Nash equilibrium in discontinuous games

Abstract

We provide several generalizations of the various equilibrium existence results in Reny (Econometrica 67:1029–1056, 1999), Barelli and Meneghel (Econometrica 81:813–824, 2013), and McLennan et al. (Econometrica 79:1643–1664, 2011). We also provide an example demonstrating that a natural additional generalization is not possible. All of the theorems yielding existence of pure-strategy Nash equilibria here are stated in terms of the players’ preference relations over joint strategies. Hence, in contrast to much of the previous work in the area, the present results for pure-strategy equilibria are entirely ordinal.

Appendix

The proof of Theorem 3.4 below is inspired by the proof in McLennan et al. (2011). A key distinction is that we must introduce a well-chosen “dominance” relation to play the role that, in McLennan et. al.’s proof, is played by the players’ utility functions, which are of course unavailable here. A secondary distinction is the presence here of a subset of players whose best-reply correspondences are well defined.

Proof of Theorem 3.4

Fix any $x^{0}\in X$ and let $J=N\backslash I.$ Letting $F_{i}$ denote the best-reply correspondence of any player $i\in J,$ the correspondence $\times _{i\in J}F_{i}(x_{I}^{0},\cdot ):\times _{i\in J}X_{i}\twoheadrightarrow \times _{i\in J}X_{i}$ is non-empty-valued, non-convex-valued, and closed, and if $x_{J}^{*}$ is any one of its fixed points, then $(x_{I}^{0},x_{J}^{*})$ is a member of $B_{I},$ the set of points in X at which players in J are simultaneously best-replying. By Glicksberg’s (1952) theorem, $B_{I}$ is non-empty. Moreover, $B_{I}$ is compact because the best-reply correspondences of the players in J are closed.

Suppose, by way of contradiction, that there is no equilibrium in $B_{I}.$ Then, by point security with respect to I,  for every $x\in B_{I}$, there is a neighborhood $U^{x}$ of x and a point $x^{x}\in X$ such that for every $y\in U^{x}\cap B_{I}$, there is a player $i\in I$ for whom

$$\begin{aligned} (x_{i}^{x},x_{-i}^{\prime })>_{i}y,\text { for every }x^{\prime }\in U^{x}\cap B_{I}. \end{aligned}$$

Thus, we have a collection of pairs $\{(U^{x},x^{x})\}_{x\in B_{I}}$ such that each $U^{x}\cap B_{I}$ is non-empty (both sets contain x) and where the $U^{x}$ form an open cover of $B_{I}.$ We may therefore extract a finite subcollection $\{(U^{k},x^{k})\}$ such that the $U^{k}$ form a finite open cover of $B_{I}$, and such that: for each k,  and for every $y\in U^{k}\cap B_{I}$, there is a player $i\in I$ for whom

$$\begin{aligned} \left( x_{i}^{k},x_{-i}^{\prime }\right) >_{i}y,\text { for every }x^{\prime }\in U^{k}\cap B_{I}. \end{aligned}$$

(*)

By construction, $U^{k}\cap B_{I}$ is non-empty for every k. Say that k i -dominates $k^{\prime },$ and write $k\trianglerighteq _{i}k^{\prime },$ if for every $x\in U^{k}\cap B_{I}$, there exists $x^{\prime }\in U^{k^{\prime }}\cap B_{I}$ such that

$$\begin{aligned} \left( x_{i}^{k},x_{-i}\right) \ge _{i}\left( x_{i}^{k^{\prime }},x_{-i}^{\prime }\right) . \end{aligned}$$

The i-dominance relation inherits from $\ge _{i},$ completeness, reflexivity, and transitivity. Let $\vartriangleright _{i}$ denote the strict relation associated with $\trianglerighteq _{i}$.²²

Because $B_{I}$ is a compact subset of a Hausdorff space, for each k we may choose a closed set $C^{k}\subset U^{k}$ such that the resulting finite collection $\{C^{k}\}$ covers $B_{I}.$ For each k and each player i,  define $W_{i}^{k}=U^{k}\backslash \cup _{j\vartriangleright _{i}k}C^{j}.$ Then, for each i,  $\{W_{i}^{k}\}_{k}$ is a finite open cover of $B_{I}$ and $\cup _{k}W_{i}^{k}=\cup _{k}U^{k}$ is independent of i.²³ Let $W=\cup _{k} W_{i}^{k}.$ By Munkres (1975, Theorem 5.1) for each i,  there is for every k a continuous function $\lambda _{i}^{k}:W\rightarrow [0,1]$ such that $\sum _{k}\lambda _{i}^{k}(x)=1$ for all $x\in W$ and such that $\lambda _{i}^{k}(x)>0$ implies $x\in W_{i}^{k}$.²⁴

