }$ is non-singularly positive.

Lemma 2

If $S$ is a non-singularly positive subset of ${\Omega }$, then for any $\omega \in S$, every $\omega ' \in {\Omega }$ that is connected to $\omega$ via a positive chain is also an element of $S$. It follows that every non-singularly positive subset $S$ can be decomposed as $S = \cup T_j$, where each $T_j$ is a non-singularly positive subset such that all of the members of $T_j$ are connected to each other by positive chains.

Proof

Suppose that $S \subseteq {\Omega }$ is non-singularly positive, and let $\omega _0 \in S$ be chosen arbitrarily. Suppose that $\omega _0$ is connected to $\omega _n$ by a positive chain $c =\omega _0\stackrel{i_0}{\rightarrow }\omega _1\stackrel{i_1}{\rightarrow }\dots \stackrel{i_{n-1}}{\rightarrow }\omega _n$. Since $\omega _0$ is in $S$ and $\omega _1$ is in the same partition element of player $i_0$ as $\omega _0$, by Definition 7, $\omega _1$ must also be in $S$, and continuing the same argument by induction, we conclude that all the elements in $c$ are members of $S$. $\square$

1.3 Type ratios

Definition 8

Let $\tau$ be a type space and $(\omega _1,\omega _2)$ an ordered pair of positive states in $\pi \in \Pi _i$. The type ratio of $(\omega _1,\omega _2)$ relative to $i$ is ⁵ $\mathrm{tr}_\tau ^i(\omega _1,\omega _2)= t_i(\pi ,\omega _1)/t_i(\pi ,\omega _2)$. If a chain $c=\omega _0\stackrel{i_0}{\rightarrow }\omega _1\stackrel{i_1}{\rightarrow }\dots \stackrel{i_{n-1}}{\rightarrow }\omega _n$ of length $n>0$ is a positive chain, the type ratio of $c$ is $\mathrm{tr}_\tau (c)=\times _{k=0}^{n-1}\,\mathrm{tr}_\tau ^{i_k}(\omega _k,\omega _{k+1})$. For a positive chain $c$ of length 0, $\mathrm{tr}_\tau (c)= 1$. Thus, if $c=c'\stackrel{i}{\rightarrow }\omega$ where $c'$ is a positive chain from $\omega _0$ to $\omega '$ and $\omega '$ is a positive state, $\mathrm{tr}_\tau (c)=\mathrm{tr}_\tau (c')\mathrm{tr}_\tau ^i(\omega ',\omega )$. ⁶

We note here for later use two equalities involving type ratios that follow immediately from the definitions:

• For any chain $c$,

  $$\begin{aligned} \mathrm{tr}(c^{-1}) = [\mathrm{tr}(c)]^{-1}, \end{aligned}$$

  (7)

• If $c =\omega _1\stackrel{i}{\rightarrow }\omega _2\stackrel{i}{\rightarrow }\omega _3$ (i.e., $\omega _2,\omega _3 \in \Pi _i(\omega _1)$), then

  $$\begin{aligned} \mathrm{tr}(c) = \mathrm{tr}^i(\omega _1,\omega _3). \end{aligned}$$

  (8)

The following proposition extends the results of Proposition 4 in Hellman and Samet (2012), from positive type spaces to general type spaces.

Proposition 2

Let $\tau$ be a type space with a connected knowledge space. Then there exists a common improper prior for $\tau$ if and only if ${\Omega }$ has a non-singularly positive subspace $S$ with respect to $\tau$, and for each $\omega _0$ and $\omega$ in $S$, and chains $c$ and $c'$ entirely contained in $S$ from $\omega _0$ to $\omega , \mathrm{tr}_\tau (c)=\mathrm{tr}_\tau (c')$.

Proof

Suppose that there exists a common improper prior $p$ for $\tau$. Let $S = \{\omega \in {\Omega }\text { } | \text { } p(\omega ) > 0\}$. $S$ is guaranteed to be positive, because $p \ne 0$. We next show that $S$ is non-singularly positive: Suppose that for arbitrary $i \in I$ and $\omega \in S, \omega ' \in \Pi _i(\omega )$. Furthermore, suppose that $\omega ' \notin S$. Then $p(\omega ') = 0$, while $p(\omega ) > 0$. Hence $p(\Pi _i(\omega )) > 0$, and by the definition of an improper prior,

$$\begin{aligned} t_i(\omega ') = \frac{p(\omega ')}{p(\Pi _i(\omega ))} = 0. \end{aligned}$$

It follows from Definition 7 that $S$ is non-singularly positive.

