Economics Noncooperative Game Theory Lectures 3. October 15, 1997 Lecture 3

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1 Economics Noncooperative Game Theory October 15, 1997 Lecture 3 Professor Andrew McLennan Nash Equilibrium I. Introduction A. Philosophy 1. Repeated framework a. One plays against dierent opponents each time, so that retaliation and rewards for prior behavior are impossible. b. One does not know the history of specic opponents, so reputation building is impossible. c. The \social norm" is observed by all agents, and is used to form expectations about opponents' behavior. d. The paper by Okuno-Fujiwara and Postlewaite (GEB 1995) presents a detailed analysis of this interpretation. 2. This interpretation depends on a notion of \similarity" between games. a. This may be culturally determined. b. Essentially this is the philosophical problem of induction. B. Importance of Nash's existence theorem 1. It is mathematically useful of course. 2. It guarantees that the solution concept is not ad hoc in the sense of applying only to some games. Insofar as solution concepts are theories of rational behavior, they should be applicable to all environments. C. Outline 1. We develop the mathematical properties of the best response correspondence. 2. Then the Kakutani xed point theorem is applied. 1

2 3. Computation is discussed. 4. An alternative proof of existence using the Brouwer xed point theorem is given. 5. The existence of symmetric equilibria is shown. II. Review of the Best Response Correspondence A. Recall the formula for agent i's expected payo: u i () = X Si = X Si i (s i )[ X S,1( j6=i j (s j )) u i (s i ;s,i )] i (s i ) u i (s i ;,i ) 1. u i is continuous. 2. u i is the restriction to (S 1 ) ::: (S n ) ofamultilinear function u i : IR S 1 :::IR S n! IR: B. BR i () =f 0 i 2(S i)ju i ( 0 i ;,i)u i ( 00 i ;,i) for all 00 i 2 (S i)g: Lemma 1: BR i () is the face of (S i ) spanned by the pure strategies in BR i (): Consequently it is nonempty and convex. Proof: This follows from the multilinearity of u i : C. Denition: A correspondence is a set valued function f : X! Y (i.e. f(x) Y ) such that f(x) 6= ; for all x 2 X: If X and Y are topological spaces, f is upper semicontinuous if its graph Gr(f) = f(x; y)jy 2 f(x)g is closed. 1. Remark: In the mathematical literature correspondences are sometimes called multifunctions and are said to be upper semicontinuous if, for every x 2 X and every neighborhoods V of f(x); there is a neighborhood U of x such that f(x 0 ) V for all x 0 2 U: If Y is compact the two notions of upper semicontinuity coincide. (Exercise) 2

3 Lemma 2: BR i () is an upper semicontinuous correspondence. Proof: Suppose f m g is a sequence in ; 0m i 2BR i ( m ) for all m; m! ; and 0m i! 0 i : If 0 i is not in BR i () then there is 00 i with u i ( 0 i ;,i)<u i ( 00 i ;,i): The continuityof u i 0m i 2 BR i ( m ): then implies that u i (i 0m ;,i m )<u i(i 00;m,i) for large m; contradicting D. Denition: The best response correspondence is BR :! dened by BR() = i2i BR i (): Corollary: BR is an upper semicontinuous and convex valued. Proof: :Obvious. III. The Existence Proof The Kakutani Fixed Point Theorem: If C is a convex compact subset of IR m ; and f : C! C is an u.s.c. convex valued correspondence, then f has a xed point, that is, there is x 2 C with x 2 f(x): A. Remarks 1. This is proved by \bootstrapping" the Brouwer xed point theorem. This is discussed later. 2. This theorem was generalized by Eilenberg and Montgomery in the late 1940's to allow, among other things, correspondences that are contractible valued. (A topological space X is contractible if there is a continuous contraction c : X [0; 1]! X; that is, c 0 = c(; 0) = Id X and c 1 (X) =fptg: Exercise { convex sets are contractible.) 3. Another generalization due to Fan and Glicksburg shows that it is enough to assume that C is a compact convex subset of a (possibly innite dimensional) topological vector space. 3

