A note on the symmetry of all Nash equilibria in games with increasing best replies

Abstract

It is well known that a symmetric game has only symmetric (pure strategy) Nash equilibria if its best-reply correspondences admit only increasing selections and its strategy sets are totally ordered. Several nonexamples of the literature show that this result is generally false when the totality condition of the relation that orders the strategy sets is simply dropped. Making use of the structure of interaction functions, this note provides sufficient conditions for the symmetry of all (pure strategy) Nash equilibria in symmetric games where best-reply correspondences admit only increasing selections, but strategy sets are not necessarily totally ordered.

Appendix: On the notion of symmetry

Let $\mathbb {N}$ denote the set $\left\{ 1,2, \ldots \right\}$ of positive integers and $\mathbb {N}_{0}$ denote the set $\left\{ 0,1,2,\ldots \right\}$ of nonnegative integers. Let $X$ be an arbitrary nonempty set.

• A permutation ⁹ of $X$ is a bijection from $X$ to $X$.

• A finite permutation $\sigma$ of $X$ is a permutation of $X$ such that

  $$\begin{aligned} \left| \left\{ x\in X:\sigma \left( x\right) \ne x\right\} \right| \in \mathbb {N}_{0}. \end{aligned}$$

• A transposition $\sigma$ of $X$ is a finite permutation of $X$ such that

  $$\begin{aligned} \left| \left\{ x\in X:\sigma \left( x\right) \ne x\right\} \right| \le 2. \end{aligned}$$

A permutation $\sigma$ of $X$ fixes an element $x$ of $X$ if $\sigma \left( x\right) =x$. A permutation $\sigma$ of $X$ moves an element $x$ of $X$ if $\sigma \left( x\right) \ne x$. The family of all permutations of $X$ (also called the symmetric group of $X$) is denoted by $\mathcal {S}_{X}$. The family of all finite permutations of $X$ is denoted by $\mathcal {F}_{X}$. The family of all transpositions of $X$ is denoted by $\mathcal {T}_{X}$. Clearly,

$$\begin{aligned} \emptyset \ne \left\{ \mathop {\mathrm{id}}\nolimits _{X}\right\} \subseteq \mathcal {T} _{X}\subseteq \mathcal {F}_{X}\subseteq \mathcal {S}_{X}\text { and, when }X \text { is finite, }\mathcal {F}_{X}=\mathcal {S}_{X}. \end{aligned}$$

Note that in this appendix a transposition of $X$ can fix all elements of $X$ (i.e., ${\hbox {id}}_{X}\in \mathcal {T}_{X}$); clearly, if a transposition of $X$ does not fix all elements of $X$, then it moves exactly two distinct elements of $X$ (interchanging them). A game $\varGamma =\left( N,\left( S_{i}\right) _{i\in N},\left( u_{i}\right) _{i\in N}\right)$ is defined as in Sect. 2, but now we only assume $N$ to be nonempty, and hence, in this appendix the set of players $N$ can well be a singleton.

Definition 1

A game $\varGamma =\left( N,\left( S_{i}\right) _{i\in N},\left( u_{i}\right) _{i\in N}\right)$ is said to be T-symmetric (resp. F-symmetric, P-symmetric) if:

• for all $\left( i,l\right) \in N\times N$,

  $$\begin{aligned} S_{i}=S_{l}; \end{aligned}$$

• for all $\left( i,s\right) \in N\times S_{N}$ and for all $\sigma \in \mathcal {T}_{N}$ (resp. $\sigma \in \mathcal {F }_{N}$, $\sigma \in \mathcal {S}_{N}$),

  $$\begin{aligned} u_{i}\left( s\right) =u_{\sigma \left( i\right) }((s_{\sigma \left( l\right) })_{l\in N}). \end{aligned}$$

This appendix shows that: (1) there exist T-symmetric games that are not P-symmetric; (2) the relations illustrated in Table 1 are true.

Table 1 Relation between the three notions of symmetry

Fact 1 below should be considered well known; however, we do not have a precise reference for its part (ii) when $N$ is not finite.

Fact 1

Let $N$ be an arbitrary nonempty set, $\tau \in \mathcal {T }_{N}$ and $\phi \in \mathcal {F}_{N}$. Then:

• (i) $\tau$ is an involution (i.e., $\tau ^{-1}=\tau$);

• (ii) there exists a finite family $\left\{ \tau _{k}\right\} _{k=1}^{m}$ of transpositions of $N$ such that

  $$\begin{aligned} \phi =\tau _{m}\circ \tau _{m-1}\circ \cdots \circ \tau _{1}. \end{aligned}$$