To complete this part of the proof, note that for any pair of states $\omega _1,\omega _2 \in S$ such that $\omega _1$ and $\omega _2$ are in the same element of $\Pi _i$ for some $i \in I, \mathrm{tr}_t^i(\omega _1,\omega _2)= p(\omega _1)/p(\omega _2)$. It then easily follows from the definition of the type ratio of a chain that for any chain $c$ entirely contained in $S$ and connecting $\omega _0$ and $\omega$, one has $\mathrm{tr}_{\tau }(c) = p(\omega _0)/p(\omega )$.

Conversely, suppose that ${\Omega }$ has a non-singularly positive subspace $S$ with respect to $\tau$, and that for each $\omega _0$ and $\omega$ in $S$, any pair of chains $c$ and $c'$ entirely contained in $S$ connecting $\omega _0$ to $\omega$ satisfy $\mathrm{tr}_\tau (c)=\mathrm{tr}_\tau (c')$. Using Lemma 2, we may assume that $S$ is connected (replacing $S$ by a connected subset of itself if necessary).

We will construct a cip $p$. For $\omega \notin S$, set $p(\omega ) = 0$. Otherwise, fix $\omega _0 \in S$ and for each $\omega \in S$, let $p(\omega )=\mathrm{tr}(c)$ for some chain $c$ from $\omega _0$ to $\omega$ contained in $S$.

To see that $p$ is a cip consider $\pi \in \Pi _i$. Suppose first that $\pi \cap S = \emptyset$. Then for all $\omega \in \pi , p(\omega ) = 0$, hence $p(\pi ) = 0$, and $p(\pi )t_i(\omega )=p(\omega )$ is satisfied.

Suppose instead that $\pi \cap S \ne \emptyset$, and that $\omega \in \pi \cap S$. Let $c$ be a chain from $\omega _0$ to $\omega$ entirely contained in $S$. For $\omega '\in \pi \cap S$, consider the chain $c'=c\stackrel{i}{\rightarrow }\omega '$. Then, by the definitions of $\mathrm{tr}$ and $p, p(\omega ')=\mathrm{tr}(c')=\mathrm{tr}(c)\mathrm{tr}^i(\omega ,\omega ')=p(\omega )t_i(\pi )(\omega ')/t_i(\pi )(\omega )$. Thus,

$$\begin{aligned} p(\pi )=\sum _{\omega '\in \pi \cap S}p(\omega ') =[p(\omega )/t_i(\pi )(\omega )]\sum _{\omega '\in \pi \cap S}t_i(\pi )(\omega ') =p(\omega )/t_i(\pi )(\omega )<\infty , \end{aligned}$$

and $p(\omega )=p(\pi )t_i(\pi )(\omega )$.

Finally, suppose that $\pi \cap S \ne \emptyset$, and that $\omega \in \pi$ is such that $\omega \notin S$. By construction, $p(\omega ) = 0$, and by the assumption that $S$ is non-singularly positive, it must be the case that $t_i(\omega ) = 0$. We have already shown that $p(\pi ) < \infty$, hence $p(\pi )t_i(\omega )=p(\omega )$ is satisfied. $\square$

Definition 9

A chain $c =\omega _0\stackrel{i_0}{\rightarrow }\omega _1\stackrel{i_1}{\rightarrow }\dots \stackrel{i_{n-1}}{\rightarrow }\omega _n$, where $\omega _{s+1}\in \Pi _{i_s}(\omega _s)$ for $s=0,\dots , n-1$, is a cycle ⁷ if $\omega _n = \omega _0$. If with respect to a cycle $c$ of length $n$ there is a pair $s,s' \in \{0,\dots , n-1\}$ such that $s' > s + 1$ and $\omega _{s'} \in \Pi _{i_s}(\omega _s)$, then we say that $c$ has a self-crossing point at $\omega _{s'}$. A cycle $c$ is a non-crossing cycle if it is alternating, and has no self-crossing points, i.e., for every pair $s,s' \in \{0,\dots , n-1\}$ such that $s' > s + 1, \omega _{s'} \notin \Pi _{i_s}(\omega _s)$.

