Random matching markets

Itai Ashlagi

Stable Matchings = Core Allocations A matching is stable if there is no man and

woman who both prefers each other over their current match.

The set of stable matchings is a non-empty lattice, whose extreme points are the Men Optimal Stable Match (MOSM) and the Women Optimal Stable Match (WOSM) Men are matched to their most preferred stable

woman under the MOSM and their least preferred stable woman under the WOSM

Multiplicity Core has a lattice structure and can be large

(Knuth) Roth, Peranson (1999) – small core in the NRMP Hitsch, Hortascu, Ariely (2010) – small core in

online dating Banerjee et al. (2009) – small core in Indian

marriage markets

Random matching markets Random Matching Market: A set of men and

a set of women Man has complete preferences over women,

drawn randomly at uniform iid. Woman has complete preferences over men,

drawn randomly at uniform iid.

Questions Average rank, or to whom should agents expect to

match? Define men’s average rank of their wives under

excluding unmatched men from the average Average rank is if all men got their most preferred wife,

higher rank is worse.

How many agents have multiple stable assignments? Agents can manipulate iff they have multiple stable

partners

Large core in random balanced marketsTheorem[Pittel 1998, Knuth, Motwani, Pittel 1990] Consider a random market with men and women.

1. With high probability, the average ranks in the MOSM are

and

2. The fraction of agents that have multiple stable partners tends to 1 as tends to infinity.

Suppose there are n+1 men and n women and the

Random markets with short preference lists have a small coreRoth, Peranson (1999) – find a small core in the NRMP

Theorem[Immorlica, Mahdian 2005]:

Consider a market with n men and n women where men have short (constant length) uniform at random preference lists and women have arbitrary preferences.

Then of men and women have multiple stable matches.

Kojima, Pathak 2009 show that there is a limited scope for manipulating a stable mechanism in such many-to-one large markets (each hospital has a constant number of positions).

Literature (multiplicity) Pittel (1989), Knuth, Motwani, Pittel (1990), Roth,

Peranson (1999) – when there are equal number of men and women Under MOSM men’s average rank of wives is , but it is under the WOSM The core is large – most agents have multiple stable partners

Roth, Peranson (1999) – small core in the NRMP Immorlica, Mahdian (2005) and Kojima, Pathak (2009)

show that if one side has short random preference lists the core is small Many (popular) agents are unmatched

Hitsch, Hortascu, Ariely (2010) – small core in online dating

Banerjee et al. (2009) – small core in Indian marriage markets

Literature Hitsch, Hortascu, Ariely (2010) – online dating Banerjee, Duflo, Ghatak, Lafortune (2009) - Indian

marriage markets Abdulkadiroglu, Pathak, Roth (2005) – NYC school choice

All use Deferred Acceptance (Gale & Shapely) to make predictions…

Crawford (1991) – comparative statics on adding men (women), but only in a given stable matching.

Large core in balanced marketsTheorem[Pittel 1998, Knuth, Motwani, Pittel 1990] Consider a random market with men and women.

1. With high probability, the average ranks in the MOSM are

and

2. The fraction of agents that have multiple stable partners tends to 1 as tends to infinity.

Question: Suppose there are n+1 men and n women and men are proposing. What is the likely average men’s rank of their wives?” [Gil Kalai’s blog]

Matching markets are very competitive

When there are unequal number of men and women:

The short side is much better off under all stable matchings;roughly, the short side “chooses” and the long side gets “chosen” sharp effect of competition despite heterogeneity

The core is small;there is little difference between the MOSM and the WOSM Small core despite long lists and uncorrelated

preferences

Question:Are there any real/natural matching markets with large cores?

Men’s average rank of wives,

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MOSMWOSM

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Men’s average rank of wives,

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Percent of matched men with multiple stable partners

20 23 26 29 32 35 38 41 44 47 50 53 56 59 62 65 68 71 74 77 800

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Theorem [Ashlagi, Kanoria, Leshno 2013]Consider a random market with men and women,for . With high probability in any stable matching

and

Moreover,

And of men and women have multiple stable

matches.

That is, with high probability under all stable matchings Men do almost as well as they would if they

choose in a random order, ignoring women’s preferences.

Women are either unmatched or roughly getting a randomly assigned man.

