Scoring rules and social choice properties: some characterizations

Abstract

In many voting systems, voters’ preferences on a set of candidates are represented by linear orderings. In this context, scoring rules are well-known procedures to aggregate the preferences of the voters. Under these rules, each candidate obtains a fixed number of points, $s_k$, each time he/she is ranked $k$th by one voter and the candidates are ordered according to the total number of points they receive. In order to identify the best scoring rule to use in each situation, we need to know which properties are met by each of these procedures. Although some properties have been analyzed extensively, there are other properties that have not been studied for all scoring rules. In this paper, we consider two desirable social choice properties, the Pareto-optimality and the immunity to the absolute loser paradox, and establish characterizations of the scoring rules that satisfy each of these specific axioms. Moreover, we also provide a proof of a result given by Saari and Barney (The Mathematical Intelligencer 25:17–31, 2003), where the scoring rules meeting reversal symmetry are characterized. From the results of characterization, we establish some relationships among these properties. Finally, we give a characterization of the scoring rules satisfying the three properties.

Appendix 1: Proofs

1.1 A.1 Pareto-optimality

By using contraposition, Theorem 1 is rewritten in an equivalent form in Theorem 6. Before stating and proving Theorem 6, we give two useful lemmas.

Lemma 1

Let $\varvec{p}$ be a profile. If the candidate $A_i$ is not Pareto-optimal, then $V_{1}^{i}=0$ and there exists another candidate $A_j$ with $V_{k}^{i} \le V_{k-1}^{j}$ for all $k \in \{2,\dots ,m\}$ and $0=V_{q}^{i} < V_{q}^{j}$, where $q=\min \bigl \{k\in \{1,\dots ,m-1\}\;\vert \; V_{k}^{j}\ne 0\bigr \}$. Moreover, $V_{k}^{i} \le V_{k}^{j}$ for all $k \in \{1,\dots ,m-1\}$.

Proof

Given a profile $\varvec{p}$, if $A_i$ is not Pareto-optimal, there exists another candidate $A_j$ that is preferred to $A_i$ by all voters, so $V_{1}^{i}=0$. On the other hand, when a voter places candidate $A_i$ in the $k$th position that voter will have placed $A_j$ in a higher scale position. Therefore, $V_{k}^{i} \le V_{k-1}^{j}$ for all $k \in \{2,\dots ,m\}$. Furthermore, if $q \in \{1,\dots ,m-1\}\,$ is the highest scale position achieved by candidate $A_j$, then $0 = V_{q}^{i} < V_{q}^{j}$. Finally, given that $V_{k-1}^{j} \le V_{k}^{j}$ for all $k \in \{2,\dots ,m-1\}$ and $V_{1}^{i}=0$, we have $V_{k}^{i} \le V_{k}^{j}$ for all $k \in \{1,\dots ,m-1\}$.$\square$

Lemma 2

Let $\varvec{S}\in \fancyscript{S}$. If $\varvec{S}\gg \varvec{0}$, then the scoring rule associated with $\varvec{S}$ is Pareto-optimal.

Proof

Given a profile $\varvec{p}$, we are going to prove that if a candidate $A_{i}$ is not Pareto-optimal, then he/she is not a winning candidate. By Lemma 1, there exist a candidate $A_j$ with $V_{k}^{i} \le V_{k}^{j}$ for all $k \in \{1,\dots ,m-1\}$ and $V_{q}^{i} < V_{q}^{j}$ for some $q \in \{1,\dots ,m-1\}$. Since $\varvec{S}\gg \varvec{0}$, we have $F_{\varvec{S}}(\varvec{V}^i) < F_{\varvec{S}}(\varvec{V}^j)$. Therefore, $A_{i}$ is not a winning candidate.   $\square$

Theorem 6

Let $\varvec{S}\in \fancyscript{S}$. The following conditions are equivalent:

• 1. The scoring rule associated with $\varvec{S}$ is not Pareto-optimal.