Definition 9 leads to an immediate corollary of Proposition 2:

Corollary 1

Let $\tau$ be a type space with a connected knowledge space. Then there exists a common improper prior for $\tau$ if and only if ${\Omega }$ has a non-singularly positive subspace $S$ with respect to $\tau$, and for each $\omega$ in $S$, every cycle $\overline{c}=\omega \rightarrow \omega$ that is entirely contained in $S$ satisfies $\mathrm{tr}(\overline{c})=1$.

Proof

It suffices to note the following: suppose that $c_1$ and $c_2$ are two distinct chains entirely contained in $S$ connecting a pair of states $\omega$ and $\omega '$. Then $\overline{c} \mathrel {\mathop :}= c_1c_2^{-1}$ is a cycle connecting $\omega$ to itself. By Eq. (7), $\mathrm{tr}(c_1) = \mathrm{tr}(c_2)$ if and only if $\mathrm{tr}(\overline{c}) = 1$. $\square$

In fact, we can do even better, and show that it suffices to check the type ratios only of non-crossing cycles, instead of all cycles:

Proposition 3

Let $\tau$ be a type space with a connected knowledge space. Then there exists a common improper prior for $\tau$ if and only if ${\Omega }$ has a non-singularly positive subspace $S$ with respect to $\tau$, and every non-crossing cycle $\overline{c}$ that is entirely contained in $S$ satisfies $\mathrm{tr}(\overline{c})=1$.

Proof

If there exists a cip, then by Corollary 1, there is a non-singularly positive $S \subseteq {\Omega }$ such that every cycle contained in $S$ has type ratio equal to $1$, hence in particular every non-crossing cycle satisfies the same property.

In the other direction, if there does not exist a cip, then for every non-singularly positive $S \subseteq {\Omega }$, there is at least one cycle $\overline{c}$ entirely contained in $S$ such that $\mathrm{tr}(\overline{c}) \ne 1$. Suppose that $\overline{c}$ is not non-crossing.

If $\overline{c}$ fails to be non-crossing because it is not alternating, this ‘flaw’ can easily be corrected: if two consecutive states $\omega _s$ and $\omega _{s+1}$ in $\overline{c}$ are identical, since that implies that $\mathrm{tr}^i(\omega _s,\omega _{s+1}) = 1$, the state $\omega _{s+1}$ is redundant and can be removed from $\overline{c}$ without changing the type ratio. Similarly, if $\omega _s, \omega _{s+1}, \omega _{s+2}$ and $\omega _{s+3}$ are consecutive states that are all members of the same partition element of player $i$, then $\mathrm{tr}^i(\omega _s,\omega _{s+1})\mathrm{tr}^i(\omega _{s+2},\omega _{s+3}) = \mathrm{tr}^i(\omega _s,\omega _{s+3})$, hence we may remove $\omega _{s+1}$ and $\omega _{s+2}$ from $\overline{c}$ without changing the type ratio.

We will therefore assume that $\overline{c}$ is alternating but not non-crossing, and that we can then write $\overline{c} =\omega _0\stackrel{i_0}{\rightarrow }\omega _1\stackrel{i_1}{\rightarrow }\dots \stackrel{i_{n-1}}{\rightarrow }\omega _n = \omega _0$, where $\omega _{s+1}\in \Pi _{i_s}(\omega _s)$ for $s=0,\dots , n-1$, where there exists at least one pair $r,k$ such that $k > r+1$, and $\omega _{k} \in \Pi _{i_r}(i_r)$.

We can ‘shorten’ $\overline{c}$ into another cycle:

$$\begin{aligned} \widehat{c} = \omega _0\stackrel{i_0}{\rightarrow }\omega _1\stackrel{i_1}{\rightarrow }\dots \omega _r\stackrel{i_r}{\rightarrow }\omega _k\stackrel{i_k}{\rightarrow }\dots \stackrel{i_{n-1}}{\rightarrow }\omega _n = \omega _0. \end{aligned}$$

If $\mathrm{tr}(\widehat{c}) \ne 1$, then we have a cycle of type ratio not equal to $1$, with a number of self-crossing points that is strictly less than the number of self-crossing points in $\overline{c}$, and we can continue by induction to apply the same process to $\mathrm{tr}(\widehat{c})$.

