A Fourier-Theoretic Perspective on the Condorcet Paradox and Arrow ’ s Theorem. By Gil Kalai,...

A Fourier-Theoretic Perspective on the Condorcet Paradox and Arrow’s Theorem.

By Gil Kalai, Institute of Mathematics, Hebrew UniversityPresented by: Ilan Nehama

2

Basic notations n players m alternatives Each player have a preference over the

alternatives Ri a >i b := Player i prefers a over b Linear order I.e.

Full and asymmetric: a, b : (a>b) XOR (a<b) Transitive

The vector of all preferences (R1, R2,…,Rn) is called a profile.

3

Basic notations

The preferences are aggregated to the society preference. a > b := The society prefers a over b

Full and asymmetric: a, b : (a>b) XOR (a<b)

We do not require it to be transitive

The aggregation mechanism is called a social choice function

4

Basic notations

Probability space For a social choice function F and a

property φ Pr[φ(RN)]:=#{Profiles

RN:φ(F)]}/#{Profiles}

5

Social choice function’s properties Social choice function is a function

between profiles to relations. The social choice function is called

rational on a specific profile RN if f(RN) is an order.

The social choice function is called rational if it is rational on every profile.

An important property of a social choice function is Pr[F is non-rational].

6

Social choice function’s properties

IIA–Independence of Irrelevant Alternatives. for any two alternatives a>b depends

only on the players preferences between a and b.

{i: a>ib} determines whether a>b

7

Social choice function’s properties

Balanced-For any two alternatives x,y : Pr[x>y]=Pr[y>x]

Neutral-The function is invariant under permutations of the alternatives

8

Social choice function’s properties Dictator

Profile-For a profile each player i that the social aggregation over the profile agrees with his opinion is called a dictator for that profile.

General-A player that is a dictator on a ‘big portion’ of the profiles is called a dictator.

Dictatorship-A social choice function that have one dictator player is called a dictatorship.

9

Main results There exists an absolute constant K

s.t.: For every >0 and for any neutral social

choice function If the probability that the function is non-

rational on a random profile < Then there exists a dictator such that for every

pair of alternatives the probability that the social choice differs from the dictator’s choice < K

10

Main results For the majority function the

probability of getting an order as result (avoiding the Condorcet Paradox) approaches (as n approaches to infinity) to G

0.9092<G<0.9192

11

Agenda Defining the mathematical base –

The Discrete Cube The probability of irrational social

choice for three alternatives The probability of the Condorcet

paradox A Fourier-theoretic proof of Arrow’s

theorem

12

Discrete Cube

Xn={0,1}n=P([n])=[2n] Uniform probability f,g:X->R

2

[ ]

2 0.5

[ ]

, : ( ) ( )2

, ( ( ))2

,

n

n

s n

s n

f g f S g S

f f f f f S

f g f g

13

An orthonormal basis: us(T)=(-1)|ST|

[ ]

, : ( ) ( )2n

s n

f g f S g S

| | 2

[ ]

| | | |

[ ]

| | | | | ( { })| | ( { })|

[ ]\{ }

| | | | | | | |

[ ]\{ }

, (( 1) ) 12

. \

, ( 1) ( 1)2

( 1) ( 1) ( 1) ( 1)2

( 1) ( 1) ( ( 1) )( 1)2

2

n

n

n

n

n

S RS S

R n

S R T RS T

R n

S R T R S R x T R x

R n x

S R T R S R T R

R n x

u u

S T x S T

u u

[ ]\{ }

0 0R n x

14

us(T)=(-1)|ST| form an orthonormal basis

2

[ ]

[ ]

2 2

[ ]

( )

( ) ,

, ( ) ( )

( , ) ( )

SS n

S

S n

S n

f f S u

f S f u

f g f S g S

f f f f S

15

For f a boolean function f:X->{0,1}. F is a characteristic function

for some AX. A2(2[n])

P[A]:=|A|/2n

2

2

2

[ ]

0

, 2 ( ) 2 | | [ ]

