the house allocation problem jordi massó spring 2010pareto.uab.es/jmasso/pdf/house allocation...

MatchingThe House Allocation Problem

Jordi Massó

International Doctorate in Economic Analysis (UAB)

Spring 2010

J. Massó (International Doctorate in Economic Analysis (UAB))Matching: The House Allocation Problem Spring 2010 1 / 29

The House Allocation Problem

References:

A. Hylland and R. Zeckhauser. �The e¢ cient allocation of individualsto positions,� Journal of Political Economy 87, 293-314 (1979).

L. Shapley and H. Scarf. �On cores and indivisibilities,� Journal ofMathematical Economics 1, 23-28 (1974).



A �nite set of n agents A = fa1, ..., ang.

A �nite set of n houses H = fh1, ..., hng (or indivisible objects likeo¢ ces, parking spaces, dormitory rooms, school seats, kidneys, etc.).

Each agent a 2 A has a strict preference ordering Pa over the set ofhouses H.

A pro�le P = (Pa)a2A is a list of preference orderings, one for eachagent.

The triple (A,H,P) is a house allocation problem.

Fix A and H. A problem is identi�ed only with a pro�le P.



A matching µ : A �! H is a one-to-one mapping that assigns toeach agent one house in such a way that no house is assigned to morethan one agent.

A housing market is a list (A,H,P, µ), where (A,H,P) is a houseallocation problem and µ is a matching of initial endowments.

De�nitionA matching η is in the core of the housing market (A,H,P, µ) if there isno coalition S � A and a matching ν such that:

ν(a) 2 µ(S) for all a 2 S ,

ν(a)Raη(a) for all a 2 S ,

ν(a)Paη(a) for some a 2 S .



Remarks:

A core matching is individually rational (every agent receives a houseat least as good as her initial house): take singleton coalitions in thede�nition of core.

A core matching is Pareto e¢ cient: take S = A in the de�nition ofcore.

Theorem(Shapley and Scarf, 1974) The core is nonempty for each housing market.

Two alternative proofs.

1 To show that an associate cooperative game is balanced (and hence ithas a nonempty core).

2 Attributed to David Gale, using an algorithm known as Gale�stop-trading cycles (TTC) algorithm.



Gales�s Top Trading Cycles Algorithm

Input: A housing market (A,H,P, µ).

Step 1:

Each agent �points to� the owner of his favorite house. Since there are�nite number of agents, there is at least one cycle.

Each agent in a cycle is assigned the house of the agent he points toand removed from the market with his assignment.

If there is at least one remaining agent, proceed with the next step.Otherwise, the output of the algorithm is the matching consisting ofthe assignment.



Step k:

Each remaining agent points to the owner of his favorite house amongthe remaining houses.

Each agent in a cycle is assigned the house of the agent he points toand removed from the market with his assignment.

If there is at least one remaining agent, proceed with the next step.Otherwise, the output of the algorithm is the matching consisting ofthe assignment done in all steps.



Let ν be the matching obtained as the result of the Gale�s TTCalgorithm.

To see that ν is in the core, let S1, ...,SK be the agents in cycles (inthe order they are removed) in Gale�s TTC algorithm.

Observe that no agent in S1 can be in a blocking coalition with thestrict preference, since they get their �rst choice under ν.

Given this, no agent in S2 can be in a blocking coalition with thestrict preference, since they get their �rst choice in Hnfν(S1)g.

Iteratively, we continue. Thus, ν is in the core.


The House Allocation Problem: Example

Example

Let jAj = jO j = 11, µ(ai ) = oi for all i = 1, ..., 11, and consider thepro�le P

Pa1 Pa2 Pa3 Pa4 Pa5 Pa6 Pa7 Pa8 Pa9 Pa10 Pa11o2 o6 o2 o2 o4 o2 o3 o3 o6 o7 o2o6 o1 o1 o8 o7 o1 o8 o6 o4 o1 o6o7 o8 o11 o4 o3 o6 o11 o1 o5 o5 o10o5 o9 o4 o7 o6 o8 o6 o2 o1 o4 o7o1 o5 o7 o3 o1 o3 o1 o10 o8 o10 o5o4 o4 o10 o9 o8 o9 o9 o7 o11 o6 o4o9 o7 o3 o6 o2 o10 o2 o5 o10 o2 o1o8 o3 o6 o10 o5 o5 o7 o4 o3 o9 o11o11 o2 o5 o1 o10 o11 o4 o11 o9 o11 o9o3 o10 o8 o5 o11 o4 o10 o8 o2 o8 o3o10 o11 o9 o11 o9 o7 o5 o9 o7 o3 o8


