Bimatrix Games. R.Chandrasekaran,

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1 Bimatrix Games R.Chandrasekaran, 1 Bimatrix Games These are two person non-zero(constant)-sum games in which each player has finitely many pure strategies. We call these non-zero(constant)- sumgamesbecausetheinterestsoftheplayersisnotrequiredtobeexactly opposed to each other. Of course, these include zero(constant)-sum games and are a true generalization of zero(constant)-sum games. But the methods usedtoanalyzethemaredifferent. Theyalsobringourmanymoredifficulties as shown inthe initialpart. Supposeplayer1hasmpure strategies andplayer2hasnpurestrategies. LetAbeanm nmatrixrepresenting payoffstoplayer1andsimilarlyletb beanm nmatrixrepresentingthe payoffmatrixtoplayer2. IfA+B=0(αJ)wheretherightsideisanm n matrix all of whose entries are zero(j is a matrix all of whose entries are equla to 1 and α is a number), then we have the zero(constant)-sum case. Else we have the non-zero(constant)-sum case. A pair of(mixed) strategies x X ={x:x 0; m i=1 x i=1}andy Y ={y:y 0; n j=1 y j =1} which satisfy the relations: (x ) t Ay x t Ay x X (x ) t By (x ) t By y Y is called a Nash equilibrium. J. Nash showed that these exist as in the following theorem. His proof uses the well known Brower fixed point theorem. [There are other fixed point theorems that are more powerful and can extend these results] SchoolofManagement,UniversityofTexasatDallas,Richardson,Texas,U.S.A.. 1

2 Theorem1 (Brower)LetT :S S beacontinuousfunctionthatmapsa compactconvexsets R n toitself. Thenthereisapointx S suchthat T(x)=x. SuchapointiscalledafixedpointforT. Theorem 2 Every bimatrix game has at least one equilibrium point. Proof. (Owen,LuceandRaiffa): Letxandybeanypairofmixedstrategies forthebimatrixgame(a,b). Define: c i =max[a i,. y x t Ay,0]; 1 i m d j =max[x t B.,j x t By,0]; 1 j n x i = x i +c i 1+ m k=1 c k 1 i m y j = y j+d j 1+ n l=1 d 1 j n l Think of the transformation T by the relation: T(x,y)=(x,y ) SuchaT isclearlycontinuousandmapsx Y toitself[showthis]. If(x,y) isanequilibriumpairofstrategies,x t Ay (e i ) t Ay=A i,. y andsoc i =0for alli. [Heree i denotestheunitvectorwitha1inthei th position.] Similarly, d j =0 forall j. Hence, (x,y )=(x,y) forsuchapair. Henceequilibrium pairs of strategies are fixed points of this transformation. Now we show the converse is also true. Proof. Since both X and Y are compact convex sets so is the Cartesian product. Hence we can apply Brower fixed point theorem and there is fixed point. Letusdenotethefixedpointby(x,y ). Nowweusetherelationsthat T(x,y )=(x,y ) to show that these are an equilibrium pair of strategies for the game. where x i = x i+c i 1+ m k=1 c k y j = y j+d j 1+ n l=1 d l 1 i m 1 j n c i =max[a i,. y (x ) t Ay,0]; 1 i m d j =max[x t B.,j (x ) t By,0]; 1 j n 2

3 Claim3 If (x,y ) is not an equilibrium pair of strategies, at least one of thevaluesofc i oroneofthevaluesofd j isstrictlypositive. Proof. Iftheclaimisnottrue,thenitfollowsthat (x ) t Ay A i,. y =(e i ) t Ay for1 i m Multiplyingthethei th oftheseinequalitiesbyx i andadding[thisispermitted sincex i 0],weget Similarly we can show that (x ) t Ay x t Ay x X (x ) t By (x ) t Ay y Y Proof. (of the main theorem continued): Suppose without loss some c i > 0 (equivalently A i,. y > (x ) t Ay ). Hence m i=1 c i > 0. But (x ) t Ay is weightedaverageof{a i,. y } m i=1. Hence,thereissomeotherindexpsuchthat A p,. y <(x ) t Ay x p >0 So,c p=0andhence x x p p= 1+ m <x k=1 c p k andthereforex x. Ifinsteadsomed j >0,wewillshowthaty y. In either casethiswill notbeafixedpointofthistransformation. Hencefixed points are equilibrium pairs and therefore equilibrium pairs of strategies exist for any bimatrix game. Proof of existence of a Nash equilibrium for n-person games is similar. There is alternate proof based on Kakutani s fixed point theorem. This isbasedonthefactthatinthesegames,thesetofbestresponseforplayeri againstastrategyprofileoftheremainingplayersisacompactconvexset( itisalospolyhedralifthesetofpurestrategieofeachplayerisfinite). This is exercise 2.4 in your homework assignments. Hereistheproof: Proof. Let (p 1,p 2,...,p n ) be a mixed strategy profile. [Note that each of thesevectorsp i isfinitedimensional.]letb i (p 1,p 2,...,p n )bethesetof"best 3

