# Tiers, Preference Similarity, and the Limits on Stable Partners

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1 Tiers, Preference Similarity, and the Limits on Stable Partners KANDORI, Michihiro, KOJIMA, Fuhito, and YASUDA, Yosuke February 7, 2010 Preliminary and incomplete. Do not circulate. Abstract We consider the one-to-one matching problem where agents on one side of the market have similar preferences over those on the other side. More specifically, we present two formalizations of the idea of preference similarity, called the tiers model and the distance model. In the tiers model, preferences can vary only within a fixed subset of agents. In the distance model, preferences are modified from the underlying common ranking within certain distance. In each model, we characterize the upperbound of the number of possible partners in stable matchings. 1 Introduction We consider the one-to-one matching problem where agents on one side of the market have similar preferences over those on the other side. More specifically, we present two formalizations of the idea of preference similarity, called the tiers model and the distance model. In the tiers model, preferences can vary only within a fixed subset of agents (which we call a tier ). In the distance model, preferences are modified from the underlying common ranking (which we call the master ranking ) within a certain distance. In each model, we characterize the upper-bound of the number of possible partners in stable matchings. Kandori: Faculty of Economics, University of Tokyo ( Kojima: Department of Economics, Stanford University ( Yasuda: National Graduate Institute for Policy Studies ( 1

2 2 Framework There are a set of students S and a set of colleges C. Each student s has a strict preference relation s over the set of colleges and being unmatched (being unmatched is denoted by ø). Each college c has a strict preference relation c over the set of students and being unmatched (being unmatched is denoted by ø). If s c, then s is said to be acceptable to c. Similarly, c is acceptable to s if c s ø. A matching µ is a function from C S to C S {ø} such that (i) µ(s) C {ø} for every s S, (ii) µ(c) S {ø} for every s, and (iii) for all s S and c C, µ(s) = c if and only if µ(c) = s. That is, a matching simply specifies the college where each student is assigned or if the student is unmatched. We say that a matching µ is blocked by a pair of student s and college c if s strictly prefers c to µ(s) and c strictly prefers s to µ(c). A matching µ is individually rational if µ(i) i for all i S C. A matching µ is stable if it is individually rational and is not blocked. 3 Tiers Model Suppose that all students agree that C C is the top K colleges (the ranking within C may vary across students). Then, C satisfies the following top tier properties: (TTP1) Stable matching is assortative at the top tier: Any c C is matched with one of the top K students of c, in any stable matching (provided that all top K students for c are better than unmatching for c). (TTP2) Existence of the top tier students: In any stable matching, the set of partners of C is always the same: S (the top tier students ) such that, for any stable matching µ, µ(c ) = S. (TTP3) The matching market can be segmented and operated sequentially: First, find a stable matching among C and any subset of students S which contains the top tier students S, defined in (TTP2). For example, S can be the set of all students, or the set of all top K students of C (this set contains S, by (TTP1)). Then find a stable matching in the remaining colleges and students. This provides a stable matching. Conversely, any stable matching can be found in this way. TTP1 says that stable matching is assortative at the top tier. The top K colleges always obtain their top Kstudents. TTP2 shows that, if there exists top tier K colleges, 2

