1. Market Design and Analysis Lecture 2 Lecturer: Ning Chen ( 陈宁 ) Email: ningc@ntu.edu.sg

2. 2 Extension One – Different # of Men and Women  When the numbers of men and women are different  Gale-Shapley algorithm still generates a men-optimal or women-optimal stable matching  all other results still hold m 2 > m 1 > m 3 \ m 1 > m 3 > m 2 \ \ \\ w 1 > w 2 w 2 > w 1 w 1 > w 2 m1m1 m2m2 m3m3 w2w2 w1w1 \

3. 3 Extension Two – Incomplete Preference Lists  The preference list of a man/woman may not be complete, i.e., it only contains a partial list of the opposite side.  If w is not on m’s preference list, then m prefers “single” (i.e. unmatched) to w  If m is not on w’s preference list, then w prefers “single” (i.e. unmatched) to m

4. 4 Extension Two – Incomplete Preference Lists  A matching f is called individually rational if for any f(m) = w, both m and w are on each other’s preference list.  All matchings considered are individually rational  otherwise, it is “blocked” by an individual who is matched to a partner that he/she doesn’t prefer.

5. 5 Extension Two – Incomplete Preference Lists  For any given matching f, if m (or w) is not matched to any one, denote f(m) = Ø (or f(m) = Ø).  For any w on m’s preference, we say w > m Ø  For any m on w’s preference, we say m > w Ø  M atching f is called stable if it does not contain any blocking pair (m,w) where w > m f(m) and m > w f(w).

6. 6 Extension Two – Incomplete Preference Lists  G-S algorithm still generates a stable matching  All other results still hold m 2 > m 3 > m 1 > m 4 \ m 4 > m 2 > m 3 \\ \ \ \ \ w2w2 w1w1 w3w3 m1m1 m2m2 m3m3 m4m4 w 1 > w 2 w 3 > w 2 w 1 > w 3 > w 2

7. 7 Decomposition Theorem  Theorem 3.1. Let f and f’ be two stable matchings of a given market. Let  M(f) be the subset of men who prefer f to f’  M(f’) be the subset of men who prefer f’ to f  W(f) be the subset of women who prefer f to f’  W(f’) be the subset of women who prefer f’ to f Then f and f’ map M(f’) onto W(f) and vice versa, and M(f) onto W(f’) and vice versa. M(f) M(f’) W(f’) W(f) f, f’ m f’(m) f’ f w = f(m) w’ =

8. 8 Decomposition Theorem  Proof.  Suppose m ∈ M(f). Then w := f(m) > m f’(m). This implies that f’(w) > w f(w) = m. That is, w ∈ W(f’), i.e., f(M(f)) ⊆ W(f’)  On the other hand, suppose w ∈ W(f’). Then m := f’(w) > w f(w). This implies that f(m) > m f’(m). That is, m ∈ M(f), i.e., f’(W(f’)) ⊆ M(f)  Hence, f and f’ map M(f) onto W(f’) m f’(m) f’ f w = f(m) f’ f(w) f m = w f’ f’(w) f f(m) f’(w)

9. 9 Application of Decomposition Theorem  Theorem 2.3. For any two stable matchings f and f’, f > M f’ if and only if f’ > W f.  Proof. If f > M f’, then M(f’) = Ø = W(f). That is, all women prefer f’, i.e., f’ > W f On the other hand, if f’ > W f, then W(f) = Ø = M(f’). That is, all men prefer f, i.e., f > M f’ M(f) M(f’) W(f’) W(f) f, f’

10. 10 Application of Decomposition Theorem  Theorem 3.2. For any given market, the set of people who are single is the same for all stable matchings.  Proof. Suppose m is matched under f’ but not f. Then m ∈ M(f’) and W(f) ≠Ø. Since f maps W(f) onto M(f’), m should be matched under f as well, a contradiction. M(f) M(f’) W(f’) W(f) f, f’

11. 11 Solution Concept – Pareto Optimality  Informally, a solution (i.e., outcome) is Pareto optimal (or Pareto efficient) if there is no other solution such that no one gets worse off and at least one gets better off.  That is, it is impossible to improve one’s benefit without hurting others.