Proof

Part (i) is an immediate consequence of the definition of a transposition. We shall prove only part (ii) of Fact 1, as follows. If $\phi$ is a transposition, there is nothing to prove. Suppose $\phi$ is not a transposition. Then, the set

$$\begin{aligned} M=\left\{ x\in N:\phi \left( x\right) \ne x\right\} \end{aligned}$$

of all elements of $N$ that are moved by $\phi$ is finite and $\left| M\right| \ge 3$. Clearly, each element of $N\backslash M$ (if any) is fixed by $\phi$ and $\phi |_{M}$ is a (finite) permutation of $M$ that moves all elements of $M$. By Proposition 2.35 in Rotman (2005), associated with the (finite) permutation $\phi |_{M}$ of the finite set $M$, there exists a finite family $\left\{ \theta _{k}\right\} _{k=1}^{m}$ of transpositions of $M$ such that

$$\begin{aligned} \phi |_{M}=\theta _{m}\circ \theta _{m-1}\circ \cdots \circ \theta _{1}. \end{aligned}$$

For each $k=1,\ldots ,m$, let $\tau _{k}$ be the transposition of $N$ that fixes all elements of $N\backslash M$ and such that $\tau _{k}|_{M}=\theta _{k}$. Now, it suffices to note that

$$\begin{aligned} \phi =\tau _{m}\circ \tau _{m-1}\circ \cdots \circ \tau _{1} \end{aligned}$$

to conclude the proof. $\square$

Proposition 1

Let $\varGamma$ be a game with an arbitrary set of players.

• (i) If $\varGamma$ is T-symmetric then $\varGamma$ is F-symmetric.

• (ii) If $\varGamma$ is F-symmetric then $\varGamma$ is T-symmetric.

• (iii) If $\varGamma$ is P-symmetric then $\varGamma$ is F-symmetric.

• (iv) If $\varGamma$ is F-symmetric and $N$ is finite then $\varGamma$ is P-symmetric.

Proof

Parts (ii), (iii) and (iv) of Proposition 1 follow immediately from the three definitions of a symmetric game provided in Definition 1. Only part (i) of Proposition 1 needs to be proved true. The proof is as follows. Suppose $\varGamma$ is T-symmetric. Pick a triple $\left( i,s,\phi \right) \in N\times S_{N}\times \mathcal {F}_{N}$. By part (ii) of Fact 1, there exists a finite family $\left\{ \tau _{k}\right\} _{k=1}^{m}$ of transpositions such that

$$\begin{aligned} \phi =\tau _{m}\circ \tau _{m-1}\circ \cdots \circ \tau _{1}. \end{aligned}$$

(11)

The T-symmetry of $\varGamma$ implies

$$\begin{aligned} u_{i}\left( s\right) =u_{\tau _{1}\left( i\right) }((s_{\tau _{1}\left( l\right) })_{l\in N}). \end{aligned}$$

(12)

On the other hand, the T-symmetry of $\varGamma$ implies also

$$\begin{aligned} u_{\tau _{k-1}\circ \cdots \circ \tau _{1}\left( i\right) }((s_{\tau _{k-1}\circ \cdots \circ \tau _{1}\left( l\right) })_{l\in N})=u_{\tau _{k}\left( \tau _{k-1}\circ \cdots \circ \tau _{1}\left( i\right) \right) }((s_{\tau _{k}\left( \tau _{k-1}\circ \cdots \circ \tau _{1}\left( l\right) \right) })_{l\in N}) \end{aligned}$$

for all $k=2,\ldots ,m$ as each $\tau _{k-1}\circ \cdots \circ \tau _{1}\left( l\right)$ is indeed a distinct player; therefore,

$$\begin{aligned} u_{\tau _{k-1}\circ \cdots \circ \tau _{1}\left( i\right) }((s_{\tau _{k-1}\circ \cdots \circ \tau _{1}\left( l\right) })_{l\in N})=u_{\tau _{k}\circ \tau _{k-1}\circ \cdots \circ \tau _{1}\left( i\right) }((s_{\tau _{k}\circ \tau _{k-1}\circ \cdots \circ \tau _{1}\left( l\right) })_{l\in N}) \end{aligned}$$

for all $k=2,\ldots ,m$, and hence,

$$\begin{aligned} u_{\tau _{1}\left( i\right) }((s_{\tau _{1}\left( l\right) })_{l\in N})= & {} u_{\tau _{2}\circ \tau _{1}\left( i\right) }((s_{\tau _{2}\circ \tau _{1}\left( l\right) })_{l\in N}) \nonumber \\= & {} \cdots =u_{\tau _{m}\circ \cdots \circ \tau _{1}\left( i\right) }((s_{\tau _{m}\circ \cdots \circ \tau _{1}\left( l\right) })_{l\in N}). \end{aligned}$$

(13)

By (12) and (13),

$$\begin{aligned} u_{i}\left( s\right) =u_{\tau _{m}\circ \cdots \circ \tau _{1}\left( i\right) }((s_{\tau _{m}\circ \cdots \circ \tau _{1}\left( l\right) })_{l\in N}). \end{aligned}$$

(14)

By (11) and (14),

$$\begin{aligned} u_{i}\left( s\right) =u_{\phi \left( i\right) }((s_{\phi \left( l\right) })_{l\in N}). \end{aligned}$$