Consequently, for each player i,  the function on W defined by $g_{i}(x)=\sum _{k}\lambda _{i}^{k}(x)x_{i}^{k}$ is continuous. The correspondence mapping the compact, convex and locally convex space $(\times _{i\in I}co\{x_{i}^{k}\}_{k})\times (\times _{i\in J}X_{i})$ into subsets of $co\{x_{i}^{k}\}_{k}$ defined by $G_{i}(x)=\{g_{i}(x)\}$ if $x\in W$ and $G_{i}(x)=co\{x_{i}^{k}\}_{k}$ if $x\in X\backslash W$ is therefore non-empty-valued, convex-valued, and closed. Hence, by Glicksberg’s (1952) theorem, there exists $y^{*}\in X$ such that $(y_{i}^{*})_{i\in J}\times (y_{i}^{*})_{i\in I}\in (\times _{i\in J}F_{i}(y^{*}))\times (\times _{i\in I}G_{i}(y^{*})).$ Consequently, (i) $y_{i}^{*}\in F_{i}(y^{*})$ for every $i\in J$ and (ii) $y_{i}^{*}\in G_{i}(y^{*})$ for every $i\in I.$ But (i) implies that $y^{*}\in B_{I}\subset W$ so that $G_{i}(y^{*})=\{g_{i}(y^{*})\}$ which together with (ii) implies that (iii) $y_{i}^{*}=\sum _{k}\lambda _{i}^{k}(y^{*})x_{i}^{k}$ for every $i\in I.$

Because the $C^{k}$ cover $B_{I},$ we may choose $\hat{k}$ such that $y^{*}\in C^{\hat{k}}\subset U^{\hat{k}}.$ Then, for any i and k with $\lambda _{i}^{k}(y^{*})>0,$ we have $y^{*}\in W_{i}^{k}\subset U^{k}$ and so $k\trianglerighteq _{i}\hat{k}$ (otherwise $\hat{k} \vartriangleright _{i}k$ implying that $y^{*}\in C^{\hat{k}}$ is not in $W_{i}^{k}=U^{k}\backslash \cup _{j\vartriangleright _{i}k}C^{j}).$ Consequently, because $y^{*}\in U^{k}\cap B_{I},$ there exists $x^{i,k}\in U^{\hat{k}}\cap B_{I}$ such that $(x_{i}^{k},y_{-i}^{*})\ge _{i}(x_{i}^{\hat{k}},x_{-i}^{i,k}).$

For each player $i\in I,$ let $k_{i}$ be the value of k in the (non-empty by (iii)) set $\{k:\lambda _{i}^{k}(y^{*})>0\}$ giving the least preferred outcome for i in all the $(x_{i}^{\hat{k} },x_{-i}^{i,k}).$ Then, for every $i\in I$ and every k such that $\lambda _{i}^{k}(y^{*})>0,$ $(x_{i}^{k},y_{-i}^{*})\ge _{i}(x_{i}^{\hat{k} },x_{-i}^{i,k})\ge _{i}(x_{i}^{\hat{k}},x_{-i}^{i,k_{i}}),$ and hence,

$$\begin{aligned} (x_{i}^{k},y_{-i}^{*})\ge _{i}(x_{i}^{\hat{k}},x_{-i}^{i,k_{i}}). \end{aligned}$$

Together with (iii) and the convexity of $\ge _{i},$ this implies that $y^{*}\ge _{i}\left( x_{i}^{\hat{k}},x_{-i}^{i,k_{i}}\right)$ for every $i\in I,$ contradicting (*) because $y^{*}$ and each $x^{i,k_{i}}$ are in $U^{\hat{k}}\cap B_{I}.$ $\square$

Proof of Theorem 4.2

Let G be correspondence secure with respect to $I\subseteq N.$ As shown in the proof of Theorem 3.4, the set $B_{I}$ is non-empty and compact. Suppose by way of contradiction that G has no Nash equilibrium in $B_{I}$. Then, for each $x\in B_{I}$, there is a neighborhood U of x and a closed correspondence $d:U\twoheadrightarrow X$ with non-empty and convex values such that the condition stated in Definition 4.1 holds. Since the collection of all such Us forms an open cover of the compact set $B_{I},$ we may extract a finite subcover $U^{1},...,U^{K}$, together with their associated correspondences $d^{1},...,d^{K}.$ Hence, for every $k=1,...,K$ and every $y\in U^{k}\cap B_{I}$, there is a player $i\in I$ for whom

$$\begin{aligned} (z_{i},x_{-i}^{\prime })>_{i}y,\text { for every }x^{\prime }\in U^{k}\cap B_{I}\text { and every }z_{i}\in d_{i}^{k}(x^{\prime }). \end{aligned}$$

(1)

We now define a surrogate game, $G^{*},$ and will obtain the desired contradiction by showing that $G^{*}$ satisfies all the hypotheses of Theorem 3.4 but has no Nash equilibrium.

