Charges and bets: a general characterisation of common priors

Abstract

We show that the equivalence of common priors and absence of agreeable bets of the famous no betting theorem can be generalised to any infinite space (not only compact spaces) if we expand the set of priors to include probability charges as priors. Going beyond the strict prior/no common prior dichotomy, we further uncover a fine-grained decomposition of the class of type spaces into a continuum of subclasses in each of which an epistemic condition approximating common priors is equivalent to a behavioural condition limiting acceptable bets.

Appendices

A Topologies Used in this Paper

We make use of the $\sup$-norm on $B(X,\mathcal {A})$, defined by $\Vert f \Vert _{\sup } = \sup _{x \in X} | f (x) |$, $f \in B(X,\mathcal {A})$. Denote by $B^*(X,\mathcal {A})$ the dual of (the set of linear functionals on) $B(X,\mathcal {A})$, noting that ${\mathrm{ba}}(X,\mathcal {A}) = B^*(X,\mathcal {A})$ (see, e.g., Dunford and Schwartz (1958) 1 Theorem p. 258). Let ${\mathrm{ba}}(X,\mathcal {A})$ be equipped with the weak* topology by $B(X,\mathcal {A})$. For any $\mu \in {\mathrm{ba}}(X,\mathcal {A})$, $\varepsilon > 0$, and $F \subseteq B (X,\mathcal {A})$ such that $| F | < \infty$, denote by

$$\begin{aligned} \mathcal {O}^*(\mu ,F,\varepsilon ) = \left\{ \nu \in {\mathrm{ba}}(X,\mathcal {A}) :\left| \int f {\mathrm{d}}\nu - \int f {\mathrm{d}}\mu \right| < \varepsilon , \ \forall f \in F \right\} , \end{aligned}$$

(4)

a neighbourhood of $\mu$. The collection of such neighbourhoods forms a subbase for the weak* topology on ${\mathrm{ba}}(X,\mathcal {A})$. $\overline{A}^*$ will denote the weak* closure of a set $A \subseteq {\mathrm{ba}}(X,\mathcal {A})$.

Notice that by the Alaoglu Compactness Theorem (see e.g. Aliprantis and Border (2006) Theorem 5.105, p. 218) the set of probability charges ${\mathrm{pba}}(X,\mathcal {A})$ is a weak* compact set.

In addition to the weak* topology, we will also make use of a different topology on ${\mathrm{ba}}(X,\mathcal {A})$, the one generated by the total variation norm. The total variation norm is defined as follows. For any $\mu \in {\mathrm{ba}}(X,\mathcal {A})$ let

$$\begin{aligned} \Vert \mu \Vert _{TV} = \sup \limits _{P \in \Pi (X,\mathcal {A})} \sum \limits _{A \in P} | \mu (A) | , \end{aligned}$$

where $\Pi (X,\mathcal {A})$ denotes the class of $\mathcal {A}$-measurable finite partitions of X. Then the collection of sets

$$\begin{aligned} \mathcal {O}_{TV} (\mu ,\varepsilon ) = \left\{ \nu \in {\mathrm{ba}}(X,\mathcal {A}) :|| \mu - \nu ||_{TV} < \varepsilon \right\} , \end{aligned}$$

$\mu \in {\mathrm{ba}}(X,\mathcal {A})$, $\varepsilon > 0$, forms a subbase of the topology by the total variation norm.

B Proofs

Proof of Lemma 1

It follows directly from Definition 4 that $\Pi _i^\varepsilon$ is convex for $\varepsilon \in [-1,1]$. Since ${\mathrm{pba}}(\Omega ,\mathcal {M})$ is a weak* compact set we have to show only that $\Pi _i = \overline{{\mathrm{conv}}(\{t_i (\omega ,\cdot ) :\omega \in \Omega \})}^*$.

It is easy to see that $\overline{{\mathrm{conv}}(\{t_i (\omega ,\cdot ) :\omega \in \Omega \})}^*\subseteq \Pi _i$. Here we show that $\Pi _i \subseteq \overline{{\mathrm{conv}}(\{t_i (\omega ,\cdot ) :\omega \in \Omega \})}^*$.

