Efficient and fair trading algorithms in market design environments
aa r X i v : . [ ec on . T H ] M a y Fractional Top Trading Cycle ∗ Jingsheng Yu † Jun Zhang ‡ May 20, 2020
Abstract
Efficiency and fairness are two desiderata in market design. Fairness requires ran-domization in many environments. As one of the few successful matching mechanisms,Top Trading Cycle (TTC) is well known for being efficient to solve deterministic al-location problems. But it is inadequate to incorporate randomization efficiently andfairly. We propose a class of Fractional TTC mechanisms to solve random allocationproblems efficiently and fairly. Dropping the graph-based definition of TTC, we use pa-rameterized linear equations to describe how agents trade endowments or priorities ateach step. Our mechanisms are ex-ante efficient. They satisfy various fairness axiomswhen parameters are properly chosen. We apply them to a couple of market designproblems and obtain efficient and fair assignments in all of them.
Keywords : Top Trading Cycle; random allocation; efficiency; fairness; balanced trade
JEL Classification : C71, C78, D71 ∗ We thank Yeon-Koo Che, Federico Echenique, Fuhito Kojima, William Thomson, Utku ¨Unver, and XiWeng for comments. We received helpful feedbacks from seminars and conferences at Beijing University,California Institute of Technology, Korea Advanced Institute of Science and Technology, Waseda University,Wuhan University, University of International Business and Economics, 2018 Social Choice and WelfareMeeting, 2018 Econometric Society Asian Meeting, 2018 Shanghai Microeconomics Workshop, and 2018International Conference on Economic Theory at Chengdu. All errors are ours. Jingsheng Yu is supportedby the Fundamental Research Funds for the Central Universities (Grant No. JBK2001072). Jun Zhang issupported by National Natural Science Foundation of China (Grant No. 71903093). † Yu: School of Economics & China Center for Behavioral Economics and Finance, Southwestern Univer-sity of Finance and Economics, 555 Liutai Road, Chengdu, 611130, China. Email: [email protected] ‡ Zhang: Institute for Social and Economic Research, Nanjing Audit University, 86 Yushan West Road,Nanjing, 211815, China. Email: [email protected]
Introduction
The past decades have witnessed the rapid development of matching theory and its applica-tions. Gale and Shapley’s (1962) Deferred Acceptance (DA) and Gale’s Top Trading Cycle(TTC; Shapley and Scarf, 1974) are the two best known matching mechanisms. DA is knownfor finding stable allocations when both sides of a market have preferences (sometimes takingthe form of priorities) over their partners, while TTC is known for finding efficient alloca-tions when resources on one side are allocated to agents on the other side. Both mechanismsperform well in solving deterministic allocation problems, but they are inadequate to solverandom allocation problems. Fairness, when imposing symmetry requirements like “equals”being treated equally, is often obtained through randomization. As Budish et al. (2013)put it, “Randomization is commonplace in everyday resource allocation. It is used to breakties among students applying for overdemanded public schools and for popular after-schoolprograms, to ration offices, parking spaces and tasks among employees, to allocate coursesand dormitory rooms amongst college students, and to assign jury and military duties amongcitizens. [...] Randomization can restore symmetry, and thus a measure of fairness.”However, it is hard to incorporate randomization into the procedures of DA and TTCwithout losing their flagship properties. When agents co-own objects or are ranked by coarsepriorities, we have to use randomized endowments or random tie breaking to restore sym-metry. For example, in school choice with coarse priorities, DA with random tie-breakingresults in ex-ante discrimination among students of equal priority, while TTC with randomtie breaking is no longer efficient from an ex-ante view. Moreover, although random tiebreaking before running TTC seems to be fair, this randomization is preference-independent.After it interacts with preferences in the trading process, the fairness of the found assignmentbecomes not transparent. This paper solves the inadequacy of TTC and proposes proba-bilistic generalizations of TTC to solve random allocation problems efficiently and fairly. When agents are ranked by priorities, we often use priority-based fairness notions (e.g., no justified envyin school choice). But if priorities are not strict, which happens in many environments, fairness also requiressymmetry among agents of equal priority. Observing this inadequacy, Kesten and ¨Unver (2015) propose probabilistic variants of DA to find con-strained efficient and ex-ante stable allocations. Yet the mechanisms have undesirable feature in computation. In the extreme case that all students have equal priorities at all schools, TTC with uniformly randompriorities is equivalent to the Random Priority mechanism (RP; Abdulkadiro˘glu and S¨onmez, 1998), whichis shown by Bogomolnaia and Moulin (2001) to be ex-ante inefficient. Following footnote 3, although the ordering of agents in RP is uniformly random so that agents seem tobe treated equally, there may exist envy among agents in the random assignment found by RP.
1e present our mechanisms in a direct extension of Shapley and Scarf’s housing marketmodel that we call fractional endowment exchange model (FEE). In the model an agentmay own fractions of several objects and the fractions of an object may be owned by sev-eral agents. The house allocation model where random allocation has been studied sinceHylland and Zeckhauser (1979) can be regarded as an FEE problem in which agents ownequal fractions of all objects, whereas the housing market model is a special case of FEEin which agents have discrete endowments. In applications we extend our mechanisms topriority-based allocation problems.To see the difficulty of solving FEE by TTC, suppose we mimic its procedure by lettingagents point to favorite objects and objects point to owners step by step, and clearing cyclesat each step. When an object has several owners, to avoid discrimination let the object pointto all of its owners. Then a problem arises: The cycles generated at each step may not bedisjoint and an agent may appear in several cycles. How to trade intersecting cycles in anefficient and fair way is unclear. Moreover, even though the cycles at some step are disjoint,clearing all of them can lead to unfair outcomes. We present examples to illustrate thesesituations in Section 3.To circumvent the above difficulty, we let agents report favorite objects step by step, butwe switch to use parameterized linear equations to describe how agents trade endowmentsat each step. The equations require that agents trade endowments in a balanced way: Ateach step the amount of favorite object obtained by each agent, which is an unknown in ourequations, be equal to the amount of endowments lost by the agent. To control fairness, weuse a set of parameters to describe how the owners of each object divide the right of using theobject to trade with the others. For example, at each step the owners of each object may useequal amounts of the object to trade with the others, or they may use amounts proportionalto their endowments to trade with the others. We use another set of parameters to controlthe amount of endowments agents trade at each step. The solution to the equations at eachstep tells us how much agents gain and how much they lose. By choosing different parametervalues, we obtain different mechanisms. So we define a class of mechanisms, and call all ofthem Fractional Top Trading Cycle (FTTC).To prove that FTTC is well-defined, we prove that the solution to the equations ateach step exists. Our equations essentially describe an equilibrium that has been studied inthe closed Leontief input-output model (Leontief, 1941). The model describes input-outputrelationships, characterized by a set of linear equations, among finite industries in a simpleeconomy. When the economy reaches an equilibrium, every industry makes zero profit. This2mplies that every column of the coefficient matrix of the input-output equations sums to one;in linear algebra such matrix is called stochastic. In this paper, balanced endowment tradeimplies that every column of the coefficient matrix of our equations sums to one. Based onexisting (linear algebra) results on the Leontief model, we prove the existence of the solutionat each step. Interestingly, by connecting to the Leontief model we obtain a description ofFTTC resembling that of TTC. In the equilibrium of the Leontief model, industries formautonomous sub-economies. Similarly, at each step of FTTC, if we generate a graph byletting agents point to favorite objects and objects point to owners, we will see that someagents and objects form disjoint parts in the graph that we call absorbing sets . The agents ineach absorbing set trade their endowments in the set among themselves, whereas the agentsand objects not in any absorbing set are not involved in any trade. Thus, FTTC is actually aprocedure of trading absorbing sets, extending the usual description of TTC. In the housingmarket model, absorbing sets reduce to cycles.Efficiency and fairness are two desiderata in market design. FTTC can be efficient and fairsimultaneously. Every FTTC is individually rational and sd-efficient (Bogomolnaia and Moulin,2001). That is, the lottery every agent obtains from FTTC first-order stochastically dom-inates his fractional endowments, and no agent can be made better off without harmingthe others. FTTC satisfies various fairness axioms when its parameters are properly cho-sen. We propose a fairness axiom called bounded envy , which imposes fairness among anytwo agents. It requires that if an agent is envied by another agent, the envy should bebounded by the former agent’s advantage in endowments. In particular, bounded envy im-plies no envy among agents of equal endowments. As examples, we present three FTTCthat have different fairness motivations but all satisfy bounded envy. To clarify their fair-ness, we relate them to three classical solution rules in the bankruptcy problem (O’Neill,1982; Aumann and Maschler, 1985; Thomson, 2003, 2015).In the FEE model there is an inherent conflict between individual rationality, efficiency,and (weak) strategy-proofness (Athanassoglou and Sethuraman, 2011; Aziz, 2018). So everyFTTC is not strategy-proof. As many papers in the literature (e.g., Che and Kojima, 2010),we consider large markets and prove that fair FTTC is asymptotically strategy-proof asmarket size increases and agents’ preferences diversify. We also examine the core notion inthe FEE model. In the housing market model, it is well known that TTC finds the uniquecore allocation. With ordinal preferences there are more than one ways to define the corenotion in the FEE model. We propose a reasonable core notion, and prove that the core canbe empty, but the weak core is always nonempty. There always exists a weak core assignment3hat treats equals equally, but sometimes all weak core assignments violate equal-endowmentno envy. So a fair FTTC may not always find a weak core assignment.This paper contributes to matching theory and its applications. We add new mechanismsto market designers’ toolkit. They provide new solutions to existing applications and makenew applications possible. As examples, we present four applications . We first apply FTTCto school choice with coarse priorities. Although coarse priorities are widely used in practice,strict priorities are commonly assumed by many papers. As well recognized by the literature,any preference-independent tie breaking rule is inefficient. Our FTTC does not explicitlybreak priority ties. At each step, we endow the students of equal highest priority at eachschool with equal rights of using the school’s seats to trade with the others. The foundassignment is efficient and fair. This solution generalizes Abdulkadiro˘glu and S¨onmez (2003)who first apply TTC to school choice by assuming strict priorities.In the second application we study time exchange markets. A prominent example is thevarious form of time banks that appear in many countries. Agents in such markets exchangetime and skills through serving each other without using payment. A time exchange marketcan be modeled as an FEE problem in which objects are services and endowments are timeunits of services that agents can provide. FTTC can be straightforwardly used to solve suchmodel and find efficient and fair assignments.In the third application we study the allocation of agents to institutions under (geograph-ically) distributional constraints, as studied by Kamada and Kojima (2015) in the contextof Japanese medical residency match. A floor (or ceiling) constraint restricts the minimum(or maximum) number of agents assigned to a group of institutions. The literature hasdemonstrated the difficulty caused by floor constraints. We develop an efficient and fairmethod to deal with both ceiling and floor constraints. By assuming institutions prefer tohire more agents when constraints are not violated, we first generate an assignment in whichagents have equal access to all institutions and the numbers of agents assigned to institutionscannot be Pareto improved. We then treat this assignment as initial endowment, and runa fair FTTC. The found assignment satisfies all constraints, ensures no envy among agents,and is Pareto efficient when the preferences of both sides are taken into account.In the last application we study the house allocation model, which we regard as a specialcase of FEE. It turns out that every FTTC coincides with a simultaneous eating algorithmdefined by Bogomolnaia and Moulin (2001), and every fair FTTC that treats agents of equalendowments equally coincides with Probabilistic Serial (PS). In an eating algorithm agents See Andersson et al. (2019) for more description of real-life time banks. In the house allocation model, our individual rationality becomes equivalent to the fairness axiom called equal-division lower bound . The coincidence between fair FTTC and PS transparently clarifies the connectionbetween PS and TTC, which was first found by Kesten (2009). .1 Related literature This paper contributes to the market design literature, and is in particular related to therandom allocation literature. Many studies on random allocation have centered around twoefficient and fair mechanisms: Hylland and Zeckhauser’s (1979) Pseudo-market mechanismand Bogomolnaia and Moulin’s (2001) PS mechanism. Both mechanisms are first proposedin the house allocation model and then generalized to various environments. Pseudo-marketrequires cardinal utilities and seeks a competitive equilibrium allocation, while PS, as othermatching mechanisms, requires only ordinal preferences and has an algorithmic definition.We survey several papers that develop these two mechanisms and compare them with ours.Yılmaz (2010) extends PS to the house allocation with existing tenants model. His ideais to minimally deviate from equal eating rate in PS without violating IR constraints ofexisting tenants. For example, if an existing tenant least prefers his private endowment,his ownership is ignored in the mechanism. The mechanism satisfies a fairness axiom calledno justified envy (NJE). Athanassoglou and Sethuraman (2011) extend Yilmaz’s mechanismto the FEE model. Their mechanism satisfies NJE but violates equal-endowment no envy(EENE). As they put it, EENE is natural because “two agents with identical endowmentsbring exactly the same resources to the group, so any differences in their final assignmentshould be explained solely by their preferences.” They wonder the existence of a mechanismsatisfying sd-efficiency, individual rationality and EENE, and regard a generalization of TTCto the FEE model as a challenging question. This paper answers both questions.Many allocation problems are subject to quantity constraints. Budish et al. (2013) char-acterize the constraint structures that preserve the Birkhoff-von Neumann theorem, which iscrucial for the implementation of random assignments. They generalize the two mechanismsmentioned above to accommodate various constraints. Akbarpour and Nikzad (forthcoming)extend their result by relaxing some constraints and considering approximate implementa-tion. In an application we use FTTC to solve geographically distributional constraints. Howto solve more general constraints is an interesting direction for future research.Echenique et al. (2019) generalize Pseudo-market to solve any constraints that pin downa set of feasible deterministic assignments. By taking the convex hull of feasible deterministicassignments as the primitive, they circumvent the implementation issue. Echenique et al.