A Quantitative Arrow Theorem

# A Quantitative Arrow Theorem

Elchanan Mossel Weizmann Institute and U.C. Berkeley. Supported by an Alfred Sloan fellowship in Mathematics, by NSF CAREER grant DMS-0548249 (CAREER), by DOD ONR grant (N0014-07-1-05-06), by BSF grant 2004105 and by ISF grant 1300/08
###### Abstract

Arrow’s Impossibility Theorem states that any constitution which satisfies Independence of Irrelevant Alternatives (IIA) and Unanimity and is not a Dictator has to be non-transitive. In this paper we study quantitative versions of Arrow theorem. Consider voters who vote independently at random, each following the uniform distribution over the rankings of alternatives. Arrow’s theorem implies that any constitution which satisfies IIA and Unanimity and is not a dictator has a probability of at least for a non-transitive outcome. When is large, is a very small probability, and the question arises if for large number of voters it is possible to avoid paradoxes with probability close to .

Here we give a negative answer to this question by proving that for every , there exists a , which depends on only, such that for all , and all constitutions on alternatives, if the constitution satisfies:

• The IIA condition.

• For every pair of alternatives , the probability that the constitution ranks above is at least .

• For every voter , the probability that the social choice function agrees with a dictatorship on at most .

Then the probability of a non-transitive outcome is at least .

Our results generalize to any number of alternatives and to other distributions over the alternatives. We further derive a quantitative characterization of all social choice functions satisfying the IIA condition whose outcome is transitive with probability at least . Our results provide a quantitative statement of Arrow theorem and its generalizations and strengthen results of Kalai and Keller who proved quantitative Arrow theorems for and for balanced constitutions only, i.e., for constitutions which satisfy for every pair of alternatives , that the probability that the constitution ranks above is exactly .

The main novel technical ingredient of our proof is the use of inverse-hypercontractivity to show that if the outcome is transitive with high probability then there are no two different voters who are pivotal with for two different pairwise preferences with non-negligible probability. Another important ingredient of the proof is the application of non-linear invariance to lower bound the probability of a paradox for constitutions where all voters have small probability for being pivotal.

## 1 Introduction

### 1.1 Background on Arrow’s Theorem

Arrow’s Impossibility theorem [1, 2] states that certain desired properties cannot hold simultaneously for constitutions on three or more alternatives. Arrow’s results were fundamental in the development of social choice theory in Economics. The most celebrated results in this area are Arrow’s Impossibility theorem and the Gibbard-Satterthwaite [8, 19] Manipulation theorem. Both results demonstrate the non-existence of ranking and voting schemes with very natural properties.

Arrow’s theorem demonstrates that IIA, Transitivity and Non-dictatorship, all of which will be defined below, cannot hold simultaneously. Quantitative versions of Arrow’s theorems prove tradeoff between being ”close to transitive” and being ”close to a dictator” assuming the IIA property holds. We proceed with a more formal discussion of Arrow’s theorem.

Consider , a set of alternatives. A transitive preference over is a ranking of the alternatives from top to bottom where ties are not allowed. Such a ranking corresponds to a permutation of the elements where is the rank of alternative .

A constitution is a function that associates to every -tuple of transitive preferences (also called a profile), and every pair of alternatives a preference between and . Some key properties of constitutions include:

• Transitivity. The constitution is transitive if is transitive for all . In other words, for all and for all three alternatives and , if prefers to , and prefers to , it also prefers to . Thus is transitive if and only if its image is a subset of the permutations on elements.

• Independence of Irrelevant Alternatives (IIA). The constitution satisfies the IIA property if for every pair of alternatives and , the social ranking of vs. (higher or lower) depends only on their relative rankings by all voters.

• Unanimity. The constitution satisfies Unanimity if the social outcome ranks above whenever all individuals rank above .

• The constitution is a dictator on voter , if , for all , or , for all , where is the ranking by reversing the ranking .

Arrow’s theorem states [1, 2] that:

###### Theorem 1.1.

Any constitution on three or more alternatives which satisfies Transitivity, IIA and Unanimity is a dictatorship.

