Transcript otws3 7284
Incentive compatibility in 2-sided matching markets Mohammad Mahdian Yahoo! Research Based on joint work with Nicole Immorlica Centralized matching markets Many examples: certain job markets match-making markets auction houses kidney exchange markets Netflix DVD rental market … The objective of the “center” is to find a matching that is optimal from individuals’ perspective. Stable Marriage Consider a set of n women and n men. Each person has an ordered list of some members of the opposite sex as his or her preference list. Let µ be a matching between women and men. A pair (m, w) is a blocking pair if both m and w prefer being together to their assignments under µ. Also, (x, x) is a blocking pair, if x prefers being single to his/her assignment under µ. A matching is stable if it does not have any blocking pair. Example Schroeder Charlie Charlie Linus Franklin Charlie Franklin Linus Schroeder Lucy Peppermint Marcie Sally Schroeder Franklin Charlie Lucy Peppermint Marcie Linus Marcie Sally Marcie Lucy Stable! Linus Franklin Peppermint Sally Marcie Deferred Acceptance Algorithms (Gale and Shapley, 1962) In each iteration, an unmarried man proposes to the first woman on his list that he hasn’t proposed to yet. A woman who receives a proposal that she prefers to her current assignment accepts it and rejects her current assignment. This is called the men-proposing algorithm. Example Schroeder Charlie Charlie Linus Franklin Charlie Franklin Linus Schroeder Lucy Peppermint Marcie Sally Schroeder Franklin Charlie Lucy Peppermint Marcie Linus Marcie Sally Marcie Lucy Stable! Linus Franklin Peppermint Sally Marcie Classical Results Theorem 1. The order of proposals does not affect the stable matching produced by the men-proposing algorithm. Theorem 2. The matching produced by the menproposing algorithm is the best stable matching for men and the worst stable matching for women. This matching is called the men-optimal matching. Theorem 3. In all stable matchings, the set of people who remain single is the same. Applications of stable matching Stable marriage algorithm has applications in the design of centralized two-sided markets. For example: National Residency Matching Program (NRMP) since 1950’s Dental residencies and medical specialties in the US, Canada, and parts of the UK. New York school match National university entrance exam in Iran Placement of Canadian lawyers in Ontario and Alberta Sorority rush Matching of new reform rabbis to their first congregation … Incentive Compatibility Question: Do participants have an incentive to announce a list other than their real preference lists? Answer: Yes! In the men-proposing algorithm, sometimes women have an incentive to be dishonest about their preferences. Example Schroeder Charlie Charlie Linus Franklin Charlie Franklin Linus Schroeder Lucy Peppermint Marcie Sally Schroeder Franklin Charlie Lucy Peppermint Marcie Linus Marcie Sally Stable! Marcie Lucy Linus Franklin Peppermint Sally Marcie Incentive Compatibility Next Question: Is there any truthful mechanism for the stable matching problem? Answer: No! Roth (1982) proved that there is no mechanism for the stable marriage problem in which truthtelling is the dominant strategy for both men and women. However, data from NRMP show that the chance that a participant can benefit from lying is slim. 1993 1994 1995 1996 # applicants 20916 22353 22937 24749 # positions 22737 22801 22806 22578 # applicants who could lie 16 20 14 21 Number of applicants who could lie can be computed using the following theorem. Theorem. The best match a woman can receive from a stable mechanism is her optimal stable husband with respect to her true preference list and others’ announced preference lists. In particular, a woman can benefit from lying only if she has more than one stable husband. Explanations (Roth and Peranson, 1999) The following limit the number of stable husbands of women: Preference lists are correlated. Applicants agree on which hospitals are most prestigious; hospitals agree on which applicants are most promising. If all men have the same preference list, then everybody has a unique stable partner, whereas if preference lists are independent random permutations almost every person has more than one stable partner. (Knuth et al., 1990) Preference lists are short. Applicants typically list around 15 hospitals. A Probabilistic Model Men choose preference lists uniformly at random from lists