An upper bound for the $\ell _1$-variation along the road to agreement

Abstract

Two agents with a common prior on the possible states of the world participate in a process of information transmission, consisting of sharing posterior probabilities of an event of interest. Aumann’s Agreement Theorem implies that such a process must end with both agents having the same posterior probability. We show that the $\ell _1$-variation of the sequence of posteriors of each agent, obtained along this process, must be finite, and provide an upper bound for its value.

Appendices

Appendix A: $\ell _1$-Variation of the conditional means of a martingale

Let $(\Omega , {\mathcal {B}}, \mathbb {P})$ be a probability space. A filtration $\mathcal {F}= (\mathcal {F}_n)_{n=1}^{\infty }$ is an increasing sequence of sub-$\sigma$-fields of ${\mathcal {B}}$. A sequence $(X_n)_{n=1}^{\infty }$ of random variables is said to be a discrete-time martingale with respect to the filtration $(\mathcal {F}_n)_{n=1}^{\infty }$ if

• (i) $\mathbb {E}|X_n| < \infty$,

• (ii) $(X_n)_n$ is adapted to $\mathcal {F}$, i.e., $X_n$ is measurable with respect to $\mathcal {F}_n$ for all $n \in \mathbb {N}$,

• (iii) $\mathbb {E}(X_{n+1}\,|\,\mathcal {F}_n)=X_n$ for all $n \in \mathbb {N}$. An important example of a discrete-time martingale that we will consider in the paper is a Doob martingale. Such a martingale arises when we consider the sequence of conditional expectations $X_n = {\mathbb {E}}(Y\,|\,\mathcal {F}_n)$ of a random variable Y satisfying ${\mathbb {E}}|Y|<\infty$, with respect to a filtration $\mathcal {F}= (\mathcal {F}_n)_{n=1}^{\infty }$. In case $Y = \mathbb {1}_C$ for some event C (i.e., $C \in {\mathcal {B}}$), we will use a standard notation and write $X_n = \mathbb {P}(C\,|\,\mathcal {F}_n)$.

We say that the discrete-time martingale sequence $X = (X_n)_{n=1}^{\infty }$ is an ${\mathcal {L}}_2$ (${\mathcal {L}}_1$) bounded martingale if $\sup _n {\mathbb {E}}(|X_n|^2)<\infty$ ($\sup _n {\mathbb {E}}|X_n|<\infty$). Let us now introduce two distinct types of variation for discrete-time martingales. The $\ell _1$-variation of the discrete-time martingale sequence $X = (X_n)_{n=1}^{\infty }$ (with respect to $\mathcal {F}$) is the random variable

$$\begin{aligned} V(X) = \sum \limits _{n=1}^{\infty } |X_{n+1}-X_{n}|. \end{aligned}$$

(13)

Next, for every event A with $\mathbb {P}(A)>0$, we define the $\ell _1$-variation of the conditional means of X on A by

$$\begin{aligned} V(X,A) = \sum \limits _{n=1}^{\infty } \big |{\mathbb {E}}(X_{n+1}\,|\,A) -{\mathbb {E}}(X_{n}\,|\,A)|, \end{aligned}$$

(14)

where

$$\begin{aligned} {\mathbb {E}}(X_{n}\,|\,A) = \frac{{\mathbb {E}} (X_n\mathbb {1}_A)}{\mathbb {P}(A)}, \ \ \forall n \in {\mathbb {N}}. \end{aligned}$$

(15)

We say that X has finite $\ell _1$-variation of the conditional means if $V(X,A)<\infty$ for all events A with $\mathbb {P}(A)>0$.

Our main technical result regarding martingales bounds the $\ell _1$-variation of the conditional means.

Theorem A1

If X is an ${\mathcal {L}}_2$-bounded martingale, then

$$\begin{aligned} V(X,A) \le \frac{{\mathbb {E}}(X_{\infty }^2)-{\mathbb {E}}(X_{1}^2)}{2\mathbb {P}(A)} + \frac{{\mathbb {E}}(Y_{\infty }^2)-{\mathbb {E}}(Y_{1}^2)}{2\mathbb {P}(A)} \end{aligned}$$

(16)

for all events A with $\mathbb {P}(A)>0$, where $X_{\infty }$ is the ${{\mathcal {L}}}_2$-limit of X, and $Y_{\infty }$ is the ${{\mathcal {L}}}_2$-limit of the ${\mathcal {L}}_2$-bounded martingale Y defined by $Y_n = \mathbb {P}(A \,|\,\mathcal {F}_n)$ for all $n \in \mathbb {N}$. In particular, X has finite $\ell _1$-variation of conditional means.

