The single crossing conditions for incomplete preferences

Abstract

We study the implications of the single crossing conditions for preferences described by binary relations. All restrictions imposed on the preferences are satisfied in the case of approximate optimization of a bounded-above utility function. In the context of the choice of a single agent, the transitivity of strict preferences ensures that the best response correspondence is increasing in the sense of a natural preorder; if the preferences are represented by an interval order, there is an increasing selection from the best response correspondence. In a strategic game, a Nash equilibrium exists and can be reached from any strategy profile after a finite number of best response improvements if all strategy sets are chains, the single crossing conditions hold w.r.t. pairs (one player’s strategy, a profile of other players’ strategies), and the strict preference relations are transitive. If, additionally, there are just two players, every best response improvement path reaches a Nash equilibrium after a finite number of steps. If each player is only affected by a linear combination of the strategies of others, the single crossing conditions hold w.r.t. pairs (one player’s strategy, an aggregate of the strategies of others), and the preference relations are interval orders, then a Nash equilibrium exists and can be reached from any strategy profile with a finite best response path.

Appendix

1.1 Proof of Theorem 3.8

We call a subset $S'\subseteq S$ an interval if $s\in S'$ whenever $s'<s<s''$ and $s',s''\in S'$. The intersection of any number of intervals is an interval too. By Proposition 3.7, the set $\{s\in S\mid x\in {\mathcal {R}}(s)\}$, for every $x\in X$, is an interval (perhaps, empty one).

The key role is played by the following recursive definition of sequences $x^k\in X$, $s^k\in S$, and $S^k\subseteq S$ $(k\in {\mathbb N})$ such that:

$$\begin{aligned}&\displaystyle s^k\in S^k;\end{aligned}$$

(20a)

$$\begin{aligned}&\displaystyle S^k\text { is an interval};\end{aligned}$$

(20b)

$$\begin{aligned}&\displaystyle \forall s\in S^k\, \bigl [x^k\in {\mathcal {R}}(s)\bigr ];\end{aligned}$$

(20c)

$$\begin{aligned}&\displaystyle \forall m<k\, \bigl [S^k\cap S^m=\emptyset \bigr ];\end{aligned}$$

(20d)

$$ \begin{aligned}&\displaystyle \forall m<k\, \bigl [[s^k<s^m\Rightarrow x^k<x^m]\mathrel { \& }[s^k>s^m\Rightarrow x^k>x^m]\bigr ];\end{aligned}$$

(20e)

$$\begin{aligned}&\displaystyle \forall m<k\, \bigl [x^k{\succ ^{\!\!\!s^k}} x^m\mathrel {\text { or }}x^m\in {\mathcal {R}}(s^k)\bigr ];\end{aligned}$$

(20f)

$$ \begin{aligned}&\displaystyle \forall s\in S\, \bigl [[x^k\in {\mathcal {R}}(s)\mathrel { \& }s\notin S^k]\nonumber \\&\displaystyle \Rightarrow \exists m<k\,\bigl (s\in S^m\mathrel {\text { or }}s<s^m<s^k\mathrel {\text { or }}s^k<s^m<s\,\bigr )\bigr ]. \end{aligned}$$

(20g)

We start with an arbitrary $s^0\in S$, pick $x^0\in {\mathcal {R}}(s^0)$, and set $S^0=\{s\in S\mid x^0\in {\mathcal {R}}(s)\}$. Now (20a), (20c), and (20g) for $k=0$ immediately follow from the definitions; (20b), from Proposition 3.7; (20d), (20e), and (20f) hold vacuously.

Let $k\in {\mathbb N}\setminus \{0\}$, and let $x^m$, $s^m$, $S^m$ satisfying (20) have been defined for all $m<k$. We define $\bar{S}^k=\bigcup _{m<k} S^m$. For every $s\in \bar{S}^k$, there is a unique, by (20d), $\mu (s)<k$ such that $s\in S^{\mu (s)}$. By (20c), $r(s)=x^{\mu (s)}$ is a selection from ${\mathcal {R}}$ on $\bar{S}^k$. The conditions (20b) and (20e) imply that $r$ is increasing. If $\bar{S}^k=S$, then we already have an increasing selection, so we stop the process.

Otherwise, we pick $s^k\in S\setminus \bar{S}^k$ arbitrarily and denote $K^-=\{m<k\mid s^m<s^k\}$, $K^+=\{m<k\mid s^m>s^k\}$ $K^*=\{m<k\mid x^m\notin {\mathcal {R}}(s^k)\}$, $m^-=\text {argmax}_{m\in K^-}s^m$, $m^+=\text {argmin}_{m\in K^+}s^m$, and $I=\{s\in S\mid s^{m^-}<s<s^{m^+}\}$. If one of $K^\pm$ is empty (both cannot be), the respective $m^\pm$ is left undefined, in which case $I=\{s\in S\mid s^{m^-}<s\}$ or $I=\{s\in S\mid s<s^{m^+}\}$.