12. 12 Pareto Optimality in Game Theory  In game theory, a strategy profile is Pareto optimal if there is no other profile such that no one gets worse off and at least one gets better off (in terms of their payoff).

13. 13 Pareto Optimality in Game Theory  Does a Pareto optimal strategy profile always exist?  Yes. Consider any strategy profile S = (s 1,…,s n )  If S is not Pareto optimal, then there is another profile S’ = (s’ 1,…,s’ n ) such that the payoff of all players in S’ is at least that in S and someone is strictly better.  Keep doing this and it will stop at some point when we could not find another “better” strategy profile, since the total number of profiles is finite.  (Use a potential function F, which is sum of total payoffs of all players, to show convergence of the above process.)

14. 14 Pareto Optimality in Game Theory  What’s the relation between Pareto optimal strategy and pure Nash equilibrium (if one exists)?  Pure Nash equilibrium may not be Pareto optimal  Pareto optimal may not be a pure Nash equilibrium cooperatedefect cooperate-1, -1-10, 0 defect0, -10-5, -5

15. 15 Pareto Optimality in Stable Marriage  In a stable marriage problem, a matching is Pareto optimal if there is no other matching such that no one gets worse off and at least one gets better off (in terms of their preference).  Does a Pareto optimal matching always exist?  What’s the relation between Pareto optimal matching and stable matching?

16. 16 Pareto Optimality in Stable Marriage  Theorem 3.3. Any stable matching is Pareto optimal.  Proof. Consider any stable matching f. If f is not Pareto optimal, then there is f’ such that no one gets worse off and at least one gets better off. Assume wlog that m get better off. Denote w = f(m) and w’ = f’(m), then m prefers w’ to w. Assume m’ = f(w’). Since no one gets worse off in f’ than f, w’ should prefer m to m’ as well, which implies that (m,w’) is a blocking pair for f, a contradiction. w’ wm m’ f f f’

17. 17 Pareto Optimality in Stable Marriage  Question. Are Pareto optimal matchings always stable? m 2 > m 1 m 1 > m 2 w2w2 w1w1 w 1 > w 2 m1m1 m2m2

18. 18 Solution Concept – Core  Given any two possible outcomes x and y of a game, we say that x dominates y if there is a coalition of players A such that  every member of the coalition A prefers x to y.  the rule of the game allows the coalition A to enforce x.  Example. Two workers  if they work separately (outcome x), then each get $10 and $5  if they work together (outcome y), then they get $16 in total (and share the profit evenly)  If the outcome is y, then first guy will choose x..  The core of a game is the set of undominated outcomes.

19. 19 Solution Concept – Core  In matching market, a matching f’ dominates another matching f if there is a subset of agents A ⊆ M ∪ W such that for any man m and woman w in A,  f’(m) ∈ A and f’(w) ∈ A  f’(m) > m f(m) and f’(w) > w f(w)  Core is “stronger” than stability in matching markets  core: no group of agents can improve  stability: no (m,w) pair can improve

20. 20 Core = Stability  Theorem 3.4. The core of the marriage market equals the set of stable matchings.  Proof. Consider any matching f.  Assume that f is in the core. If f is blocked by (m,w) with w > m f(m) and m > w f(w), then it is dominated via the coalition S = {m,w} by any matching f’ with f’(m) = w, a contradiction. Hence, f is stable.  Assume that f is not in the core. Then f is dominated by a matching f’ via coalition A. Consider any m ∈ A, let w = f’(m) ∈ A. By the definition of domination, w > m f(m) and m > w f(w) That is, (m,w) is a blocking pair of f, i.e., f is not stable.

21. 21 G-S Algorithm – Strategic Consideration  In Gale-Shapley algorithm, individuals provide their preferences, and the algorithm gives a stable matching. That is,  input: preferences lists (private info of every individual)  output: stable matching (via match maker)  different inputs give different outputs  Can any individual get a better assignment if they lie on their preference?