(2020) extend Yilmaz’s NJE to a general environment with participation constraints, anduse Pseudo-market with price-dependent incomes to find desirable allocations. Both papersassume cardinal utilities and find market equilibrium allocations. So their methodology israther different than ours. 6n an unpublished paper Aziz (2015) makes an attempt to generalize TTC to the FEEmodel. To solve the difficulty caused by non-disjoint cycles (see Section 3), Aziz uses ex-ogenous rankings of agents and objects to prioritize cycles, and essentially selects a subsetof cycles at each step to trade. This makes his mechanism unfair. In Online Appendix wedescribe his mechanism as an FTTC, and explain that it does not treat equal agents equally.Similarly, Altuntas and Phan (2017) define probabilistic variants of TTC by letting objectsuse strict priorities to rank agents. So their mechanisms are straightforward extensions ofTTC. Priorities introduce asymmetry and thus cause artificial unfairness.Many papers are related to our applications, but we only discuss the most related here. Inthe school choice model with coarse priorities, Kesten and ¨Unver (2015) propose two ex-antestability notions as well as two probabilistic variants of DA to find constrained efficient andex-ante stable allocations. He et al. (2018) generalize Pseudo-market to the same model. Byusing priority-dependent prices, their mechanism finds ex-ante stable allocations. FollowingAbdulkadiro˘glu and S¨onmez (2003), in our application of FTTC we do not take stability asa desideratum.Andersson et al. (2019) and Manjunath and Westkamp (2019) propose two different timeexchange models. In the former model each agent provides a distinct service and has dichoto-mous preferences over services in the market. In the latter model agents are endowed withdisjoint sets of shifts and have dichotomous preferences over the other agents’ shifts. Bothpapers recommend priority mechanisms because fairness is not their concern. Differently,in our model an agent can provide several services and several agents can provide a sameservice. FTTC lets agents trade services, and finds fair assignments.Considering ceiling constraints, Kamada and Kojima (2015) adapt DA to the Japanesemedical residency match model. Their mechanism satisfies a priority-based fairness notion.Akin (2019) follows their approach and generalizes their fairness notion to accommodate floorconstraints. Kamada and Kojima (2018) relax their fairness notion but find that absenceof floor constraints is still crucial for the existence of desirable matchings. When transfersare allowed, Kojima et al. (2019) find that only ceiling and floor constraints on capacitiesof hospitals preserve the gross substitutes property of hospitals’ demands and guarantee theexistence of competitive equilibria. Our method can solve both ceiling and floor constraints.By using random allocation, our method abandons priorities and treats doctors equally.TTC has been intensively studied in deterministic allocation environments. Alcalde-Unzu and Molis(2011) use the notion of absorbing sets to deal with weak preferences in TTC (also seeJaramillo and Manjunath (2012)). By identifying absorbing sets at each step, they de-7ermine when it is safe to remove a group of agents and their assignments without los-ing efficiency. Differently, we use absorbing sets to provide a description of FTTC. Torun FTTC, we do not need to identify absorbing sets at each step. TTC has been ex-tended by P´apai (2000), Pycia and ¨Unver (2017), Morrill (2015), and Hakimov and Kesten(2018), among others. TTC has been applied to allocation problems including school choice(Abdulkadiro˘glu and S¨onmez, 2003), house allocation (Abdulkadiro˘glu and S¨onmez, 1999),kidney exchange (Roth et al., 2004), and tuition and worker exchange (Dur and ¨Unver,2019). TTC has been characterized in different ways by Ma (1994), Morrill (2013), Dur(2013), Fujinaka and Wakayama (2018), Dur and Morrill (2018), and Abdulkadiroˇglu et al.(2017). A fractional endowment exchange (FEE) problem is a four-tuple ( I, O, ≻ I , ω ) in which • I is a finite set of agents; • O is a finite set of objects; • ≻ I = {≻ i } i ∈ I is the preference profile of agents; • ω = ( ω i,o ) i ∈ I,o ∈ O is the endowment matrix.For each agent i , ω i = ( ω i,o ) o ∈ O denotes i ’s endowments, with ω i,o ∈ [0 , being the amount(probability share) of object o owned by i . Let q o = P i ∈ I ω i,o denote the total amount ofobject o , which is an integer. We require P o ∈ O ω i,o ≤ for all i ∈ I . Each i has a strictpreference relation ≻ i , which is a linear order on O , with % i being the at least as good asrelation; that is, o % i o ′ if and only if o ≻ i o ′ or o = o ′ . We present our results by treatingobjects as indivisible since this is the case in most market design problems. But our modeland mechanisms can apply to environments in which objects are divisible. Our applicationto time exchange markets is an example. The FEE model reduces to the housing marketmodel if | I | = | O | and ω is a permutation matrix.A lottery is a vector l ∈ ℜ | O | + such that P o ∈ O l o ≤ . A lottery l weakly (first-order)stochastically dominates another lottery l ′ for agent i , denoted by l % sdi l ′ , if P o ′ % i o l o ′ ≥ When each object represents a service type, if one unit of a service means one day or one hour, then ahalf of the service means a half day or a half hour. o ′ % i o l ′ o ′ for all o ∈ O . If the inequality is strict for some o , l strictly stochastically dominates l ′ for agent i , denoted by l ≻ sdi l ′ .An assignment is a matrix p = ( p i,o ) i ∈ I,o ∈ O ∈ ℜ | I |×| O | + such that P i ∈ I p i,o ≤ P i ∈ I ω i,o forall o ∈ O and P o ∈ O p i,o ≤ for all i ∈ I . Each p i,o is the amount of o assigned to i . The rowvector p i = ( p i,o ) o ∈ O is the lottery assigned to i . When objects are indivisible, if all elements of p are integers, then p is a deterministic assignment. The Birkhoff-von Neumann theorem andits generalization (Birkhoff, 1946; Von Neumann, 1953; Kojima and Manea, 2010) guaranteethat every assignment is a convex combination of deterministic assignments. An assignment p strictly stochastically dominates another assignment p ′ , denoted by p ≻ sdI p ′ , if p i % sdi p ′ i for all agent i and p j ≻ sdj p ′ j for some agent j . An assignment p is sd-efficient if it is neverstrictly stochastically dominated. It is individually rational (IR) if p i % sdi ω i for all i ∈ I . IRimplies that P o ∈ O p i,o = P o ∈ O ω i,o for all i .An assignment p satisfies equal treatment of equals (ETE) if for all i, j ∈ I such that ω i = ω j and ≻ i = ≻ j , p i = p j ; p satisfies equal-endowment no envy (EENE) if for all i, j ∈ I such that ω i = ω j , p i % sdi p j and p j % sdj p i . EENE is stronger than ETE.We denote an FEE problem by its preference profile when its other elements are fixed.Let P denote the set of all linear orders on O . A mechanism ϕ finds an assignment ϕ ( ≻ I ) foreach ≻ I ∈ P | I | . The lottery assigned to each i in ϕ ( ≻ I ) is denoted by ϕ i ( ≻ I ) . A mechanismsatisfies an efficiency or fairness axiom if its found assignments satisfy the axiom.An agent i weakly manipulates a mechanism ϕ at ≻ I by reporting ≻ ′ i ∈ P\{≻ i } if ϕ i ( % I ) % sdi ϕ i ( % ′ i , % − i ) . Agent i strongly manipulates ϕ at ≻ I by reporting ≻ ′ i if ϕ i ( % ′ i , % − i ) ≻ sdi ϕ i ( % I ) . ϕ is (weakly) strategy-proof if it is never (strongly) manipulated. In the housing market model, every agent owns a distinct object. The procedure of TTC issimple: At each step, first let agents point to favorite objects and objects point to owners,and then let the agents in each resulting cycle exchange endowments. Clearly, at each stepcycles must exist and be disjoint. To extend TTC to the FEE model, a straightforward ideais to first let agents point to favorite objects and objects point to all of their owners (toavoid discrimination among owners), and then clear cycles. However, two examples in thissection illustrate the problem with this idea.Example 1 shows that the cycles generated as above can be complicated and not disjoint.It is unclear how to clear cycles to obtain an efficient and fair assignment.9 xample 1.
Consider five agents { , , , , } and five objects { o , o , o , o , o } . Agentshave endowments and preferences shown in Table 1. Letting agents point to favorite objectsand objects point to all of their owners, we obtain Figure 1. The weight of every edge o → i denotes the amount of o owned by i . o o o o o / / / / / / /
44 0 0 1 / / /
45 0 0 1 / / (a) Endowments ≻ ≻ ≻ ≻ ≻ o o o o o o o o o o o o o o o o o o o o o o o o o (b) Preferences Table 1 o o o o o
12 34 5 / / / / / / / / / / / / Figure 1
There are five cycles in the figure: o → → o → → o → → o , o → → o → → o , o → → o → → o , o → → o , and o → → o . These cycles arenot disjoint. Some share edges and some nest the others. These cycles cannot be clearedsimultaneously as in TTC. Which cycles to clear and how to clear them are crucial forefficiency and fairness. One idea is to find some way to clear all cycles simultaneously andfairly. Another idea is to select disjoint cycles by some rule and then clear them fairly. In A cycle is nested by another cycle if every node in the former cycle is involved in the latter cycle. nline Appendix E, we discuss several ways to implement these ideas but show that they areundesirable in fairness. Example 2 shows that even though the cycles generated as above are disjoint at somestep, clearing all of them does not give us a fair assignment.
Example 2.
Consider four agents { , , , } and three objects { o , o , o } . Agent 1 owns / o and most prefers o . Agent 2 owns / o and prefers o to o and o to o . Agent 3owns / o and most prefers o . Agent 4 owns / o and most prefers o . The remainingamounts of objects are owned by other agents omitted for simplicity. Letting agents pointto favorite objects and objects point to all of their owners, we obtain Figure 2. o o o / / / / Figure 2
There are two cycles and they are disjoint. If we clear both cycles, 1 obtains his endow-ment / o , 3 and 4 exchange endowments so that 3 obtains / o and 4 obtains / o ,and 2 remains with his endowment / o . This assignment is unfair for 2. Because 1 mostprefers o , in any IR assignment 1 must obtain / o . After removing 1 and his endowment,2 and 3 essentially have equal preferences. But letting 3 and 4 exchange endowments before2 points to o , 2 has no chance to exchange endowments with to obtain any amount of o .This makes the assignment violate EENE among 2 and 3. As TTC, our FTTC lets agents report favorite objects step by step. But FTTC uses linearequations to define how agents trade endowments at each step. The key to our equations is Without loss of generality, let omitted agents most prefer their endowments. d , let I ( d ) and O ( d ) respectively denote theset of remaining agents and the set of remaining objects; let ω ( d ) = ( ω i,o ( d )) i ∈ I,o ∈ O denoteremaining endowment matrix; let p ( d ) = ( p i,o ( d )) i ∈ I,o ∈ O denote the found assignment bystep d , where p i,o ( d ) is the amount of object o obtained by agent i ; let o i ( d ) denote agent i ’sfavorite object among O ( d − . At step d , let x i ( d ) denote the amount of o i ( d ) obtained by i ∈ I ( d − , and let x o ( d ) denote the total amount of o ∈ O ( d − assigned to agents. Soby definition, for all o ∈ O ( d − , x o ( d ) = X i ∈ I ( d − o i ( d )= o x i ( d ) . If no agents report an object o ∈ O ( d − as favorite at step d , x o ( d ) = 0 .Because all objects are owned by agents, x o ( d ) is also the total amount of o lost by itsowners from their endowments at step d . Different owners may lose different amounts of o .We use a parameter λ i,o ( d ) to denote the ratio of the amount lost by agent i to the totalamount x o ( d ) . In other words, i loses the amount λ i,o ( d ) x o ( d ) of o from his endowments atstep d . We put these parameters into a matrix λ ( d ) = (cid:0) λ i,o ( d ) (cid:1) i ∈ I ( d − ,o ∈ O ( d − , and call it ratio matrix . λ ( d ) satisfies that, for all i ∈ I ( d − and all o ∈ O ( d − , λ i,o ( d ) ∈ [0 , , P i ∈ I ( d − λ i,o ( d ) = 1 , and λ i,o ( d ) > only if ω i,o ( d − > . The total amount of endowmentsthat each i ∈ I ( d − loses at step d is P o ∈ O ( d − λ i,o ( d ) x o ( d ) . So the balanced trade conditionrequires that, for all i ∈ I ( d − , x i ( d ) = X o ∈ O ( d − λ i,o ( d ) x o ( d ) . We use another parameter β i,o ( d ) to control the maximum amount of o ∈ O ( d − that i ∈ I ( d − can lose at step d . That is, λ i,o ( d ) x o ( d ) ≤ β i,o ( d ) . For all i ∈ I ( d − and all o ∈ O ( d − , ≤ β i,o ( d ) ≤ ω i,o ( d − . We call β ( d ) = (cid:0) β i,o ( d ) (cid:1) i ∈ I ( d − ,o ∈ O ( d − quota matrix .Now we are ready to present the procedure of FTTC. Fractional Top Trading CycleInitialization : I (0) = I , O (0) = O , ω (0) = ω , and p (0) = . Step d ≥ : Each i ∈ I ( d − reports his favorite remaining object o i ( d ) . Given apair of λ ( d ) and β ( d ) , let x ∗ ( d ) = ( x ∗ a ( d )) a ∈ I ( d − ∪ O ( d − denote the maximum solution12o the equation system x o ( d ) = P i ∈ I ( d − o i ( d )= o x i ( d ) for all o ∈ O ( d − ,x i ( d ) = P o ∈ O ( d − λ i,o ( d ) x o ( d ) for all i ∈ I ( d − , (1)that satisfies the constraints λ i,o ( d ) x o ( d ) ≤ β i,o ( d ) for all i ∈ I ( d − and o ∈ O ( d − . (2)For all i ∈ I and all o ∈ O , let p i,o ( d ) = p i,o ( d −
1) + x ∗ i ( d ) if i ∈ I ( d − and o = o i ( d ) ;otherwise let p i,o ( d ) = p i,o ( d − . Let ω i,o ( d ) = ω i,o ( d − − λ i,o ( d ) x o ( d ) if i ∈ I ( d − and o ∈ O ( d − ; otherwise let ω i,o ( d ) = ω i,o ( d − . Let I ( d ) = { i ∈ I : P o ∈ O ω i,o ( d ) > } and O ( d ) = { o ∈ O : P i ∈ I ( d ) ω i,o ( d ) > } . If O ( d ) is empty, stop the procedure;otherwise go to step d + 1 .In the next section we prove that the solutions to the equation system (1) exist. Con-straints (2) pin down the maximum solution: for any other solution x ( d ) that satisfies (2), x ∗ a ( d ) ≥ x a ( d ) for all a ∈ I ( d − ∪ O ( d − and x ∗ a ( d ) > x a ( d ) for some a . The linearequations (1) can be solved by many well-developed computation methods and computerprograms. In simple examples, they can be solved by hand.Each step of FTTC takes a pair of λ ( d ) and β ( d ) as inputs. By choosing different λ ( d ) and β ( d ) , we obtain different mechanisms. So we define a class of FTTC mechanisms. Thechoice of λ ( d ) is usually related to fairness. It determines how the owners of each objectdivide the right of using the object to trade with the others. We discuss fairness in Section6. The choice of β ( d ) is usually related to the speed of FTTC (yet it may also be used tocontrol fairness as shown in Section 7). For example, by choosing β i,o ( d ) = ω i,o ( d − , welet agents trade endowments as much as possible at each step. It means that at each stepat least one agent uses up his endowment of some object, so FTTC must stop in at most | I | × | O | steps. In general, as long as a (fixed) minimum amount of objects are traded at eachstep, FTTC stops in finite steps. To illustrate the procedure defined above, in Appendix Dwe use a specific FTTC presented in Section 7 to solve Example 1. The procedure to solveExample 2 is straightforward. For example, if we want and to have equal rights to use o to trade with the others, we will only clear the cycle o → → o in Figure 2.13 Leontief model and absorbing set
The procedure of FTTC needs to solve linear equations at each step. The coefficient matrixdoes not have full rank, but we prove that its solutions exist. Then the constraints pin downthe maximum solution, which is nonnegative.
