Mechanism Design without
Money
Lecture 9
1
Debt – Israel psychology market
• I was asked to give more examples for
markets.
• Clinical psychology
– Offers are made in three rounds
– If you agreed in round 2, you are not allowed to
back off in round 3.
• What are the problems?
2
Debt - A linear fraction of couples
• The number of married people is proportional
to the number of people
• Simulations
decent
success
for an
Thm : considershow
a random
market,
with rates
n singles,
constant
fraction
of couples
couples and
more than
20n hospitals.
With
constant
is no stable
• Is
there
a wayprobability,
to insert there
the couples,
to get a
outcome
stable
matching?
Proof idea – isolated markets
1. Find a small structure, which prevents a
stable outcome
•
A few hospitals and doctors, which (if left alone)
can not form a stable outcome
2. Show that this small structure exists with
constant probability
3. Show that no one outside the structure ever
enters a hospital in the isolated market
Instable structure
• For a single s and a
couple c, with
probability O(1/n2) we
have the structure
• If the structure occurs
– no “local” stable
outcome
• There are n singles and
n couples, so with
constant probability
this structure will occur
Doctor Preferences
Hospital Preferences
Capacity is 1
c=(c1,c2)
S
H1
H2
H1,H2
H2
…
…
whatever
H1
s
c2
…
…
…
c1
s
…
…
Isolated market
• The only solution is to insert someone else to h1,h2
thus avoiding the problem
• There is an excess of positions, so if a doctor goes to
h1,h2 there are hospitals which are left free. We need
to show that the doctor prefers them
• A quantitative version of the Rural Hospital Theorem
– Define a probabilistic process, show it’s a martingale, use
Azuma’s inequality
l Pessimistic deferred acceptance
• l-Pessimistic DA: At each step t=1; 2 …
– either a single doctor s or a couple c that has less than
l temporary assignments are chosen at random and
applies to the most preferred hospital or pair of
hospitals on their list respectively that they haven't
applied so far.
– Each hospital h assigns a doctor d if and only if no
other doctor is currently assigned to h and no other
doctor applied at this step to h. If some doctor d
applies to h and some doctor d’ is temporarily
assigned to h, h rejects both d and d’ and no doctor is
never allowed to apply to h anymore.
7
Halting condition
• With constant probability no more than n
hospitals are visited in the process 3pessimistic DA for some  < 20.
• Let Ai(t) be the number of doctors who have i
positions
• Vt number of hospitals visited until time t
• Follow the variable
Xt = t + Vt + 15A0(t) + 10A1(t) + 5A2(t)
• Show it is a super martingale, E(Xt+1|Xt) < Xt
8
3-pessimistic DA bounds the number
of hospitals visited
• For each entity d (single or couple) , let p1(d),
p2(d) and p3(d) denote the hospitals they get
• A doctor is poor, if in the real DA it gets
something which is outside of pi(d). Let k denote
the number of poor entities
• But that means that the hospitals there are filled
with doctors.
– And only poor doctors can fill them
• But we need to fill 3k hospitals with at most 2k
doctors
9
What’s in today’s class?
• School choice
– Case studies: NYC and Boston
– Algorithms: the Boston mechanism, Top Trading
Cycles
– More on tie-breaking
• The signaling mechanism
10
What is different about school choice?
•
•
•
•
11
Schools and students as strategic players
True indifferences
The concept of justified envy
Stability vs. optimality
The NYC High School Match
• Until 2002:
–
–
–
–
Decentralized applications and admissions
Only five choices allowed
Three rounds of waiting lists, waiting lists run by mail
Congestion (out of over 90,000 kids every year, 30,000
administratively assigned, and 17,000 receiving
multiple offers)
– 30,000 students assigned to schools not on their
choice list
– Schools take students’ ranking into account
– Withholding of capacity
12
The NYC High School Match
• Are NYC schools really two-sided matching problem?
– Schools conceal capacities
– EdOpt schools have different preferences (high scores,
attendance records, etc.)