Suppose by contradiction that there exists $\mu \in \Pi _i$ such that $\mu \notin \overline{{\mathrm{conv}}(\{t_i (\omega ,\cdot ) :\omega \in \Omega \})}^*$. Then $\mu$ is separated from $\overline{{\mathrm{conv}}(\{t_i (\omega ,\cdot ) :\omega \in \Omega \})}^*$ in the weak* topology, meaning that there exists $f \in B (\Omega ,\mathcal {M})$, $f \ne 0$, and $\varepsilon > 0$ such that

$$\begin{aligned} \langle f ,\mu \rangle {:}{=} \int f {\mathrm{d}}\mu \ge \alpha + \varepsilon \text { and } \langle f ,\nu \rangle {:}{=} \int f {\mathrm{d}}\nu \le \alpha - \varepsilon \end{aligned}$$

(5)

for all $\nu \in \overline{{\mathrm{conv}}(\{t_i (\omega ,\cdot ) :\omega \in \Omega \})}^*$ (where $\alpha$ defines the separating hyperplane). Let K be a bound of f, that is, $| f | \le K$.

There exists a partition $B_1,\ldots ,B_n \in \mathcal {M}$ of $\Omega$ and a simple function $s_f$ on it such that for all $\omega \in \Omega$

$$\begin{aligned} \left| f (\omega ) - s_f (\omega ) \right| < \frac{\varepsilon }{2} \,. \end{aligned}$$

(6)

Since $\mu \in \Pi _i$, for each $B_m$, $m=1,\ldots ,n$

$$\begin{aligned} \mu (B_m) = \mu (B_m \cap \Omega ) = \int \limits _\Omega t_i (\cdot ,B_m) \, d \mu \, . \end{aligned}$$

(7)

Since $t_i (\cdot ,B_m)$ is bounded and $\mathcal {M}_i$-integrable for each $B_m$, there exist a partition $C_1^m,\ldots ,C_{n_m}^m \in \mathcal {M}_i$ of $\Omega$ and a simple function $s_{t_i (\cdot ,B_m})$such that for each $\omega \in \Omega$

$$\begin{aligned} \left| t_i (\omega ,B_m) - s_{t_i (\cdot ,B_m)} (\omega ) \right| < \frac{\varepsilon }{2Kn} \,. \end{aligned}$$

(8)

Let $C_1,\ldots ,C_l$ be the common refinement of the partitions $B_1,\ldots ,B_n$, $C_1^m,\ldots ,C_{n_m}^m$, $m=1,\ldots ,n$. It is easy to see that $C_1,\ldots ,C_l \in \mathcal {M}$ and that $\{C_1,\ldots ,C_l\}$ forms a partition of $\Omega$. Let $\omega _k \in C_k$, $k=1,\ldots ,l$ be arbitrarily fixed. Then by Equations (7) and (8) for all $m=1,\ldots ,n$

$$\begin{aligned} \left| \mu (B_m) - \sum \limits _{k=1}^l t_i (\omega _k,B_m) \mu (C_k) \right| < \frac{\varepsilon }{2Kn} \, , \end{aligned}$$

that is,

$$\begin{aligned} \left| \sum \limits _{m=1}^n s_f (B_m) \mu (B_m) - \sum \limits _{m=1}^n s_f (B_m) \sum \limits _{k=1}^l t_i (\omega _k,B_m) \mu (C_k) \right| < \frac{\varepsilon }{2} \, . \end{aligned}$$

