A Bayesian Game without epsilon equilibria

TL;DR

This paper constructs a three-player Bayesian game demonstrating the non-existence of epsilon equilibria in measurable strategies for small epsilon, while showing that non-measurable equilibria do exist, highlighting limitations of epsilon equilibrium concepts.

Contribution

It introduces a specific Bayesian game where epsilon equilibria in measurable strategies do not exist, revealing new insights into equilibrium existence in Bayesian games.

Findings

No epsilon equilibria in measurable strategies for small epsilon

Existence of non-measurable equilibria in the constructed game

Highlights limitations of epsilon equilibrium concepts in Bayesian games

Abstract

We present a three player Bayesian game for which there is no epsilon equilibria in Borel measurable strategies for small enough epsilon, however there are non-measurable equilibria.

Equations

$$R_1\begin{array}{ccc}& {\mathbf{a}}_0& {\mathbf{a}}_1\\ {\mathbf{b}}_0& 300& 0\\ {\mathbf{b}}_1& 0& 100\end{array}\text{mbox}and{R}_2\begin{array}{ccc}& {\mathbf{a}}_0& {\mathbf{a}}_1\\ {\mathbf{b}}_0& 100& 0\\ {\mathbf{b}}_1& 0& 300\end{array}.$$

$$R_1\begin{array}{ccc}& {\mathbf{a}}_0& {\mathbf{a}}_1\\ {\mathbf{b}}_0& 100& 0\\ {\mathbf{b}}_1& 0& 300\end{array}\text{mbox}and{R}_2\begin{array}{ccc}& {\mathbf{a}}_0& {\mathbf{a}}_1\\ {\mathbf{b}}_0& 300& 0\\ {\mathbf{b}}_1& 0& 100\end{array}.$$

$${\mathbf{b}}_0\begin{array}{ccc}& {\mathbf{a}}_0& {\mathbf{a}}_1\\ {\mathbf{a}}_0& 1000& 0\\ {\mathbf{a}}_1& 0& 2000\end{array}{\mathbf{b}}_1\begin{array}{ccc}& {\mathbf{a}}_0& {\mathbf{a}}_1\\ {\mathbf{a}}_0& 0& 1000\\ {\mathbf{a}}_1& 2000& 0\end{array}$$

$${\mathbf{b}}_0\begin{array}{ccc}& {\mathbf{a}}_0& {\mathbf{a}}_1\\ {\mathbf{a}}_0& 0& 1000\\ {\mathbf{a}}_1& 2000& 0\end{array}{\mathbf{b}}_1\begin{array}{ccc}& {\mathbf{a}}_0& {\mathbf{a}}_1\\ {\mathbf{a}}_0& 2000& 0\\ {\mathbf{a}}_1& 0& 1000\end{array}$$

$$N=\{x\in X:U(x)=V(x),\text{mbox}forsomedistinctU,V\in G^{+}\}.$$

Full text

A Bayesian Game without $\epsilon$-equilibria

Robert Samuel Simon and Grzegorz Tomkowicz

Robert Simon

London School of Economics

Department of Mathematics

Houghton Street

London WC2A 2AE

email: R.S.Simon@lse.ac.uk

Grzegorz Tomkowicz

Centrum Edukacji $G^{2}$

ul.Moniuszki 9

41-902 Bytom

Poland

e-mail: gtomko@vp.pl

Abstract

We present a three player Bayesian game for which there is no $\epsilon$-equilibria in Borel measurable strategies for small enough positive $\epsilon$, however there are non-measurable equilibria.

