of the row player, and B = (bij) is the payoff matrix of the column player. ... R (the row player), and y a mixed strategy of player C (the column player). ...... [1] K. C. Border, Fixed Point Theorems with Applications to Economics and Game Theory.
Computational Aspects of Game Theory
Bertinoro Spring School 2011
Lecture 4: Nash Equilibria Lecturer: Bruno Codenotti
The concept of Nash Equilibrium embodies the fundamental idea of economics that people act according to their incentives, and thus Nash equilibria turn out to be relevant in every area where incentives are important, i.e., all of Economic Theory.
4.1
Introduction
In this lecture, we first analyze two-player zero sum games, and then move on to consider two-player nonconstant sum games. For zero-sum games, we first look at the cases where equilibria in the pure strategies do exist. These cases are characterized by the presence of a saddle point in the payoff matrix. Since saddle points do not exist in general, we then consider the case where players can use mixed strategies. In this scenario, we show the equivalence of zero-sum games and linear programming. More precisely, we prove that: • Given a zero-sum game, the optimal strategies of the players can be represented by two linear programs, which are the dual of one another; we can then use duality theory to show the existence of equilibria. • Given a pair of primal-dual linear programs, we construct a zero sum game, and show that the optimal solutions to the linear programs can be easily extracted from the mixed strategies which give a NE for the game. In the second part of the lecture, we first show that NE for two-player non-constant sum games can be described as the solution to a computational problem known as linear complementarity, and then prove the Nash Theorem, which shows the existence of NE. Before starting with our analysis, we recall some basic definitions. We consider two-player games in normal form. These games are described by a pair of matrices (A, B), whose entries are the payoffs of the two players, called row and column player. A = (aij ) is the payoff matrix of the row player, and B = (bij ) is the payoff matrix of the column player. The rows (resp. columns) of A and B are indexed by the row (resp. column) player’s pure strategies. The entry aij is the payoff to the row player, when she plays her i-th pure strategy and the opponent plays his j-th pure strategy. Similarly, bij is the payoff to the column player, when he plays his j-th pure strategy and the opponent plays her i-th pure strategy. A mixed strategy is a probability distribution over the set of pure strategies which indicates how likely it is that each pure strategy is played. More precisely,∑ in a mixed strategy a player associates to her i-th pure strategy a quantity pi between 0 and 1, such that i pi = 1, where the sum ranges over all pure strategies. Let us consider the game (A, B), where A and B are m × n matrices. In such a game the row player has m pure strategies, while the column player has n pure strategies. Let x be a mixed strategy of player R (the row player), and y a mixed strategy of player C (the column player). Strategy x is the m-tuple 4-1
4-2
Lecture 4: Nash Equilibria
x = (x1 , x2 , . . . , xm ), where xi ≥ 0, and ∑ n i=1 yi = 1.
∑m i=1
xi = 1. Similarly, y = (y1 , y2 , . . . , yn ), where yi ≥ 0, and
Let A = (aij ) be the payoff matrix of player R. Consider the table below: R/C x1 x2 x3 ... xm
y1 a11 a21 a31 ... am1
y2 a12 a22 a32 ... am2
y3 a13 a23 a33 ... am3
... ... ... ... ... ...
yn a1n a2n a3n ... amn
When the pair of mixed strategies x and y is played, the entry aij contributes to the expected payoff of the row player with weight xi yj . The expected payoff of the row player∑ can be evaluated by adding up all the entries of A weighted by the corresponding entries of x and y, i.e., ij xi yj aij . This can be rewritten ∑ ∑ as i xi j aij yj , which can be expressed in matrix terms as1 xT Ay. Similarly, the expected payoff of the column player is xT By. ∑k Definition 4.1 (Stochastic vector). A k-vector z = (zi ) with real nonnegative entries such that i=1 zi = 1 is called a stochastic k-vector. Definition 4.2 (Nash Equilibrium). A pair of mixed strategies (x, y) is in Nash equilibrium if xT Ay ≥ T x′ Ay, for all stochastic m-vectors x′ , and xT By ≥ xT By ′ , for all stochastic n-vectors y ′ . We say that x (resp. y) is a Nash equilibrium strategy for the row (resp. column) player. The set of indices such that xi > 0 (resp. yi > 0) is called the support of the Nash equilibrium strategy x (resp. y). The definition of Nash equilibrium can be immediately extended to multiplayer non-cooperative games.