: ( ) ( 1) 1

( ) , [ ]

n n

S n

f f f f S A P A

R u R

f f u P A

[ ]

, : ( ) ( )2n

s n

f g f S g S

Boolean functions over X

16





theorem

17

Domain definition

F is a social choice function < = F(<1, <2,…,<n) F is not necessarily rational Three alternatives – {a,b,c}

F is IIA {i: a>ib} determines whether a>b

18

Each player preference can be described by 3 boolean variables xi=1 <=> a>ib yi=1 <=> b>ic zi=1 <=> c>ia

Domain definition

19

F can be described by three boolean functions of 3n variables

f(x1,..,xn,y1,..,yn,z1,..,zn)=1 <=> a>b g(x1,..,xn,y1,..,yn,z1,..,zn)=1 <=> b>c h(x1,..,xn,y1,..,yn,z1,..,zn)=1 <=> c>a

Domain definition

20

F is IIA {i: a>ib} determines whether

a>b

f,g,h are actually functions of n variables

f(x)=f(x1,..,xn) g(y)=g(y1,..,yn) h(z)=h(z1,..,zn)

21

Define

1

2

3

p =P[{x| f(x)=1}] = f ( )

p =P[{y| g(y)=1}] = g( )

p =P[{z| h(z)=1}] = h( )

F will be called balanced when p1=p2=p3=½

22

The domain of F is: Ψ = {all (x,y,z) that correspond to

rational profiles}= {(x,y,z) | i (xi,yi,zi) {(0,0,0),

(1,1,1)}

P[Ψ] = (6/8)n

23

W=W(F)=W(f,g,h) is defined to be The probability of obtaining a non-rational

outcome (from rational profile) f(x)g(y)h(z)+(1-f(x))(1-g(y))(1-h(z))=1

<=> F(x,y,z) is non-rational

( , , )

( ( ) ( ) ( ) (1 ( ))(1 ( ))(1 ( ))

| |x y z

f x g y h z f x g y h z

W

W- Probability of a non-rational outcome

24

Theorem 3.1

| | 1

[ ]

1

2

3

, : ( ) ( )( 1/ 3)

p =P[{x| f(x)=1}] = f ( )

p =P[{y| g(y)=1}] = g( )

p =P[{z| h(z)=1}] = h( )

S

s n

f g f S g S

1 2 3 1 2 3(1 )(1 )(1 ) ( , , , ) / 3W p p p p p p f g g h h g

25

Proof of Thm. 3.1 A,B are

boolean functions on 3n variables Subsets of 23n

A=ΧΨ

B=f(x)g(y)h(z)

3

3

( , , ) 2

2 ( ) ( ) ( ) , ( ) ( )n

n

x y z S

f x g y h z A B A S B S

26

32

, ( ) ( )

A=

B=f(x)g(y)h(z)

nS

A B A S B S

3

3

3 | |

2

| || | | |3

( , , )

( )

2 : ( , , )

( ) , 2 ( )( 1)

| | | | | | | |

2 ( ) ( ) ( )( 1) ( 1) ( 1)

( ) ( ) ( )

n

y yx x z z

x y z

nx y z x k y k z k

n S RS

R

x x y y z z

S RS R S Rnx y z

R R R

x y z

B S

S S S S S S x S y S z

B S B u B R

S R S R S R S R

f R g R h R

f S g S h S

Proof of Thm. 3.1

27

32

, ( ) ( )

A=

B=f(x)g(y)h(z)

nS

A B A S B S

3

1

3 | |

2

1

| |3

1 1( )

( )

: { , , }

1 S is a rational profile( ) {

0

( , , )

0 ' ' '( ', ', ') {

1

( ) , 2 ( )( 1)

| | | |

2 ( ) ( 1)

n

i i

i i

i i i i

n

i i ii

n S RS

R

n

i ii

n nS Rn

i ii iR

A S

F F x y z

A SOtherwise

A S S S

x y zA x y z

otherwise

A S A u A R

S R S R

A R

1

1

( )

n

n

i ii

A S

Proof of Thm. 3.1

28

32

, ( ) ( )