4 B. Denition: A Nash equilibrium is a xed point of BR: Theorem: Every normal form game has a nonempty set of Nash equilibria. IV.Computation A. In general the computation of the set of Nash equilibria can be very tedious. B. A Nash equilibrium has the following two parts: 1. The sets of pure best responses for the various agents. 2. The probabilities assigned to the pure best responses. C. The algorithm 1. For each collection of subsets ; 6= T i S i compute the solution of the equilibrium conditions: a. all pure strategies in T i have the same expected payo; and b. T i i(s i )=1: c. This system of equations is algebraic of degree n, 1: 2. Given a solution of the rst stage, check whether: a. all probabilities are nonnegative; and b. all pure strategies in S i, T i are inferior. 3. Often logic allows a drastic reduction in the number of cases that need to be considered. D. An Example 1n2 ` c r 0 U (8; 6) (3; 1) (4; 8) M (1; 7) (5; 5) (9; 3) D (6; 2) (7; 9) (2; 4) 1 C A 1. Pure strategy equilibria a. We look at the best responses i. U ) r ) M ) ` ) U 4

5 ii. D ) c ) D b. Thus (D; c) is the only pure equilibrium. 2. Equilibria with mixtures over two strategies a. We look at the consequences in the best response correspondence of not playing assigning positive probability to a pure strategy. i. U ) r )M ) `) U ii. Thus T 1 = fm; Dg or fu; Dg and T 2 = f`; cg or fc; rg are not possible. iii. D ) c ) D; so T 1 = fu; Mg and T 2 = f`; rg must be checked out. The system 8q` +4q r =1q`+9q r ;q`+q r =1; 8p U +p M =4p U +9p M ;p U +p M =1; has the solution (p U ;p M )=(2=3;1=3); (q`; q r )=(5=12; 7=12): Since the probabilities are positive and D ) c ) D; this is an equilibrium. 3. Totally mixed equilibria. a. In principle we should solve two systems of three equations and three unknowns. b. Since I cleverly chose the payos so that the payo matrix for each agent is a magic square { the sums of all rows, columns, and diagonals are the same { (p U ;p M ;p D )=(1=3;1=3;1=3) and (q`; q c ;q r )=(1=3;1=3;1=3) is the unique totally mixed equilibrium. V. More on Fixed Point Theory A. It is very important to understand xed point theorems. 1. Nash equilibrium, and the techniques used to prove existence, are central. 5

6 2. More sophisticated aspects of xed point theory are used in developing some solution concepts. 3. We will do the following: a. Argue that Brouwer's xed point theorem is plausible; b. Sketch the argument passing from Brouwer's theorem to Kakutani's; c. Show how the existence of Nash equilibria can be demonstrated using Brouwer's theorem. B. The Brouwer Theorem Brouwer's Fixed Point Thoerem: If C is a compact convex subset of IR m (or homeomorphic to one), and g : C! C is a continuous function, then g has a xed point. 1. For economists the best source of a proof is Topology from the Dierentiable Viewpoint by John Milnor, pp. 1 { 19, since one also learns Sard's Theorem and the avor of dierentiable topology. 2. We present one step in this argument. Any compact convex subset of IR m is homeomorphic to D` = fx 2 IR` jkxk1g for some `: (Exercise) Let S`,1 = fx 2 IR` jkxk =1g be the (`, 1){sphere. If g : D`! D` has no xed point, let h : D`! S`,1 be the function that maps each x to the intersection of S`,1 with the ray from g(x) through x: (Draw a picture.) Then h is continuous, and h(x) =x for all x 2 S`,1 : This is intuitively impossible (a drumhead cannot be stretched onto the rim of the drum without tearing), but it is nonetheless not at all easy to prove. 3. The argument passing from Brouwer's theorem to Kakutani's is a matter of approximating the graph of the correspondence with the graphs of functions. [Illustration] The limit points of a sequence of xed points of a sequence of ner and ner 6