• 2. $\min \nolimits _{k\in \{1,\dots ,m-1\}} S_k=0\,$ and

  $$\begin{aligned} \sum _{k=1}^{p-1} (m+k) S_k + \sum _{k=p+1}^{m-1} k S_k \le m, \end{aligned}$$

  where $p=\min \bigl \{k\in \{1,\dots ,m-1\}\;\vert \; S_{k}=0\bigr \}$.

Proof

$1\;\Rightarrow \; 2:$ If the scoring rule associated with $\varvec{S}$ is not Pareto-optimal, then, by Lemma 2, we have $\min \nolimits _{k\in \{1,\dots ,m-1\}} S_k=0$. Let $p=\min \bigl \{k\in \{1,\dots ,m-1\}\;\vert \; S_{k}=0\bigr \}$. Moreover, we can find a profile $\varvec{p}$ of $n$ voters such that one winning candidate, $A_i$, is not Pareto-optimal. By Lemma 1, there exists a candidate $A_j$ with $V_{k}^{i} \le V_{k}^{j}$ for all $k\in \{1,\dots ,m-1\}$. Therefore, $F_{\varvec{S}}(\varvec{V}^i) \le F_{\varvec{S}}(\varvec{V}^j)$. Since $A_i$ is a winning candidate, we have $F_{\varvec{S}}(\varvec{V}^i) = F_{\varvec{S}}(\varvec{V}^j)$; that is,

$$\begin{aligned} \sum _{k=1}^{m-1} S_{k}V_{k}^{i} = \sum _{k=1}^{m-1} S_{k}V_{k}^{j}. \end{aligned}$$

Given that $S_{k}\ge 0$ and $V_{k}^{i} \le V_{k}^{j}$ for all $k\in \{1,\dots ,m-1\}$, the previous equality is satisfied if and only if $S_{k}V_{k}^{i} = S_{k}V_{k}^{j}$ for all $k\in \{1,\dots ,m-1\}$. As $S_{k} > 0$ for all $k<p$, we get $V_{k}^{i} = V_{k}^{j}$ for all $k<p$. If $q=\min \bigl \{k\in \{1,\dots ,m-1\}\;\vert \; V_{k}^{j}\ne 0\bigr \}$, then $0=V_{q}^{i} < V_{q}^{j}$. Therefore, $q\ge p$ and, consequently, $V_{k}^{i} = 0$ for all $k\in \{1,\dots ,p\}$.

On the other hand, since $A_i$ is a winning candidate, we have

$$\begin{aligned} \sum _{k=1}^{m-1}S_k V_{k}^{l} \le \sum _{k=1}^{m-1}S_k V_{k}^{i}, \end{aligned}$$

for all $l\in \{1,\dots ,m\}$. When we add, member to member, all the inequalities we have

$$\begin{aligned} \sum _{l=1}^{m} \sum _{k=1}^{m-1} S_k V_{k}^l = \sum _{k=1}^{m-1} S_k \sum _{l=1}^{m} V_{k}^l = \sum _{k=1}^{m-1} S_k nk = n\sum _{k=1}^{m-1} k S_k \le m \sum _{k=1}^{m-1} S_k V_{k}^i. \end{aligned}$$

Since $V_{k}^{i} = 0$ for all $k\in \{1,\dots ,p\}$, and $V_{k}^{i} \le n$ for all $k\in \{p+1,\dots ,m-1\}$, we have

$$\begin{aligned} n \sum _{k=1}^{m-1} k S_k \le m n \sum _{k=p+1}^{m-1} S_k, \end{aligned}$$

that is,

$$\begin{aligned} \sum _{k=1}^{p-1} k S_k + \sum _{k=p+1}^{m-1} k S_k \le m \sum _{k=p+1}^{m-1} S_k = m\left( 1-\sum _{k=1}^{p-1} S_k\right) . \end{aligned}$$

So, we get

$$\begin{aligned} \sum _{k=1}^{p-1} (m+k) S_k + \sum _{k=p+1}^{m-1} k S_k \le m. \end{aligned}$$