By Proposition 2.2 applied to ${\succ ^{\!\!\!s^k}}$, we can pick $x^k\in {\mathcal {R}}(s^k)$ such that $x^k{\succ ^{\!\!\!s^k}} x^m$ for each $m\in K^*$, hence (20f) holds. Finally, we define $S^k=\{s\in S\setminus \bar{S}^k\mid x^k\in {\mathcal {R}}(s)\}\cap I$. Now the conditions (20a), (20c), and (20d) immediately follow from the definitions; (20b) and (20g), from Proposition 3.7.

Checking (20e) needs a bit more effort. If we assume that $x^{m^-}\in {\mathcal {R}}(s^k)$, then the condition (20g) for $m^-$ and $s^k$ implies the existence of $m<m^-$ such that $s^{m^-}<s^m<s^k$, contradicting the definition of $m^-$; therefore, $x^k{\succ ^{\!\!\!s^k}}x^{m^-}$ by (20f), hence $x^k>x^{m^-}$ by Proposition 3.1. Therefore, $x^k>x^m$ for all $m\in K^-$. A dual argument shows that $x^k<x^{m^+}\le x^m$ for all $m\in K^+$. Thus, (20e) holds.

To summarize, either we obtain an increasing selection on some step, or our sequences are defined [and satisfy (20)] for all $k\in {\mathbb N}$.

Lemma 7.1

If conditions (20) hold for all $k\in {\mathbb N}$, then there exists an increasing sequence $\langle k_h\rangle _{h\in {\mathbb N}}$ such that $s^{k_h}$ is either monotone increasing or monotone decreasing in $h$, and $x^{k_{h+1}}{\succ ^{\!\!\!s^{k_{h+1}}}}x^{k_h}$ for each $h\in {\mathbb N}$.

Proof

We denote ${\mathbb N}^{\scriptscriptstyle \downarrow }$, respectively, ${\mathbb N}^{\scriptscriptstyle \uparrow }$, the set of $k\in {\mathbb N}$ such that $s^m<s^k$, or $s^m>s^k$, holds for an infinite number of $m\in {\mathbb N}$. Clearly, ${\mathbb N}={\mathbb N}^{\scriptscriptstyle \downarrow }\cup {\mathbb N}^{\scriptscriptstyle \uparrow }$; without restricting generality, ${\mathbb N}^{\scriptscriptstyle \downarrow }\ne \emptyset$. We consider two alternatives.

Let there exist $\min \{s^k\mid k\in {\mathbb N}^{\scriptscriptstyle \downarrow }\}=s^*$; then the set $\{m\in {\mathbb N}\mid s^m<s^k\}$ is finite for every $s^k<s^*$, hence the set $\{m\in {\mathbb N}\mid s^k<s^m<s^*\}$ is infinite. We define $k_0=\min \{k\in {\mathbb N}\mid s^k<s^*\}$, and then recursively define $k_{h+1}$ as the least $k\in {\mathbb N}$ for which $s^{k_h}<s^k<s^*$. The minimality of $k_h$ ensures that $k_{h+1}>k_h$. Whenever $s^{k_h}<s^m<s^{k_{h+1}}$, we have $m>k_{h+1}$ by the same minimality; therefore, $x^{k_h}\notin {\mathcal {R}}(s^{k_{h+1}})$ by (20g), hence $x^{k_{h+1}}{\succ ^{\!\!\!s^{k_{h+1}}}}x^{k_h}$ by (20f).

Let $\min \{s^k\mid k\in {\mathbb N}^{\scriptscriptstyle \downarrow }\}$ not exist; then the set $\{m\in {\mathbb N}^{\scriptscriptstyle \downarrow }\mid s^m<s^k\}$ is nonempty (actually, infinite) for every $k\in {\mathbb N}^{\scriptscriptstyle \downarrow }$. We set $k_0=\min {\mathbb N}^{\scriptscriptstyle \downarrow }$, and then recursively define $k_{h+1}$ as the least $k\in {\mathbb N}^{\scriptscriptstyle \downarrow }$ for which $s^k<s^{k_h}$. The minimality of $k_h$ again ensures that $k_{h+1}>k_h$. Whenever $s^{k_{h+1}}<s^m<s^{k_h}$, we have $m\in {\mathbb N}^{\scriptscriptstyle \downarrow }$, hence $m>k_{h+1}$; therefore, $x^{k_h}\notin {\mathcal {R}}(s^{k_{h+1}})$ by (20g), hence $x^{k_{h+1}}{\succ ^{\!\!\!s^{k_{h+1}}}}x^{k_h}$ by (20f).$\square$