Suppose, therefore, that $\mathrm{tr}(\widehat{c}) = 1$. Denote:

$$\begin{aligned}&c_0 =\omega _0\stackrel{i_0}{\rightarrow }\omega _1\stackrel{i_1}{\rightarrow }\dots \stackrel{i_{r-1}}{\rightarrow }\omega _r,\\&c_k =\omega _k\stackrel{i_k}{\rightarrow }\omega _{k+1} \stackrel{i_{k+1}}{\rightarrow }\dots \stackrel{i_{n-1}}{\rightarrow }\omega _n , \end{aligned}$$

and

$$\begin{aligned} c_{l} =\omega _r\stackrel{i_r}{\rightarrow }\omega _{r+1} \stackrel{i_{r+1}}{\rightarrow }\dots \stackrel{i_{k-1}}{\rightarrow }\omega _{k}. \end{aligned}$$

Then $\overline{c} = c_0c_{l}c_k$, and $\widehat{c} = c_0(\omega _r,\omega _k)c_k$. By assumption, $1 = \mathrm{tr}(\widehat{c}) = \mathrm{tr}(c_0)\mathrm{tr}^{i_r}(\omega _r,\omega _k)\mathrm{tr}(c_k)$. It follows that $[\mathrm{tr}^{i_r}(\omega _r,\omega _k)]^{-1} = \mathrm{tr}^{i_r}(\omega _k,\omega _r) = \mathrm{tr}(c_0)\mathrm{tr}(c_k)$. We also assumed that $\mathrm{tr}(\overline{c}) \ne 1$, so $1 \ne \mathrm{tr}(c_0c_{l}c_k) = \mathrm{tr}(c_0)\mathrm{tr}(c_k)\mathrm{tr}(c_l) = \mathrm{tr}^{i_r}(\omega _k,\omega _r)\mathrm{tr}(c_l)$.

Writing out the last inequality in full yields

$$\begin{aligned} \mathrm{tr}(\omega _k\stackrel{i_r}{\rightarrow }\omega _r\stackrel{i_r}{\rightarrow }\omega _{r+1}\stackrel{i_{r+1}}{\rightarrow }\dots \stackrel{i_{k-1}}{\rightarrow }\omega _{k})\ne 1. \end{aligned}$$

But by Eq. (8), $\mathrm{tr}(\omega _k\stackrel{i_r}{\rightarrow }\omega _r\stackrel{i_r}{\rightarrow }\omega _{r+1}) = \mathrm{tr}(\omega _k\stackrel{i_r}{\rightarrow }\omega _{r+1})$, hence

$$\begin{aligned} \mathrm{tr}(\omega _k\stackrel{i_r}{\rightarrow }\omega _{r+1}\stackrel{i_{r+1}}{\rightarrow }\dots \stackrel{i_{k-1}}{\rightarrow }\omega _{k})\ne 1. \end{aligned}$$

We deduce then that the cycle $\widetilde{c}=\omega _k\stackrel{i_r}{\rightarrow }\omega _{r+1}\stackrel{i_{r+1}}{\rightarrow }\dots \stackrel{i_{k-1}}{\rightarrow }\omega _{k}$ satisfies both that $\mathrm{tr}(\widetilde{c}) \ne 1$, and that it has a number of self-crossing points that is strictly less than the number of self-crossing points in $\overline{c}$. We can continue by induction to apply the same process to $\mathrm{tr}(\widetilde{c})$.