$2\;\Rightarrow \; 1:$ Let $\varvec{S}\in \fancyscript{S}$ such that $\min \nolimits _{k\in \{1,\dots ,m-1\}} S_k=0$ and $\sum \nolimits _{k=1}^{p-1} (m+k) S_k + \sum \nolimits _{k=p+1}^{m-1} k S_k \le m$, where $p=\min \bigl \{k\in \{1,\dots ,m-1\}\;\vert \; S_{k}=0\bigr \}$. Consider $m-2$ voters and let $\varvec{p}$ be a profile obtained by considering the forward cyclic list of orders ⁸ generated by $A_{3}\,A_{4}\cdots A_{m}$ and by placing candidates $A_{1}$ and $A_{2}$ in the $p$th and $(p+1)$th positions, respectively; that is,

$$\begin{aligned}&1 \hbox {voter}: A_{3}A_{4}\ldots A_{p+1}A_{1}A_{2}A_{p+2}\ldots A_{m-1}A_{m}\\&1 \hbox {voter}: A_{4}A_{5}\ldots A_{p+2}A_{1}A_{2}A_{p+3}\ldots A_{m}A_{3}\\&1 \hbox {voter}: A_{5}A_{6}\ldots A_{p+3}A_{1}A_{2}A_{p+4}\ldots A_{3}A_{4}\\&.........................................................................\\&1 \hbox {voter}: A_{m}A_{3}\ldots A_{p}A_{1}A_{2}A_{p+1}\ldots A_{m-2}A_{m-1}. \end{aligned}$$

As we can see in Table 1, candidates $A_{3},\dots ,A_{m}$ have the same cumulative standings; so they get the same score. The cumulative standings of candidates $A_1$ and $A_2$ differ only in the $p$th cumulative standing, but $S_{p}=0$. Therefore, $A_1$ and $A_2$ obtain the same score. We calculate the scores of $A_{2}$ and $A_{3}$:

$$\begin{aligned} F_{\varvec{S}}(\varvec{V}^2)&= \sum _{k=p+1}^{m-1} (m-2)S_{k} = (m-2)\sum _{k=p+1}^{m-1} S_{k} = (m-2)\left( 1-\sum _{k=1}^{p-1}S_{k}\right) ,\\ F_{\varvec{S}}(\varvec{V}^3)&= \sum _{k=1}^{p-1} k S_{k} + \sum _{k=p+1}^{m-1} (k-2) S_{k}. \end{aligned}$$

We now compare both scores:

$$\begin{aligned} F_{\varvec{S}}(\varvec{V}^2)&\ge F_{\varvec{S}}(\varvec{V}^3) \;\Leftrightarrow \; m-2 - (m-2)\sum _{k=1}^{p-1}S_{k} \ge \sum _{k=1}^{p-1} k S_{k} + \sum _{k=p+1}^{m-1} (k-2) S_{k} \\&\Leftrightarrow \; m-2 \ge \sum _{k=1}^{p-1} (m+k) S_{k} + \sum _{k=p+1}^{m-1} k S_{k} - 2\left( \sum _{k=1}^{p-1} S_{k} + \sum _{k=p+1}^{m-1} S_{k} \right) \\&\Leftrightarrow \; m \ge \sum _{k=1}^{p-1} (m+k) S_{k} + \sum _{k=p+1}^{m-1} k S_{k}, \end{aligned}$$

where the last inequality is satisfied by hypothesis. Therefore, $A_{2}$ is a winning candidate but he/she is not Pareto-optimal because $A_{1}$ is preferred to $A_{2}$ by all voters. So, the scoring rule associated with $\varvec{S}$ is not Pareto-optimal.$\square$

Table 1 Cumulative standings of profile $\varvec{p}$ (Proof of Theorem 6)

Proof of Corollary 1

If $\varvec{S}\in \fancyscript{S}$ is the scoring vector corresponding to the $k$-approval voting, the expression

$$\begin{aligned} \sum _{l=1}^{p-1} (m+l) S_l + \sum _{l=p+1}^{m-1} l S_l, \end{aligned}$$

where $p=\min \bigl \{l\in \{1,\dots ,m-1\}\;\vert \; S_{l}=0\bigr \}$, is equal to $m+1$ when $k=1$ and equal to $k$ when $k\ge 2$. From Theorem 1 we get the result.$\square$