The core is small

Corollary 1: One women makes a differenceIn a random market with men and women, with high probability

and

in all stable matches, and a vanishing fraction of agents have multiple stable partners.

(n=1000, log n= 6.9, n/logn = 145)

(n=100000,log n =11.5, n/logn =8695)

Corollary 2: Large UnbalanceConsider a random market with men and women for . Then for we have that, with high probability, in all stable matchings

and

Ashlagi, Braverman, Hassidim (2011) also establish a small core in this setting.

Strategic implications Men proposing DA (MPDA) is strategyproof for men,

but no stable mechanism is strategyproof for all agents.

A woman can manipulate MPDA only if she has multiple stable husbands Misreport truncated preferences.

In unbalanced matching market a diminishing number of woman have multiple stable husbands Under full information, a diminishing fraction can

manipulate At the interim, under mild assumptions on utilities,

all agents reporting truthfully is an -Nash

Intuition In a competitive assignment market with

homogenous buyers and homogenous sellers the core is large, but the core shrinks when there is one extra seller.

In a matching market the addition of an extra woman makes all the men better off Every man has the option of matching with the single

woman But only some men like the single woman Changing the allocation of some men requires changing

the allocation of many men. If some men are made better off, some women are made worse off creating more options for men. All man benefit, and the core is small.

Proof overviewCalculate the WOSM using: Algorithm 1: Men-proposing Deferred

Acceptance gives MOSM Algorithm 2: MOSM→WOSMBoth algorithms use a sequence of proposals by men

Stochastic analysis by sequential revelation of preferences.

Algorithm 1: Men-proposing DA (Gale & Shapley)

Everyone starts unmatched. Each man one at a time, begins a ‘chain’.

Chain beginning with : Set to be the proposer. The proposer proposes to his (next) most

preferred woman If is unmatched, end chain and start a new one

with Otherwise, rejects her less preferred man

between and her current partner. Repeat, with rejected man proposing.

Algorithm 2: MOSM → WOSM We look for stable improvement cycles for women.

We iterate:Phase: For candidate woman , reject her match starting a chain.Two possibilities for how the chain ends:

(Improvement phase) Chain reaches new stable match.

(Terminal phase) Chain ends with unmatched woman is ’s best stable match. is no longer a candidate.

Algorithm 2: MOSM → WOSM Initialize 1. Choose in if non-empty.2. Phase: Record the current match as . Woman

rejects her partner, man , starting a chain where accept a proposal only if the proposal is preferred to .

3. Two possibilities for how the chain ends: (Improvement phase) If the chain ends with

acceptance by , we have found a new stable match. Return to Step 2.

(Terminal phase) Else the chain ends with acceptance by . Woman has found her best stable partner. Roll the match back to . Add to S and return to Step1.

prefers to rejects to start a chain

𝑤1

𝑚1

Illustration of Algorithm 2: MOSM → WOSM

𝑤2

𝑚2

𝑤1

𝑚1

Illustration of Algorithm 2: MOSM → WOSM

𝑤2

𝑚2

𝑤3

𝑚3

prefers to

𝑤1

𝑚1

Illustration of Algorithm 2: MOSM → WOSM

𝑤2

𝑚2

𝑤3

𝑚3

𝑤1

𝑚1

Illustration of Algorithm 2: MOSM → WOSM

𝑤2

𝑚2

𝑤3

𝑚3

prefers to

𝑤1

𝑚1

Illustration of Algorithm 2: MOSM → WOSM

𝑤2

𝑚2

𝑤3

𝑚3

New stable match found (Convince yourself that there is no blocking pair),Update match and continue.