It is possible to give a characterization of all constitutions satisfying IIA and Transitivity. Results of Wilson [20] provide a partial characterization for the case where voters are allowed to rank some alternatives as equal. In order to obtain a quantitative version of Arrow theorem, we give an explicit and complete characterization of all constitutions satisfying IIA and Transitivity in the case where all voters vote using a strict preference order. Write for the set of all constitutions on alternatives and voters satisfying IIA and Transitivity. For the characterization it is useful write for the statement that for all it holds that ranks all alternatives in above all alternatives in . We will further write for the constitution restricted to the alternatives in . The IIA condition implies that depends only on the individual rankings of the alternatives in the set . The characterization of we prove is the following.

###### Theorem 1.2.

The class consist exactly of all constitutions satisfying the following: There exist a partition of the set of alternatives into disjoint sets such that:

•  A1>FA2>F…>FAr,
• For all s.t. , there exists a voter such that is a dictator on voter .

• For all such that , the constitution is an arbitrary non-constant function of the preferences on the alternatives in .

We note that for all all elements of are not desirable as constitutions. Indeed elements of either have dictators whose vote is followed with respect to some of the alternatives, or they always rank some alternatives on top some other. For related discussion see [20].

The main goal of the current paper is to provide a quantitative version of Theorem 1.2 assuming voters vote independently and uniformly at random. Note that Theorem 1.2 above implies that if then . However if is large and the probability of a non-transitive outcome is indeed as small as , one may argue that a non-transitive outcome is so unlikely that in practice Arrow’s theorem does not hold.

The goal of the current paper is to establish lower bounds on the probability of paradox in terms that do not depend on . Instead our results are stated in terms of the statistical distance between and the closet element in . Thus our result establishes that the only way to avoid non-transitivity is by being close to the family .

In the following subsections we introduce the probabilistic setup, state our main result, discuss related work and give an outline of the proof.

### 1.2 Notation and Quantitative Setup

We will assume voters vote independently and uniformly at random so each voter chooses one of the possible rankings with equal probability. We will write for the underlying probability measure and for the corresponding expected value. In this probabilistic setup, it is natural to measure transitivity as well as how close are two different constitutions.

• Given two constitutions on voters, we denote the statistical distance between and by , so that:

 D(F,G)=P[F(σ)≠G(σ)].
• Given a constitution , we write for the probability that the outcome of is transitive and for the probability that the outcome of is non-transitive so ( stands for paradox):

 T(F)=P[F(σ) is transitive],P(F)=1−T(F).

### 1.3 Main Result

In our main result we show that

###### Theorem 1.3.

For every number of alternatives and , there exists a , such that for every , if is a constitution on voters and alternatives satisfying:

• IIA and

• ,

then there exists satisfying . Moreover, one may take:

 δ=exp(−Cϵ21), (1)

for some absolute constant .

We therefore obtain the following result stated at the abstract:

###### Corollary 1.4.

For any number of alternatives and , there exists a , such that for every , if is a constitution on voters and alternatives satisfying:

• IIA and

• is far from any dictator, so for any dictator ,

• For every pair of alternatives and , the probability that ranks above is at least ,

then the probability of a non-transitive outcome, , is at least , where may be taken as in (1).

###### Proof.

Assume by contradiction that . Then by Theorem 1.3 there exists a function satisfying . Note that for every pair of alternatives and it holds that:

 P[G ranks a above b]≥P[F ranks a % above b]−D(F,G)>0.

Therefore for every pair of alternatives there is a positive probability that ranks above . Thus by Theorem 1.2 it follows that is a dictator which is a contradiction. ∎

###### Remark 1.5.

Note that if and is any constitution satisfying then .

###### Remark 1.6.

The bounds stated in Theorem 1.3 and Corollary 1.4 in terms of and is clearly not an optimal one. We expect that the true dependency has which is some fixed power of . Moreover we expect that the bound should be improved to .