Proof of Corollary 2

Let $\varvec{s}\in \fancyscript{S}^{*}$ and consider the scoring vector $\varvec{S}\in \fancyscript{S}$ associated with it. Given that $\varvec{S}\gg \varvec{0}$ if and only if $1 > s_{2} > \cdots > s_{m-1} > 0$, $S_{p}=0$ if and only if $s_{p}=s_{p+1}$, and

$$\begin{aligned}&\sum _{k=1}^{p-1} (m+k) S_k + \sum _{k=p+1}^{m-1} k S_k > m\\&\qquad \Leftrightarrow \sum _{k=1}^{p-1} (m+k) (s_{k} - s_{k+1}) + \sum _{k=p+1}^{m-1} k (s_{k} - s_{k+1}) > m\\&\qquad \Leftrightarrow \sum _{k=1}^{p-1} (m+k) s_{k} - \sum _{k=2}^{p} (m+k-1) s_{k} + \sum _{k=p+1}^{m-1} k s_{k} - \sum _{k=p+2}^{m} (k-1) s_{k} > m\\&\qquad \Leftrightarrow (m+1)s_{1} + \sum _{k=2}^{p-1} s_{k} - (m+p-1) s_{p} \\&\qquad \quad \,\,\,+ (p+1) s_{p+1} + \sum _{k=p+2}^{m-1} s_{k} - (m-1)s_{m} > m\\&\qquad \Leftrightarrow (m+1) + \sum _{k=2}^{p-1} s_{k} - (m-2) s_{p} + \sum _{k=p+2}^{m-1} s_{k} > m \\&\qquad \Leftrightarrow 1 + \sum _{k=2}^{p-1} s_{k} + \sum _{k=p+2}^{m-1} s_{k} > (m-2) s_{p}\\&\qquad \Leftrightarrow \sum _{k=1}^{m-1} s_{k} > m\cdot s_{p}, \end{aligned}$$

the result is obvious from Theorem 1.$\square$

1.2 A.2 Absolute loser paradox

By using contraposition, Theorem 2 is rewritten in an equivalent form as follows.

Theorem 7

Let $\varvec{S}\in \fancyscript{S}$. The following conditions are equivalent:

• 1. The scoring rule associated with $\varvec{S}$ is vulnerable to the absolute loser paradox.

• 2. $\sum _{k=1}^{m-1} k S_{k} < \displaystyle \frac{m}{2}$.

Proof

$1\;\Rightarrow \; 2:$ If the scoring rule associated with $\varvec{S}$ is vulnerable to the absolute loser paradox, we can find a profile $\varvec{p}$ such that one winning candidate, $A_{i}$, is the absolute loser. Therefore, he/she is ranked last by an absolute majority of voters and, consequently,

$$\begin{aligned} V_{k}^{i}\le n-\big (\lfloor n/2\rfloor +1\big ) = \left\lfloor (n-1)/2\right\rfloor , \end{aligned}$$

for all $k\in \{1,\dots ,m-1\}$. On the other hand, since $A_i$ is a winning candidate, we have

$$\begin{aligned} \sum _{k=1}^{m-1}S_k V_{k}^{l} \le \sum _{k=1}^{m-1}S_k V_{k}^{i}, \end{aligned}$$

for all $l\in \{1,\dots ,m\}$. When we add, member to member, all the inequalities we have

$$\begin{aligned} \sum _{l=1}^{m} \sum _{k=1}^{m-1} S_k V_{k}^l&= \sum _{k=1}^{m-1} S_k \sum _{l=1}^{m} V_{k}^l = \sum _{k=1}^{m-1} S_k nk = n\sum _{k=1}^{m-1} k S_k \le m \sum _{k=1}^{m-1} S_k V_{k}^i\\&\le m \left\lfloor (n-1)/2\right\rfloor . \end{aligned}$$

Therefore,

$$\begin{aligned} \sum _{k=1}^{m-1} k S_k \le \frac{m}{n} \left\lfloor (n-1)/2\right\rfloor < \frac{m}{2}. \end{aligned}$$

$2\;\Rightarrow \; 1:$ Since $\sum \nolimits _{k=1}^{m-1} k S_{k} < m/2$, there exists $r\in \mathbb {N}$ such that

$$\begin{aligned} \sum _{k=1}^{m-1} k S_{k} < \frac{m}{2}-\frac{1}{2r}. \end{aligned}$$