Note that since $\mu (C_k) \ge 0$, $k=1,\ldots ,l$ and $\sum _{k=1}^l \mu (C_k) =1$ the term $\sum \nolimits _{k=1}^l t_i (\omega _k,B_m) \mu (C_k)$ is a convex combination of $t_i (\omega _k,B_m)$, $k=1,\ldots ,l$, $m = 1,\ldots ,n$. Recall that for each $\omega$, $t_i(\omega , \cdot ) \in {\mathrm{pba}}(\Omega ,\mathcal {M})$ satisfies the conditions for being a prior of player i. Hence, writing $\nu {:}{=}\sum \nolimits _{k=1}^l \mu (C_k) t_i (\omega _k,\cdot )$, we have $\nu \in \Pi _i$. Finally, by Eq. (6)

$$\begin{aligned} \left| \int f {\mathrm{d}}\mu - \int f {\mathrm{d}}\nu \right| < \varepsilon \, , \end{aligned}$$

which contradicts Equation (5). $\square$

Proof of Proposition 1

If: Suppose by contradiction that $\cap _{m=1}^n {\mathrm{cone}}(K_m ) \ne \{ 0 \}$ and that at the same time there exist linear functionals $f_1,\ldots ,f_n$ such that $f_m (x) \ge \alpha > 0$, for each $m=1\ldots ,n$ and for each $x \in K_m$, with $\sum _{m=1}^n f_m = 0$. Since $\cap _{m=1}^n {\mathrm{cone}}(K_m) \ne \{0\}$, there is an $x \ne 0$ such that $x \in \cap _{m=1}^n {\mathrm{cone}}(K_m)$; such an x then satisfies the property that there exist $\beta _1,\ldots ,\beta _n > 0$ such that $\beta _m x \in K_m$ for $m=1,\ldots ,n$, and $\sum f_m (x) \ge \alpha \sum _{m=1}^n \frac{1}{\beta _m}$. This contradicts $\sum f_m(x) = 0$.

Only if: Let $\hat{K} = \{ x \in K_1^{n-1} :x_1 = \ldots = x_{n-1},\ x_1 \in K_1)$ (so that $\hat{K}$ is an $n-1$ copy of $K_1$), and let $\tilde{K} = {\mathrm{cone}}(K_2) \times \ldots \times {\mathrm{cone}}(K_n)$. It is clear that $\tilde{K}$ is weakly* closed and convex and that $\hat{K}$ is weakly* compact and convex.

Suppose that $\cap _{m=1}^n {\mathrm{cone}}(K_m) = \{ 0 \}$. Then it follows from the definitions that $\tilde{K} \cap \hat{K} = \emptyset$, which implies that there exists a continuous linear functional g, a real number $\beta$, and $\varepsilon > 0$ such that $g(x) \ge \beta + \varepsilon$ for all $x \in \tilde{K}$ and $g(x) \le \beta - \varepsilon$ for all $x \in \hat{K}$. Since $\tilde{K}$ contains the origin, it must be the case that $\beta + \varepsilon \le 0$, which implies that $\beta < 0$. Moreover, by definition of $\tilde{K}$, $g = g_2 + \ldots + g_n$, and $g_m (x) \ge 0$ for each $x \in K_m$ and each $m=2,\ldots ,n$.

Since $K_m$ is weak* compact and $0 \notin K_m$, $m =2,\ldots ,n$, there exist $\alpha _m > 0$ and $\delta _m$ continuous functionals such that $\alpha _m \le \delta _m (x) \le \frac{-\beta }{2 (n-1)}$, $x \in K_m$, $m = 2,\ldots ,n$. Let $\alpha {:}{=}\min \{ \alpha _2, \ldots ,\alpha _n \} > 0$.

Then $(g_m + \delta _m) (x) \ge \alpha > 0$ for each $m=2,\ldots ,n$ and for each $x \in K_m$. Furthermore, $\sum _{m=2}^n (g_m + \delta _m) (x) \le \frac{\beta }{2}$ for all $x \in \hat{K}$. Note that $K_1$ and $\hat{K}$ are isomorphic, hence $\sum _{m=2}^n (g_m + \delta _m) (x) \le \frac{\beta }{2}$ for all $x \in K_1$.

