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Noam Nisan Non-price Equilibria in Markets of Discrete goods Avinatan Hassidim, Haim Kaplan, Yishay Mansour, Noam Nisan
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Noam Nisan Market Equilibrium x ij - fraction of good j that player i gets p j - price of good j Each player gets his “demand” Bundle that he prefers most under current prices Market clears demand=supply for each good goodsBuyers / sellers x ij pjpj
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Noam Nisan Main Dogma of Economics Market equilibria exist (…, Arrow-Debreu, …): theorem if “convexity” A Market equilibrium gives an efficient allocation “First welfare theorem” Convexity is a big assumption Does not hold for indivisible goods What happens without it?
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Noam Nisan Simple Market Model for this talk m heterogeneous indivisible items to be allocated among n bidders. Each bidder i has a valuation v i, where v i (S) is his real value for the set S of items. Walrasian equilibrium: prices + allocation 2 items1item 139 2 items1item 127 $4.5
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Noam Nisan Lack of equilibrium 2 items1item 80 2 items1item 66 AND bidder OR bidder AND Wins? one of the prices ≤ 4 OR wants it OR Wins an item? other item has price 0 OR prefers other item So what “happens”?
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Noam Nisan The Market as a Game Every player i makes an offer b ij for each item j Complete information Highest bidder on each item wins it and pays his offer Ties are handled according to a pre-defined tie-breaking rule Utility of player i = v i (S i ) - ∑ j Si b ij
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Noam Nisan Single item case Other equilibria exist (e.g. 6 7 and/or 4 5) Assuming that ties are broken in player 1’s favor Otherwise, no pure equilibrium exists But (6+ , 6, 4) is an -equilibrium V 1 =8 V 2 =6 V 3 =4 b 1 =6 b 2 =6 b 3 =4 An equilibrium A game
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Noam Nisan Pure Equilibria Theorem: Pure Nash equilibria of the game correspond to the Walrasian equilibria. Comment: Exactly so for some tie-breaking rule, - Nash for all tie-breaking rules. Corollary: Any pure equilibrium is efficient (PoA=1). Proof: (W N) Everyone bids eq. prices; winner + (N W) with winning prices 2 items1item 139 2 items1item 127 $4.5
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Noam Nisan Is there always a mixed-Nash equilibrium? Nash theorem does not apply: Continuum of strategies, discontinuous utilities For some games, for some tie breaking rules, there is no exact mixed Nash equilibrium. Simon&Zame ’90 implies that for some (randomized) tie breaking rule a mixed Nash equilibrium exists. Utility functions are continuous except for ties Conjecture: -Nash for every tie-breaking rule
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Noam Nisan Mixed Nash for AND-OR game Proof (for symmetric deviations): Any 0≤t≤1/2 is best-reply for OR: Expected utility = Pr[AND’s-bid ≤ t](v-t) = (v-1/2) = constant Any 0≤t≤1/2 is best-reply for AND: Expected utility = Pr[OR’s-bid ≤ t](1-t) – t = 0 = constant 2 items1item 10 2 items1item vv AND bidder OR bidder (v>0.5( Bids y for each item, with 0≤y≤1/2 according to cumulative distribution: Pr[bid ≤y]=(v-1/2)/(v-y) (atom at 0: Pr[y = 0] = 1-1/(2v)) Bids x for random item, with 0≤x≤1/2 according to cumulative distribution: Pr[bid≤x]=x/(1-x)
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Noam Nisan Price of Anarchy and Stability Mixed Nash equilibria are not always efficient How inefficient? Previous work about PoA of 2 nd price auctions: Christodoulou, Kovacs & Schapira 2008 Lucier & Borodin 2010 Bhawalkar & Roughgarden 2011
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Noam Nisan First Non-Welfare Theorem Consider an AND-OR game with m items AND player has value 1 for the bundle of m items OR player has value v = 1/√m for any single item Theorem: Any equilibrium has welfare ≤ O(lg m/√m) Proof: 1. In eq., AND player never bids a total of more than 1. 2. OR can get utility v-O(lg m/m) by bidding 2/m on random log m items 3. Pr[OR looses in equilibrium] ≤ O(lg m/√m) 4. Social welfare ≤ v + Pr[AND wins] ≤ O(lg m/√m)
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Noam Nisan Approximate First Welfare Theorem Theorem: for every game the social welfare of any mixed Nash equilibrium is at least fraction of the optimum, where: ≤ m, in general ≤ log m, for sub-additive valuations ≤ 2, for sub-modular and XOS valuations This is also true in the Bayesian setting, for Bayes-Nash equilibria.
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Noam Nisan Proof of ≤ 2 for sub-modular case Let us look at a mixed Nash equilibrium EQ. Consider the following deviation for player i: for each item j OPT i bid the median value of the highest other bid for j in EQ. This would win each item with probability ½. Expected value is ≥ v i (OPT i )/2 Uses sub-modularity (or fractional sub-additivity) Expected payment is ≤ ∑ j OPTi E EQ [price j ]
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Noam Nisan Proof (cont.) Since the deviation from last slide cannot be profitable: E EQ [value i ] - E EQ [payment i ] ≥ v i (OPT i )/2 - ∑ j OPTi E EQ [price j ] Summing over all players i: E EQ [SW] - E EQ [Revenue] ≥ SW(OPT)/2 - E EQ [Revenue]
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Noam Nisan Conclusions Looked at Nash equilibria in markets Pure-Nash corresponds to price-based equilibrium Mixed-Nash exists even when no price eq. exist Analyzed mixed-Nash equilibria in some basic cases Unlike pure equilibria, Mixed equilibria may have an efficiency loss We can bound the efficiency loss
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Noam Nisan Further work Within our model: Existence of (ε-) mixed equilibrium for all tie breaking rules? Characterization of all mixed equilibria in our games PoA lower bounds for sub-classes of valuations General program Other market models (budgets, two-sided….) Other “auction” rules More on non-complete information models
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Noam Nisan Thank You!
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