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Mechanism Design without Money. Lecture 7. M-optimal and W-optimal matchings.

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m optimal and w optimal matchings
M-optimal and W-optimal matchings

A stable matching is M-optimal if every man likes it at least as well as any other stable matching; that is, a stable matching is M-optimal if for every other stable matching we have for all (from now on we will write ).

Similarly, is W-optimal if for every other stable matching we have .

m optimal and w optimal matchings1
M-optimal and W-optimal matchings

Theorem (GS 1962): When all men and women have strict preferences, the matching produced by the DA algorithm with men-proposing is the M-optimal stable matching. Symmetrically, the matching produced by the DA algorithm with women-proposing is the W-optimal stable matching.

Proof: Look at the first man being rejected by an “achievable” woman . If is unacceptable to , we reach contradiction. If not, and the man that chose over in that step block any stable matching in which and are matched.

conflicting interests
Conflicting interests

Theorem (Knuth, 1976): When all agents have strict preferences, the common preferences of the two sides of the markets are opposed on the set of stable matchings. That is, If and are both stable matchings then:

if and only if .

Proof: Let and be two stable matchings such that . Suppose for some woman . Then . But for man we also have . So the pair block , which was supposedly stable.

conflicting interests1
Conflicting interests

When preferences are strict, for any two matchings and let be the matching such that if and otherwise, and if and otherwise. In an opposite way define the matching .

The Lattice theorem (Conway): When and are stable matchings, then and are both stable matchings.

the lattice structure
The lattice structure

Simple example:

,

The stable matchings are:

A more complicated example with 10 matches in a lattice in Roth and Sotomayor (1990), pp. 37-38.

pareto efficiency
Pareto efficiency

Pareto – no one gets hurt, and someone improves their situation

Basic optimality criterion

Obviously, both and are optimal with respect to the entire population (otherwise, there would be a blocking pair).

Thm: is Pareto efficient with respect to the men That is, no other matching (even non stable) is weakly preferred by all men.

many to one matchings
Many-to-one matchings

Also known as the “college admissions” model.

A market is a finite set of hospitals , a finite set of doctors , and preferences .

For every doctor , is a strict order over .

For every hospital , is a strict order over .

many to one matchings1
Many-to-one matchings
  • A matching is such that:
    • if and only if .
  • A matching is blocked by agent if .A matching is individually rational if it is not blocked by any .
  • A matching is blocked by the pair and if and either or there exists such that . A matching is pairwise stable if it is not blocked by any pair .
  • A matching is stable if it is both individually rational and pairwise stable.
many to one matchings2
Many-to-one matchings
  • A matching is blocked by the coalition if there exists a matching such that:
  • A matching is group-stable if it is not blocked by any coalition.
many to one matchings3
Many-to-one matchings

A hospital has responsive preferences with quota if its preference between any two sets of doctors that differ by only one doctor is determined by some constant order over , and it prefers any set of size or less to any set of size or more.

Lemma: When hospitals have responsive preferences, a matching is stable if and only if it is group-stable.

many to one matchings4
Many-to-one matchings

When hospitals have responsive preferences we can consider the related marriage marketin which each hospital is split into mini-hospitals with demand of one doctor (quota of 1).

Lemma: A matching is stable if and only if the corresponding matchings of the related marriage market are stable.

many to one matchings5
Many-to-one matchings

Some, but not all, of the results still hold.

For example, a stable matching still exists (modification of DA algorithm), and it takes some work, but the lattice structure is also there.

However, weak Pareto optimality is lost. For example:

,

Here the only stable matching is given by:

, ,

All hospitals prefer the (non-stable) matching :

, ,

timeline
Timeline

~1900: Internships introduced

1944: Contracts signed two years before internship begins.

1946: Transcripts released only at the end of junior year.

1945-1951: Students holding offers. Offers with 12 hours deadline.

1951: Trial run of a centralized matching algorithm.

1952: First use of the NIMP algorithm to preform a match.

1962: Gale and Shapley (first?) prove existence of stable matching.

Late 1980’s and early1990’s: The market gradually becomes more complex. Lack of confidence in the algorithm.

1995: NRMP board decides to design a new algorithm.