A=

B=f(x)g(y)h(z)

nS

A B A S B S

1 2 31 2

1

| || 3|x y z

1

( )

( ) ( )

3 / 4

By direct computation: ( ) { 1/ 4 | | 2

0

( ) ( 1/ 4) (3 / 4) : belongs to two or none of S ,S ,and S( )

0

n

i ii

i

i i i

nn S S SS S S

i ii

A S

A S A S

S

A S S

otherwise

A S i iA S

Otherwise

Proof of Thm. 3.1

29

| | | |x y z

1

x y z x y z

3

: belongs to two or none of S ,S ,and S( 1/ 4) (3 / 4)( ) ( ) {

0

We'll denote by Q the triplets (S ,S ,S ) for which (S ,S ,S ) 0

( ) ( ) ( ) ( )

2 ( ) (

x y z x y zS S S n S S Sn

i ii

x y z

n

i iA S A S

Otherwise

A

B S f S g S h S

f x g y

3( , , ) 2

| |

) ( ) , ( ) ( )

(1 )( ) 2 (1 )( ) ( )

2 ( ) 2 ( ) ( )

2 ( 1) ( )

( ){1 ( )

nx y z S

nS

R

n nS S

R R

n R S

R

h z A B A S B S

f S f R u R

u R f R u R

f S

f S S

Sf S

Proof of Thm. 3.1

30

1

( , , )

3

( , , ) ( , , )

| | | |3 3

( , , )

| | ( ( ) ( ) ( ) (1 ( ))(1 ( ))(1 ( ))

6 2 ( ) ( ) ( ) (1 ( ))(1 ( ))(1 ( ))

2 ( ) ( ) ( )( 1/ 4) (3 / 4) 2 (1 )( )(1x y z x y z

x y z

x y z

n n

x y z x y z

S S S n S S Sn nx y z x

S S S Q

W f x g y h z f x g y h z

W f x g y h z f x g y h z

f S g S h S f S

| | | |

( , , )

| | | |3

( , , )

3

)( )(1 )( )( 1/ 4) (3 / 4)

2 ( 1/ 4) (3 / 4) [ ( ) ( ) ( ) (1 )( )(1 )( )(1 )( )]

2 (3 / 4) [ ( ) ( ) ( ) (1 )( )(1 )( )(1

x y z x y z

x y z

x y z x y z

x y z

S S S n S S Sy z

S S S Q

S S S n S S Snx y z x y z

S S S Q

n n

g S h S

f S g S h S f S g S h S

f g h f g h

3 | | | | | | | | | | | |

| |1 2 3 1 2 3

)( )] 2 ( 1/ 4) (3 / 4) ( ) ( ) ( 1/ 4) (3 / 4) ( ) ( ) ( 1/ 4) (3/ 4) ( ) ( )

[ (1 )(1 )(1 )] ( 1/ 3) [ ( ) ( ) ( ) ( ) ( ) ( )

x y x z y

yz x

n S n S S n S S n S

S S S S S S Sz S SSS S

S

f S g S f S h S h S g S

W p p p p p p f S g S g S h S f S h S

3 3 3 3

1 2 3 1 2 3

1 2 3

2 | | 1

]

, , ,(1 )(1 )(1 )

3

Note that if f=g=h then we get ( )

(1 ) , (1 ) ( )( 1/ 3)

S

S

S

f g g h h fp p p p p p

p p p p

W p p f f p p f S

Proof of Thm. 3.1

31




paradox A Fourier-theoretic prosof of

Arrow’s theorem

32

The Condorcet Paradox

There are cases that the majority voting system (which seems natural) yields irrational results.

Three voters, three alternatives 1) a>1b>1c 2) b>2c>2a 3) c>3a>3b

Result: a>b>c>a

Marie Jean Antoine Nicolas Caritat, marquis de Condorcet

33

Computing the probability of the Condorcet Paradox

3 alternatives n=2m+1 voters f=g=h are the majority function

G(n,3):=The probability of a rational outcome.