4 responses"forplayeriagainstp i. RecallfromHW2.4,thatthesesetsB i arecompactconvexsets. HenceB 1 B 2... B n isalsoacompactconvex set. Consider the correspondence (p 1,p 2,...,p n ) B 1 B 2... B n Itmapsastrategyprofiletoacompactconvexsubsetofstrategyprofiles. It is also upper-hemi-continuous (see definition below). Hence by Kakutani s theroem, a strategy profile (p 1,p 2,...,p n ) which is a fixed point of this correspondence. This clearly is a Nash equilibrium. Definition4 AcorrespondenceΓ:A Bissaidtobeupperhemicontinous at the point x A if for any open neighborhood V of Γ(x), there exists a neighborhoodu ofxsuchthatγ(y)isasubsetofv forally U. Definition5 ThegraphofacorrespondenceΓ:A B istheset{(x,y): y Γ(x);x A,y B}. Thisgraphissaidtobeclosedifthesetisclosed. If the sets A, B are compact and Γ(x) is closed for each x this concept is equivalenttoupperhemicontinuity. Itisthisthatisusedtoassurethatthe conditions of kakutani s theorem are vlaid for our case. Pairsofstrategies(x,y )thatsolve min(x ) t Ay=max min x X y x t Ay maxxay =max y Y min x X xt Ay do not necessarily form Nash equilibrium pairs for a bimatrix game. The computational problem of finding these equilibrium pairs for a bimatrixgamewas"solved"byformulatingitasaspecialcaseofthewellknown "Linear Complementarity Problem" proposed by Lemke and Howson. The general LCP has many other applications (see book by R.W. Cottle, J.S. Pang and R.E. Stone). Lemke- Howson algorithm is a finite algorithm but it is not guaranteed to be polynomially bounded indeed no such algorithm is known. 4

5 1.0.1 Linear Complementarity Problem(LCP)& Bimatrix Games Given a matrix M R n n, a vector q R n, the linear complementarity problemistofindvectorsw,z R n satisfyingtherelations: w=mz+q w 0;z 0 w t z=0 Pleasenotethat[w 0;z 0;w t z =0] [w i =0;z i =0foralli];hence, atmostoneofeachpair{w i,z i }canbe nonzero. Thisisthereasonforthe term"complementary" in the name of the problem. For this formulation, it is more convenient to think of the payoff matrices as "loss" matrices. Let [a i,j,b i,j ]representthelosstoplayer1and2respectivelyifplayer1usespure strategyiandplayer2usespurestrategyj. Then,apairofmixedstrategies (x,y )areanashequilibriumpairiftheysatisfytherelations: This is equivalent to the system: (x ) t Ay x t Ay x X (x ) t By (x ) t By y Y [(x ) t Ay ]e m Ay [(x ) t By ]e n B t x wheree m ande n refertovectorofall1 sofsizemandnrespectively. We can show that this reduces to the linear complementarity problem: or equivalently the problem: u=ay e m v=b t x e n x 0;y 0;u 0;v 0 x t u=0;y t v=0 w=mz+q w 0;z 0 w t z=0 5

6 with M = 0 A B t 0 ;q= e m e n w= u v ;z= x y Toshowtheequivalence,suppose(x,y )isanequilibriumpairofstrategies. Soclearly, x 0;y 0. Suppose(andwewillshowthatthisiswithout lossofgenerality)thata>0;b>0. Let x x= (x ) t By y y= (x ) t Ay u=ay e m v=b t x=e n Since m i=1 x i =1and n j=1 y j =1,itfollowsthat(x ) t By >0;(x ) t Ay > 0byourassumptionthatA>0,B>0. Thus,x 0;y 0. [[(x ) t Ay ]e m Ay ] [Ay e m ] and [[(x ) t By ]e n B t x ] [B t x e n ]. Hence u 0;v 0. x t u= (x ) t [A y e (x ) t Ay m ] (x ) t By 1 1 = (x ) t By (x ) t By =0 Similarlyy t v=0. Hence,(x,y,u,v)isasolutiontotheLCP. Todotheconversesupposex 0,y 0,u 0,v 0 solvesthelinearcomplementarity problem. Clearly,neitherx 0 nory 0 canbeequaltothe0vectorsinceu 0,v 0 arenonnegative. Hence m i=1 x0 i >0; n j=1 y0 j >0. Nowlet x = y = x 0 m i=1 x0 i y 0 n j=1 y0 j 6