Gale�s TTC Algorithm: Step 1

Pa1 Pa2 Pa3 Pa4 Pa5 Pa6 Pa7 Pa8 Pa9 Pa10 Pa11o2 o6 o2 o2 o4 o2 o3 o3 o6 o7 o2o6 o1 o1 o8 o7 o1 o8 o6 o4 o1 o6o7 o8 o11 o4 o3 o6 o11 o1 o5 o5 o10o5 o9 o4 o7 o6 o8 o6 o2 o1 o4 o7o1 o5 o7 o3 o1 o3 o1 o10 o8 o10 o5o4 o4 o10 o9 o8 o9 o9 o7 o11 o6 o4o9 o7 o3 o6 o2 o10 o2 o5 o10 o2 o1o8 o3 o6 o10 o5 o5 o7 o4 o3 o9 o11o11 o2 o5 o1 o10 o11 o4 o11 o9 o11 o9o3 o10 o8 o5 o11 o4 o10 o8 o2 o8 o3o10 o11 o9 o11 o9 o7 o5 o9 o7 o3 o8,



��

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9 10 11

7 8

6

4 5

1 2 3-

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6

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HHHH

HHHY

BBBBBBBBBBBM

(a1, o1) (a2, o2) (a3, o3)

(a4, o4) (a5, o5)

(a6, o6)

(a7, o7) (a8, o8)

(a9, o9) (a10, o10) (a11, o11)



η(a2) = o6η(a6) = o2

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9 10 11

7 8

6

4 5

1 2 3-

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HHHH

HHHY

BBBBBBBBBBBM

(a1, o1) (a2, o2) (a3, o3)

(a4, o4) (a5, o5)

(a6, o6)

(a7, o7) (a8, o8)

(a9, o9) (a10, o10) (a11, o11)



Pa1 Pa3 Pa4 Pa5 Pa7 Pa8 Pa9 Pa10 Pa11o4 o3 o3 o7

o1 o8 o7 o8 o4 o1o7 o11 o4 o3 o11 o1 o5 o5 o10o5 o4 o7 o1 o4 o7o1 o7 o3 o1 o1 o10 o8 o10 o5o4 o10 o9 o8 o9 o7 o11 o4o9 o3 o5 o10 o1o8 o10 o5 o7 o4 o3 o9 o11o11 o5 o1 o10 o4 o11 o9 o11 o9o3 o8 o5 o11 o10 o8 o8 o3o10 o9 o11 o9 o5 o9 o7 o3 o8



��

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7 8

4 5

1 3

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AAAAAAAK

AAAAAAAK

HHHH

HHHY

�

(a1, o1) (a3, o3)

(a4, o4) (a5, o5)

(a7, o7) (a8, o8)

(a9, o9) (a10, o10) (a11, o11)



η(a1) = o7η(a2) = o6η(a3) = o1η(a6) = o2η(a7) = o3

��

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7 8

4 5

1 3

��

�

XXXXXXXXXXXXXXXXz

�

��3

AAAAAAAK

AAAAAAAK

HHHH

HHHY

�

(a1, o1) (a3, o3)

(a4, o4) (a5, o5)

(a7, o7) (a8, o8)

(a9, o9) (a10, o10) (a11, o11)



Pa4 Pa5 Pa8 Pa9 Pa10 Pa11o4

o8 o4o4 o5 o5 o10

o4o10 o8 o10 o5

o9 o8 o11 o4o5 o10

o10 o5 o4 o9 o11o10 o11 o9 o11 o9

o5 o11 o8 o8o11 o9 o9 o8



��

��

�� 9 10 11

8

4 5XXXXXXXXXXXXXXXXz

�

��AAAAAAAK

��

�

(a4, o4) (a5, o5)

(a8, o8)

(a9, o9) (a10, o10) (a11, o11)



η(a1) = o7η(a2) = o6η(a3) = o1η(a4) = o8η(a5) = o4η(a6) = o2η(a7) = o3η(a8) = o10η(a10) = o5

��

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�� 9 10 11

8

4 5XXXXXXXXXXXXXXXXz

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��

�

(a4, o4) (a5, o5)

(a8, o8)

(a9, o9) (a10, o10) (a11, o11)



Pa9 Pa11

o11

o11o9 o9



�� 9 11-

(a9, o9) (a11, o11)



η(a1) = o7η(a2) = o6η(a3) = o1η(a4) = o8η(a5) = o4η(a6) = o2η(a7) = o3η(a8) = o10η(a10) = o5η(a11) = o11

�� 9 11-

(a9, o9) (a11, o11)



Agent a9 is asigned to his own object o9.