### 1.4 Generalizations and Small Paradox Probability

Theorem 1.3 and Corollary 1.4 extend to more general product distributions. We call a distribution over the permutations of elements , symmetric if for all . We will write for . We will write and for the probability and expected value according to the product measure .

###### Theorem 1.7.

Theorem 1.3 and Corollary 1.4 extend to the following setup where voters vote independently at random according to a symmetric distribution over the permutations of elements. In this setup it suffices to take

 δ=exp(−C1αϵC2(α)), (2)

where . In particular one may take .

The dependency of on in (1) and (2) is a bad one. For values of it is possible to obtain better dependency, where is polynomial in . In Section 4 we prove the following.

###### Theorem 1.8.

Consider voting on alternatives where voters vote uniformly at random from . Let

 1324>ϵ>0. (3)

For every , if is a constitution on voters satisfying:

• IIA and

•  P(F)<136ϵ3n−3, (4)

then there exists satisfying . If each voter follows a symmetric voting distribution then with minimal probability then the same statement holds where (3) is replaced with and (4) is replaced with

 P(F)<α2ϵ−12αn−3.

### 1.5 Related Work

The first attempt at getting a quantitative version of Arrow’s theorem is Theorem 1.2 in a beautiful paper by Kalai [10] which we state in our notation as follows.

###### Theorem 1.9.

There exists a such that the following holds: Consider voting on alternatives where voters vote uniformly at random from . Assume is a balanced constitution, i.e., for every pair of alternatives, it holds that the probability that ranks above is exactly . Then if , then for some dictator .

Comparing Kalai’s result to Theorem 1.3 we see that

• Kalai obtains better dependency of in terms of .

• Kalai’s result holds only for alternatives, while ours hold for any number of alternatives.

• Kalai’s result holds only when is balanced while ours hold for all .

The approach of [10] is based on ”direct” manipulation of the Fourier expression for probability of paradox. A number of unsuccessful attempts (including by the author of the current paper) have been made to extend this approach to a more general setup without assuming balance of the functions and to larger number of alternatives.

A second result of [10] proves that for balanced functions which are transitive the probability of a paradox is bounded away from zero. Transitivity is a strong assumption roughly meaning that all voters have the same power. We do not assume transitivity in the current paper. A related result [14, 15] proved a conjecture of Kalai showing that among all balanced low influence functions, majority minimizes the probability of a paradox. The low influence condition is weaker than transitivity,but still requires that no single voter has strong influence on the outcome of the vote.

Keller [11] extended some of Kalai’s result to symmetric distributions (still under the balance assumption). Keller [11] also provides lower bounds on the probability of a paradox in the case the functions are monotone and balanced.

We want to note of some natural limitation to the approach taken in [10] and [11] which is based on ”direct” analysis of the probability of a paradox in terms of the Fourier expansion. First, this approach does not provide a proof of Arrow theorem nor does it ever use it (while our approach does). Second, it is easy to see that one can get small paradox probability by looking at constitutions on alternatives which almost always rank one candidates at the top. Thus a quantitative version of Arrow theorem cannot be stated just in terms of distance to a dictator. Indeed an example in [11] (see Theorem 1.2) implies that for non-balanced functions the probability of a paradox cannot be related in a linear fashion to the distance from dictator or to other functions in .

As noted in [10], there is an interesting connection between quantitative Arrow statements and the concept of testing introduced in [18, 9] which was studied and used extensively since. Roughly speaking a property of functions is testable if it is possible to perform a randomized test for the property such that if the probability that the function passes the test is close to , then the function has to be close to a function with the property (say in the hamming distance). In terms of testing, our result states that among all functions satisfying the IIA property, the Transitivity property is testable. Moreover, the natural test ”works”: i.e., in order to test for transitivity, one can pick a random input and check if the outcome is transitive.

We finally want to note that the special case of the quantitative Arrow theorem proved by Kalai [10] for balanced functions has been used to derive the first quantitative version of the Gibbard-Satterthwaite Theorem [8, 19] in [7]. The results of [7] are limited in the sense that they require neutrality and apply only to 3 candidates. It is interesting to explore if the full quantitative version of Arrow theorem proven here will allow to obtain stronger quantitative version of the Gibbard-Satterthwaite Theorem.