The final step of the proof consists in showing that the existence of a sequence described in Lemma 7.1 contradicts the strong acyclicity assumption. If $s^{k_h}$ is increasing, the relations $x^{k_{h+1}}{\succ ^{\!\!\!s^{k_{h+1}}}}x^{k_h}$ “translate,” by (6a), to $x^{k_{h+1}}{\succ ^{\!\!\!\max S}}x^{k_h}$ for each $h\in {\mathbb N}$. If $s^{k_h}$ is decreasing, we obtain $x^{k_{h+1}}{\succ ^{\!\!\!\min S}}x^{k_h}$ for each $h\in {\mathbb N}$ by (6b).

1.2 Proof of Theorem 6.1

Since each $X_i$ contains its maximum and minimum, the same holds for each $S_i$. Applying Theorem 3.8, we obtain an increasing selection $r_i:S_i\rightarrow X_i$ from the best responses such that $r_i(S_i)$ is finite for each $i\in N$. Now we define an admissible best response quasi-improvement relation by (15) with

$$\begin{aligned} {\mathcal {R}}_i^*(x_{-i})=\{r_i(\sigma _i(x_{-i}))\}. \end{aligned}$$

Before showing the impossibility of an infinite admissible best response quasi-improvement path, we develop quite a number of auxiliary constructions.

For each $i\in N$, we denote $r_i(S_i)=\{x_i^1,\dots ,x_i^{m_i}\}$, assuming $x_i^{h+1}>x_i^h$ for all relevant $h$. For every $x_i\in X_i$, we define $\eta _i(x_i)\in \{0,1,\dots ,m_i\}$ as the minimal $h$ such that $x_i^{h+1}>x_i$; if no such $h$ exists, we set $\eta _i(x_i)=m_i$. Then we define $s_i^0=\min S_i$ and $s_i^h=\sup \{s_i\in S_i\mid r_i(s_i)=x_i^h\}$ for $h\in \{1,\dots ,m_i\}$; clearly, $s_i^{m_i}=\max S_i$. Then we denote $\Delta _i^h=s_i^h-s_i^{h-1}\,[\ge 0]$ for $h=1,\dots ,m_i$. Note that $\Delta _i^h=0$ is possible for some $h$ (if $r_i(s_i)=x_i^h$ for a unique $s_i=s_i^h=s_i^{h-1}$), in which case $\Delta _i^{h'}>0$ for (typically, both) adjacent $h'$.

For every $x_N\in X_N$, we define a set $N^0(x_N)=\{i\in N\mid x_i\in r_i(S_i)\}$ and a function

$$\begin{aligned} P(x_N)=\sum _{i\in N}\Bigl [-x_i\cdot s_i^{\eta _i(x_i)} + \sum _{j\in N:\,j\ne i} \frac{1}{2} {\,} a_{ij}\cdot x_i\cdot x_j + \sum _{h=1}^{\eta _i(x_i)} x_i^h\cdot \Delta _i^h\Bigr ]. \end{aligned}$$

(21)

For each $i\in N$, we define a binary relation $\triangleright _i$ on $r_i(S_i)$ by setting (for each $h\in \{1,\dots ,m_i-1\}$ such that $s_i^h\in S_i$) $x_i^{h+1}\triangleright _i x_i^h$ if $r_i(s_i^h)=x_i^{h+1}$, and $x_i^h\triangleright _i x_i^{h+1}$ if $r_i(s_i^h)=x_i^h$. Clearly, $x_i^h\triangleright _i x_i^{h+1}$ and $x_i^h\triangleright _i x_i^{h-1}$ whenever $\Delta _i^h=0$ (provided $h<m_i$ and $h>0$, respectively).

Lemma 7.2

Let $i\in N$, $z_i,y_i,x_i\in r_i(S_i)$, and $z_i\triangleright _i y_i\triangleright _i x_i$. Then either $z_i>y_i>x_i$ or $z_i<y_i<x_i$.

Immediately follows from the definitions.

Then we extend $\triangleright _i$ to the whole $X_i$, setting $y_i\triangleright _i x_i$ whenever $x_i\notin r_i(S_i)\ni y_i$, and define ${\triangleright \!\triangleright }_i$ as the transitive closure of $\triangleright _i$ on $X_i$.