Proposition 1.
At each step d of FTTC, the maximum solution x ∗ ( d ) to (1) subject to (2)exists and is nonnegative. The proof relies on an observation that the equation system (1) essentially describes anequilibrium that has been studied in the closed Leontief input-ouput model (Leontief, 1941).Specifically, the coefficient matrix of (1) is stochastic ; that is, its every column sums to one.This is also the key feature of the Leontief model. Existing studies on the Leontief model(Peterson and Olinick, 1982; Leon, 2015) imply Proposition 1. In Appendix A we provide aself-contained proof.The Leontief model describes an economy consisting of n industries that produce n dif-ferent products. Each industry requires input of the products from the other industries andpossibly also of its own. Let a uv ∈ [0 , denote the amount of input from the u -th indus-try that necessary to produce one unit of output in the v -th industry (by a unit we meanone dollar’s worth). Let x u denote the units of output produced by the u -th industry. Bydefinition, x = ( x , x , . . . , x n ) satisfies the set of linear equations: x u = a u x + a u x + · · · + a un x n for all u = 1 , , . . . , n. That is, all output of the u -th industry becomes input into the other industries and possiblyits own. When the economy reaches an equilibrium, every industry makes a least zero profit.Eventually, all industries make zero profits, and thus P nu =1 a uv = 1 for all v = 1 , , . . . , n . Similarly, at each step d of FTTC, suppose I ( d −
1) = { i , . . . , i n } and O ( d −
1) = { o , . . . , o m } . Then the equation system (1) can be written as " B n × m C m × n x i ( d ) ... x in ( d ) x o ( d ) ... x om ( d ) = x i ( d ) ... x in ( d ) x o ( d ) ... x om ( d ) , (3) In equilibrium, the cost of producing one unit of output in every v -th industry is no greater than onedollar. That is, P nu =1 a uv ≤ . By definition, P nv =1 a uv x v = x u . Thus, P nu =1 P nv =1 a uv x v = P nu =1 x u .Rearranging terms, we have P nv =1 ( P nu =1 a uv ) x v = P nu =1 x u . So P nu =1 a uv = 1 . B n × m = ( b u,v ) and C m × n = ( c v,u ) are such that, for all u ∈ { , . . . , n } and all v ∈{ , . . . , m } , b u,v = λ i u ,o v ( d ) , c v,u = , if o v = o i u ( d ) , , otherwise . In words, B n × m coincides with λ and C m × n indicates agents’ favorite objects. Let A denotethe coefficient matrix of (3). The definition of λ and agents’ strict preferences imply thatevery column of A sums to one.In the equilibrium of the Leontief model, industries form autonomous sub-economies.Similar happens at each step of FTTC. In the proof of Proposition 1, we show that someagents and objects form disjoint groups that we call absorbing sets . The agents in eachabsorbing set trade their endowments in the absorbing set among themselves, and the agentsand objects not in any absorbing set are not involved in any trade. In this sense, absorbingsets are extensions of cycles in TTC, and FTTC can be regarded as a procedure of clearingabsorbing sets. In the housing market model, absorbing sets reduce to cycles and FTTCreduces to TTC.Formally, at each step d , we generate a graph by letting agents point to favorite objectsand letting every o ∈ O ( d − point to every i ∈ I ( d − such that λ i,o ( d ) > . We saythere is a directed path from a node v to another node v ′ if there exists a sequence of nodes v , v , . . . , v z such that v = v , v z = v ′ , and for every ℓ ∈ { , . . . , z − } , v ℓ points to v ℓ +1 . Definition 1.
A subset V ⊂ I ( d − ∪ O ( d − is an absorbing set at step d if1. V is inside connected: within V there is a directed path from every node to every othernode;2. V has no outgoing edge: there is no directed path from any node in V to any nodeoutside of V . If an agent i belongs to an absorbing set V , his favorite object o i ( d ) belongs to V ; ifan object o belongs to V , every i ∈ I ( d − with λ i,o ( d ) > belongs to V . If we restrictthe equations (1) and constraints (2) to the nodes in each absorbing set V and denote themaximum solution by x ∗ V ( d ) , then x ∗ V ( d ) tells us how the agents in V trade their endowmentsin V among themselves at step d . Let V , . . . , V k denote the absorbing sets at step d , andlet U denote the set of agents and objects not in any absorbing set. Then the maximumsolution x ∗ ( d ) at step d is simply the combination of x ∗ V ( d ) : x ∗ ( d ) = ( x ∗ V ( d ) , . . . , x ∗ V k ( d ) , U ) .
15n the two examples in Section 3, if λ i,o (1) > if and only if ω i,o > , then the nodes otherthan o , o constitute the only absorbing set in Figure 1, and agent and his endowment o constitute the only absorbing set in Figure 2. Remark 1.
We can directly prove the existence of absorbing sets, their disjointness, and thefact that agents in each absorbing set trade endowments among themselves. • Existence is implied by the following facts: first, the whole graph has no outgoing edge;second, if a set of nodes has no outgoing edge but is not inside connected, then it mustcontain a strict subset that has no outgoing edge; last, every singleton set is insideconnected. • Suppose two absorbing sets share a node. Then there must be a directed path fromevery node in one absorbing set to the shared node, and another directed path from theshared node to every node in the other absorbing set. This is a contradiction. • Suppose an agent outside of an absorbing set obtains a positive amount of some objectin the absorbing set. Because the agents in the absorbing set do not obtain any amountof objects outside of the absorbing set, the aggregate balanced trade condition for theagents in the absorbing set is violated.
FTTC can satisfy efficiency and fairness simultaneously. We first prove that FTTC is indi-vidually rational and sd-efficient. It is the obvious consequence of stepwise balanced trade.
Proposition 2.
FTTC is individually rational and sd-efficient.
We then present conditions on the parameters in FTTC to ensure various fair axioms.ETE and EENE are standard axioms that require fairness among “equal” agents. In the houseallocation model, EENE is known as envy-freeness because all agents have equal endowments.In the FEE model agents may own different endowments. This motivates us to define a newaxiom that requires fairness among agents of unequal endowments. The axiom allows for If V ⊂ I ( d − ∪ O ( d − has no outgoing edge but is not inside connected, then there exist v, v ′ ∈ V such that there is no directed path from v to v ′ . Let V be the set of nodes in V that can be reached from v through directed paths. Then v ′ / ∈ V . Since V has no outgoing edge, there is no directed path from anynode in V to any node outside of V . The definition of V implies that there is no directed path from anynode in V to any node in V \ V . So V has no outgoing edge. i envies another agent j in anassignment p , we use max o ∈ O (cid:2) P o ′ % i o p j,o ′ − P o ′ % i o pi i,o ′ (cid:3) to measure i ’s envy towards j . Definition 2.
A mechanism ϕ satisfies bounded envy if, for all ≻ I ∈ P | I | and all i, j ∈ I , max o ∈ O (cid:2) X o ′ % i o ϕ j,o ′ ( ≻ I ) − X o ′ % i o ϕ i,o ′ ( ≻ I ) (cid:3) ≤ X o ∈ O : ω j,o >ω i,o (cid:0) ω j,o − ω i,o (cid:1) . It is clear that bounded envy implies EENE.Now we present fairness conditions on the procedure of FTTC.
Definition 3.
An FTTC satisfies(1) stepwise equal treatment of equals (stepwise ETE) if at each step d , ω i ( d −
1) = ω j ( d − and o i ( d ) = o j ( d ) imply λ i ( d ) = λ j ( d ) ;(2) stepwise equal-endowment equal treatment (stepwise EEET) if at each step d , ω i ( d −
1) = ω j ( d − implies λ i ( d ) = λ j ( d ) ;(3) bounded advantage if at each step d , ω i,o ( d − ≥ ω j,o ( d − implies that λ i,o ( d ) ≥ λ j,o ( d ) and that ω i,o ( d ) ≥ ω j,o ( d ) . The first two conditions apply the ideas of ETE and EENE to the procedure of FTTC.Note that λ i ( d ) = λ j ( d ) implies x i ( d ) = x j ( d ) and ω i ( d − − ω i ( d ) = ω j ( d − − ω j ( d ) .Stepwise ETE means that if two agents have equal remaining endowments at the beginningof a step and they most prefer the same object at the step, then they obtain equal amountsof the favorite object and lose equal amount of each endowment at the step. Stepwise EEETmeans that if two agents have equal remaining endowments at the beginning of a step, theyobtain equal amounts of their respective favorite objects and lose equal amount of eachendowment at the step.In the third condition, if ω i,o ( d − ≥ ω j,o ( d − , bounded advantage implies that ≤ λ i,o ( d ) x o ( d ) − λ j,o ( d ) x o ( d ) = (cid:2) ω i,o ( d − − ω i,o ( d ) (cid:3) − (cid:2) ω j,o ( d − − ω j,o ( d ) (cid:3) = (cid:2) ω i,o ( d − − ω j,o ( d − (cid:3) − (cid:2) ω i,o ( d ) − ω j,o ( d ) (cid:3) ≤ ω i,o ( d − − ω j,o ( d − .
17o if i owns more of an object o than j at the beginning of step d , then i uses more of o than j to trade with the others at step d , but the extra amount of o that i uses is boundedby the difference between their remaining endowments of o . Proposition 3. (1) An FTTC satisfying stepwise ETE satisfies ETE;(2) An FTTC satisfying stepwise EEET satisfies EENE;(3) An FTTC satisfying bounded advantage satisfies bounded envy.
The proofs for this section are presented in Appendix B.
By now we have not presented any specific FTTC mechanism. In this section we presentthree fair FTTC. They have different fairness motivations, but all of them satisfy boundedadvantage defined in Section 6. So they satisfy bounded envy, and thus EENE. We regardthem as good choices for applications that plan to use an FTTC mechanism.The first mechanism is called equal-FTTC and denoted by T e . At each step, the remain-ing owners of each object use equal amounts of the object to trade with the others. Formally,at each step d , T e uses the following λ e ( d ) and β e ( d ) : λ ei,o ( d ) = { ω i,o ( d − > } P j ∈ I ( d − : { ω j,o ( d − > } ,β ei,o ( d ) = ω i,o ( d − . Here is the indicator function. If two agents i and j both own an object o and ω i,o > ω j,o ,then they use equal amounts of o to trade at each step of T e until j uses up his endowmentof o earlier than i . In Appendix D we use T e to solve Example 1.The second mechanism is called proportional-FTTC and denoted by T p . At each step,the remaining owners of each object use amounts proportional to their endowments of theobject to trade with the others. Formally, at each step d , T p uses the following λ p ( d ) and β p ( d ) : λ pi,o ( d ) = ω i,o ( d − P j ∈ I ( d − ω j,o ( d − ,β pi,o ( d ) = ω i,o ( d − . If two agents own different amounts of an object, they use different amounts of the object totrade at each step of T p , but they use up their endowments of the object at the same step.18he third mechanism is called ordered-FTTC and denoted by T o . At each step, amongthe remaining owners of each object, only those who own the most amount can use theobject to trade with the others, and the amount they can use is no more than the differencebetween the most amount and the second most amount. Formally, T o uses the following λ o ( d ) and β o ( d ) : λ oi,o ( d ) = { ω i,o ( d −
1) = ω { } o ( d − } P j ∈ I ( d − : { ω j,o ( d −
1) = ω { } o ( d − } ,β oi,o ( d ) = { ω i,o ( d −
1) = ω { } o ( d − } (cid:2) ω { } o ( d − − ω { } o ( d − (cid:3) , where ω { } o ( d −
1) = max { ω i,o ( d −
1) : i ∈ I ( d − } , and ω { } o ( d −
1) = max { ω i,o ( d −
1) : i ∈ I ( d − , ω i,o ( d − = ω { } o ( d − } . So if two agents i and j both own some object o and ω i,o > ω j,o , then i uses o to trade earlierthan j in T o , but once they use o at the same step, they use equal amounts, and they useup their endowments of o at the same step.Example 3 explains the difference between the three mechanisms. Example 3.
