Post on 15-Jan-2016
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
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
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
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 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
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
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
<f,uS>=0 S: |S|>1f is not constant=> f depends on one variable
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
<f,uS>=0 S: |S|>1f is not constant=> f depends on one variable
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
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
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
W p p p p p p f g g h h g
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