After applying this reasoning as often as necessary, we arrive at the existence of a cycle entirely contained in $S$ with no self-crossing points, i.e., a non-crossing cycle, whose type ratio is not equal to $1$, which is what we needed to show. $\square$

Note that it follows from the definitions that if $(\omega _1,\omega _2)$ is an ordered pair of positive states in $\pi \in \Pi _i^X$ then

$$\begin{aligned} \mathrm{tr}_{\tau ^X}^i(\omega _1,\omega _2)= \frac{t_i^X(\omega _1)}{t_i^X(\omega _2)} = \frac{t_i(\omega _1)}{\tau _i(\pi )}\frac{\tau _i(\pi )}{t_i(\omega _2)} = \frac{t_i(\omega _1)}{t_i(\omega _2)} = \mathrm{tr}_{\tau }^i(\omega _1,\omega _2), \end{aligned}$$

(9)

from which it further immediately follows that for any chain $c$ of $\tau$ whose elements are entirely contained in $X$,

$$\begin{aligned} \mathrm{tr}_{\tau ^X}(c) = \mathrm{tr}_{\tau }(c). \end{aligned}$$

(10)

We need one more definition.

Definition 10

Let $\tau$ be a type space over $({\Omega },\mathbf{\Pi })$, let $X \subseteq {\Omega }$ be a positive subset of ${\Omega }$, and let $f$ be a random variable. A state $\omega \in X$ is a surplus state for player $i$ relative to $f$ and $\tau ^X$ if $E_i^X(f|\omega ) > 0$. In the context of a sequence $f = \{f_1, \ldots , f_m\}$ of r.v., we will say that $\omega$ is an $i$ -surplus state if $f_i$ is a surplus state for player $i$.

1.4 Constructing disagreements

Proposition 4

Let $\tau$ be a type space over $({\Omega },\mathbf{\Pi })$, let $S$ be a finite connected subset of positive states in ${\Omega }$, and let $X \subseteq S$. Suppose that there exists an agreeable bet relative to $\tau ^X$. Then there exists an agreeable bet relative to $\tau ^S$.

Proof

Let $f$ be an agreeable bet relative to $\tau ^X$. If $X = S$, there is nothing to prove. If $X \subset S$, then by the assumption of the connectedness of $S$, we can find at least one player $i$ and a point $\omega ' \notin X$ such that $\Pi _i(\omega ') \cap X \ne \emptyset$. By the assumption of positivity, $t_i(\omega ') > 0$, and by the assumption that $f$ is an agreeable bet, every state $\omega \in \Pi _i(\omega ')$ is an $i$-surplus state relative to $f$.

Denote $Y \mathrel {\mathop :}= X \cup \omega '$, and let $\varepsilon$ be the (by the $i$-surplus state assumption) positive value

$$\begin{aligned} \varepsilon \mathrel {\mathop :}= \sum _{\omega '' \in \Pi _i(\omega ') \cap X}f_i(\omega '')t_i^X(\omega ''). \end{aligned}$$

(11)

Next, let $\overline{f}_i(\omega ')$ be a negative real number satisfying

$$\begin{aligned} 0 > \overline{f}_i(\omega ') > \frac{-(1 - t_i^Y(\omega '))}{t_i^Y(\omega ')}\varepsilon \text { ,} \end{aligned}$$

(12)

and for $j \ne i$, set $\overline{f}_j(\omega ') \mathrel {\mathop :}= -\overline{f}_i(\omega ')/(m-1) > 0$, where $m = |I|$. Clearly, by construction, $\sum _{j \in I} \overline{f}_j(\omega ') = 0$. Complete the definition of $\overline{f}$ by letting $\overline{f}(\omega '') \mathrel {\mathop :}= f(\omega '')$ for all $\omega '' \in X$.

It is straightforward to check that $\overline{f}$ is an agreeable bet relative to $\tau ^Y$. Now simply repeat this procedure as often as necessary to extend the agreeable bet to every state in the finite set $S$. $\square$

Lemma 3

Let $\tau$ be a type space over $({\Omega },\mathbf{\Pi })$, and let $X$ be a non-crossing cycle such that $\mathrm{tr}(X) \ne 1$. Then there exists a random variable $f$ that is an agreeable bet relative to $\tau ^X$.