22. 22 Example m 1 > m 2 > m 3 m 1 > m 3 > m 2 w 1 > w 2 > w 3 w 1 > w 3 > w 2 m 1 > m 2 > m 3 m2m2 m1m1 m2m2 m3m3 w2w2 w1w1 w3w3

23. 23 Strategic Consideration  Is honesty the best policy?  Gale-Shapley algorithm, no!  What exactly are the incentives to behave untruthfully? Which agents?  Are there other ways of organizing the market so that honesty would always be the best policy?

24. 24 Formal Strategic Model  Players: M and W  Private information: preference list P(m) for m ∈ M and P(w) for w ∈ W  Strategy space: all possible preferences  Matching mechanism F  players submit preferences Q = (Q(x)) x ∈ M ∪ W (Note that Q(x) can be either same or different as P(x))  mechanism decides on a matching f = F(Q)

25. 25 Best Response and Nash Equilibrium  For any agent x and fixed strategies of other agent s (denoted by Q -x ), the best response of x is a preference Q(x) such that for any other preference Q’(x), we have f(x) ≥ x f’(x), where  f(x) = F(Q -x,Q(x))  f’(x) = F(Q -x,Q’(x))  A strategy profile (Q(x)) x ∈ M ∪ W is called a Nash equilibrium if it is a best response for everyone.

26. 26 Dominant Strategies  A dominant strategy for agent x is a strategy Q(x) that is a best response to all possible sets of strategies of other agent s.  A matching mechanism is called truthful (or incentive compatible, or strategy proof) if it is a dominant strategy to state true preference for all agents.

27. 27 Solution Concepts  Provide a benchmark for fairness, efficiency etc.  Nash equilibrium: a stable state where no single individual is willing to change  Stable matching: a stable state where no pair of man and woman are willing to change  Core: a stable state where no subset of individuals is willing to change  Pareto optimality: a state where all individuals as a whole do not want to change (i.e., for any other state where someone is better off, then at least one is worse off)

28. 28 Solution Concepts  Provide a tool to analyze selfish behaviors of individuals  best response: a best possible “move” given that all others have selected their strategies always exists  dominant strategy: a best possible “move” for any strategies selected by others may not exist  truthfulness: reporting true private info is a dominant strategy for all individuals  there is a dominant strategy for all individuals design truthful protocol

29. 29 Matching Mechanisms  F is called a stable matching mechanism if for all submitted strategies Q, f = F(Q) is a stable matching with respect to Q.  Are there any truthful stable matching mechanisms?  F is called a Pareto-optimal matching mechanism if for all submitted strategies Q, f = F(Q) is Pareto- optimal with respect to Q.  Are there any truthful Pareto-optimal matching mechanisms?

30. 30 Truthful Pareto-Optimal Matching Mechanism  Mechanism: given submitted preferences Q from all men and women  all women are free  order all men arbitrarily, say m 1,…,m n  for i = 1,…,n match m i his most preferred available woman according to Q(m i )

31. 31 Truthful Pareto-Optimal Matching Mechanism  Theorem 3.5. The above mechanism is truthful and generates a Pareto optimal matching  Proof. Assume all men are ordered by m 1,…,m n  Truthfulness. For any fixed reported preferences of other agents, consider any man m i The mechanism first decides matches for m 1,…,m i-1 When the mechanism reaches m i, the best possible result is to match him with the most preferred free woman. Reporting true preference will have such match.

32. 32 Truthful Pareto-Optimal Matching Mechanism  Theorem 3.5. The above mechanism is truthful and generates a Pareto optimal matching  Proof. Assume all men are ordered by m 1,…,m n  Pareto optimality. For m 1, there is no other matching that m 1 can get a better partner Given assignment of m 1 (i.e., without hurting m 1 ), there is no other matching that m 2 can get a better partner Given assignment of m 1 and m 2, there is no other matching that m 3 can get a better partner … Given assignment of m 1,m 2,…,m n-1, there is no other matching that m n can get a better partner

33. 33 Truthful Pareto Optimal Matching Mechanism  Mechanism: given submitted preferences Q from all men and women  all women are free  order all men arbitrarily, say m 1,…,m n  for i = 1,…,n match m i his most preferred available woman according to Q  Is the mechanism stable?  Any truthful stable mechanism? m 2 > m 1 m 1 > m 2 w 1 > w 2 m1m1 m2m2 w2w2 w1w1

34. 34 Truthful Stable Matching Mechanism  Theorem 3.6. [Roth] (Impossibility theorem) No stable matching mechanism exists for which stating the true preferences is a dominant strategy (i.e. truthful) for very man and woman.