Key words: Bayesian games, non-amenable semi-group action, equilibrium existence

1 Introduction

Game theory is about strategies and the expected payoffs resulting from these strategies. The fundamental concept of game theory is that of an equilibrium, strategies for each player such that no player prefers to switch to another strategy, given that the strategies of the other players remain fixed. An $\epsilon$-equilibrium for any $\epsilon\geq 0$ is defined in the same way – no player prefers by more than $\epsilon$ to switch to another strategy. An equilibrium is a [math]-equilibrium. What do we mean by a player preferring some strategy over another by at least $\epsilon$? Implied by a preference is an evaluation in an ordered field, usually the real numbers. Conventionally we assume that there is a topological probability space on which the game takes place, that each player is assigned a Borel sigma field corresponding to its knowledge of the space, and that that the players choose strategies that are measurable with respect to their respective Borel sigma fields. Given fixed measurable strategies of the other players, the evaluation of a strategy for a player is global, an integration of a function over the whole probability space. When the players attain an $\epsilon$-equilibrium by this criteria, a global or Harsanyi $\epsilon$-equilibrium is obtained. On the other hand, a player could orient itself locally to some minimal sets in its sigma algebra, known as information sets (should they exist), and maximise its payoff with regard to these sets and its knowledge of them (which we presume is consistent with its knowledge of the whole probability space through regular conditional distributions). When each player at each such minimal set cannot obtain more than $\epsilon$ with the payoff evaluated locally at this set, a Bayesian $\epsilon$-equilibrium is obtained. Bayesian games are ancient, most card games being good examples. J. Harsanyi (1967) introduced a global theoretical perspective to Bayesian games, the origin of the term Harsanyi equilibrium. P. Milgrom and R. Weber (1985) asked implicitly the question whether Bayesian games always have measurable equilibria after proving existence for a special class of Bayesian games and analysing a game which did not belong to that class. R. Simon (2003) demonstrated an example of a three player Bayesian game for which there is no Borel measurable equilibrium however there are many non-measurable equilibria. This example was made possible by a structure of knowledge for which always the players have common knowledge of a countable set of measure zero. These sets of measure zero give no orientation for compiling the many local equilibria toward Borel measurable strategies. Indeed, elementary applications of ergodic theory showed that Borel measurable equilibria could not exist. A significant advance was performed by Z. Hellman (2014). He showed that there is a two-player Bayesian game with Bayesian equilibria but no Bayesian $\epsilon$-equilibrium that is also Borel measurable for small enough positive $\epsilon$. This discovery was advanced further by Z. Hellman and J. Levy (2016), in which it was demonstrated that a broad class of knowledge structures support games for which the same holds. This paper serves well as a general source to the structure, problems, and history of Bayesian games, especially in their relation to countable equivalence relations that are amenable. It is required of a Bayesian $\epsilon$-equilibrium that throughout the space each player cannot gain locally more than $\epsilon$ through an alternative strategy. However in a Harsanyi $\epsilon$ -equilibrium there could be a player who, according to the strategies defining the $\epsilon$ equilibrium, could improve its payoff by as much as $B>0$ at a set of measure no more than $\epsilon/B$. A Borel measurable Bayesian $\epsilon$-equilibrium is a Harsanyi $\epsilon$-equilbrium, but not vice versa. Indeed it is not difficult to show that for every $\epsilon>0$ there are Harsanyi $\epsilon$-equilibria to the Hellman game cited above. One can perceive sets of very small measure where a player can act foolishly as a kind of firewall, absorbing the problems of the measurability requirement. Amenable structures tend to allow for such firewalls; for example with the closely related topic of Borel colouring; see Kechris, et all (1999). Therefore we would not have expected to find a game example lacking Harsanyi $\epsilon$ equilibria (yet possessing Bayesian equilibria) without utilizing a non-amenable structure.

In this paper we demonstrate that there is a Bayesian game played on a topological probability space $\Omega$ for which there are no Harsanyi $\epsilon$-equilibria for all $\epsilon\leq\frac{1}{1000}$, and yet there are non-measurable Bayesian equilibria that employ pure strategies almost everywhere (pure meaning all weight given to one action). Because $\Omega$ is a Cantor set and the concern is the existence of approximate equilibria, moving from Borel to Lesbegue measurability does not alter the result.

As long as a game has an Harsanyi $\epsilon$-equilibrium for every positive $\epsilon$ there is an equilibrium payoff, namely a cluster point of payoffs corresponding to the $\epsilon$-equilibria as $\epsilon$ goes to [math]. By this interpretation of an equilibrium payoff, ours is a Bayesian game that has equilibria, but no equilibrium payoff.

With our example, there is no proper subset of the probability space for which the players have common knowledge, hence the arguments used are different from that of previous Bayesian games that lack Harsanyi equilibria but have Bayesian equilibria (which do utilize countable equivalence relations). Nevertheless we use a non-amenable semi-group action that contains some similarities to the structure of the Hellman example. The rest of this paper is organised as follows. In the next section we define the game, followed in the third section by a proof that this game has no Harsanyi $\frac{1}{1000}$-equilibrium. In the fourth section we show that it does have non-measurable Bayesian equilibria. The concluding fifth section is a presentation of open problems.

2 The Game:

Let $G^{+}$ be the free semi-group generated by the non-negative powers of the two generators $T_{1}$ and $T_{2}$, with $e$ the identity included in $G^{+}$. Let $X$ be the space $\{0,1\}^{G^{+}}$, with an $x\in X$ a collection of the form $(x^{e},x^{T_{1}},x^{T_{2}},x^{T_{1}T_{2}},x^{T_{2}T_{1}},$

$x^{T_{1}^{2}},x^{T_{2}^{2}},\dots)$ with $x^{U}\in\{0,1\}$ for every $U\in G^{+}$. For both $i=1,2$ define $T_{i}:X\rightarrow X$ to be the shift: $T_{i}(x)^{V}=x^{T_{i}V}$ for all $V\in G^{+}$, and for every $U,V\in G^{+}$ define $U:X\rightarrow X$ by $U(x)^{V}=x^{UV}$. This defines a right action of $G^{+}$ on $X$, meaning that $UV(x)=V\circ U(x)$. For every $x\in X$ define $G^{+}(x)$ to be the countable set $\{U(x)\ |\ U\in G^{+}\}$. Very important to the structure of $X$ is that for every $x,y\in X$ there are two $z_{1},z_{2}\in X$ such that $T_{1}(z_{i})=x$ and $T_{2}(z_{i})=y$ for both $i=1,2$. We call the points the twins determined by $x$ and $y$. We place the canonical probability distribution $m$ on $X$ which gives $\frac{1}{2}$ to each [math] or $1$ placed in each position of $G^{+}$ and independently, so that a cylinder defined by $n$ positions is given the probability $2^{-n}$. With this probability distribution, we see that all $U\in G^{+}$ are measure preserving actions, meaning that $m(U^{-1}A)=m(A)$ for all Borel subsets $A$. Of special importance, the indendence probability assumption on each position implies that the distribution $m$ on $X$ can be reconstructed from the measure preserving property of $T_{1}:X\rightarrow X$ and $T_{2}:X\rightarrow X$, its product distribution $m^{2}$, combined with the equal probability given to both twins.