For each $i\in N\backslash I,$ let $F_{i}:X\twoheadrightarrow X_{i}$ denote i’s best-reply correspondence. For each $k=1,...,K,$ extend $d^{k}$ to X by defining $d^{k}(x)=X$ whenever $x\notin U^{k}.$ Each $d^{k}:X \twoheadrightarrow X$ is then closed with non-empty and convex values.

Introduce two new players, A and C. The surrogate game $G^{*}$ has player set $\{A,C\}\cup I.$ Player A chooses $a\in X,$ player C chooses $c\in X$, and each player $i\in I$ chooses $\alpha _{i}\in \Delta =\{\lambda \in [0,1]^{K}:\sum _{k=1}^{K}\lambda _{k}=1\},$ the unit simplex in $\mathbb {R}^{K}.$

Players A and C have preferences on $X\times X\times \Delta ^{I}$ that are represented by the own-strategy quasi-concave utility functions $u_{A}$ and $u_{C},$ respectively, where $u_{A}(a,c,\alpha )=1$ if $a=c$ and 0 otherwise, and $u_{C}(a,c,\alpha )=1$ if $c\in d(a,\alpha )$ and 0 otherwise, where $d(a,\alpha ):=\left( \times _{i\in I}\sum _{k=1}^{K}\alpha _{ik}d_{i}^{k}(a)\right) \times \left( \times _{i\in N\backslash I}F_{i}(a)\right) .$

Player $i\in I$ has preferences $\ge _{i}^{*}$ on $X\times X\times \Delta ^{I}$ defined by $(a,c,\alpha )\ge _{i}^{*}(a^{\prime },c^{\prime },\alpha ^{\prime })$ if and only if $\forall z_{i}\in d_{i}(a,\alpha ),$ $\exists z_{i}^{\prime }\in d_{i}(a^{\prime },\alpha ^{\prime })$ such that $(z_{i},a_{-i})\ge _{i}(z_{i}^{\prime },a_{-i}^{\prime })$.²⁵ Each relation $\ge _{i}^{*}$ is complete, reflexive, transitive, and convex.²⁶

So defined, the game $G^{*}$ satisfies A.1–A.3, and players A and C (all players in fact) have locally convex strategy spaces. Note also that players A and C have closed best-reply correspondences whose values are non-empty and convex. We next show that $G^{*}$ is point secure with respect to I, which will allow us to apply Theorem 3.4.

So as not to confuse the original game, G,  with the surrogate game, $G^{*},$ let $B_{I}^{*}$ denote the set of strategies $(a,c,\alpha )$ such that players A and C are simultaneously best-replying in $G^{*}$ . Hence, $B_{I}^{*}=\{(a,c,\alpha ):a=c$ and $c\in d(a,\alpha )\}.$ Observe that if $(a,c,\alpha )$ is in $B_{I}^{*},$ then $a\in d(a,\alpha )$ and so $a\in B_{I}.$

Consider any $(a,c,\alpha )\in B_{I}^{*}.$ Then, as just observed, $a\in B_{I}.$ Hence, there is $\hat{k}$ such that $a\in U^{\hat{k}}$ and so $(a,c,\alpha )\in (U^{\hat{k}}\times U^{\hat{k}}\times \Delta ^{I})\cap B_{I}^{*}.$ For every $(y,w,\beta )\in (U^{\hat{k}}\times U^{\hat{k} }\times \Delta ^{I})\cap B_{I}^{*},$ since (similar to a) $y\in U^{ \hat{k}}\cap B_{I},$ condition (1) implies that there is a player $i\in I$ for whom $(z_{i},a_{-i}^{\prime })>_{i}y$ for every $a^{\prime }\in U^{\hat{k}}\cap B_{I}$ and every $z_{i}\in d_{i}^{\hat{k} }(a^{\prime }).$ But then, because $y_{i}\in d_{i}(y,\beta )$ ,

$$\begin{aligned} (a^{\prime },c^{\prime },e_{i}^{\hat{k}},\alpha _{-i}^{\prime })>_{i}^{*}(y,w,\beta ) \end{aligned}$$

(2)

holds for every $(a^{\prime },c^{\prime },\alpha ^{\prime })\in (U^{\hat{k} }\times U^{\hat{k}}\times \Delta ^{I})\cap B_{I}^{*},$ where $e_{i}^{ \hat{k}}$ is the $\hat{k}$-th unit vector in $\Delta$.

