Economics Letters 156 (2017) 65–67
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Economics Letters journal homepage: www.elsevier.com/locate/ecolet
Stable matchings and fixed points in trading networks: A note Hiroyuki Adachi 1
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Article history: Received 24 February 2017 Accepted 23 March 2017 Available online 18 April 2017 JEL classification: C78 D47
abstract Building on the work of Ostrovsky (2008) on supply chain networks, Fleiner et al. (2016) showed that, in possibly cyclic trading networks under fully substitutable preferences, the set of fully trail-stable matchings is nonempty by characterizing it using the set of the fixed points of an increasing function. I characterize the fully trail-stable matchings using a different fixed point (increasing) function, rendering generalized T-algorithm for such environments. © 2017 Elsevier B.V. All rights reserved.
Keywords: The gross substitutes condition Full substitutability
1. Introduction Ostrovsky (2008) analyzed supply chain networks (acyclic trading networks) by extending two-sided matching models such as Gale and Shapley (1962), Crawford and Knoer (1981), Kelso and Crawford (1982), Roth (1984), Blair (1988), Adachi (2000), Alkan (2002), Fleiner (2003), Echenique and Oviedo (2004) and Hatfield and Milgrom (2005) among others. He introduced a set of conditions on preferences now called full substitutability, generalizing the gross substitutes condition by Kelso and Crawford (1982). His solution concept is chain stability and he showed that under full substitutability the chain-stable matchings is characterized using the fixed points of an increasing function. Existence of chain-stable matchings is proved by invoking Tarski’s fixed point theorem, a technique used in Adachi (2000), Fleiner (2003), Echenique and Oviedo (2004) and Hatfield and Milgrom (2005) among others. Hatfield and Kominers (2012) showed that a stable matching (in their sense) always exists in an acyclic trading network but it may fail to exist in a cyclic network. See Hatfield et al. (2013) for trading networks under quasilinear preferences. Fleiner et al. (2016) analyzed possibly cyclic trading networks and introduced new solution concepts such as fully trail-stable matchings. Under full substitutability, they characterized the set of fully trail-stable matchings using the set of fixed points of an increasing function and showed that a fully trail-stable matching always exists. Full trail stability is stronger than chain stability and weaker than the stability concept in Hatfield and Kominers (2012).
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[email protected]. 1 Any errors are the author’s responsibility. http://dx.doi.org/10.1016/j.econlet.2017.03.032 0165-1765/© 2017 Elsevier B.V. All rights reserved.
I characterize the set of fully trail-stable matchings using a different fixed point (increasing) function. It provides a different proof to existence and a generalization of T-algorithm by Ostrovsky (2008). As in the prior matching literature, my analysis relies on the irrelevance of rejected contracts condition. It is explicitly assumed by Alkan (2002) and Fleiner (2003) and implicitly assumed by Hatfield and Milgrom (2005). Its importance is emphasized by Aygun and Sonmez (2013). 2. A trading network 2.1. Preliminary There is a finite set F of firms. A typical firm is denoted by f . There is a finite set K of all possible contracts. For each contract x ∈ K , there exist a unique seller s(x) and a unique buyer b(x) associated with contract x. It is not allowed that s(x) = b(x). For any contract set Y ⊆ K , Yf → = {x ∈ Y : f = s(x)} is the set of contracts in Y for which f is the seller, and Y→f = {x ∈ Y : f = b(x)} is the set of contracts in Y for which f is the buyer. We say that Yf → is f ’s downstream contracts in Y and Y→f is f ’s upstream contracts in Y . Yf = Yf → ∪ Y→f is the set of contracts pertinent to f in Y . Each firm f ’s preferences are expressed by its choice function C f (·) such that C f (Yf ) ⊆ Yf for any subset Y of K . We write C f (Y ) = C f (Yf ) for any Y ⊆ K . We assume that C f (·) satisfies the following irrelevance of rejected contracts (IRC) condition: For any X and Y , C f (X ) ⊆ Y ⊆ X
implies C f (X ) = C f (Y ).