We modify Example 2 by removing agent 1 and object o . Now agents 2 and3 respectively own / o and / o , and they prefer o to o . Agent 4 owns ω ,o of o andprefers o to o . o o / / ω ,o (a) Preferences and endowments p ,o p ,o / / / / , / / , / / , / T e T o T p (b) Assignments found by the three FTTC Figure 3
Figure 3a shows agents’ preferences and endowments. When we vary ω ,o from to / ,the three lines in Figure 3b show the amounts of o obtained by 2 and 3 respectively in thethree mechanisms. The gray area is the set of IR assignments satisfying bounded envy. ake ω ,o = 1 / for example. In T e , 2 and 3 each use / o to trade with 4. In T p , 2and 3 respectively use / o and / o to trade with 4. In T o , at step one 2 uses / o totrade with 4; at step two 2 and 3 each use / o to trade with 4. The three nodes in the threelines in Figure 3b denote the assignments found by the three mechanisms. The fairness motivations behind the three mechanisms are related to solution rules inthe classical bankruptcy problem. The bankruptcy problem describes a situation in which agroup of agents I claim on a resource, each claiming c i ∈ R + , but the amount of the resource E ∈ R + is not enough to satisfy all claims, that is, P i ∈ I c i ≥ E . A solution rule is an awardsvector x ∈ R n + such that ≤ x i ≤ c i for all i ∈ I and P i ∈ I x i = E . Different fairness criteriamotivate different rules. We list three classical rules in Table 2.Solution rule DefinitionConstrained equal awards rule (CEA) ∃ α ∈ R + such that x i = min { c i , α } Constrained equal losses rule (CEL) ∃ α ∈ R + such that x i = max { α − c i , } Propositional rule (P) ∃ α ∈ R + such that x i = αc i Table 2: Three rules in the bankruptcy problemIn a rule x , x i and c i − x i are the award and loss of agent i . CEA equalizes the awardsof agents as much as possible. It first gives agents amounts of resource equal to the smallestclaim. It then removes satisfied agents and repeats the operation for the remaining agents.In a symmetric manner, CEL equalizes the losses of agents as much as possible. The P rulesimply gives agents awards and losses proportional to their claims.In Example 3, when ω ,o < / , the example can be regarded as a bankruptcy problemin which 2 and 3 respectively claim / o and / o , but there is no enough of o to satisfyboth claims. The three mechanisms find assignments equal to the solutions found by thethree rules. In this sense the three mechanisms are analogous to the three rules. Moreover,the bounded advantage condition in Section 6 is analogous to the order preservation propertyin the bankruptcy problem (Aumann and Maschler, 1985). Order preservation requires thatthe awards and losses of any two agents should be ordered as their claims are; that is, if c i ≥ c j , then x i ≥ x j and c i − x i ≥ c j − x j . Many solution rules including the abovethree satisfy order preservation. In bounded advantage, if ω i,o ( d − ≥ ω j,o ( d − , the firstcondition λ i,o ( d ) ≥ λ j,o ( d ) essentially requires that i receive more awards (more amount ofobjects) than j , and the second condition ω i,o ( d ) ≥ ω j,o ( d ) essentially requires that i havemore loss (more unused endowments) than j .20 Core and incentive
TTC finds the unique core assignment in the housing market model (Roth and Postlewaite,1977). It means that no coalition of agents can obtain better objects than their TTC assign-ments by reassigning endowments among themselves. In the FEE model, because stochasticdominance is an incomplete relation over lotteries, there are more than one ways to defineblocking coalition and core. One way is to say that a coalition block an assignment if theirlotteries obtained by reassigning endowments stochastically dominate their lotteries in thegiven assignment. Another way is to require that their lotteries obtained by reassigning en-dowments should not be stochastically dominated by their lotteries in the given assignment.The former definition is stronger and thus the induced core is bigger. We choose a definitionsomehow in between. We believe that individuals are easier to block an assignment than non-singleton coalitions since the latter requires coordination. So in our definition every agent canblock an assignment that is not IR for him (i.e., p i % sdi ω i ), but every non-singleton coalitioncan block an assignment if their lotteries obtained by reassigning endowments stochasticallydominate their lotteries in the given assignment. Definition 4.
A non-singleton coalition I ′ ⊂ I block an assignment p via another p ′ if P i ∈ I ′ p ′ i = P i ∈ I ′ ω i , p ′ i % sdi p i for all i ∈ I ′ , and p ′ j ≻ sdj p j for some j ∈ I ′ . If p ′ i ≻ sdi p i forall i ∈ I ′ , I ′ strongly block p . The core consists of IR assignments that are not blockedby non-singleton coalitions. The weak core consists of IR assignments that are not stronglyblocked by non-singleton coalitions. The above definitions reduce to the familiar ones in the housing market model.
Proposition 4. (1) There exists an FEE problem in which the core is empty.(2) In every FEE problem there exists a weak core assignment satisfying ETE.(3) There exists an FEE problem in which every weak core assignment violates EENE,and every FTTC satisfying stepwise ETE fails to find a weak core assignment.
Appendix B presents two FEE problems that respectively satisfy the first and the thirdstatements. To prove the second statement, we endow agents with cardinal utilities consistentwith their ordinal preferences, and in particular endow “equal” agents with equal utilities.We prove the existence of a Walrasian equilibrium with slack by using the main theoremof Mas-Colell (1992). We then construct a sequence of Walrasian equilibria with slack andprove that their limit points are weak core assignments satisfying ETE.21lthough we do not answer whether there exists some FTTC always finding a weak coreassignment, from the third statement we know that if it exists, it violates stepwise ETE,which is a weak fairness condition on the procedure of FTTC.
TTC is the unique individually rational, Pareto efficient and strategy-proof mechanism in thehousing market model (Ma, 1994). However, a couple of impossibility theorems have shownthe conflict between efficiency, fairness, IR, and strategy-proofness in random allocationproblems. Specifically, in the house allocation model, Bogomolnaia and Moulin (2001) provethat any sd-efficient mechanism satisfying ETE is not strategy-proof. In the FEE model,Athanassoglou and Sethuraman (2011) prove that any IR and sd-efficient mechanism is notstrategy-proof, while Aziz (2018) prove that such a mechanism is not even weakly strategy-proof. So every FTTC is not weakly strategy-proof in the FEE model. Example 4 from Aziz(2018) explains this.
Example 4.
There are two agents i, j and three objects o , o , o . i owns / o and / o ,while j owns / o . We omit the agents holding remaining amounts of objects for simplicity.Consider the following preferences. ≻ i ≻ j o o o o o o In any IR assignment p , i must obtain an amount of objects equal to and j must obtainan amount of objects equal to / . If p is sd-efficient, i must obtain / o from j and i must give an amount / of endowment to j . There are two cases. If p j,o > , then p i,o < / . Now if i pretends to have the preference relation ≻ ′ i : o , o , o , then the onlyIR and sd-efficient assignment is the one such that i obtains / o and / o . So i stronglymanipulates the mechanism. If p j,o = 0 , suppose j pretends to have the preference relation ≻ ′ j : o , o , o , then the only IR and sd-efficient assignment is the one such that j obtains / o . So j strongly manipulates the mechanism. So in any case one agent can stronglymanipulate an IR and sd-efficient mechanism. As many papers in the literature, we evaluate the incentive property of FTTC in largemarkets. We identify agents by their endowments and preferences, and classify them into22ypes. A fair FTTC does not distinguish between agents of equal type. In a large FEEproblem in which each type consists of a large number of agents, if any agent misreportshis preferences, the distribution of the population of each type changes a little bit. If a fairFTTC is insensitive to such a small change, which we formalize as a continuity property,then the assignment found by the FTTC changes a little bit. If the FTTC satisfies EENE,then the benefit from manipulation will be approximately zero as the market size increases.In this sense we say the FTTC is asymptotically strategy-proof. We formalize this idea inAppendix C, and simply state our result here.
Proposition 5.
Any continuous FTTC that satisfies EENE is asymptotically strategy-proof.
The three mechanisms presented in Section 7 are continuous and satisfy EENE.
Since the seminal paper Abdulkadiro˘glu and S¨onmez (2003), school choice has been inten-sively studied under the assumption that schools use strict priorities, though real-life pri-orities are often coarse. A few papers study the consequence of explicit tie breaking andshow its efficiency loss. For example, Erdil and Ergin (2008) show that tie breaking in-troduces artificial stability constraints and makes deferred acceptance (DA) no longer con-strained efficient. In the extreme case that all students have equal priority at all schools(Abdulkadiro˘glu et al., 2011), TTC with ties broken uniformly at random is equivalent to RP(Abdulkadiro˘glu and S¨onmez, 1998), which is ex-ant inefficient (Bogomolnaia and Moulin,2001). In this subsection we adapt FTTC to solve school choice without explicitly breakingties.Formally, the school choice model consists of a set of students I with preferences ≻ I and a set of schools O with priority rankings ( % o ) o ∈ O . Each % o is a complete and transitiverelation on I . Let ≻ o and ∼ o respectively denote the asymmetric and symmetric componentsof % o . Two students i, j have equal priority at school o if i ∼ o j . Intuitively, equal prioritymeans that the two students have equal rights over o . By interpreting equal priority asequal ownerships, we extend the equal-FTTC mechanism T e to school choice. At eachstep d , each remaining student reports his favorite remaining school. Among the remainingstudents I ( d − , let I o ( d − be the set of those who have highest priority at school o . Two mechanisms are equivalent if they always find equal assignments. i ∈ I o ( d − , we choose λ i,o ( d ) = 1 / | I o ( d − | . That is, students of equal highestpriority at a school use equal amounts of the school’s seats to join the trading process. Ateach step students trade school seats as many as possible until some student is satisfied orthe seats of some school are exhausted. So we no longer introduce β i,o ( d ) . Extension of T e to school choiceNotations : Let I ( d ) , O ( d ) , o i ( d ) , p ( d ) and x ∗ ( d ) have same definitions as before. Step d ≥ : Let each i ∈ I ( d − report his favorite school o i ( d ) . Let x ∗ ( d ) be themaximum solution to the equations x o ( d ) = P i ∈ I ( d − o i ( d )= o x i ( d ) for all o ∈ O ( d − ,x i ( d ) = P o ∈ O ( d − i ∈ I o ( d − x o ( d ) | I o ( d − | for all i ∈ I ( d − , that satisfies the constraints x o ( d ) ≤ q o − P d − k =1 x ∗ o ( k ) for all o ∈ O ( d − ,x i ( d ) ≤ − P d − k =1 x ∗ i ( k ) for all i ∈ I ( d − . Then for all i ∈ I and o ∈ O , p i,o ( d ) = p i,o ( d −
1) + x ∗ i ( d ) if i ∈ I ( d − and o = o i ( d ) ,p i,o ( d − otherwise.Let I ( d ) = { i ∈ I : P dk =1 x ∗ i ( k ) < } and O ( d ) = { o ∈ O : P dk =1 x ∗ o ( k ) < q o } . If O ( d ) or I ( d ) is empty, stop the procedure. Otherwise, go to step d + 1 .As before, the found assignment is sd-efficient. By extending T e we also intend to makethe assignment fair. But to formalize the fairness here, we need to define hierarchical en-dowments implied by coarse priorities and other notions. We omit the details because theintuition has been implied by our main results. An obvious implication of the fairness hereis that, in the found assignment no student envies another student who has weakly lowerpriority at all schools. In particular, there is no envy among students of equal priority at allschools. We summarize this discussion as follows. Result 1.
In school choice with coarse priorities, the extension of T e is sd-efficient and fair. .2 Time exchange Many people around the world choose to exchange time and skills on centralized platforms(e.g., time banks) without using transfers. Participants trade a wide range of services,including legal assistance, babysitting, medical care, and many others. A time exchangemarket can be described as an FEE problem. Objects are types of services that agents canprovide. Conceptually, service types can be as abstract as contracts in the matching theoryliterature (Hatfield and Milgrom, 2005). Usually, a service type specifies at least the timeand location to finish a certain task. Each agent i ’s endowment ω i,o ∈ [0 , is the amountof service o that i can provide. One unit of a service could mean one hour, one day, or oneweek, which depends on the context. Services can be regarded as divisible because time isdivisible (e.g., a half of a service could mean 12 hours).In applications of our model in which objects are indivisible, we assume that each agentdemands one object. A difference in a time exchange market is that an agent may demandseveral services and his demand over each service can be strictly between 0 and 1. So weintroduce d i,o ∈ [ ω i,o , and d i,o ∈ [0 , ω i,o ] to denote the upper bound and lower bound of i ’sdemand over o . d i,o can regarded as the amount of service o that i reserves for himself. Anassignment p is called feasible if, for all i ∈ I and all o ∈ O , p i,o ∈ [ d i,o , d i,o ] . Agents mayhave complex preferences over bundles of services, but it may be impractical to elicit suchcomplex preferences. In our solution we let agents report ordinal preferences on services.FTTC solves a time exchange market with the following modifications:1. At each step d , let each remaining agent i report his favorite remaining service type o that satisfies p i,o ( d − < d i,o .2. For all i ∈ I ( d − and o ∈ O ( d − , λ i,o ( d ) > only if ω i,o ( d − > d i,o .3. The constraints (2) are replaced by λ i,o ( d ) x o ( d ) ≤ ω i,o ( d − − d i,o and x i ( d ) ≤ d i,o i ( d ) − p i,o i ( d ) ( d − for all i ∈ I ( d − and o ∈ O ( d − .4. After each step d , an agent i remains if ω i,o ( d ) > d i,o for some o ∈ O ; a service type o remains if ω i,o ( d ) > d i,o for some i ∈ I .To obtain a fair assignment, we can choose a fair FTTC mechanism. The three mecha-nisms in Section 7 are candidates. Result 2.