3 there is the top tier K students who are always matched with the top tier colleges at any stable matching. We now prove that those properties follow form a condition weaker than the unanimity condition (all students agree that C C is the top K colleges). Proposition 1 Let C be a subset of colleges such that (I) C = K and (II) for each c C, the top K students for c agree that C is the set of the top K colleges for them. Then the top tier properties (TTP1)-(TTP3) hold. Proof. (TTP1): Assume that c C is not matched with one of its top K students at stable matching µ. Then, at least one of c s top K students, say s 0, is matched with a college outside of C. This follows from the fact that it is impossible to match all of c s top K students to C {c}, because C {c} = K 1. Given this observation µ(s 0 ) / C, the premise (II) of Proposition shows that a top K student s 0 for c must prefer c C to µ(s 0 ) / C. In addition, since c C is not matched with one of its top K students, µ(c) (which can be c) is worse than s 0, who is a top K student of c. Hence, (c, s 0 ) can block µ, a contradiction. (TTP2): This is equivalent to: For any stable matching µ, µ(c ) µ CP (C ), where µ CP is the college-pessimal stable matching. (Note that, by feasibility we have µ(c ) = µ CP (C ), and therefore µ(c ) µ CP (C ) implies µ(c ) = µ CP (C )). We must show that, whenever µ(c) µ CP (c) for some c C, s 0 µ(c) is matched with a college in C at µ CP (µ(c) µ CP (c) and the rural hospitals theorem shows that µ(c) is indeed a student (i.e., not equal to c)). Suppose that this fails so that µ CP (s 0 ) / C. The fact that s 0 is matched to c in a stable matching µ and (TTP1) imply that s 0 is a top K student for c. Then the premise (II) of Proposition implies that s 0 prefer any college in C to any college outside C. In particular, c s 0 µ CP (s 0 ). On the other hand, since µ CP is college-pessimal stable matching, s 0 µ(c) c µ CP (c). Hence µ CP is blocked by (s 0, c), a contradiction. (TTP3) Part 1: A stable matching can be segmented. Let µ be a stable matching, and decompose it as µ = (µ C, µ C ), which means the following. (a) µ C : C S C S, where S is any subset of students such that S S (b) µ C (c) = µ(c) for c C (c) µ C (s) = µ(s) for s µ(c ) (d) µ C (s) = s for s / µ(c ) (e) µ C : C C S µ(c ) C C S µ(c ) (f) µ C (c) = µ(c) for c / C (g) µ C (s) = µ(c) for s S µ(c ). Then TTP3 claims that (i) µ C is a stable matching among C and all students S, and (ii) µ C is a stable matching among C C and S µ C (C ) = S µ(c ). Claim (ii) is 3

4 trivial, because the stability of µ directly implies no blocking pair for µ C. In contrast, the same is not true for (i): Because of (d), there can be potential blocking pairs to µ C whose blocking conditions are different under µ and µ C. Those are pairs (c C, s / µ(c )), where s is unmatched in µ C but matched with a college in µ (= µ(s) C ). However, any such pair satisfies the property µ C (c) c s so that it cannot block µ C. The reason is the following. Suppose on the contrary that s c µ C (c) = µ(c). By (TTP1), s must be a top K student for c, and the premise (II) of Proposition implies that s prefers c C to µ(s) C. This means that µ is blocked by (s, c), a contradiction. For any other blocking pair to µ C, the no blocking condition is identical under µ and µ C, so that the stability of µ implies (i). (TTP3) Part 2: Segmented matchings form a stable matching. Let µ C be a stable matching among C and S, and let µ C be a stable matching among C C and S µ C (C ) = S µ(c ). Then, (TTP3) claims that µ = (µ C, µ C ), defined by (c),(d),(f) and (g) is a stable matching. We show this in the following way. Given the stability of µ C and µ C, the only pairs who can possibly block (µ C, µ C ) are (c C, s / µ C (C )) and (c / C, s µ C (C )). First, we show that a pair (c C, s / µ C (C )) cannot block (µ C, µ C ). Suppose on the contrary such a pair (c, s) blocks (µ C, µ C ) so that s c µ C (c) and c s µ C (s). By the stability (individual rationality) of µ C, we have µ C (s) s s. Then, we have s c µ C (c) and c s s, and this implies that µ C (where s is unmatched ) is not stable. Second, we show that a pair (c / C, s µ C (C )) cannot block (µ C, µ C ). By (TTP1), s µ C (C ) implies that s is a top K student for some college in C. Then, the premise (II) of Proposition implies that s prefer µ C (s) (which is in C ) to any c / C. Hence (c / C, s µ C (C )) cannot form a blocking pair of (µ C, µ C ). Corollary 1 Suppose the set of colleges is partitioned in to C 1,..., C K and all students agree that C k is the set of kth tier colleges (their preferences with in a tier C k may vary). Then, the matching market (to find a stable matching) can operate sequentially, starting from the top tier colleges. First, a stable matching can be found for the top tier colleges and all students. No matter which stable matching is obtained the set of students the top tier colleges obtain is always the same. Call it the top-tier students S 1. Then, a stable matching can be found for the second tier colleges and the remaining students S S 1 and so on. Any stable matching is assortative: the top tier (with K 1 C 1 colleges) obtain their top K 1 students, the second tier obtains their top K 2 (= C 2 ) students in the remaining students S S, and so on. Corollary 2 Suppose the set of colleges is partitioned in to C 1,..., C K and let K k = C k. Suppose also that all top K k students for C k agree that C k is the set of their best colleges. 4

