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 atomFootnote 5A. 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 \(\omega \in \Omega \).