Since $(a,c,\alpha )\in B_{I}^{*}$ was arbitrary, (2) holds in particular when $(a,c,\alpha )\in B_{I}^{*}$ is not a Nash equilibrium and so we have shown that $G^{*}$ is point secure with respect to I. But (2) also shows that $G^{*}$ has no Nash equilibrium since any Nash equilibrium $(a,c,\alpha )$ must be in $B_{I}^{*}$ and so we may set $(a^{\prime },c^{\prime },\alpha ^{\prime })=(y,w,\beta )=(a,c,\alpha )$. This contradicts Theorem 3.4 and completes the proof. $\square$

Proof of Theorem 5.5

Follow the steps of the proof of Theorem 3.4, except that (a) $( \hat{x}_{i}^{x},x_{-i}^{\prime })>_{i}y$ in the display preceding (*) should be replaced with $y_{i}\notin co\{w_{i}:(w_{i},y_{-i})\ge _{i}(\hat{x}_{i}^{x},x_{-i}^{\prime })\},$ (b) $(\hat{x}_{i}^{k},x_{-i}^{\prime })>_{i}y$ in (*) should be replaced with $y_{i}\notin co\{w_{i}:(w_{i},y_{-i})\ge _{i}(\hat{x} _{i}^{k},x_{-i}^{\prime })\},$ and (c) $y^{*}\ge _{i}(x_{i}^{\hat{k} },x_{-i}^{i,k_{i}})$ in the final sentence should be replaced with $y_{i}^{*}\in co\{w_{i}:(w_{i},y_{-i}^{*})\ge _{i}(\hat{x} _{i}^{k},x_{-i}^{i,k_{i}})\}.$ $\square$

Proof of Theorem 5.6

The proof follows the steps of the proof of Theorem 4.2 except that (a) the correspondences $d^{k}$ satisfying (1) are co-closed, even when extended to all of X,  (b) player $B\,$’s payoff is defined by $u_{B}(a,c,\alpha )=1$ if $c\in cod(a,\alpha )$ and 0 otherwise, (c) $\ge _{i}^{*}$ is not necessarily convex and so $G^{*}$ satisfies only A.1 and A.2, and (d) the last three paragraphs of the proof are replaced with the following four paragraphs:

So as not to confuse the original game, G,  with the surrogate game, $G^{*},$ let $B_{I}^{*}$ denote the set of strategies $(a,c,\alpha )$ such that players A and C are simultaneously best-replying in $G^{*}$ . Hence, $B_{I}^{*}=\{(a,c,\alpha ):a=c$ and $c\in cod(a,\alpha )\}.$ Observe that if $(a,c,\alpha )$ is in $B_{I}^{*},$ then $a\in cod(a,\alpha )$ and so $a\in B_{I}.$

Consider any $(a,c,\alpha )\in B_{I}^{*}.$ Then, as just observed, a is in $B_{I}.$ Hence, there is $\hat{k}$ such that $a\in U^{\hat{k}}$ and so $(a,c,\alpha )\in (U^{\hat{k}}\times U^{\hat{k}}\times \Delta ^{I})\cap B_{I}^{*}.$ For every $(a^{1},c^{1},\alpha ^{1})\in (U^{\hat{k}}\times U^{\hat{k}}\times \Delta ^{I})\cap B_{I}^{*},$ since (similar to a) $a^{1}\in U^{\hat{k}}\cap B_{I},$ condition (1) implies that there is a player $i\in I$ for whom

$$\begin{aligned} a_{i}^{1}\notin co\{w_{i}:(w_{i},a_{-i}^{1})\ge _{i}(z_{i},a_{-i}^{\prime })\} \end{aligned}$$

(3)

holds for every $a^{\prime }\in U^{\hat{k}}\cap B_{I}$ and every $z_{i}\in d_{i}^{\hat{k}}(a^{\prime }).$ Because $(a^{1},c^{1},\alpha ^{1})\in B_{I}^{*}$, we have $a_{i}^{1}\in cod_{i}(a^{1},\alpha ^{1})=\sum _{k=1}^{K}\alpha _{ik}^{1}cod_{i}^{k}(a^{1})=co\sum _{k=1}^{K}\alpha _{ik}^{1}d_{i}^{k}(a^{1}),$ and so $a_{i}^{1}$ is a convex combination of $a_{i}^{1j}$’s such that each $a_{i}^{1j}=\sum _{k=1}^{K}\alpha _{ik}^{1}\delta _{i}^{kj}$ and each $\delta _{i}^{kj}\in d_{i}^{k}(a^{1}).$