Remark A1

The conditional mean of the $\ell _1$-variation of an ${\mathcal {L}}_2$-bounded martingale can be infinite on any event of positive probability. For instance, let $(d_n)_{n=1}^{\infty }$ be a sequence of independent random variables distributed according to the law $\mathbb {P}\left( d_n = \frac{1}{n}\right) = \mathbb {P}\left( d_n = -\frac{1}{n}\right) = \frac{1}{2}$. Define the martingale $M = (M_n)_{n=1}^{\infty }$ by $M_n = d_1 + \cdots + d_n$ for all $n \in {\mathbb {N}}$. Then M is an ${\mathcal {L}}_2$-bounded martingale satisfying

$$\begin{aligned} {\mathbb {E}}\left( \sum \limits _{n=1}^{\infty } |M_{n+1}-M_n|\,\big |\,A\right) = \sum \limits _{n=1}^{\infty } \frac{1}{n} = \infty , \end{aligned}$$

for every event A with $\mathbb {P}(A)>0$.

As a corollary of Theorem A1 we obtain the following result.

Corollary A1

Suppose that X is a discrete-time ${\mathcal {L}}_2$-bounded martingale. Suppose that the probability space $(\Omega , {\mathcal {B}}, \mathbb {P})$ contains an atom⁵A. Let $(a_n)_{n=1}^{\infty }$ be the (a.s. fixed) values of $(X_n)_{n=1}^{\infty }$ on A. Note that $a_n = {\mathbb {E}}(X_n\,|\,A)$ for all $n \in {\mathbb {N}}$, thus implying that $V(X) = V(X,A)$ a.s. on A. With the help of Theorem A1 we may deduce that

$$\begin{aligned} V(X) \le \frac{{\mathbb {E}}(X_{\infty }^2)-{\mathbb {E}}(X_{1}^2)}{2\mathbb {P}(A)} + \frac{{\mathbb {E}}(Y_{\infty }^2)-{\mathbb {E}}(Y_{1}^2)}{2\mathbb {P}(A)}, \end{aligned}$$

(17)

a.s. on A.

This result is strongly related to the following theorem of Burkholder (1966), which holds for a wider class of martingales.

Theorem A2

(Burkholder, 1966) Suppose that X is an ${\mathcal {L}}_1$-bounded martingale. If A is an atom of the probability space, then

$$\begin{aligned} V(X)<\infty , \end{aligned}$$

almost everywhere on A.

Let us now proceed to the proof of the Theorem A1.

Proof of Theorem A1

By conditioning on ${\mathcal {F}}_{n+1}$, one has

$$\begin{aligned} {\mathbb {E}}\left( X_{n+1}\mathbb {1}_{A}\right) = {\mathbb {E}}\left( X_{n+1}Y_{n+1}\right) ,\quad \forall n \ge 0. \end{aligned}$$

(18)

Similarly, by conditioning on ${\mathcal {F}}_{n}$ we have ${\mathbb {E}}\left( X_{n+1}Y_{n}\right) = {\mathbb {E}}\left( X_{n}Y_{n}\right)$ for all $n \in {\mathbb {N}}$. Therefore, with the use of Eq. (18) we obtain

$$\begin{aligned} {\mathbb {E}}\left( X_{n}\mathbb {1}_{A}\right) = {\mathbb {E}}\left( X_{n}Y_{n}\right) = {\mathbb {E}}\left( X_{n+1}Y_{n}\right) ,\, \ \ \forall n \in {\mathbb {N}}. \end{aligned}$$

(19)

Hence by combining Eqs. (18) and (19) we have

$$\begin{aligned} \sum \limits _{n=1}^{\infty } |{\mathbb {E}} (X_{n+1}\mathbb {1}_A)-{\mathbb {E}} (X_n\mathbb {1}_A)|= & {} \sum \limits _{n=1}^{\infty }\big |{\mathbb {E}}X_{n+1}(Y_{n+1}-Y_n)\big | \nonumber \\= & {} \sum \limits _{n=1}^{\infty }\big |{\mathbb {E}}(X_{n+1}-X_n)(Y_{n+1}-Y_n)\big | \nonumber \\\le & {} \sum \limits _{n=1}^{\infty }{\mathbb {E}}\big |(X_{n+1}-X_n)(Y_{n+1}-Y_n)\big | \nonumber \\\le & {} \sum \limits _{n=1}^{\infty }{\mathbb {E}}\left( \frac{(X_{n+1}-X_n)^2+(Y_{n+1}-Y_n)^2}{2}\right) \nonumber \\= & {} \frac{1}{2} \sum \limits _{n=1}^{\infty } {\mathbb {E}}(X_{n+1}-X_n)^2 + \frac{1}{2}\sum \limits _{n=1}^{\infty }{\mathbb {E}}(Y_{n+1}-Y_n)^2 \nonumber \\= & {} \frac{1}{2}\left( {\mathbb {E}}(X_{\infty }^2)-{\mathbb {E}}(X_{1}^2)\right) + \frac{1}{2}\left( {\mathbb {E}}(Y_{\infty }^2)-{\mathbb {E}}(Y_{1}^2)\right) , \end{aligned}$$