Proof

Write the non-crossing cycle as $X =\omega _1\stackrel{i_1}{\rightarrow }\omega _2\stackrel{i_2}{\rightarrow }\dots \omega _n\stackrel{i_{n}}{\rightarrow }\omega _{n+1} = \omega _0$, where $\omega _{s+1}\in \Pi _{i_s}(\omega _s)$ for $s=1,\dots , n$. Assume without loss of generality that $\mathrm{tr}(X) < 1$ (otherwise simply reverse the ordering of states in the cycle). To cut down on notational clutter, write $r_s \mathrel {\mathop :}= \mathrm{tr}^{i_{s}}(\omega _{s},\omega _{s+1})$, hence $\mathrm{tr}(X) = r_1r_2 \cdots r_{n}$. Furthermore, denote by $P$ the set such that $i \in P$ if and only if $i$ is one of $i_1, i_2, \ldots i_n$ used in the presentation above of the cycle $X$.

We will several times make use of the following simple technical observation: suppose that $i$ is a player, $\pi$ is a partition element of $\Pi _1$, and $\omega ',\omega '' \in \pi$. Furthermore, suppose that $g$ is a random variable satisfying the property that $g(\omega ) = 0$ for all $\omega \in \pi$ such that $\omega \ne \omega ',\omega ''$. Then:

$$\begin{aligned} \left\{ \begin{array}{l@{\quad }l@{\quad }l} E_i(g|\omega ') = 0 &{} \ \ \ \ \ &{} \text { if } \quad g(\omega '') = -\mathrm{tr}^i(\omega ',\omega '')g(\omega ') ,\\ E_i(g|\omega ') > 0 &{} \ \ \ \ \ &{} \text { if } \quad g(\omega '') > -\mathrm{tr}^i(\omega ',\omega '')g(\omega ') ,\\ E_i(g|\omega ') < 0 &{} \ \ \ \ \ &{} \text { if } \quad g(\omega '') < -\mathrm{tr}^i(\omega ',\omega '')g(\omega '). \end{array}\right\} \end{aligned}$$

(13)

We now proceed to the construction of an agreeable bet, in stages.

First note that by the assumption that $r_1r_2 \cdots r_{n} < 1$, we may choose a $\delta _n > 1$ such that $r_1r_2 \cdots r_{n}\delta _n < 1$. We may then further define $\delta _{n-1}$ such that $\delta _n > \delta _{n-1} > 1$ and so on, to yield a sequence $\delta _n > \delta _{n-1} > \ldots > \delta _2 > 1$. Use this to define $\overline{f}$ by:

$$\begin{aligned} \begin{array}{l@{\quad }l@{\quad }l} \overline{f}_{i_1}(\omega _1) = -1 &{} \ \ \ &{} \\ \overline{f}_{i_n}(\omega _1 = \omega _{n+1}) = 1 &{} \ \ \ &{} \\ \overline{f}_{i_{s-1}}(\omega _{s}) = \delta _{s}r_1r_2 \cdots r_{s-1} &{} \ \ \ &{} \text {for} \quad 2 \le s \le n \\ \overline{f}_{i_s}(\omega _s) = -\delta _{s}r_1r_2 \cdots r_{s-1} &{} \ \ \ &{} \text {for} \quad 2 \le s \le n \\ \overline{f}_j(\omega ) = 0 &{} \ \ \ &{} \text { for all other } \omega \text { and } j. \end{array} \end{aligned}$$

Using Eq. (13) repeatedly (here is also where we use the assumption that $X$ is non-crossing, which ensures that in each partition element of every player $i$ there are at most two states at which $\overline{f}_i$ takes on non-zero values) we deduce that $E_{i_s}(\overline{f}_{i_s}|\omega _s) > 0$ for $1 \le s \le n+1$.

We still need to ensure that the players who are not in $P$ have positive expectations at the states participating in $X$. To do so, note the following: since $r_1r_2 \cdots r_{n}\delta _n < 1$, we can choose an $\varepsilon _{n+1}$ such that $1 - \varepsilon _{n+1} > r_1r_2 \cdots r_{n}\delta _n$. Similarly, since for any $2 \le s \le n, \delta _sr_1r_2 \ldots r_{s-1} > \delta _{s-1}r_1r_2 \ldots r_{s-1}$, we can choose $\varepsilon _{s}$ such that

$$\begin{aligned} \delta _sr_1r_2 \ldots r_{s-1} - \varepsilon _{s} > \delta _{s-1}r_1r_2 \ldots r_{s-1}. \end{aligned}$$

