Appendix B. Proof of Theorem 1
We split up the proof in short propositions.
Lemma A1. is a complete lattice, with a largest and smallest element.
Proof. That is a partially ordered set follows from the definition of . Take a subset of . We need to show that has the least upper bound and the greatest lower bound in to complete the proof.
For every
, define
. By definition, for every
,
. Assign a new index
over
such that
Let . Define inductively as follows: .
Note that and . We define z as a -vector, , with respect to the original index. Clearly, and , so .
Claim 1. zis the least upper bound of.
Proof. By definition, which is greater than for all and . Therefore, for all , which means that z is an upper bound.
Suppose that is another upper bound of . Therefore, for all and . If we take the supremum of the right hand side, then we get for all . On the other hand, . The last two impressions imply for all i, so . Thus, z is the least upper bound of . □
Similarly, we can construct the greatest lower bound of as follows: . Define inductively: . The proof that u is the greatest lower bound of of is similar to the proof that z is the least upper bound of , so it is omitted. □
For each
m, let the choice
be defined as follows. For a vector
, let
be the vector in
that is maximal for
. In other words, if
x represents the quantities of type of women available for
m,
chooses according to
from the best choice downwards until filling quota
. Note that if
then
, and if
then
. Define
analogously.
Proposition A1. is a nonempty and complete lattice.
Proof. A man pre-matching is a matrix such that and . A woman pre-matching is a matrix such that and .
We consider pairs
, where
A is a man prematching, and
B is a woman prematching, ordered by a partial order ≤. The order ≤ is defined as
if
The order ≤ is a product order of complete lattices by Lemma A1, so that the set of all pairs ordered by ≤ is a complete lattice.
We define a function C, mapping pairs of prematchings into pairs of prematchings. Fix : For a type-m man, the number of type-w women who are willing to match with m at B is . Let , i.e., the -matrix such that entry is the number of type-w women who are willing to match with type m men at B. Similarly, let be the -matrix for which entry is the number of type-m men who are willing to match with type-w women at A. Now let where and . Note that is a man prematching and is a woman prematching.
We now prove that C is isotone. Assuming that , we prove that .
Fix
m. For any
w, we have
because
. As a consequence, a type-
m man has weakly more women of each type willing to match with him in
than in
B: i.e.,
. Thus
. Similarly, for a type
w woman,
. It follows that
. By Tarski’s fixed point theorem, there is a fixed point of
C, and the set of fixed points of
C is a complete lattice when ordered by ≤.
Let be a fixed point of C, i.e., . Assume that for some m and w. and implies that although number of type-m men were available, only of them are chosen by . Therefore, all nonnegative entries in are at least as good as m with respect to . This implies that . Since , we get which contradicts . Therefore, for all m and w. Hence the fixed point has the property that , and they are not only a prematching but a matching as well since a man prematching that is also woman prematching is a matching.
Finally, we prove that the set of fixed points of C is the set of stable matchings. More precisely, is a fixed point of C if and only if A is a stable matching.
Suppose that a fixed point
of
C is not stable. By construction of
and
,
A is individually rational. Then there is a blocking pair
. By definition of blocking pairs, there exist
and
such that
,
,
, and
. Now, the number of type-
w women who are willing to match with
m at
A is
as
. However,
and
contradicts that there is
with
and
.
Suppose that A is a stable matching. We fix m and show that where . Suppose, for contradiction, that . Denote as the most preferred type of women (with respect to ) such that . For all preferred to , . Thus, implies either or , neither of which is possible. On the other hand, contradicts that A is stable. Although there are type women available more than , some type m men are matched to less preferred women. Then, there is such that and , so is a blocking pair. Similarly, we can show that , and therefore . □
Proposition A2. Suppose that X and Y are stable matchings, then if and only if .
Proof. Suppose that X and Y are stable matchings such that , we show that . The other direction of the claim can be proved analogously.