𝑤1

Illustration of Algorithm 2: MOSM → WOSM

𝑚3

𝑤1

𝑚3

Illustration of Algorithm 2: MOSM → WOSM

rejects to start a chain prefers to

𝑤1

𝑚3

Illustration of Algorithm 2: MOSM → WOSM

𝑤4

𝑚4

𝑤1

𝑚3

Illustration of Algorithm 2: MOSM → WOSM

𝑤4

𝑚4

𝑤5

𝑚5

prefers to

ends with a proposal to unmatched woman

is ’s best stable partner

W and

𝑤1

𝑚3

Illustration of Algorithm 2: MOSM → WOSM

𝑤4

𝑚4

𝑤5

𝑚5

𝑤

𝑤1

𝑚3

Illustration of Algorithm 2: MOSM → WOSM

𝑤4

𝑚4

𝑤5

𝑚5

𝑤

ends with a proposal to unmatched woman

is ’s best stable partner

and similarly and already had their best stable partner

𝑤1

𝑚3

Illustration of Algorithm 2: MOSM → WOSM

𝑤4

𝑚4

𝑤5

𝑚5

𝑤

ends with a proposal to unmatched woman

is ’s best stable partner

and similarly and already had their best stable partner

𝑤1

𝑚3

Illustration of Algorithm 2: MOSM → WOSM

𝑤4

𝑚4

𝑤5

𝑚5

𝑤

𝑤1

𝑚3

Illustration of Algorithm 2: MOSM → WOSM

𝑤4

𝑚4

𝑤5

𝑚5

𝑤

𝑤8

𝑚8

rejects to start a new chain

𝑤1

𝑚3

Illustration of Algorithm 2: MOSM → WOSM

𝑤4

𝑚4

𝑤5

𝑚5

𝑤

𝑤8

𝑚8

But is already matched to her best stable partner

prefers to

𝑤1

𝑚3

Illustration of Algorithm 2: MOSM → WOSM

𝑤4

𝑚4

𝑤5

𝑚5

𝑤

𝑤8

𝑚8

prefers to

But is already matched to her best stable partner

is ’s best stable partner

𝑤1

𝑚3

Illustration of Algorithm 2: MOSM → WOSM

𝑤4

𝑚4

𝑤5

𝑚5

𝑤

𝑤8

𝑚8

prefers to

But is already matched to her best stable partner

is ’s best stable partner

Algorithm 2: MOSM → WOSM Initialize 1. Choose in if non-empty.2. Phase: Record the current match as . Woman

rejects her partner, man , starting a chain where accepts a proposal only if the proposal is preferred to .

3. Two possibilities for how the chain ends: (Improvement phase) If the chain ends with

acceptance by , we have found a new stable match. Return to Step 2.

(Terminal phase) Else the chain ends with acceptance by . Woman has found her best stable partner. Roll the match back to . Add to S and return to Step1.

Algorithm 2: MOSM → WOSM Initialize (S set of women matched to best stable partners)1. Choose in if non-empty.2. Phase: Record the current match as . Woman

rejects her partner, man , starting a chain. A proposal is accepted if the woman prefers the proposer to her match under .

3. Two possibilities: (Improvement cycle) If there is an acceptance by a

woman in the chain, we have found a new stable match. Update , erase the women with new stable matches from the chain and continue.

(Terminal phase) Else the chain ends with acceptance by . Woman has found her best stable partner. Roll the match back to . Add and all women who accepted proposals in this chain to S and return to Step 1.

Proof idea:

Analysis of MPDA is similar to that of Pittel (1989) Coupon collectors problem

Analysis of Algorithm 2: MOSM → WOSM more involved. S grows quickly (set of woman that are already

matched to best stable partner) Once S is large improvement phases are rare Together, in a typical market, very few agents

participate in improvement cycles.

𝑤1

𝑚1

Why does the set S (women matched to best stable partners) grows quickly?

𝑤2

𝑚2

𝑤5

𝑚5

𝑤

• After finding the MOSM, men haven’t made and women haven’t received many proposals.

• Proposals to or with probability ~1/n

=> After eliminating cycles first terminal phase will include women!

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Further questions

Can we allow correlation in preferences? Perfect correlation leads to a unique core. But “short side” depends on more than

Tiered market: 15 men, 20 women: 10 top, 10 mid What is a general balance condition?

Are there any real/natural matching markets with large cores?