G(3):=limn→∞G(n,3)

34

Computing the probability of the Condorcet Paradox It is known that

3 3 1(3) arcsin 0.91226

4 2 3G

We will prove 1 1 2 8 1

0.9092 1 ( ) (3) 1 ( ) 0.91924 2 9 9 2

G

35

2 1 2

1

2 1 2

(2 1) |{ } |

[2 1] [2 1]\{ } [2 1]\{ }| | 1 | { }| 1 | | 1

2 2

1

: ({ })

2( 2 ) (2 1)

2 ({ }) ( 1) ( 1) 1

2 2 2

m

mk

mm

m k S

S m S m k S m kS m S k m S m

m m

i m i m

d f k

md m

m

f k

m m m

i i m

36

2 1 2

1

2 1 2

({ })

2( 2 ) (2 1)

m

mk

m

d f k

mm

m

m

2 2 1 2 2 2 1 22

m+1

2 1 22 1 2m2

2 24

4

d is a decreasing sequence

2 2 (2 2)!( 2 ) (2 3) ( 2 ) (2 3)1d ( 1)!

(2 )!2d ( 2 ) (2 1)( 2 ) (2 1)( !)

(2 1) (2 2) (2 3) (2 1)(2 3)2

( 1) (2 1) (2 2)

m m

mm

m mm mm m

mm mmmm

m m m m m

m m m

2

2 2

2 1 2 2 1 2m 2

2 22 1 2

2

22 1 2

4 8 31

4 8 4

1lim

2

2 (2 )!d ( 2 ) (2 1) ( 2 ) (2 1)

( !)

Striling's approximation : ( !) 2

(2 ) 2 2( 2 ) (2 1)

( 2 )

2 2 1( 2 ) (2 1)

4

mm

m m

k k

m mm

m m

mm

m

m m

m m

d

m mm m

m m

k k e k

m e mm

m e m

mm

mm

1

2

37

1 2 n 1 2 n

|R S| |R S| |R S|+1

R [n] R [n]\{x}

|R S|

R [n]\{x}

f is the majority function

f(1-x ,1-x ,...,1-x )=1-f(x ,x ,...,x )

f (S)=0 S , |S| is even

x S

f (S)= f(R)(-1) f(R)(-1) f(R {x})(-1)

(-1) [f(R)-f(R {x})]=

|R S|

R [n]\{x}|R|=m

| | 1i

0(|R S|=i)

| | 1

2i | | 1

0(|R S|=i)

(-1) [0-1]=

| | 1 2 1 | |(-1)

| | 1 2 1 | | | | 1 2 1 | |(-1) (-1)

| | 1 (| | 1 )

|

S

i

S

S i

i

S m S

i m i

S m S S m S

i m i S i m S i

S

| | 1

2i 1

0(|R S|=i)

| 1 2 1 | | | | 1 2 1 | |(-1) (-1) 0

)

S

i

i

m S S m S

i m i i m i

38

3 32

| | 1

2 2| | 1 | | 1

|S| is odd |S| is odd

2 | | 1

|S| is odd

2

m

|S| =1

1 ( ,3) (1 ) ( )( 1/ 3)

¼- ( )( 1/ 3) ¼- ( )(1/ 3)

1 11 ( ) (3)

4 2

1 ( ,3) ¼- ( )(1/ 3)

1¼- ( ) ¼-d ¼

2

S

S

S S

S S

S

S

S

G n W p p f S

f S f S

G

G n f S

f S

1 1 2 8 1Proving 1 ( ) (3) 1 ( )

4 2 9 9 2G

39

2 2 2 22

| | 3 |S|=1 | | 2

2

2| | 1

|S| is odd

2 2| | 1

| | 3 | | 3

2 8 1(3) 1 ( ) 0.9192

9 9 2

( ) ( ) ( ) ( )

0 ¼-

1 ( ,3) ¼- ( )(1/ 3)

¼- ( )(1/ 3) ¼- 1/ 9 ( )

¼- 1/ 9(¼- ) 2 / 9 8 / 9

1

S S

m m

S

S

Sm m

S S

m m m

G

f S f f f S f S

p p d d

G n f S

d f S d f S

d d d

G

( ,3) 2 / 9 8 / 9

2 8 11 ( )

9 9 2

m

m

n d

G n

1 1 2 8 1Proving 1 ( ) (3) 1 ( )

4 2 9 9 2G

40





theorem

41

Arrow’s Theorem At least three alternatives Let f be a social choice function which

is: unanimity respecting / Pareto optimal independent of irrelevant alternatives

Then f is a dictatorship.