Consider the construction of C in the proof of Proposition A1: Let and correspond to the matrices for X and Y as in the proof. Since X and Y are stable matchings we have and . Since for all m, for all m and w. Therefore, for all w, which implies that . □
Proposition A3. Suppose that X and Y are two stable matchings. Then for any men or women type a, either or . Consequently, and .
Proof. We only prove the first part that either or , in three steps depending on whether X and Y have integer, rational, and real entries. The second part follows immediately from the first part.
Case 1 (Integer Entries): We first start with the case when X and Y have integer entries as well as and . From , we create a many-to-one matching market (of colleges and students) as follows.
The set of types of men remains the same; interpreted as the set of colleges. A college m has a capacity of . Whereas, each type-w woman is split into copies, all of which have the same preferences over colleges; women types are interpreted as students. On the other hand, m’s preferences replace w in with her copies enumerated from 1 to , in increasing order. Denote ’s copy by . So if and only if . In addition, each college has responsive preferences over groups of students. The new matching market with colleges and students is a many-to-one matching markets where an outcome for a college is a group of students, and an outcome for a student is either a college or being single.
Now, we construct a new matching, , in the new market from X. (The new matching is a many-to-one matching, which is not an aggregate matching. Nevertheless, we use the aggregate matching notation to stress the relations between and X, and and Y.) It is enough to describe the matches of students in . Rank woman w’s outcomes in X in decreasing order according to her preference . Let the copy of , , match to the highest outcome of in X. Similarly, construct from Y.
We claim that and are stable matchings in the new market. Suppose for contradiction that is not a stable matching. Since is individually rational by construction, there exists a blocking pair . This means that ’s match is worse than m. Similarly, m’s match includes a student worse than : this agent cannot be where by definition of , and it cannot be where because by construction ’s match is worse than ’s match with respect to . Hence, one of m’s matches is worse than , and not a copy of . This means that forms a blocking pair in X: A contradiction to the stability of X. Therefore, must be stable. Similarly, is also stable.
Now, by Theorem 5.26 of [
12], for any college
m the outcomes in
and
are comparable. This means that the responsive preferences over groups of students inherited from
, which is equivalent to the first-order stochastic dominance, can compare the outcomes of
m in these two stable matchings. Therefore,
can compare the outcomes in
X and
Y since
is a coarser order than
.
An analogous argument shows that can compare the matching outcomes in X and Y.
Case 2 (Rational Entries): Suppose for now that all entries of X and Y are rational numbers as well as and . Define a new matching market from as follows.
M, W, and P are the same. We change the capacities as follows. Find a common factor of denominators in all entries in X and Y, say r, and multiply all capacities by r. Therefore, the new market is . Define and with non-negative integer entries. By the argument above, for any agent a, the matchings in and can be compared with respect to which implies that the outcomes in X and Y can also be compared.
Case 3 (Real Entries): Suppose now that entries of X and Y are real numbers. We construct two sequences of matrices, and , as follows:
and have rational entries,
and for all ,
the sum of entries in row i and column j is the same for and , and
and as for all .
By construction, stability of X and Y imply stability of and for the same market with adjusted capacities. By the argument above, for each type a, and can be compared with respect to . Take a subsequence such that the ordering is the same for all entries. Therefore, for all l or for all l. By taking l to ∞ we get that either in the former case, or in the latter since implies and as for all . □
We have shown that is a lattice in Proposition A1. Using the proposition above, we show that the lattice, , is distributive.
Proposition A4. is a distributive lattice.
Proof. Suppose that
X,
Y, and
Z are stable matchings. We are going to prove that type
a has the same matching in
and
for all agent types
a:
where min and max operators are defined with respect to
and where we repeatedly use Proposition A3.
The proof that is analogous. □
Proposition A5. Suppose that X and Y are stable matchings. Then for all i and similarly for all j.
Proof. The proof has the same structure as in the proof of Proposition A3: it has three steps depending on whether X and Y have integer, rational, and real entries.