We claim that

$$\begin{aligned} \alpha _{i}^{1}\notin co\{\gamma _{i}:(a^{1},c^{1},\gamma _{i},\alpha _{-i}^{1})\ge _{i}^{*}(a^{\prime },c^{\prime },e_{i}^{\hat{k}},\alpha _{-i}^{\prime })\} \end{aligned}$$

(4)

holds for every $(a^{\prime },c^{\prime },\alpha ^{\prime })\in (U^{\hat{k} }\times U^{\hat{k}}\times \Delta ^{I})\cap B_{I}^{*},$ where $e_{i}^{ \hat{k}}$ is the $\hat{k}$-th unit vector in $\Delta$. Otherwise, for some such $(a^{\prime },c^{\prime },\alpha ^{\prime })$, $\alpha _{i}^{1}$ would be a convex combination of $\alpha _{i}^{1n}$’s such that for each n, $(a^{1},c^{1},\alpha _{i}^{1n},\alpha _{-i}^{1})\ge _{i}^{*}(a^{\prime },c^{\prime },e_{i}^{\hat{k}},\alpha _{-i}^{\prime }).$ Defining $z_{i}^{nj}:=\sum _{k=1}^{K}\alpha _{ik}^{1n}\delta _{i}^{kj},$ we have $z_{i}^{nj}\in d_{i}(a^{1},\alpha _{i}^{1n},\alpha _{-i}^{1}).$ Hence, because $(a^{1},c^{1},\alpha _{i}^{1n},\alpha _{-i}^{1})\ge _{i}^{*}(a^{\prime },c^{\prime },e_{i}^{\hat{k}},\alpha _{-i}^{\prime }),$ there exists for each n and j,  $\tilde{z}_{i}^{nj}\in d_{i}^{\hat{k} }(a^{\prime })$ such that $(z_{i}^{nj},a_{-i}^{1})\ge _{i}(\tilde{z} _{i}^{nj},a_{-i}^{\prime }).$ Let $\tilde{z}_{i}$ denote a $\tilde{z} _{i}^{nj}$ that makes $(\tilde{z}_{i}^{nj},a_{-i}^{\prime })$ the least desirable for i as n and j vary. Then, $\tilde{z}_{i}\in d_{i}^{\hat{k} }(a^{\prime })$ and for every n and j, we have $(z_{i}^{nj},a_{-i}^{1}) \ge _{i}(\tilde{z}_{i},a_{-i}^{\prime }).$ But, because $a_{i}^{1}$ is evidently a convex combination of the $z_{i}^{nj}$ and because $(a^{\prime },c^{\prime },\alpha ^{\prime })\in (U^{\hat{k}}\times U^{\hat{k}}\times \Delta ^{I})\cap B_{I}^{*}$ implies $a^{\prime }\in U^{\hat{k}}\cap B_{I},$ this contradicts (3) and so establishes (4).

Since $(a,c,\alpha )\in B_{I}^{*}$ was arbitrary, (4 ) holds in particular when $(a,c,\alpha )\in B_{I}^{*}$ is not a Nash equilibrium and so we have shown that $G^{*}$ is point secure with respect to I. But (4) also shows that $G^{*}$ has no Nash equilibrium since any Nash equilibrium $(a,c,\alpha )$ must be in $B_{I}^{*}$ and so we may set $(a^{\prime },c^{\prime },\alpha ^{\prime })=(y,r,\beta )=(a,c,\alpha )$. This contradicts Theorem 3.4 and completes the proof. $\square$

Proof of Theorem 5.8

The proof follows that of Theorem 3.4. One need only replace (*) and the display immediately preceding (*) with their weak point-security counterparts. Then, at the end of the proof of Theorem 3.4, let V be the neighborhood of $y^{*}$ defined by $V=\cap _{j,k}W_{j}^{\hat{k}},$ where the intersection is over all j, k such that $\lambda _{j}^{k}(y^{*})>0.$ Continuing now as in the proof of Theorem 3.4, it is not difficult to see that for every player i and every $y^{\prime }\in V\cap B_{I},$ there exists $x^{i}\in U^{\hat{k}}\cap B_{I}$ s.t. $(y_{i}^{*},y_{-i}^{\prime })\ge _{i}(\hat{x}_{i}^{k},x_{-i}^{i}),$ contradicting the adjusted (*) because $y^{*}\in U^{\hat{k}}\cap B_{I}.$ $\square$