Lemma 7.3

Each relation ${\triangleright \!\triangleright }_i$ is irreflexive and transitive.

Immediately follows from Lemma 7.2.

Finally, we define a potential:

$$ \begin{aligned}&y_N {\succ \!\!\!\succ }x_N\!\iff \!\bigl [N^0(y_N)\!\supset \! N^0(x_N)\!\mathrel {\text { or }}\! [N^0(y_N)\!=\!N^0(x_N)\!\mathrel { \& }\! P(y_N)\!>\!P(x_N)] \mathrel {\text { or }}\nonumber \\&\quad \bigl (N^0(y_N)=N^0(x_N)\mathrel { \& }P(y_N)=P(x_N)\mathrel { \& }\nonumber \\&\quad \forall i\in N\,[y_i=x_i\mathrel {\text { or }}y_i\,{\triangleright \!\triangleright }_i\, x_i]\mathrel { \& }\exists i\in N\, [y_i\,{\triangleright \!\triangleright }_i\, x_i]\bigr )\bigr ]. \end{aligned}$$

(22)

Lemma 7.4

The relation ${\succ \!\!\!\succ }$ is irreflexive and transitive.

Immediately follows from the definition.

Lemma 7.5

If $y_N{\triangleright _{\!i}^{\mathrm{[BR]^*}}}x_N$, then $y_N{\succ \!\!\!\succ }x_N$.

Proof

By definition, $y_i=r_i(\sigma _i(x_{-i}))$ and $y_{-i}=x_{-i}$, hence $i\in N^0(y_N)$. If $x_i\notin r_i(S_i)$, then we have $N^0(x_N)\subset N^0(y_N)$ since $y_j=x_j$ for all $j\ne i$; therefore, $y_N{\succ \!\!\!\succ }x_N$ by the first lexicographic component in (22). Otherwise, we have $N^0(x_N)=N^0(y_N)$; let us compare $P(y_N)$ and $P(x_N)$.

Let $y_i=x_i^{h''}$ and $x_i=x_i^{h'}$; we denote $\bar{s}_i=\sigma _i(x_{-i})$. Since $y_i=r_i(\bar{s}_i)$, we have $s_i^{h''-1}\le \bar{s}_i\le s_i^{h''}$. Since $\sum _{j\in N:j\ne i}a_{ij}\cdot x_j=\bar{s}_i$, we have

$$\begin{aligned} P(y_N)=y_i\cdot (\bar{s}_i-s_i^{h''}) +\sum _{h=1}^{h''}x_i^h\cdot \Delta _i^h + C(x_{-i}) =y_i\cdot (\bar{s}_i-s_i^{h''-1}) +\sum _{h=1}^{h''-1}x_i^h\cdot \Delta _i^h + C(x_{-i}). \end{aligned}$$

(23)

Let us consider two alternatives.

1. Let $y_i>x_i$, i.e., $h''>h'$. Similarly to (23), we have

$$\begin{aligned} P(x_N)&= x_i\cdot \left( \bar{s}_i-s_i^{h'}\right) +\sum _{h=1}^{h'}x_i^h\cdot \Delta _i^h + C(x_{-i})\nonumber \\&= x_i\cdot \left( \bar{s}_i-s_i^{h''-1}\right) +x_i\cdot \left( s_i^{h''-1}-s_i^{h'}\right) +\sum _{h=1}^{h'}x_i^h\cdot \Delta _i^h + C(x_{-i}).\nonumber \\ \end{aligned}$$

(24)

Note that $C(x_{-i})$ is indeed the same.

Subtracting (24) from (23), we obtain

$$\begin{aligned} P(y_N)-P(x_N)=(y_i-x_i)\cdot \left( \bar{s}_i-s_i^{h''-1}\right) +\sum _{h=h'+1}^{h''-1}\left( x_i^h-x_i\right) \cdot \Delta _i^h\ge 0. \end{aligned}$$

(25)

If $P(y_N)>P(x_N)$, we are home by the second lexicographic component in (22). Let $P(y_N)=P(x_N)$; then both terms, the product and the sum, in (25) must equal zero. Since $y_i>x_i$, we have $\bar{s}_i=s_i^{h''-1}$, hence $y_i=x_i^{h''}\triangleright _i x_i^{h''-1}$. If $h''=h'+1$ (so the sum is empty), then $y_i\,{\triangleright \!\triangleright }_i\, x_i$ and hence $y_N{\succ \!\!\!\succ }x_N$ by the third lexicographic component in (22).