Finally, for each $m=2,\ldots ,n$ let $f_m = g_m + \delta _m$. We now have all the ingredients for defining a zero-sum agreeable bet: let $f_1 = - \sum _{m=2}^n f_m$. Then $f_m (x) \ge \alpha > 0$ for each $m=1,\ldots ,n$ and each $x \in K_m$, and $\sum f_m = 0$. $\square$

Lemma 4

Let $A \subseteq {\mathrm{ba}}(\Omega ,\mathcal {M})$, $\alpha \in \mathbb {R}$, $\varepsilon \in \mathbb {R}_+$, and $f \in B(\Omega ,\mathcal {M})$. Then

$$\begin{aligned} f (x) \ge \alpha + \varepsilon || f ||_{\sup } \; \forall x \in A \iff f (x) \ge \alpha \;\forall x \in \overline{A + \mathcal {O}_{TV} (0,\varepsilon )}^*, \end{aligned}$$

and equivalently

$$\begin{aligned} f (x) \ge \alpha \;\forall x \in A \iff f (x) \ge \alpha - \varepsilon || f ||_{\sup } \;\forall x \in \overline{A + \mathcal {O}_{TV} (0,\varepsilon )}^*. \end{aligned}$$

Proof of Lemma 4

In this proof, for $f \in B(\Omega , \mathcal {M})$ and $x \in {\mathrm{ba}}(\Omega ,\mathcal {M})$ we will write $\langle f,x \rangle$ to denote f(x); this underscores the duality between $B(\Omega , \mathcal {M})$ and ${\mathrm{ba}}(\Omega ,\mathcal {M}) = B^*(\Omega , \mathcal {M})$.

Let $(x_n) \subseteq \Omega$ satisfy $| f (x_n) | \nearrow || f ||_{\sup }$.

For each $x \in A \subset {\mathrm{ba}}(\Omega ,\mathcal {M})$, it is the case that $x + \varepsilon \delta _{\{x_n\}}, x - \varepsilon \delta _{\{x_n\}} \in \overline{A + \mathcal {O}_{TV} (0,\varepsilon )}^*$ for all n.

Suppose that $\lim f (x_n) \ge 0$. Note that $x - \varepsilon \delta _{\{x_n\}} \in \overline{A + \mathcal {O}_{TV} (0,\varepsilon )}^*$ for all n and for all $x \in A$, and that in addition $\langle f, x - \varepsilon \delta _{\{x_n\}} \rangle = \langle f, x \rangle - \varepsilon \langle f, \delta _{\{x_n\}} \rangle = \langle f, x \rangle - \varepsilon f (x_n)$ for all $x \in A$. It follows that $\langle f, x \rangle - \varepsilon \lim f(x_n) = \langle f, x \rangle - \varepsilon || f ||_{\sup }$ for all $x \in A$. From this we can deduce that $\langle f, x \rangle \ge \alpha$ for all $x \in \overline{A + \mathcal {O}_{TV} (0,\varepsilon )}^*$ if and only if $\langle f, x \rangle \ge \alpha + \varepsilon || f ||_{\sup }$ for all $x \in A$.

Suppose that $\lim f (x_n) \le 0$. We have that $x + \varepsilon \delta _{\{x_n\}} \in \overline{A + \mathcal {O}_{TV} (0,\varepsilon )}^*$ for all n and for all $x \in A$ and that $\langle f, x + \varepsilon \delta _{\{x_n\}} \rangle = \langle f, x \rangle + \varepsilon f (x_n)$ for all n and for all $x \in A$, hence $\langle f, x \rangle + \varepsilon \lim f(x_n) = \langle f, x \rangle - \varepsilon || f ||_{\sup }$ for all $x \in A$. We can deduce that $\langle f, x \rangle \ge \alpha$ for all $x \in \overline{A + \mathcal {O}_{TV} (0,\varepsilon )}^*$ if and only if $\langle f, x \rangle \ge \alpha + \varepsilon || f ||_{\sup }$ for all $x \in A$. $\square$

Lemma 5

For any $\varepsilon > 0$

$$\begin{aligned} \bigcap \limits _{i \in N} \Pi _i^\varepsilon = \emptyset \iff \bigcap \limits _{i \in N} \overline{\Pi _i + \mathcal {O}_{TV} (0,\varepsilon )}^*= \emptyset . \end{aligned}$$

Proof of Lemma 5

Since $\Pi _i^\varepsilon = \overline{\Pi _i + \mathcal {O}_{TV} (0,\varepsilon )}^*\cap {\mathrm{pba}}(\Omega ,\mathcal {M})$, $i \in N$, if $\cap _{i \in N} \overline{\Pi _i + \mathcal {O}_{TV} (0,\varepsilon )}^*= \emptyset$, we deduce $\cap _{i \in N} \Pi _i^\varepsilon = \emptyset$.