66
H. Adachi / Economics Letters 156 (2017) 65–67
2.2. Fully trail-stable matchings
For any given contracts sets Y and Z , define f CB
(Y |Z ) = C (Y→f f CS (Z |Y ) = C f (Y→f
f
∪ Zf → ) →f and ∪ Zf → ) f → .
These are the set of contracts f chooses as a buyer (as a seller, respectively) when upstream contracts in Y and downstream contracts in Z are available to f . Also define f
f
CB (Y |Z ) = ∪f ∈F CB (Y |Z ) and CS (Z |Y ) = ∪f ∈F CS (Z |Y ). f
f
Note that because CB (Y |Z ) ⊆ Y→f and CS (Z |Y ) ⊆ Zf → , CB (Y |Z ) ⊆ Y and CS (Z |Y ) ⊆ Z . Ostrovsky (2008) introduced a set of conditions on firms’ choice functions called full substitutability (same-side substitutability and cross-side complementarity) to analyze supply chain networks or acyclic trading networks. The preferences of f are same-side substitutable (SSS) if for all f Y ′ ⊆ Y ⊆ K and Z ′ ⊆ Z ⊆ K , (a) k ̸∈ CB (Y ′ ∪ {k}|Z ) implies k ̸∈ f
f
f
CB (Y ∪ {k}|Z ) and (b) k ̸∈ CS (Z ′ ∪ {k}|Y ) implies k ̸∈ CS (Z ∪ {k}|Y ). The preferences of f are cross-side complementary (CSC) if for all f f Y ′ ⊆ Y ⊆ K and Z ′ ⊆ Z ⊆ K , (a) CB (Y |Z ′ ) ⊆ CB (Y |Z ) and (b) f
f
CS (Z |Y ′ ) ⊆ CS (Z |Y ). The preference of f are fully substitutable if they are same-side substitutable and cross-side complementary. If each firm f ’s choice function C f (·) satisfies the IRC condition, CB (·|·) satisfies the following condition, which we call the sameside irrelevance of rejected contracts condition. (We refer to it as the IRC condition, too.) The similar property holds for CS (·|·). Lemma 1. If C (·) satisfies the IRC condition, for any subsets X , Y and Z of K , we have that CB (X |Z ) ⊆ Y ⊆ X implies CB (X |Z ) = CB (Y |Z ). f
f
Proof. Because CB (·|·) = ∪f ∈F CB (·|·), we just need to prove f that each CB (·|·) satisfies the similar [C f (X→f ∪ f property. Suppose Zf → )]→f ⊆ Y→f ⊆ X→f . Then C (X→f ∪ Zf → ) →f ⊆ (Y→f ∪ Zf → ) ⊆ (X→f ∪ Zf → ). Also, if C f (X→f ∪ Zf → ) f → ⊆ Zf → ⊆ (X→f ∪ Zf → ), then C f (X→f ∪ Zf → ) →f ⊆ (Y→f ∪ Zf → ) ⊆ (X→f ∪ Zf → ).
From the IRC condition of C f (·), C f (X→f ∪ Zf → ) = C f (Y→f ∪ Zf → ). f
f
This means CB (X |Z ) = CB (Y |Z ).