1998: First run of the Roth-Peranson algorithm.

strategic behavior
Strategic behavior

We mentioned that the trial-run algorithm was rejected because it was prone to strategic behavior.

What can we say about strategic behavior when using stable mechanisms?

Direct revelation mechanisms: doctors and hospitals reveal their private information (their preferences over matchings), and the mechanism selects a matching accordingly.

We will be interested in direct revelation mechanisms that output stable matchings.

strategic behavior1
Strategic behavior

Reminder 1: a matching is called stable if it is individually rational and pairwise stable.

Reminder 2: for every set of preferences, a stable matching exists. Furthermore, we can point to two special matchings: the doctor-optimal stable matching and the hospital-optimal stable matching.

Reminder 3: One way for constructing those two matchings is by running (e.g. on a computer) a deferred-acceptance algorithm.

strategic behavior2
Strategic behavior

When using the doctor-optimal stable mechanism, a hospital may be able to manipulate the outcome by submitting different preferences.

Example:

,

The doctor-optimal stable matching is .

If submits instead the preference , then the only stable matching is .

strategic behavior3
Strategic behavior

Theorem: There exists no stable mechanism for which truthful revelation of the preference is dominant strategy.

Proof: Use the last example. If the stable matching that the mechanism selects for truthful revelation is and not , then one of the doctors can manipulate.

strategic behavior4
Strategic behavior

Theorem: When using the doctor-optimal stable mechanism, it is a (weakly) dominant strategy for the doctors to state their true preferences.

Proof: Follow the doctor that can cheat. Use independence of order.

strategic behavior5
Strategic behavior

Theorem: If hospitals all have quota of one (that is, in one-to-one matching) and the doctor-optimal stable mechanism is being used, all outcomes of Nash equilibria in weakly undominated strategies are stable with respect to the true preferences. Furthermore, every stable matching is the outcome of some Nash equilibrium.

Proof: The second part of the theorem is easy (doctors submit preference truthfully), and hospitals according to the stable matching. The proof of the first part is omitted.

strategic behavior6
Strategic behavior

Theorem: When using the hospital-optimal stable mechanism, truthful revelation is not a dominant strategy for the hospitals.

Intuition: Hospital with more than one position is like a group of men and they can deviate together in order to induce a weakly better allocation to all of them.

Theorem: If some hospitals have quota above one and the mechanism is either the doctor-optimal stable mechanism or the hospital optimal stable mechanism, not every outcome that corresponds to a Nash equilibrium in weakly undominated strategies is necessarily stable with respect to the true preferences.

Proof: Omitted.

strategic behavior where next
Strategic behavior - where next?

It seems like many of the results are negative. Why is it then that the NIMP algorithm “survived” and the Roth-Peranson algorithm (to be discussed later) is also considered a success?

Answer 1: Imperfect information makes truthful revelation more appealing (Roth and Rothblum, 1999).

Answer 2: NRMP data and computational experiments show that in fact the set of stable matchings is small, and only very few participants can manipulate the results.

why the differences are small
Why the differences are small?

Answer 1: Correlation. Note that if one side has perfectly correlated preferences, then there exists only one stable matching, and there is no incentive to misrepresent the true preference.

Answer 2: Short ROLs (rank order lists) do not allow long rejection chains (in DA algorithm).

It turns out the second answer is quite sufficient to explain the results.

immorlica and mahdian 2003
Immorlica and Mahdian (2003)
  • Settings
    • One-to-one market of size governed by men-optimal stable mechanism.
    • Women have arbitrary preferences over men.
    • Men have rank ordered lists of size at most , independently distributed by repeated drawing from (common) distribution over women. satisfies for every .
immorlica and mahdian 20031
Immorlica and Mahdian (2003)

Theorem:

Intuition: Rejecting a man will cause a rejection cycle that might or might not return to the original woman. When no woman is popular enough there is an increasingly big chance that the rejection cycle will end with a man who stays single, or with a woman who was single before.

immorlica and mahdian 20032
Immorlica and Mahdian (2003)

Corollary 1: The game induced by the men-proposing stable mechanism has an equilibrium in which, in expectation, fraction of the strategies are truthful.

