Strict Nash equilibria in non-atomic games with strict single crossing in players (or types) and actions

Abstract

In this paper, we study games where the space of players (or types, if the game is one of incomplete information) is atomless and payoff functions satisfy the property of strict single crossing in players (types) and actions. Under an additional assumption of quasisupermodularity in actions of payoff functions and mild assumptions on the player (type) space—partially ordered and with sets of uncomparable players (types) having negligible size—and on the action space—lattice, second countable and satisfying a separation property with respect to the ordering of actions—we prove that every Nash equilibrium is essentially strict. Furthermore, we show how our result can be applied to incomplete information games, obtaining the existence of an evolutionary stable strategy, and to population games with heterogeneous players.

Appendices

Appendix 1: Lemma 1 and its proof

A key result for the Proof of Theorem 1 is that any set of weakly best responders is a countable union of sets having measure zero. Lemma 1 below provides such result.

The logic of the Proof of Lemma 1 goes as follows. The joint use of quasisupermodularity in actions (AU1) and strict single crossing in players and actions (AU2) is similar to that in Theorem 4 of Milgrom and Shannon (1994), and it allows to arrange multiple best replies of different players in a linear order. The crucial economic assumption is the strict single crossing property in players and actions, which implies that the sets of weakly best replies of any two distinct players intersect at most at an extreme point and hence are—roughly speaking—rather separated one from the other. The technical assumptions on countability (AA3) and separation (AA4) complete the job, allowing at most a countable number of such sets (see Sect. 4.3 for a discussion on the importance of the countability and separation properties). Therefore, there can exist only a countable number of comparable players that are weakly best responders; for any such player, there can be many (even uncountable) players that are all uncomparable and weakly best responders, but for the comparability assumption (AT2) their measure is null. This leads to conclude that the set of weakly best responders is formed by countably many sets having measure zero, and hence, its measure is zero as well.

Preliminarily, we define $R_{i,t}(f)$ as the set of best replies to f for $t \in T_i$, namely $R_{i,t} (f) = \{a \in A_i :\; u_i(t,a,f_{-t}) \ge u(t,a',f_{-t}) \text { for all } a' \in A_i\}$.

Lemma 1

Let $\varGamma$ be a game that satisfies AT, AA, and AU. Then, for every $i \in I$, $\{t \in T_i:\; ||R_{i,t}(f)|| > 1\}$ is a countable union of sets having measure zero.

Proof

This is the outline of the proof. For a generic $i \in I$, first we define a function $g_i$ that maps every $t \in \{t \in T_i:\; ||R_{i,t}(f)|| > 1\}$ into a pair $(a,a')$ of her best replies, then we define a function $h_i$, and we use it to assign $(a,a')$ to a base set. We show that function $h_i$ is injective and that function $g_i$ is such that any set of players assigned to the same pair of actions has measure zero. Finally, we invoke the fact that there exists only a countable number of base sets to obtain the desired result.

For each $i \in I$, we consider the partial orders assumed in AA1 (lattice structure) and AT1 (partial ordering) and we take a function $g_i: \{t \in T_i:\; ||R_{i,t}(f)|| > 1\} \rightarrow A_i^2$ such that $g_i(t) = (g_{i,0}(t), g_{i,1}(t))$ with $g_{i,0}(t), g_{i,1}(t) \in R_{i,t}(f)$, $g_{i,0}(t) <^A_i g_{i,1}(t)$, and $g_{i,1}(t) \le ^A_i g_{i,0}(t')$ for $t' >^T_i t$. The following two arguments show that such a function exists for each $i \in I$. First, $a \in R_{i,t}(f)$ and $a' \in R_{i,t}(f)$ imply $a \vee a' \in R_{i,t}(f)$, so that we can set $g_{i,0}(t) = a$ and $g_{i,1}(t) = a \vee a'$, with $a \vee a'$ existing thanks to AA1 (lattice structure). In fact, $u_i(t,a,f_{-t}) \ge u_i(t,a \wedge a',f_{-t})$ since $a \in R_{i,t}(f)$, and hence, $u_i(t,a \vee a',f_{-t}) \ge u_i(t,a',f_{-t})$ by AU1 (quasisupermodularity in actions), which in turn implies that $u_i(t,a \vee a',f_{-t}) = u_i(t,a,f_{-t}) = u_i(t,a',f_{-t})$ since $a \in R_{i,t}(f)$ and $a' \in R_{i,t}(f)$. Second, $a \in R_{i,t}(f)$ and $a' \in R_{i,t'}(f)$ for $t' >^T_i t$ imply $a \le ^A_i a'$. This is true since $u_i(t,a,f_{-t}) \ge u_i(t,a \wedge a',f_{-t})$ due to $a \in R_{i,t}(f)$, and hence, $u_i(t,a \vee a',f_{-t}) \ge u_i(t,a',f_{-t})$ by AU1 (quasisupermodularity in actions), and therefore, $u_i(t',a \vee a',f_{-t}) > u_i(t',a',f_{-t})$ by AU2 (strict single crossing in players and actions), with $a \wedge a'$ existing thanks to AA1 (lattice structure).