Consider twice the forward cyclic list of orders generated by $A_{2}\,A_{3}\ldots A_{m}$. In the first forward cyclic list, we place candidate $A_{1}$ in the first position and in the second forward cyclic list we place $A_{1}$ in the last position. Now we consider a profile $\varvec{p}$ of $2(m-1)r-1$ voters where each order is considered $r$ times but the first one, which is considered $r-1$ times:

$$\begin{aligned}&r-1 \,\hbox {voters}: A_{1}\,A_{2}\,A_{3}\ldots A_{m-1}\,A_{m}\\&r \,\hbox {voters}\qquad \,: A_{1}\,A_{3}\,A_{4}\ldots A_{m}\,A_{2}\\&r \,\hbox {voters}\qquad \,: A_{1}\,A_{4}\,A_{5}\ldots A_{2}\,A_{3}\\&....................................................\\&r \,\hbox {voters}\qquad \,: A_{1}\,A_{m}\,A_{2}\ldots A_{m-2}\,A_{m-1}\\&r \,\hbox {voters}\qquad \,: A_{2}\,A_{3}\ldots A_{m-1}\,A_{m}\,A_{1}\\&r \,\hbox {voters}\qquad \,: A_{3}\,A_{4}\ldots A_{m}\,A_{2}\,A_{1}\\&r \,\hbox {voters}\qquad \,: A_{4}\,A_{5}\ldots A_{2}\,A_{3}\,A_{1}\\&.....................................................\\&r \,\hbox {voters}\qquad \,: A_{m}\,A_{2}\ldots A_{m-2}\,A_{m-1}\,A_{1} \end{aligned}$$

In Table 2, we show the cumulative standings of candidates according to the profile $\varvec{p}$. Given that $\varvec{V}^{m}\ge \varvec{V}^{i}$ for all $i\in \{2,\dots ,m-1\}$, we have $F_{\varvec{S}}(\varvec{V}^{m}) \ge F_{\varvec{S}}(\varvec{V}^{i})$ for all $i\in \{2,\dots ,m-1\}$. We now compare the scores of $A_{1}$ and $A_{m}$:

$$\begin{aligned} F_{\varvec{S}}(\varvec{V}^1) > F_{\varvec{S}}(\varvec{V}^m)&\Leftrightarrow \sum _{k=1}^{m-1} \bigl ((m-1)r-1\bigr ) S_{k} > \sum _{k=1}^{m-1} (2k-1)r S_{k}\\&\Leftrightarrow (m-1)r-1 > r \sum _{k=1}^{m-1} (2k-1) S_{k} \\&\Leftrightarrow m - 1 - \frac{1}{r} > \sum _{k=1}^{m-1} (2k-1) S_{k}\\&\Leftrightarrow m - \frac{1}{r} > \sum _{k=1}^{m-1} (2k-1) S_{k} + \sum _{k=1}^{m-1} S_{k}\\&\Leftrightarrow m - \frac{1}{r} > 2\sum _{k=1}^{m-1} k S_{k} \Leftrightarrow \frac{m}{2} - \frac{1}{2r} > \sum _{k=1}^{m-1} k S_{k}, \end{aligned}$$

where the last inequality is satisfied by the choice of $r$. Since $A_{1}$ is at the same time the winner and the absolute loser, $F_{\varvec{S}}$ is vulnerable to the absolute loser paradox.$\square$

Table 2 Cumulative standings of profile $\varvec{p}$ (Proof of Theorem 2)

Proof of Corollary 3

• 1. If $\varvec{S}\in \fancyscript{S}$ is the scoring vector corresponding to the Borda rule, we have

  $$\begin{aligned} \sum _{k=1}^{m-1} k S_{k} = \frac{1}{m-1} \sum _{k=1}^{m-1} k = \frac{1}{m-1} \frac{m(m-1)}{2}=\frac{m}{2}. \end{aligned}$$

  So, the result is obvious from Theorem 2.