4.2
Zero-sum games: Equilibria in the Pure Strategies
In non-cooperative two person zero-sum games, if A denotes the payoff matrix of R, then the payoff matrix of C is −A. For this reason, these games can be described by a single matrix, and are thus called matrix games, as opposed to general two-person bimatrix games. We will just focus on the row player’s payoff matrix A, and think of both players acting on this matrix. Player C (the minimizer) will select a column with the goal of minimizing the maximum payoff of the row player (because of the zero-sum property, this coincides with maximizing his own payoff). Conversely, player R (the maximizer) selects the row which maximizes her minimum payoff. We represent each pure strategy i by a vector whose entries are all zero, except for the i-th entry, which is set to one. If x∗ and y ∗ are vectors representing the pure strategies of R and C, respectively, then we can assume that they are chosen so that x∗ Ay ∗ is either maxx miny xAy or miny maxx xAy. Note that maxx miny xAy is the payoff that R can guarantee for herself independently of what C does, while miny maxx xAy is the “ payment” that C is guaranteed not to exceed, independently of what R does. 1 We use the notation xT to denote the transpose of vector x. When the context does not lead to ambiguities, we omit the superscript, and use the same notation for row and column vectors.
Lecture 4: Nash Equilibria
4-3
In the Fair Division Example of Lecture 2, R cuts the cake into two equal pieces, and thus maximizes the minimum share he receives. If C had declared his strategy first, the outcome would have been the same, since he would have chosen the bigger piece, thus minimizing the maximum share received by R. The trouble is that, in the pure strategies, maximin (maxx miny xAy) and minimax (miny maxx xAy) need not be the same, as the following example shows. [ Example 4.3 (minimax vs maximin). Let A =
1 −1 −1 1
] .
The minimum of the row maxima of A is 1, whereas the maximum of the column minima is −1, so that maxx miny xAy is different from miny maxx xAy. Definition 4.4 (Saddle Point). Given a matrix A = (aij ), the entry akl is called a saddle point for A if akl = maxi ail and akl = minj akj . It is easy to see that a saddle point is a NE in the pure strategies. Theorem 4.5. Let A describe a matrix game. We have maxx miny xAy ≤ miny maxx xAy. Proof. Let i∗ ∈ arg maxi minj aij , and j ∗ ∈ arg minj maxi aij . There is a column ¯j such that ai∗ ¯j = minj ai∗ j , i.e., ai∗ ¯j is the maximin value. We have that ai∗ ¯j ≤ ai∗ j ∗ . Similarly there is a row ¯i such that a¯ij ∗ = maxi aij ∗ , i.e., a¯ij ∗ is the minimax value. We have that a¯ij ∗ ≥ ai∗ j ∗ . Thus a¯ij ∗ ≥ ai∗ j ∗ ≥ ai∗ ¯j . Corollary 4.6. A has a saddle point if and only if maxx miny xAy = miny maxx xAy. Therefore a zero-sum game described by a matrix A has a pure strategy NE if and only if A has a saddle point. The gap of Example 4.3 arises because A does not have saddle points. On the contrary in the Fair Division example the order of play could not influence the outcome, since the minimum of the row maxima was equal to the maximum of the column minima. It is easy to see that if (i, j) and (k, l) are saddle points, then (i, l) and (k, j) are saddle points as well.