For all $i \in I$, by AA2 (topology structure), $A_i$ has a topology and by AA3 (second countability) we can take a countable base $\mathcal {B}_i$ for such a topology. For each $i \in I$, we take a function $h_i: g_i(\{t \in T_i:\; ||R_{i,t}(f)|| > 1\}) \rightarrow \mathcal {B}_i$ such that $a_1 \in h_i(a_0,a_1)$ and $a \notin h_i(a_0,a_1)$ for all $a \le ^A_i a_0$. To see that such a function $h_i$ exists, note that by AA4 (order separation) for each $(a_0,a_1) \in g_i(\{t \in T_i:\; ||R_{i,t}(f)|| > 1\})$ there exists some open set $S_{a_1} \subset A_i$ such that $a_1 \in S$ and $a \notin S$ for all $a \le ^A_i a_0$; since $\mathcal {B}_i$ is a base, there must exist some $B_{a_1} \in \mathcal {B}_i$ such that $a_1 \in B_{a_1}$ and $B_{a_1} \subseteq S_{a_1}$. We set $h_i(a_0,a_1) = B_{a_1}$.

We check that, for all $i \in I$, $g_i$ is such that, for all $(a,a') \in A^2_i$, $g_i^{-1}(a,a')$ has measure zero. For all $t,t' \in \{t \in T_i:\; ||R_{i,t}(f)|| > 1\}$, $t <^T_i t'$, we have that $g_{i,0}(t) < g_{i,1}(t) \le g_{i,0}(t') < g_{i,1}(t')$ from the definition of function $g_i$. Therefore, $t, t' \in g_i^{-1}(a,a')$ implies $t \not \le ^T_i t'$ and $t' \not \le ^T_i t$, and AT2 (negligibility of sets of uncomparable players) guarantees that $\tau _i(g_i^{-1}(a,a'))=0$.

We check that, for all $i \in I$, $h_i$ is injective. For all $(a_0,a_1), (a'_0,a'_1) \in g_i(\{t \in T_i:\; ||R_{i,t}(f)|| > 1\})$, $(a_0,a_1) \ne (a'_0,a'_1)$, we know that either $a_0 < a_1 \le a'_0 < a'_1$ or $a'_0 < a'_1 \le a_0 < a_1$. Suppose, without loss of generality, that $a_0 < a_1 \le a'_0 < a'_1$. Then, by the definition of function $h_i$, we know that $a_1 \in h_i(a_0,a_1)$, $a'_1 \in h_i(a'_0,a'_1)$, and $a_1 \notin h_i(a'_0,a'_1)$ since $a_1 \le a'_0$. Hence, $h_i(a_0,a_1) \ne h_i(a'_0,a'_1)$.

Therefore, $g \circ h$ maps $\{t \in T_i:\; ||R_{i,t}(f)|| > 1\}$ into $\mathcal {B}_i$ in such a way that for every $B \in \mathcal {B}_i$ such that there exists $t \in T_i$ with $h(g(t)) = B$, we have that $\tau _i(\{t \in T_i: h(g(t))=B\})=0$. Since $\mathcal {B}_i$ is countable, we can conclude that $\{t \in T_i:\; ||R_{i,t}(f)|| > 1\}$ is the countable union of sets having measure zero. $\square$

Appendix 2: Proof of Proposition 2

We start by checking that Theorem 1 can be applied to $\varGamma ^{IS}$. Clearly, $\varGamma ^{IS}$ is a special case of $\varGamma ^{I}$. First, we note that $\varGamma ^{I}$ is a specific instance of $\varGamma$. To see this, set i’s type space $T_i = [0,1]^h$, with associated probability space $(T_i,\mathcal {T}_i,\tau _i)$ where $\mathcal {T}_i$ is the sigma algebra of all Lebesgue measurable subsets of $T_i$ and measure $\tau _i$ is the one induced by $\phi _i$, implying that $\tau _i$ is atomless since $\phi _i$ is bounded. Furthermore, set i’s action space $A_i = A$. Finally, note that utility $u^I_i$ is a special case of $u_i$ where the utility of type t does not depend on the actions chosen by other types of the same player role.