• 2. If $\varvec{S}\in \fancyscript{S}$ is the scoring vector corresponding to the $k$-approval voting, we have

  $$\begin{aligned} \sum _{l=1}^{m-1} l S_{l} = k. \end{aligned}$$

  Therefore, the result is obvious from Theorem 2.$\square$

Proof of Corollary 4

Given $\varvec{s}\in \fancyscript{S}^{*}$, consider the scoring vector $\varvec{S}\in \fancyscript{S}$ associated with it. Since

$$\begin{aligned} \sum _{k=1}^{m-1} k S_{k}&= \sum _{k=1}^{m-1} k (s_{k}-s_{k+1}) = \sum _{k=1}^{m-1} k s_{k} - \sum _{k=2}^{m} (k-1) s_{k} = \sum _{k=1}^{m-1} s_{k} - (m-1)s_{m}\\&= \sum _{k=1}^{m-1} s_{k}, \end{aligned}$$

the result is obvious from Theorem 2.$\square$

1.3 A.3 Reversal symmetry

Before we give the proof of Theorem 3, we previously establish two characterizations and a technical lemma.

Proposition 1

The scoring rule associated with $\varvec{S}$ satisfies reversal symmetry if and only if

$$\begin{aligned} F_{\varvec{S}}(V_{1}^{i},\dots ,V_{m-1}^{i})&> F_{\varvec{S}}(V_{1}^{j},\dots ,V_{m-1}^{j})\\&\Leftrightarrow F_{\varvec{S}}(n-V_{m-1}^{i},\dots ,n-V_{1}^{i}) < F_{\varvec{S}}(n-V_{m-1}^{j},\dots ,n-V_{1}^{j}) \end{aligned}$$

for all profile $\varvec{p}$ and all pair of candidates $A_{i}$ and $A_{j}$.

Proof

Given a profile $\varvec{p}$, let $(V_{1}^{i},\dots ,V_{m-1}^{i})$ and $(V_{1}^{j},\dots ,V_{m-1}^{j})$ be the cumulative standings of candidates $A_{i}$ and $A_{j}$, respectively. It is easy to check that, when the profile $\varvec{p}$ is reversed, the cumulative standings of $A_{i}$ and $A_{j}$ for this new profile are $(n-V_{m-1}^{i},\dots ,n-V_{1}^{i})$ and $(n-V_{m-1}^{j},\dots ,n-V_{1}^{j})$, respectively. From this, the result is obvious.$\square$

Proposition 2

The scoring rule associated with $\varvec{S}$ satisfies reversal symmetry if and only if

$$\begin{aligned} F_{\varvec{S}}(\varvec{V}^{i}) > F_{\varvec{S}}(\varvec{V}^{j}) \;\Leftrightarrow \; F_{\varvec{S}^{d}}(\varvec{V}^{i}) > F_{\varvec{S}^{d}}(\varvec{V}^{j}) \end{aligned}$$

for all profile $\varvec{p}$ and all pair of candidates $A_{i}$ and $A_{j}$.

Proof

The proof is obvious from Proposition 1 and the following equivalence:

$$\begin{aligned} F_{\varvec{S}}(n-V_{m-1}^{i},\dots ,n-V_{1}^{i})&< F_{\varvec{S}}(n-V_{m-1}^{j},\dots ,n-V_{1}^{j})\\&\Leftrightarrow \displaystyle \sum _{k=1}^{m-1} S_{k}(n-V_{m-k}^{i}) < \displaystyle \sum _{k=1}^{m-1} S_{k}(n-V_{m-k}^{j})\\&\Leftrightarrow n - \displaystyle \sum _{k=1}^{m-1} S_{k}V_{m-k}^{i} < n - \displaystyle \sum _{k=1}^{m-1} S_{k}V_{m-k}^{j}\\&\Leftrightarrow \displaystyle \sum _{k=1}^{m-1} S_{k}V_{m-k}^{i} > \displaystyle \sum _{k=1}^{m-1} S_{k}V_{m-k}^{j}\\&\Leftrightarrow \displaystyle \sum _{k=1}^{m-1} S_{m-k}V_{k}^{i} > \displaystyle \sum _{k=1}^{m-1} S_{m-k}V_{k}^{j}\\&\Leftrightarrow F_{\varvec{S}^{d}}(\varvec{V}^{i}) > F_{\varvec{S}^{d}}(\varvec{V}^{j}). \end{aligned}$$