Let $D$ be the set $D:=\{r,g\}$, $r$ for red $r$ and $g$ for green. The probability space on which the game is played will be $\Omega:=D\times X$. We define the topology on $\Omega$ to be that induced by the clopen (closed and open) sets defined by the set $D$ and the cylinder sets of $X$, so that $\Omega$ is a Cantor set. We define the canonical probability distribution $\mu$ on $\Omega$ so that for each choice of $d\in D$ and [math] or $1$ in $n$ distinct positions the probability for this cylinder set will be $\frac{1}{2}\ 2^{-n}$. For example, $\mu$ gives the set $\{(r,x)\ |\ x^{e}=0,x^{T_{1}}=1\}$ the probability $\frac{1}{8}$. The measure $\mu$ is the common prior for the game, meaning the Borel probability measure by which the game is defined.

There are three players, labelled $G_{0},R_{1},R_{2}$. The information sets of each player are defined as follows. For each $x\in X$ Player $G_{0}$ considers $(g,x)$ and $(r,x)$ possible, and with equal $\frac{1}{2}$ probability, and these two points constitute its information set. For each $i=1,2$ and each $x$ Player $R_{i}$ consider $(r,x)$ and $\{g\}\times T_{i}^{-1}(x)$ possible, with the point $(r,x)$ and the set $\{g\}\times T_{i}^{-1}x$ of equal $\frac{1}{2}$ probability, and this pairing of a point with the corresponding Cantor set is its information set. Notice that this belief by the player $R_{i}$ is consistent with the probability distribution on $\Omega$, as the measure preserving property of the $T_{i}$ implies that $m(T_{i}^{-1}(A))=m(A)$ for all Borel subsets $A$ of $X$. If $B$ is the information set of a player, it means that this player cannot distinguish between any two points of this set and therefore has to conduct the same behaviour throughout the set. The Borel sigma algebra defining the knowledge of a player is that induced by these information sets, meaning the collection of all Borel sets $B$ such that every information set of this player is either inside of $B$ or disjoint from $B$. All players have only two actions. The red players $R_{1}$ and $R_{2}$ have the actions ${\bf a_{0}},{\bf a_{1}}$ and the green player $G_{0}$ has the actions ${\bf b_{0}}$, ${\bf b_{1}}$. For either player $R_{i}$ the only payoff that matters is that obtained at those states labelled $r$, and for the player $G_{0}$ the same is true for those states labelled $g$. There are two equivalent approaches to be taken, illustrated for a player $R_{i}$. Either the payoff obtained at $(r,x)$, described below, is duplicated at all the other points in the same information set, namely the set $\{g\}\times T^{-1}_{i}(x)$, or the payoffs obtained at $(r,x)$, described below, is multiplied by $2$ and at all other points in the same information set the payoff is [math]. Though the latter interpretation may be better suited to some theoretical approaches, as it employs the probability distribution $\mu$ on $\Omega$, we will assume throughout the former equivalent interpretation (and for Player $G_{0}$ as well). This will allow us to focus on the set $X$ and its probability distribution $m$.

The ${\bf a_{0}}$ and ${\bf a_{1}}$ pertain to actions of Player $G_{0}$ at both $(g,x)$ and $(r,x)$. If $x^{e}=0$ then the payoff matrices for the players $R_{i}$ at the states $(r,x)$ are

[TABLE]

If $x^{e}=1$ then the payoff matrices at $(r,x)$ are reversed:

[TABLE]

More complex are the payoffs of the player $G_{0}$ at a state labelled $g$. The matrix is three dimensional, meaning that it is a $2\times 2\times 2$ matrix. We need only to describe a $2\times 2$ matrix corresponding to each action of the $G_{0}$ player. The rows and columns stand for the actions of the $R_{1}$ and $R_{2}$ players, respectively. Those actions ${\bf a_{0}}$ and ${\bf a_{1}}$ are performed by the $R_{1}$ player at both $(g,x)$ and $(r,T_{1}x)$ and by the $R_{2}$ player at both $(g,x)$ and $(r,T_{2}x)$. First we describe the payoff matrices if $x^{e}=0$:

[TABLE]

On the other hand, if $x^{e}=1$ then the structure of payoffs is reversed:

[TABLE]

A strategy of a player is a function from its collection of information sets to the probability distributions on its two actions (a one dimensional simplex). The strategy is Borel measurable if that function is measurable with respect to its Borel sigma algebra (which is defined canonically as above from its information sets).

Notice that however the $G_{0}$ player acts at some $(g,x)$, that action is copied at $(r,x)$ (because the $G_{0}$ player cannot distinguish between these two points). However the $R_{i}$ players respond at $(r,x)$, those actions are copied at the sets $\{g\}\times T_{i}^{-1}(x)$ respectively (as the $R_{i}$ player cannot distinguish between $(r,x)$ and $\{g\}\times T_{i}^{-1}(x)$). The behaviour of a player at $(g,x)$ or $(r,x)$ will influence inductively the behaviour of all players at an uncountable subset leading upward through repetitive applications of the $T_{i}^{-1}$. However the behaviour of players that influences inductively a player’s payoff at $(g,x)$ or $(r,x)$ lie entirely within the countable set $D\times G^{+}(x)$. With regard to this latter aspect of influence, our game shares similarity with those defined by countable equivalence relations.

