Unversty of Illnos Fall 08 ECE 586GT: Problem Set : Problems and Solutons Unqueness of Nash equlbra, zero sum games, evolutonary dynamcs Due: Tuesday, Sept. 5, at begnnng of class Readng: Course notes, Sectons.5-.7 and Chapter. [Exstence and unqueness of NE for games wth contnuous type strateges] Ths problem concerns an n player game wth strategy space S = (0, + ) for all players, and space of strategy profles S = S S n. (a) Consder the payoff functons u (s) = (ln s ) s s tot, where s tot = s + + s n. Does there exst a pure strategy Nash equlbrum? If so, s t unque? Can you fnd an explct expresson for t? Soluton: For [n], for any s fxed, u s a strctly concave functon of s wth lmt at 0 and +. Hence, for each s fxed there s a unque best response whch s obtaned by settng u s = 0. Note that s tot = s + s, where s = [n], s. Wrtng u (s) = ln s s s s we fnd u s = 0 s equvalent to s s = 0. () s (Although t sn t needed, we note () mples that the best response functon for any player s B (s ) = s + s +8 4.) Usng the fact s tot = s + s, () can be rewrtten as whch can be used to determne s as a functon of s tot : s s s tot = 0, () s = s tot + s tot + 4 We can conclude that any Nash equlbrum (n pure strateges) s symmetrc. Therefore, for a Nash equlbrum, s tot = ns and so s = ns + n s + 4 Thus, there s a unque Nash equlbrum gven by s = n+ for all [n]. (b) Consder the payoff functons u (s) = (ln s ) s s tot + ɛ cos(s ) for some ɛ > 0. Fnd a postve constant ɛ such that there exsts a Nash equlbrum f 0 ɛ < ɛ. To avod some tedous steps, you may assume that the strategy space for each player s restrcted to the closed nterval [δ, /δ], such that δ can be chosen so small that any best response for any player s n the nteror of the nterval. Soluton: Note that u (s) ( s = ɛ cos(s ) s ) < 0 f ɛ. so u (s) s a concave functon of s for s fxed, f ɛ. Also, the payoff functons are contnuous and the strategy spaces [δ, /δ] are compact, so there exsts a pure strategy NE f ɛ by the Debreu, Glcksberg, and Fan exstence theorem.
(c) For the payoff functons n part (b), fnd a postve constant ɛ such that there exsts a unque pure strategy Nash equlbrum f 0 ɛ < ɛ. (Hnt: A dagonal matrx wth negatve entres on the dagonal s negatve defnte, and the sum of a negatve defnte matrx and a negatve sem-defnte matrx s negatve defnte.) Soluton: We use the suffcent condtons for unqueness n the notes based on the followng matrx: U(s) = u (s) ( s ) u (s)... s s. u (s) s s By drect ( calculaton we) fnd U(s) = D(s) J, where D(s) s the dagonal matrx dag ɛ cos(s s ) and J s the all ones n n matrx. In partcular, U s a symmetrc matrx so we don t need to consder the symmetrzaton U(s) + U T (s). If 0 ɛ then D(s) s negatve defnte. Also, J s negatve semdefnte. Thus, f 0 ɛ, U(s) s negatve defnte, and there exsts a unque Nash equlbrum,. [Market power n terrtory control game] Recall the terrtory control game for a graph G = (V, E) and n players, defned n problem set. The game s symmetrc n the sense that the payoff functons are nvarant under permutaton of the players. A strategy profle (n ether pure or mxed strateges) s symmetrc f every player uses the same strategy. (a) Prove that for any symmetrc fnte player game, there exsts a symmetrc Nash equlbrum n mxed strateges. Thus, for the terrtory control game based on a connected graph G = (V, E), there s a Nash equlbrum such that the expected payoff of each player s V n. Soluton: Ths follows from the same proof as Nash s theorem. (b) Show that a two-player game such that the sum of payoffs s constant for all strategy pars s equvalent to a zero sum two-player game. In partcular, there exsts a saddle pont and a value of the game for each player. Soluton: Let u tot denote the sum of payoffs, whch s assumed to be ndependent of the strategy profle. We could mpose a tax n the amount u tot to player, or mpose a tax of u tot /n to every player, to make the sum of payoffs always equal to zero. (c) Suppose there are three players, but players and 3 team up and splt ther payoffs wth each other. Thus, players and 3 together can be vewed as a super player; from the perspectve of player, the game becomes a two-player game, wth constant sum of payoffs V. Player could be placng a Burger Kng and players and 3 together place two McDonalds. Show that the value of the modfed game for player s greater than or equal to V /4 for any connected graph G. (Hnt: What f there were two super players, each selectng nodes for two restaurants?) Soluton: If there were two super players, each selectng two nodes usng a mxed strategy, then by symmetry the value of the game would be V / for each of the players. Transform the strategy of one of those super players by havng the player frst select ts two nodes (possbly the same one twce), and then flppng a far con to determne whch of those choces to keep, and whch to dscard. The mean payoff generated by the randomly retaned node was at least V /4 before the other randomly selected node
