Economics and Computation Week 6: Assignment games Prof. Pingzhong Tang kenshin@tsinghua.edu.cn http://iiis.tsinghua.edu.cn/~kenshin/ec Fall 2016 9/22/2018 Pingzhong Tang

References Two-sided matching A study in game theoretical modeling and analysis Alvin Roth & Marilda Sotomayor Chapters 7 (one seller and many buyers) and 8 (assignment games) 9/22/2018 Pingzhong Tang

One seller and many buyers A market (thought of as an auction) One seller s Indivisible item for sale Values at rs: will not sell the item below price rs Interpretation: has an outside option that worth rs A set of buyers Each buyer b values it at rb: will not buy it above rb Transaction happens between b and s at price p If, no other monetary transfers are made Then, s gains p (not p- rs), b gains rb -p 9/22/2018 Pingzhong Tang

Transferable utility model Money is freely transferable Not only possible between seller and buyers But also possible among buyers! Utility of a buyer can be anything A rigorous game-theoretical analysis is in need Cooperative game theory Stability, core Spoiler alert: At stable outcomes, no money transfer for losers Losers always get utility 0 9/22/2018 Pingzhong Tang

Formulation A cooperative game N={1, …, n, n+1} 1 to n are buyers, n+1 is the seller ri is the valuation of i A subset S of N is called a coalition Coalition function: v(S), for each S The maximum amount of utility S can create v(S)=0, if n+1 not in S, otherwise v(S)=max{ri|i in S}, v(N)=max{ri|i in N} Knowing v(S) summarize possible ways S divides the money Money transferable after transaction 9/22/2018 Pingzhong Tang

Outcome(=Payoff) of the game Payoff vectors: The amount of money each player ends up with Notation: x=(x1,x2,…,xn+1) x(S)=∑xi|i in S Feasibility constraints X={x=(x1,x2,…,xn+1) in Rn+1 |x(N)≤v(N)} Question: why don’t we have x(S)≤v(S) for each S? 9/22/2018 Pingzhong Tang

Core Define a payoff vector x dominates y If there exists a coalition S, such that xi>yi, for all i in S, and x(S)≤v(S) In other words, every member of S prefer x to y S can afford x (can create x on their own) Two vectors can dominate each other Core is the set of undominated payoff vectors Theorem: Core={x in X|x(S)≥v(S) for all S } If for some S, we have a <, easily construct y that dominates x Can be empty 9/22/2018 Pingzhong Tang

Stability A payoff vector is stable, if It is individually rational (nonnegative) There does not exist a buyer i and price p such that p > xn+1 ri-p > xi if such i and p exists, they will both deviate and block x i and n+1 will make transaction, both better off than at x IOW, stability =core constraints applied to |S|=1, 2 9/22/2018 Pingzhong Tang

Main theorem Renaming N: {1*, 2*, …, (n+1)*} Theorem: so that ri*≥ri+1* Theorem: 1. The core is always non-empty 2. If n+1 ≠ 1*, and r1*≥r2*, Core={x in X| such that xn+1=p and x1*=r1*-p for r1*≥ p ≥r2*, and xi=0 for other} Otherwise, Core={(0, …,0, v(N))} 3. The set of stable vectors equals core 9/22/2018 Pingzhong Tang

Statement 1 Non-emptyness of the core (0, …,0, v(N)) is always in the core Easy to check: every S, x(S) no less than v(S) 9/22/2018 Pingzhong Tang

Statement 2 If n+1 ≠ 1*, and r1*≥r2*, Core={x in X| such that xn+1=p and x1*=r1*-p for r1*≥ p ≥r2*, and xi=0 for loser}. Otherwise, Core={(0, …,0, v(N))} If unique highest buyer, core divides the benefit between seller and this buyer. Losers gets nothing. Intuition of proof If non-zero for loser, S={1*, n+1} dominated If p less than r2*, S={2*, n+1} dominated 9/22/2018 Pingzhong Tang

Statement 3 Core  Stability Not core  not stable By statement 2, when the price is above the second highest price, no other buyer wants to buy the item Not core  not stable Then either not individually rational Or {1*, n+1} dominated (transfer losers’ payoff to {1*, n+1} ) Thus, unstable 9/22/2018 Pingzhong Tang

Truthfulness Define a stable matching mechanism to be a function h that selects, for every valuation input r, a payoff vector h(r) in the core Theorem: No stable matching mechanism exists for which truthful report of valuation is a dominant strategy for every agent 9/22/2018 Pingzhong Tang

Proof If exists unique highest buyer By previous theorem, core = continuum set Let h(r) be in the core, and is not buyer optimal buyer 1* obtains the item at p>r2* 1* could have reported p>r’>r2* Since h selects member in the core, in h(r’), 1* obtains the item at price less or equal to r’. Better off for 1* Let h(r) be in the core and is not seller optimal buyer 1* obtains the item at p<r1* Seller could have reported p<r’≤r1* 9/22/2018 Pingzhong Tang

