Robust Version of Arrow’s Theorem

Pr

ICA[rational outcome] = 1 if and only if W

1_{[f] = 1.}

However, going back to the calculation of the upper bound immediately after Equation (4.1), by examining the numeric values of the coefficients we see that in fact we have something more, namely:

Pr

ICA[rational outcome]≈1 if and only if W

1_[f]≈1.

This can be made precise by using Lemma 4.6, as we will do below.

What does it mean for f : {−1,1}n _{→ {−1,1} to satisfy W}1_{[f] = 1−ε}

with ε >0 small? Does it then necessarily have to be “close” (in the sense of Definition 2.6) to a dictator function? Or is it the case that there are functions which are unlike a dictatorship but that still have degree-one Fourier weight very close to one? This is not an elementary question, but the answer is known: the Friedgut-Kalai-Naor (FKN) Theorem.

Theorem 4.9

(FKN Theorem) Let f : {−1,1}n → {−1,1} with W1[f] ≥ 1−ε.

Then there exists ani∈[n] such that f is O(ε)-close to χi or −χi.

To make the statement more precise: given an f and someε, “f is O(ε)-close to χi or −χi” means that there is a universal constant M > 0 such that

dist(f,±χi)≤Mε.

Before we show the FKN Theorem using Theorem 3.19, it is worth noting that for any Boolean-valued function h:{−1,1}n_{→ {−1,1} we have}

CHAPTER 4. ARROW’S THEOREM 64

if and only if

|bh(i)| ≥1−O(ε).

This is easy to see using Theorem 2.8 and Equation (2.1). We will use this observation in what follows.

Proof of Theorem 4.9. Without loss of generality we may assume W1[f] = 1−ε.

To make the below proof easier, we first show that without loss of generality we may assume that f does not have a constant coefficient in its Fourier expansion, i.e., W0_{[f] = 0. So suppose f : {−1,1}}n _{→ {−1,1} satisfies}

W1_{[f] = 1−ε. Consider the functionf}0 _:{−1,1}n+1 _{→ {−1,1} defined by}

f0_(x

0, x1, . . . , xn)

def

=x0f(x0x1, . . . , x0xn).

If the Fourier expansion of f is b

f(∅) +fb({1})x₁+. . .+fb({n})x_n+fb({1,2})x₁x₂+. . . ,

then by looking at the definition off0 _{we see that the Fourier expansion of f}0

is of the form b

f(∅)x0+fb({1})x₁+. . .+fb({n})x_n+fb({1,2})x₀x₁x₂+. . . . Note that here we have used the fact that x2

0 = 1 for any x0 ∈ {−1,1}. We

did not write the higher-order terms since they are not relevant for us. It is important to notice thatf0 _{does not have a degree-zero coefficient so, assuming}

the theorem has been proven for such functions, from W1_[f0_{] =}

b

f(∅)2_{+ W}1_[f]≥W1_{[f] = 1−ε}

we deduce thatf0 _{isO(ε)-close toχ}

i or−χi, for somei∈[n]. By the comment

preceding this proof, |fb0(i)| ≥1−O(ε) for some i∈ {0} ∪[n]. However since b

f0_{(0) =} b

f(∅) and W1_{[f] = 1−ε, it cannot be thati= 0. Thus, i∈[n] and}

|fb0(i)| ≥1−O(ε); but from the Fourier expansions we directly read off that b

f0_{(i) =} b

f(i), so by again applying the comment preceding this theorem we are done.

The previous argument shows that we can assume without loss of generality thatf does not have a constant coefficient in its Fourier expansion. Therefore f is of the form f(x) = n X i=1 b f(i)xi+ X |S|>1 b f(S)xS. By hypothesis,Pn

i=1fb(i)2 = 1−ε, so Parseval’s Theorem implies P |S|>1fb(S)2 = ε. Let us define g, h:{−1,1}n_→ R by g(x) def = n X i=1 b f(i)xi, h(x) def = X |S|>1 b f(S)xS,

CHAPTER 4. ARROW’S THEOREM 65

so that f =g+h.

Since the square of a Boolean-valued funtion is always 1, we have f2 _{= 1,}

where 1 denotes the constant one function. Thus, (g+h)2_{= 1, so}

g2_{+ 2gh+h}2 _{= 1,}

g2_{+h(2g+h) = 1.}

Since f =g+h this can be rewritten as

g2_{+h(2f −h) = 1. (4.3)}

We now consider these two terms separately. • First we analyze the termg2_{. We have}

g(x)2₌ n X i=1 b f(i)2_x2 i + X i6=j b f(i)fb(j)x_ix_j = n X i=1 b f(i)2₊X i6=j b f(i)f(j)xb _ix_j, so g2_{= 1−ε+q (4.4)} where q:{−1,1}n_→

Ris defined as q(x)def= Pi6=jfb(i)fb(j)x_ix_j.

