Voting and Mechanism Design

Voting and Mechanism Design


Jose M Vidal

Department of Computer Science and Engineering, University of South Carolina

March 26, 2010

Abstract

Voting, Mechanism design, and distributed algorithmicsmechanism design. Chapter 8.


Voting

1 VotingThe ProblemSolutionsSummary

2 Centralized Mechanism DesignProblem DescriptionAn Example Problem and SolutionThe Groves-Clarke MechanismThe Vickrey-Clarke-Groves Mechanism

3 Distributed Mechanism Design

4 Conclusion


Voting

Why Vote?

Common way of aggregating agents’ preferences.

Well understood.

But, centralized.


Voting

The Problem




4 Conclusion


Voting

The Problem

The Voting Problem

MilkWineBeer

WineBeerMilk

BeerWineMilk

Beer Wine Milk

Plurality

5 4 6

Runoff

5,9 4 6,6

Pairwise

1 2 0


Voting

The Problem

The Voting Problem

MilkWineBeer

WineBeerMilk

BeerWineMilk

Beer Wine Milk

Plurality 5 4 6Runoff

5,9 4 6,6

Pairwise

1 2 0


Voting

The Problem

The Voting Problem

MilkWineBeer

WineBeerMilk

BeerWineMilk

Beer Wine Milk

Plurality 5 4 6Runoff 5,9 4 6,6

Pairwise

1 2 0


Voting

The Problem

The Voting Problem

MilkWineBeer

WineBeerMilk

BeerWineMilk

Beer Wine Milk

Plurality 5 4 6Runoff 5,9 4 6,6

Pairwise 1 2 0


Voting

Solutions




4 Conclusion


Voting

Solutions

Symmetry

Reflectional symmetry: If one agent prefers A to B andanother one prefers B to A then their votes should cancel eachother out.

Rotational symmetry: If one agent prefers A,B,C and anotherone prefers B,C,A and another one prefers C,A,B then theirvotes should cancel out.

Plurality vote violates reflectional symmetry, so does runoffvoting.

Pairwise comparison violates rotational symmetry.


Voting

Solutions

Symmetry

Reflectional symmetry: If one agent prefers A to B andanother one prefers B to A then their votes should cancel eachother out.

Rotational symmetry: If one agent prefers A,B,C and anotherone prefers B,C,A and another one prefers C,A,B then theirvotes should cancel out.

Plurality vote violates reflectional symmetry, so does runoffvoting.

Pairwise comparison violates rotational symmetry.


Voting

Solutions

Borda Count

Jean-Charles deBorda. 1733–1799.

1 With x candidates, each agent awards xto points to his first choice, x − 1 pointsto his second choice, and so on.

2 The candidate with the most points wins.

Borda satisfies both reflectional and rotationalsymmetry.


Voting

Solutions

Formalization

There is a set of A agents, and O outcomes.

Each agent i has a preference function >i over the set ofoutcomes.

Let >∗ be the global set of social preferences. That is, whatwe want the outcome to be.


Voting

Solutions

Definition (Desirable Voting Outcome Conditions)

1 >∗ exists for all possible inputs >i

2 >∗ exists for every pair of outcomes

3 >∗ is asymmetric and transitive over the set of outcomes

4 >∗ should be Pareto efficient.

5 The scheme used to arrive at >∗ should be independent ofirrelevant alternatives.

6 No agent should be a dictator in the sense that >∗ is alwaysthe same as >i , no matter what the other >j are.


Voting

Solutions









Voting

Solutions









Voting

Solutions









Voting

Solutions









Voting

Solutions









Voting

Solutions

Kenneth Arrow

Theorem (Arrow’s Impossibility)

There is no social choice rule that satisfies thesix conditions.

Plurality voting relaxes 3 and 5. Adding athird candidate can wreak havoc.

Pairwise relaxes 5.

Borda violates 5.


Voting

Solutions

Kenneth Arrow

Theorem (Arrow’s Impossibility)

There is no social choice rule that satisfies thesix conditions.

Plurality voting relaxes 3 and 5. Adding athird candidate can wreak havoc.

Pairwise relaxes 5.

Borda violates 5.


Voting

Solutions

Borda Example

1 a > b > c > d

2 b > c > d > a

3 c > d > a > b

4 a > b > c > d

5 b > c > d > a

6 c > d > a > b

7 a > b > c > d

1 c gets 20 points

2 b gets 19 points

3 a gets 18 points

4 d gets 13 points


Voting

Solutions

Borda Example

1 a > b > c > d

2 b > c > d > a

3 c > d > a > b

4 a > b > c > d

5 b > c > d > a

6 c > d > a > b

7 a > b > c > d

1 c gets 20 points

2 b gets 19 points

3 a gets 18 points

4 d gets 13 points


Voting

Solutions

Borda Example

Let’s get rid of d .