### 1.6 Proof Ideas

We first recall the notion of influence of a voter. Recall that for , the influence of voter is given by

 Ii(f)=P[f(X1,…,Xi−1,0,Xi+1,…,Xn)≠f(X1,…,Xi−1,1,Xi+1,…,Xn)],

where are distributed uniformly at random. The notion of influence is closely related to the notion of pivotal voter which was introduced in Barabera’s proof of Arrow’s Theorem [3]. Recall that voter is pivotal for at if . Thus the influence of voter is the expected probability that voter is pivotal.

We discuss the main ideas of the proof for the case . By the IIA property that pairwise preference and are decided by three different functions and depending on the pairwise preference of the individual voters.

• The crucial and novel step is showing that for every , there exists , such that if two different voters satisfy and , then the probability of a non-transitive outcome is at least , for some . The proof of this step uses and generalizes the results of [16], which are based on inverse-hyper-contractive estimates [5]. We show that if and then with probability at least , over all voters but and , the restricted and , have and pivotal. We show how this implies that with probability we may chose the rankings of and , leading to a non-transitive outcome. And therefore the probability of a paradox is at least . This step may be viewed as a quantitative version of a result by Barbera [3]. The main step in Barbera’s proof of Arrow theorem is proving that if two distinct voters are pivotal for two different pairwise preferences that the constitution has a non-rational outcome.

• The results above suffice to establish a quantitative Arrow theorem for . This follows from the fact that all influences of a function are bounded by then the function is close to a constant function. The probability of paradox obtained here is of order .

• Next, we show that the statement of the theorem holds when is large and all functions are symmetric threshold functions. Note that in this case, since symmetric thresholds functions are low influence functions, the conclusion of the theorem reads: if non of the alternatives is ranked at top/bottom with probability , then the probability of a paradox is at least .

• Using the Majority is stablest result [15] (see also [14]) in the strong form proven in [13] (see also [12]) we extend the result above as long as for any pair of functions say there exist no variable for which both and is large.

• The remaining case is where there exists a single voter , such that is large for at least two of the functions and all other variables have low influences. By expanding the paradox probability in terms of the possible ranking of voter and using the previous case, we obtain the conclusion of the theorem, i.e., that in this case either there is a non-negligible probability of a paradox, or the function close to a dictator function on voter .

Some notation and preliminaries are given in Section 2. The proof for the case where two different functions have two different influential voters is given in Section 3. This already allows to establish a quantitative Arrow theorem in the case where the functions is very close to an element of in Section 4. The proof of the Gaussian Arrow Theorem is given in Section 5. Applying ”strong” non-linear invariance the result is obtained for low influence functions in Section 6. The result with one influential variable is the derived in Section 7. The proof of the main result for alternatives is then given in Section 8. Section 9 concludes the proof by deriving the proof for any number of alternatives. The combinatorial Theorem 1.2 is proven in Section 10. Section 11 provides the adjustment of the proofs needed to obtain the results for symmetric distributions.

### 1.7 Acknowledgement

Thanks to Marcus Issacson and Arnab Sen for interesting discussions. Thanks to Salvador Barbera for helpful comments on a manuscript of the paper.

## 2 Preliminaries

For the proof we introduce some notation and then follow the steps above.

### 2.1 Some Notation

The following notation will be useful for the proof. A social choice function is a function from a profile on permutation, i.e., an element of to a binary decision for every pair of alternatives which one is preferable. The set of pairs of candidates is nothing but . Therefore a social choice function is a map where means

 F ranks a above b if ha>b(σ)=1,F % ranks b above a if ha>b(σ)=−1.

We will further use the convention that .