• Second we analyze the termh(2f−h). Ash does not have a constant coefficient in its Fourier expansion, from of the comment after Definition 2.12 we get Ex[h(x)] = 0; also, Ex[h(x)2] =εby Parseval’s Theorem, so

we have Var[h(x)] =ε. Chebychev’s inequality implies that Pr[|h(x)| ≥10√ε]≤ ₁₀1₂ = ₁₀₀1 .

So with probability at least 99% we have

|h(x)(2f(x)−h(x))| ≤ |h(x)|(2|f(x)|+|h(x)|)<10√ε(2 + 10√ε). Notice that in the term 10√ε(2+10√ε) = 20√ε+100ε, the nonnegligible one is 20√ε. It is therefore easy to see that for εsmall enough we have that the this term is smaller than or equal to 21√ε. To be very concrete: letting ε0 _{=ε/10000, sinceε <1 this bound will hold.}

From Equation (4.3) and Equation (4.4) we deduceh(2f−h) =ε−q. Therefore we obtain that |ε−q(x)|=|h(x)(2f(x)−h(x))| ≤21√εwith probability at least 99%. Since it holds thatε+ 21√ε≤22√ε, using the triangle inequality we obtain

Pr[|q(x)| ≤22√ε]≥ ₁₀₀99 . (4.5) The idea is now to find an upper bound for E[q(x)2_{] in the form of a (possible}

CHAPTER 4. ARROW’S THEOREM 66

“innocent” form being a multilinear polynomial composed of random bits x1, . . . , xn, this will be possible; here is where the Hypercontractive Inequality

will come into play.

Claim. Let q(x1, . . . , xn) be a random variable which is of multilinear poly-

nomial form, where x1, . . . , xn are independent and uniformly distributed

on {−1,1}. Then if we have Pr[|q(x)| ≤ 22√ε] ≥ 99/100, it follows that E[q(x)2_]≤5000ε.

We prove this claim. First, it is clear we may assume without any loss of generality that Pr[|q(x)| ≤ 22√ε] = 99

100. Let us say E[q(x)

2_{] =Kε. Taking}

conditional expectations, we have Kε= E[q2_]≤ 99

100(22 √

ε)2₊ 1

100E[q2|q2 >484ε].

By the Hypercontractive Inequality (Theorem 3.19) we have E[q4_]≤92_E[q2_]2₌

92_K2_ε2_{.Therefore 9}2_K2_ε2 _{is at least} E[q4_]≥ 1 100E[q4|q2 >484ε]≥ 1 100E[q2|q2>484ε]2 ≥ 1 100(100K−99·484)2ε2. Analyzing the quadratic equation inKgives us a bound: we getK ∈]252,4791[. In particular, K≤5000. This ends the proof of the claim.

Now it is quite easy to finish the proof of the FKN Theorem. By Parseval’s Theorem we get E[q(x)2_{] =}X i6=j b f(i)2 b f(j)2 ₌ n X i=1 b f(i)22₋ n X i=1 b f(i)4_{= (1−ε)}2₋ n X i=1 b f(i)4_.

By the above claim we get

n X i=1 b f(i)4 _≥(1−ε)2_{−5000ε= 1−(5002ε−ε}2_{) = 1−O(ε).} Also, Pn

i=1fb(i)4 is at most max_jfb(j)2 Pn

i=1fb(i)2. But this is just equal to maxjf(j)b 2, by Parseval’s Theorem. Thus there is an i ∈ [n] such that

b

f(i)2 _{≥1−O(ε).}

There are two possibilities. If fb(i) ≥ p

1−O(ε), then by Equation (2.1) we have dist(f, χi) = 1− hf, χii 2 = 1−f(i)b 2 ≤ 1−p 1−O(ε) 2 .

By a Taylor expansion we know √1−δ = 1−1

2δ+O(δ2) for each−1< δ <1,

so we get dist(f, χi)≤ 1₄O(ε) =O(ε). Iffb(i)≤ − p

1−O(ε), we can similarly deduce dist(f,−χi)≤O(ε). We conclude thatf isO(ε)-close to either χi or

CHAPTER 4. ARROW’S THEOREM 67

Finally we get the robust version of Arrow’s Theorem: Theorem 4.10

(Robust Version of Arrow’s Theorem) Let f : {−1,1}n → {−1,1} be unanimous. Furthermore, suppose f is such that when doing a 3-candidate Condorcet election based on f, the probability that the outcome is rational under the impartial culture assumption is 1−ε. Then there exists ani∈[n] such that f is O(ε)-close to χi.

Proof. From Lemma 4.6 we get 1−ε≤ 7₉+2₉W1[f], i.e., W1[f]≥1−9₂ε. The

FKN Theorem then implies that there is ani∈[n] such thatf isO(ε) =O(9 2ε)-

close to χi or−χi. However, unanimity of f implies that the latter of these

two possibilities is not viable.

Our proof is based on O’Donnell [97], which is in turn based on Kalai [68].