7 approximations must be xed points of the correspondence. C. A Proof of Existence of Nash Equilibria Using Brouwer's Theorem 1. For 2 and s i 2 S i let ' i (; s i ) = maxf0; u i (s i ;,i ),u i ()g; and dene the better response function for i to be i :!(S i ) given by i ()(s i )=( i (s i )+' i (; s i ))=(1 + S i ' i(; s 0 i )): Lemma 3: i :!(S i ) is continuous. Proof: Obvious. Lemma 4: i () = i if and only if i 2 BR i (): Proof: If i 2 BR() then ' i ( ; s i ) = 0 for all s i 2 S i ; so i () = i : Conversely, if i ()= i then there is 0 such that ' i (; s i )= i (s i ) for all s i 2 S i ; and if >0 then it must be the case that all the pure strategies receiving positive probability in i must have an expected payo greater than i ; an impossibility. 2. Let () =( 1 (); :::; n ()): Then our results above imply that :! is a continuous function whose xed points are precisely the Nash equilibria. The existence of Nash equilibria now follows from Brouwer's theorem. VI. Symmetry A. Introduction 1. In some games there are agents who are clones of each other, and one may want equilibria in which they all play the same way. 2. There may be rearrangements of the strategies of the agents that leave the payo matrices unchanged, and one may wish to consider equilibria in which the strategic probabilities are invariant with respect to such rearrangements. 3. There are reasons to regard symmetric equilibria with suspicion. 7

8 a. I know only one instance in which this notion appears in the literature, and the symmetric equilibrium for that game is, in my opinion, the least plausible. b. For the following simple example the only symmetric equilibrium is (p L ;p R ) = (1=2; 1=2); (q`; q r )=(1=2;1=2); and it is easy to argue that it is implausible. 0 1n2 ` r L (1; 1) (0; 0) R (0; 0) (1; 1) 4. On the other hand Nash was very smart, in fact smarter in some ways than was apparent when he wrote, so it is advisable to pay attention to his thought. B. Denition: A symmetry of N is a collection of bijections =( :I!I; ( i : S i! S (i)) i2i ) with the property that u i (s) =u (i)(( i (s i )) i2i ) for all i and s: We abbreviate this equation with the notation u i (s) =u (i) ((s)): 1. If is a symmetry, we extend to a map :! dened by () (i) ((s i )) = i (s i ): 2. Let Sym be the set of strategy vectors such that () = for all symmetries : 1 C A Lemma 5: Sym is nonempty, closed, and convex. Proof: For all symmetries and all i we have # (S i )= # (S (i) ); so 2 dened by i (s i )=1= # (S i ) is an element of Sym : Since :! is continuous, if ( m )= m for m =1;2; ::: and m! ; then () =; so Sym is closed. Also, it is easy to compute that ( +(1,) 0 )=()+(1,)( 0 ); so if and 0 are symmetric, so is +(1,) 0 : 8

9 Lemma 6: (()) = (()) for all 2 : Proof: We compute that u i () = S ( j2i j (s j ))u i (s) = S ( j2i () j ((s j )))u (i) ((s)) = u (i) (()); '(; s i ) = maxf0; u i (s i ;,i ),u i ()g = maxf0; u (i) ((s i ;,i )), u (i) (())g = ' (i) ((); (s i )); and i ((s i )=( i (s i )+' i (; s i ))=(1 + S i ' i(; s 0 i)) = (() (i) ((s i )) + ' (i) ((); (s i ))) (1 + S(i) ' (i) ((); s 0 (i) )) = (i) (())((s i )): Combining this with the denition of (()) yields the equation (()) (i) ((s i )) = i ()(s i )= (i) (())((s i )): Corollary: If 2 Sym ; then () 2 Sym. Proof: For any 2 Sym and any symmetry we have (()) = (()) = (): Theorem: There is a Nash equilibrium in Sym : Proof: : Sym! Sym satises the assumptions of the Brouwer xed point theorem. VII. Innite Spaces of Pure Strategies A. For i 2 I; let S i be a compact metric space of pure strategies, let S = S 1 :::S n ; and let u i : S! IR be a continuous function. 9

10 B. Let (S i ) be the set of Borel probability measures on S i endowed with the weak topology. (This won't be on the exam!) Let = i2i (S i ): C. Denition: A Nash equilibrium is 2 such that for each i; Z S 1 ::: Z Sn u i (s)d 1 ::: d i ::: d n Z S 1 ::: Z Sn u i (s)d 1 ::: d i ::: d n for all i 2 (S i ): Theorem: Under the assumptions above, there are Nash equilibria. Proof: The proof is the same as the one given above using Kakutani's xed point theorem, executed at a higher level of generality. One shows that each u i is continuous, that (S i ) is compact and convex, and that the best response correspondence is upper semicontinuous and convex valued. The desired conclusion follows from the Fan-Glicksburg xed point theorem, provided one knows the requisite functional analysis. 10