The exisTence of nash equilibrium for all finiTe GamesXintong Tian

Swarthmore College, Department of Mathematics & Statistics

IntroductionSuppose two players are playing a game where they each choose either 0 or 1. The payoff table for this game is shown below.

The notation (x, y) indicates that x represents the profits of Player 1 and y represents the profits of Player 2. In this game, choosing 1 will result in profits of either $0 or $5, while choosing 0 will result in profits of either $1 or $6. Hence both players have strong incentive to choose 0, though they would benefit from playing cooperatively and choosing 1 together. The action of both players choosing 0 is known as a Nash equilibrium.

What is a game?A finite game played with n people is defined by the tuple (X, u) where• X = (X1 , X2 ,..., Xn), where Xi is the finite set of all possible actions (or pure strategies) available to player i. The vector (x1 , x2 ,..., xn), an element of X, is called an action profile; player 1 chooses action x1, player 2 chooses action x2, and so on.

• u = (u1 , u2 ,..., un), where ui is the payoff function for player i. Different action profiles will result in different payoffs for each player depending on the payoff function.

A mixed strategy is a probability distribution of pure strategies. The equivalent to an action profile for mixed strategies is a strategy profile s. A game can be more generally defined with the strategy profiles S.

Definition of a Nash EquilibriumLet G = (S, u) be a game. We say a strategy profile s* in X is a Nash equilibrium if si* is an element of

argmax {u(si , s-i )} over si for all i = 1, ..., n, where s-i is the strategy profiles of all players except player i. In other words, a Nash equilibrium occurs when no player has any incentive to change his or her strategy, assuming that the strategies of all other players remain fixed.

The Existence of Nash equilibriumEvery game with a finite number of players and action profiles has at least one Nash equilibrium. To prove this, we must first learn about fixed points. A fixed point of a function is an element of the function’s domain that is mapped to itself by the function.The proof for the existence of at least one Nash equilibrium involves three steps:1. Construct a function that maps the set of all mixed strategies for all players to itself.

2. Prove that this function has a fixed point.3. Show that the fixed point is a Nash equilibrium. It turns out that steps 1 and 3 are relatively simple, but step 2 requires proving another theorem.

Application of Nash EquilibriumLet a > 0, and consider the game G = ([0, a], u) with two players where

u1(p1 , p2) = p1x2(p1 , p2) - cx1(p1 , p2),with x1 defined as x1(p1 , p2) = a - p1 if p1 < p2, x1(p1 , p2) = 0 if p1 > p2 , and x(p1 , p2) = (1/2)a - p1 if p1 = p2. The payoff function u2 is defined similarly. This game models two firms engaging in price competition. Here a stands for the maximum possible price level in the market (where demand is 0), c is the constant marginal cost, and xi(p1 , p2) denotes the output sold by firm i at the price profile (p1 , p2). The payoff functions reflect the hypotheses that consumers always buy the cheaper good, and the firms share the market equally in case of a tie. Since this is a finite game, we know this game has a Nash equilibrium (p1* , p2*). We can easily show that p1* , p2* ≥ c and p1* , p2* ≤ c, so the Nash equilibrium is (c , c).

References• Efe Ok. Real analysis with economic applications. Princeton University Press, Princeton, New Jersey, 2007.

• John Nash (1950). Equilibrium points in n-person games. Proceedings of the National Academy of Sciences USA, 36, 48-49. Reprinted in Harold Kuhn, Classics in Game Theory. Princeton University Press, Princeton, New Jersey, 1997.

• John Nash (1951). Non-cooperative games. Annals of Mathematics, 54, 286-295.

• Kim Border. Fixed point theorems with applications to economics and game theory. Cambridge University Press, New York, New York, 1985.

Brouwer Fixed Point TheoremLet S be a nonempty, compact, and convex set. Let f be a continuous function that maps S to itself. Then there exists at least one fixed point of f in S. Once we establish this theorem, step 2 only requires proving that the function created in step 1 is continuous and the domain of it is compact and convex.

Contact InformationEmail: xtian1@swarthmore.edu | xintong.tian@gmail.comPlease feel free to contact me if you have any questions.