35. 35 Truthful Stable Matching Mechanism  Idea of the proof.  We construct an example, where there are only two stable matchings f 1 and f 2  Any stable matching mechanism must output either f 1 or f 2 (given that everyone tells truth)  If it outputs f 1, there is a person who wants to lie  If it outputs f 2, there is a person who wants to lie  Hence, no matter what the matching that the mechanism outputs, there always exists a person who wants to lie

36. 36 Truthful Stable Matching Mechanism  Proof. Consider the following setting: There are two stable matchings f 1 and f 2 w 1 > w 2 w 2 > w 1 m 2 > m 1 m 1 > m 2 m1m1 m2m2 w2w2 w1w1 m1m1 m2m2 w2w2 w1w1 m1m1 m2m2 w2w2 w1w1

37. 37 Truthful Stable Matching Mechanism  So any mechanism that outputs a stable matching must output either f 1 or f 2, if everyone reports its true preference.  Suppose it outputs f 1. Note that if w 1 unilaterally changes her stated preference to m 2 (i.e. w 1 only prefers m 2 ), then the only stable matching is f 2 and w 1 will get m 2, a better assignment.  Similarly, if the table matching algorithm outputs f 2, m 2 can unilaterally increase his assignment by lying. w 1 > w 2 w 2 > w 1 m 2 > m 1 w2w2 w1w1 m1m1 m2m2 m 1 > m 2 m2m2

38. 38 Men-Opt SM Mechanism – for Men  Recall that in G-S algorithm with men proposing, it returns the unique men-optimal stable matching (for the given stated preferences).  Theorem 3.7. [Roth]  Men-optimal stable matching mechanism (i.e.,. G-S algorithm with men proposing) is truthful for every man.  Women-optimal stable matching mechanism is truthful for every woman.

39. 39 Men-Opt SM Mechanism – for Men  Theorem 3.8. [Dubins and Freedman] Assume that f is the men-optimal stable matching when all men and women state their true preferences. Let f’ be the men-optimal stable matching when a coalition of men S ⊆ M lie on their preferences. Then there is m ∈ S such that f(m) ≥ m f’(m).  That is, for any coalition of men, it cannot be the case that all men in the coalition get strictly improved assignment in men-optimal stable matching mechanism.

40. 40 Men-Opt SM Mechanism – for Men m 1 > m 2 > m 3 m 3 > m 1 > m 2 m 1 > m 2 > m 3 w 2 > w 1 > w 3 w 1 > w 2 > w 3 w2w2 w1w1 m1m1 m3m3  For any coalition S of men,  impossible that all men in S are better off (by Theorem 3.8)  It is possible that some in S get better and others get the same w3w3 m2m2 w3w3 w 1 > w 2 > w 3 w 2 > w 1 > w 3

41. 41 Men-Opt SM Mechanism – for Men  Theorem 3.8. [Dubins and Freedman] Assume that f is the men-optimal stable matching when all men and women state their true preferences. Let f’ be the men-optimal stable matching when a coalition of men S ⊆ M lie on their preferences. Then there is m ∈ S such that f(m) ≥ m f’(m).  Theorem 3.8 implies Theorem 3.7 if we consider any man m ∈ M and let S = {m} in the statement of Theorem 3.8 (then we know that f(m) ≥ m f’(m), i.e., m cannot get a better assignment by lying)

42. 42 Blocking Lemma  Blocking Lemma. For a given instance with set of men M and set of women W, let f be the men-optimal stable matching. Let f’ be any unstable matching and S be the subset of men who prefer f’ to f. If S ≠ Ø, then there is a blocking pair (m,w) for f’ such that m ∈ M-S and w ∈ f’(S). S m M-S f’(S) w f’

43. 43 Men-Opt SM Mechanism – for Men  Theorem 3.8. [Dubins and Freedman] Assume that f is the men-optimal stable matching when all men and women state their true preferences. Let f’ be the men-optimal stable matching when a coalition of men S ⊆ M lie on their preferences. Then there is m ∈ S such that f(m) ≥ m f’(m).