Finally, let us show that the equality $P(y_N)=P(x_N)$ is incompatible with the inequality $h''>h'+1$. If $h''>h'+2$, then the sum in (25) contains at least one strictly positive term. The only remaining possibility is $h''-1=h'+1$ and $\Delta _i^{h''-1}=0$, i.e., $s_i^{h''-2}=s_i^{h''-1}=\bar{s}_i$. But then we must have $r_i(\bar{s}_i)=x_i^{h''-1}$, which contradicts $r_i(\bar{s}_i)=y_i=x_i^{h''}$.

2. Let $y_i<x_i$, i.e., $h''<h'$. We ignore what follows the second equality sign in (23), and replace (24) with

$$\begin{aligned} P(x_N)&= x_i\cdot \left( \bar{s}_i-s_i^{h'}\right) +\sum _{h=1}^{h'}x_i^h\cdot \Delta _i^h + C(x_{-i})\nonumber \\&= x_i\cdot \left( \bar{s}_i-s_i^{h''}\right) +x_i\cdot \left( s_i^{h''}-s_i^{h'}\right) +\sum _{h=1}^{h'}x_i^h\cdot \Delta _i^h + C(x_{-i}). \end{aligned}$$

(26)

Subtracting (26) from (23), we obtain

$$\begin{aligned} P(y_N)-P(x_N)=(x_i-y_i)\cdot \left( s_i^{h''}-\bar{s}_i\right) +\sum _{h=h''+1}^{h'}\left( x_i-x_i^h\right) \cdot \Delta _i^h. \end{aligned}$$

(27)

Again, $P(y_N)\ge P(x_N)$, and an equality is only possible if $\bar{s}_i=s_i^{h''}$ and $h'=h''+1$, which means that $y_i\triangleright _i x_i$, hence $y_i\,{\triangleright \!\triangleright }_i\, x_i$ as well. In other words, $y_N\,{\succ \!\!\!\succ }x_N$ again.$\square$

Finally, let $\langle x^k_N\rangle _{k=0,1,\dots }$ be an admissible best response quasi-improvement path, i.e., whenever $x_N^{k+1}$ is defined, there holds $x_N^{k+1}{\triangleright _{\!i}^{\mathrm{[BR]^*}}}x_N^k$ for some (unique) $i\in N$. By Lemma 7.5, we have $x_N^{k+1}{\succ \!\!\!\succ }x_N^k$. We set $N^*=\{i\in N\mid \exists k\,[x_i^{k+1}=r_i(x_{-i}^k)]\}$. If $i\in N\setminus N^*$, then $x_i^k$ is the same for all $k$. Thus, our path moves upwards (in the sense of ${\succ \!\!\!\succ }$) in a finite set $\prod _{i\in N^*}r_i(S_i)$, hence it cannot be infinite.

1.3 Proof of Theorems 6.2 and 6.3

The two proofs are so similar that we do not have to distinguish almost to the very end.

Exactly as in the proof of Theorem 6.1, we apply Theorem 3.8, obtaining an increasing selection from the best responses $r_i:S_i\rightarrow X_i$ such that $r_i(S_i)$ is finite for each $i\in N$. Then we again define an admissible best response quasi-improvement relation by (15) with ${\mathcal {R}}_i^*(x_{-i})=\{r_i(\sigma _i(x_{-i}))\}$.

Denoting $X_i^0=r_i(S_i)$ and ${\mathcal {X}}=\bigcup _{i\in N}X_i^0\subset {\mathbb R}$, we define strictly increasing mappings $\rho :{\mathcal {X}}\rightarrow {\mathbb N}$ by $\rho (x)=\#\{y\in {\mathcal {X}}\mid y<x\}$ (rank function) and $\varphi :{\mathcal {X}}\rightarrow {\mathbb R}$ by $\varphi (x)=n^{\rho (x)}$, where $n=\#N$.

Lemma 7.6

Let $I\subset N$, $y_I,x_I\in X_I^0$, and $\max _{i\in I}y_i>\max _{i\in I}x_i$. Then $\sum _{i\in I}\varphi (y_i)>\sum _{i\in I}\varphi (x_i)$.

Proof

Let $\max _{i\in I}x_i=\mu$. Then $\sum _{i\in I}\varphi (y_i)\ge n^{\mu +1}$, while $\sum _{i\in I}\varphi (x_i)\le \#I\cdot n^\mu <n^{\mu +1}$.

Supposing, to the contrary, that $\langle x_N^k\rangle _{k\in {\mathbb N}}$ is an infinite admissible best response quasi-improvement path, we denote $N^*=\{i\in N\mid \exists k\in {\mathbb N}\,[x_i^{k+1}=r_i(x_i^k)]\}$ and consider two alternatives.