• Solution: Deferred Acceptance
– Only 12 options allowed (breaks truthful revelation,
Haeringer and Klijn, JET, 2009)
– Due to historical rules about specialized schools –
matching is done in 3 rounds (round 3 for unmatched kids)
13
The NYC High School Match
• Single tie-breaking vs. multiple tie-breaking
(for schools’ indifferences)?
– Multiple tie-breaking increases number of
instabilities, and it therefore constrains the
efficiency
– NYC DOE saw simulations, and tried both tiebreaking rules, and decided on single tie-breaking
rule
14
The NYC High School Match
• Outcome:
– Only 3,000 did not receive any school they chose (compare to
30,000 the previous year)
– The reasons: relieving congestion (many offers and acceptances,
instead of only three rounds), giving each student a single offer
(instead of people getting multiple offers), allowing ranking of
12 instead of 5 schools, but also…
– The results continued to be better and better each year
(comparing rankings), even though there were no changes to
the algorithm… hmmm…
– The answer: schools have learned to stop withholding capacity!
– Open question: how to do appeals? (TTC? but that’s later)
15
Boston Public Schools (BPS)
• About 4000 kids in each cohort. Four cohorts
are making choices: K, 1, 6, and 9.
• Priorities (= schools’ preferences) come from
walking zones, siblings, and random tiebreaking
16
Boston Public Schools (BPS)
• Until 2006 the mechanism used is “The Boston
Mechanism” (but also used in many other places):
– Step 𝑘. 1: Each student that is still unmatched applies to her
most preferred school
– Step 𝑘. 2: Each school fills its quota as much as possible with
those applicants that it prefers the most, and rejects the rest
• Problems with the “Boston Mechanism”:
1.
2.
3.
4.
17
Does not produce stable matchings
Truth-telling is not dominant (far from it)
Not immediately clear that something is wrong…
Those who do not play strategically get hurt
Boston Public Schools (BPS)
• Unlike NYC – unclear that the market is twosided:
– No gaming by schools
– Lots of people in each priority class, and looks like
priorities are meant to help parents select schools
– If the market is actually one-sided, then stable
matchings are not Pareto optimal (it is better for
people to trade priorities)
18
Top Trading Cycles
• Introduced in Shapley and Scarf (1974), but
attributed to David Gale.
• Draw a graph where each agent is a node, with
each agent pointing to his/her/its most preferred
match.
• Remove a cycle, and redraw the edges, now each
agent points to most preferred match among
those remaining.
• Repeat until all nodes are removed.
19
Top Trading Cycles
20
Top Trading Cycles
Theorem (Shapley and Scarf, 1974): the
outcome of TTC is in the core.
Theorem (Roth, 1982): TTC is strategyproof.
21
Boston Public Schools (BPS)
• So there were two options for Boston:
– DA – Strategyproof, stable, selects student-optimal
matching (except for tie-breaking issues)
– TTC – Strategyproof, Pareto efficient for the students
• The most important thing: that the algorithm will be
strategyproof. This levels the playing field and allows
gathering data about actual preferences over schools.
• The DA algorithm was chosen because it is more
transparent and easier to explain to parents.
22
A bit more on tie-breaking
Proposition: For any set of strict preferences for
students and weak preferences for schools, any
matching that can be produced by deferred acceptance
with multiple tie-breaking, but not by deferred
acceptance with single tie-breaking is not a studentoptimal stable matching.
DA-MTB
23
DA-STB
Studentoptimal
stable
matchings
A bit more on tie-breaking
• Example:
There are three schools 𝑆 = 𝑠1 , 𝑠2 , 𝑠3 and three students 𝐼 = 𝑖1 , 𝑖2 , 𝑖3 .