In a time exchange market, a fair FTTC is feasible, individually rational, (feasibility-constrained) sd-efficient, and fair. .3 Geographically distributional constraints Distributional constraints appear widely when a market designer wants to control the num-ber of agents assigned to institutions. A prominent example is the Japanese medical resi-dency match (Kamada and Kojima, 2015). Because hospitals in rural regions suffer doctorshortages, the Japanese government imposes constraints on the number of doctors that thehospitals in each geographic region should hire. The literature has been successful at dealingwith ceiling constraints, yet it demonstrates the difficulty of dealing with floor constraints.Consequently, the Japanese government uses a modification of DA by imposing artificialregional caps, which is proved to be inefficient by Kamada and Kojima. We use FTTC toprovide a different solution to the problem.Formally, I is a set of doctors and O is a set of hospitals. Each constraint ( S, q S , q S ) ischaracterized by a subset of hospitals S ⊂ O and a pair of integers ( q S , q S ) with ≤ q S ≤ q S .An assignment p satisfies the constraint if q S ≤ P o ∈ S P i ∈ I p i,o ≤ q S . We require the collec-tion of constraints { ( S, q S , q S ) } form a hierarchy (Budish et al., 2013; Kamada and Kojima,2014); that is, for any two constraint sets S and S ′ , if S ∩ S ′ = ∅ , then either S ⊂ S ′ or S ′ ⊂ S . We also require that the collection include a capacity constraint ( { o } , q o , q o ) for eachhospital o ∈ O . We assume that all constraints are compatible. Assignments that satisfy allconstraints are called feasible .Hospitals often prefer to hire more doctors as long as constraints are not violated. Sowe assume that every hospital o weakly prefers a feasible assignment p ′ to another feasibleassignment p if P i ∈ I p ′ i,o ≥ P i ∈ I p i,o , and the preference is strict if the inequality is strict.Taking account of all players’ preferences, we say a feasible assignment p is two-sided efficient if there does not exist another feasible assignment p ′ such that p ′ strictly stochasticallydominates p for agents and P i ∈ I p ′ i,o ≥ P i ∈ I p i,o for all o ∈ O . Doctors do not have endowments in this problem. To apply FTTC, we choose a feasibleassignment ω and treat it as doctors’ endowment. It reflects doctors’ rights/obligations toattend each hospital. We let doctors exchange their rights/obligations by running an FTTCmechanism. Because the number of doctors assigned to each hospital does not change inthe procedure of FTTC, the found assignment must be feasible. To enhance efficiency andfairness, we choose a fair ω and a fair FTTC as follows:1. Let ω be a feasible assignment such that ω i,o = ω j,o for all i, j ∈ I and all o ∈ O , and Combe et al. (2018) study a teacher (re)assignment problem and define a similar two-sided efficiencynotion to take account of preferences of both teachers and schools. p ′ such that X i ∈ I p ′ i,o ≥ X i ∈ I ω i,o for all o ∈ O, and X i ∈ I p ′ i,o ′ > X i ∈ I ω i,o ′ for some o ′ ∈ O.
2. Regard ω as endowment and run an FTTC mechanism satisfying stepwise EEET.Let p denote the found assignment. Because ω is feasible, p is also feasible . Because agentshave equal endowments in ω and we impose stepwise EEET, p is envy-free for doctors. Thisis different than the priority-based fairness used by Kamada and Kojima (2015). Becausethe numbers of doctors assigned to hospitals in ω cannot be Pareto improved, and giventhat, p is sd-efficient for doctors. So p is two-sided efficient . Result 3.
In the Japanese medical residency match model, our method is feasible, envy-freefor agents, and two-sided efficient.
The initial assignment ω determines the numbers of doctors assigned to hospitals. So ω determines the fairness among hospitals. We do not discuss how to evaluate such fairness andhow to choose ω . We leave this to the market designer. But note that our method cannotsolve constraints that distinguish between agents’ types, such as those in controlled schoolchoice (Abdulkadiro˘glu and S¨onmez, 2003; Ehlers et al., 2014; Echenique and Yenmez, 2015;Fragiadakis and Troyan, 2017). To solve such constraints, our method will start with afeasible assignment and then let agents of equal type trade endowments among themselves.But the found assignment is not ensured to be efficient. How to solve such constraints is aninteresting direction for future research.
The literature since Hylland and Zeckhauser (1979) often studies random allocation in thehouse allocation model. The model has a simple setup: a number of objects are assigned toan equal number of agents, and there is no endowment or priority structure that distinguishesbetween agents. With ordinal preferences, the earliest well-known fair mechanism to solvethe model is Random Priority (RP; Abdulkadiro˘glu and S¨onmez (1998)). RP first drawsan ordering of agents uniformly at random and then lets agents sequentially choose favoriteobjects. Observing the ex-ante inefficiency of RP, Bogomolnaia and Moulin (2001) propose See Kamada and Kojima (2015) for some related discussion about fairness among hospitals. ω i,o = 1 / | I | for all i ∈ I and all o ∈ O .Then the IR notion coincides with the fairness axiom called equal-division lower bound inthe model (Thomson, 2011). In FTTC, agents also obtain favorite objects step by step. Soit is easy to make the following observations. Result 4.
In the house allocation model:1. Every FTTC coincides with an SEA satisfying equal-division lower bound;2. Every FTTC satisfying stepwise EEET coincides with PS.
The second result makes the relation between PS and TTC transparent. The relation isfirst discovered by Kesten (2009), but Kesten shows the relation in a different way. Kestendefines a probabilistic variant of TTC in which agents are endowed with equal divisions ofobjects and decomposed into pseudo-agents who trade endowments on behalf of them. Ateach step, to avoid redundancy and ensure fairness, Kesten carefully selects short cycles toclear. These features make his mechanism complicated and not coincide with PS, though itis equivalent to PS in outcomes.In the house allocation with existing tenants model (HET; Abdulkadiro˘glu and S¨onmez,1999), in addition to social endowments O , a subset of agents called existing tenants furtherown private endowments. This endowment structure can be translated into a coarse prioritystructure in which social endowments rank all agents equally, and private endowments rankowners highest and rank the others equally. Then our extension of T e in Section 9.1 becomesan appealing solution to HET because of its intuitive fairness implied by the coarse prioritystructure. Actually, the mechanism can be described as an extension of PS that incorporatesthe feature of TTC: at any time t ∈ [0 , , • If a subset of existing tenants demand each other’s private endowments so that theyform a cycle, let them trade (fractional) endowments immediately. • If there do not exist cycles among existing tenants, let agents consume their favoriteobjects with following rates. First, all agents have basic eating rates of one. Second, if28n exsiting tenant’s private endowment is being consumed by other agents, his eatingrate is instantly increased by an amount that is equal to the total rates of the agentswho are consuming his private endowment. We call this rule “you request my house -I get your rate”.The formal definition of this mechanism is presented in Online Appendix F. This mech-anism is first presented in the third chapter of Zhang (2017). Using the technique ofChe and Kojima (2010), Zhang (2017) shows that this mechanism is asymptotically equiva-lent to the “you request my house - I get your turn” mechanism (Abdulkadiro˘glu and S¨onmez,1999) in large markets. This result resembles the asymptotic equivalence between RP andPS proved by Che and Kojima.
10 Conclusion
Top Trading Cycle is one of the few successful mechanisms in market design. Fairness invarious environments requires the use of randomization, yet TTC has to take deterministicendowments or strict priorities as inputs. This restricts its application to random allocationproblems without losing its flagship efficiency property. To generalize TTC to random allo-cation problems, we consider a direct extension of the housing market model and proposethe class of FTTC mechanisms. We drop the graph-based definition of TTC and use pa-rameterized linear equations to describe balanced endowment exchange among agents. Theequations essentially describe an equilibrium that has been studied in the Leontief input-output model. This enables us to prove that FTTC in our description is well-defined. FTTCis individually rational and efficient. By choosing different parameter values in its definition,FTTC satisfies different fairness axioms. As examples, we present three fair FTTC and con-nect their fairness to three classical rules in the bankruptcy problem. We apply FTTC to acouple of market design problems and obtain efficient and fair assignments in all of them. Anotable application is school choice with coarse priorities. The literature has demonstratedthat preference-independent tie breaking causes efficiency loss. Our application of FTTCto school choice is efficient and fair, and does not explicitly breaking ties. In a companionpaper Yu and Zhang (2020), we extend FTTC to the full preference domain in which agentsmay be indifferent between objects, preserving the efficiency and fairness properties.29 eferences
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A Proof of Proposition 1
In Section 5 we have explained that the equation system (1) can be written as (3), and if weuse A = ( a ij ) m + ni,j =1 = " B n × m C m × n
34o denote the coefficient matrix, then (3) can be written as Ax ( d ) = x ( d ) , or ( I − A ) x ( d ) = . (4)Let N = { , . . . , m + n } be the index set of A . So every i ∈ N denotes an agent or anobject (here we use i as an index in N ). A satisfies that, for all i, j ∈ N , a ii = 0 , a ij ∈ [0 , ,and P m + ni =1 a ij = 1 . So A is a stochastic matrix. Because the sum of all columns of I − A iszero, I − A is singular. So (4) must have nonzero solutions. We prove that there must exista maximum solution that is nonnegative and satisfies the constraints (2).We first introduce some definitions. For any N ′ ⊂ N , let A N ′ = ( a ij ) i,j ∈ N ′ be therestriction of A to the index subset N ′ . For any N ′′ ⊂ N ′ ⊂ N , A N ′′ is called a submatrix of A N ′ ; if N ′′ ( N ′ , A N ′ is called a strict submatrix of A N ′ . A N ′′ is called isolated in A N ′ if a ij = 0 for all i ∈ N ′ \ N ′′ and j ∈ N ′′ . For any N ′ ⊂ N , A N ′ is called isolated if it is isolatedin A ; otherwise it is called unisolated . A N ′ is called reducible if it has a strict nonemptysubmatrix that is isolated in A N ′ ; otherwise, it is called irreducible . For convenience, we say N ′ is (un)isolated or (ir)reducible if A N ′ is (un)isolated or (ir)reducible. Lemma 1. N can be uniquely partitioned into disjoint sets N , N , . . . , N k , N k +1 such that(1) for all ℓ = 1 , . . . , k , N ℓ is nonempty, isolated and irreducible;(2) N k +1 , which can be empty, is either unisolated or reducible, and it does not containany strict nonempty subset that is isolated and irreducible.Proof. We prove the lemma by constructing the partition. We first construct N . If N isirreducible, let N = N and we are done by letting k = 1 and N = ∅ . Otherwise, N containsa strict nonempty subset N ′ that is isolated. If N ′ is irreducible, let N = N ′ . Otherwise, N ′ contains a strict nonempty subset N ′′ that is isolated in N ′ . Since N ′ is isolated, N ′′ isisolated. If N ′′ is irreducible, let N = N ′′ . Otherwise, N ′′ also contains a strict nonemptysubset N ′′′ that is isolated in N ′′ . So N ′′′ is also isolated. Note that every submatrix thatconsists of a single element must be irreducible. So by repeating the above argument wemust be able to find an irreducible and isolated nonempty subset. Let the subset be N . Wethen construct N .If N \ N is irreducible and isolated, we are done by letting N = N \ N , k = 2 , and N = ∅ . Otherwise, if N \ N does not contain any strict nonempty subset that is irreducibleand isolated, we are done by letting k = 1 and N = N \ N . If N \ N contains a strictnonempty subset N ′ that is irreducible and isolated, let N = N ′ . Then we construct N and possibly N , . . . , N k +1 from N \ ( N ∪ N ) in a similar way. Since N is a finite set, wemust stop in finite steps. 35emma 1 implies that by permuting rows and columns, we can write A as the followingblock form: A = A N Z A N . . . A N k A N k +1 (5)For every ℓ = 1 , . . . , k + 1 , let I N ℓ denote the identity matrix of dimension | N ℓ | × | N ℓ | . Lemma 2. (1) Rank ( I N ℓ − A N ℓ ) = | N ℓ | − for all ℓ = 1 , . . . , k ;(2) Rank ( I N k +1 − A N k +1 ) = | N k +1 | .Proof. (1) For every ℓ = 1 , . . . , k , since N ℓ is isolated, it must be that P i ∈ N ℓ a ij = 1 for every j ∈ N ℓ . Therefore, the sum of all columns of I N ℓ − A N ℓ is zero. So det ( I N ℓ − A N ℓ ) = 0 andRank ( I N ℓ − A N ℓ ) < | N ℓ | . Claim 1.
For every nonempty N ′ ( N ℓ , there exists j ∈ N ′ such that P i ∈ N ′ a ij < . Proof : Suppose P i ∈ N ′ a ij = 1 for all j ∈ N ′ . Since P i ∈ N ℓ a ij = 1 for all j ∈ N ℓ , it meansthat a ij = 0 for all i ∈ N ℓ \ N ′ and j ∈ N ′ . So A N ′ is isolated in A N ℓ , which contradicts theassumption that A N ℓ is irreducible.Corollary 3.3 of Peterson and Olinick (1982) states that for a general matrix D = ( d ij ) ki,j =1 such that d ij ∈ [0 , and P ki =1 d ij ≤ for all j ∈ X = { , . . . , k } , if det ( I k × k − D ) = 0 anddet ( I ( k − × ( k − − D X \ j ) = 0 for all j ∈ X , then every column of D sums to one. This resultand Claim 1 imply the following claim. Claim 2.