3 No Harsanyi $\frac{1}{1000}$ -equilibria

Before we show that the game has no Harsanyi $\frac{1}{1000}$-equilibrium, we focus in on the subset $\{g\}\times X$.

Let $A_{0}$ be the subset of $X$ such that the probability that Player $G_{0}$ at $\{g\}\times A_{0}$ chooses ${\bf b_{0}}$ is at least $\frac{19}{20}$. Let $A_{1}$ be the corresponding subset of $X$ such that the probability that Player $G_{0}$ chooses ${\bf b_{1}}$ is at least $\frac{19}{20}$. Let $A_{M}$ be the subset $X\backslash(A_{0}\cup A_{1})$.

As a general rule, from the above payoff matrices and the assumption that players are following their interests (the interests of the $R_{i}$ players at $(r,x)$ being that of conveying the choice of the $G_{0}$ player at $(g,x)$), we would expect that if $T_{1}(x)\in A_{i}$ and $T_{2}(x)\in A_{j}$, and $x^{e}=k$ then $x\in A_{i+j+k}$ where $i+j+k$ is represented modulo two. We call this the parity rule, and say that this rule holds for a point $(g,x)$ whenever these three containments are true. We say that the parity rule holds for any $x\in X$ when it holds for $(g,x)$. If any player chooses both actions at some point with strictly more than $\frac{1}{20}$ we say that the player is mixing at that point (meaning in $A_{M}$ when this player is $G_{0}$). If there is a player and a set $A$ of measure at least $w>0$ where that player prefers one strategy over another by at least $r>0$ and either that player is mixing or choosing the non-preferred action, then that player can gain at least $\frac{rw}{20}$ by choosing a different strategy. Therefore in an $\epsilon$ equilibrium it follows that $w$ is at most $\frac{20\epsilon}{r}$. This simple fact is the bridge between the equilibrium concept and the semi-group action on $X$. With respect to an $\epsilon$-equilibrium for sufficiently small enough $\epsilon$, there are two aspects of the game very important to our following arguments, First, where the strategies in approximate equilibrium are not mixing, they tend to fall into the parity rule and stay there. Second, mixing is strongly discouraged by the structure of the payoffs. Looking at the payoffs of the $R_{1}$ and $R_{2}$ players at $(r,x)$, it is not possible for $G_{0}$ at $(g,x)$ and $(r,x)$ to make both other players indifferent to their two different actions. And then if $z_{0}$ and $z_{1}$ are twins, namely $T_{i}(z_{0})=T_{i}(z_{1})$ for both $i=1,2$, if the $R_{i}$ player is not mixing at $(r,T_{i}(z_{j}))$ for at least one of $i=1,2$, it is not possible for the $G_{0}$ player at $(g,z_{j})$ to be indifferent to its two actions at both $j=0,1$. This dynamic is formalised in the next lemma. Lemma 1: For every $x\in X$, either one or the other corresponding Player $R_{1}$ or $R_{2}$ at $(r,x)$ has an incentive of at least $80$ to choose either ${\bf a_{0}}$ or ${\bf a_{1}}$ over the other action. Let $x,y\in X$ be any two points in $X$ and let $z_{0}$ and $z_{1}$ be the two twins where $T_{1}(z_{i})=x$ and $T_{2}(z_{i})=y$ for $i=0,1$ and $z_{0}^{e}=0$ and $z_{1}^{e}=1$. If one of $R_{1}$ or $R_{2}$ is mixing at $(r,x)$ or $(r,y)$, respectively, and the other is not, then Player $G_{0}$ at either $(g,z_{0})$ or at $(g,z_{1})$ has an incentive of at least $80$ to choose either ${\bf b_{0}}$ or ${\bf b_{1}}$ over the other strategy. Proof: Without loss of generality assume that $x^{e}=0$ and that the Player $G_{0}$ at $(g,x)$ chooses ${\bf b_{0}}$ with probability at least $\frac{1}{2}$. By choosing ${\bf a_{1}}$ the $R_{1}$ player would get no more than $50$ and by choosing ${\bf a_{0}}$ the $R_{1}$ player would get at least $150$. On the other hand, if the Player $G_{0}$ at $(g,x)$ chooses ${\bf b_{1}}$ with probability at least $\frac{1}{2}$ then the $R_{2}$ player would get no more than $50$ by choosing ${\bf a_{0}}$ and at least $150$ by choosing ${\bf a_{1}}$. Next, due to symmetries, it suffices to consider the two cases of the $R_{2}$ player choosing ${\bf a_{0}}$ with probability no more than $\frac{1}{20}$ and the $R_{2}$ player choosing ${\bf a_{1}}$ with probability no more than $\frac{1}{20}$. Let $w\leq\frac{1}{20}$ be the probability that the $R_{2}$ player chooses $\bf a_{0}$. We break this case into two subcases, where Player $R_{1}$ chooses ${\bf a_{0}}$ with at least $\frac{3}{5}$ and where Player $R_{1}$ chooses ${\bf a_{0}}$ with at most $\frac{3}{5}$. If Player $R_{1}$ chooses ${\bf a_{0}}$ with at least $\frac{3}{5}$ then the $G_{0}$ player at $(g,z_{1})$ gets at least $570$ for playing ${\bf b_{0}}$ and no more than $400(1-w)+2000w$ for playing ${\bf b_{1}}$, which reaches a maximum of $480$ at $w=\frac{1}{20}$. If Player $R_{1}$ chooses ${\bf a_{1}}$ with at least $\frac{2}{5}$ then the $G_{0}$ player at $(g,z_{0})$ gets at least $760$ from choosing ${\bf b_{0}}$ and no more than $600(1-w)+2000w$ for playing ${\bf b_{1}}$, which reaches a maximum of $670$ at $w=\frac{1}{20}$. Now let $w\leq\frac{1}{20}$ be the probability that the $R_{2}$ player chooses $\bf a_{1}$. We break this case into two subcases, where Player $R_{1}$ chooses ${\bf a_{0}}$ with at least $\frac{3}{5}$ and where Player $R_{1}$ chooses ${\bf a_{0}}$ with at most $\frac{3}{5}$. If Player $R_{1}$ chooses ${\bf a_{0}}$ with at least $\frac{3}{5}$ then the $G_{0}$ player at $(g,z_{1})$ get at least $1140$ from choosing ${\bf b_{1}}$ and no more than $820$ by choosing ${\bf b_{0}}$. If Player $R_{1}$ chooses ${\bf a_{1}}$ with at least $\frac{2}{5}$ then the $G_{0}$ player at $(g,z_{0})$ gets at least $780$ from choosing ${\bf b_{1}}$ and no more than $600(1-w)+2000w$ from choosing ${\bf b_{0}}$, which reaches a maximum of $670$ at $w=\frac{1}{20}$. $\Box$The consequence of Lemma 1 is that the players are hardly ever mixing in an approximate equilibrium. That is formalized in the next lemma. Lemma 2: In any $\frac{1}{1000}$ Borel measurable equilibrium of the game, the $G_{0}$ player mixes with probability less than $\frac{16}{10,000}$ and the parity rule holds for all but at most $\frac{4}{1000}$ of the space $X$. Proof: Let $B_{1}$ be the subset of $z\in A_{M}$ such that the corresponding $R_{1}$ player at $(r,T_{1}z)$ is mixing and let $B_{2}$ be the subset of $z\in A_{M}$ such that the corresponding $R_{2}$ Player at $(r,T_{2}z)$ is mixing. Let $c=m(A_{M})$, $a=m(B_{1})$ and $b=m(B_{2})$. As the $T_{1}z$ and $T_{2}z$ are distributed independently as variables of $z$, in an $\frac{1}{1000}$ equilibrium the following holds: c\leq\frac{1}{4000}+ab+\frac{1}{2}\big{(}a+b\big{)}, where the $\frac{1}{4000}$ refers to the maximum probability for the $G_{0}$ player to choose a strategy that is suboptimal by a quantity of at least $80$, the $ab$ refers to the probability that both $R_{i}$ are mixing at both $T_{1}z$, and $\frac{a+b}{2}$ refers to the probability that Player $G_{0}$’s actions are within $80$ of each other for one but not both of the twins $z_{0}$ and $z_{1}$ (where one or the other of $R_{1}$ at $T_{1}z_{i}$ or $R_{2}$ at $T_{2}z_{i}$ are mixing, but not both). But as the sets $B_{i}$ are the sets $T^{-1}_{i}$ applied to where $R_{i}$ is mixing in $\{r\}\times X$ and the $T_{i}$ are measure preserving, $a$ is also the probability throughout $\{r\}\times X$ that $R_{1}$ is mixing and the same holds for $b$ and $\{r\}\times X$. By Lemma 1, for any $z\in X$ the probability of both $R_{1}$ mixing at $(r,z)$ and $R_{2}$ mixing at $(r,z)$ cannot exceed $\frac{1}{4000}$ (from $\frac{20}{80\cdot 1000}=\frac{1}{4000}$). From this we conclude that $a+b\leq c+\frac{1}{1000}$, since where $(g,z)$ is mixing at most $\frac{1}{2000}$ of the points following in both $(r,z)$ and $(r,z)$ can be mixing, (with the other $\frac{1}{2000}$ referring to the possibility that $G_{0}$ is not mixing at $(g,z)$ nevertheless one of the $R_{i}$ players at $(r,z)$ is mixing). From $ab\leq\frac{1}{4}(a+b)^{2}$, and the above, we get the quadratic $0\leq c^{2}-\frac{999}{500}c+\frac{3,001}{1,000,000}$. After completing the square we get that $|c-\frac{999}{1000}|\geq\sqrt{.995}$. Since $c$ cannot be greater than $1$ we are left with $c<.999-.9974=.0016$. The probability that the parity rule is not followed for an $x\in X$ is no more than the probability of the $G_{0}$ player mixing at either $(g,T_{1}x)$ or $(g,T_{2}x)$ plus the probability that the $R_{1}$ player at $(r,T_{1}x)$, the $R_{2}$ player at $(r,T_{2}x)$ or the $G_{0}$ player at $(g,x)$ is not properly responding to the corresponding non-mixing behaviour. These probabilities sum to $.00395$. . $\Box$ Next we show it is impossible for there to exist a $\frac{1}{1000}$ equilibrium Borel measurable equilibrium, using the regularity of the measure. Let ${\cal C}_{n}$ be the set of cylinders of depth $n$, where the two cylinders defined by the values $x^{e}=0$ and $x^{e}=1$ have depth [math]. With $2^{n+1}-1$ words of length $n$ or less the cardinality of ${\cal C}_{n}$ is $2^{2^{n+1}-1}$ and $m(c)=2^{-2^{n+1}+1}$ for all $c\in{\cal C}_{n}$. Recall the definition of $A_{0}$ and $A_{1}$ as the sets where either [math] or $1$ is played by $G_{0}$ with probability at least $\frac{19}{20}$. For every $c\in{\cal C}_{n}$ and $i=0,1$ let $w_{i}(c)$ be the conditional probability of action ${\bf b_{i}}$ at the cylinder $c$, in other words $m(A_{i}\cap c)/m(c)$. For every cylinder $c$ define $\eta(c):=\min_{i=0,1}w_{i}(c)$ and let $r(c)$ be the conditional probability of belonging in the set where the parity rule does not hold. In the next lemma, we show that the parity rule is a powerful force to equalize the probabilities for both actions ${\bf b_{0}}$ and ${\bf b_{1}}$. This cannot be guaranteed for all cylinders, due to the small probability that the parity rule doesn’t hold. But it does hold in general for most cylinders, regardless of the depth. Two free generators and the dual causation implicit in the parity rule force this equalization.

