Mechanism Design Without Money

1
Mechanism Design Without Money

• James Schummer and Rakesh V. Vohra

presented by Adriaan ter Mors
2
Contents

• Single-peaked preference functions
• Motivation
• Restricted domains
• Strategy-proof allocation rules
• House Allocation problem
• Stable Marriage problem

3
Arrows Impossibility Result
Theorem 9.8 (Gibbard-Satterthwaite) Let f be an
incentive compatible social choice funtion onto
A, where A 3, then f is a dictatorship.

• Alter preference functions with money (e.g.
  Vickrey-Clarke-Groves mechanism)
• Restrict possible preference functions

4
A Restricted Domain single-peaked preference
functions
Utility ?
Alternatives ?
5
Allocation Rule Definition

• f R1 x x Rn ? A
• Strategy proof truth telling is a dominant
  strategy
• Restriction to onto functions for all x ? A,
  there is a multi-agent preference function such
  that f chooses x

6
Equivalence Lemma

• Onto all x ? A can be chosen
• Unanimity if all agents most prefer x, f will
  choose x
• Pareto-optimality for all ? Rn, there exists
  no point x ? 0,1 such that x i f() for all i
  ? N

Lemma 10.1 Suppose that f is strategy-proof. Then
f is onto if and only if it is unanimous if and
only if it is Patero-optimal.
7
Proof of Lemma 10.1 (1)
Onto ? Unanimity

• Fix x ? 0,1
• Because f is onto, there exists a joint profile
  such that f() x
• Consider the unanimous joint profile such that
  every agents peak is at x
• Since f is strategy proof, we have
• f(1, 2, , n) x
• and consequently f(1, 2, , n) x

8
Proof of Lemma 10.1 (2)
Onto ? Pareto-optimality

• Proof by contradiction suppose that f is onto
  but not Pareto-optimal.

9
Proof of Lemma 10.1 (3)
Onto ? Pareto-optimality

• Tranform preference function of agent with second
  left-most peak
• Idea replacing 2 with 2 in will not affect
  social choice

a2
x
10
Proof of Lemma 10.1 (4)
Onto ? Pareto-optimality

• Tranform preference function of agent with second
  left-most peak
• Idea replacing 2 with 2 in will not affect
  social choice.
• After replacing all i, a unanimous joint profile
  emerges, with a joint peak p gt x
• A contradiction has been derived f is onto and
  strategy proof, but the unanimous peak is not
  chosen
• Still to prove transformed joint profile has
  same social choice

11
Proof of Lemma 10.1 (5)
Onto ? Pareto-optimality
y f(2, 1, 3, , n)
1) agent a2 y ? x, p2
2) agent a2 y ? (x, p2
x
p1
p2
y1
y2
12
The Median Voter Rule
13
The Median Voter Rule
14
The Average Voter Rule?
15
Fairness in Median Voting
0
1
16
Median Voter Theorem
Theorem 10.2 A rule f is strategy-proof, onto,
and anonymous if and only if there exist y1, y2,
, yn-1 ? 0,1 such that for all ? Rn, f()
medp1, p2, , pn, y1, y2, , yn-1.
Proof
Maybe some other time
17
Individual Rationality opting out

• Example setting labourers (agents) that
  simultaneously have to work on some machine, and
  jointly enjoy the machines output.
• Each agent has a single-peaked labour function
  relating work-hours to value of the machines
  output.
• An individually rational rule selects an outcome
  that is weakly preferred to no labour, no reward.
• Unique rule choosing the minimum peak

18
Final Thoughts single-peaked preferences

• Median (kth-order statistics-) rules only the
  most preferred point matters
• Allocations are easy to compute

19
House Allocation Problem

• There are n agents, each of which owns a house
• Each agent has a preference order over the n
  houses
• Object to find an appropriate reallocation of
  houses
• Agents are indifferent to allocations that give
  him the same house, so Gibbard-Satterthwaite does
  not apply (it applies only for totally ordered
  preference functions)