For every nonempty N ′ ( N ℓ , if det ( I N ′ − A N ′ ) = 0 , then there exists some j ∈ N ′ such that det ( I N ′ \{ j } − A N ′ \{ j } ) = 0 . Proof : Suppose det ( I N ′ \{ j } − A N ′ \{ j } ) = 0 for all j ∈ N ′ . By Corollary 3.3 of Peterson and Olinick(1982), every column of A N ′ sums to one. But it contradicts Claim 1.Now we prove that for all j ∈ N ℓ , det ( I N ℓ \{ j } − A N ℓ \{ j } ) = 0 . Suppose towards acontradiction that det ( I N ℓ \{ j } − A N ℓ \{ j } ) = 0 for some j ∈ N ℓ . By Claim 2, there exists j ∈ N ℓ \{ j } such that det ( I N ℓ \{ j,j } − A N ℓ \{ j,j } ) = 0 . By Claim 2 again, there further exists j ∈ N ℓ \{ j, j } such that det ( I N ℓ \{ j,j ,j } − A N ℓ \{ j,j ,j } ) = 0 . By repeatedly using Claim 2,we must find a submatrix consisting of only one element i ∈ N ℓ such that − a ii = 0 , which36ontradicts the fact that a ii = 0 for all i ∈ N . So det ( I N ℓ \{ j } − A N ℓ \{ j } ) = 0 for all j ∈ N ℓ .This implies that Rank ( I N ℓ − A N ℓ ) = | N ℓ | − .(2) We first prove that A N k +1 is unisolated. Suppose it is isolated, then by definitionit must be reducible. So A N k +1 contains a strict nonempty submatrix A N ′ that is isolatedin A N k +1 . Since A N k +1 is isolated, A N ′ is also isolated. Still by the definition of A N k +1 , A N ′ must be reducible. So A N ′ also contains a strict nonempty submatrix A N ′′ that isisolated in A N ′ . Since A N ′ is isolated, A N ′′ is also isolated. By the definition of A N k +1 again, A N ′′ must be reducible. By repeating this argument, we finally find a submatrixconsisting of a single element and conclude that it is reducible, which is a contradiction.So A N k +1 is unisolated. It means that not every column of A N k +1 sums to one. Then byCorollary 3.3 of Peterson and Olinick (1982) and same arguments as Claim 1 and Claim 2,if det ( I N k +1 − A N k +1 ) = 0 , then there must exist some i ∈ N k +1 such that − a ii = 0 , whichis a contradiction. So det ( I N k +1 − A N k +1 ) = 0 , which implies that Rank ( I N k +1 − A N k +1 ) = | N k +1 | .Given the block form (5), Lemma 2 implies thatRank ( I − A ) = k +1 X ℓ =1 Rank ( I N ℓ − A N ℓ ) = m + n − k. So ( I − A ) x ( d ) = has k linearly independent solutions. Below we construct the k solutions.For all every ℓ = 1 , . . . , k , we consider the equation system ( I N ℓ − A N ℓ ) x N ℓ ( d ) = .Since det ( I N ℓ − A N ℓ ) = 0 , is an eigenvalue of A N ℓ . Since A N ℓ is irreducible, by FrobeniusTheorem (see Section 6.8 of Leon, 2015), has a positive eigenvector ˜ x N ℓ ( d ) that is a solutionto ( I N ℓ − A N ℓ ) x N ℓ ( d ) = . Recall that A can be written in the block form (5). So ˜ x ℓ ( d ) = ( , . . . , | {z } ℓ − , ˜ x N ℓ ( d ) , , . . . , ) is a nonnegative solution to ( I − A ) x ( d ) = . It is clear that the k solutions ˜ x ( d ) , ˜ x ( d ) , . . . , ˜ x k ( d ) are linearly independent. Therefore, every solution to ( I − A ) x ( d ) = 0 is a linear combinationof the k solutions. That is, there exist y , . . . , y k ∈ R such that x ( d ) = y ˜ x N ( d ) ... + y ˜ x N ( d ) ... + · · · + y k ... ˜ x N k ( d ) . ℓ = 1 , . . . , k , define y ∗ ℓ = min (cid:26) β i,o ( d ) λ i,o ( d )˜ x o ( d ) : o ∈ N ℓ , i ∈ N ℓ , λ i,o ( d ) > (cid:27) . Then, x ∗ ( d ) = k X ℓ =1 y ∗ ℓ ˜ x ℓ ( d ) is the maximum solution that satisfies the constraints: λ i,o ( d ) x o ( d ) ≤ β i,o ( d ) for all i ∈ I ( d − and o ∈ O ( d − . In the remaining part of this section, we show that { N ℓ } ≤ ℓ ≤ k defined in Lemma 1 is theset of absorbing sets at step d . Lemma 3.
Any N ′ ⊂ N is an absorbing set if and only if A N ′ is irreducible and isolated.Proof. (Only if) Let N ′ be an absorbing set. So every node j ∈ N ′ does not effectively pointto any node i ∈ N \ N ′ . In the definition of A , it means that a ij = 0 for all i ∈ N \ N ′ and j ∈ N ′ . So A N ′ is isolated. Suppose A N ′ is reducible. Then there is a strict subset N ′′ of N ′ such that A N ′′ is isolated in N ′ . It means that a ij = 0 for all i ∈ N ′ \ N ′′ and all j ∈ N ′′ . Inthe generated graph it means that there is no path from every node in N ′ to every node in N ′ \ N ′′ , which contradicts the assumption that N ′ is an absorbing set. So A N ′ is irreducibleand isolated.(If) Let A N ′ be irreducible and isolated. Being isolated directly implies that there is nopath from every node in N ′ to every node not in N ′ . Suppose there exist two nodes i, j ∈ N ′ such that there is no path from i to j . Let N be the subset of N ′ such that there is a pathfrom i to every k ∈ N . Let N be the set of remaining nodes in N ′ . Obviously, i ∈ N and j ∈ N . Then there must be no path from every node in N to every node in N ; otherwise,there would be a path from i to every node in N , which contradicts the definition of N . Itmeans that a ij = 0 for all i ∈ N and all j ∈ N . So A N is isolated in A N ′ , which contradictsthe assumption that A N ′ is irreducible. So N ′ must be an absorbing set. B Proofs of Proposition 2, 3, and 4
Proof of Proposition 2 . (Individual rationality) At each step d , for all i ∈ I ( d − and all o ∈ O ( d − with ω i,o ( d − > , o i ( d ) % i o . Suppose the assignment p found by FTTC is not38ndividually rational. Then there exist o ∗ ∈ O and i ∈ I such that P o % i o ∗ p i,o < P o % i o ∗ ω i,o .Let d be the earliest step after which all objects in { o ∈ O : o % i o ∗ } are exhausted. Thatis, { o ∈ O : o % i o ∗ } ∩ O ( d ) = ∅ and { o ∈ O : o % i o ∗ } ∩ O ( d − = ∅ . Then i ’s favoritesobjects from step one to step d must belong to { o ∈ O : o % i o ∗ } , and P o % i o ∗ p i,o is the totalamount of objects obtained by i at the end of step d . But given P o % i o ∗ p i,o < P o % i o ∗ ω i,o ,the balanced trade condition implies that there must exist o ′ % i o ∗ such that ω i,o ′ ( d ) > .This is a contradiction.(Sd-efficiency) Still let p denote the assignment found by FTTC. Define a binary relation ⊲ on O : o ⊲ o ′ if, for some i ∈ I , o ≻ i o ′ and p i,o ′ > . Bogomolnaia and Moulin (2001) andChe and Kojima (2010) have shown that p is sd-efficient if and only if ⊲ is acyclic. Since inFTTC an agent reports an object only if all better objects are exhausted, for any o ≻ i o ′ and p i,o ′ > , o must be exhausted earlier than o ′ . So ⊲ must be acyclic. Proof of Proposition 3 . (1) For any FTTC that satisfies stepwise ETE, if any two agents i, j have equal endowments and equal preferences, then at step one they report the samefavorite object and λ i (1) = λ j (1) . So they obtain equal amounts of the favorite object andtheir remaining endowments after step one are equal. By induction it is easy to see that thisholds for all remaining steps. So the two agents must obtain equal lotteries.(2) Similarly, for any FTTC that satisfies stepwise EEET, if any two agents i, j have equalendowments, then at each step d it must be that λ i ( d ) = λ j ( d ) . It means that the two agentsobtain equal amounts of their respective favorite objects at each step. Suppose i envies j in the found assignment p . Then there exists o ∗ ∈ O such that P o % i o ∗ p i,o < P o % i o ∗ p j,o .Let d be the earliest step after which all objects in { o ∈ O : o % i o ∗ } are exhausted.Then i ’s favorite objects from step one to step d must belong to { o ∈ O : o % i o ∗ } , and P o % i o ∗ p i,o is the total amount of objects obtained by i at the end of step d . So we shouldhave P o % i o ∗ p j,o ≤ P dd ′ =1 x j ( d ′ ) = P o % i o ∗ p i,o . This is a contradiction.(3) Let p denote the assignment found by any FTTC satisfying bounded advantage. Forany distinct i, j ∈ I , let o ∗ be the solution to max o ∈ O (cid:2) P o ′ % i o p j,o ′ − P o ′ % i o p i,o ′ (cid:3) . Let d bethe earliest step after which all o % i o ∗ are exhausted. That is, { o ∈ O : o % i o ∗ } ∩ O ( d ) = ∅ and { o ∈ O : o % i o ∗ } ∩ O ( d − = ∅ . By the balanced trade condition, X o % i o ∗ p i,o = X o ∈ O (cid:0) ω i,o − ω i,o ( d ) (cid:1) and X o % i o ∗ p j,o ≤ X o ∈ O (cid:0) ω j,o − ω j,o ( d ) (cid:1) . For all o ∈ O such that ω i,o ≥ ω j,o , bounded advantage implies that for all ≤ d ′ ≤ d , ω i,o ( d ′ ) ≥ ω j,o ( d ′ ) and λ i,o ( d ′ ) ≥ λ j,o ( d ′ ) . So ω j,o − ω j,o ( d ) = P dd ′ =1 λ j,o ( d ′ ) x ∗ o ( d ′ ) ≤ dd ′ =1 λ i,o ( d ′ ) x ∗ o ( d ′ ) = ω i,o − ω i,o ( d ) . Equivalently, (cid:0) ω j,o − ω j,o ( d ) (cid:1) − (cid:0) ω i,o − ω i,o ( d ) (cid:1) ≤ . For all o ∈ O such that ω i,o < ω j,o , bounded advantage implies that for all ≤ d ′ ≤ d , ω i,o ( d ′ ) ≤ ω j,o ( d ′ ) and λ i,o ( d ′ ) ≤ λ j,o ( d ′ ) ; in particular, ω i,o ( d ) ≤ ω j,o ( d ) . So (cid:0) ω j,o − ω j,o ( d ) (cid:1) − (cid:0) ω i,o − ω i,o ( d ) (cid:1) ≤ ω j,o − ω i,o . Therefore, X o % i o ∗ p j,o − X o % i o ∗ p i,o ≤ X o ∈ O (cid:20)(cid:0) ω j,o − ω j,o ( d ) (cid:1) − (cid:0) ω i,o − ω i,o ( d ) (cid:1)(cid:21) = X o ∈ O : ω i,o ≥ ω j,o (cid:20)(cid:0) ω j,o − ω j,o ( d ) (cid:1) − (cid:0) ω i,o − ω i,o ( d ) (cid:1)(cid:21) + X o ∈ O : ω i,o <ω j,o (cid:20)(cid:0) ω j,o − ω j,o ( d ) (cid:1) − (cid:0) ω i,o − ω i,o ( d ) (cid:1)(cid:21) ≤ X o ∈ O : ω i,o <ω j,o (cid:0) ω j,o − ω i,o (cid:1) . Thus, p satisfies bounded envy. Proof of Proposition 4 . We first prove the second statement (2). For any FEE prob-lem, we construct a sequence of Walrasian equilibria with slack and prove that the limitpoints are weak core assignments satisfying ETE. We endow each agent i with von Neu-mann–Morgenstern utilities ( u i,o ) o ∈ O ∈ R | O | + such that o ≻ i o ′ if and only if u i,o > u i,o ′ . Twoagents of equal endowments and equal preferences are endowed with equal utilities. Theutility of obtaining nothing is normalized to zero. So every i ’s expected utility of obtaininga lottery p i is u i ( p i ) = P o ∈ O u i,o p i,o . Agents compare lotteries by expected utilities.For any ε > , define i ’s consumption space to be X εi = { x i is a lottery : X o ′ % i o x i,o ′ ≥ X o ′ % i o ω i,o ′ − ε for all o ∈ O } . It is clear that X εi is nonempty, closed and convex. Then define the space of price vectorsto be ∆ = { P ∈ R | O | + : k P k ≤ } . Basteck (2018) uses a similar method to prove the existence of weak core from equal division in thehouse allocation problem. ε > ensures that for every nonzero P ∈ ∆ , P · ω i > inf x i ∈ X εi P · x i . By Mas-Colell’s Theorem 1, for any ε > , there exists a Walrasian equilibrium with slack (WE-slack) ( x ε , P ε , α ε ) such that1. P ε ∈ ∆ and α ε ≥ ;2. P i ∈ I x εi,o = P i ∈ I ω i,o for all o ∈ O ;3. x εi ∈ arg max { u i ( x i ) : x i ∈ X εi , P · x i ≤ P · ω i + α ε } for all i ∈ I .The facts that x εi ∈ X εi for all i and that P i ∈ I x εi,o = P i ∈ I ω i,o for all o imply that x ε isan assignment. Actually x ε can be chosen to satisfy ETE. To see it, let { I , · · · , I K } bethe partition of agents such that each I ℓ consists of agents of equal endowments and equalpreferences. Given any WE-slack ( x ε , P ε , α ε ) , for every I ℓ and every i ∈ I ℓ , define y εi = P j ∈ I ℓ x εj | I ℓ | . It is clear that y εi ∈ X εi , P ε · y εi ≤ P · ω i + α ε , and y εi solves i ’s expected utility maximization(since expected utility functions are linear). It is also obvious that P i ∈ I y εi,o = P i ∈ I ω i,o forall o . So ( y ǫ , P ε , α ε ) is a WE-slack and y ǫ satisfies ETE.Let ( ε k ) k ∈ N be a sequence of positive numbers such that ε k → as k → ∞ . For each ε k , let ( x k , P k , α k ) be a WE-slack that satisfies ETE. Since ( x k ) k ∈ N is bounded, it has limitpoints (the limits of convergent subsequences). Let x ∗ be any limit point. Since each x k isan assignment satisfying ETE, x ∗ must be an assignment satisfying ETE. Moreover, for all i , x ki ∈ X ε k i implies that x ∗ i ∈ X i , which means that x ∗ is individually rational. We prove that x ∗ is a weak core assignment. Suppose towards a contradiction that for some non-singletoncoalition I ′ there exists a different assignment x ′ such that P i ∈ I ′ x ′ i = P i ∈ I ′ ω i and x ′ i ≻ sdi x ∗ i for all i ∈ I ′ . So for every i ∈ I ′ , x ′ i ∈ X ε k i for all k and u i ( x ′ i ) > u i ( x ∗ i ) . Since x ∗ is a limitpoint of ( x k ) k ∈ N , there exists a large enough n ∈ N such that u i ( x ′ i ) > u i ( x ni ) for all i ∈ I ′ .It implies that P n · x ′ i > P n · ω i + α n ≥ P n · ω i for all i ∈ I ′ . But it contradicts the fact that P n · P i ∈ I ′ x ′ i = P n · P i ∈ I ′ ω i .Below we present two FEE problems that respectively satisfy (1) and (3). Example 5.