Lemma 3: In any $\frac{1}{1000}$ equilibrium of the game, the average $q_{i}=\frac{\sum_{c\in{\cal C}_{i}}\eta(c)}{|{\cal C}_{i}|}$ is at least $\frac{1}{3}$ for every $i$. Proof: The proof is by induction. There are two elements in ${\cal C}_{0}$ and eight elements in ${\cal C}_{1}$. Let $c_{0}$ and $c_{1}$ be the two elements of ${\cal C}_{0}$. Both cylinders $c_{0}$ and $c_{1}$ are composed of four elements of ${\cal C}_{1}$, created (by means of $T_{1}$ and $T_{2}$) from the combination of the two elements $c_{0},c_{1}$ of ${\cal C}_{0}$ and the same two elements $c_{0},c_{1}$ of ${\cal C}_{0}$ along with a value of $x^{e}=0$ for all $x\in c_{0}$ and $x^{e}=1$ for all $x\in c_{1}$. Let $x_{0}$ and $x_{1}$ be two points such that $x_{0}^{e}=0$, $x_{1}^{e}=1$, and $x_{0}^{U}=x_{1}^{U}$ for every other $U\not=e$. However membership in $A_{0}$ or $A_{1}$ is determined by $T_{1}x_{i}$ and $T_{2}x_{i}$, the parity rule requires opposite memberships for $i=0,1$. As the parity rule must hold in a set of size at least $1-\frac{1}{250}$, it follows that in the whole space the probability given to both $A_{0}$ and $A_{1}$ must be approximately the same, more precisely these probabilities must be at least $\frac{124}{250}$ for both $A_{0}$ and $A_{1}$. Now let $c$ be either $c_{0}$ or $c_{1}$. As $c$ is created by either the $e$ position being [math] or $1$ and the four combinations of $c_{0}$ and $c_{1}$ in both direction $T_{1}$ and $T_{2}$, whatever are the probabilities given for the two $w_{i}(c_{0})$ and the two $w_{i}(c_{1})$, the fact that $\frac{w_{i}(c_{0})+w_{i}(c_{1})}{2}\geq\frac{124}{250}$ for both $i=0,1$, implies that the conditional probability given to both $A_{0}$ and $A_{1}$ at $c$ must be at least $2(\frac{124}{250})^{2}-\frac{1}{125}\geq.48$. We assume the claim is true for every $t\in{\cal C}_{i}$. Every $t\in{\cal C}_{i+1}$ is created through the combination of a pair $c,d\in{\cal C}_{i}$ with a determination of [math] or $1$ (though this determination will play no rule in the following argument). Let $i_{c}$ be the action that is less frequent at $c$, and define $i_{d}$ the same way. Let $j$ be the action following from the parity rule determined by the value of $t^{e}$ and the combination of $i_{c}$ with $i_{d}$ (however that is determined by the $e$ position). If $r(c)=r(d)=r(t)=0$ then the parity rule would give $j$ exactly $\eta(c)(1-\eta(d))+\eta(d)(1-\eta(c))$, as it would give the other action the greater quantity $(1-\eta(c))(1-\eta(d))+\eta(c)\eta(d)$. Due to the influence of the quantities $r(c),r(d),r(t)$ we cannot say for sure that $j$ is the action less taken at $t$. But we can say that $\eta(t)\geq-r(t)+\eta(c)(1-\eta(d)-r(d))+\eta(d)(1-\eta(c)-r(c))\geq\eta(c)+\eta(d)-2\eta(c)\eta(d))-r(t)-\frac{r(c)+r(d)}{2}$. But with $\sum_{c\in{\cal C}_{j}}r(c)\leq\frac{1}{250}|{\cal C}_{j}|$ for all $j$ it follows that $q_{i+1}\geq-\frac{1}{125}+\frac{1}{|{\cal C}_{i}|^{2}}\sum_{c,d\in{\cal C}_{i}}\eta(c)+\eta(d)-2\eta(c)\eta(d)=-\frac{1}{125}+\frac{1}{|{\cal C}_{i}|}\sum_{c\in{\cal C}_{i}}\eta(c)+\frac{1}{|{\cal C}_{i}|}\sum_{d\in{\cal C}_{i}}\eta(d)+\frac{1}{|{\cal C}_{i}|^{2}}(\sum_{c\in{\cal C}_{i}}\eta(c))(\sum_{d\in{\cal C}_{i}}\eta(d))=-\frac{1}{125}+2q_{i}-2q_{i}^{2}$. By induction we conclude that $q_{i+1}\geq\frac{4}{9}-\frac{1}{125}>\frac{1}{3}$. $\Box$