In particular, if we substitute Y = CB (X |Z ) in the above Lemma, we have CB (X |Z ) = CB (CB (X |Z )|Z ). This property is sometimes called idempotency. If each f ’s choice function C f (·) satisfies the SSS condition along with the IRC condition, then CB (·|·) satisfies the following (sameside) path independence (PI). The same condition holds for CS (·|·). Lemma 2. If C f (·) satisfies the SSS condition and the IRC condition, for any subsets X , Y and Z of K , it holds that CB (Y ∪ X |Z ) = CB (CB (Y |Z ) ∪ X |Z ). Proof. It suffices to prove that f
f CB
(·|·) satisfies the similar property f
f
for each f . To see CB (Y ∪ X |Z ) ⊆ CB (CB (Y |Z ) ∪ X |Z ), suppose k ∈
f CB
(Y ∪ X |Z ) and k ̸∈ f
f
f CB
f CB
( (Y |Z ) ∪ X |Z ). If k ∈ f
f CB
(Y |Z ) ∪ X ,
then k ̸∈ CB (CB (Y |Z ) ∪ X |Z ) implies k ̸∈ CB (Y ∪ X |Z ) by SSS. A
A contract set A is an individually rational matching if Af = C f (A) for each f ∈ F . In particular, an empty set ∅ of contracts is an individually rational matching. Fleiner et al. (2016) introduced full trail-stability. A finite set of different contracts W = {x1 , . . . , xM } is called a trail if b(xm ) = s(xm+1 ) holds for all m ∈ {1, . . . , M − 1}. An individually rational matching A is a fully trail-stable matching if there is no trail W = {x1 , . . . , xM } with W ∩ A = ∅ such that 1. x1 ∈ C f (A ∪ {x1 }) where f = s(x1 ), 2. {xm−1 , xm } ⊆ C f (A ∪ {xm−1 , xm }) where f = b(xm−1 ) = s(xm ) for m ∈ {2, . . . , M }, and 3. xM ∈ C f (A ∪ {xM }) where f = b(xM ). Such a trail W is called a locally blocking trail to A.2 Ostrovsky (2008) introduced the model of supply chain networks (acyclic trading networks) and considered chainstability, which is weaker than full trail stability. A chain is a trail in which the trail does not cross the same firm more than once. An individually rational matching A is a chain-stable matching if there is no blocking chain. We characterize the fully trail-stable matchings using the fixed points of an increasing function, different from the one in Fleiner et al. (2016). Define a pair of functions, TB (Z , Y ) and TS (Z , Y ), by TB (Z , Y ) := {x ∈ K : x ∈ CB (Y ∪ {x}|Z )} and
(1)
TS (Z , Y ) := {x ∈ K : x ∈ CS (Z ∪ {x}|Y )}.
(2)
Then consider a pair of equations Z = TB (Z , Y )
(3)
Y = TS (Z , Y ).
(4)
Then (Z , Y ) is a solution to the pair of Eqs. (3) and (4) if and only if it is a fixed point of function T := (TB , TS ). So we refer to (3) and (4) as the fixed point conditions. This fixed point function is more similar to those of Ostrovsky (2008) and Hatfield and Kominers (forthcoming) than to those of Fleiner (2003) and Hatfield and Milgrom (2005). Under SSS and CSC, Lemma 2 of Ostrovsky (2008) showed that, with respect to the usual set inclusion order, TB (Z , Y ) is increasing in Z and decreasing in Y and TS (Z , Y ) is decreasing in Z and increasing in Y . We define a partial order ⊒ on K × K by (Z ′′ , Y ′′ ) ⊒ (Z ′ , Y ′ ) if Z ′′ ⊇ Z ′ and Y ′′ ⊆ Y ′ . The same partial order is previously used by Fleiner (2003) and Hatfield and Milgrom (2005). We restate Lemma 2 of Ostrovsky (2008). Lemma 3. T (Z , Y ) is increasing in (Z , Y ) with respect to partial order ⊒. Because there are a finite number of contracts, ⟨K × K , ⊒⟩ is a complete lattice with a largest element (K , ∅) and a smallest element (∅, K ). Since T (Z , Y ) is increasing in (Z , Y ) with respect to ⊒, the fixed point theorem of Tarski (1955) implies Lemma 4. The set of fixed points of T (or the set of solutions to Eqs. (3) and (4)) is a nonempty lattice. The next theorem states that the set of fixed points of T corresponds to the set of fully trail-stable matchings in a trading network. This theorem allows us to extend T-algorithm of Ostrovsky (2008). The proof generalizes Ostrovsky’s.
f
contradiction. So k ̸∈ CB (Y |Z ) ∪ X . We know k ∈ Y ∪ X . So k ∈ Y and k ̸∈ f CB
f CB
(Y |Z ). We have assumed k ∈ CBf (Y ∪ X |Z ). By SSS,
(Y |Z ). A contradiction. f f f So we have CB (Y ∪ X |Z ) ⊆ CB (CB (Y |Z )∪ X |Z ) ⊆ Y ∪ X . Using the f f f f same-side IRC and idempotency, CB (Y ∪ X |Z ) = CB (CB (CB (Y |Z ) ∪ f f X |Z )|Z ) = CB (CB (Y |Z ) ∪ X |Z ).