Suppose there are four agents I = { , , , } and four objects O = { o , o , o , o } .Consider two FEE problems in which agents have same endowments but they respectively havethe preferences ≻ I and ≻ ′ I in Table 3. o o o / / / / / /
24 0 1 / / (a) Endowments ≻ ≻ ≻ ≻ o o o o o o o o o o o o o o o o (b) ≻ I ≻ ′ ≻ ′ ≻ ′ ≻ ′ o o o o o o o o o o o o o o o o (c) ≻ ′ I Table 3 (Core can be empty) Let p be an IR assignment for the problem ( ≻ I ) . IR requires that,for all i ∈ I , p i,o + p i,o = 1 / and p i,o + p i,o = 1 / ; in particular, agent 4 must obtain hisendowment in p . Then p is not blocked by { , } through exchanging endowments if and onlyif p = ω and p = ω . Similarly, p is not blocked by { , } through exchanging endowmentsif and only if p = ω and p = ω . So p is a core assignment only if p = p = ω and p = ω = ω , which is impossible given p = ω . So p is not a core assignment.(Weak core violates EENE) Let p be an IR assignment that satisfies EENE for the problem ( ≻ ′ I ) . IR requires that, for all i ∈ I , p i,o + p i,o = 1 / and p i,o + p i,o = 1 / ; in particular, p ,o = 1 / and p ,o = 1 / . EENE requires that p ,o = p ,o and p ,o = p ,o . So p ,o = p ,o ≤ / and p ,o = p ,o ≤ / . It implies that p ,o = p ,o ≥ / and thus p ,o ≤ / .Then it means that { , } can strongly block p by exchanging endowments. So p is not aweak core assignment.(FTTC satisfying stepwise ETE cannot find weak core) It is easy to verify that all FTTCsatisfying stepwise ETE must find an IR assignment that satisfies EENE for the problem ( ≻ ′ I ) . So any such FTTC cannot find weak core assignments. C Asymptotic strategy-proofness
For any problem M = ( I, O, ≻ I , ω ) , we assume that each i ∈ I has von Neumann–Morgensternutilities ( u i,o ) o ∈ O ∈ R | O | + such that o ≻ i o ′ if and only if u i,o > u i,o ′ . The utility of receiving ∅ is normalized to zero. So i ’s expected utility of obtaining a lottery p i is u i ( p i ) = P o ∈ O u i,o p i,o .We assume that agents use expected utilities to compare lotteries. Define Ω M = { ˆ ω : ∃ i ∈ I, ˆ ω = ω i } to be the space of endowment types. We say an agent is type- (ˆ ω, ≻ ) if his42ndowment is ˆ ω and his preference relation is ≻ . For all (ˆ ω, ≻ ) ∈ Ω M × P , define A (ˆ ω, ≻ ) = |{ i ∈ I : ω i = ˆ ω, ≻ i = ≻}|| I | to be the proportion of type- (ˆ ω, ≻ ) agents in the problem.Fixing the set of object types O , we say a sequence of economies ( M [ n ] ) ∞ n =1 , where M [ n ] =( O, I [ n ] , ≻ I [ n ] , ω [ n ] ) , is regular if(1) For all n ≥ , | I [ n ] | = n | I [1] | and Ω M [ n ] = Ω M [1] .(2) For all (ˆ ω, ≻ ) ∈ Ω M [1] × P , there exists A [ ∞ ] (ˆ ω, ≻ ) ∈ (0 , such that A [ n ] (ˆ ω, ≻ ) → A [ ∞ ] (ˆ ω, ≻ ) as n → ∞ . (3) There exits K > such that for all n ≥ , all i ∈ I [ n ] , and all distinct lotteries p i and p ′ i , | u i ( p i ) − u i ( p ′ i ) | < K · max o ∈ O | p i,o − p ′ i,o | .Condition (1) says that as the market size increases, the number of agents increases but allagents still hold endowment types same as the agents in the base economy M [1] . Condition(2) says that as the market size increases, for every (ˆ ω, ≻ ) ∈ Ω M [1] × P , the proportion oftype- (ˆ ω, ≻ ) agents converges to a positive fraction A [ ∞ ] (ˆ ω, ≻ ) . Condition (3) essentially saysthat the vNM utilities of all agents in the sequence of economies have a common finite upperbound.A mechanism ϕ is asymptotically strategy-proof if, given any regular sequence ( M [ n ] ) ∞ n =1 ,for any ε > , there exists n ∗ ∈ N such that for any n > n ∗ , if any i ∈ I [ n ] reports any ≻ ′ i ∈ P\{≻ i } in the economy M [ n ] , then the utility gain from manipulation is bounded by ε : u i (cid:0) ϕ i ( ≻ ′ i , ≻ I [ n ] \{ i } ) (cid:1) < u i (cid:0) ϕ i ( ≻ I [ n ] ) (cid:1) + ε. For any FTTC that satisfies ETE and any regular sequence ( M [ n ] ) ∞ n =1 , let p [ n ] denotethe assignment found for each M [ n ] , and, for every (ˆ ω, ≻ ) ∈ Ω M [1] × P , let p [ n ](ˆ ω, ≻ ) denotethe lottery assigned to type- (ˆ ω, ≻ ) agents in M [ n ] . We say two regular sequences ( M [ n ] ) ∞ n =1 and ( ˜ M [ n ] ) ∞ n =1 converge to each other if Ω M [1] = Ω ˜ M [1] , and, for all (ˆ ω, ≻ ) ∈ Ω M [1] × P , lim n →∞ A [ n ] (ˆ ω, ≻ ) = lim n →∞ ˜ A [ n ] (ˆ ω, ≻ ) . Then we say an FTTC satisfying ETE is continuous if, for any two regular sequences ( M [ n ] ) ∞ n =1 and ( ˜ M [ n ] ) ∞ n =1 that converge to each other andany (ˆ ω, ≻ ) ∈ Ω M [1] × P , both lim n →∞ p [ n ](ˆ ω, ≻ ) and lim n →∞ ˜ p [ n ](ˆ ω, ≻ ) exist, and lim n →∞ p [ n ](ˆ ω, ≻ ) = lim n →∞ ˜ p [ n ](ˆ ω, ≻ ) . roof of Proposition 6 . Given any regular sequence of problems ( M [ n ] ) ∞ n =1 , there exists N > such that for all n ≥ N and all (ˆ ω, ≻ ) ∈ Ω M [1] × P , the number of type- (ˆ ω, ≻ ) agentsin M [ n ] are more than one. Suppose for some (ˆ ω, ≻ ) ∈ Ω M [1] × P , some type- (ˆ ω, ≻ ) agent i reports ≻ ′ i ∈ P\{≻ i } in the economy M [ n ] . Then we obtain a new economy ˜ M [ n ] such that |{ i ∈ ˜ I [ n ] : ˜ ω [ n ] i = ˆ ω, ≻ i = ≻}| = |{ i ∈ I [ n ] : ω [ n ] i = ˆ ω, ≻ i = ≻}| − , |{ i ∈ ˜ I [ n ] : ˜ ω [ n ] i = ˆ ω, ≻ i = ≻ ′ }| = |{ i ∈ I [ n ] : ω [ n ] i = ˆ ω, ≻ i = ≻ ′ }| + 1 , and the numbers of agents of other types in ˜ M [ n ] and M [ n ] are equal. Then ( M [ n ] ) n ≥ N and ( ˜ M [ n ] ) n ≥ N are two regular sequences that converge to each other.Let ( p [ n ] ) n ≥ N and (˜ p [ n ] ) n ≥ N be the sequences of assignments found by a continuous FTTCsatisfying EENE for the two sequences of economies. By continuity, lim n →∞ p [ n ](ˆ ω, ≻ ′ ) = lim n →∞ ˜ p [ n ](ˆ ω, ≻ ′ ) . So for any ε > , there exists n ∗ (ˆ ω, ≻ , ≻ ′ ) > N such that, for all n > n ∗ (ˆ ω, ≻ , ≻ ′ ) , max o ∈ O | p [ n ](ˆ ω, ≻ ′ ) ,o − ˜ p [ n ](ˆ ω, ≻ ′ ) ,o | < ε/K. Thus, | u i ( p [ n ](ˆ ω, ≻ ′ ) ) − u i (˜ p [ n ](ˆ ω, ≻ ′ ) ) | < ε. Since the FTTC satisfies EENE, p [ n ](ˆ ω, ≻ ) % sd p [ n ](ˆ ω, ≻ ′ ) . So u i ( p [ n ](ˆ ω, ≻ ) ) ≥ u i ( p [ n ](ˆ ω, ≻ ′ ) ) > u i (˜ p [ n ](ˆ ω, ≻ ′ ) ) − ε. Define n ∗ = max { n ∗ (ˆ ω, ≻ , ≻ ′ ) : (ˆ ω, ≻ , ≻ ′ ) ∈ Ω M × P × P , ≻6 = ≻ ′ } . For all n > n ∗ , if any i ∈ I [ n ] in M [ n ] misreports any ≻ ′ i ∈ P\{≻ i } , then i ’s utility gain is bounded by ε . So theFTTC is asymptotically strategy-proof. 44 Using T e to solve Example 1 • At step 1, we solve the equations x (1) = x (1) = 1 / x o (1) + 1 / x o (1) ,x (1) = x (1) = 1 / x o (1) + 1 / x o (1) + 1 / x o (1) ,x (1) = 1 / x o (1) + 1 / x o (1) ,x o (1) = x (1) ,x o (1) = x o (1) = 0 ,x o (1) = x (1) + x (1) ,x o (1) = x (1) + x (1) , subject to the constraints / x o (1) ≤ / , / x o (1) ≤ / , / x o (1) ≤ / , / x o (1) ≤ / , / x o (1) ≤ / . The maximum solution is x ∗ (1) = x ∗ (1) x ∗ (1) x ∗ (1) x ∗ (1) x ∗ (1) x ∗ o (1) x ∗ o (1) x ∗ o (1) x ∗ o (1) x ∗ o (1)1 / / / / / / / ! . The nodes other than o , o constitute the only absorbing set in Figure 1. • At step 2, because o is exhausted, agents and report o as favorite. We omit theequations and constraints. The maximum solution is x ∗ (2) = x ∗ (2) x ∗ (2) x ∗ (2) x ∗ (2) x ∗ (2) x ∗ o (2) x ∗ o (2) x ∗ o (2) x ∗ o (2)1 /
24 1 /
24 1 /
12 1 /
12 1 /
12 1 /
12 0 3 /
12 0 ! . At this step the nodes other than o , o constitute the only absorbing set. • At step 3, the maximum solution is the one such that obtains / o . { o , } is theonly absorbing set. • At step 4, the maximum solution is the one such that and each obtain / o . { , , o } is the only absorbing set. • At step 5, the maximum solution is the one such that and each obtain / o , and , , and each obtain / o . { , , o } and { , , , o } are two absorbing sets. We omit the constraints / x o (1) ≤ / and / x o (1) ≤ / obtained from agent ’s endowment. At step 6, agent is the only remaining agent who owns the remaining endowment / o . So obtains / o . • So T e finds the following assignment. o o o o o / / / / / /
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12 2 / /
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45 0 0 1 / / nline Appendix (not for publication) E Possible ideas to solve FEE by trading cycles
We discuss several ideas to solve the FEE model through clearing cycles, and apply them toExample 1 in our paper. We show that these ideas are undesirable in fairness.Recall that in Example 1, by letting agents point to favorite objects and objects point toall owners, we obtain Figure 1. o o o o o
12 34 5 / / / / / / / / / / / / Figure 1There are five cycles in the graph:cycle 1: → o → cycle 2: → o → cycle 3: → o → → o → cycle 4: → o → → o → cycle 5: → o → → o → → o → These cycles are not disjoint and some nest the others. For example, cycle 3 and cycle 4share the edge → o , and cycle 1 and cycle 3 are nested by cycle 5. If we want to use theidea of clearing cycles to solve Figure 1, we need to answer (1) whether all cycles are cleared,or only a subset of cycles are selected to clear, and (2) how much of objects is traded inevery (selected) cycle. A1 .1 Clearing non-redundant cycles Because cycles are not disjoint and some nest the others, it seems difficult to clear all cyclessimultaneously and reasonable to select a subset of cycles to clear. The first idea observesthat some long cycles seem redundant in the presence of short cycles, because the agents insuch long cycles can obtain same objects by clearing short cycles. In Figure 1, cycle 5 isredundant in the presence of cycle 1 and cycle 3, because every agent in cycle 5 can obtainthe object he points to by clearing cycle 1 or cycle 3. So clearing non-redundant cycles seemsto be a reasonable idea. There can be many definitions of redundancy. Here we say a cycleis redundant if it nests multiple shorter cycles and every agent in the cycle is involved in oneof the shorter cycles. Otherwise, we say the cycle is non-redundant .To obtain a mechanism, we need to specify how much of objects is traded in every non-redundant cycle. In every cycle the maximum amount of objects that can be traded is equalto the smallest remaining endowment (the smallest weight of the edges o → i ) in the cycle. However, if multiple cycles share an edge o → i , we need to divide the weight of o → i amongthe multiple cycles. How to choose the division is related to fairness. o o o o o / / / / / / / /
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45 0 0 1 / / Table 4: Assignment found by the first ideaYet even without choosing the division rule, we are already able to use the first idea tosolve Example 1. In Figure 1, only cycle 5 is redundant; cycle 3 and cycle 4 only sharethe edge → o . So we can trade objects as much as possible in every cycle. We tradethe amount / in cycle 1, the amount / in cycle 2, the amount / in cycle 3, and theamount / in cycle 4. So obtains / o , obtains / o , obtains / o , obtains / o ,and obtains / o . This finishes the first step of the mechanism. The following steps arestraightforward because only self-cycles ( i → o → i ) appear. In particular, at step two, obtains / o ; at step three, obtains / o and obtains / o ; at step four, and each We do consider agents’ demands because their remaining endowments are no greater than their remainingdemands.
