1 Review of Zero-Sum Games
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1 COS 5: heoreical Machine Learning Lecurer: Rob Schapire Lecure #23 Scribe: Eugene Brevdo April 30, 2008 Review of Zero-Sum Games Las ime we inroduced a mahemaical model for wo player zero-sum games. Any such game can in principle be formulaed via is Game Marix M. We saw wo possible models for such games: deerminisic play and randomized play. In deerminisic play, players make decisions according o pure sraegies, and in randomized play, players ac according o mixed (random) sraegies. A he end of each game we have an oucome. For pure sraegy games, he oucome is a scalar. For mixed sraegy games, we can calculae he expecaion of he oucome (he expeced oucome). Deerminisic Game: Game marix M Row player (min) chooses row i (he min oucome) Column player (max) chooses column j (he max oucome) oucome = M(i, j) Randomized Game: Game marix M Row chosen according o disribuion over he rows Column chosen according o disribuion over he columns (expeced) oucome = i,j (i)m(i, j)(j) = M where for randomized games, we use he noaion M(, ) = M o denoe he expeced oucome as a funcion of he disribuions chosen by he players. here are many connecions beween game heory and machine learning, and in oday s class we will ry o bring ogeher opics from boh of hese fields and unify hem.. A Basic Inequaliy for Randomized Games Las ime we also inroduced a basic analysis for he game of Rock-aper-Scissors (R,,S). In his game, boh sides play simulaneously. Suppose he game is modified slighly, and insead play is sequenial. Mindy (he row player) goes firs, and Max (he column player) can decide wha o do. In a deerminisic seing his game is no very ineresing. Le s consider randomized play: min chooses firs
2 max chooses knowing Since Max knows ha Mindy has chosen and knows ha for a given, he oucome is M(, ), hen Max will wan o choose o maximize M(, ), giving oucome max M(, ). Since Mindy knows wha Max will do and will know he given oucome, she chooses o minimize his oucome. hus, he final oucome will be min max M(, ) where again, in his case Mindy has gone firs. he minimizing above is called he min max sraegy. Example: Consider he game where Mindy prefers rock and has chosen = (/2, /4, /4): R wih probabiliy /2 wih probabiliy /4 S wih probabiliy /4 hen Max can always choose paper and = (0,, 0). he game marix M for his game is: R S R /2 0 0 /2 S 0 /2 and hus we have M(, ) = 5/8. In fac, if Mindy chooses a which is no uniform, he payoff will always be greaer han /2. Now, if Max chooses he column disribuion firs, and Mindy chooses he o minimize he oucome, we would have an ou come of max min M(, ) where he seleced is also known as he max min sraegy. We have discussed his in oher conexs, and noe ha he player going second has he advanage. More specifically, i is sraighforward o show min max M(, ) max min M(, ). 2 Fundamenal heorem of Zero-Sum Games I urns ou ha for all zero-sum games wih finie moves, he above inequaliy can be urned ino an equaliy. hus, regardless of who goes firs, in a game of opimal players, he expeced oucome is always he same. We denoe he oucome value v, his is he Value of he game. So, v = min max M(, ) = max min M(, ). his heorem was proved by von Neumann (who was a IAS in rinceon), and is called he von Neumann min max heorem. We will aim o prove his heorem via an online learning algorihm and argumens we have seen previously. 2
3 heorem 2. (von Neumann min max heorem) For randomized zero-sum games of wo players: min max M(, ) = max min M(, ). he heorem implies ha even if Max knows Mindy s sraegy, Max canno ge a beer oucome han v; v is he bes possible value: min max sraegy such ha : M(, ) v. I also implies ha no maer wha sraegy Mindy uses, he oucome is a wors v: max min sraegy such ha : M(, ) v. and are he opimal sraegies, assuming ha he opponen is also using an opimal sraegy. hus for a wo person zero sum game agains a good opponen, your bes be is o find he min max sraegy and o always play i. his is no he end of he sory. here are many oher issues o consider in such games, usually due o limied informaion abou he game or he opponen. Here are some examples:. We don always know M. 2. M can be very large (and compuing compuaionally difficul). 3. he sraegy is only he bes when playing agains an opimal opponen; we migh do beer agains a non-opimal opponen (i.e., dumb, no mean, ec...). Bar Simpson, for example, always plays Rock insead of choosing he uniform disribuion. hus you can play aper and always bea Bar, insead of only abou /3 of he ime. As mos games are played over and over again, here is an opporuniy for learning eiher he game rules M or he opponen s sraegy, or boh, even wihou any knowledge of eiher a he beginning. Le s consider an online version of ieraions of he game: n = # rows for =,..., row (learner) chooses column (environmen) chooses learner observes M(i, ) for each row i learner suffers loss M(, ) and we define he oal loss = = M(, ). Noe ha a each ieraion, he learner is able o observe par of he game marix M. Here, M(i, ) is he oucome of M given ha Mindy chose row i and he disribuion over he columns is. Similarly M(, j) is he oucome of M given ha max chose column i and he disribuion over he rows was. If is concenraed on a paricular column, hen he learner sees ha enire column. he opposie exreme is when he learner can see only a paricular elemen of he game marix M. Such cases are more complicaed o 3