was dropped. The droppng of the other choce by the same player cannot decrease the expected ponts s generated by the retaned node, whch s thus at least V /4. For example, f the player had two Burger Kng restaurants, then closng one of them could generate more traffc for the other. (d) Consder the scenaro of part (c) for the specal case of a lne graph V = {,,..., m}, such that m = 4b + for some nonnegatve nteger b, and such that E = {[, + ] : m }. Show that the value of the modfed game for player s equal to V /4. (Hnt: By part(c) only an upper bound on the average payoff of player s needed.) Soluton: Suppose player selects b + and player 3 selects 3b +. Then the payoff of player s less than or equal to m/4 for all possbltes. Specfcally, t ranges from (k + )/ to V /4 as s ncreases from 0 to b +. It remans flat at V /4 for b + s 3b +. 3. [Dual of a transport problem] Let n and let W be an n n matrx wth nonnegatve elements. Consder the followng lnear optmzaton problem: mn X, [n] X, W, subect to: X, = [n] X, = [n] X, 0 for [n] for [n] for, [n] An nterpretaton s that there are n sources of some good, n destnatons, W, s the cost of transportng a unt of good from to, and X, s the amount of good transported from to. Each source provdes one unt of good and each destnaton receves one unt of good. The problem n vector matrx form can be wrtten as: mn X X, W subect to: X =, X T =, X 0. (a) Derve the dual problem by frst fndng the Lagrangan functon. Fnd a smple verson of the dual problem. (Hnt: You can ether elmnate the Lagrange multplers for the constrants X, 0, or don t handle those constrants wth Lagrange multplers n the frst place. 3
Soluton: Usng strong dualty for lnear programs, we fnd mn X, W X:X=, X T =, X 0 = mn X:X 0 µ (),µ () mn X:X 0 max X, W + ( X) T µ () + ( X T ) T µ () µ (),µ () X, W + ( X) T µ () + ( X T ) T µ () µ (),µ () mn X:X 0 T (µ () + µ () ) + X, W µ () T (µ () ) T µ (),µ () :µ () T +(µ () ) T W In other words the dual problem s max µ () µ (),µ () + µ () [n] [n] T (µ () + µ () ) subect to: µ () + µ () W, for, [n] (An nterpretaton of the dual varables s that µ () s a transport charge per unt good at source and µ () s a transport charge per unt good at destnaton. The dual problem entals maxmzng the sum of payments subect to a constrant showng the charges are ustfed by the transport costs. The equalty constrants n the prmal problem could be replaced by nequalty constrants: X, X T, and that would lead to nonnegatvty constrants on the multplers n the dual problem. ( ) 3 8 (b) Fnd the common value of the prmal and dual, and solutons, for W =. 4 ( ) ( ) ( ) 0 3 0 Soluton: X =, µ 0 () =, µ () =, wth value 7 each. (The ( ) soluton to the dual problem s not unque. Addng any constant multple of to µ () and subtractng t from µ () yelds another soluton.) 4. [Evolutonarly stable strateges and states] Consder the followng symmetrc, two-player game::,,, 0,0 (a) Does ether player have a (weakly or strongly) domnant strategy? Soluton: No. (b) Identfy all the pure strategy and mxed strategy Nash equlbra. Soluton: (,) and (,) are pure strategy NE. As usual, there s no NE n whch only one strategy s pure. If (p, q) s an NE such that both p and q are nondegenerate mxed ( strateges, ether acton of player one must be a best response to q, so = q, or q =, ) (. Smlarly, p =, ) ((. Thus,, ) (,, )) s the unque NE n nondegenerate mxed strateges. 4
(c) Identfy all evolutonarly stable pure strateges and all evolutonarly stable mxed strateges. Soluton: Recall that f p s an ESS, then (p, p) s an NE. The pure strategy NEs found n part (b) are not symmetrc, so there are no pure ESSs. It remans to see whether the mxed strategy NE p gven by p = (, ) s an ESS. By defnton, we need to check whether for any p p, ether () u(p, p) < u(p, p), or () (u(p, p) = u(p, p) and u(p, p ) > u(p, p )). Snce u(p, p) = u(p, p) for all choces of p, the queston comes down to whether u(p, p ) > u(p, p ) for all p p. That s, whether + ()p > p + ( p )(p ) for all p p. Or, equvalently, whether ( p ( ) > 0 for p p. Ths condton s true, so, ) s a mxed ESS. (d) The replcator dynamcs based on ths game represents a large populaton consstng of type and type ndvduals. Show that the evoluton of the populaton share vector θ(t) under the replcator dynamcs for ths model reduces to a one dmensonal ordnary dfferental equaton for θ t (), the fracton of the populaton that s type. Soluton: The replcator dynamcs for the populaton share vector θ t are gven by θ t (a) = θ t (a)(u(a, θ t ) u(θ t, θ t )) for a {, }. Although there are two equatons, ths system s actually one dmensonal because θ t () = θ t (). Let x t = θ t (). Then u(, θ t ), and u(θ t, θ t ) = x t + 3x t ( x t ) = 3x t x t. So the replcator dynamcs become ẋ t = x t ( 3x t + x t ), or, equvalently, ẋ t = x t ( x t )( x t ). (3) (e) Identfy the steady states of the replcator dynamcs. Soluton: The rght hand sde of (3) s zero for x t { 0,, }, so there are three steady states for the replcator dynamcs: (,0), (, ), and (,0). (f) Of the steady states dentfed n the prevous part, whch are asymptotcally stable states of the replcator dynamcs? Justfy your answer. Soluton: Of the three steady states, only (, ) s asymptotcally stable. If x0 = ɛ for an arbtrarly small but postve ɛ, then ẋ t > 0 and x wll converge monotoncally up to, so 0 s not even a stable steady state (so t s not asymptotcally stable). Smlarly, s not a stable steady state. However, s an asymptotcally stable steady state of x because the rght hand sde of (3) has a down crossng of zero at. Hence (, ) s an asymptotcally stable state for the replcator dynamcs. (Another ustfcaton for ths problem can be gven by applyng general facts about ESSs and replcator dynamcs. States (, 0) and (0, ) can t be asymptotcally stable, or even stable, states for the replcator dynamcs, because, when played aganst themselves, they don t gve NEs. The mxed state (, ) s an asymptotcally stable state of the replcator dynamcs because t s an ESS.) 5