Correlation -

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MOSM

WOSM

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Pct Multiple Stable

Correlation in Men's preferences - β(N+K)

Men

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Correlation -

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WOSM

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Correlation in Men's preferences - β(N+K)

Men

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Men’s average rank of wives

diff -10 -1 0 +1 +5 +10

100MOSM 29.5 20.3 5.0 4.1 3.0 2.6

WOSM 30.1 23.6 20.3 4.9 3.2 2.6

200MOSM 53.6 35.3 5.7 4.8 3.7 3.1

WOSM 54.7 41.0 35.5 5.7 3.8 3.2

500MOSM 115.8 75.9 6.7 5.7 4.5 3.9

WOSM 118.0 86.6 76.2 6.7 4.7 4.0

1000MOSM 203.8 136.2 7.4 6.4 5.2 4.6

WOSM 207.5 155.1 137.3 7.4 5.4 4.7

2000MOSM 364.5 249.6 8.1 7.1 5.9 5.3

WOSM 370.8 280.7 249.1 8.1 6.1 5.4

5000MOSM 793.1 560.0 9.1 8.1 6.8 6.2

WOSM 804.7 622.5 560.2 9.1 7.0 6.3

Percent of men with multiple stable matches

Diff: -10 -1 0 +1 +5 +10

100 2.1 15.1 75.3 15.4 4.5 2.3

200 2.2 14.6 83.6 14.6 4.1 2.1

500 2.0 12.6 91.0 13.1 3.6 2.0

1000 1.9 12.3 94.5 12.2 3.4 2.0

2000 1.8 11.1 96.7 11.1 2.9 1.7

5000 1.5 10.1 98.4 10.2 2.8 1.5

Large Simulations

  Men's rank under % Men with Men's rank under% Men with

  MOSM WOSMMultiple Stable MOSM WOSM

Multiple Stable

101.98

(0.45)2.29

(0.60)13.84 (18.82)

1.31 (0.20)

1.33 (0.21) 1.19 (5.13)

1004.09

(0.72)4.89

(1.08)15.16 (12.98)

2.55 (0.26)

2.61 (0.27) 2.30 (3.15)

1,0006.47

(0.79)7.44

(1.28)11.9

(10.17)4.59

(0.30)4.69

(0.31) 1.95 (2.03)

10,0008.80

(0.79)9.80

(1.30) 9.45 (8.30)6.88

(0.30)6.98

(0.32) 1.46 (1.47)

100,00011.11 (0.83)

12.09 (1.31) 7.66 (6.60)

9.16 (0.31)

9.26 (0.32) 1.08 (1.02)

1,000,000

13.40 (0.80)

14.41 (1.27) 6.62 (6.04)

11.46 (0.30)

11.56 (0.32) 0.85 (0.80)

Summary

Random unbalanced matching markets are very competitive:

The short side chooses in all stable matchings The core is small – most agents have a single

stable partner

Do matching markets generically have small cores?

Recent developments and future directions

1. Why do we observe short preference lists in practice?

2. Efficiency vs Stability (Lee & Yairv 14, Che & Tricieux 14 – consider cardinal utilities)

3. Surplus in random markets with transfers (Romm & Hassidim 2014 – law of one price in random Shapley Shubik model)

Topics

Unbalanced matching markets

Matching markets with couples

57

Source: https://www.aamc.org/download/153708/data/charts1982to2011.pdf59

Two-body problems

Couples of graduates seeking a residency program together.

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In the 1970s and 1980s: rates of participation in medical clearinghouses decreases from ~95% to ~85%. The decline is particularly noticeable among married couples.

1995-98: Redesigned algorithm by Roth and Peranson (adopted at 1999)

Decreasing participation of couples

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Couples’ preferences

The couples submit a list of pairs. In a decreasing order of preferences over pairs of programs – complementary preferences!

Example:

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Alice BobNYC-A NYC-XNYC-A NYC-Y

Chicago-A Chicago-XNYC-B NYC-X

No Match NYC-X

Couples in the match (n≈16,000)

Source: http://www.nrmp.org/data/resultsanddata2010.pdf

63

No stable match [Roth’84, Klaus-Klijn’05]

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C12

1

AC

2

CB

BA1 2

Option 1: Match AB

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C12

1

AC

2

CB

C-2 is blocking

BA1 2

BA1 2

Option 2: Match C2

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C12

1

AC

2

CB

C-1 is blocking

BA1 2

C12

Option 3: Match C1

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C12

1

AC

2

CB

AB-12 is blocking

BA1 2

C12

Stability with couples

But: In the last 12 years, a stable match has

always been found. Only very few failures in other markets.