At each state in $\omega _s \in X$, therefore, we intuitively can take away a positive part of the positive value of $\overline{f}_{i_{s-1}}(\omega _s)$ and ‘redistribute’ it among the other players. This enables the following construction:

$$\begin{aligned} \begin{array}{lll} f_{i_s}(\omega _s) = \overline{f}_{i_s}(\omega _s) &{} \ \ \ &{} \text {for} \quad 1 \le s \le n \\ f_{i_{s-1}}(\omega _s) = \overline{f}_{i_{s-1}}(\omega _s) - \varepsilon _{s }&{} \ \ \ &{} \text {for} \quad 1 \le s \le n \\ f_{j}(\omega _s) = \varepsilon _{s}/(m-2) &{} \ \ \ &{} \text { for } 1 \le s \le n, \text {for}\quad j \in I, j \ne i_s, i_{s-1} \\ f_j(\omega ) = 0 &{} \ \ \ &{} \text { for all other } \omega \text { and } j. \end{array} \end{aligned}$$

By construction, $f = \{f_1, \ldots , f_m\}$ is an agreeable bet relative to $\tau ^X$, which is what we needed to show. $\square$

The following proposition, one half of the no betting characterisation for finite type spaces, has several proofs in the literature, all of which ultimately rely on convex separation theorem, or equivalents thereof. The proof presented here (building on the previous lemmas) is, in contrast, constructive and entirely combinatorial.

Proposition 5

Let $\tau$ be a type space over $({\Omega },\mathbf{\Pi })$, where ${\Omega }$ is a finite state space. If $\tau$ has no common prior then there exists an agreeable bet relative to $\tau$.

Proof

Recalling that in a finite state space every improper prior can be normalised, and is therefore a proper prior, we may refer to all the previous lemmas and apply them restricting attention to the special case of common priors, rather than the more general common improper prior.

Let $S \subseteq {\Omega }$ be the set of non-singularly positive states in ${\Omega }$, and (using Lemma 2) decompose $S$ disjointly as $S = \cup _{j=1}^k T_j$, where each $T_j$ is non-singularly positive and connected. By Proposition 3, for each $1 \le j \le k$, there is a non-crossing cycle $c_{T_j}$ contained in $T_j$ such that $\mathrm{tr}(c_{T_j}) \ne 1$. Lemma 3 together with Proposition 4 imply that there exists $f^{T_j} = \{f^{T_j}_1, \ldots , f^{T_j}_k\}$ such that $f^{T_j}_i(\omega ) = 0$ for all $\omega \notin T_j$ and $i \in I$, and $f^{T_j}$ is an agreeable bet relative to $\tau ^{T_j}$.

Let $Z$ be the set of zero states, and denote by $Q$ the complement of $S \cup Z$. Define $f^Q$ in stages as follows.

In stage $0$, let $W_0$ be the set of singular states in $Q$. For every $\omega \in W_0$ decompose $I$ as $I = J \cup K$, where $\tau _j(\omega ) > 0$ for all $j \in J$ and $\tau _i(\omega ) = 0$ for $i \in K$, and choose an arbitrary $i' \in K$. Then define $g_j^0(\omega ) = 1/|J|$ for every $j \in J$, with $g^0_{i'}(\omega ) = -1$, and $g^0_i(\omega ) = 0$ for all other $i \in K, i \ne i'$. Set $g_k^0(\omega ) = 0$ for all states $\omega$ not in $W_0$ and all players $k \in I$.

Note, for the rest of this proof, that by definition every state in $Q \setminus W_0$ is a positive state.

In stage $1$, let $W_1$ be the set of all states $\omega \in Q \setminus W_0$ satisfying the property that there is at least one player $i$, and a state $\omega ' \in W_0$, such that $\omega \in \pi _i(\omega ')$. For each $\omega \in W_1$, choose $\omega '$ as just described. If $t_i(\omega ') > 0$, then by construction $\omega '$ is an $i$-surplus state relative to $g^0$. Hence we can apply a process similar to that described in the paragraph preceding and following Eq. (12) to extend $g^0$ to a function $g^1$ such that $g^1_i(\omega ) < 0$, but for all $j \ne i, g^1_j(\omega ) > 0, \sum _{j \in I}g^1_j(\omega ) = 0$, and is a surplus state for all players.