Since $\Pi _i^\varepsilon \subseteq {\mathrm{pba}}(\Omega ,\mathcal {M})$ if $x \in \cap _{i \in N} \overline{\Pi _i + \mathcal {O}_{TV} (0,\varepsilon )}^*$ we deduce (by the triangle inequality) $\cap _{i \in N} \Pi _i^\varepsilon \ne \emptyset$, i.e., if $\cap _{i \in N} \Pi _i^\varepsilon = \emptyset$ then $\cap _{i \in N} \overline{\Pi _i + \mathcal {O}_{TV} (0,\varepsilon )}^*= \emptyset$. $\square$

Proof of Theorem 1

This follows immediately from Theorem 2 in the special case that $\varepsilon = 0$. $\square$

Proof of Lemma 3

If: Since $\overline{P_i}^{TV} \subseteq \overline{P_1}^*$, $i = 1,2$, if $\overline{P_1}^*$ and $\overline{P_2}^*$ are strongly separable by a weak* continuous linear functional, then the very same linear functional is norm continuous and it strongly separates $\overline{P_1}^{TV}$ and $\overline{P_2}^{TV}$.

Only if: Suppose that $\overline{P_1}^{TV}$ and $\overline{P_2}^{TV}$ are strongly separable by a norm continuous linear function (over $\ell _1$), and let f be the non-trivial strongly separating norm continuous linear functional, i.e, there exist $c \in \mathbb {R}$ and $\varepsilon > 0$ such that

$$\begin{aligned} f (x) \ge c + \varepsilon \quad \text {and} \quad c - \varepsilon \ge f (y) , \end{aligned}$$

for all $x \in \overline{P_1}^{TV}$ and for all $y \in \overline{P_2}^{TV}$.

Since the dual of $\ell _1$ is $\ell _\infty$, $f \in \ell _\infty$ and it holds that

$$\begin{aligned} f (x) \ge c + \varepsilon \quad \text {and} \quad c - \varepsilon \ge f (y) , \end{aligned}$$

for all $x \in \overline{P_1}^*$ and for all $y \in \overline{P_2}^*$. $\square$

Proof of Theorem 2

By Lemma 1, for each $i \in N$ and $\varepsilon \in \mathbb {R}$ the set $\Pi _i^\varepsilon$ is a weak* closed subset of ${\mathrm{pba}}(\Omega ,\mathcal {M})$, which is itself weak* compact, hence $\cap _{i \in N} \Pi _i^\varepsilon = \emptyset$, i.e., there is no $\varepsilon$-common prior, if and only if there exists a finite set of indices $i_1,\ldots ,i_n \in N$ such that $\cap _{m=1}^n \Pi _{i_m}^\varepsilon = \emptyset$.

If $\varepsilon = 0$ then let $K_m = \Pi _{i_m}$; if $1> \varepsilon > 0$ then let $K_m = \overline{\Pi _{i_m} + \mathcal {O}_{TV} (0,\varepsilon )}^*$ (it is clear that there always exists 1-common prior but there cannot exist 1-agreeable bet); and if $\varepsilon < 0$ then let $K_m = \Pi _{i_m}^\varepsilon$, $m=1,\ldots ,n$, as in the statement of Proposition 1. By Lemma 1 the sets $K_m$, $m=1,\ldots ,n$, are convex, weak* compact subsets of ${\mathrm{ba}}(\Omega ,\mathcal {M})$, $0 \notin K_m$, $m=1,\ldots ,n$, and $\cap _{m=1}^n {\mathrm{cone}}(K_m )= \{ 0 \}$. This last property holds if and only if $\cap _{m=1}^n K_m = \emptyset$. We can therefore apply Proposition 1 to deduce $\cap _{m=1}^n K_m = \emptyset$ if and only if there exist continuous linear functionals $f_1,\ldots ,f_n$ satisfying $\sum f_m = 0$ as in the statement of that theorem.