20
Blocking Coalitions

• Agents own the houses, so they can opt out of any
  allocations they dont like
• Blocking coalition subset of agents in which at
  least one is better off if they trade amongst
  themselves
• Set of allocations that is not blocked by any
  subset of agents is called the core

Theorem 10.6 The core of the house allocation
problem consists of exactly one matching.
21
Top Trading Cycle Algorithm
A1
A1 2 3 4 1
A2 1 4 3 2
A3 1 2 3 4
A4 3 4 1 2
A2
A3
A4
3
4
22
Uniqueness of TTCA

• In iteration 1, each agent receives his favourite
  house. Therefore, in any other coalition, N1
  would form a blocking coalition.
• With the agents from N1 gone, the same argument
  can be applied in iteration 2, etc.
• Therefore, if an allocation is in the core, it is
  the one determined by TTCA.

23
Stable Matchings
24
Stable Matchings

• Set M of men, set W of women
• Each man has a strict preference order over the
  women, and vice versa
• Matching assignment of men to women, such that
  everyone ends up with at most one partner
• A matching S is unstable if (m1, w1) ? S, (m2,
  w2) ? S, but m1 and w2 prefer each other to their
  current partners

25
G-S Algorithm (from Wikipedia)
function stableMatching Initialize all m ?
M and w ? W to free while there is a free man
m who still has a woman w to propose to
w m's highest ranked such woman if w is
free (m, w) become engaged else
some pair (m', w) already exists if w
prefers m to m' (m, w) become engaged
m' becomes free else
(m', w) remain engaged
26
Properties of Gale-Shapley Algorithm
Theorem 10.10 The male-proposal algorithm
terminates in a stable matching.
Theorem 10.11 The stable matching produced by the
(male-proposal) algorithm is male-optimal.
27
Proof of Stability

• Suppose instability
• (m1, w1) ? S, (m2, w2) ? S
• w2 gtm1 w1 and m1 gtw2 m2
• Man m1 has proposed to w2 first
• Since (m1, w2) ? S, w2 received a better offer
  from m
• m gt m1
• (w2, m2) ? S, hence m2 w2 m
• (transitivity) m2 gtw2 m1
• Contradiction m2 gtw2 m1 and m1 gtw2 m2

28
Male Optimality

• w1 is a valid partner of m1 there exists a
  stable matching S in which (m1, w1) ? S
• w1 is the best valid partner of m1 there is no
  other valid partner that m1 prefers
• Male-optimality theorem every execution of the
  G-S algorithm couples each man with his best
  valid partner

29
Proof of Theorem 10.11

• In some execution ?, let m1 be the first man to
  be rejected by a valid partner w1
• Men propose in decreasing order of preference ?
  w1 is m1s best valid partner
• Let m2 be the man that w1 chose over m1 in ?
• Since w1 is a valid partner of m1, there exists a
  stable matching S containing (m1, w1)
• Question who is paired to m2 in S?

30
Proof of Theorem 10.11 (2)

• Let (m2, w2) ? S
• In execution ?, m1s rejection by w1 was the
  first rejection of a man by a valid partner
• Hence, when m2 and w1 got engaged in ?, m2 had
  not been rejected by any valid partner
• Hence, m2 prefers w1 to w2
• Also, w1 prefers m2 to m1 (whom she had rejected)
• In S, the pair (m2, w1) forms an instability

31
Womens woes
Corollary 1.8 (Kleinberg Tardos) In the stable
matching S, each woman is paired with her worst
valid partner.

• Suppose (m1, w1) ? S, s.t. m1 is not the worst
  valid partner of w1
• Let (m2, w1) ? S, s.t. m1 gtw1 m2
• Let (m1, w2) ? S
• (Theorem 10.11) w1 is best valid partner of m1,
  so m1 prefers w1 over w2
• The pair (m1, w1) forms an instability in S

32
Discussion

• Questions for me
• Questions for you occurrence of single-peaked
  preferences (or matching) in your research