If $t_i(\omega ') = 0$, then note that since $\omega$ is positive but not contained in any non-singularly positive set, there must be a chain $c =\omega \stackrel{i_0}{\rightarrow }\omega _1\stackrel{i_1}{\rightarrow }\dots \stackrel{i_{n-1}}{\rightarrow }\omega _n$ entirely contained in $Q \setminus W_0$ such that there is a player $i$ and a state $\omega ' \in X_0$ satisfying $\omega _n \in \pi _i(\omega ')$ and $t_i(\omega ') > 0$. But then we can apply the same argument as in the previous paragraph to extend $g^0$ to $g^1$ by induction over all the states $\omega _n, \omega _{n-1}, \ldots , \omega$, yielding a function such that $\sum _{j \in I}g^1_j(\omega ) = 0$, and is a surplus state for all players.

In all stages $l > 1$, in stage $l - 1$, a function $g^{l-1}$ has been defined such that each state $\omega \in W_{l-1}$ is an $i$-surplus state for every player $i$ relative to $g^{l-1}$. Denote by $W_l$ the set containing every state $\omega \in Q \setminus W_{l-1}$ satisfying the property that there is at least one $i \in I$ and $\omega \in W_{l-1}$ such that $\omega ' \in \pi _i(\omega )$. Since $\omega$ is an $i$-surplus state relative to $g^{l-1}$, we can again apply the same technique as in the previous paragraphs to extend $g^{l-1}$ to $g^l$.

By the finiteness of ${\Omega }$, this iterative process ends after a finite number of stages $r$. Finally, set $f^Q \mathrel {\mathop :}= g^r$, and define

$$\begin{aligned} f \mathrel {\mathop :}= f^{T_1} + f^{T_2} + \ldots f^{T_k} + f^{Q}. \end{aligned}$$

It is straightforward to check that $f$, by construction, is an agreeable bet. $\square$

As mentioned in the introduction, putting together the elements of the proofs in this section yields an algorithm that can be applied in finite type spaces. The algorithm determines whether the space has a common prior, by listing the connected non-singularly positive subspaces and checking whether there is such a subspace such that every non-crossing cycle contained in it satisfies $\mathrm{tr}(c) = 1$ (see Proposition 3). If it does have a common prior, the algorithm then constructs the common prior, by the method in the proof of Proposition 2. If the space does not have a common prior, the algorithm constructs an agreeable bet by the method in the proof of Proposition 5, thus finding a sequence of random variables about whose expected values the players ‘agree to disagree’.

Fig. 3

The knowledge space in Example 3

Example 3

This simple example illustrates how to compute an acceptable bet given a knowledge space without a common prior. The space of states is${\Omega }= \{w,x,y,z\}$. Player $1$’s partition of ${\Omega }$ is $\{\{w,x\}, \{y,z\}\}$, with corresponding posteriors $\{\{1/8,7/8\}, \{4/5,1/5\}\}$ and Player $2$’s partition is $\{\{w,y\}, \{x,z\}\}$, with corresponding posteriors $\{\{1/2,1/2\}, \{1/4,3/4\}\}$. See Fig. 3.

The sequence $w, x, y, z$ forms a cycle, with the corresponding type ratios $\frac{1}{7}, \frac{1}{3}, \frac{1}{4}$ and $1$. Multiplying them together gives $\frac{1}{84}$, hence the type ratio of the cycle is not equal to one, and therefore this type space has no common prior, hence there must exist an agreeable bet.

To compute an agreeable bet, choose real numbers $1 < \delta _2 < \delta _3 < \delta _4 < 84$. For example, let $\delta _2 = 2, \delta _3 = 6$ and $\delta _4 = 8$. Use these to define a function $f(w) \mathrel {\mathop :}= -1, f(x) \mathrel {\mathop :}= \delta _2\frac{1}{7} = \frac{2}{7}, f(y) \mathrel {\mathop :}= -\delta _3 \frac{1}{7}\frac{1}{3}= \frac{6}{21}$ and $f(z) \mathrel {\mathop :}= \delta _4 \frac{1}{7}\frac{1}{3} \frac{1}{4} = \frac{8}{84} = \frac{2}{21}$. $\{f, -f\}$ is an agreeable bet.