Appeals to Lemmata 4 and 5 then complete the argument leading to our desired result: if there is no $\varepsilon$-common prior, then there exists an $\varepsilon$-agreeable bet, and if there exists an $\varepsilon$-agreeable bet, then there is no $\varepsilon$-common prior. $\square$

Proof of Theorem 5

If: If the $\sigma$-type space admits a strong common $\varepsilon$-prior $p^\varepsilon$ for all $\varepsilon > 0$, then by that ${\mathrm{pba}}(\Omega ,\mathcal {M})$ is weak* compact $(p^{1/n})_{n \in \mathbb {N}}$ has a cluster point. Any cluster point of $(p^{1/n})_{n \in \mathbb {N}}$ is a common prior for the $\sigma$-type space.

Only if: Suppose that there exists $\varepsilon > 0$ such that the $\sigma$-type space does not admit a strong common $\varepsilon$-prior. Then by Lemma 3 it holds that the $\sigma$-type space does not admit a common prior either. $\square$

C Why bets must be bounded away from zero

Theorem 1 characterises the existence of common priors using agreeable bets (Definition 5) which not only give rise to positive expectations at each state but have expectations bounded away from zero. We show here by an example why this presumption is necessary.

There are two players, Anne and Ben. The state space and partition is the basic partition space, that is, $\Omega = \{1, 2, \ldots \}$, Anne’s knowledge partition, $\Pi ^A$, is given by

$$\begin{aligned} \left\{ \{1\}, \{2,3\}, \{4,5\}, \{6,7\}, \ldots \right\} \end{aligned}$$

and Ben’s knowledge partition, $\Pi ^B$, is given by

$$\begin{aligned} \{\{1,2\}, \{3,4\}, \{5,6\}, \ldots \}. \end{aligned}$$

The epistemic events: $\mathcal {M}$ is the field generated by the singleton sets, $\mathcal {M}_i$ is the field generated by the partition $\Pi ^i$, $i=A,B$.

Anne’s type function, $t_A$, is given by

$$\begin{aligned} t_A(n,\{n\}) = \left\{ \begin{array}{l l} 1 &{} \quad \text {if}\, n = 1\\ \frac{1}{2} &{} \quad \text {if}\, n\,\text { is even}\\ \frac{1}{2} &{} \quad \text {if}\, n\,\text { is odd,}\, n > 1.\\ \end{array} \right. \end{aligned}$$

Ben’s type function, $t_B$, is given by

$$\begin{aligned} t_B(n,\{n\}) = \left\{ \begin{array}{l l} \frac{1}{2} &{} \quad \text {if}\, n\,\text { is odd}\\ \frac{1}{2} &{} \quad \text {if}\, n\,\text { is even.}\\ \end{array} \right. \end{aligned}$$

Fig. 4

The type space of the example in this section

Notice that the probability charge which assigns zero to each finite set is a common prior for this type space (Fig. 4).

Suppose that $f_A$, with $f_B = -f_A$, is an agreeable bet, with $|f_A|$ a bounded and strictly increasing function and $f_A (n) = (-1)^{n+1} | f_A (n)|$, $n \in \Omega$. For $i = A,B$, for each $n \in \Omega$,

$$\begin{aligned} \int f_i \, \mathrm{d} t_i (n,\cdot ) > 0 , \end{aligned}$$

but there does not exist $\alpha \in \mathbb {R}$ such that for each $n \in \Omega$

$$\begin{aligned} \int f_i \, \mathrm{d} t_i (n,\cdot ) \ge \alpha > 0. \end{aligned}$$

In words, this type space admits an agreeable bet in the sense of Lehrer and Samet (2014) but it does not do so according to Definition 5.