Case 1 (Integer Entries): We first start with the case when X and Y have integer entries as well as for all m and for all w. From , we create a many-to-one matching market and also stable matchings and in this new market as in the proof of Proposition A3.
Now, by Theorem 5.12 of [
12], the set of positions filled for any
m in
and
are the same. Therefore,
for all
i. Similarly,
for all
j.
Case 2 (Rational Entries): Suppose for now that all entries of X and Y are rational numbers as well as for all a. Define a new matching market from as follows.
M, W, and P are the same. We change the capacities as follows. Find a common factor of denominators in all entries in X and Y, say r, and multiply all capacities by r. Therefore, the new market is . Define and with non-negative integer entries.
By the argument above, for all i and for all j. The conclusion follows.
Case 3 (Real Entries): Suppose now that entries of X and Y are real numbers. We construct two sequences of matrices, and , as follows:
and have rational entries,
and for all ,
the sum of entries in row i and column j is the same for and , and
and as for all .
By construction, stability of X and Y imply stability of and for the same market with adjusted capacities. By the argument above, for all i and . If we take the limit of these equalities as , we get the desired equalities. □
Appendix C. Proof of Theorem 3
Let X be a rationalizable integral complete matching. Suppose that all cycles of the associated graph are balanced. Direct the edges of such that each cycle is oriented as follows: if is a cycle, then the edge is oriented such that , which we denote by . For each path , direct the edges in a similar way. If the matching X is rationalizable, then such an orientation of the edges exists and defines a rationalizing preference profile P (Theorem 2, Part 1). The rationalizing preferences have the property that if and then if , and if .
First, if
X has no cycles, then it is rationalizable as the unique stable matching (Theorem 2, Part 2), so there is nothing to prove, as a unique stable matching is also the median stable matching. Suppose then that
X has at least one cycle
. Enumerate the vertexes of the cycle such that
in the orientation (directed graph) of
above, and
lies in the same row as
. Let
Let be the set of all matrices E of integer numbers such that
if is not in the cycle c;
for all i and j, and ; and
.
We want to make two observations about the matrices in . First, , so is a stable matching for all . Second, if and only if ; and is such that if and only if . Similarly, is such that if and only if .
We need to prove that any stable matching in the resulting market must be obtained from X through matrices in : Then X is a median stable matching.
First, we prove that only if is a vertex in a minimal cycle of X. The number of single agents of each type is the same in X as in Y (Proposition A5; in this case it is zero, as X has no single agents). So, if then there is and such that and . Similarly, if then there is and such that and . Hence, we can apply this observation repeatedly, starting from any with and obtain a sequence with such that for each n ():
.
Claim 2. For, .
Proof. Suppose for contradiction that . Without loss of generality, assume that . By definition of the rationalization P, we have that , as and . We can now show that if and differ in i, then prefers to ; and that if they differ in j, then prefers to . This fact, which we prove in the next paragraph, establishes the contradiction: , but by definition of P and because .
To prove the fact, we reason by induction. We have already established that . Suppose that . By Property 1 of the sequence , either and ; or and (or both hold). Then the stability of X and Y implies that . The proof for the case when is similar. □
The claim implies that the sequence is a cycle in . Thus a stable matching Y can only differ from X in vertexes that are part of a cycle of .
Second, we prove that . We established above that only if is a vertex in a cycle. We now prove that Clearly, . We show that if then there is h such that .
If for all then there is and such that some type- men of type who are married to women of type and in X are married to women of type in Y. Then we can define two cycles, and would be a vertex in both of them. The first cycle has as the first edge, and the remaining edges defined inductively, by the definition of above. The second cycle has as the first edge, and the remaining edges defined inductively. The resulting two cycles would be connected, which contradicts the hypothesis that X is rationalizable. So there must exist some h with . An analogous argument applied to implies that ; so . Then, , as .