1 a > b > c > d

2 b > c > d > a

3 c > d > a > b

4 a > b > c > d

5 b > c > d > a

6 c > d > a > b

7 a > b > c > d

1 a gets 15 points

2 b gets 14 points

3 c gets 13 points


Voting

Solutions

Borda Example


1 a > b > c

2 b > c > a

3 c > a > b

4 a > b > c

5 b > c > a

6 c > a > b

7 a > b > c

1 a gets 15 points

2 b gets 14 points

3 c gets 13 points


Voting

Solutions

Borda Example


1 a > b > c

2 b > c > a

3 c > a > b

4 a > b > c

5 b > c > a

6 c > a > b

7 a > b > c

1 a gets 15 points

2 b gets 14 points

3 c gets 13 points


Voting

Summary




4 Conclusion


Voting

Summary

Voting Summary

1 Use Borda count whenever possible.

2 Practically, Borda requires calculating all preferences: oftencomputationally hard.


Centralized Mechanism Design




4 Conclusion



Problem Description




4 Conclusion



Problem Description

Painting the House

Name Wants house painted?

Alice YesBob NoCaroline YesDonald YesEmily Yes



Problem Description

Formal Definition

Each agent i has a type θi ∈ Θi which is private.

θ = {θ1, θ2, . . . , θA}.The protocol results in some outcome o ∈ O.

Each agent i gets a value vi (o, θi ) for outcome o.

The social choice function f (θ) tells us the outcome we wantto achieve. For example,

f (θ) = arg maxo∈O

n∑i=1

vi (o, θi ) (1)



Problem Description

Formal Definition


θ = {θ1, θ2, . . . , θA}.

The protocol results in some outcome o ∈ O.




n∑i=1

vi (o, θi ) (1)



Problem Description

Formal Definition






n∑i=1

vi (o, θi ) (1)



Problem Description

Formal Definition






n∑i=1

vi (o, θi ) (1)



Problem Description

Formal Definition






n∑i=1

vi (o, θi ) (1)



Problem Description

Painting the House

Name Type (θi ) vi (Paint, θi ) vi (NoPaint, θi )

Alice WantPaint 10 0Bob DontNeedPaint 0 0Caroline WantPaint 10 0Donald WantPaint 10 0Emily WantPaint 10 0



Problem Description

Painting the House



Try this: Everyone votes Y/N. If majority votes Y then painthouse. All pay 1/5 of cost (4 each).

Bob must pay for a paint job he didn’t want.



Problem Description

Painting the House



Try this: Everyone votes Y/N. If majority votes Y then painthouse. All pay 1/5 of cost (4 each).

Bob must pay for a paint job he didn’t want.



Problem Description

Painting the House



Try this: Everyone votes Y/N. Split cost among those whovoted Y.

There is an incentive for all but Bob to lie.



Problem Description

Painting the House



Try this: Everyone votes Y/N. Split cost among those whovoted Y.

There is an incentive for all but Bob to lie.



Problem Description

Definition (g Implements f )

A mechanism g : S1 × · · · × SA → O implements social choicefunction f (·) if there is an equilibrium strategy profile(S∗1 (·), . . . ,S∗A(·)) of the game induced by g such thatg(S∗1 (θ1), . . . ,S∗A(θA)) = f (θ1, . . . , θA) for all θ ∈ Θ.

Where we let Si (θi ) be agent i ’s strategy given that it is of type θi .



Problem Description

Definition (Dominant Strategy Equilibrium)

A strategy profile (S∗1 (·), . . . ,S∗A(·)) of the game induced by g is adominant strategy equilibrium if for all i and all θi ,

vi (g(s∗i (θi ), s−i ), θi ) ≥ vi (g(s ′i , s−i ), θi )

for all s ′i ∈ Si and all s−i ∈ S−i .



Problem Description

Definition (g Implements f )

A mechanism g : S1 × · · · × SA → O implements social choicefunction f (·) in dominant strategies if there is a dominant strategyequilibrium strategy profile (S∗1 (·), . . . ,S∗A(·)) of the game inducedby g such that g(S∗1 (θ1), . . . ,S∗A(θA)) = f (θ1, . . . , θA) for allθ ∈ Θ.



Problem Description

Definition (Strategy-Proof)

The social choice function f (·) is truthfully implementable indominant strategies (or strategy-proof) if s∗i (θi ) = θi (for allθi ∈ Θi and all i) is a dominant strategy equilibrium of the directrevelation mechanism f (·). That is, if for all i and all θi ∈ Θi ,

vi (f (θi , θ−i ), θi ) ≥ vi (f (θi , θ−i ), θi )

for all θi ∈ Θi and all θ−i ∈ Θ−i .



Problem Description

Theorem (Revelation Principle)

If there exists a mechanism g that implements the social choicefunction f in dominant strategies then f is truthfullyimplementable in dominant strategies.



An Example Problem and Solution




4 Conclusion




Selling Example

θi ∈ <: types are the valuations.

o ∈ {1, . . . , n}: index of agent who gets the item.

vi (o, θi ) = θi if o = i , and 0 otherwise.

f (θ) = arg maxi (θi )

Each agent gets a pi (o) so that ui (o, θi ) = vi (o, θi ) + pi (o).