The binary notation above is also useful to encode the individual preferences as follows. Given we define binary vectors in the following manner:

 xa>b(i)=1,if voter i ranks a above b;xa>b(i)=−1,if voter i ranks a above b

The IIA condition implies that the pairwise preference between any pair of outcomes depends only on the individual pairwise preferences. Thus, if satisfies the IIA property then there exists functions for every pair of candidates and such that

 F(σ)=((fa>b(xa>b):{a,b}∈(k2))

We will also consider more general distributions over . We call a distribution on symmetric if for all . We will write for .

### 2.2 The Correlation Between xa>b and xb>c

For some of the derivations below will need the correlations between the random variables and . We have the following easy fact:

###### Lemma 2.1.

Assume that voters vote uniformly at random from . Then:

1. For all and all the variables and are independent.

2. If are distinct then .

For the proof of part 2 of the Lemma, note that the expected value depends only on the distribution over the rankings of which is uniform. It thus suffices to consider the case . In this case there are permutations where and permutations where .

We will also need the following estimate

###### Lemma 2.2.

Assume that voters vote uniformly at random from . Let and let . Then

 |Tf|2=1/√3.
###### Proof.

There are two permutations where determine . For all other permutations is equally likely to be and conditioned on and . We conclude that and therefore . ∎

### 2.3 Inverse Hyper-contraction and Correlated Intersections Probabilities

We will use some corollaries of the inverse hyper-contraction estimates proven by Borell [4]. The following corollary is from [16].

###### Lemma 2.3.

Let be distributed uniformly and are independent. Assume that for all and that . Let be two sets and assume that

 P[B1]≥e−α2,P[B2]≥e−β2.

Then:

 P[x∈B1,y∈B2]≥exp(−α2+β2+2ραβ1−ρ2).

We will need to generalize the result above to negative and further to different value for different bits.

###### Lemma 2.4.

Let be distributed uniformly and are independent. Assume that for all and that . Let be two sets and assume that

 P[B1]≥e−α2,P[B2]≥e−β2.

Then:

 P[x∈B1,y∈B2]≥exp(−α2+β2+2ραβ1−ρ2).

In particular if and , then:

 P[x∈B1,y∈B2]≥ϵ21−ρ. (5)
###### Proof.

Take so that are independent and and . It is easy to see there exists independent of with s.t. the joint distribution of is the same as , where . Now for each fixed we have that

 P[x∈B1,z⋅w∈B2]=P[x∈B1,z∈w⋅B2]≥exp(−α2+β2+2ραβ1−ρ2),

where . Therefore taking expectation over we obtain:

 P[x∈B1,y∈B2]=EP[x∈B1,z⋅w∈B2]≥exp(−α2+β2+2ραβ1−ρ2)

as needed. The conclusion (5) follows by simple substitution (note the difference with Corollary 3.5 in [16] for sets of equal size which is a typo). ∎

Applying the CLT and using [5] one obtains the same result for Gaussian random variables.

###### Lemma 2.5.

Let be with independent. Assume that . Let be two sets and assume that

 P[B1]≥e−α2,P[B2]≥e−β2,

Then:

 P[N∈B1,M∈B2]≥exp(−α2+β2+2ραβ1−ρ2).

In particular if and , then:

 P[N∈B1,M∈B2]≥ϵ21−ρ. (6)
###### Proof.

Fix the values of and and assume without loss of generality that is obtained for . Then by [5] (see also [13]), the minimum of the quantity under the constraints on the measures given by and is obtained in one dimension, where and are intervals . Look at random variables , where and . Let and . Then the CLT implies that

 P[Xn∈I1]→P[N1∈B1],P[Yn∈I2]→P[M1∈B2],

and

 P[Xn∈I1,Yn∈I2]→P[N1∈B1,M1∈B2].

The proof now follows from the previous lemma. ∎

## 3 Two Influential Voters

We begin the proof of Arrow theorem by considering the case of candidates named and two influential voters named and . Note that for each voter there are legal values for . These are all vector different from and . Similarly constitution given by and has a non-transitive outcome if and only if

 (fa>b(xa>b),fb>c(xb>c),fc>a(xc>a))∈{(−1,−1,−1),(1,1,1)}.

### 3.1 Two Pivots Imply Paradox

We will use the following Lemma which as kindly noted by Barbera was first proven in his paper [3].