$\square$

Lemma 3

Given $l\in \bigl \{1,\dots ,\lfloor (m-1)/2\rfloor \bigr \}$, there exist a profile $\varvec{p}$ and candidates $A_{i}$ and $A_{j}$ such that $V_{l}^{i} = V_{l}^{j}+1$, $V_{m-l}^{i} = V_{m-l}^{j}-1$ and $V_{k}^{i} = V_{k}^{j}$ for all $k\ne l,m-l$.

Proof

Consider $m$ voters and the forward cyclic list of orders generated by the order $A_{1}\,A_{2}\cdots A_{m}$; that is,

$$\begin{aligned}&1\hbox {st voter}\,\,: A_{1}\,A_{2}\ldots A_{m-1}\,A_{m}\\&2\hbox {nd voter}: A_{2}\,A_{3}\ldots A_{m}\,A_{1}\\&...................................................\\&m\hbox {th voter}: A_{m}\,A_{1}\ldots A_{m-2}\,A_{m-1}. \end{aligned}$$

Now, in the first voter’s preferences, we first change the candidates of positions $l$ and $l+1$ and, after that, we change the candidates of positions $m-l$ and $m-l+1$. Let $\varvec{p}$ be the resulting profile. We distinguish two cases:

• 1. If $m-l \ne l+1$, then the first voter’s order is

  $$\begin{aligned} 1\hbox {st voter}: A_{1}\ldots A_{l-1}\,A_{l+1}\,A_{l}\,A_{l+2}\ldots A_{m-l-1}\,A_{m-l+1}\,A_{m-l}\,A_{m-l+2}\ldots A_{m}, \end{aligned}$$

  and the cumulative standings of profile $\varvec{p}$ are given in Table 3. As we can see, candidates $A_{l+1}$ and $A_{m-l+1}$ satisfy the thesis of the theorem.

• 2. If $m-l = l+1$, then the first voter’s order is

  $$\begin{aligned} 1\hbox {st voter}: A_{1}\ldots A_{l-1}\,A_{l+1}\,A_{l+2}\,A_{l}\ldots A_{m}, \end{aligned}$$

  and the cumulative standings of profile $\varvec{p}$ are given in Table 4. As we can see, candidates $A_{l+1}$ and $A_{l+2}$ satisfy the thesis of the theorem.$\square$

Table 3 Cumulative standings of profile $\varvec{p}$ (Proof of Lemma 3, case $l+1 \ne m-l$)

Table 4 Cumulative standings of profile $\varvec{p}$ (Proof of Lemma 3, case $l+1 = m-l$)

Proof of Theorem 3

$1\;\Rightarrow \; 2:$ We are going to prove that if $\varvec{S}$ is not self-dual, then the scoring rule associated with $\varvec{S}$ does not satisfy reversal symmetry. If $\varvec{S}$ is not self-dual, then there exists $l\in \bigl \{1,\dots ,\lfloor (m-1)/2\rfloor \bigr \}$ such that $S_{l}\ne S_{m-l}$. By Lemma 3, we can find a profile $\varvec{p}$ and candidates $A_{i}$ and $A_{j}$ such that $V_{l}^{i} = V_{l}^{j}+1$, $V_{m-l}^{i} = V_{m-l}^{j}-1$ and $V_{k}^{i} = V_{k}^{j}$ for all $k\ne l,m-l$. For this profile and these candidates, we have

$$\begin{aligned} F_{\varvec{S}}(\varvec{V}^{i}) > F_{\varvec{S}}(\varvec{V}^{j})&\Leftrightarrow \displaystyle \sum _{k=1}^{m-1} S_{k}V_{k}^{i} > \displaystyle \sum _{k=1}^{m-1} S_{k}V_{k}^{j}\\&\Leftrightarrow S_{l}V_{l}^{i} + S_{m-l}V_{m-l}^{i} > S_{l}V_{l}^{j} + S_{m-l}V_{m-l}^{j}\\&\Leftrightarrow S_{l} > S_{m-l}, \end{aligned}$$

