The key is that there are two disjoint populations, and everyone in the market is on either one side or the other

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1 Econ 805 Advanced Micro Theory I Dan Quint Fall 2009 Lecture 17 So... two-sided matching markets. First off, sources. I ve updated the syllabus for the next few lectures. As always, most of the papers are available online. But if you re interested in the topic, I highly recommend picking up the Roth and Sotomayor book, Two-Sided Matching it s a paperback, it s under $40, and it covers most of what we ll be doing the next few weeks. Also, if you re interested in the topic, Al Roth s website (just google Al Roth ) has a staggering amount of material papers, references, overviews, applications, and so on. Check it out. My plan, roughly, is to spend today and Thursday on one-to-one markets; next Tuesday on extensions of the model; and next Thursday on applications. Then student presentatiosn start. Motivation The topic generally is two-sided markets Can be men and women, as in the Gale and Shapley paper Can be firms and workers in a labor market Or schools and applicants in a college admissions model (Classic examples: matching newly-minted MDs to residency programs at hospitals, which uses a centralized matching process; and matching law school graduates to clerk jobs with judges, which does not; New York and Boston public schools; econ PhDs and academic jobs.) The key is that there are two disjoint populations, and everyone in the market is on either one side or the other (We can model buyers and sellers this way, but it must be that buyers are only buyers, sellers are only sellers. We ll see an example later today of why we can t use these techniques to analyze one-sided markets, that is, settings where any player can match with any other player.) For today and Thursday, we ll focus on the one-to-one matching case, where each player wants to match with at most one player from the other side of the market. We ll refer to the two sides as men and women. 1

2 Start off with A set M of men m 1,..., m a A set W of women w 1,..., w b (generally a = b, but doesn t matter) Each man m i has strict preferences over mates W Allowing for indifferences actually complicates things a lot for the baseline model, we ll assume preferences are all strict It s customary to indicate matching with no one as matching with yourself, so a man i s preferences might be w 3 m1 w 1 m1 m1 m i m1 w 5 m1 w 2 The Roth and Sotomayor book states preferences by just listing the options in order of preference, e.g., P (m i ) = w 3, w 1,, m i, w 5, w 2 Similarly, each woman has strict preferences over potential mates (including staying alone) We refer to potential mates as acceptable if they re preferable to being alone (matching to yourself) So that s the basic environment. As a first step, we ll put aside the mechanism design problems of eliciting peoples preferences and mapping them to outcomes, and look only at outcomes. The types of outcomes we ll be looking at are called matchings. A matching is just a pairing-up of men and women Formally, a matching is a mapping µ : M W M W such that for any m M, µ(m) W {m} for any w W, µ(w) M {w} for any m M, w W, m = µ(w) if and only if w = µ(m) If µ(m) = m, man m matches to nobody; if µ(m) = w W, m matches to w We assume that preferences are only over mates, that is, each player cares only about who they are matched to, not who everyone else is matched to 2

3 Next, a couple of characteristics that might be reasonable to expect of an outcome. We ll say a matching is individually rational if everyone s mate is acceptable, that is, nobody matches to someone who they like less than being alone We say a pair (m, w) block a matching if m and w are not matched to each other (µ(m) w), but m prefers w to his mate µ(m) and w prefers m to her mate µ(w) That is, a pair blocks a matching if they d both be happier leaving their mate and matching with each other And finally, we say a matching is stable if it is individually rational and it is not blocked by any pair The name stable matching is pretty descriptive obviously, if a matching is blocked by an individual or a pair, you would expect it to be unstable Stable matchings in matching models roughly play the role of equilibria in other models we haven t yet said whether they always exist, there may be more than one, we haven t said how you arrive at one, but once you re at one, it s reasonable to think you might stay there As it happens, in two-sided matching markets, stable matchings do always exist, and there is a nice algorithm for picking out at least one or two of them But first, an example to remind ourselves that this result is not obvious, and also to show why economists like two-sided markets more than one-sided markets: the roommate problem The roommate problem is a one-sided market, that is, a game where people need to end up in pairs, but anyone can match with anyone And it s very easy to generate an example of a roommate problem with no stable matching Suppose there are four players, a, b, c, and d Preferences: P (a) : b c d, P (b) : c a d, P (c) : a b d, P (d) : doesn t matter There is no stable matching. Easiest way to see it: d is everyone s last choice, but still preferred to being alone. d must match with someone. Whoever he matches with is someone else s first choice, so d s roommate, and whoever likes that guy most, block the matching. Next, a simple example that there can be more than one stable matching: Two men and two women, everyone acceptable to everyone P (m 1 ) : w 1 w 2, P (m 2 ) : w 2 w 1, P (w 1 ) : m 2 m 1, P (w 2 ) : m 1 m 2 Two stable matchings: w 1 m 1, w 2 m 2, and w 1 m 2, w 2 m 1. (logic) 3

