Flexible level-1 consensus ensuring stable social choice: analysis and algorithms

Abstract

Level-1 consensus is a recently-introduced property of a preference-profile. Intuitively, it means that there exists a preference relation which induces an ordering of all other preferences such that frequent preferences are those that are more similar to it. This is a desirable property, since it enhances the stability of social choice by guaranteeing that there exists a Condorcet winner and it is elected by all scoring rules. In this paper, we present an algorithm for checking whether a given preference profile exhibits level-1 consensus. We apply this algorithm to a large number of preference profiles, both real and randomly-generated, and find that level-1 consensus is very improbable. We support these empirical findings theoretically, by showing that, under the impartial culture assumption, the probability of level-1 consensus approaches zero when the number of individuals approaches infinity. Motivated by these observations, we show that the level-1 consensus property can be weakened while retaining its stability implications. We call this weaker property Flexible Consensus. We show, both empirically and theoretically, that it is considerably more probable than the original level-1 consensus. In particular, under the impartial culture assumption, the probability for Flexible Consensus converges to a positive number when the number of individuals approaches infinity.

Appendices

Appendices

Appendix A: To verify Condition 1, it is sufficient to check adjacent preferences

This section provides a formal proof to the following lemma used in Sect. 3.

Lemma 1 Suppose that the preference relations in a profile $\pi$ are ordered by two criteria: first by frequency $\mu _\pi (\succ _{i})$ , then by distance $d(\succ _{i},\succ _0)$ , where $\succ _0$ is a fixed preference. In this ordering, if Condition 1 is violated for any pair of preference relations in $\pi$, then it is violated for an adjacent pair $\succ _{i},\succ _{i+1}$ for some i.

Proof

Suppose that there exist indices $i<j$ such that Condition 1 is violated for the pair $\succ _i,\succ _j$, i.e, $\mu _\pi (\succ _{i})>\mu _\pi (\succ _{j})$, yet $d(\succ _{i},\succ _0)\ge d(\succ _{j},\succ _0)$. We now prove the lemma by induction on the difference of indices, $j-i$.

Base: If $j-i=1$, then $\succ _i$ and $\succ _j$ are already adjacent, so we are done.

Step: Suppose that $j-i>1$. We prove that there exists a pair with a smaller difference that violates Condition 1. We consider several cases.

Case 1 :

i is not the largest index in its equivalence class. i.e, there exists $i'> i$ such that $\mu _\pi (\succ _{i'})=\mu _\pi (\succ _{i})$. Then, by the secondary ordering criterion, $d(\succ _{i'},\succ _0)\ge d(\succ _{i},\succ _0)$, Condition 1 is violated for the pair $\succ _{i'}$ and $\succ _{j}$.

Case 2 :

j is not the smallest index in its equivalence class. i.e, there exists $j'<j$ such that $\mu _\pi (\succ _{j'})=\mu _\pi (\succ _{j})$. Then, by the secondary ordering criterion, $d(\succ _{j},\succ _0)\ge d(\succ _{j'},\succ _0)$, Condition 1 is violated for the pair $\succ _{i}$ and $\succ _{j'}$. Otherwise, i is the largest index in its equivalence class, j is the smallest index in its equivalence class, but still $i+1<j$. This means that the equivalence classes of i and j are not adjacent, i.e, $\mu _\pi (\succ _{i})>\mu _\pi (\succ _{i+1})>\mu _\pi (\succ _{j})$. Now there are two remaining cases:

Case 3 :

$d(\succ _{i},\succ _0)\ge d(\succ _{i+1},\succ _0)$, in which case Condition 1 is violated for the adjacent pair $\succ _{i}$ and $\succ _{i+1}$ and we are done.

Case 4 :

$d(\succ _{i+1},\succ _0) > d(\succ _{i},\succ _0)$. This implies $d(\succ _{i+1},\succ _0) > d(\succ _{j},\succ _0)$, so Condition 1 is violated for the pair $\succ _{i+1}$ and $\succ _{j}$ and we are done. $\square$

Appendix B: The effect of a switch on the inversion-distance

This section provides a formal proof to an intuitive claim made within the proof of Lemma 3. Let a, b be two fixed alternatives. Let $C(a>b)$ be the set of preferences by which $a\succ b$ and $C(b>a)$ the set of preferences by which $b\succ a$. Let $w^{ab}:C(b>a)\rightarrow C(a>b)$ be a function that takes a preference-relation and creates a new preference-relation by switching the position of a and b in the ranking.