5 Then matching market (to find a stable matching) can clear in any order:c σ(1),..., C σ(k), where σ is a permutation. Those corollaries describe some aspects of junior recruiting market. As for the second corollary suppose that C 1 = strong department in theory and C 2 = strong department in labor. If the markets are segmented at the top (the top theory students prefer C 1 while the top labor candidates prefer C 2 ), the markets can clear in any order. Now let us turn to our model where players have preference relations which are similar to some master orders, and examine when we have a top tier. Suppose that colleges preferences are k c different from a master order s 1 s 2 s M. This means that s i c s j if i + k c < j. Proposition 2 Suppose that all students s 1, s 2,..., s K+kc agree that C is the set of their top K colleges (in particular, C = K). Then C satisfies the top tier properties (TTP1)- (TTP2). Proof. This is just a corollary to Proposition: The k c difference assumption implies that the top K students for any college in C are contained in {s 1, s 2,..., s K+kc }, and therefore the premise (II) of Proposition is satisfied. We now consider how likely it is that the assumption of Proposition 2 is satisfied. To address this question, let us assume that each college s preference ordering is 1-different from a master order s 1 s 2 s M, each student s preference ordering is also 1-different from a master order c 1 c 2 c N, and those preference orderings are generated by the following random utility model. We explain students preferences but a symmetric description holds for colleges. Assume the master utilities for the student are equally spaced u(c i ) u(c i+1 ) = 1, and assume that student s s utility is given by u s (c) = u(c) + η s (c), where η s (c) is a random noise x with probability r η s (c) = 0 with probability 1 2r x with probability r, 5

6 for some 1/2 < x < 1. The random nose η s (c) is i.i.d. across s and c. The requirement 1/2 < x implies that the ranking of c i and c i+1 can be flipped, while x < 1 ensures that the ranking of c i and c j, j > i + 1 cannot be flipped (so that student s preference relation is 1-different from the master order). Under those assumptions, we evaluate the probability that students s 1, s 2,..., s K+1 agree that C = {c 1,..., c K } is the set of their top K colleges. (Then Proposition 2 says that under this event C satisfies the top tier properties (TTP1)-(TTP3).) Now let c(s, K) = (c 1, c 2,..., c K ) be the list of top K colleges for student s, in the decreasing order of preference (c k is s s top k college), and let C(s, K) be the set of top K colleges for student s. We evaluate the probability that the top K colleges of student s (= C(s, K)) coincide with the top K colleges in the master order {c 1, c 2,..., c K }, conditional on the ordering of top K 1 colleges c(s, K 1). That is, we evaluate (find a lower bound of) Pr(C(s, K) = {c 1,..., c K } c(s, K 1)). To this end we are going to find an upper bound of Pr(C(s, K) {c 1,..., c K } c(s, K 1)). First, note that C(s, K) {c 1,..., c K } if and only if the ranking of c K and c K+1 is flipped. Hence Pr(C(s, K) {c 1,..., c K } c(s, K 1)) By Bayes Rule, this is equal to = Pr(η s (c K ) = x and η s (c K+1 ) = x c(s, K 1)). Pr(c(s, K 1) η s (c K ) = x and η s (c K+1 ) = x) Pr(η s (c K ) = x and η s (c K+1 ) = x) Pr(c(s, K 1)) = Pr(c(s, K 1) η s(c K ) = x and η s (c K+1 ) = x)r 2. Pr(c(s, K 1)) Let us first evaluate Pr(c(s, K 1) η s (c K ) = x and η s (c K+1 ) = x). Given the negative shock to the utility of c K (η s (c K ) = x ), colleges c K, c K+1,... can never be in the top K 1. Hence, Pr(c(s, K 1) η s (c K ) = x and η s (c K+1 ) = x) is equal to the probability 6

7 that student s s ranking of {c 1,..., c K 1 } is given by c(s, K 1). Denote this probability by Pr(Rank s {c 1,..., c K 1 } = c(s, K 1)). Hence we have obtained Pr(C(s, K) {c 1,..., c K } c(s, K 1)) Lastly, we evaluate the denominator as Pr(c(s, K 1)) = Pr(Rank s{c 1,..., c K 1 } = c(s, K 1))r 2. Pr(c(s, K 1)) = Pr(c(s, K 1) η s (c K ) = 0 or x)(1 r) + Pr(c(s, K 1) η s (c K ) = x)r. By the same argument as above, given that η s (c K ) = 0 or x, colleges c K, c K+1,... can never be in the top K 1 and therefore Pr(c(s, K 1) η s (c K ) = 0 or x) = Pr(Rank s {c 1,..., c K 1 } = c(s, K 1)). Hence Pr(c(s, K 1)) and we obtained an upper bound This implies the following. Pr(Rank s {c 1,..., c K 1 } = c(s, K 1))(1 r), Pr(C(s, K) {c 1,..., c K } c(s, K 1)) Pr(Rank s{c 1,..., c K 1 } = c(s, K 1))r 2 Pr(Rank s {c 1,..., c K 1 } = c(s, K 1))(1 r) = r2 1 r. Lemma 1 Let C(s, K) be the set of top K colleges of student s, and let c(s, K 1) be the vector of top K colleges of student s, in the decreasing order of s s preference. Then we have Pr(C(s, K) = {c 1,..., c K } c(s, K 1)) 1 r2 1 r. Given this lemmma, the probability that students s 1, s 2,..., s K+1 agree that c 1, c 2,..., c K are their top K colleges, given any information c(s k, K 1), k = 1,..., K + 1, is at least (1 r2 1 r )K+1. 7