4 The Gale and Shapley paper, despite being quite brief and very nontechnical, was extremely influential it basically launched the matching literature (although it took a long time to take off). Probably the major result of the paper was this: Theorem. In a marriage market as we ve defined it, a stable matching always exists, and here s a way to find one. The algorithm they introduced: the deferred-acceptance algorithm To run the deferred-acceptance algorithm, one side of the market is chosen as the proposer ; we ll let this be the men To begin, each man proposes to his first choice Each woman tentatively accepts her favorite mate out of the proposals she got (if any of them is acceptable), and rejects all the others That s the first round In the second round, every man who was rejected in round one, proposes to his second choice Now each woman who got new proposals compares them to the one she s holding onto (or to being alone), tentatively accepts her favorite, and rejects all the others In the third round, every man who was rejected in round one, proposes to his favorite woman he hasn t yet proposed to Each woman keeps her best option, rejects the others, and the process continues (Once a man has proposed to, and been rejected by, every woman he finds acceptable, he stops making proposals) At some point, every man is either tentatively accepted by some woman, or rejected by every woman he finds acceptable; at that point, the women all marry the man they re with, and the algorithm ends Note to self: come up with a good 3x2 example of the deferred acceptance algorithm (in past years, used p 29 of Roth and Sotomayor, but too long) 4

5 Now, what s nice about this algorithm: It has to end after finite steps if there s a period without a rejection, the algorithm ends, but since a man never proposes to the same woman twice, there are only a finite number of possible rejections And it s actually very easy to prove that when the algorithm ends, the resulting matching is stable! The reason: suppose man m and woman w block the matching that results from the deferred acceptance algorithm This means they must not be matched with each other, and each prefers the other to the mate they re paired with There are two possibilities: either w rejected m at some point while the algorithm was running, or she didn t If she did, it s because she had a better offer either m is not acceptable to her, or she had a proposal from someone she liked more than m But men never take back proposals; so if she had a proposal she liked more than m, call it m, then either she ended up with m, or she rejected m for someone she liked even more, and so on So if she rejected m, she had to end up with someone she liked more than m, so w and m couldn t block the outcome On the other hand, suppose she never rejected m. Since they didn t end up together, that must mean m never proposed to her. But m started out at his top choice, and kept working down his list each time he got rejected So if m never proposed to w, then either she s not acceptable to him, or he ended up with someone he liked more So w and m can t block the outcome Since no men propose to women who aren t acceptable, and women always immediately reject proposals from men who aren t acceptable to them, no individual can block the outcome either So the result of the algorithm is a stable match Next, return to our example from before, with two men and two women and crossed preferences (m i prefers w i, but w i prefers m 1 i ) We said before that there were two stable matchings: either both men get their first choice, or both women get their first choice So the men agree on which stable matching is better; and similarly, the women agree on which stable matching is better 5