Lemma 5

If $a \succ _0 b$, then for every preference $\succ _1\in C(b>a)$:

$$\begin{aligned} d(\succ _1,\succ _0) ~ > ~ d(w^{ab}(\succ _1 ),\succ _0) \end{aligned}$$

Proof

For every preference $\succ$, define $D(\succ ,\succ _0)$ as the set of pairs-of-alternatives $\{i,j\}$ that are ranked differently in $\succ$ and in $\succ _0$. By definition, the inversion distance is the cardinality of this set, $d(\succ ,\succ _0) = |D(\succ ,\succ _0)|$. Therefore, it is sufficient to show that there are more pairs in $D(\succ _1,\succ _0)$ than in $D(w^{ab}(\succ _1),\succ _0)$. To show this, we consider all possible pairs-of-alternatives; for each pair, we calculate its contribution to the difference in cardinalities $|D(\succ _1,\succ _0)|-|D(w^{ab}(\succ _1),\succ _0)|$, and show that the net difference is positive.

• The pair $\{a,b\}$ is in $D(\succ _1,\succ _0)$ but not in $D(w^{ab}(\succ _1),\succ _0)$, so this pair contributes $+1$ to the difference.

• Any pair that contains neither a nor b is not affected by the switch. I.e, each pair $\{c,e\}$ where $c,e\ne a,b$ is in $D(\succ _1,\succ _0)$ if-and-only-if it is in $D(w^{ab}(\succ _1),\succ _0)$, so it contributes 0 to the difference.

• Let c be an alternative that is ranked by $\succ _1$ above b or below a, i.e, either $c\succ _1 b\succ _1 a$ or $b\succ _1 a\succ _1 c$. Then, the order between c to a or b in $\succ _1$ is not affected by the switch, so $\{c,a\}$ is in $D(\succ _1,\succ _0)$ if-and-only-if it is in $D(w^{ab}(\succ _1),\succ _0)$, so it contributes 0 to the difference. The same is true for $\{c,b\}$.

• Finally, let c be an alternative that is ranked by $\succ _1$ between a and b, i.e, $b\succ _1 c\succ _1 a$. Then, the switch $w^{ab}$ changes the direction of both the pair $\{c,a\}$ and the pair $\{c,b\}$:

  • If $c \succ _0 a \succ _0 b$, then the pair $\{c,a\}$ is in $D(\succ _0,w^{ab}(\succ _1))$ but not in $D(\succ _0,\succ _1)$, and the pair $\{c,b\}$ is in $D(\succ _0,\succ _1)$ but not in $D(\succ _0,w^{ab}(\succ _1))$, so these pairs contribute $+1-1=0$ to the difference.

  • If $a \succ _0 c \succ _0 b$, then both the pair $\{c,a\}$ and the pair $\{c,b\}$ are in $D(\succ _0,\succ _1)$ but not in $D(\succ _0,w^{ab}(\succ _1))$, so these pairs contribute $+1+1=2$ to the difference.

  • If $a \succ _0 b \succ _0 c$, then the pair $\{c,a\}$ is in $D(\succ _0,\succ _1)$ but not in $D(\succ _0,w^{ab}(\succ _1))$, and the pair $\{c,b\}$ is in $D(\succ _0,w^{ab}(\succ _1))$ but not in $D(\succ _0,\succ _1)$, so these pairs contribute $+1-1=0$ to the difference.

We proved that the contribution of each pair of alternatives is at least 0, and the contribution of the pair $\{a,b\}$ is $+1$. Therefore, the difference $|D(\succ _1,\succ _0)|-|D(w^{ab}(\succ _1),\succ _0)|$ is positive and the lemma is proved. $\square$

Appendix C: Probability that binomial random variables are equal

This section justifies the following approximation, used in Sect. 4.

