This post is a sequel to my previous ad of a joint paper with Yaron, in which we prove existence of pure ${\epsilon}$-equilibria in certain games. I am now going to make a fuss over the fact that our result is essentially a collection of logical relations between linear inequalities, yet its proof seems to require Brouwer’s fixed point theorem.

I start with emphasizing game theory’s reliance on Brouwer’s Theorem to prove Nash’s Theorem, an outrage with which I have more or less already learned to live. I call this an outrage because Nash’s Theorem is essentially a theorem about logical relationships between algebraic inequalities. More formally, fix integers ${n,m}$ and let ${\xi_{n,m}}$ be the assertion that every ${n}$-player game with ${m}$ strategies to every player admits a mixed Nash Equilibrium. Then ${\xi_{n,m}}$ is a first order sentence in the language of ordered fields. The signature of this language is ${(0,1,<,+,\times)}$ where ${0,1}$ are constant symbols, ${\leq}$ is a relation symbol and ${+,\times}$ are binary function symbols. (Skip the next paragraph if this is obvious to you).

Indeed, let ${\bar x}$ be a set of ${(m^n)\cdot n}$ variables, representing the payoff matrix of an ${n}$-player game with ${m}$ strategies for every player, and let ${\bar p}$ be a set of ${n\cdot m}$ variables representing a mixed strategies profile. Then $\displaystyle \xi_{n,m}=\forall \bar x~\exists \bar p ~\phi(\bar x,\bar p)$

where ${\phi(\bar x,\bar p)}$ is a formula that says that ${\bar p}$ is a mixed Nash equilibrium in a game with payoff matrix ${\bar x}$. This is a conjunction of formulas that assert that ${\bar p}$ is indeed a mixed strategy profile (nonnegative elements which sum to ${1}$), and that if player ${i}$ plays action ${j}$ with a positive probability under this profile then player ${i}$ doesn’t gain from deviating to any pure strategy. The last assertion involved a somewhat unappealing term (the payoff for player ${i}$ under profile ${p}$ when the payoff matrix is ${\bar x}$), but this term is just products and additions of variables.

Now since by Tarski’s Theorem all real closed fields satisfy the same first order sentences, it follows that Nash’s Theorem is true in every real closed field ! Here is an interesting corollary of this conclusion: Every game in which the payoffs are algebraic numbers has an equilibrium in which the probabilities are algebraic numbers. Here is a more interesting corollary: In discounted stochastic games, an equilibrium strategy can be expressed as a fractional laurent series of the discount factor. This appears in a seminal paper (jstor) of Bewley and Kohlberg, who are to my knowledge the first to make this observation. I presented this paper in a students seminar back in grad school, probably the first paper in game theory I have ever read, and it is still one of my favorites.

Anyway, back to pure ${\epsilon}$-equilibrium. Yaron and I defined the Lipschitz Constant of a game as the maximal change in a player’s payoff when one of her opponents changes his strategy. Theorem 2.2 in our paper says that if ${\delta < \epsilon/\sqrt{8n\log (2mn)}}$ then every game with Lipschitz constant smaller than ${\delta}$ admits pure ${\epsilon}$-equilibrium.

Fix now integers ${n,m}$ and rational numbers ${\epsilon,\delta}$. Let ${\zeta_{n,m,\epsilon,\delta}}$ be the assertion that every ${n}$-player game with ${m}$ strategies for every player and Lipschitz constant ${<\delta}$ admits an ${\epsilon}$-equilibrium. Then ${\zeta_{n,m,\epsilon,\delta}}$ is a first order sentence in the language of ordered fields without multiplication. (Again, skip the next paragraph if this is obvious)

Indeed, let ${\bar x}$ be a set of ${(m^n)\cdot n}$ variables indexed ${x_a^i}$ for every strategy profile ${a}$ and player ${i}$, representing the payoff matrix of an ${n}$-player game with ${m}$ strategies for every player. Then $\displaystyle \zeta_{n,m,\epsilon,\delta}=\forall \bar x\bigl(\wedge_\psi\psi(\bar x)\rightarrow\vee_{a}\wedge_\phi\phi(\bar x)\bigr)$

The antecedent is a conjunction of linear inequalities ${\psi(\bar x)}$, each asserts that the payoff of a certain player under a certain profile does not change by more than ${\delta}$ if a certain opponent changes his strategy. Thus, each such ${\psi}$ is of the form ${|x_{a'}^i-x_{a''}^i|\leq\delta}$, where ${a',a''}$ are two strategy profiles that differs in one of player ${i}$-s opponents. The consequent is a disjunction over all strategy profiles ${a}$, and for every such ${a}$ a conjunction of linear inequalities ${\phi(\bar x)}$, that assert the fact that under ${a}$ a certain player cannot gain more than ${\epsilon}$ by a certain deviation. Thus, each of these ${\phi}$ is of the form ${x_{a'}^i-x_{a}^i\leq\epsilon}$, where ${a'}$ is a strategy profile that differs from ${a}$ only in the ${i}$-th coordinate.

Now if ${\delta < \epsilon/\sqrt{8n\log (2mn)}}$ then ${\zeta_{n,m,\epsilon,\delta}}$ is valid in ${\mathbb{R}}$. Our proof for this theorem relies on Nash’s Theorem and so, again, on uncle Brouwer’s theorem, which seems even more outrageous than the mixed strategies case, since this time there is not even multiplication involved. It’s just a theorem about relationships between finite number of linear inequalities.

At this point I hope you are telling to yourself something like `Brouwer’s Theorem to show that a bunch of linear inequalities imply a bunch of other linear inequalities !? These guys are killing a mite with a cruise missile.’ But in fact we have an elementary argument that our theorem, which we prove by Nash’s Theorem, also implies Nash’s Theorem. The idea is very simple, and is based on replacing each player in a game by a large number of agents, such that a pure equilibrium in the game played by the agents translates to a mixed equilibrium in the original game. The details are in Section 8 of our paper(pdf).

Open Problem: We know that Nash’s Theorem is still valid when payoffs are in some real algebraic field. But in order to speak about pure ${\epsilon}$-equilibrium, one only needs payoffs to be in an ordered commutative group. So let ${n,m,\epsilon,\delta}$ be such that ${\delta<\epsilon/\sqrt{8n\log (2mn)}}$ (or some other similar inequality). Is it true that every game with payoffs in some ordered commutative group and Lipschitz Constant smaller than ${\delta}$ admits pure ${\epsilon}$ equilibrium ?