44. 44 Men-Opt SM Mechanism – for Women  In men-optimal stable matching mechanism  stating true preference is dominant strategy for all men (Theorem 3.7)  it is possible that women get better assignment when lying  How will women behave in such mechanism?  Next, when considering men-optimal stable matching mechanism, assume all men state their true preferences.  For simplicity, only consider Gale-Shapley model (i.e., same number of men and women, complete true preferences). All claims still hold for general settings.

45. 45 Men-Opt SM Mechanism – for Women  For a given matching mechanism, a set of stated preferences Q = (Q(x)) x ∈ M ∪ W is a Nash equilibrium if for very individual x ∈ M ∪ W, Q(x) is his/her best response, given that other individuals will choose their strategies according to Q.  In men-optimal stable matching mechanism  any Nash equilibrium exists for women?  if yes, is it a stable matching w.r.t. true preferences?

46. 46 M-Opt SM Mechanism – for Women  Theorem 3.9. Let f be any stable matching for the instance (M,W; P), where P is true preferences. Suppose that for each woman w ∈ W, she lists only f(w) on her stated preference list, i.e., Q(w) = {f(w)}. Then this is a Nash equilibrium induced by men-optimal stable matching mechanism on Q and f is the only stable matching with respect to Q.  The theorem implies all women can get the women- opt stable matching with respect to (M,W; P) in the men-opt stable matching mechanism f(w)w f P(w)  f(w)w f Q(w): f(w)

47. 47 M-Opt SM Mechanism – for Women  Proof. We first show f is the men-optimal stable matching for Q  f is a stable matching for Q (o/w, f is not stable for P)  f is the only stable matching for Q o/w, there is another stable matching f’ there is a women in w ∈ W who is free in f’ (since every woman only lists Q(w) = {f(w)}) By Theorem 3.2 (set of singles is the same for all stable matchings), w should be free in f as well, a contradiction  Hence, f is the men-optimal stable matching for Q

48. 48 M-Opt SM Mechanism – for Women  Proof. Next we show Q is a Nash equilibrium (with output f)  all men state their true preferences (dominant strategy)  if not Nash equilibrium for all women, there is w such that stating Q’(w) (denote the resulting preferences by Q’) gives her better partner m’. That is, in the men optimal stable matching f’ of Q’, f’(w) = m’ and m’ > w f(w).  Let w’ = f(m’). Then w’ > m’ w and w’ is free in f’. Thus (m’,w’) is a blocking pair for f’, a contradiction. f(w) m’ w’ w f’ f f

49. 49 M-Opt SM Mechanism – for Women  Given strategic behaviors of women, when men- optimal stable matching mechanism is employed, will the resulting matching always stable w.r.t. the true preferences?  Theorem 3.10. In men-optimal stable matching mechanism, suppose that all women choose an equilibrium strategy Q = (Q(w)) w ∈ W, and let f be the resulting men-optimal stable matching with respect to Q. Then f is a stable matching for the instance (M,W; P), where P is their true preferences.

50. 50 M-Opt SM Mechanism – for Women  Proof. Assume o/w that f is not stable for P and let (m,w) be a blocking pair.  Consider running G-S algorithm on preferences Q (which returns f): since (m,w) is a blocking pair, then w > m f(m) m must have proposed to w in the algorithm w rejects m at some point

51. 51 M-Opt SM Mechanism – for Women  Proof. Assume o/w that f is not stable for P and let (m,w) be a blocking pair.  If w reports another preference Q’(w) = {m} (denote the resulting preferences by Q’), consider running G-S algorithm again: the process will be the same until the point when m proposes to w but w will never reject m so in the returned matching, w is matched to m  Since m > w f(w), this contradicts to the fact that Q is a Nash equilibrium

52. 52 M-Opt SM Mechanism – for Women  Theorem 3.9 says any stable matching w.r.t. the true preferences can be obtained by an equilibrium set of strategies of women  Theorem 3.10 says any equilibrium procudes a matching that is stable w.r.t. the true preferences  That is, the women cannot get too greedy, the best possible assignment that every woman can get in an equilibrium is whom?