8 Actually, for K = 1, we have better bound. The probability that students s 1 and s 2 agree that c 1 is their best, is exactly equal to 2 (1 Pr(u si (c 1 ) < u si (c 2 )) = (1 r 2 ) 2. None of i=1 those probabilities may not be a large number, but we show that the probability that at least one effective top k tier exists for k = 1,..., K is large, for a reasonably small K. Our argument above implies that the probability that students s 1, s 2,..., s K+1 do not agree that c 1, c 2,..., c K are their top K colleges, given any information c(s k, K 1), k = 1,..., K + 1, is at most 1 (1 r2 1 r )K+1. for K 2 and equal to 1 (1 r 2 ) 2 when K = 1. Hence, the probability that s 1, s 2,..., s k+1 do not agree that {c 1,..., c k } is the set of their top k colleges, for all k = 1,..., K is at most K q(k) = (1 (1 r 2 ) 2 ) (1 (1 r2 1 r )k+1 ). Hence, the probability that k=2 {c 1,..., c k } satisfies the top tier properties for at least one k = 1, 2,..., K is at least 1 q(k). We calculate this number q(k) for r = 0.1 = (so that ranking of c i and c i+1 is flipped with probability r 2 = 0.1). q(1) = 1 (1 r 2 ) 2 = 0.19 = 0.19 q(2) = (1 (1 r 2 ) 2 )(1 (1 r2 1 r )3 ) = = q(3) = (1 (1 r 2 ) 2 )(1 (1 r2 1 r )3 )(1 (1 r2 1 r )4 ) = = q(4) = (1 (1 r 2 ) 2 )(1 (1 r2 1 r )3 )(1 (1 r2 1 r )4 )(1 (1 r2 1 r )5 ) = = q(4) = (1 (1 r 2 ) 2 )(1 (1 r2 1 r )3 )(1 (1 r2 1 r )4 )(1 (1 r2 1 r )5 )(1 (1 r2 1 r )6 ) = =

9 Hence, the probability that {c 1,..., c k } satisfies the top tier properties for at least one k = 1,..., 5, or the probability that the top tier is of size less than 5 is at least 1 q(5) = Now suppose r = 0.3 = (so that ranking of c i and c i+1 is flipped with probability r 2 = 0.3). q(1) = 1 (1 r 2 ) 2 = 0.51 = q(2) = (1 (1 r 2 ) 2 )(1 (1 r2 1 r )3 ) = = q(3) = (1 (1 r 2 ) 2 )(1 (1 r2 1 r )3 )(1 (1 r2 1 r )4 ) = = q(4) = (1 (1 r 2 ) 2 )(1 (1 r2 1 r )3 )(1 (1 r2 1 r )4 )(1 (1 r2 1 r )5 ) = = q(4) = (1 (1 r 2 ) 2 )(1 (1 r2 1 r )3 )(1 (1 r2 1 r )4 )(1 (1 r2 1 r )5 )(1 (1 r2 1 r )6 ) = = Hence, the probability that {c 1,..., c k } satisfies the top tier properties for at least one k = 1,..., 5, or the probability that the top tier is of size less than 5 is at least 1 q(5) = Not so good!! And the sequence seems to converge to a number less than one. Try something in between! suppose r = 0.2 = (so that ranking of c i and c i+1 is flipped with probability r 2 = 0.2). q(1) = 1 (1 r 2 ) 2 = 0.36 = q(2) = (1 (1 r 2 ) 2 )(1 (1 r2 1 r )3 ) : = q(3) = (1 (1 r 2 ) 2 )(1 (1 r2 1 r )3 )(1 (1 r2 1 r )4 ) = = q(4) = (1 (1 r 2 ) 2 )(1 (1 r2 1 r )3 )(1 (1 r2 1 r )4 )(1 (1 r2 1 r )5 ) = = q(4) = (1 (1 r 2 ) 2 )(1 (1 r2 1 r )3 )(1 (1 r2 1 r )4 )(1 (1 r2 1 r )5 )(1 (1 r2 1 r )6 ) : = Hence, the probability that {c 1,..., c k } satisfies the top tier properties for at least one k = 1,..., 5, or the probability that the top tier is of size less than 5 is at least 1 q(5) = The below is what we originally had (Model 1). 9