6 This turns out to be a much more general phenomenon: there is always a stable matching that all the men agree is the best, and a stable matching that all the women agree is the best For a given marriage market, define a stable matching to be M-optimal if every man weakly prefers it to every other stable matching; similarly, define a stable matching to be W-optimal if every woman weakly prefers it to every other stable matching Theorem. When all preferences are strict, there exists a M-optimal and a W-optimal stable matching. The deferred acceptance algorithm with men proposing results in the M-optimal stable matching. The deferred acceptance algorithm with women proposing results in the W-optimal stable matching. Proof. We ll show that the algorithm with men proposing leads to a M-optimal stable matching, which proves one exists. We say a woman w is achievable to a man m if there is some stable matching in which m matches with w We will prove that no man is ever rejected by an achievable woman during the deferred acceptance algorithm Proof is by induction on the number of rounds. Obviously, at the start of round 1, no man has yet been rejected by an achievable woman. Suppose that, up to the start of round k, no man has been rejected by an achievable woman, but that in round k, some man m is rejected by some woman w who is achievable. If she rejects him because he s unacceptable, then she couldn t be achievable; so she must reject him because she has a better offer. Call the better offer m. Now, m proposed to w before or during round k. And up to round k, no man had been rejected by an achievable woman. So m was not rejected by an achievable woman before proposing to w. But the men propose in order of preference; so this says that there is no achievable woman that m prefers to w. Now, we said w is achievable to m. So there must be some stable matching µ which pairs w to m. Under µ, m must either be alone, or matched to someone who s achievable. Which means he prefers w to his match under µ. But w prefers m to m. So w and m would block this match, so it isn t stable; so w was never achievable to m to begin with. So there s the inductive step: if no man has been rejected by an achievable woman before round k, no man can be rejected by an achievable woman during round k. So no man is ever rejected by an achievable woman; since men propose in order, this means that in the men-proposing deferred acceptance algorithm, every man ends up with his favorite achievable woman 6

7 So every man agrees that this is the best possible stable matching! (Note that we did require strict preferences we needed m to not have any other achievable women that he liked just as much as w. If we allow for indifferences, we are no longer guaranteed M- and W-optimal stable matchings the book gives an example where this fails.) Now, in the two-men, two-women example, the men agreed on a stable matching they both preferred, and the women agreed on a stable matching they both preferred, but the preferences were opposed: the match the men liked more was worse for the women, and vice versa. This turns out to generalize as well. Theorem. Take any marriage market with strict preferences, and take two stable matchings µ and µ which are not identical. If all the men weakly prefer µ to µ, then all women weakly prefer µ to µ. Proof. Suppose some woman w strictly preferred µ to µ. She must match with different men in the two matchings; let m = µ(w) be the one she matches with under µ. We know that m, being a man, weakly prefers µ to µ. And since w matches with someone else under µ, and his preferences are strict, he must strictly prefer w to his match under µ. And we know that w strictly prefers m to her match under µ. But then under the matching µ, w and m are not matched to each other, but prefer each other to their mates, so µ can t be stable. I ll give one more, fairly surprising result today You may recall, when we introduced affiliation, the concepts of the meet and the join the pairwise max and pairwise min of two points in R n Another way to think of the meet is as a least upper bound. Think of R n as a set with a partial order x y if it s weakly bigger in every component. Then for x and y, think of the set of points which are bigger than both of them; the meet is the smallest element of this set. (DRAW IT) Similarly, the join is the greatest lower bound Now, a set with a partial order that is closed under meet and join that is, a set that, if it contains two points x and y, also contains their meet and their join is referred to as a lattice 7