Set p(o) such that the agent who wins must pay a tax equal to thesecond highest valuation. No one else pays/gets anything.

ui (o, θi ) =

{θi −maxj 6=i θj if o = i

0 otherwise.




Selling Example







ui (o, θi ) =


0 otherwise.




Selling Example







ui (o, θi ) =


0 otherwise.




Selling Example







ui (o, θi ) =


0 otherwise.




Selling Example







ui (o, θi ) =


0 otherwise.




Selling Example







ui (o, θi ) =


0 otherwise.




Truth-Telling is Dominant in Vickrey Payments Example.

1 Let bi (θi ) be i ’s bid given that his true valuation is θi .

2 Let b′ = maxj 6=i bj(θj) be the highest bid amongst the rest.

3 If b′ < θi then any bid bi (θi ) > b′ is optimal since

ui (i , θi ) = θi − b′ > 0

4 If b′ > θi then any bid bi (θi ) < b′ is optimal since

ui (i , θi ) = 0

5 Since we have that if b′ < θi then i should bid > b′ and ifb′ > θi then i should bid < b′, and we don’t know b′ then ishould bid θi .








ui (i , θi ) = θi − b′ > 0


ui (i , θi ) = 0









ui (i , θi ) = θi − b′ > 0


ui (i , θi ) = 0









ui (i , θi ) = θi − b′ > 0


ui (i , θi ) = 0









ui (i , θi ) = θi − b′ > 0


ui (i , θi ) = 0




The Groves-Clarke Mechanism




4 Conclusion




Theorem (Groves-Clarke Mechanism)

If


n∑i=1

vi (o, θi )

then calculating the outcome using


n∑i=1

vi (o, θi )

(where θ are reported types) and giving the agents payments of

pi (θ) =∑j 6=i

vj(f (θ), θj)− hi (θ−i )

(where hi (θ−i ) is an arbitrary function) results in a strategy-proofmechanism.




Groves-Clarke Payments for House Painting

Name vi (o, θ) vi (o, θ) +∑

j 6=i vj(θ)

Alice 10− 204 = 5

Bob 0− 0 = 0

Caroline 10− 204 = 5

Donald 10− 204 = 5

Emily 10− 204 = 5

Assuming all tell the truth.






j 6=i vj(θ)

Alice 10− 204 = 5 5 + 15 = 20

Bob 0− 0 = 0 0 + 20 = 20

Caroline 10− 204 = 5 5 + 15 = 20

Donald 10− 204 = 5 5 + 15 = 20

Emily 10− 204 = 5 5 + 15 = 20

Assuming all tell the truth.






j 6=i vj(θ)

Alice 0− 0 = 0 10 + (103 · 3) = 20

Bob 0− 0 = 0 0 + (103 · 3) = 10

Caroline 10− 203 = 10

3 10 + (103 · 2) = 10

Donald 10− 203 = 10

3 10 + (103 · 2) = 10

Emily 10− 203 = 10

3 10 + (103 · 2) = 10

Alice lies.



The Vickrey-Clarke-Groves Mechanism




4 Conclusion




Theorem (Vickrey-Clarke-Groves (VCG) Mechanism)

If


n∑i=1

vi (o, θi )

then calculating the outcome using


n∑i=1

vi (o, θi )

(where θ are reported types) and giving the agents payments of

pi (θ) =∑j 6=i

vj(f (θ−i ), θj)−∑j 6=i

vj(f (θ), θj)

(where hi (θ−i ) is an arbitrary function) results in a strategy-proofmechanism.




VCG Payments for House Painting

Name vi (o, θ)∑

j 6=i vj(f (θ−i ), θj)∑

j 6=i vj(·)−∑

j 6=i vj(·)

Alice 10− 204 = 5 (10− 20

3 ) · 3 = 10 10− 15 = 5

Bob 0− 0 = 0 (10− 204 ) · 4 = 20 20− 20 = 0

Caroline 10− 204 = 5 (10− 20

3 ) · 3 = 10 10− 15 = 5

Donald 10− 204 = 5 (10− 20

3 ) · 3 = 10 10− 15 = 5

Emily 10− 204 = 5 (10− 20

3 ) · 3 = 10 10− 15 = 5

Alice tells the truth.


Distributed Mechanism Design




4 Conclusion



Distributed Algorithmic Mechanism Design

Algorithmic mechanism design: make mechanism polynomialtime

Distributed algorithmic mechanism design: make mechanismdistributed



Inter-Domain Routing Problem

AcA = 5

Xcx = 2

Zcz = 4 D

cd = 1

Bcb = 2

Ycy = 3


Conclusion




4 Conclusion


Conclusion

Conclusion

GC and VCG payment equations are useful ready-madesolutions to many problems.

But, we still need more research into how to distribute themechanisms and how to make their calculationcomputationally tractable.


Conclusion

Conclusion

GC and VCG payment equations are useful ready-madesolutions to many problems.

But, we still need more research into how to distribute themechanisms and how to make their calculationcomputationally tractable.

Documents

Voting and Mechanism Design