Marriage, Honesty, & Stability

Marriage, Honesty, & Stability

NICOLE IMMORLICA, NORTHWESTERN UNIVERSITY

Math &

Romance

Love, marriage, & bipartite graphs

the boys

and girls

Jake

Elwood

Curtis

Ray

Twiggy

Claire

Jill

Holly

Holly > Claire > Twiggy > Jill

Claire > Jill > Twiggy > Holly

Twiggy > Jill > Holly > Claire


Jake > Elwood > Curtis > Ray

Jake > Curtis > Elwood > Ray

Ray > Curtis > Elwood > Jake

Ray > Jake > Elwood > Curtis

LoveMarriage

A pair (m,w) is a blocking pair for a matching if m prefers w to his match and w prefers m to her match.

When is marriage stable?

“blocking pair”

Jake

Elwood

Curtis

Ray

Twiggy

Claire

Jill

Holly









When is marriage stable?

“blocking pair”A pair (m,w) is a blocking pair for a matching if m

prefers w to his match and w prefers m to her match.

“stability”A matching is stable if there is no blocking pair.

The courtship ritual

Jake

Elwood

Curtis

Ray





Twiggy

Claire

Jill

Holly





What’s math say about romance?

Question 1. Do stable matchings always exist?Question 2. Are they easy to find?Question 3. Does the courtship ritual work?Question 4. Are stable matchings unique?Question 5. If not, who benefits?

What would Cupid prove?

Theorem. Courtship algorithm terminates.Theorem. Resulting marriage is stable.Corollary. Stable marriages always exist and are

easy to find.

We’ll prove this using potential functions.

Step 1: proof of termination

Theorem.The courtship ritual terminates.

Proof.1. Define a potential function: the number of

names on boys’ lists.2. Note this is strictly decreasing and always

non-negative.

Step 2a: defining invariant

The girls’ options only ever improve!

Invariant.For every girl G and boy B, if G is crossed off B’s list it is because she’s married to him or has a suitor she prefers.

… because B only crosses off G after proposing, and G only rejects proposals when she finds someone better.

Step 2b: proof of stability

Theorem.The resulting matching is stable.

Proof.Consider boy B and girl G that are not married to each other.

1. Suppose G was crossed off B’s list. Then G prefers husband to B, so won’t elope with B.

2. Suppose G is on B’s list. Then B didn’t propose to G yet, so B prefers wife to G, so won’t elope with G.

What can math say on romance?

Question 1. Do stable matchings always exist?Question 2. Are they easy to find?Question 3. Does the courtship ritual work?

Question 4. Are stable matchings unique????Question 5. If not, who benefits????

Yes, yes, yes.

Recall …

Jake

Elwood

Curtis

Ray





Twiggy

Claire

Jill

Holly





In an alternate universe

Jake

Elwood

Curtis

Ray





Twiggy

Claire

Jill

Holly





Conclusion: not unique

Jake

Elwood

Curtis

Ray

Twiggy

Claire

Jill

Holly









Stable spousesIn general, there are many stable marriages.

Boy B

Girl G

Girl G’

A person P is a stable spouse of a person P’ if P is married to P’ in some stable matching.

“stable spouse”

Boys are happier than girls

Theorem.Boys marry their most preferred stable wife….… and girls get their least preferred stable husband!

way

Proof.

Proof.Of 1st claim, by contradiction. Suppose some boy isn’t married to favorite stable spouse.

1. He must have proposed to her and been refused.2. Let B be 1st boy to lose his favorite stable wife G.3. Then G must have had a proposal from a boy B’

she preferred to B.4. Since B’ has not yet crossed off his favorite stable

wife, B’ must love G more than any stable wife.5. But then B’ and G will elope in marriage which

matches B to G, contradicting stability of wife G.

Proof.Of 2nd claim, by contradiction. Suppose there is a stable matching in which girl G gets worse husband.

1. Let B be her husband when boys propose and B’ be her husband in worse matching.

2. Then G prefers B to B’ by assumption.3. Furthermore, by 1st claim, G is favorite stable wife

of B, so B prefers G to wife in worse matching.4. But then B and G will elope in worse matching.

But of course … symmetry

If girls propose, then they will get their favorite stable husbands.

Applications of Stable Matching

Centralized two-sided markets.

– National Residency Matching Program (NRMP) since 1950’s – Dental residencies and medical specialties in the US,

Canada, and parts of the UK– National university entrance exam in Iran – Placement of Canadian lawyers in Ontario and Alberta – Sorority rush – Matching of new reform rabbis to their first congregation – Assignment of students to high-schools in NYC– …

Theorem 1. The matching produced by the men-proposing algorithm is the best stable matching for men and the worst stable matching for women.

This matching is called the men-optimal matching.

Theorem 2. The order of proposals does not affect the stable matching produced by the men-proposing algorithm.

Theorem 3. In all stable matchings, the set of people who remain single is the same.

Classical Results

NRMP

National Residency Matching Program• After finishing med school in the US, students

must complete internships at hospitals prior to receiving their degree

• Students interview at hospitals and rank their preferences

• Hospitals rank students they interviewed • A hospital-optimal deferred acceptance algorithm

determines placement

Question. Do participants have an incentive to announce a list other than their real preference lists?