A2btain / o , and and each obtain / o . The found assignment is shown in Table 4. Itis sd-efficient but violates EENE. Agents , have equal endowments, but ’s lottery doesnot weakly stochastically dominate 2’s lottery for ’s preferences. E.2 Clearing all cycles equally
Given the failure of selecting cycles in the first idea, the second idea is to clear all cyclessimultaneously. The difficulty is how to do so in a feasible and fair way. We choose to treatall cycles equally by trading an equal amount of objects in all cycles. At each step d , forevery edge o → i in the generated graph, we count the number of cycles that involve theedge and denote it by n o → i ( d ) . Since the weight of the edge is ω io ( d − , at most ω io ( d − n o → i ( d ) of objects can be traded in each of the n o → i ( d ) cycles. So we define the trading quota of allcycles at step d to be min o → i : n o → i ( d ) > ω io ( d − n o → i ( d ) . This mechanism seems to be fair. If an edge o → i is involved in multiple cycles, i trades an equal amount of o in each of the multiple cycles. It also allows agents of differentendowments to obtain different benefits from their endowments. If an agent’s endowmentis demanded by more others, he is involved in more cycles and thus obtains more of hisfavorite object. However, this mechanism is undesirable. If we run it in Example 1, at stepone the trading quota of all cycles is / . So obtains / o , obtains / o , obtains / o , obtains / o , and obtains / o . The following steps are straightforward becauseonly self-cycles appear. At step two, obtains / o and obtains / o ; at step three, obtains / o and obtains / o ; at step four, and each obtain / o , and and each obtain / o . The found assignment is shown in Table 5. It is sd-efficient but violatesEENE. Agents , have equal endowments, but ’s lottery does not weakly stochasticallydominate ’s lottery for ’s preferences. o o o o o / / / / / / /
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45 0 0 1 / / Table 5: Assignment found by the second ideaA3 .3 Decomposing agents
The third idea is to mimic Kesten (2009) to decompose every agent into multiple pseudo-agents and let pseudo-agents trades endowments on behalf of real agents. Specifically, every i is decomposed into pseudo-agents { i o } o ∈ O such that every i o holds i ’s endowment of o . Thebenefit of this idea is that in the generated graph, every pseudo-agent is pointed by only oneobject and points to only one object. So the graph seems to be simplified. However, somereflection should convince the reader that this simplification is illusory. In the generatedgraph an object may point to several pseudo-agents and every pseudo-agent may be involvedin several cycles, which is same as before. If we use the former two ideas to solve Example1, we obtain same assignments as before.Aziz (2015) has used this idea to extend TTC to the FEE model. To solve the difficultywe have discussed, Aziz uses exogenous rankings of agents to select cycles. At each step d ,let I ′ ( d −
1) = { i ∈ I ( d −
1) : ω io i ( d ) > } be the set of agents whose remaining endowmentscontain their favorite objects. If I ′ ( d − is nonempty, his mechanism lets every i ∈ I ′ ( d − obtain his endowment of o i ( d ) immediately. Otherwise, an exogenous ranking determines anagent that each remaining object should point to. We describe his mechanism as a memberof FTTC, and denote it by T a . T a uses the following λ a ( d ) and β a ( d ) : • If I ′ ( d − = ∅ , for all i ∈ I ′ ( d − , λ ai,o i ( d ) ( d ) = ω i,o i ( d ) ( d − P j ∈ I ′ ( d − o j ( d )= o i ( d ) ω j,o i ( d ) ( d − ,β ai,o i ( d ) ( d ) = ω i,o i ( d ) ( d − . For all i ∈ I ′ ( d − and all o ∈ O ( d − \{ o i ( d ) } , and all i ∈ I ( d − \ I ′ ( d − and all o ∈ O ( d − , λ ai,o ( d ) is arbitrary and β ai,o ( d ) = 0 . • If I ′ ( d −
1) = ∅ , let ⋗ be an exogenous ranking of agents. For all i ∈ I ( d − and all o ∈ O ( d − , λ ai,o ( d ) = if i = arg ⋗ - max { j ∈ I ( d −
1) : ω j,o ( d − > } , otherwise ,β ai,o ( d ) = ω i,o ( d − . Aziz considers weak preferences. For simplicity, we only discuss his mechanism under strict preferences. A4 a is undesirable in fairness. It violates ETE. In Example 3 of the paper, suppose both2 and 3 own / b whereas 4 owns / c . If 2 is ranked highest in the exogenous ranking, thenat step one, b only points to 2. So 2 and 4 form a cycle. By trading endowments, 2 obtains / c and 4 obtains / b . At step two, 3 obtains his endowment / b . So 3 envies 2.Aziz proves that T a always finds a weak core assignment. Through the following examplewe show that this statement is incorrect. Example 6.
Consider four agents { , , , } and four objects { o , o , o , o } . Suppose T a uses the exogenous ranking ⋗ ⋗ ⋗ . Table 6 shows agents’ endowments and prefer-ences, and the assignment found by T a . The assignment is strongly blocked by { , } throughexchanging their endowments. o o o o / / / / / /
24 0 1 / / (a) Endowments ≻ ≻ ≻ ≻ o o o o o o o o o o o o o o o o (b) Preferences o o o o / / / /
23 0 1 / / / / (c) Assignment Table 6Aziz’s mistake is due to an incorrect lemma stating that a weak core assignment in thedecomposed problem is also a weak core assignment in the original problem. This lemmais incorrect because the blocking incentive of an agent in the original problem is not equalto the “sum” of the blocking incentives of its pseudo-agents in the decomposed problem. InExample 6, T a finds a weak core assignment in the decomposed problem: the pseudo-agent o does not want to form a blocking coalition with the pseudo-agent o , and the pseudo-agent o does not want to form a blocking coalition with the pseudo-agent o . But in theoriginal problem, and want to block the assignment by exchanging their endowments. F Extension of T e to HET The extension of T e to HET is first proposed by Zhang (2017) and denoted by P S E . Wefirst present Zhang’s definition that describes P S E as an extension of PS and TTC. We thenexplain that P S E coincides with the extension of T e to HET when HET is described by acoarse priority structure as stated in the paper.A5n HET, we denote the set of existing tenants by I E and the set of their private endow-ments by O E . Let π : I E → O E be a bijection such that π ( i ) is the private endowment of i ∈ I E . Let ¯ O = O ∪ O E denote the set of all objects. After each step d of P S E , let I ( d ) bethe set of remaining agents, ¯ O ( d ) the set of remaining objects, r i ( d ) the remaining demandof each i ∈ I ( d ) , and r o ( d ) the remaining amount of each o ∈ ¯ O ( d ) .The P S E algorithm • Initialization : I (0) = I , ¯ O (0) = O ∪ O E , and r a (0) = 1 for all a ∈ I (0) ∪ O (0) . Set theclock t = 0 . • Step d ≥ : Let every i ∈ I ( d − point to his favorite object o i ( d ) among ¯ O ( d − .Let every o ∈ ¯ O ( d − ∩ O E point to its owner π − ( o ) if π − ( o ) ∈ I ( d − . – If there are cycles among existing tenants, let I ( c ) and H ( c ) be the set of agentsand the set of objects involved in each cycle c . The agents in each cycle trade theamount min { r a ( d − } a ∈ I ( c ) ∪ H ( c ) of private endowments instantly. Remove agentswhose demands are satisfied and objects that are exhausted. If the remainingagents I ( d ) or the remaining objects ¯ O ( d ) become empty, stop; otherwise, go tostep d + 1 . – If there are no cycles, let agents simultaneously consume their favorite objectswith rates (denoted by s i ( d ) ) that satisfy the “you request my house - I get yourrate” rule. For all i ∈ I ( d − \ I E , s i ( d ) = 1 ; for all j ∈ I ( d − ∩ I E , s j ( d ) = s π ( j ) ( d ) + 1 where s π ( j ) ( d ) = P i ∈ I ( d − o i ( d )= π ( j ) s i ( d ) . This step stops when someagent’s demand is satisfied or some object is exhausted. If I ( d ) or ¯ O ( d ) becomeempty, stop; otherwise, go to step d + 1 .As said, HET can be described by a coarse priority in which social endowments O give allagents equal priority, and every private endowment o ∈ O E gives its owner highest priorityand the other agents equal priority. In the extension of T e , let I ( d ) , ¯ O ( d ) , and o i ( d ) bedefined as above. Let O E ( d ) = { o ∈ O E : π − ( o ) ∈ I ( d ) } be the set of private endowmentswhose owners remain in the procedure after step d . Then at each step d of the extension of T e , agents trade endowments according to the maximum solution x ∗ ( d ) to the equations x o ( d ) = P i ∈ I ( d − o i ( d )= o x i ( d ) for all o ∈ ¯ O ( d − ,x i ( d ) = P o ∈ ¯ O ( d − \ O E ( d − x o ( d ) | I ( d − | for all i ∈ I ( d − \ I E ,x j ( d ) = P o ∈ ¯ O ( d − \ O E ( d − x o ( d ) | I ( d − | + x π ( j ) ( d ) for all j ∈ I ( d − ∩ I E . A6ubject to the constraints x o ( d ) ≤ − P d − k =1 x ∗ o ( k ) for all o ∈ ¯ O ( d − ,x i ( d ) ≤ − P d − k =1 x ∗ i ( k ) for all i ∈ I ( d − . To prove that
P S E coincides with the extension of T e , we prove that agents’ consumptionat each step d of P S E is exactly the maximum solution x ∗ ( d ) at step d of T e . At each step d of P S E , if there are cycles among existing tenants, for each existing tenant j involvedin a cycle, it is obvious that x j ( d ) = x π ( j ) ( d ) , whereas for each agent i and each object o not involved in any cycle, x i ( d ) = x o ( d ) = 0 . So x ( d ) satisfies the above equationsand constraints. If there are no cycles, let t ( d ) be the duration of step d . Then for any o ∈ ¯ O ( d − , x o ( d ) = P i ∈ I ( d − o i ( d )= o x i ( d ) , which holds by definition. For any i ∈ I ( d − \ I E , x i ( d ) = s i ( d ) t ( d ) = t ( d ) , whereas for any j ∈ I ( d − ∩ I E , x j ( d ) = s j ( d ) t ( d ) = P i ∈ I ( d − o i ( d )= π ( j ) s i ( d ) t ( d ) + t ( d ) = P i ∈ I ( d − o i ( d )= π ( j ) x i ( d ) + t ( d ) = x π ( j ) ( d ) + t ( d ) . We sayan agent i is linked to an object o if there exist distinct existing tenants j , j , . . . , j ℓ suchthat i points to π ( j ) , j points to π ( j ) , j points to π ( j ) , . . . , j ℓ points to o . Because thereare no cycles, every remaining agent must be linked to some object in ¯ O ( d − \ O E ( d − .Then the “you request my house - I get your rate” rule implies that the total rates of theagents who consume ¯ O ( d − \ O E ( d − is equal to the number of remaining agents. So t ( d ) = P o ∈ ¯ O ( d − \ O E ( d − x o ( d ) / | I ( d − | . It means that agents’ consumption at step d of P S E still satisfies the above equations and constraints. Since at each step of P S E agentstrade endowments or consume objects as much as possible until some agent is satisfied orsome object is exhausted, P S E coincides with the extension of T ee