(20)

where the second inequality holds since $|ab| \le \frac{a^2+b^2}{2}$ for every $a,b \in \mathbb {R}$, and the last equality follows from Theorem 12.1 in Williams (1991). Thus, combining Eq. (20) together with Eqs. (14) and (15) we deduce that

$$\begin{aligned} V(X,A) \le \frac{{\mathbb {E}}(X_{\infty }^2)-{\mathbb {E}}(X_{1}^2)}{2\mathbb {P}(A)} + \frac{{\mathbb {E}}(Y_{\infty }^2)-{\mathbb {E}}(Y_{1}^2)}{2\mathbb {P}(A)} \end{aligned}$$

as desired. $\square$

The following example shows that Theorem A1 cannot be extended to ${{\mathcal {L}}}_1$-bounded martingales, by providing an ${{\mathcal {L}}}_1$-bounded martingale M and an event A for which $V(M,A) = \infty$.

Example A1

Consider the i.i.d. random variables $(d_n)_{n=1}^{\infty }$ distributed according to the law

$$\begin{aligned} \mathbb {P}\left( d_n = 0\right) =\mathbb {P}\left( d_n = 2\right) = \frac{1}{2}. \end{aligned}$$

Let ${\mathcal {D}} \subseteq {\mathcal {G}}$ be the smallest $\sigma$-field on which the random variables $(d_n)_{n=1}^{\infty }$ are measurable. Define the martingale $M = (M_n)_{n=1}^{\infty }$ on $(\Omega ,{\mathcal {D}},\mathbb {P})$ with respect to the natural filtration induced by $(d_n)_{n=1}^{\infty }$ by $M_n = \prod _{k=1}^{n} d_k,\ \ \forall n \in \mathbb {N}.$ Since $M_n\ge 0$ for every $n \in \mathbb {N}$, we have $\mathbb {E}|M_n| = \mathbb {E}M_n = 1$, implying that M is bounded in ${{\mathcal {L}}}_1$. For each $n \in \mathbb {N}$ define the event $A_n = \big \lbrace M_1 =2 , M_2 =4,\ldots ,M_n=2^n,M_{n+1}=0\big \rbrace ,$ and let $A = \bigcup _{n=1}^{\infty }A_n$. We have

$$\begin{aligned} \mathbb {P}(A)V(M,A) = \sum \limits _{n=1}^{\infty } \mathbb {P}(A_n)V(M,A_n)= \sum \limits _{n=1}^{\infty } 2^{-(n+1)}(2+4+\cdots +2^n)=\infty , \end{aligned}$$

and so $V(M,A) = \infty$.

Remark A2

The events $(A_n)_{n=1}^{\infty }$ in Example A1 are disjoint atoms of the probability space $(\Omega ,{\mathcal {D}},\mathbb {P})$, and thus the result of Burkholder (Theorem A2) cannot be extended to an infinite union of atoms of ${{\mathcal {L}}}_1$-bounded martingales.

Appendix B: Complements to Aumann’s Bayesian dialogue

Proof of Proposition 1

We begin by proving (i). It is easily verified that $\mathbb {P}(A) = Q^1_1$. Since the partition element of agent i is determined by the pair $(n,x) \in \mathbb {N}\times \{0,\ldots ,n\}$ of tosses he was allotted and the number of H outcomes he observed, following the notation of Theorem 2 we have

$$\begin{aligned} \sum \limits _{s \in {\hat{Y}}_i} \frac{\mathbb {P}(F_i(s)\cap A)^2}{\mathbb {P}(F_i(s))}= & {} \sum \limits _{n=1}^{\infty } \sum \limits _{x=0}^n \left[ \left( t^i_n \left( {\begin{array}{c}n\\ x\end{array}}\right) \int _0^1 \theta ^{x+1} (1-\theta )^{n+1-(x+1)} dQ(\theta )\right) ^2 \bigg /(t^i_n Q^n_x) \right] \nonumber \\= & {} \sum \limits _{n=1}^{\infty } \sum \limits _{x=0}^n \left[ \left( t^i_n \frac{(x+1)Q^{n+1}_{x+1}}{n+1}\right) ^2 \bigg /(t^i_n Q^n_x)\right] \nonumber \\= & {} \sum \limits _{n=1}^{\infty } t^i_n \sum \limits _{x=0}^n \frac{(Q^{n+1}_{x+1})^2}{Q^n_x}\left( \frac{x+1}{n+1}\right) ^2. \end{aligned}$$