(15)

As $\left( i,s,\phi \right)$ has been chosen arbitrarily in $N\times S_{N}\times \mathcal {F}_{N}$, (15) implies the F-symmetry of $\varGamma$. $\square$

Note that, by part (i) of Fact 1, a game is T-symmetric if and only if $S_{i}=S_{l}$ for all $\left( i,l\right) \in N\times N$ and

$$\begin{aligned} u_{i}\left( s\right) =u_{\tau \left( i\right) }((s_{\tau ^{-1}\left( l\right) })_{l\in N})\quad \text {for all }\left( i,s,\tau \right) \in N\times S_{N}\times \mathcal {T}_{N}. \end{aligned}$$

Thus, any game that satisfies the notion of symmetry as enunciated—for games with finitely many players—in part (ii) of Definition 1 of Amir et al. (2008) is T-symmetric. (Note that the latter equality is not generally satisfied by T-symmetric games when $\tau$ is a permutation but not a transposition: e.g., see the last footnote in the proof of Claim 1).

Claim 1

A T-symmetric game need not be P-symmetric. Indeed:

• (i) there exists a game with a countably infinite set of players which is T-symmetric but not P-symmetric;

• (ii) there exists a game with an uncountably infinite set of players which is T-symmetric but not P-symmetric.

Proof

(i) Consider the following game $\varGamma =\left( N,\left( S_{i}\right) _{i\in N},\left( u_{i}\right) _{i\in N}\right)$. Put

$$\begin{aligned} N=\mathbb {N}. \end{aligned}$$

For each $i\in N$, let

$$\begin{aligned} S_{i}=\left\{ 0,1\right\} . \end{aligned}$$

Say that $s\in S_{N}$ is a 1-even-infinite joint strategy if

$$\begin{aligned} \left| \left\{ i\in \mathbb {N}:s_{2i}=1\right\} \right| =\left| \mathbb {N}\right| , \end{aligned}$$

and¹⁰ say it is 1-even-finite otherwise. For all $i\in N$, let

$$\begin{aligned} u_{i}:s\mapsto \left\{ \begin{array}{l@{\quad }l} 900 &{} \text {if }s_{i}=1\text { and }s\text { is 1-even-infinite,} \\ 800 &{} \text {if }s_{i}=0\text { and }s\text { is 1-even-infinite,} \\ 600 &{} \text {if }s_{i}=1\text { and }s\text { is 1-even-finite,} \\ 700 &{} \text {if }s_{i}=0\text { and }s\text { is 1-even-finite.} \end{array} \right. \end{aligned}$$

Noting that the “1-even-infiniteness” and the “1-even-finiteness” of a joint strategy are preserved under any transposition of $N$, one easily concludes that $\varGamma$ is T-symmetric. We cannot infer the same conclusion about the P-symmetry. Let $\overline{s}$ be a 1-even-infinite joint strategy¹¹ such that

$$\begin{aligned} \overline{s}_{i}=1\text { for infinitely many even }i\in \mathbb {N}\text { and }\overline{s}_{i}=0\text { for all odd }i\in \mathbb {N}. \end{aligned}$$

We have

$$\begin{aligned} u_{1}\left( \overline{s}\right) =800. \end{aligned}$$

Let $\overline{\sigma }$ be a permutation of $N$ that interchanges $i$ with $i+1$ for each odd integer $i\ge 3$ and fixes each $i\in N$ such that $i<3$. Then

$$\begin{aligned} u_{1}\left( \overline{s}\right) =800\ne 700=u_{1}((\overline{s}_{\overline{ \sigma }\left( i\right) })_{i\in N})=u_{\overline{\sigma }\left( 1\right) }(( \overline{s}_{\overline{\sigma }\left( i\right) })_{i\in N}) \end{aligned}$$

and hence, $\varGamma$ is not P-symmetric.¹²

(ii) Exactly the same proof of part (i) of Claim 1, but now replace the defining equality $N=\mathbb {N}$ with $N=\left[ -4/5,-3/5\right] \cup \mathbb {N}$. $\square$

Note that the two games constructed in the proof of Claim 1 satisfy the conditions of Lemma 1 (when each strategy set $S_{i}$ is endowed with the partial order inherited from $\mathbb {R}$), and hence all Nash equilibria are symmetric. It can be easily checked that the two joint strategies where all players implement 0 and 1 are Nash equilibria. Since these two joint strategies are the only possible symmetric joint strategies, by Lemma 1 they are the only Nash equilibria for $\varGamma$. Finally, note that there exists a bijection from $\mathbb {N}$ to a compact subset of $\left[ 0,1\right]$ and there exists a bijection from $\left[ -4/5,-3/5\right] \cup \mathbb {N}$ to the compact interval $\left[ 0,1\right]$; therefore, it is immaterial that in the proof of Claim 1 the set $N$ is unbounded.