We next check that all hypotheses of Theorem 1 are satisfied.

AU is satisfied by assumption.

We check AT. Since $[0,1]^h$ is a partial order, AT1 is satisfied. Take a set $\widehat{T} \subseteq [0,1]^h$ which is made of types that are all uncomparable. For any $(t_1, t_2, \ldots , t_{h-1}) \in [0,1]^{h-1}$, there exists at most one $t_h \in [0,1]$ such that $(t_1, t_2, \ldots , t_{h-1}, t_h) \in \widehat{T}$; otherwise, we would have two elements belonging to $\widehat{T}$ that are comparable. This shows that $\widehat{T}$ is contained in the graph of a function from $[0,1]^{h-1}$ to [0, 1], which constitutes an hypersurface in $[0,1]^h$. We know that an hypersurface has Lebesgue measure equal to zero and hence $\widehat{T}$ as well. Therefore, the measure of $\widehat{T}$ according to the marginal density function $\phi _i$ is null, since the integration of $\phi _i$ over a zero-measure set is zero. So, AT2 is satisfied.

We check AA. If A is a finite lattice, then AA1–AA4 hold trivially. If $A = [0,1]^k$, then AA1 and AA2 are satisfied by considering, respectively, the standard order and the Euclidean topology on $[0,1]^k$. It is well known that the Euclidean space (and any of its subsets) is second countable (it is enough to consider as base the set of all open balls with rational radii and whose centers have rational coordinates). So AA3 is also satisfied. Finally, consider $a, a' \in [0,1]^k$ such that $a'_i \ge a_i$, $a' \ne a$. Then take an open ball centered at $a'$ with radius lower than the Euclidean distance between $a'$ and a; clearly, $a'$ belongs to the ball, while every $a'' \in [0.1]^k$ such that $a''_i \le a_i$ does not belong to the ball. This shows that AA4 is satisfied.

So, we can apply Theorem 1 to conclude that every pure-strategy Nash equilibrium must be essentially strict and monotone in types and actions.

Consider now a symmetric pure-strategy Nash equilibrium where every player chooses strategy $\alpha$. Consider also any strategy $\alpha '$, with $\alpha ' \ne \alpha$. We have already shown, by exploiting Theorem 1, that $\alpha$ is essentially strict, and so $u^I(t,\alpha (t),\varvec{\alpha }_{-i}(\mathbf {t}_{-i})) > u^I(t,\alpha ' (t),\varvec{\alpha }_{-i}(\mathbf {t}_{-i}))$ for almost all $t \in [0,1]^{h}$. Therefore,

$$\begin{aligned} \int _{[0,1]^{h}} \left( u(t, \alpha (t),\varvec{\alpha }'_{-i}(\mathbf {t}_{-i})) \right) \phi _i(t) \mathrm {d} t > \int _{[0,1]^{h}} \left( u(t,\alpha ' (t),\varvec{\alpha }_{-i}(\mathbf {t}_{-i})) \right) \phi _i(t) \mathrm {d} t, \end{aligned}$$

(1)

which means that $V(\alpha , \alpha ) > V(\alpha ', \alpha )$. Hence, for $\epsilon$ small enough, we can conclude that $(1 - \epsilon ) V(\alpha , \alpha ) + \epsilon V(\alpha , \alpha ') > (1 - \epsilon ) V(\alpha ', \alpha ) + \epsilon V(\alpha ', \alpha ')$. We have so established that $\alpha$ is an ESS.

Finally, to show that an ESS exists, we can rely on Theorem 1 in McAdams (2003) that can be applied since AU2 implies the single crossing condition—which is required by the Theorem. Such theorem, if applied to symmetric games, establishes the existence of a symmetric pure-strategy Nash equilibrium.²⁴ By the previous argument, we conclude that the strategy played in such equilibrium must be an ESS.