4.3
Zero-sum games: Equilibria in the Mixed Strategies
First of all, it is easy to see that the equivalence between NE and minimax equilibria continue to hold in the mixed strategies. The celebrated von Neumann Minimax Theorem states that, if players are allowed to use mixed strategies, then maxx miny xAy = miny maxx xAy, so that the situation of Example 4.3 can be overcome in the mixed strategies. Before proving von Neumann Minimax Theorem, we recall some basic results from Linear Programming Duality. Definition 4.7 (The Primal Linear Program). The general form of the primal Linear Program (P) is maxx Subject to
cT x Ax ≤ b, x ≥ 0.
4-4
Lecture 4: Nash Equilibria
The primal can be also expressed in terms of equality constraints, and corresponding slack variables as follows: maxx Subject to
cT x Ax + z = b, x, z ≥ 0.
Definition 4.8 (The Dual Linear Program). The general form of the dual Linear Program (D) is yT b
miny
AT y ≥ c, y ≥ 0.
Subject to
The dual can be also expressed in terms of equality constraints, and corresponding slack variables as follows: miny Subject to
yT b AT y + w = c, y, w ≥ 0.
Definition 4.9 (Feasible Point). We say that x is feasible for P if Ax ≤ b and x ≥ 0, and that y is feasible for D if AT y ≥ c, and y ≥ 0. Theorem 4.10 (Weak Duality). If x is feasible for P and y is feasible for D, then 1. cT x ≤ y T b. 2. If cT x = y T b, x is an optimal solution to P, and y is an optimal solution to D. Definition 4.11 (Complementary Slackness). yi (b − Ax)i = 0, for all i, and (AT y − c)j xj = 0, for all j. Theorem 4.12. If x is feasible for P and y is feasible for D, then 1. x and y satisfy complementary slackness if and only if cT x = y T b. 2. if x and y satisfy complementary slackness, then x is an optimal solution to P, and y is an optimal solution to D. Theorem 4.13 (Strong Duality). If P and D are both feasible, then there exist feasible x and y satisfying complementary slackness, and so (equivalently) cT x = y T b. We are now ready to prove the von Neumann Minimax Theorem, and show the equivalence between optimality in linear programming and minimax equilibria. Theorem 4.14 (von Neumann Minimax Principle). Let A be an n × m matrix. There exist a stochastic n-vector x∗ and a stochastic m-vector y ∗ such that max min xT Ay = min max xT Ay = x∗ T Ay ∗ , x
y
y
x
where the minimum is taken over all stochastic m-vectors y, and the maximum over all stochastic n-vectors x. The quantity x∗ T Ay ∗ is called the value of the zero-sum game described by A. Proof. First of all, note that if R picks the mixed strategy x, then her average payoff will be at least miny xT Ay. Similarly, if C picks the mixed strategy y, then he will have to pay at most maxx xT Ay.
Lecture 4: Nash Equilibria
4-5
It is easy to see that: min xT Ay y
max xT Ay x
= min
n ∑
j
= max i
aij xi
i=1 m ∑
aij yj .
j=1
Therefore the problem faced by R can be written as max min x
j
n ∑
aij xi ,
i=1
which is equivalent to the linear program
maximize
z
Subject to z −
n ∑
aij xi ≤ 0, for 1 ≤ i ≤ n,
(4.1)
i=1
and
n ∑
xi = 1 , xi ≥ 0.
i=1
Similarly, the problem of C can be written as
minimize
w
Subject to w −
m ∑
aij yj ≥ 0, for 1 ≤ j ≤ m,
(4.2)
j=1
and
m ∑
yj = 1 , yj ≥ 0.
j=1
Problem 4.1 and Problem 4.2 are feasible linear programs that are the dual of one another. Therefore z ∗ = w∗ . z ∗ = w∗ is the value of the game. The proof of von Neumann Theorem gives one direction of the equivalence between zero-sum two person games and Linear Programming; more precisely it shows that the Strong Duality Theorem implies the minimax Theorem. We now sketch the path that leads to the equivalence. We need to show that the minimax Theorem implies the Strong Duality Theorem. A special family of zero-sum games which is of considerable interest consists of the so called symmetric games, in which the payoff matrix is skew-symmetric, i.e., A = −AT , and AT denotes the transpose of the matrix A. These games are in a sense the most fair. Indeed the (i, j)-th entry of A coincides with the (j, i)-th entry of −A, so that the payoff to R when the combination of strategies (i, j) is played is equal to the payoff of C when the strategies (j, i) are played. Lemma 4.15. The value of symmetric games is zero.