Kenneth Arrow

42

Lemma 6.1: For f a boolean function:If <f,uS>=0 S: |S|>1

Then exactly one of the following holds f is constant

f=1 or f=0 f depends on one variable (xi)

f(x1, x2,…,x1)=xi or f(x1, x2,…,x1)=1-xi

43

<f,uS>=0 S: |S|>1f is not constant=> f depends on one variable

S S

2

S [n]

2 2 2

1 S [n] S

p [ ( ) 1]. With no loss of generality p ½

(otherwise we will prove for (1-f) , <1-f,u >= -<f,u >)

p= f ( ) f ( )

Assume that f is not constant and hence p [½,1)

f ({ }) f ( ) f ( )n

i

P f x

S

i S S

22

[n]|S|>1

2 2

1 1

f ( )

|f ({ }) | |f ({ }) | !i f ({ }) 0n n

i i

p p

i p p i p p i

Proof of Lemma 6.1

44

i

{ }

2{ }

S [n] 1 1

2

1

1 f ({ }) 0Define x as: x {

0

01( ) {

11

1 ( ) f ( ) ( ) f ({ }) ( ) |f ({ }) | 1

½ |f ({ }) | ! f ({ }) 0, | f ({ }) | ½

ii

i

n n

s ii i

n

i

i

Otherwise

xu x

x

f x S u x p i u x p i p p p

p p i p p i i


Proof of Lemma 6.1

45

|{ } |{ }

[ ] [ ] [ ]|

i i

i i

i

f ({ }) , 2 ( )( 1) 2 ( )(1) ( )( 1)

½Pr[x =0 f(x)=1] - ½Pr[x =1 f(x)=1]

| Pr[x =0 f(x)=1] - Pr[x =1 f(x)=1] | = 1

Then one of the two cases:

x =0Pr[

n k S ni

S n S n S ni S i S

i f u f S f S f S

i

i i

x =1 ]=0 , Pr[ ]=1 ( )

f(x)=1 f(x)=1

x =0 x =1 Pr[ ]=1 , Pr[ ]=0 ( )

f(x)=1 f(x)=1

i

i

f x x

f x x


Proof of Lemma 6.1

46

Proof of Arrow’s theorem (assuming neutrality)

From lemma 6.1 one can prove Arrow’s theorem for neutral social choice function

Instead we will use a generalization of this lemma to prove a generalization of Arrow’s theorem.

47

1 2 3

1 2 3 1 2 3

| | 1

[ ]

2| | 1

[ ]

F is neutral

f=g=h p p p ½

(1 )(1 )(1 ) ( , , , ) / 3

, : ( ) ( )( 1/ 3)

0 ¼ , ¼ ( )( 1/ 3)

¼ ( )

S

s n

S

s n

p

W p p p p p p f g g h h g

f g f S g S

W f f f S

f S

| | 1

2 2

| | 1

2 2

2 | | 1 2

, ( 1/ 3) ( )

¼ ( ) * ( 1/ 3) ( )

( ) ¼ , ( 1/ 3) ( ) ¼

¼ ¼ *¼ 0

SS S

s s

SS S

s s

SS S

s s

u f S u

f S u f S u

f S u p f S u p

Proof of Arrow’s theorem using lemma 6.1

48

| | 1

i

: ( ) ( 1/ 3) ( )

( ) 0 S:|S|>1

( ) 0 S:|S|>1

f is not constant, f(0,...,0)=0. Hence i f(x)=x

SS S

s s

f S u f S u

f S

f S

49

2

2

| | 1

2

1 2 n i 1

Theorem 7.1: There exists constants K and K' s.t.