Thus, the assignment η in the Core obtained by the Gale�s TTCalgorithm can be represented by

η =

�a1 a2 a3 a4 a5 a6 a7 a8 a9 a10 a11o7 o6 o1 o8 o4 o2 o3 o10 o9 o5 o11

�.



References:

J. Ma. �Strategy-proofness and the strict core in a market withindivisibilities,� International Journal of Game Theory 23, 75-83 (1994).

A. Roth. �Incentive compatibility in a market with indivisible goods,�Economics Letters 9, 127-132 (1982).

A. Roth and A. Postlewaite. �Weak versus strong domination in amarket with indivisible goods,� Journal of Mathematical Economy 4,131-137 (1977).

Theorem(Roth and Postelwaite, 1977) The core of each housing market containsonly one matching.



Proof

Let ν be the output of the Gale�s TTC algorithm and assume thatη 6= ν for some matching η.

Let b be the �rst agent who satis�es η(b) 6= ν(b) (according to theorder of the cycles S1, ...,SK , if there are multiple agents in a cyclelike b, then choose one of them arbitrarily).

Let b be in the cycle Sk . Observe that for every agent a assignedbefore the cycle Sk we have η(a) = ν(a).

Thus, for every a 2 Sk , ν(a)Raη(a) and ν(a) 2 µ(Sk ). Moreover,v(b)Pbη(b).

Thus, Sk blocks matching η with ν.



Theorem(Roth, 1982) The core, as a mechanism, is strategy-proof.

Proof Let ϕ : P �!M be the core mechanism. Let P be a pro�le. LetS1, ...,SK be the agents removed in Gale�s TTC algorithm in theconstruction of ν = ϕ[P ]. The proof of the Theorem is by iteration oncycles.

Each agent in S1 receives her �rst choice under ν with respect to P.Hence, none of these agents will bene�t by reporting a di¤erentpreference ordering. Moreover, observe that S1 will form as it is inStep 1, regardless of any agent in AnS1 submits a di¤erent preferenceordering.

...



Each agent in Sk (for k � 2) receives her �rst choice under ν inHn(ν(S1) [ ...[ ν(Sk�1)) with respect to P. Since the previouscycles are una¤ected by them reporting di¤erent preference orderings,if one agent changes her preference ordering still the same houses willbe assigned to the agents in ν(S1) [ ...[ ν(Sk�1). Therefore, thisagent will not bene�t by reporting a di¤erent preference orderingunder Gale�s TTC algorithm, completing the proof.



Let (A,H,P, µ) be a housing market.

Prices of houses is a vector p = (p1, ..., pn).

A house hj is a¤ordable for an agent ai at p if pj � pµ(ai ).

A matching ν and a price vector p is a competitive equilibrium if forevery agent a, ν(a) is the best house she can a¤ord at prices p.

Theorem(Roth and Postlewaite, 1977) There is a unique competitive equilibriummatching for a housing market which is given by the core matching.



Theorem(Ma, 1994) A mechanism φ : P �!M is strategy-proof, Pareto e¢ cient,and individually rational if and only if it is the core mechanism.

Remark: The three axioms are independent.

A serial-dictatorship is strategy-proof, Pareto e¢ cient but it is notindividually rational.

The constant mechanism (φ[P ] = µ for all P 2 P) is strategy-proof,individually rational but it is not Pareto e¢ cient.

Consider the following example.


Example

Consider a housing market with n = 3, µ(ai ) = hi for i = 1, 2, 3, andPa1 : h2, h3, h1, Pa2 : h1, h3, h2, and Pa3 : h1, h3, h2. Then,

ϕ[P ] =�a1 a2 a3h2 h1 h3

�and φ[P ] =

�a1 a2 a3h2 h3 h1

�satisfy individual rationality and Pareto e¢ ciency at P. Complete thede�nition of the mechanism φ by setting, for all P 0 6= P, φ[P 0] = ϕ[P 0].Thus, φ is individually rational and Pareto e¢ cient. To see that themechanism φ is not strategy-proof consider P 0 = (Pa1 ,Pa2 ,P

0a3) with

P 0a3 : h2, h1, h3. Then,

φ[Pa1 ,Pa2 ,Pa3 ] =�a1 a2 a3h2 h3 h1

�and φ[P 0] =

�a1 a2 a3h2 h1 h3

�.

Hence, φ[P ](a3) = h1P 0a3h3 = φ[P 0](a3). Thus, a3 manipulates φ at P 0 bydeclaring Pa3 .


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