Theorem 1: There can be no Borel ( $\mu$) measurable $\frac{1}{1000}$-equilibrium. Proof: With $\eta(c)$ defined as in the proof of Lemma 3, for the mutually exclusive measurable sets $A_{0},A_{1}$ of $X$ it follows from the regularity of the measure $\mu$ that $\lim_{n\rightarrow\infty}q_{n}=\lim_{n\rightarrow\infty}\frac{\sum_{c\in{\cal C}_{n}}\eta(c)}{|{\cal C}_{n}|}=0$. But by Lemma 3 it never falls below $\frac{1}{3}$. $\Box$ Notice that where the players are obeying the parity rule, even approximately so, the location where the payoff to Player $G_{0}$ is close to $2000$ or to $1000$ is determined by one or the other of the two other players, by Player $R_{1}$ in the half $\{x\in X\ |\ x^{e}=0\}$ and by Player $R_{2}$ in the half $\{x\in X\ |\ x^{e}=1\}$. Where the parity rule holds the Lesbegue measurability of the payoff of Player $G_{0}$ implies the Lesbegue measurability of the strategies of the $R_{i}$ players and hence also the Lesbegue measurability of the strategy of the $G_{0}$ player in response. For the $G_{0}$ player to have an “equilibrium payoff” by some interpretation one must define that concept quite differently from the existence of Harsanyi $\epsilon$-equilibria.

4 Bayesian equilibria:

In this section we will show that we can colour the space $X=\{0,1\}^{G^{+}}$ modulo a null set $N$ using only two colours: $1$ and [math] or red and blue, respectively, so that the parity rule is obeyed, and furthermore extend this to an equilibrium of the whole space $\Omega$. Recall that the parity rule is a function $c:X\rightarrow X$ such that $c(x)=c(T_{1}(x))+c(T_{2}(x))+x^{e}$ (modulo 2). Notice that such a colouring on such a set defines a Bayesian equilibrium on the subset $\{g,r\}\times(X\backslash N)$. We show then how to extend these strategies to a Bayesian equilibrium on all of the space $\Omega=\{g,r\}\times X$.

Recall the definition of the twins determined by some $x,y\in X$. We say that a subset $A$ of $X$ is closed if for every pair $x,y$ in $A$ the twins determined by $x$ and $y$ are also in $A$. By the closure $\overline{A}$ of a set $A\subseteq X$ we mean the smallest closed set containing $A$. We say that $A\subseteq X$ is pyramidic if $x\in A$ implies that $U(x)\in A$, where $U\in G^{+}$. The main example of pyramidic set is $G^{+}(x)$, where $x$ is an arbitrary element of $X$. Notice that whenever a set is pyramidic that its closure is also pyramidic.

Define now the set

[TABLE]

This set is a null set with respect to the product measure $m$ on $X$. Indeed, for any given two distinct words $U,V\in G^{+}$, the equality implies an agreement on infinitely many coordinates, and there are only countably many words $U\in G^{+}$.

We are ready to prove the following lemma:

Lemma 4: Let $X_{1}=X\setminus N$ and let $c$ be the parity rule defined on $X$. Then there exists a colouring of $X_{1}$ using the two colours $\{0,1\}$ which is consistent with the rule $c$.

Proof: We will proceed by transfinite induction, assuming the Axiom of Choice and thus Zorn’s Lemma. Let $x_{0}$ be any element of $X_{1}$ and obtain the set $P_{0}:=G^{+}(x_{0})$. We define now the colouring of $P_{0}$ of $G^{+}(x_{0})$ as follows:

$(i)$ colour all the points $T_{1}U(x_{0})$ in red, where $U\in G^{+}(x_{0})$ and $U$ is the identity or begins on the right with $T_{1}$; $(ii)$ colour all the points $T_{2}V(x_{0})$ in blue, where $V\in G^{+}(x_{0})$ is the identity or begins on the right with $T_{1}$; $(iii)$ colour the remaining points of the pyramid $P_{0}$ in the way that they satisfy the rule $c$. After colouring all the points of $P_{0}$, extend the colouring to all the points in the closure $\overline{P_{0}}$ of $P_{0}$.

Next create a partial ordering on colourings of pyramidic and closed subsets of $X$ that obey the parity rule, with one colouring greater than another if the subset is larger and their colourings agree on their common intersection (the smaller subset). Any tower of such colourings will define a colouring that obeys the parity rule. As Zorn’s Lemma implies that there is a maximal element (as a tower defines its own least upper bound), it suffices to show that maximality implies that all of $X_{1}$ has been coloured. Let $P$ be any pyramidic and closed subset of $X_{1}$ with a colouring that obeys the parity rule. Assume that $P$ is not all of $X_{1}$. Let $x$ be the first member of $X_{1}$ that is not in $P$. We say that $x$ has a *hitting point in * $P$ if $U(x)\in P$ for some $U\in G^{+}$ and whenever $U=VW$ and $W$ is not the identity then $V(x)\not\in P$.

Now we have the two cases:

Case 1) $x$ has no hitting point with respect to $P$. Then we colour the closure $G^{+}(x)$ in the same way as the initial pyramidic set $P_{0}$.