Truthfulness There does exist a stable matching mechanism in which truthful report of valuation is a dominant strategy for every buyer Why? Exercise, check whether third price auction satisfies this theorem. 9/22/2018 Pingzhong Tang

Assignment games Similar to the previous model But with multiple sellers P: set of m buyers; Q: set of n sellers αij: how much buyer i values item of seller j Seller has valuation 0 of her own item Thus, the coalition function v is given by V({i,j})= αij for i in P and j in Q. V(S)=0, if S consists only of P agents or Q agents V(S)= value of maximum matching in S 9/22/2018 Pingzhong Tang

Payoff vectors Payoff vector (u,v) in Rm x Rn If no other monetary transfer are made or received by i or j: ui= αij-p vj=p Buyer i purchased item from seller j at price p Seller’s valuation is normalized to 0. WLOG. Again, money is freely transferable Question: find core/stable outcome of this game 9/22/2018 Pingzhong Tang

Worth of P∪Q v(P∪Q) Find via LP Maximize ∑ αijxij Maximum total payoff achievable At core, ∑ui+∑vi= v(P∪Q) Find via LP Maximize ∑ αijxij Subject to (a) ∑i xij≤1 (b) ∑j xij≤1 (c) xij ≥ 0 9/22/2018 Pingzhong Tang

About the LP Guarantee integer solution xij is either 0 or 1 Denote whether there is a transaction between (i, j) v(P∪Q) = ∑ αijxij where x is the solution of LP 9/22/2018 Pingzhong Tang

Feasible and optimal assignment Feasible: integer x that satisfies a), b), c) Optimal: feasible and is solution of the LP Optimal assignment always exists, not unique 9/22/2018 Pingzhong Tang

Feasible payoff vectors A payoff vector (u,v) is feasible, if there exist a feasible assignment x such that ∑ui+∑vi= ∑ αijxij In this case, we say (u,v) and x are compatible and ((u,v),x) is a feasible outcome 9/22/2018 Pingzhong Tang

Stable outcomes Definition: A feasible outcome ((u,v),x) is stable, if for all (i, j): 1) ui, vj ≥ 0 2) ui+vj ≥ αij Lemma: Let ((u,v),x) be a stable outcome, then 1) ui+vj = αij for all pairs matched in x (i.e., xij=1) 2) ui, vj = 0 for unassigned i,j Proof ∑ui+∑vi= ∑ (ui+vj) xij|xij=1 + ∑i unmachedui + ∑j unmachedvj = ∑ αijxij Now apply stability. QED. Again, losers get 0 payoff in stable matching First =, by decomposition, second = by feasibility ∑ui+∑vi= ∑ αijxij 9/22/2018 Pingzhong Tang

Core outcomes Dual of the maximum matching LP: Minimize ∑ui+∑vi Subject to a) ui, vj ≥ 0 b) ui+vj ≥ αij By strong duality: If x is an optimal assignment (solution of primal) and (u,v) is a solution of the dual, we have ∑ui+∑vi = ∑ αijxij = v(P∪Q) Thus, 1) ((u,v) ,x) is feasible and optimal. 2) ((u,v) ,x) is stable. 9/22/2018 Pingzhong Tang

The core - continued Minimize ∑ui+∑vi Subject to a) ui, vj ≥ 0 b) ui+vj ≥ αij b) says ui+vj ≥ v({i,j})  for any S: ∑ui+∑vi ≥ v(S) This is exactly the constraints of core! i) ∑ui+∑vi = ∑ αijxij = v(P∪Q) ensures feasibility ii)∑ui+∑vi ≥ v(S) ensures non-dominance for any S Or simply by the theorem… Conversely, any payoff in the core is a solution of the dual 9/22/2018 Pingzhong Tang

To summarize… Theorem (Shapley and Shubik, 1972). In any assignment game, The set of stable outcomes and the core are the same The core is the set of solutions of the dual At the core/stable outcomes, primal maximized, dual minimized, sum of payoffs equals the value of maximum matching In other words, any algorithmically optimal solution satisfies desirable economic properties! 9/22/2018 Pingzhong Tang

Corollaries Corollary: If x is an optimal assignment, it is compatible with any stable/core payoff. In stable payoff, ∑ui+∑vi = ∑ αijxij = v(P∪Q), for any optimal x Corollary: if ((u,v),x) is a stable outcome, then x is optimal. Again, ∑ui+∑vi = v(P∪Q)= ∑ αijxij First =, feasible, second feasible 9/22/2018 Pingzhong Tang

More corollaries Corollary: Let ((u,v),x) and ((u’,v’),x’) be stable outcomes of an assignment game. Then if x’ij=1, ui’>ui implies vj’<vj. Interpretation: opposed interests between seller and buyer. Proof: Suppose otherwise vj’≥vj. Then, αij = u’i+v’j > ui+vj = αij Corollary: The core forms a lattice. Exists a P optimal stable payoff. Exists a Q optimal stable payoff. = is from a previous Lemma. The order on the lattices is defined by relations on the two stable outcomes. 9/22/2018 Pingzhong Tang