7 Thesearemixedstrategiesforthetwoplayers. Nowweshowthat(x,y )is anashequilibriumpair. Todothis,weneedtoshowthat which is equivalent to showing; y 0 [(x ) t Ay ]e m Ay Ay =A[ ) t Ay ]e (e n ) t y 0] [(x m = Since(x 0,y 0,u 0,v 0 )solvesthelcp, andso Similarly, 0=(x 0 ) t u 0 =(x 0 ) t [Ay 0 e m ] y 0 (e m ) t x 0 =(x 0 ) t Ay 0 (e n ) t y 0 =(x 0 ) t By 0 Ay =A[ ) t Ay ]e (e n ) t y 0] [(x m = (x 0 ) t Ay 0 [(e m ) t x 0 ] [(e n ) t y 0 ] (x 0 ) t Ay 0 [(e m ) t x 0 ] [(e n ) t y 0 ] Sinceu 0 =Ay 0 e m 0,itfollowsthatAy 0 e m. Therefore, Ay =A y0 (e n ) t y 0 [(x0 ) t Ay 0 (e m ) t x 0 ][ 1 (e n ) t y 0]e m=[(x ) t Ay ]e m whichiswhatwesetouttoprove. Similarlywecanshowthat B t x [(x ) t By ]e n Thuswehaveshownthat(x,y )isanequilibriumpairofstrategies. Lemma6 Let A and B be "loss" matrices in a bimatrix game. Let k be a scalar and let J be a matrix of all 1 s of the same size as A and B. Let C = A+kJ;D = B+kJ. Any equilibrium pair of strategies for the pair (A,B)isalsoanequilibriumpairofstrategiesfor thepair(c,d). Proof of this lemma is quite straightforward and left to the reader. If k isasufficientlylargepositivevalue,thenmatricesc andd willbepositivematricesanditisinthissensethatthisassumptioniswithoutlossof generality. It will be a digression to discuss the algorithm of Lemke-Howson here;sufficeittosaythatitisafiniteprocedurebutnotguaranteedtobe polynomially bounded. 7

8 1.1 Papadimitriou-Daskalakis Formulation: HerewecontinuetothinkofmatricesAandB as"gain"orpayoffmatrices asuslaingametheory. Then, a pair of mixed strategies (x,y ) are a Nash equilibrium pair if they satisfy the relations: This is equivalent to the system: (x ) t Ay x t Ay x X (x ) t By (x ) t By y Y [(x ) t Ay ]e m Ay [(x ) t By ]e n B t x wheree m ande n refertovectorofall1 sofsizemandnrespectively. Wecan show that this reduces to the"modified linear complementarity problem": or equivalently the problem: with u+ay=e m v+b t x=e n x 0;y 0;u 0;v 0 x t u=0;y t v=0 w+mz=q w 0;z 0 w t z=0 M = 0 A B t 0 ;q= e m e n w= u v ;z= x y Toshowtheequivalence,suppose(x,y )isanequilibriumpairofstrategies. So clearly, x 0;y 0. Suppose without loss of generality that A > 8

9 0;B>0. Let x x= (x ) t By 0 y y= (x ) t Ay 0 u= Ay+e m 0 v= B t x+e n 0 Since m i=1 x i =1and n j=1 y j =1,itfollowsthat(x ) t By >0;(x ) t Ay > 0byourassumptionthatA>0,B>0. Thus,x 0;y 0. [[(x ) t Ay ]e m Ay ] [Ay e m ] and [[(x ) t By ]e n B t x ] [B t x e n ]. Hence u 0;v 0. x t u= (x ) t [ A y +e (x ) t Ay m ] (x ) t By 1 1 = (x ) t By + (x ) t By =0 Similarlyy t v =0. Hence, (x 0,y 0,u 0,v 0) isasolutiontothe LCP. Todotheconversesupposex 0 0,y 0 0,u 0 0,v 0 0solvesthelinear complementarityproblem. Hence m i=1 x0 i >0; n j=1 y0 j >0. Nowlet x = y = x 0 m i=1 x0 i y 0 n j=1 y0 j Thesearemixedstrategiesforthetwoplayers. Nowweshowthat(x,y )is anashequilibriumpair. Todothis,weneedtoshowthat which is equivalent to showing; y 0 [(x ) t Ay ]e m Ay Ay =A[ ) t Ay ]e (e n ) t y 0] [(x m = 9 (x 0 ) t Ay 0 [(e m ) t x 0 ] [(e n ) t y 0 ]