Case 2) $x$ has a hitting point in $P$. Then colour the elements of $G^{+}(x)$ taking into an account the colours of the hitting points. Notice that the closure of $P$ implies that if $Ux$ is not in $P$ then one of $T_{i}Ux$ is also not in $P$, so that if $T_{i}Ux$ is a hitting point then $T_{j}Ux$ is not a hitting point, for $i\not=j$. This allows us to colour $x$ arbitrarily and then move downward in a consistent way, with the colouring of $T_{i}Ux$ determined already only if $T_{j}Ux$ is a hitting point or had just been just coloured arbitrarily (for $j\not=i$).

And then we colour the closure of $G^{+}(x)\cup P$ according to the parity rule $c$ for a larger set that is closed, pyramidic, and consistent with both the parity rule and the pre-existing colouring. $\Box$

Theorem 2: There exists a Bayesian equilibrium on all of $\Omega$.

Proof: Following on from the proof of Lemma 4, let $\overline{X_{1}}$ be the closure of $X_{1}$ and extend the colouring of $X_{1}$ to one on $\overline{X_{1}}$. For all three players $G_{0}$, $R_{1}$ and $R_{2}$ define pure strategies on $D\times\overline{X_{1}}$ accordingly. With $x$ any point in $X\backslash X_{1}$, let $\Gamma_{x}$ be the game defined on $D\times G^{+}(x)$ such that the strategies on $D\times(G^{+}(x)\cap\overline{X_{1}})$ are already fixed by the above colouring. As the game has only countably many positions, by Simon (2003) there is a Nash equilibrium defined on the game $\Gamma_{x}$. But notice that it defines an equilibrium when including those strategies on $D\times(G^{+}(x)\cap\overline{X_{1}})$ that are already fixed. Extend this equilibrium to an equilibrium on the set $D\times(\overline{G^{+}(x)\cap\overline{X_{1}}})$ through best reply responses (noticing that nothing done at a point $y$ has any influence on any player at points $Uy$ for any $U\not=e$ - just consider $x,T_{1}x,T_{2}x$ and follow by induction). We can even assume that these best reply responses are pure strategies. As with the proof of Lemma 4, we define a partial ordering on pyramidic and closed subsets $P$ of $X$ and the equilibria defined on $D\times P$. In the same way, we show that an equilibrium can be defined on all of $\Omega$, using critically that any extension of an equilibrium from a closed and pyramidic set $P$ will not disturb the pre-existing equilibrium property on $P$. $\Box$ There are some points in $X$ for which any equilibrium requires a mixed strategy. Let $x,y$ be the two points defined by $x^{e}=0$, $y^{e}=1$, $T_{1}(x)=y$, $T_{2}(x)=x$, $T_{1}(y)=x$ and $T_{2}(y)=x$. No matter how $x$ is coloured, because $T_{1}(y)=T_{2}(y)=x$ and $y^{e}=1$, $y$ must be coloured with $1$. But then $T_{1}(x)=y$, $T_{2}(x)=x$ and $x^{e}=0$ forces $x$ to be coloured differently from itself.

5 Conclusion: open questions

Is there an example of an ergodic game (Simon 2003) that has no Harsanyi $\epsilon$-equilibrium for some positive $\epsilon$? The examples of Simon (2003) and of Hellman (2014) were ergodic games, and ergodic games have Bayesian equilibria (Simon 2003). We believe the answer is yes and that it can be done through the action of a non-amenable group which defines the information structure of the players. With the example of this paper, there was a very strong mixing structure that kept the probability high for both actions at all cylinders. We believe that the weaker mixing structure from a group action would be sufficient to obtain the same result. In the example of this paper, there are three players. Can the same result be accomplished with two players? We believe that the answer is yes. The structure of our example is similar to that of a free action of the free product $G=C_{2}*C_{3}$ on $\{0,1\}^{G}$. We believe it can be done through associating $C_{3}$ with the knowledge of one player and $C_{2}$ with the other player, (or with two other finite groups whose free product is non-amenable). Lastly, what is the relationship between Bayesian equilibria and the Banach-Tarski paradox? Let $G$ be a group acting in a measure preserving way on a probability space $X$ and for every player $i$ let $G_{i}$ be a finite subgroup of $G$ so that the information sets of Player $i$ are the orbits of $G_{i}$ and $G$ is generated by the $G_{i}$. Is there a Bayesian game so defined such that every Bayesian equilibrium defines through its fibres a paradoxical decomposition of the space where the group elements demonstrating the paradox belong to $G$? On this question we are agnostic.

6 References

Harsanyi, J. (1967)

, Games with Incomplete Information Played by Bayesian Players, Management Science, Vol. 14, pp. 159-182.

Hellman, Z. (2014)

, A Game with No Bayesian Approximate Equilibria, Journal of Economic Theory, Vol. 153, pp. 138-151.

Hellman, Z. and Levy, J. (2016)

, Bayesian Games over Continuum Many States, Theoretical Economics, forthcoming.

Kechris, A., Solecki, S. Todorcevic, S. (1999)

, Borel Chromatic Numbers, Vol 141, Issue 1, pp. 1-44.

Milgrom, P. and Weber, R. (1985)

, Distribution Strategies for Games with Incomplete Information, Mathematics of Operations Research, Vol. 10, No. 4, pp. 619-632.

Simon, R. S. (2003)

, Games of Incomplete Information, Ergodic Theory, and the Measurability of Equilibria, Israel J. Math., Vol. 138, No. 1, pp. 73-92.