4-6
Lecture 4: Nash Equilibria
Symmetric games capture duality in a natural way, as shown by the proof of the Skew Symmetric Theorem below. Theorem 4.16 (The Skew Symmetric Theorem). To any pair P and D of dual linear programs there corresponds a skew-symmetric matrix game such that the optimal solution to the linear programs can be extracted from any optimal mixed strategy of the game. Proof. Given P and D in the form of Definitions 4.7 and 4.8, we construct the following payoff matrix
y x z y 0 -A b x AT 0 -c z −bT cT 0 By Lemma 4.15 we know that the value of the game has to be zero. Therefore the optimal mixed strategies (y ∗ , x∗ , z ∗ ) must satisfy the inequalities −Ax + bz ≥ 0, AT y + cz ≥ 0, −bT y + cT x ≥ 0, x, y ≥ 0. ∗
If now z ∗ > 0 we can see that (x = xz∗ , y = needs additional work, which we omit here.
4.4
y∗ z∗ )
are the optimal solutions to P and D. If z ∗ = 0, the proof
Non constant-sum games
Consider a bimatrix game (A, B), where A and B are m × n matrices with rational entries. Let (x, y) be a pair of mixed strategies. We say that y is a best response to x if it maximizes the expected payoff xT By. Similarly we say that x is a best response to y if it maximizes the expected payoff xT Ay. The pair (x, y) constitutes a NE for the game (A, B) if xT Ay ≥ x′T Ay, for all stochastic m-vectors x′ , and xT By ≥ xT By ′ , for all stochastic n-vectors y ′ . Thus a NE is a pair of mutual best responses. The following lemma provides a combinatorial characterization of best responses. Lemma 4.17. Let (x, y) be a pair of mixed strategies for a bimatrix game (A, B). x is a best response to y if and only if all the pure strategies in the support of x are best responses to y. Proof. Let us consider the action of y on A. Ay =
a1,1 a2,1 a3,1 ... am,1
a1,2 a2,2 a3,2 ... am,2
. . . a1,n . . . a2,n . . . a3,n ... ... . . . am,n
y1 y2 y3 ... yn
∑ a y ∑j 1,j j j a2,j yj = . ... . . . ∑ j am,j yj
∑ ∑ ∑ ∑ Let M = maxi j ai,j yj . xT Ay = i xi (Ay)i = M − i (xi (M − (Ay)i ). Since i (xi (M − (Ay)i ) ≥ 0, we ∑ have that xT Ay ≤ M , and that xT Ay achieves its maximum M if and only if i (xi (M − (Ay)i ) = 0. But this equality holds if and only if xi > 0 implies that (Ay)i = M .
Lecture 4: Nash Equilibria
4-7
We now derive a useful formulation for the problem of finding a NE. For a pair (x, y) in NE, we must have • xT Ay ≥ x′T Ay, which implies that xT Ay ≥ • xT By ≥ xT By ′ , which implies that xT By ≥
∑ j
ai,j yj , i = 1, 2, . . . , m.
∑
i bi,j xi ,
j = 1, 2, . . . , n.
Note that, without loss of generality, we may assume that both xT Ay and xT By are strictly greater than zero2 . ∑ ∑ Let M = maxi j ai,j yj , and N = maxj i bi,j xi . We have (Ay)i ≤ M , and (B T x)j ≤ N . From the best response property, we also have that xi > 0 implies (Ay)i = M , and that yi > 0 implies (B T )j = N . Now we introduce the vectors u and v, whose components are uj =
yj , xT Ay
and vi =
xi , xT By
respectively.