10 Since(x 0,y 0,u 0,v 0 )solvesthelcp, andso Similarly, 0=(x 0 ) t u 0 =(x 0 ) t [ Ay 0 +e m ] y 0 (e m ) t x 0 =(x 0 ) t Ay 0 (e n ) t y 0 =(x 0 ) t By 0 Ay =A[ ) t Ay ]e (e n ) t y 0] [(x m = (x 0 ) t Ay 0 [(e m ) t x 0 ] [(e n ) t y 0 ] Sinceu 0 =Ay 0 e m 0,itfollowsthatAy 0 e m. Therefore, Ay =A y0 (e n ) t y 0 [(x0 ) t Ay 0 (e m ) t x 0 ][ 1 (e n ) t y 0]e m=[(x ) t Ay ]e m whichiswhatwesetouttoprove. Similarlywecanshowthat B t x [(x ) t By ]e n Thuswehaveshownthat(x,y )isanequilibriumpairofstrategies. 1.2 Symmetric Games and Symmetrization Ann-playergameissaidtobesymmetricifeachplayer ssets={1,2,...,s} ofpurestrategiesisthesameandpayofffunctionisthesameandequalto π(σ;n 1,n 2,...,n s )wheren i isthenumberofplayersusingpurestrategyi. Definition7 Asymmetric equilibriumforasymmetricgameisanash equilibrium where all players use the same strategy.. Theorem 8 Nash: There exists a symmetric equilibrium for a symmetric game. Proof. Inusingthefixedpointtheroemusethesetwhichistheintersection ofproductofthesetsforeachplayerintersectedwiththesetthatrequiresthe choice of strategies across players to be the same. 10

11 Theorem 9 Every two person game is equivalent to two person symmetric game. Proof. LetthebimatrixgamehavepayoffmatricesAandB. Withoutlossof generality,weamyassumethata>0andb>0. Considerthesymmetric gamewithmatricesc andc t where C= 0 A B t 0 Bytheoremabove,thisgamehasasymmetricNashequilibriumz = [ x y ] Claim 10 x 0;y 0 Proof. [ ] Suppose x = 0 or y = 0. Consider Player I using strategy z = x withx 0;y 0againstPlayerIIusingz y =(x,y ). (z ) t Cz =(x ) t Ay +(x ) t B t y =0 <(x) t Ay +(x ) t B t y The last ofthese follows from A>0,B >0 and x 0,y 0. This would leadtoacontradictionthatz isanashequilibrium. Hencetheclaim. Proof. (continued) Let x= x (e m ) t x ȳ= y z 1 = (e n ) t y [ x 0 ] ;z 2 = Usingthefactthatz isthebestresponsetoz,wecanobtainthatz 1 isthe best response to z 2 and hence ( x,ȳ) is a Nash equilibrium for the original game. [ 0 ȳ ] 11

12 1.3 Computing Equilibrium in Two-person Symmetric Game[P,D] LetM bethesquarematrixc intheabovediscussion; forthisdiscussion, assumethatsizeofthismatrixisn n. Considerthesystem; Mz e n z 0 LetM i, denotethei th rowofthismatrix. Withoutloss,wemayassumethat elements of this matrix are nonnegative with no zero rows or zero columns. ThisimpliesthatP ={z R n :z 0;Mz e n }isnonemptyandbounded and closed. P is said to be nondegenerate if at each vertex exactly n of these 2n inequalities are tight. By small perturbation, we can make any such polyhedron nondegenerate and we will assume this is the case from now on. Eachvertexofthispolytopeisdescribedbyitstightconstraintsinthe followingmanner: Weusetheindexiasmanytimesasthenumberoftight constraintsamong{m i, z=1;z i =0}. Forexample,ifM 1, z=1andz 1 =0 andz 2 =0butM 2, z<1,thisvertexwillbedenotedby{1,1,2}. Soevery vertexhasncomponents. Ifavertexhasthelabel{1,2,...,n}(i.e. allindices occuronce),thenoneofthefollwoingistrue: (a)z=0;or(b)thenormalized vector z is a symmetric Nash equilibrium. To show the existence(and a computation mechanism) for a Nash equilibrium we do the following: Select an index a (1,2,...,n). Consider the set of all vertices whose indexsetscontainallindicesfrom{1,2,..,n}exceptpossiblytheindexa;call thissetofverticesv. Eachvertexinthepolytopehasexactlynneighbors obtainedbyrelaxingoneofthetightconstraints. SoforavertexinV that containsallindices,thereisonlyoneneighborinv;thisobtainedbyrelaxing oneofthetwoconstraintsfortheindexathatistightinthisvertex. Characterize other vertices in V by the index that is repeated say b. There are two constraints correspondingtoindex b;m b, z = 1 and z b = 0. HencethevertexinV thathasindexbrepeatedhasexactlytwoneighbors inv. TheadjacencygraphinV hasdegree2forverticesthatdonotahve allindicesanddegreeequalto1forthosethathaveallindicesrepresented. So if start at z = 0 which has all indices represented and move along this graph we must reach another vertex with all indices represented which must correspond to a symmetric Nash equilibrium! The process is finite because thesizeofv isfinite. ThisisthecruxofC.E.Lemke salgorithm. 12