(21)

Moreover,

$$\begin{aligned} \mathbb {P}(C(\omega ))= & {} t(n_1,n_2)\left( {\begin{array}{c}n_1\\ x_1\end{array}}\right) \left( {\begin{array}{c}n_2\\ x_2\end{array}}\right) \int _0^1 \theta ^{x_1+x_2} (1-\theta )^{n_1+n_2-(x_1+x_2)}dQ(\theta )\nonumber \\= & {} t(n_1,n_2)\gamma (n_1,x_1,n_2,x_2) Q_{x_1+x_2}^{n_1+n_2}. \end{aligned}$$

(22)

Combining the latter with Eqs. (7), and (21) we obtain Eq. (11), thus showing (i). Let us now move on and assume the Q is the uniform distribution on [0, 1]. For each $n \in \mathbb {N}$, and $x \in \{0,\ldots ,n\}$ we have

$$\begin{aligned} Q_x^n = \left( {\begin{array}{c}n\\ x\end{array}}\right) \int _0^1 \theta ^x (1-\theta )^{n-x}d\theta= & {} \left( {\begin{array}{c}n\\ x\end{array}}\right) \mathbf{B }(x+1,n-x+1)\nonumber \\= & {} \left( {\begin{array}{c}n\\ x\end{array}}\right) \frac{\Gamma (x+1)\Gamma (n-x+1)}{\Gamma (n+2)}\nonumber \\= & {} \frac{1}{n+1}, \end{aligned}$$

(23)

where $\mathbf{B }$ is the beta-function, for which we used the identity $\mathbf{B }(\alpha ,\beta ) = \frac{\Gamma (\alpha )\Gamma (\beta )}{\Gamma (\alpha +\beta )}$. In turn, Eq. (23) implies that

$$\begin{aligned}&Q^1_1-\sum \limits _{n=1}^{\infty } t^i_n \sum \limits _{x=0}^n \frac{(Q^{n+1}_{x+1})^2}{Q^n_x}\left( \frac{x+1}{n+1}\right) ^2 \\&\quad = \frac{1}{2} - \sum \limits _{n=1}^{\infty } t^i_n \sum \limits _{x=0}^n \frac{n+1}{(n+2)^2}\left( \frac{x+1}{n+1}\right) ^2 \\&\quad = \frac{1}{2} - \sum \limits _{n=1}^{\infty } t^i_n \frac{1}{(n+1)(n+2)^2} \sum \limits _{x=0}^n (x+1)^2 \\&\quad = \frac{1}{2} - \sum \limits _{n=1}^{\infty } t^i_n \frac{(n+1)(n+2)(2n+3)}{6(n+1)(n+2)^2} \\&\quad = \frac{1}{2} - \sum \limits _{n=1}^{\infty } t^i_n \frac{2n+3}{6n+12} \le \frac{1}{2} - \frac{1}{4} \sum \limits _{n=1}^{\infty } t^i_n = \frac{1}{4}. \end{aligned}$$

Item (i) in Proposition 1 yields the bound

$$\begin{aligned} \sum \limits _{n=1}^{\infty } |p_{n+1}^{i}(\omega )-p_{n}^{i}(\omega )| \le \frac{1}{2} + \frac{n_1+n_2+1}{8t(n_1,n_2)\gamma (n_1,x_1,n_2,x_2)}, \end{aligned}$$

(24)

proving the second item of Proposition 1. $\square$

Remark B1

In the proof of Proposition 1 we did not fully utilize the bound presented in Theorem 2. In fact, we did not subtract the quantity $\mathbb {P}(C(\omega ))/2\mathbb {P}(F_i(\omega ))$. The reason for this is that $\mathbb {P}(C(\omega ))/2\mathbb {P}(F_i(\omega )) \le \frac{1}{2}$ $\forall \omega \in \Omega$, making it negligible compared to the right-hand side of Eq. (11), which need not be bounded across different values of $\omega \in \Omega$.