𝑠2 ≻𝑖1 𝑠1 ≻𝑖1 𝑠3 𝑖1 ~𝑠1 𝑖2 ~𝑠1 𝑖3
𝑠1 ≻𝑖2 𝑠2 ≻𝑖2 𝑠3 𝑖2 ≻𝑠2 𝑖1 ≻𝑠2 𝑖3
𝑠1 ≻𝑖3 𝑠2 ≻𝑖3 𝑠3 𝑖3 ≻𝑠3 𝑖1 ≻𝑠3 𝑖2
Three stable matchings from student-proposing DA with different tie-breaking
rules:
𝑖
𝑖
𝑖
𝑖
𝑖
𝑖
𝑖
𝑖
𝑖
𝜇1 = ( 1 2 3 )
𝜇2 = ( 1 2 3 )
𝜇3 = ( 1 2 3 )
𝑠1 𝑠2 𝑠3
𝑠2 𝑠1 𝑠3
𝑠3 𝑠2 𝑠1
Note that while all are stable, 𝜇1 is not student-optimal, because 𝜇2
dominates 𝜇1 .
24
Stable improvement cycles
• Based on Erdil and Ergin (AER, 2008)
• Given a stable matching 𝜇, strict preferences for students and priorities for
the schools, a stable improvement cycle consists of students 𝑖1 , … , 𝑖𝑛 = 𝑖0
such that:
1.
2.
3.
𝜇 𝑖𝑘 ∈ 𝑆 (every student is matched to a school)
𝜇 𝑖𝑘+1 ≻𝑖𝑘 𝜇 𝑖𝑘 (every student prefers the school the next student is
currently allocated)
𝑖𝑘 ∈ arg max 𝑗 𝜇 𝑖𝑘+1 ≻𝑗 𝜇 𝑗 , where the argmax is taken with respect to
school 𝜇(𝑖𝑘+1 )’s priorities.
• Given a stable improvement cycle create a new matching:
𝜇 𝑗
𝑗 ∉ 𝑖1 , … , 𝑖𝑛
𝜇′ 𝑗 =
𝜇 𝑖𝑘+1
𝑗 = 𝑖𝑘
Proposition: 𝜇′ is stable and it (weakly) Pareto dominates 𝜇.
25
Stable improvement cycles
Theorem: Fix the preferences and priorities, and
let 𝜇 be a stable matching. If 𝜇 is (weakly) Pareto
dominated by another stable matching, then 𝜇
admits a stable improvement cycle.
Corollary: In order to find a student-optimal
stable matching, we can run deferred
acceptance, and then find and implement stable
improvement cycles until none are left.
26
Stable improvement cycles
27
Stable improvement cycles
Theorem (Abdulkadiroglu, Pathak and Roth, 2008):
For any tie breaking rule 𝜏, there is no mechanism
that is strategy-proof and dominates 𝐷𝐴𝜏 .
Furthermore, when considering stable
improvement cycles, it is kind of clear what kind of
manipulations might be profitable. It is worthwhile
to list schools that are over-demanded and in which
you might have priority in order to replace them
with people who have priority in other schools that
you actually want.
28
Stable improvement cycles
Example (Azavedo and Leshno, 2010):
Four students, 𝐼 = 𝑖1 , 𝑖2 , 𝑖3 , 𝑖4 and two schools 𝑆 = 𝑠1 , 𝑠2 with quotas 𝑞𝑠1 = 1 and 𝑞𝑠2 = 2.
𝑠2 ≻𝑖1 𝑠1 ≻𝑖1 ∅
𝑖1 ≻𝑠1 𝑖2 ~𝑠1 𝑖3 ~𝑠1 𝑖4
𝑠2 ≻𝑖2 ∅
𝑖3 ~𝑠2 𝑖4 ≻𝑠2 𝑖1 ~𝑠2 𝑖2
𝑠1 ≻𝑖3 ∅
𝑠1 ≻𝑖4 ∅
1
2
1
2
Assume utility from first choice is 1, from staying single is 0, and that 𝑢𝑖3 𝑠2 > − and 𝑢𝑖4 𝑠2 > − .