1. Let $N^*=N$. We pick $\bar{k}\in {\mathbb N}$ such that $x_i^k\in X_i^0$ whenever $k\ge \bar{k}$. Clearly, $\langle x_N^k\rangle _{k\ge \bar{k}}$ is an infinite admissible best response quasi-improvement path in a subgame where each player is restricted to strategies from $X_i^0$. On the other hand, Lemma 7.6 implies that the subgame can be perceived as generated by the aggregation rules $\sigma _i^*(x_{-i})=\sum _{j\in I(i)}x_j$ in the case of Theorem 6.2, or $\sigma _i^*(x_{-i})=\sum _{j\in I(i)}(-x_j)$ in the case of Theorem 6.3. Therefore, it is covered by Theorem 6.1 in either case. The contradiction proves both theorems.

2. Let $N^*\subset N$. For each $i\in N\setminus N^*$, we have $x_i^k=x_i^0$ for all $k$. Therefore, we may consider a reduced game with the set of active players $N^*$, and each $i\in N\setminus N^*$ always choosing $x_i^0$. The game satisfies all assumptions of our theorem and $\langle x_N^k\rangle _{k\in {\mathbb N}}$ remains an infinite admissible best response quasi-improvement path; besides, Alternative 1 holds. Now the argument of the previous paragraph applies.

1.4 Counterexamples

Example 7.1

Without the transitivity of preferences, even the existence of a Nash equilibrium does not follow from the single crossing conditions, even in a two-person game with finite chains $X_i$.

Let $N=\{1,2\}$, $X_1=\{0,1,2,3,4\}$ and $X_2=\{5,6\}$ (both with natural orders); let preference relations ${\succ ^{\!\!x_{-i}}_{\!i}}$ be defined by: $2{\succ ^{\!\!5}_{\!1}}4{\succ ^{\!\!5}_{\!1}}0{\succ ^{\!\!5}_{\!1}}1{\succ ^{\!\!5}_{\!1}}3$; $1{\succ ^{\!\!6}_{\!1}}3{\succ ^{\!\!6}_{\!1}}2{\succ ^{\!\!6}_{\!1}}4{\succ ^{\!\!6}_{\!1}}0$; $5{\succ ^{\!\!x_1}_{\!2}}6$ whenever $x_1\le 1$; $6{\succ ^{\!\!x_1}_{\!2}}5$ whenever $x_1\ge 2$. The preferences of player 1 are intransitive, but single crossing conditions (16) are easy to check: (16a) is nontrivial only for $4{\succ ^{\!\!5}_{\!1}}0$; (16b), only for $1{\succ ^{\!\!6}_{\!1}}3$ and $2{\succ ^{\!\!6}_{\!1}}4$. Player 2’s preferences are described by a family of total orders; (16) are obvious. There is no Nash equilibrium: ${\mathcal {R}}_1(5)=\{2\}$ and ${\mathcal {R}}_1(6)=\{1\}$, whereas ${\mathcal {R}}_2(2)=\{6\}$ and ${\mathcal {R}}_2(1)=\{5\}$. It may be noted that ${\mathcal {R}}_1$ admits no increasing selection.

Remark

The game in Example 7.1 admits no infinite best response improvement path, but does not have the FBRP, due to the second condition in the definition. However, it does admit infinite best response quasi-improvement paths.

Example 7.2

Theorem 3.8 cannot be extended to strongly acyclic and transitive preference relations.

Let $X=[-2,2]$, $S=[-1,1]$ (both with natural orders), and relations ${\succ ^{\!\!\!s}}$ be defined by

$$ \begin{aligned} y{\succ ^{\!\!\!s}}x\iff [u_1(y,s)>u_1(x,s)\mathrel { \& }u_2(y,s)>u_2(x,s)], \end{aligned}$$

(28)

where $u:X\times S\rightarrow {\mathbb R}^2$ is this: $u(1,s)=\langle 5,2\rangle$ and $u(-1,s)=\langle 2,5\rangle$ for all $s\in S$; $u(2,s)=u(-2,s)=u(x,s)=\langle 0,0\rangle$ for all $x\in {]-1,1[}$ and $s\in S$; whenever $x\in {]1,2[}$ and $s\ge 0$,

$$\begin{aligned} u_1(x,s)= {\left\{ \begin{array}{ll} x+s-1,&{} \quad \text {if }\quad x+s\le 2,\\ x+s+4,&{} \quad \text {if }\quad x+s>2, \end{array}\right. } \end{aligned}$$

while $u_2(x,s)=6-x-s$; whenever $x\in {]-2,-1[}$ and $s\ge 0$, $u(x,s)=\langle x+6,-1-x\rangle$; finally, $u_i(x,s)$ for all $s<0$, $i=1,2$, and $x\in {]-2,-1[}\cup {]1,2[}$ is such that the equality

$$\begin{aligned} u_i(x,s)=u_{3-i}(-x,-s) \end{aligned}$$

(29)

holds for all $s\in S$, $i=1,2$, and $x\in X$.