Answer. Yes! In the men-proposing algorithm, sometimes women have an incentive to be dishonest about their preferences.

Incentive Compatibility

Recall …

Jake

Elwood

Curtis

Ray





Twiggy

Claire

Jill

Holly





A small white lie

Jake

Elwood

Curtis

Ray





Twiggy

Claire

Jill

Holly





Next Question. Is there any truthful mechanism for the stable matching problem?

Answer. No! Roth (1982) proved that there is no mechanism for the stable marriage problem in which truth-telling is the dominant strategy for both men and women.


Next Question. When can people benefit from lying?

Theorem. The best match a woman can receive from a stable mechanism is her optimal stable husband with respect to her true preference list and others’ announced preference lists.

In particular, a woman can benefit from lying only if she has more than one stable husband.


Data from NRMP show that the chance that a participant can benefit from lying is slim.

1993 1994 1995 1996

# applicants 20916 22353 22937 24749

# positions 22737 22801 22806 22578

# applicants who could lie 16 20 14 21

Explanations (Roth and Peranson, 1999)

The following limit the number of stable husbands of women:

Preference lists are correlated. Applicants agree on which hospitals are most prestigious; hospitals agree on which applicants are most promising.

If all men have the same preference list, then everybody has a unique stable partner, whereas if preference lists are independent random permutations almost every person has more than one stable partner. (Knuth et al., 1990)

Preference lists are short.Applicants typically list around 15 hospitals.

A Probabilistic Model

Men. Choose preference lists uniformly at random from lists of at most k women.

Women. Randomly rank men that list them.

Conjecture (Roth and Peranson, 1999). Holding k constant as n tends to infinity, the fraction of women who have more than one stable husband tends to zero.

Our Results

Theorem. Even allowing women arbitrary preference lists in the probabilistic model, the expected fraction of women who have more than one stable husband tends to zero.

Economic Implications Corollary 1. When other players are truthful, almost surely a

random player’s best strategy is to tell the truth.

Corollary 2. The stable marriage game has an equilibrium in which in expectation a (1-o(1)) fraction of the players are truthful.

Corollary 3. In stable marriage game with incomplete information there is a (1+o(1))-approximate Bayesian Nash equilibrium in which everybody tells the truth.

Structure of proof

Step 1. An algorithm that counts the number of stable husbands of a given woman.

Step 2. Bounding the probability of having more than one stable husband in terms of the number of singles

Step 3. Bounding the number of singles by the solution of the occupancy problem.

Step 1. Finding stable husbands of g 1. Use men-proposing algorithm to find a stable

matching.2. Whenever the algorithm finds a stable matching,

have g divorce her husband and continue the men-proposing algorithm (but now g has a higher standard for accepting new proposals).

3. Terminate when either – a man who is married in the men-optimal matching runs

through his list, or – a woman who is single in the men-optimal matching

receives a proposal.

Question. If each woman has an arbitrary complete preference list, and each man has a random list of k women, what is the probability that this algorithm returns more than one stable husband for g?

The main tool that we will use to answer this question is the principle of deferred decisions:

Men do not pick the list of their favorite women in advance; Instead, every time a man needs to propose, he picks a woman at random and proposes to her. A man remains single if he gets rejected by k different women.

Step 2. Bounding the probability• Consider moment when algorithm finds first (i.e.,

men-optimal) matching. Call this matching μ.

• Let A denote the set of women who are single in μ, and X denote |A|.

• Fix random choices before the algorithm finds μ, and let probabilities be over random choices that are made after that.

Step 2, cont’d.

• Look at sequence of women who receive a proposal.

• Probability algorithm finds another stable husband for g is bounded by probability g comes before all members of A in sequence, i.e., 1/(X+1).

• Therefore, the probability that g has more than one stable husband is at most E[1/(X+1)].

Step 3. Number of singles

We need to compute E[1/(X+1)], where X is the number of singles in the men-optimal matching.

Simple Observation. The probability that a woman remains single is at least the probability that she is never named by men.

Step 3, cont’d.Let Ym,n denote the number of empty bins in an experiment where m balls are dropped independently and uniformly at random in n bins.

Lemma.

Proof Sketch. Assume (without loss of generality!) that men are amnesiacs and might propose to a woman twice. The total number of proposals (bins) is at most (k+1)n w.h.p.

1X+1

1Y(k+1)n, n + 1

k2

nE[ ] < E[ ] +

E[ ] <

The occupancy problem

Lemma.

Proof sketch. – Use the principle of inclusion and exclusion to

compute E[1/(Ym,n+1)] as a summation.– Compare this summation to another (known)

summation term-by-term.

1Ym, n + 1

em/n

n

Putting it all together…

Theorem. In the model where women have arbitrary complete preference lists and men have random lists of size k, the probability that a fixed woman has more than one stable husband is at most

ek+1 + k2

n

Future Directions

Preference Elicitation. How do students choose hospitals to interview at? Can we increase the efficiency of the market?

Settings with correlation. What if students have geographic preferences?

Stable matching with couples. Why has the NRMP algorithm found a matching every year?

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