k∈
2 In a trail, no contracts are repeated. If repeated contracts are allowed, such a sequence of contracts is called a walk. One can think of a blocking walk and a walkstable matching, but one will soon realize that a fully trail-stable matching and a walk-stable matching are equivalent under general preferences because one can make a locally blocking trail out of a blocking walk. This fact is used in the proof of Theorem 5.
H. Adachi / Economics Letters 156 (2017) 65–67
Theorem 5. (a) If (Z , Y ) is a fixed point of T , then Z ∩ Y is a fully trail-stable matching and Z ∩ Y = CB (Y |Z ) = CS (Z |Y ). (b) Suppose A is a fully trail-stable matching. Then there is a pair of sets (Z , Y ) such that (Z , Y ) is a fixed point of T and A = Z ∩ Y = CB (Y |Z ) = CS (Z |Y ). Proof. Part (a). Let (Z , Y ) be a fixed point of T . First we claim CB (Y |Z ) = Z ∩ Y . To see CB (Y |Z ) ⊇ Z ∩ Y , suppose x ∈ Z ∩ Y . Then x ∈ CB (Y ∪ {x}|Z ) = CB (Y |Z ). To show CB (Y |Z ) ⊆ Z ∩ Y , suppose x ∈ CB (Y |Z ). Then x ∈ Y and x ∈ CB (Y ∪{x}|Z ). From the fixed point conditions, x ∈ Z . So CB (Y |Z ) = Z ∩ Y . Similarly, CS (Z |Y ) = Z ∩ Y . These conditions mean that, for each f , C f (Y→f ∪ Zf → ) = (Z ∩ Y )f ⊆ Y→f ∪ Zf → . By IRC, C f ((Z ∩ Y )f ) = C f (Y→f ∪ Zf → ) = (Z ∩ Y )f . Z ∩ Y is an individually rational matching. Next we show that A := Z ∩ Y is fully trail-stable. Suppose for a contradiction that there is a locally blocking trail W = {x1 , . . . , xM } with W ∩ A = ∅. Since xM ∈ CB (A ∪ {xM }|A) ⊆ CB (A ∪ {xM }|Z ) = CB (CB (Y |Z ) ∪ {xM }|Z ) = CB (Y ∪ {xM }|Z ), we have xM ∈ Z . Symmetrically, we have x1 ∈ Y . Since A = Z ∩ Y and x1 ̸∈ A, we must have x1 ̸∈ Z . Since x1 ̸∈ Z and (Z , Y ) is a fixed point, x1 ̸∈ CB (Y ∪ {x1 }|Z ) = CB (CB (Y |Z ) ∪ {x1 }|Z ) = CB (A ∪ {x1 }|Z ). By assumption, x1 ∈ b(x ) b(x ) CB 1 (A ∪ {x1 }|A ∪ {x2 }). If x2 ∈ Z , by CSC, x1 ∈ CB 1 (A ∪ {x1 }|Z ). A contradiction. So x2 ̸∈ Z . By proceeding in the same way, we have xM ̸∈ Z . A contradiction. Next we prove part (b). Suppose A is a fully trail-stable matching. Let Z (0) = Y (0) = A and for n ≥ 1 define Z (n) = {x ∈ K : x ∈ CB (A ∪ {x}|Z (n − 1))} and Y (n) = {x ∈ K : x ∈ CS (A ∪ {x}|Y (n − 1))}. By individual rationality of A, Z (1) ⊇ A and Y (1) ⊇ A. By CSC, Z (n) ⊇ Z (n − 1) and Y (n) ⊇ Y (n − 1) for all n ≥ 1. Because there are a finite number of contracts, there exists an N ∗ such that Z (N ∗ ) = Z (N ∗ − 1) and Y (N ∗ ) = Y (N ∗ − 1). Let us denote such sets by Z ∗ and Y ∗ . We will argue that A = Z ∗ ∩ Y ∗ = CB (Y ∗ |Z ∗ ) = CS (Z ∗ |Y ∗ ) and (Z ∗ , Y ∗ ) is a fixed point of T . We show by induction that for, all n ≥ 1, Z (n) ∩ Y (n) = A. We know Z (1) ∩ Y (1) ⊇ A. To see Z (1) ∩ Y (1) ⊆ A, suppose x ∈ Z (1)∩ Y (1). Then x ∈ CB (A ∪{x}|A) and x ∈ CS (A ∪{x}|A). Since A has no locally blocking trail, x ∈ A. Thus Z (1)∩ Y (1) = A. Suppose that Z (n) ∩ Y (n) = A holds up to n − 1. We will show it holds for n. Assume otherwise. Take x1 ∈ (Z (n) ∩ Y (n)) \ A and let f = b(x1 ). f We will construct a blocking walk.3 If x1 ∈ CB (A ∪ {x1 }|A), stop. (x1 f
will be the end of a blocking walk.) Assume x1 ̸∈ CB (A ∪ {x1 }|A). f CB
Since x1 ∈ Z (n), x1 ∈ (A ∪ {x1 }|Z (n − 1)). Then Z (n − 1)f → must be a strict superset of Af → . Let {z1 , . . . , zm } = Z (n − 1)f → \ A. We claim that there exists at least one zj which satisfies zj ∈ C f (A ∪ {x1 } ∪ {z1 , . . . , zm }). (Otherwise, for all i ∈ {1, . . . , m}, zi ̸∈ C f (A ∪{x1 }∪{z1 , . . . , zm }). By IRC, x1 ∈ C f (A ∪{x1 }), leading to a contradiction.) By SSS, zj ∈ C f (A ∪ {x1 } ∪ {zj }). It must also be the case that x1 ∈ C f (A∪{x1 }∪{zj }). (Otherwise, by IRC, zj ∈ C f (A∪{zj }) and so zj ∈ Y (1) ⊆ Y (n − 1). Then zj ∈ Z (n − 1) ∩ Y (n − 1) = A and this is a contradiction.) Therefore, {x1 , zj } ⊆ C f (A ∪ {x1 } ∪ {zj }).
3 See footnote 2.
67 g
Denote zj by x2 and let g = b(x2 ). If x2 ∈ CB (A ∪ {x2 }|A), stop. (x2 g will be the end of a blocking walk.) Assume x2 ̸∈ CB (A ∪ {x2 }|A). g Since x2 ∈ Z (n − 1)\ A, x2 ∈ CB (A ∪{x2 }|Z (n − 2)). As in the previous step, let {w1 , . . . , wl } = Z (n − 2)g → \ A. Then there exists at least one wi such that {x2 , wi } ⊆ C g (A ∪ {x2 } ∪ {wi }). Denote wi by x3 . This process eventually stops because, at n, xn ∈ Z (1) \ A and xn ∈ b(x ) CB n (A ∪ {xn }|A). Symmetrically, starting from x1 , we can stretch a walk in the other direction. The resulting walk {yL , . . . , x1 , . . . , xM } is a blocking walk to A. A contradiction. Next we show CB (Y ∗ |Z ∗ ) = A. If x ∈ CB (Y ∗ |Z ∗ ), we have x ∈ Y ∗ and x ∈ CB (A ∪ {x}|Z ∗ ). So CB (Y ∗ |Z ∗ ) ⊆ Z ∗ ∩ Y ∗ = A. Applying IRC to CB (Y ∗ |Z ∗ ) ⊆ A ⊆ Y ∗ , A = CB (A|Z ∗ ) = CB (Y ∗ |Z ∗ ) ⊆ A. So CB (Y ∗ |Z ∗ ) = A. To see Z ∗ = TB (Z ∗ , Y ∗ ), note that x ∈ Z ∗ ⇐⇒ x ∈ CB (A ∪ {x}|Z ∗ )
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