The very form of (28) ensures that every ${\succ ^{\!\!\!s}}$ is irreflexive and transitive. Whenever $x\in \{-2\}\cup {]{-1},1[}\cup \{2\}$ and $y\in {]-2,-1]}\cup {[1,2[}$, $y{\succ ^{\!\!\!s}}x$ for every $s\in S$. Whenever $x,y\in {]-2,-1[}$ or $x,y\in {]1,2[}$, $y{\succ ^{\!\!\!s}}x$ does not hold for any $s\in S$. Let $s\ge 0$; if $-2<x<-1$, then $u_1(x)<5$ and $u_2(x)\le 1$, hence $1{\succ ^{\!\!\!s}}x$; if $1<x\le 2-s<2$, then $u_1(x)\le 1$ and $u_2(x)<5$, hence $-1{\succ ^{\!\!\!s}}x$; if $2-s<y<2$, then $u_1(y)>6$ and $u_2(y)>3$, hence $y{\succ ^{\!\!\!s}}1$. “Dually,” by (29), $y{\succ ^{\!\!\!s}}-1{\succ ^{\!\!\!s}} x$ whenever $s<0$, $-2<y<-2-s$, and $1<x<2$; $1{\succ ^{\!\!\!s}}x$ whenever $s<0$ and $-2-s\le x<-1$. Thus, ${\mathcal {R}}(s)=\{-1\}\cup ]2-s,2[$ for $s>0$, ${\mathcal {R}}(s)=\{1\}\cup ]-2,-2-s,2[$ for $s<0$, and ${\mathcal {R}}(0)=\{-1,1\}$ (Fig. 1). We see that every relation ${\succ ^{\!\!\!s}}$ is strongly acyclic: no more than three consecutive improvements can be made from any starting point (e.g., $2-s/2{\succ ^{\!\!\!s}}1{\succ ^{\!\!\!s}}-1.5{\succ ^{\!\!\!s}}-2$ when $s>0$). Single crossing conditions (6) are also easy to check.

Suppose there is an increasing selection $r$ from ${\mathcal {R}}$. If $r(s)>-1$ for some $s>0$, then $2>r(s)>2-s$; defining $s'=2-r(s)>0$, we have $s'<s$, hence $r(s')\le r(s)$, hence $r(s')<2-s'$, hence $r(s')\in {\mathcal {R}}(s')$ is only possible if $r(s')=-1$. Therefore, $r(s)=-1$ for some $s>0$; dually, $r(s)=1$ for some $s<0$. We have a contradiction, i.e., there is no increasing selection.

Fig. 1

Best responses in Example 7.2

Example 7.3

If discontinuous utility functions are replaced with their “upper semicontinuous closures,” there may be no $\varepsilon$-Nash equilibrium of the original game close to a Nash equilibrium of the modified game.

Let $N=\{1,2\}$, $X_1=X_2=[0,1]$ (with the natural order), and the preferences of the players be defined by “isomorphic” utility functions

$$ \begin{aligned} u_i(x_i,x_j)= {\left\{ \begin{array}{ll} (1+x_j)x_i,&{}\quad \text {if } \quad 0\le x_i<1/2\mathrel { \& }0\le x_j\le 1,\\ (1+x_j)x_i-2,&{}\quad \text {if } \quad 1/2\le x_i<3/4\mathrel { \& }0\le x_j<3/4,\\ (1+x_j)x_i,&{}\quad \text {if } \quad 1/2\le x_i<3/4\mathrel { \& }3/4\le x_j\le 1,\\ (x_j-2)x_i,&{} \quad \text {if } \quad 3/4\le x_i\le 1\mathrel { \& }0\le x_j<3/4,\\ (x_j-2)x_i+2,&{} \quad \text {if } \quad 3/4\le x_i\le 1\mathrel { \& }3/4\le x_j\le 1. \end{array}\right. } \end{aligned}$$

It is easily checked that $u_i(x_i,x_j)$ thus defined is supermodular: whenever it increases in $x_i$, the rate of increase goes up as $x_j$ increases; whenever it decreases in $x_i$, the rate of decrease goes down as $x_j$ increases. For every $x_j$, the set ${\mathcal {R}}_i(x_j)$ is empty; therefore, there is no Nash equilibrium.