10 Model 1: Colleges have tiers: there exist tiers t 1, t 2,..., t n such that c s c for any student s and c t i, c t i+1 Claim 1 In Model 1, each c t has at most t different students as stable partners. Proof. First, consider the best tier t 1 an let S 1 be the set of students that are matched to some college in t 1 in the college-pessimal stable matching µ. Then, in any stable matching µ and any college c t 1, µ (c) S 1 since, if not, c and µ (c) blocks µ, contradicting stability of µ. By an inductive argument we can show that for any tier t i and S i = µ(t 1 ) and any stable matching µ, µ (t 1 ) = S i. This completes the proof. Remark 1 Although the statement of the claim is given in terms of colleges partners, the claim has an implication on partners of students. More specifically, if t = k for all tiers t, then the number of stable partners for each student is at most k. Remark 2 A connection with the slope condition exists. Consider the college final offers game. Given two equilibria x x, it is easy to see that x c x c k if t = k for all tier t. The proof is, however, via an argument that is essentially the same as the above proof, so we re not very confident that our slope interpretation is particularly illuminating. 4 Distance Model Suppose that colleges and students preferences are generated from the underlying common ranking (called master ranking ) in the following way (Model 2). Model 2: Colleges and students rankings are modified no more than k S and k C : there exist master rankings c 1 > S c 2 > S > S c C and s 1 > C s 2 > C > C s S such that for any s S, if j > i + k S then c i s c j, (1) and for any c C, if j > i + k C then s i c s j. (2) We denoted a college (student) with master number l by c l (s l ). Lemma 2 In Model 2, for all m {1, 2,..., C }, c m is matched with (m + k S )th best student or above in its preference order, and for all n {1, 2,..., S }, s n is matched with her (n + k C )th best or better college in any stable matching. 10

11 Proof. We will only show the first part. The exactly symmetric proof can establish the second part. Suppose there is a stable matching in which c m is matched with its (m+k)th best student for k > k S. Since each student can be matched with at most one college, there must exist some student s whom c m ranks higher than its current partner, and is either unmatched or matched with c j for j > m + k S. By 1, c m is preferred to c j by all students. Clearly in both cases c m and s constitute a blocking pair, and hence we obtain a contradiction. Proposition 3 In any stable matching, for all m {1, 2,..., C }, c m s partner s i must be such that m (k S + k C ) i m + (k S + k C ). (3) Similarly, for all n {1, 2,..., S }, s n s partner c j must satisfy n (k S + k C ) j n + (k S + k C ). (4) Proof. By the assumption on preference and Lemma 2, Rearranging (5), we obtain i m + (k S + k C ) and j n + (k S + k C ). (5) i (k S + k C ) m and j (k S + k C ) n. Note the above inequalities should also hold when c m is matched with s n, i.e., for m = j and n = i. Thus, the following is true Combining (5) and (6) gives (3) and (4). n (k S + k C ) j and m (k S + k C ) i. (6) Proposition 4 For each college (resp. student), the number of stable partners is at most 2k S + k C + 1 (resp. 2k C + k S + 1). Proof. We will only consider colleges stable partners. The argument for students stable partners is completely analogous, so is skipped. By (3), in any stable matching c m is not matched with student s j for j < m (k S +k C ). There are m (k S + k C ) 1 such students. By (2), s j must be ranked (j + k C )th best or above by all colleges. This implies that s j is always included in the (m + k S )th best students in c m preference order (but not available to c m ). Since, c m is matched with one 11

12 of these (m + k S )th best students (Lemma 2), the maximum number of stable partners is given by m + k S [m (k S + k C ) 1] = 2k S + k C + 1, which concludes the proof. The followings are possible future works. Remark 3 Is Proposition 4 tight? It is clearly not when k S or k C (but not both) becomes 0, since stable matching is unique in such cases. Remark 4 What happens if k S and k C are heterogeneous? For instance, we can consider k i s such that for each s S and i C, if j > i + k i s then c i s c j. References To be added. 12

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