We can use this substitution to derive the inequalities ∑
bi,j vi
≤ 1 ∀j
ai,j uj
≤ 1 ∀i,
i
∑ j
where uj > 0 implies that
∑
i bi,j vi
= 1, and vi > 0 implies that
∑ j
ai,j uj = 1.
We can now add two vectors of slack variables, r and t, and transform these inequalities into equalities, i.e., {
B T v + t = 1n , t ≥ 0 Au + r = 1m , r ≥ 0,
where 1k denotes the k-vector whose components are all equal to 1. Now we have to add the constraint that characterizes the support of a NE in the mixed strategies in terms of best response. Such property translates into an orthogonality condition, i.e., r · v = t · u = 0, which ∑ expresses the fact that if ri > 0, i.e., if j aij yj < xAy, then xi = 0, otherwise x is not a best response. The same argument shows the other orthogonality requirement. Note that the pair of stochastic vectors which constitute the NE can be recovered by normalizing u and v. This process leads to the Linear Complementarity formulation of the NE conditions (LC-NASH): Let w be the (n + m)-vector obtained concatenating v and u, and z be the (n + m)-vector obtained concatenating t and r. LC-NASH: Find nonnegative (n + m)-vectors w and z such that
Hw + z T
w z
= 1(n+m) =
0,
2 We can add to all the entries of both payoff matrices a large positive constant without changing the nature of the game under investigation.
4-8
Lecture 4: Nash Equilibria
where ( H=
4.5
0 BT
A 0
) .
Nash Equilibria for Multiplayer Games
The definition of NE can be immediately extended to multi-player games. Consider a k-player game, and let uj (s1 , s2 , . . . , sk ) denote the payoff to player j, when the players use strategies s1 , . . . , sk (si is the strategy of player i, i = 1, . . . , k). A strategy profile u = (u1 , u2 , . . . , uk ) can be denoted by (ui , u−i ), i ∈ {1, 2, . . . , k}, where u−i = (u1 , u2 , . . . , ui−1 , ui+1 , . . . , uk ). Definition 4.18 (NE for k-player games). An NE is a strategy profile x∗ such that no player can increase her payoff by switching her strategy unilaterally. Formally, for each player i, and for every mixed strategy yi , ui (x∗ ) ≥ ui (x∗−i , yi ). Consider games where the payoffs are rational numbers. A major difference between two- and multi-player games is that if the number of players is at least three, then all the NE can be irrational numbers. On the contrary, all the NE in two-player games are rational. Example 4.19 (Three-player game with irrational NE). Let a1 , a2 , b1 , b2 , c1 , c2 , be the strategies available to player 1, 2, and 3, respectively. Consider the payoff matrices
a1 a2
b1 b2 (0,0,1) (1,0,0) (1,1,0) (2,0,8)
a1 a2
b1 b2 (2,0,9) (0,1,1) (0,1,1) (1,0,0)
where the matrix on the left gives the payoffs when player 3 chooses c1 , and the matrix on the right gives the payoffs when player 3 chooses c2 . Each entry of the matrices is a triple, whose i-th component is the payoff to player i, i = 1, 2, 3. This game has a unique NE given by (p, 1 − p), (q, 1 − q), (r, 1 − r), where p = √ r = 9−12 51 .
4.6
√ 30−2 51 , 29
q =
√ 2 51−6 , 21
and
Brouwer’s Fixed Point Theorem
Let f be a function mapping a set S into itself. A fixed point of f is a point x ∈ S such that f (x) = x. We now prove a basic theorem on fixed points, i.e., Brouwer’s Fixed Point Theorem. There are several proofs of this theorem. We will prove it via Sperner’s Lemma (see Lecture 3 for the statement and the proof of this lemma). We first need to extend to the simplex the definition of completely labeled triangle of Lecture 3.
