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There are two equivalent ways to understand the best response property of a Nash Equilibrium strategy. First, we can say that the player plays a mixed strategy whose expected payoff is maximal among all possible mixed strategies. Second, we can say that the player randomly chooses a pure strategy from the set of pure strategies whose expected payoff is maximal among all possible pure strategies.

So far so good, and every student of game theory is aware of this equivalence. What I think is less known is that the two perspectives are not identical for ${\epsilon}$-best response and ${\epsilon}$-equilibrium: A mixed strategy whose expected payoff is almost optimal might put some positive (though small) probability on a pure strategy which gives a horrible payoff. In this post I am going to explain why I used to think the difference between the two perspectives is inconsequential, and why, following a conversation with Ayala Mashiah-Yaakovi about her work on subgame perfect equilibrium in Borel games, I changed my mind.

This is the most frustrating part in academic career: You come up with a cool idea, google around a bit for references, and discover that the Simpsons did it twenty years ago. It happened to Ronen and I recently when we were talking about computability of Nash equilibrium. Only thing left is to blog about it, so here we are.

A good starting point is the omitted paragraph from John Nash’ Thesis (scanned pdf), in which Nash motivates his new idea. The paragraph is not included in the published version of the thesis, it is not clear whether because of editorial intervention or Nash’ own initiative.

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. Read the rest of this entry »

Among game theoretic concepts, mixed strategy is arguably the most difficult to digest. We don’t see people tossing coins all the time, and it’s difficult to justify rational decision as based on Lady Fortuna’s unpredictable caprices. The case of Nash Equilibrium is especially disturbing — if you are indifferent between a couple of strategies then why bother randomizing between them according to the mixture prescribed by the equilibrium. Just pick one of these strategies arbitrary and get it over with.

I know of two types of answers that game theory gives for this conundrum. One, which may be called `interpretation of mixed strategies’ is arguing that the mixed strategies in Nash equilibrium do not reflect an actual randomization performed by the players: Epistemic game theory interprets mixed strategies as opponent’s beliefs about a player’s (non-randomized) strategy; Harsanyi’s Purification approach embeds a normal form game in a larger game with incomplete information and pure equilibrium. The other type of answers is identifying classes of games which admit pure equilibrium, such as games with strategic complementarity and potential games.

In my new paper with Yaron(pdf) we suggest another class of games which admit pure ${\epsilon}$-equilibrium, which means that no player gains more than ${\epsilon}$ from deviating. These are games in which a player’s payoff does not change much if one of her opponents changes his strategy:

Math and open problems below the fold…

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