of at most k women. Women randomly rank men that list them. Conjecture (Roth and Peranson, 1999): Holding k constant as n tends to infinity, the fraction of women who have more than one stable husband tends to zero. Our Results Theorem. Even allowing women arbitrary preference lists in the probabilistic model, the expected fraction of women who have more than one stable husband tends to zero. Economic Implications Corollary 1. When other players are truthful, almost surely a given player’s best strategy is to tell the truth. Corollary 2. The stable marriage game has an equilibrium in which in expectation a (1-o(1)) fraction of the players are truthful. Corollary 3. In stable marriage game with incomplete information there is a (1+o(1))-approximate Bayesian Nash equilibrium in which everybody tells the truth. Structure of proof Step 1: An algorithm that counts the number of stable husbands of a given woman. Step 2: Bounding the probability of having more than one stable husband in terms of the number of singles Step 3: Bounding the number of singles by the solution of the occupancy problem. Step 1: Finding stable husbands of g Use men-proposing algorithm to find a stable matching. Whenever the algorithm finds a stable matching, have g divorce her husband and continue the men-proposing algorithm (but now g has a higher standard for accepting new proposals). Terminate when either a man who is married in the men-optimal matching runs through his list, or a woman who is single in the men-optimal matching receives a proposal. Question: If each woman has an arbitrary complete preference list, and each man has a random list of k women, what is the probability that this algorithm returns more than one stable husband for g? The main tool that we will use to answer this question is the principle of deferred decisions: Men do not pick the list of their favorite women in advance; Instead, every time a man needs to propose, she picks a woman at random and proposes to her. A man remains single if he gets rejected by k different women. Stable! End! Schroeder Charlie Franklin Linus Charlie Linus Franklin Schroeder Charlie Schroeder Linus Franklin Lucy Peppermint Marcie Sally Schroeder Franklin Charlie Lucy Peppermint Linus Marcie Sally Sally Marcie Lucy Linus Schroeder Franklin Charlie Marcie Sally Lucy Step 2: Bounding the probability Consider the moment when the algorithm finds the first (i.e., men-optimal) matching. Call this matching μ. Let A denote the set of women who are single in μ, and X denote |A| . Fix random choices before the algorithm finds μ, and let probabilities be over random choices that are made after that. Step 2, cont’d. Look at the sequence of women who receive a proposal. The probability that the algorithm finds another stable husband for g is bounded by the probability that g comes before all members of A in this sequence. This probability is 1/(X+1). Therefore, the probability that g has more than one stable husband is at most Step 3: Number of singles We need to compute E[1/(X+1)], where X is the number of singles in the menoptimal matching. Simple Observation: The probability that a woman remains single is at least the probability that she is never named by men. Step 3, cont’d. Let Ym,n denote the number of empty bins in an experiment where m balls are dropped independently and uniformly at random in n bins. Lemma. Proof Sketch: Assume (without loss of generality!) that men are amnesiacs and might propose to a woman twice. The total number of proposals (bins) is at most (k+1)n w.h.p. The occupancy problem Lemma. Proof sketch: Use the principle of inclusion and exclusion to compute E[1/(Ym,n+1)] as a summation. Compare this summation to another (known) summation term-by-term. Putting it all together… Theorem. In the model where women have arbitrary complete preference lists and men have random lists of size k, the probability that a fixed woman has more than one stable husband is at most Generalizations More general classes of distributions: Arbitrary non-uniform distribution instead of the uniform distribution: still we can prove that the probability tends to zero. Many-to-one matchings: [Kojima & Pathak]: result generalizes. Open Questions Stable matching with couples: Why has the NRMP algorithm found a matching every year? Restricting to complete preference lists: There are similar observations about the probability that a participant can benefit from lying. (Teo, Sethuraman, Tan, 2001)