To obtain the upper semicontinuous closure of $u_i$, we have to modify it at $x_i=1/2$ for $0\le x_j<3/4$ and at $x_i=3/4$ for all $0\le x_j\le 1$. The resulting best responses are depicted as thick lines in Fig. 2: $\bar{\mathcal {R}}_i(x_j)=\{1/2\}$ for $0\le x_j<3/4$; $\bar{\mathcal {R}}_i(x_j)=\{3/4\}$ for $3/4<x_j\le 1$; $\bar{\mathcal {R}}_i(3/4)=\{1/2,3/4\}$. That new game has two Nash equilibria: $(1/2,1/2)$ and $(3/4,3/4)$. Finally, if we switch from the original utilities $u_i$ to preferences defined by (4) with a small enough $\varepsilon >0$, then the “best” responses are depicted in the same figure by small arrows. We see that there is no $\varepsilon$-Nash equilibrium of the original game near the point $(3/4,3/4)$, so any search in that vicinity would be futile.

Fig. 2

“Best” responses in Example 7.3

Remark

In accordance with Theorem 5.2, it is impossible to make more than three consecutive $\varepsilon$-best response improvements in this game, whatever the starting point.

Example 7.4

The FBRP cannot be asserted in Theorem 6.1, even for a finite game with the preferences described by utility functions.

Let $N=\{1,2,3\}$, $X_1=\{0,1,2,3,4\}$, $X_2=\{0,1,2,3,4,5\}$, $X_3=\{0,1\}$; let the preferences of the players be defined by utility functions $u_i(x_N)=U_i(x_i,-\sum _{j\ne i}x_j)$. Clearly, we have an aggregative game as in Theorem 6.1 with $a_{ij}=-1$, hence $S_1=\{-6,-5,\dots ,0\}$, $S_2=\{-5,-4,\dots ,0\}$, $S_3=\{-9,-8,\dots ,0\}$. Let the utilities be:

$$\begin{aligned} \begin{array}{l@{\quad }l@{\quad }l@{\quad }l@{\quad }l@{\quad }l} U_1: &{} \left[ \begin{array}{lllll} 2&{} \quad 3&{} \quad 3&{}\quad 3&{}\quad 4\\ 2&{}\quad 3&{}\quad 3&{}\quad 3&{}\quad 2\\ 2&{}\quad 3&{}\quad 3&{}\quad 3&{}\quad 2\\ 2&{}\quad 3&{}\quad 3&{}\quad 3&{}\quad 2\\ 2&{}\quad 3&{}\quad 3&{}\quad 3&{}\quad 2\\ 2&{}\quad 3&{}\quad 3&{}\quad 3&{}\quad 2\\ 2&{}\quad 1&{}\quad 1&{}\quad 1&{}\quad 0 \end{array}\right] &{} U_2: &{} \left[ \begin{array}{llllll} 1&{} \quad 1&{}\quad 1&{}\quad 1&{}\quad 1&{}\quad 2\\ 1&{} \quad 1&{}\quad 1&{}\quad 1&{}\quad 1&{}\quad 2\\ 1&{} \quad 1&{}\quad 1&{}\quad 1&{}\quad 1&{}\quad 2\\ 1&{} \quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 1&{} \quad 0&{}\quad 0&{}\quad 0&{}\quad 0&{}\quad 0\\ 1&{} \quad 0&{}\quad 0&{}\quad 0&{} \quad 0&{}\quad 0 \end{array}\right] &{} U_3:&{} \left[ \begin{array}{ll} 1&{}\quad 2\\ 1&{}\quad 2\\ 1&{}\quad 2\\ 1&{}\quad 2\\ 1&{}\quad 2\\ 1&{}\quad 0\\ 1&{}\quad 0\\ 1&{}\quad 0\\ 1&{}\quad 0\\ 1&{}\quad 0\\ 1&{}\quad 0 \end{array}\right] \end{array} \end{aligned}$$

where own choice, $x_i$, is on the abscissae axis, and $s_i$ (= minus the sum of the partners’ choices), on the ordinates axis. Conditions (6), even (7), are easy to check. By Theorem 6.1, the game has a restricted F[BR]P; actually, even a restricted FBRP. However, it does not have the FBRP since there is a best response improvement cycle:

Remark

The question of whether such an example is possible, was left open in Kukushkin (2004). If we retained the same sets $X_i$ and utilities $U_i$, but redefined $\sigma _i$, setting $a_{ij}=1$ for all $i,j\in N$, $j\ne i$, then there could be no best response improvement cycle (Kukushkin 2004, Theorem 1).