Lecture 4: Nash Equilibria
4-9
Definition 4.20 (Completely Labeled Subsimplex). Given a simplicial subdivision of the unit (n − 1)simplex, a subsimplex labeled with all the n labels is called completely labeled subsimplex. Theorem 4.21 (Brouwer). Let S n be the unit (n − 1)-simplex. Let us consider a continuous mapping x → f (x) of S n into itself. There is a point z ∈ S n such that f (z) = z. Proof. Let us consider a continuous mapping x → f (x) of the unit (n − 1)-simplex into itself. This mapping is specified by the n∑one-dimensional projections of f , fi (x), i = 1, 2, · · · , n, which satisfy fi (x) ≥ 0, for n i = 1, 2, · · · , n, and i=1 fi (x) = 1. Let v1 , v2 , · · · , vn be the vertices of the simplex, and vn+1 , vn+2 , · · · , vn+k be additional points so that altogether v1 , v2 , · · · , vn , · · · , vn+k form a simplicial subdivision of the simplex. We assume that none of the above points is a fixed point of the mapping f , otherwise the theorem would trivially follow. Thus for each vertex u we can find an index i such that ui > fi (u). We label vertex ∑ u with∑such an i. Indeed it cannot be that all the ui ’s satisfy ui ≤ fi (u) because then the conditions i ui = i fi = 1 would force a fixed point. Now notice that if u lies on the boundary of the simplex, with, e.g., uj = 0, then the label of u will be different from j. Therefore this labeling satisfies the assumptions of Sperner’s Lemma, and the subdivision contains a completely labeled subsimplex. Consider now a sequence of simplicial subdivisions whose mesh size tends to zero. Each of these subdivisions yields a completely labeled simplex. From the compactness of the unit simplex, it follows that there is a convergent subsequence of completely labeled simplices all of whose vertices tend to a single point x∗ . Such a point is∑the limit of all the labels. Since f is continuous, we must have x∗i ≥ fi (x∗ ), for all ∑npoints bearing n ∗ ∗ i. Since i=1 xi = i=1 fi (x ), we must have x∗i = fi (x∗ ), for all i. Note that there is no loss of generality to consider functions mapping the unit simplex into itself. If f were defined in a more general, convex and compact set C, we could embed C in a larger simplex S, and define the function g : S → S as follows: 1. if x ∈ C then g(x) = f (x); / C then g(x) = f (y), where y is the unique point in C which is closest to x. 2. if x ∈
4.7
Existence of Nash Equilibria
We will prove the existence of a Nash equilibrium for a two-person game. The proof can be easily extended to handle multi-person games. Let (A, B) denote a bimatrix game, and p and q denote the mixed strategies of R and C, respectively. The expected payoffs to the two players can be written as u1 (p, q) = pT Aq and u2 (p, q) = pT Bq. Let Ai and Bj denote the i-th row of A and the j-th column of B, respectively. Then Ai q is the expected payoff of R when she plays her i-th pure strategy against the mixed strategy q. Similarly, pT Bj is the expected payoff of C when he plays his j-th pure strategy against the mixed strategy p3 . Theorem 4.22 (Nash). Every two person bimatrix game has a Nash Equilibrium in the mixed strategies. Proof. Let us define 3 For
simplicity, sometimes we omit the superscript T when denoting row vectors.