Let $B_1,\ldots ,B_t$ be i.i.d. random variables distributed binomially with m trials and success-probability $p < 1/2$. Let $q:=1-p$. Define:

$$\begin{aligned} P^{\text {equal}}(m,p,t) := \Pr [B_1 = \cdots = B_t] \end{aligned}$$

Then, for every $t\ge 1$, when m is sufficiently large,

$$\begin{aligned} P^{\text {equal}}(m,p,t)&\approx {1 \over (\sqrt{2\pi p q m})^{t-1} } \end{aligned}$$

Proof

The value of each of the variables $B_i$ can be any integer between 0 and m. Therefore we can present $P^{\text {equal}}$ as a sum of probabilities of disjoint events:

$$\begin{aligned} P^{\text {equal}}(m,p,t) = \sum _{i=0}^m \Pr [B_1 = \cdots = B_t = i] \end{aligned}$$

Since the $B_i$ are i.i.d:

$$\begin{aligned} P^{\text {equal}}(m,p,t)&= \sum _{i=0}^m \left( \Pr [B_1 = i]\right) ^t \\&= \sum _{i=0}^m \left( {m\atopwithdelims ()i} p^i q^{m-i}\right) ^t \end{aligned}$$

where $q:=1-p$. Using Stirling’s approximation, when $m,i,m-i$ are sufficiently large:

$$\begin{aligned} {m\atopwithdelims ()i}&\approx \sqrt{m\over 2 \pi i (m-i)}\cdot {m^m\over i^i (m-i)^{m-i}} \end{aligned}$$

Substitute this in $P^{\text {equal}}$ and approximate the sum by an integral:

$$\begin{aligned} P^{\text {equal}}(m,p,t)&\approx \int _{y=0}^m \left( \sqrt{m\over 2 \pi y (m-y)}\cdot {m^m\over y^y (m-y)^{m-y}}\cdot p^y q^{m-y}\right) ^t dy \end{aligned}$$

Substitute $y = m x$ and $dy = m dx$:

$$\begin{aligned} P^{\text {equal}}(m,p,t)&\approx m\int _{x=0}^1 \left( \sqrt{1 \over 2 \pi m x (1-x)}\cdot {p^{m x} q^{m-mx}\over x^{m x} (1-x)^{m-mx}}\right) ^t dx \\&= m^{1-{t/ 2}}\int _{x=0}^1 \left( \sqrt{1 \over 2 \pi x (1-x)}\cdot {p^{m x} q^{m-mx}\over x^{m x} (1-x)^{m-mx}}\right) ^t dx \end{aligned}$$

The integral can be approximated by Laplace’s method. Define:

$$\begin{aligned} h(x)&:= \left( \sqrt{1 \over 2 \pi x (1-x)}\right) ^t \\ g(x)&:= t \cdot \bigg [ x \ln ({p\over x}) + (1-x) \ln ({q\over 1-x}) \bigg ] \end{aligned}$$

Then:

$$\begin{aligned} P^{\text {equal}}(m,p,t)&\approx m^{1-{t / 2}} \int _{x=0}^1 h(x) e^{m g(x)} dx \end{aligned}$$

The function g(x) is twice continuously differentiable on (0, 1) and has a unique maximum at $x_0={p\over p+q} = p$; the maximum value is $g(x_0) = 0$. Moreover, $g''(x_0) = -{1 \over p q}< 0$. Therefore, by Laplace’s method:

$$\begin{aligned} P^{\text {equal}}(m,p,t)&\approx m^{1-t/2} \sqrt{ 2\pi \over - m g''(x_0) } \cdot h(x_0) \cdot e^{m g(x_0)} \end{aligned}$$

where the symbol $\approx$ means that the ratio between the expressions in its two sides goes to 1 as $m\rightarrow \infty$. Substituting the functions g and h gives:

$$\begin{aligned} P^{\text {equal}}(m,p,t)&\approx m^{1-t/2} \sqrt{ 2\pi \over m / (p q) } \cdot \left( {1\over 2\pi p q}\right) ^{t/2}\cdot e^0 \\&= m^{(1-t)/2} \cdot (2\pi p q)^{(1-t)/2} \\&= (2\pi p q m)^{(1-t)/2} \\ P^{\text {equal}}(m,p,t)&\approx {1 \over (\sqrt{2\pi p q m})^{t-1} } \end{aligned}$$

$\square$

Note that $P^{\text {equal}}(m,p,1) = 1$, which is trivially true, since a single random variable always equals itself. When $t\ge 2$, $P^{\text {equal}}(m,p,1)\rightarrow 0$ as $m\rightarrow \infty$.