###### Proposition 3.1.

Consider a social choice function on candidates and and two voters denoted and . Assume that the social choice function satisfies that IIA condition and that voter is pivotal for and voter is pivotal for . Then there exists a profile for which is non-transitive.

For completeness we provide a proof using the language of the current paper (the proof of [3] like much of the literature on Arrow’s theorem uses binary relation notation).

###### Proof.

Since voter is pivotal for and voter is pivotal for there exist such that

 fa>b(0,y)≠fa>b(1,y),fb>c(x,0)≠fb>c(x,1).

Look at the profile where

 xa>b=(x∗,y),xb>c=(x,y∗),xc>a=(−x,−y).

We claim that for all values of this correspond to transitive rankings of the two voters. This follows from the fact that neither nor belong to the set . Note furthermore we may chose and such that

 fc>a(−x,−y)=fa>b(x∗,y)=fb>c(x,y∗).

We have thus proved the existence of a non-transitive outcome as needed. ∎

### 3.2 Two influential Voters Implies Joint Pivotality

Next we establish the following result.

###### Lemma 3.2.

Consider a social choice function on candidates and and voters denoted . Assume that the social choice function satisfies that IIA condition and that voters vote uniformly at random. Assume further that and . Let

 B={σ:1 is pivotal for fa>b(xa>b(σ)) and 2% is pivotal for fb>c(xb>c(σ))}.

Then

 P[B]≥ϵ3.
###### Proof.

Let

 B1={σ:1 is pivotal for fa>b},B2={σ:2 % is pivotal for fb>c}.

Then and , and our goal is to obtain a bound on . Note that the event is determined by and the event is determined by . Further by Lemma 2.1 it follows that and The proof now follows from Lemma 2.4. ∎

### 3.3 Two Influential Voters Imply Non-Transitivity

We can now prove the main result of the section.

###### Theorem 3.3.

Let and . Consider the uniform voting model on . Let be a constitution on voters satisfying:

• IIA and

• There exists three distinct alternatives and and two distinct voters and such that

 Ii(fa>b)>ϵ,Ij(fb>c)>ϵ.

then .

###### Proof.

We look at restricted to rankings of and . Note that in the uniform case each permeation has probability Without loss of generality assume that and and consider first the case of the uniform distribution over rankings. let be the event from Lemma 3.2. By the lemma we have . Note that if satisfies that , then fixing we may apply Proposition 3.1 to conclude that there are values of and leading to a non-transitive outcome. Therefore:

 P[P(F)]≥P[(σ∗(1),σ∗(2),σ(3),…,σ(n)):σ∈B]≥136P[B]≥136ϵ3.

## 4 Arrow Theorem for Almost Transitive Functions

In this section we prove a quantitative Arrow Theorem in the case where the probability of a non-transitive outcome is inverse polynomial in . In this case it is possible to obtain an easier quantitative proof which does not rely on invariance. We will use the following easy and well known Lemma.

###### Lemma 4.1.

Let and assume for all . Then there exist a constant function such that .

Similarly, let and assume for all . Then there exists a function such that .

###### Proof.

For the first claim, use

 12min(P[f=1],P[f=−1])≤P[f=1]P[f=−1]=Var[f]≤n∑i=1Ii(f)≤ϵ. (7)

For the second claim assume WLOG that . Let and . Apply (7) to chose so that

 D(f1,s1)≤∑i>1Ii(f1).

Similarly, let be chosen so that

 D(f,s−1)≤∑i>1Ii(f−1).

Let and . Then:

 2D(f,g)=D(f1,s1)+D(f−1,s−1)≤∑i>1Ii(f1)+∑i>1Ii(f−1)=2∑i>1Ii(f)≤2ϵ.

The proof follows. ∎

###### Theorem 4.2.

Consider voting on alternatives where voters vote uniformly at random from . Let

 1324>ϵ>0. (8)

For every , if is a constitution on voters satisfying:

• IIA and

•  P(F)<136ϵ3n−3, (9)

then there exists satisfying .