With DA-STB with random tie-breaking the equilibrium is truthful revelation, and allocation is
𝑖3
𝑖4
𝑖1 𝑖2
1
1
1
1
𝑠2 𝑠2
𝑠1 + ∅
𝑠1 + ∅
2
2
2
2
29
Stable improvement cycles
If, however, both 𝑖3 and 𝑖4 report the preference 𝑠1 ≻ 𝑠2 ≻ ∅ then the DA-STB
allocation is
𝑖3
𝑖1
𝑖4
𝑖2
1
1
1
3
1
3
𝑠1 + 𝑠2 ∅
𝑠1 + 𝑠2
𝑠1 + 𝑠2
2
2
4
4
4
4
and the unique Pareto efficient assignment (with respect to reported preferences)
that dominates DA-STB is
𝑖3
𝑖4
𝑖1 𝑖2
1
1
1
1
𝑠2 ∅
𝑠1 + 𝑠2
𝑠1 + 𝑠2
2
2
2
2
And this is equilibrium.
Corollary: Consider any mechanism that is Pareto efficient with respect to reported
preferences, and Pareto dominates DA-STB. In the economy described, this
mechanism has a unique equilibrium assignment which is Pareto dominated by the
DA-STB assignment, and is unstable with respect to the true preferences.
30
Signaling
• Based on Coles, Kushnir and Niederle (AEJ: Micro,
forthcoming).
• Two types of signaling:
– Quality signaling
– Preference signaling
• Examples of relevant markets:
–
–
–
–
31
Job market for new Ph.D. economists (Coles et al., 2010)
Informal preference signaling (Roth and Xing, 1994)
Internet dating markets (Lee and Niederle, 2011)
College admissions (Avery and Levin, 2009).
Signaling
• Simple example:
– Two firms and two workers
– Preference on matching with other side are drawn at random (each order with
1
probability 2).
– Utilities: top choice ⇒ 1, second choice ⇒ 𝑥, unmatched ⇒ 0.
– One offer from each firm, then workers choose among offers.
• Without signals (eq. with anonymous strategies): each firm offers to best
3
3
1
1
worker, and gets it with probability , 𝑢𝑓 = , 𝑢𝑤 = + 𝑥.
4
4
2
4
• Now suppose workers simultaneously send signals before firms offer
positions: each worker sends to better firm, firm proposes to a worker
1
1
3
who is going to accept, matching probability is 1, 𝑢𝑓 = + 𝑥, 𝑢𝑤 =
2
2
4
1
+ 𝑥.
4
• Note that there are also other equilibria.
32
Signaling
• In congested markets, a signaling mechanism:
– Increases the expected number of matches
– Increases the welfare of workers
– Ambiguously changes the welfare of firms
• The value of signaling mechanism:
– is high for balanced markets
– decreases when there are additional periods of
interaction
33
More on the AEA job market
• Why is it necessary in this market? (schools not reading all
applications, students submit a lot of enthusiastic
applications, truncation from the top and coordination
problem)
• How many signals to provide?
• Signaling mechanism is open until December
• Signals are scarce, credible and equitable
• Where should people send signals:
– Candidate 1 already has already attracted some interest from
very competitive universities…
– Candidate 2 is a strong but not flashy candidate…
– Candidate 3 has an unusual background… departments don’t
normally recruit from his university…
34
More on the AEA job market
35
More on the AEA job market
• Where are signals effective?
– Liberal arts colleges
– Departments in towns (population < 50,000)
– “Unranked” schools (including internationals!)
– For non-current PhDs
– Departments that don’t receive many signals
36
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)
37
Related topics
• Roth and Vande Vate (1990) – Random paths
to stability
• Jackson and Watts (2002)
• Ausubel and Milgrom (2000) on package
bidding
38
Questions?
39
Extra Slides
40
Chicken
41
Road example
1 hour
A
N minutes
N minutes
B
1 hour
• 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
42
Nash in road example
1 hour
A
N minutes
N minutes
B
1 hour
• In the Nash equilibrium, 25 people would take
each route, for a travel time of 85 minutes
43
Braess’ paradox
1 hour
A
N minutes
N minutes
Free
B
1 hour
• Now suppose someone adds an extra road which takes
no time at all. Travel time goes to 100 minutes
44