Corollary 2: For every , if is large enough the game has an -approximate Nash equilibrium in which everybody is truthful.

other results on large markets
Other results on large markets

Kojima and Pathak (2009) – Extension to many-to-one markets.

Azevedo and Leshno (working paper, 2011) – A model with continuum of students.

Lee (working paper, 2013) – With bounded underlying cardinal utility, utility gains from preference misrepresentation are less than for fraction of the population.

Ashlagi, Kanoria, and Leshno (working paper, 2013) – Small number of stable matches in large uniform and uneven markets.

rural hospitals
Rural hospitals

The problem with allocation of doctors to rural hospitals.

Theorem: When preferences are strict, the set of doctors employed and positions filled is the same at every stable matching.

Theorem: When preferences are strict, any hospital that does not fill its quota at some stable matching is assigned precisely the same set of students at every stable matching.

couples
Couples

With time and progress, couples became a problem for the NRMP.

The “leading member” adjustment didn’t work very well…

This was one of the main reasons the original NRMP algorithm had to be replaced.

complementarities and stable matching
Complementarities and stable matching

It turns out that when couples are present, the set of stable matchings may be empty.

Example:

, ,

Even when stable matchings do exist, there does not have to be a side-optimal stable matching.

roth and peranson 1999
Roth and Peranson (1999)
  • The new algorithm works as follows:
    • First do doctor-proposing deferred-acceptance, with only single doctors involved.
    • Then add the couples one by one (in random order) and using similar “proposing” mechanism.
    • If any cycles are detected, start over.
  • If the algorithm terminates, the resulting matching is stable.
  • And while supposedly there is no good reason for that, this algorithm always terminates (when using data from previous years or when using random preferences).
  • Furthermore, it is generally considered a success story, and it was adopted for other programs as well…
roth and peranson 19991
Roth and Peranson (1999)

Clearinghouses currently using the algorithm (Taken from Roth’s slides):

some questions
Some questions

Why does the Roth-Peranson algorithm works?

When is the set of stable matchings (with couples) non-empty?

large markets again
Large markets again…

Kojima, Pathakand Roth (working paper) – Using a model similar to Immorlica and Mahdian (2003) and Kojima and Pathak (2009), if the number of couples is then a stable matching exists (and can be reached by the Roth and Peranson algorithm).

Intuition: Instead of bounding using the probability that there is no rejection cycle (and all rejection chains end with a hospital that had a vacant position). For couples bound using also that there is no rejection path from one member of the couple to the other.

large markets again1
Large markets again…

Ashlagi, Bravermanand Hassidim (working paper) – Better bounds and slightly more sophisticated algorithm insure existence of stable matching even if the number of couples is for any .

They also provide a counterexample for the case in which the number of couples is .

extensions
Extensions

Roommate problems, multi-sided matching

Many-to-one with discrete money and substitutable preferences (Crawford and Knoer, 1981; Kelso and Crawford, 1982)

Many-to-many with responsive preferences (Roth, 1984)

Matching with contracts (Hatfield and Milgrom, 2005)

Many-to-many matching with contracts (Echenique and Oveido, 2006)

Matching in supply chains (Ostrovsky, 2008)

Matching in networks with bilateral contracts (Hatfield, Kominers, Nichifor, Ostrovsky and Westkamp, working paper)

Matching with minimum quotas, regional caps, etc. (Biro, Fleiner, Irving and Manlove, 2010, Kamada and Kojima, 2013)

related topics
Related topics

Roth and VandeVate (1990) – Random paths to stability

Jackson and Watts (2002)

Ausubel and Milgrom (2000) on package bidding

road example
Road example
  • 50 people want to get from A to B
  • There are two roads, each one has two segments. One takes an hour, and the other one takes the number of people on it

1 hour

N minutes

A

B

1 hour

N minutes

nash in road example
Nash in road example
  • In the Nash equilibrium, 25 people would take each route, for a travel time of 85 minutes

1 hour

N minutes

A

B

1 hour

N minutes

braess paradox
Braess’ paradox
  • Now suppose someone adds an extra road which takes no time at all. Travel time goes to 100 minutes

1 hour

N minutes

A

B

Free

1 hour

N minutes