###### Proof.

We prove the theorem for the uninform case. The proof for the symmetric case is identical. Let be the three pairwise preference functions. Let .

Consider three cases:

1. Among the functions , there exist two different functions and and s.t: and .

2. There exists a voter such that for all and all , it holds that .

3. There exists two different functions such that for all it holds that and .

Note that each satisfies one of the three conditions above. Note further that in case I. we have by Theorem 3.3 which contradicts the assumption (9). So to conclude the proof is suffices to obtain assuming (9).

In case II. it follows from Lemma 4.1 that there exists functions and of voter only such that

 D(fa>b,ga>b)<2ϵ,D(fb>c,gb>c)<2ϵ,D(fc>a,gc>a)<2ϵ.

Letting be the constitution defined by the ’s we therefore have and .

Furthermore if this implies that . So which is a contradiction.

In the remaining case III. assume WLOG that and have all influences small. By Lemma 4.1 if follows that and are far from a constant function. There are now two subcases to consider. In the first case there exists an such that and . Note that in the case letting

 ga>b=s,gb>c=−s,gc>a=fc>a,

and be the constitution defined by the ’s, we obtain that and .

We finally need to consider the case where and for some . Let be the set of where and similarly for and . Then and . Furthermore by transitivity

 P[A(a,c)]≤P[A(a,b)]+P[A(b,c)]+P(F)≤6ϵ.

We thus conclude that . Letting and the constitution defined by we have that . A contradiction. The proof follows

It is now easy to prove Theorem 1.8 for the uniform distribution. The adaptations to symmetric distributions will be discussed in Section 11.

###### Proof.

The proof follows by applying Theorem 4.2 to triplets of alternatives. We give the proof for the uniform case. Assume .

Note that if are two different function each of which is either a dictator or a constant function than . Therefore for all it holds that for at most one function which is either a dictator or a constant function. In case there exists such function we let , otherwise, we let .

Let be the social choice function defined by the functions . Clearly:

 D(F,G)<10(k2)ϵ<10k2ϵ.

The proof would follow if we could show and therefore .

To prove that is suffices to show that for every set of three alternatives, it holds that . Since , Theorem 4.2 implies that there exists a function s.t. . There are two cases to consider:

• is a dictator. This implies that is close to a dictator for each and therefore for all pairs , so .

• There exists an alternative (say ) that always ranks at the top/bottom. In this case we have that and are at most far from the constant functions and (or and ). The functions and have to take the same constant values and therefore again we have that .

The proof follows.

## 5 The Gaussian Arrow Theorem

The next step is to consider a Gaussian version of the problem. The Gaussian version corresponds to a situation where the functions can only ”see” averages of large subsets of the voters. We thus define a dimensional normal vector . The first coordinate of is supposed to represent the deviation of the number of voters where ranks above from the mean. The second coordinate is for ranking above and the last coordinate for ranking above .

Since averaging maintain the expected value and covariances, we define:

 E[N21]=E[N22]=E[N23]=1, (10) E[N1N2]=E[xa>b(1)xb>c(1)]:=−1/3, E[N2N3]=E[xb>c(1)xc>a(1)]:=−1/3, E[N3N1]=E[xc>a(1)xa>b(1)]:=−1/3.

We let be independent copies of . We write and for we write . The Gaussian version of Arrow theorem states:

###### Theorem 5.1.

For every there exists a such that the following hold. Let . Assume that for all and all it holds that

 P[fi(Ni)=u,fi+1(Ni+1)=−u]≤1−ϵ (11)

Then with the setup given in (10) it holds that:

 P[f1(N1)=f2(N2)=f3(N3)]≥δ.

Moreover, one may take .

We note that the negation of condition (11) corresponds to having one of the alternatives at the top/bottom with probability at least . Therefore the theorem states that unless this is the case, the probability of a paradox is at least . Since the Gaussian setup excludes dictator functions in terms of the original vote, this is the result to be expected in this case.

###### Proof.

We will consider two cases: either all the functions satisfy , or there exists at least one function with