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Large random market n doctors, k=n1-ε couples λn residency spots, λ>1 Up to c slots per hospital Doctors/couples have uniformly random

preferences over hospitals (can also allow “fitness” scores)

Hospitals have arbitrary responsive preferences over doctors.

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Stable match with few couples

Theorem [Kojima,Pathak, Roth 10]: In a large random market with n doctors and n0.5-ε couples, with probability →1• a stable match exists• truthfulness is an approximated Bayes-Nash equilibrium

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Remark: Kojima-Pathak-Roth actually model this when there are n doctors and n slots and each doctor has a short preference list

Theorem [Ashlagi, Braverman, Hassidim 2012]: In a large random market with at most n1-ε couples, with probability →1: a stable match exists, and we find it using a

new Sorted Deferred Acceptance (SoDA) algorithm

truthfulness is an approximated Bayes-Nash equilibrium

Existence in large markets when the number of couples is not too large

The main idea of the proof We would like to run deferred acceptance in

the following order: singles; couples: singles that are evicted apply down their

list before the next couple enters. If no couple is evicted in this process, it

terminates in a stable matching.

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What can go wrong?

Alice evicts Charlie. Charlie evicts Bob. H1 regrets letting

Charlie go.

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C12

1

AC

2

CB

BA1 2

Solution

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Find some order of the couples so that no previously inserted couples is ever evicted.

The couples (influence) graph

Is a graph on couples with an edge from AB to DE if inserting couple AB may displace the couple DE.

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BA1 2

C12

BA1 2

The couples graph

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A B

C D

E F

GA B

E F

The couples graph

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A B

C D

E F

GA B

E F

The SoDA algorithm The Sorted Deferred Acceptance algorithm

looks for an insertion order where no couple is ever evicted.

This is possible if the couples graph is acyclic.

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A BC D

E FG H

Insert the couples in the order:AB, CD, EF, GH

orAB, CD, GH, EF

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A BC D

E FG H

Sorted Deferred Acceptance (SoDA)

Set some order π on couples.Repeat: Deferred Acceptance only with singles. Insert couples according to π as in DA:

If AB evicts CD: move AB ahead of CD in π. Add the edge AB→CD to the influence graph.

If the couples graph contains a cycle: FAIL If no couple is evicted: Fantastic!

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Couples graph is acyclic The probability of a couple AB influencing a

couple CD is bounded by (log n)c/n≈1/n. With probability →1, the couples graph is

acyclic.

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Influence trees and the couples graph

If:1. (h,d’) IT(cj,0)2. (h,d) IT(ci,0)3. Hospital h prefers d to d’

ci

cj

IT(ci,0) - set of hospitals-doctor pairs ci can affect if it was inserted as the first couple

cjcih

dd’

IT(ci,r) - similar but allow r adversarial rejections (to capture that other couples may have already applied)

Influence trees and the couples graph

The influence tree of c consists of all the hospitals-doctors that are likely to be part of the rejection chain due to c ‘s presence.

Key steps:1. Influence tree of each couple is “small” .2. There are no directed cycles in the couples graph.3. If c influences some hospital h, then h will belong

to that influence tree.

Linear number of couples

Theorem [Ashlagi, Braverman, Hassidim 12]: in a

random market with n singles, αn couples and large enough λ>1, with constant probability no stable matching exists.

Idea:1. Show that a small submarket with no stable

outcome exists2. No doctor outside the submarket ever enters a

hospital in this submarket market

Results from the APPIC data

Matching of psychology postdoctoral interns.

Approximately 3000 doctors and 20 couples.

Years 1999-2007. SoDA was successful in all of them. Even when 160 “synthetic” couples are

added.

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SoDA: the couples graphs

In years 1999, 2001, 2002, 2003 and 2005 the couples graph was empty.

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2008 2004 2006 2007

number of doctors

SoDA: simulation results

Success Probability(n) with number of couples equal to n. 4% means that ~8% of the individuals participate as couples.

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808 per 16,000 ≈ 5%

probability of success

Summary and some directions for research1. More structure on couples preferences (some

“cities” structure is given in Ashlagi, Braverman, Hassidim). Relax preferences of hospitals.

2. What to do if there is no stable matching? Roth and Peranson make decisions in the algorithm when there is no stable matching.

3. What would be a good strategy for an employer in a big city? In a rural area?