4 analyze. Now, he learner wishes o minimize he oal loss, as compared o he bes loss possible, had he learner jus chosen he bes fixed sraegy for he ieraions and suck wih i. We wan: M(, ) min M(, ) + small. = Noe ha min = M(, ) v, wih equaliy if =. his happens when he opponen is an opimal opponen, leading o =. In general his doesn have o be he case and we can do significanly beer. hus we can do almos as well as if we had known ahead of ime. 2. Muliplicaive Updaes Le s consider a simple muliplicaive weigh updae algorihm, which oupus disribuions over rows, hus elling us how o play: (i) = n i + (i) = (i)β M(i, ) Z where 0 < β < and Z is a normalizing consan. Noe ha M(i, ) is he loss for exper i. We can prove he following bound for his algorihm (which is a direc generalizaion of he weighed majoriy voe algorihm we sudied earlier): heorem 2.2 (erformance of Muliplicaive Weigh Updaes) Using he algorihm above, we have M(, ) a β min M(, ) + c β ln n = where a β and c β are funcions of β. We are no going o prove his heorem, bu we have seen similar poenial-based argumens of his sor of proof before. Corollary 2.3 I is possible o se β so ha, when dividing boh sides by o calculae he average per-round loss: ( ) ln n where = O. M(, ) min = M(, ) + Noe ha he small error erm falls off o 0 as, for n fixed. hus he average per-round loss approaches he bes possible. 4
5 In he heorem above, depends on, bu we keep his same on he righ hand side. hus, we don ge everyhing we waned; however i is sufficien o prove he min max heorem. o prove he heorem, firs assume is chosen adversarially o maximize he loss: and define = arg max M(, ) = arg max M(, j) j = =. Noe ha he average of several disribuions is also a disribuion. Now, we know ha max min M(, ) min max M(, ). We wish o show ha max min M(, ) min max M(, ). Consider he following sring of inequaliies: min max M max M () = max = min M (2) max M (3) M (4) M + (5) = = min M + (6) max M +. (7) Here, () is by definiion of he minimum, (2) is by definiion of, (3) is by convexiy: max avg avg max, (4) is by definiion of he adversary, (5) is by he bound of he online algorihm, (6) is by definiion of, (7) is because we maximize over. hus we have ha he min max max min +, wih 0 as. his proves he von Neumann min max heorem. We also learn somehing useful from he algorihm. If we skip he firs inequaliy, hen we see ha: max M(, ) v + where v = max min M(, ). hus, aking he average of he s compued during he algorihm, we ge a disribuion ha is wihin of opimal. If = 0, hen is opimal. 5 (8)
6 hus we ge as close o he maximum as we wish by running he game for more seps. is called he approximae min max sraegy. Similarly we can skip he las inequaliy and show ha is he approximae max min sraegy. As an aside, games ha are no zero-sum are much harder o analyze. In general, one wishes o find a sraegy for finding he Nash equilibrium. he sraegies we are considering are relaed o a class of sraegies ha find anoher equilibrium poin called he correlaed equilibrium. 3 Relaion o Online Learning Le us consider he following learning problem: for =,..., observe x X predic ŷ {0, } observe he rue label c(x ) (we made a misake if ŷ c(x )) Suppose ha we associae expers wih each hypohesis h H, and we wish o do almos as well as he bes hypohesis. We consider X and H o be finie ses. And we would like, similarly o wha we ve seen in he pas: # misakes # misakes of bes h + small. For he given learning problem, we can se up an equivalen game by formulaing a game marix M: M = h h 2. x x 2 where {x i }, i =,..., X are simply indices ino he se of possible insances (no he observaions ha we see in any given round of learning). Similarly we have {h j }, j =,..., H. We fill in his misake marix: { if h(x) c(x) M(h, x) = 0 oherwise and apply he game playing algorihm o he game marix. Given a paricular observaion x, he algorihm chooses a disribuion from he rows. his is a disribuion on he hypoheses. he algorihm chooses a random hypohesis according o his disribuion and applies i o he example given. For he analysis, we have: M(, X ) } {{ } E[# misakes] min M(, x ) + small = min M(h, x ) +small h }{{} # misakes of bes h which, afer all he definiions are properly plugged in, is he same bound as we achieved in he online learning model. 6
7 4 Relaion o Boosing We can consider he basic Boosing learning problem as a game beween wo players: he Boosing algorihm and he weak learning algorihm. for =,..., H = {weak hypoheses} X = raining examples he booser chooses a disribuion D over X he WL algorihm selecs h H such ha: r x D [h (x) c(x)] 2 γ. We canno urn his procedure ino a game readily using he M we derived in he las secion, because he hypohesis space is no longer over he examples. However, by flipping he game board, convering he row player ino he min player, and renormalizing so ha he resuling M values are in [0, ], we can apply he game playing algorihm o he Boosing problem. hus, we use he game marix Now M (x, h) = M = M { if h(x) = c(x) 0 oherwise Our reducion becomes, for boosing, o le D = and is he disribuion concenraed on he h given o us: all he weigh is on one paricular column. By applying he game playing algorihm, we have M (, ) = M (, h ) = r x D [h (x) = c(x)] + γ. (0) 2 And in hree seps we have he guaranees of a Boosing algorihm:. Using he bound given in corollary (2.3): 2 + γ M (, h ) min (9) M (x, h ) + () where he firs inequaliy is due o (0) applied o he individual ieraions. 2. Since he second inequaliy in () applies o he minimum over all (all rows), i applies o all rows: x : = M (x, h ) 2 + γ > /2 where he second inequaliy is rue provided is large enough (so ha < γ). Noe ha M (x, h ) is he fracion of weak hypoheses ha correcly classify x. 7
8 3. Because for any x, he fracion of hypoheses ha correcly classify x approaches a value sricly greaer han /2 we have ha: on all x. MAJ(h (x),, h (x)) = c(x) hus he game formulaion M (x, h) as given in (9) solves he problem of choosing sample weighs in he simple, non-adapive case of Boosing. 8
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