Monotone bargaining is Nash-solvable

Abstract

Given two finite ordered sets $A = \{a_1, \ldots, a_m\}$ and $B = \{b_1, \ldots, b_n\}$, introduce the set of $m n$ outcomes of the game $O = \{(a, b) \mid a \in A, b \in B\} = \{(a_i, b_j) \mid i \in I = \{1, \ldots, m\}, j \in J = \{1, \ldots, n\}$. Two players, Alice and Bob, have the sets of strategies $X$ and $Y$ that consist of all monotone non-decreasing mappings $x: A \rightarrow B$ and $y: B \rightarrow A$, respectively. It is easily seen that each pair $(x,y) \in X \times Y$ produces at least one {\em deal}, that is, an outcome $(a,b) \in O$ such that $x(a) = b$ and $y(b) = a$. Denote by $G(x,y) \subseteq O$ the set of all such deals related to $(x,y)$. The obtained mapping $G = G_{m,n}: X \times Y \rightarrow 2^O$ is a game correspondence. Choose an arbitrary deal $g(x,y) \in G(x,y)$ to obtained a mapping $g : X \times Y \rightarrow O$, which is a game form. We will show that each such game form is tight and, hence, Nash-solvable, that is, for any pair $u = (u_A, u_B)$ of utility functions $u_A : O \rightarrow \mathbb R$ of Alice and $u_B: O \rightarrow \mathbb R$ of Bob, the obtained monotone bargaining game $(g, u)$ has at least one Nash equilibrium in pure strategies. Moreover, the same equilibrium can be chosen for all selections $g(x,y) \in G(x,y)$. We also obtain an efficient algorithm that determines such an equilibrium in time linear in $m n$, although the numbers of strategies $|X| = \binom{m+n-1}{m}$ and $|Y| = \binom{m+n-1}{n}$ are exponential in $m n$. Our results show that, somewhat surprising, the players have no need to hide or randomize their bargaining strategies, even in the zero-sum case.

In this version we extend significantly Section 4. We add more classes of dual hypergraphs and show that for some of these classes the proof of the main theorem becomes much simpler than in general