4-10
Lecture 4: Nash Equilibria
ci (p, q) = max{Ai q − pAq, 0}, i = 1, 2, · · · , m, dj (p, q) = max{pBj − pBq, 0}, j = 1, 2, · · · , n. Note that ci (dj , resp.) represents the gain obtained by switching from the mixed strategy p (q, resp.) to the pure strategy i (j, resp.). We can now define the functions P (p, q) = (P1 (p, q), . . . , Pm (p, q)) and Q(p, q) = (Q1 (p, q), . . . , Qn (p, q)), where
Pi (p, q) =
pi + ci (p, q) ∑m , 1 + k=1 ck (p, q)
Qj (p, q) =
qj + dj (p, q) ∑n . 1 + k=1 dk (p, q)
Ai q − pT Aq and pT Bj − pT Bq are continuous functions. Since the maximum of two continuous functions is a continuous function, we have that ∑m ∑nci and dj are continuous. Therefore Pi and Qj are continuous, too. Furthermore i=1 Pi (p, q) = 1, and j=1 Qj (p, q) = 1. Now let us define a function T mapping the Cartesian product between the sets of the mixed strategies of R and C into itself, as follows: T (p, q) = (P (p, q), Q(p, q)). Since T is continuous, then we can apply Brouwer’s fixed point theorem to establish the existence of a point (p∗ , q ∗ ) such that T (p∗ , q ∗ ) = (p∗ , q ∗ ). ∑m ∑m We claim that k=1 ck (p∗ , q ∗ ) = 0. Suppose that the claim is not true. Then we would have i=1 ck (p∗ , q ∗ ) > 0. Since (p∗ , q ∗ ) is a fixed point, we have pi ∗ =
pi ∗ + ci (p∗ , q ∗ ) ∑m 1 + k=1 ck (p∗ , q ∗ )
for every i. This equality implies that pi ∗ = 0 whenever ci (p∗ , q ∗ ) = 0. Let I = {i : pi ∗ > 0}. Let u1 ∗ = u1 (p∗ , q ∗ ), an recall that u1 (p, q) = pAq. ∑ Note that I ̸= ∅ because i∈I pi ∗ = 1. For i ∈ I, we have ci (p∗ , q ∗ ) > 0, i.e., Ai q ∗ > u1 ∗ . Multiplying by pi ∗ > 0 we obtain pi ∗ Ai q ∗ > pi ∗ u1 ∗ . Summing over all i’s, we obtain u1 ∗ =
m ∑ i=1
pi ∗T Ai q ∗ ≥
∑ i∈I
pi ∗T Ai q ∗ > (
∑ i∈I
pi ∗ )u1 ∗ = u1 ∗ ,
Lecture 4: Nash Equilibria
4-11
∑m which is a contradiction. Therefore k=1 ck (p∗ , q ∗ ) = 0, i.e., ck (p∗ , q ∗ ) = 0, ∀k, which means that p∗ is a best response to q ∗ . Similarly we can prove that q ∗ is a best response to p∗ . Since p∗ and q ∗ are mutual best responses they are a Nash Equilibrium pair.
Bibliographic notes For the presentation of the results on zero-sum games, we have mainly followed [3]. The first proof of the minimax theorem by von Neumann appeared in 1928 [7]. A completely different proof was then presented in 1944 [8]. Brouwer’s fixed point theorem was presented in [2]. For the proof of this result, we have followed [1]. Nash Theorem was first shown in [5]. Nash then showed how to reduce his theorem to Brouwer’s fixed-point theorem in [6].
References [1] K. C. Border, Fixed Point Theorems with Applications to Economics and Game Theory. Cambridge, UK: Cambridge University Press (1985). [2] L. Brouwer, Uber Abbildung von Mannigfaltikeiten, Mathematische Annalen, 71, pp. 97-115 (1912). [3] V. Chandru, M.R. Rao, 175 Years of Linear Programming. Part 4. Minimax and the Cake. Resonance, Vol. 4, No.7, July 1999, pp.4-13. [4] R.W. Cottle, J.S. Pang, R.E. Stone, The Linear Complementarity Problem, Academic Press, New York, 1992. [5] J. Nash, Equilibrium points in N-Person Games, Proceedings of the National Academy of Sciences (USA), 36(1):48–49, 1950. [6] J. Nash, Non-Cooperative Games, Annals of Mathematics, 54(2):286–296, 1951. [7] J. von Neumann, Zur Theorie der Gesellschaftsspiele, Mathematische Annalen, 100, 1928, pp. 295320. [8] J. von Neumann, O. Morgenstern, Theory of Games and Economic Behavior. Princeton University Press, Princeton, 1944.