8 And it turns out that with a natural partial ordering, the set of stable matchings of a given marriage market will turn out to be a lattice (The partial order we use will basically be one side s preferences) Formally, take some marriage market, and let µ and µ be two stable matchings. Define a new mapping λ : M W M W, such that λ(m) is whichever of µ(m) and µ (m) is preferred by m λ(w) is whichever of µ(w) and µ (w) is least-preferred by w So we can think of λ as the pairwise max of µ and µ, from the mens perspective (and the pairwise min from the womens perspective). Then it turns out that λ is not just an arbitrary map, but it s a matching (not obvious); and it turns out to be a stable matching. (By construction, all men unanimously prefer it (weakly) to both µ and µ, and all women like it weakly less than either.) We could do the same thing with a new mapping ν which matches men to the worse of µ(m) and µ (m), and women to the better of µ(w) and µ (w), and this would also be a stable matching This means that the set of stable matchings is closed under meet and join, and is therefore a lattice which gives us a bunch of nice mathematical structure First, let s show λ is a matching, not just some random meaningless mapping between two sets We need to show two things: w = λ(m) m = λ(w) and m = λ(w) w = λ(m) If w = λ(m), this means that w is m s favorite out of µ(m) and µ (m). Suppose w = µ(m). We need to show that m = λ(w), which would require that m be w s less-preferred out of µ(w) and µ (w), that is, w must prefer µ (w) to m. But this has to be true; if not, then w would prefer m to µ (w), and m would prefer w to µ (m), so m and w would block the matching µ. So now we know w = λ(m) implies m = λ(w), and we need to prove the other direction. Let M be the set of men who are not single under λ, that is, M = {m λ(m) W } = { m µ(m) W or µ (m) W } Now, for any w and any m M, w = λ(m ) implies m = λ(w); so λ(m ) {w λ(w) M} = { w µ(w) M and µ (w) M } W so W is at least as big as λ(m ) 8

9 But by definition, M = { m µ(m) W or µ (m) W } {m µ(m) W } { m µ(m) W } = µ(w ) Since µ is a matching, µ(w ) is the same size as W ; so M is at least as big as W And λ is one-to-one on M, because if w = λ(m) = λ(m ) then m = m = λ(w), so M and λ(m ) are the same size, so λ(m ) is at least as big as W So now λ(m ) and W are the same size, and one contains the other, so λ(m ) = W Now, what we needed to show was that m = λ(w) w = λ(m), so suppose m = λ(w) Then w W = λ(m ), so w = λ(m ) for some m M But for m M, we know w = λ(m ) implies m = λ(w); and since λ(w) is unique, this can t hold for any m m, but since it has to hold for some m, w = λ(m) and we re done So now we know λ is a matching. Next, we need to show it s stable. THIS IS WHERE WE RAN OUT OF TIME WE LL PICK IT UP HERE ON THURSDAY 9

10 Individual rationality follows from individual rationality of µ and µ if nobody matches to anybody unacceptable in µ or µ, then the better of those two, or the worse of those two, must still be acceptable Suppose there was a pair (m, w) that blocked λ. If m prefers w to λ(m), then by definition, he prefers w to both µ(m) and µ (m). And if w prefers to m to λ(w), then she prefers to m to at least one of µ(w) and µ (w) If she prefers m to µ(w), and he prefers her to µ(m), then they would have blocked µ; similarly with µ in the other case So λ is a stable matching One more result for today. Roth and Sotomayor give this as a corollary of a different result, which we ll see Thursday, but it s apparent from the proof we just did. Take two matching µ and mu, and let M and M be the set of men who match to some woman (don t end up alone) under each, and W and W likewise Let λ be the meet of µ and µ, as defined above; let M λ and W λ be the men, and women, who end up matched up λ Now, if m matches under µ, it s because he prefers his mate to being alone; so if he matches under µ, he matches under λ. Similarly if he matches under µ. So M λ = M M But by the same logic, if a woman w ends up alone at either µ or µ, that s her worse result, so she ends up alone at λ, so W λ = W W This means that M λ is at least as big as the bigger of M and M, and W λ is at least as small as the smaller of W and W But the men who match under µ, match to the women who match under µ, so M and W are the same size; same with M and W So M λ is at least as big as the bigger of M and M, and W λ is at least as small as the smaller of them But the men who match under λ, match to the women who match under λ, so M λ and W λ are the same size 10

11 So M and M are the same size so any two stable matchings have the same number of couples But we re not done yet Since M λ = M M and W λ = W W, M λ is at least as big as M, and W λ is at least as small as M; since they re the same size, this means they re both the size of M But since M λ = M M, that means M = M So the set of men who end up matched to women is the same under any two stable matchings, same with the women 11

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