For every f a boolean function and every , if

f =p

( )

Then one of the following cases holds

p<K' or p>1-K'

i f(x , x ,..., x ) x <K or f(x , x

S

f S

2

2 n i,..., x ) (1 x ) <K

Notice that for =0 we get Lemma 6.1

Proofs of this thorem are the issue of

"Boolean Functions whose Fourier Transform is Concentrated on

the First Two Levels" / Friedgut, Kalai and

Naor.

50

Generalized Arrow’s Theorem Theorem 7.2: For every ε>0 and for

every neutral social choice function on three alternatives:

If the probability the social choice function if non-rational≤ε

Then there is a dictator such that the probability that the social choice differs from the dictator’s choice is smaller than Kε

Notice that for ε=0 we get Arrow’s theorem.

51

1 2 3

1 2 3 1 2 3

| | 1

[ ]

2| | 1

[ ]

2

F is neutral

f=g=h p p p ½

(1 )(1 )(1 ) ( , , , ) / 3

, : ( ) ( )( 1/ 3)

¼ , ¼ ( )( 1/ 3)

¼ (

S

S n

S

S n

p


f g f S g S

W f f f S

f S

2| | 1

[ ] | | 1

22 | | 1

| | 1

2 2 2| | 1 3 1

| | 1 | | 1 | | 1

) ( )[( 1/ 3) 1]

¼ (½-½ ) ( )[( 1/ 3) 1]

( )[1 ( 1/ 3) ] ( )[1 (1/ 3) ] 8 / 9 ( )

S

S n S

S

S

S

S S S

f S

f S

f S f S f S

Proof of theorem 7.2 using theorem 7.1

52

2

| | 1

2 2

1 2 n i 1 2 n i

2

1 2 n i

( ) 9 / 8

1/ 2

i f(x , x ,..., x ) x <K9 / 8 or f(x , x ,..., x ) (1 x ) <9 / 8K

(If we assume pareto optimality f(x , x ,..., x ) x <9 / 8K )

S

f S

p

53

Corollary

For fm a balanced social choice family on m alternatives For every ε>0, as m tends to infinity,

If for every pair of alternatives there is no dictator with probability (1- ε)

Then, the probability for a rational outcome tends to zero

54

The End

55

Proposition 5.2

If the social choice function is neutral then the probability of a rational outcome is at least 3/4

56

Proof of Proposition 5.2

1 2 3

1 2 3 1 2 3

| | 1

[ ]

2

[ ]

F is neutral

f=g=h p p p ½ f(1-x)=f(x)

(1 )(1 )(1 ) ( , , , ) / 3

, : ( ) ( )( 1/ 3)

¼ , ¼ ( )( 1/ 3)

S

S n

S n

p


f g f S g S

W f f f S

C

| | 1

|R S| |R S| C |R S|

R [n] R [n]

|R S| |S|-| S|

R [n]

|R S| | S| |R S|

R [n] R [n]

f (S)=0 S : |S| is even

x S

f (S)= f(R)(-1) ½ f(R)(-1) f(R )(-1)

½ f(R)(-1) (1 f(R))(-1)

½ f(R)(-1) (1 f(R))(-1) ½ (-1)

½ (-

S

R

R

|R S| |R {x} S| |R S| |R S|

R [n]\{x} R [n]\{x}

1) +(-1) ½ (-1) -(-1) 0

57

Proof of Proposition 5.2

2 | | 1

[ ]

2 | | 1

[ ] |S| is odd

2 | | 1

[ ] |S| is odd

¼ ( )( 1/ 3)

¼ ( )( 1/ 3)

¼ ( )1/ 3 ¼

S

S n

S

S n

S

S n

W f S

f S

f S

A Fourier-Theoretic Perspective on the Condorcet Paradox and Arrow ’ s Theorem. By Gil Kalai,...

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Transcript of A Fourier-Theoretic Perspective on the Condorcet Paradox and Arrow ’ s Theorem. By Gil Kalai,...