and

$$\begin{aligned} F_{\varvec{S}^{d}}(\varvec{V}^{i}) > F_{\varvec{S}^{d}}(\varvec{V}^{j})&\Leftrightarrow \displaystyle \sum _{k=1}^{m-1} S_{m-k}V_{k}^{i} > \displaystyle \sum _{k=1}^{m-1} S_{m-k}V_{k}^{j}\\&\Leftrightarrow S_{m-l}V_{l}^{i} + S_{l}V_{m-l}^{i} > S_{m-l}V_{l}^{j} + S_{l}V_{m-l}^{j}\\&\Leftrightarrow S_{m-l} > S_{l}. \end{aligned}$$

Therefore, according to Proposition 2, the scoring rule associated with $\varvec{S}$ does not satisfy reversal symmetry.

$2\;\Rightarrow \; 1:$ Obvious by Proposition 2.$\square$

Proof of Corollary 5

Given $\varvec{s}\in \fancyscript{S}^{*}$, consider the scoring vector $\varvec{S}\in \fancyscript{S}$ associated with it. Since

$$\begin{aligned} \varvec{S}=\varvec{S}^{d}&\;\Leftrightarrow \; S_{k}=S_{m-k}\quad \text {for all } k\in \{1,\dots ,\lfloor (m-1)/2\rfloor \}\\&\;\Leftrightarrow \; s_{k} - s_{k+1} = s_{m-k} - s_{m-k+1} \quad \text {for all } k\in \{1,\dots ,\lfloor (m-1)/2\rfloor \}\\&\;\Leftrightarrow \; s_{k} + s_{m-k+1} = s_{k+1} + s_{m-k}\quad \text {for all } k\in \{1,\dots ,\lfloor (m-1)/2\rfloor \}\\&\;\Leftrightarrow \; 1 = s_{k+1} + s_{m-k} \quad \text {for all } k\in \{1,\dots ,\lfloor (m-1)/2\rfloor \}, \end{aligned}$$

the result is obvious from Theorem 3.$\square$

1.4 A.4 Relationships among the properties

Proof of Theorem 4

By Theorem 1, it is sufficient to prove that if $\min \nolimits _{k\in \{1,\dots ,m-1\}} S_k=0$, then

$$\begin{aligned} \sum _{k=1}^{p-1} (m+k) S_k + \sum _{k=p+1}^{m-1} k S_k \le m, \end{aligned}$$

where $p=\min \bigl \{k\in \{1,\dots ,m-1\}\;\vert \; S_{k}=0\bigr \}$. Since the scoring rule associated with $\varvec{S}$ satisfies reversal symmetry, then, by Theorem 3, $S_{k}=S_{m-k}$ for all $k\in \{1,\dots ,m-1\}$. Therefore, $S_{m-p}=S_{p}=0$ and $p\le \lfloor m/2 \rfloor$. We distinguish two cases:

• 1. If $m-1$ is even, then

  $$\begin{aligned}&\sum _{k=1}^{p-1} (m+k) S_k + \sum _{k=p+1}^{m-1} k S_k \\&\qquad = \sum _{k=1}^{p-1} (m+k) S_k + \sum _{k=p+1}^{\frac{m-1}{2}} k S_k + \sum _{k=\frac{m+1}{2}}^{m-p-1} k S_{m-k} + \sum _{k=m-p+1}^{m-1} k S_{m-k}\\&\qquad = \sum _{k=1}^{p-1} (m+k) S_k + \sum _{k=p+1}^{\frac{m-1}{2}} k S_k + \sum _{k=p+1}^{\frac{m-1}{2}} (m-k) S_{k} + \sum _{k=1}^{p-1} (m-k) S_{k}\\&\qquad = 2m \sum _{k=1}^{p-1} S_k + m \sum _{k=p+1}^{\frac{m-1}{2}} S_{k} \le 2m \sum _{k=1}^{\frac{m-1}{2}} S_k = m \sum _{k=1}^{m-1} S_k = m. \end{aligned}$$

• 2. If $m-1$ is odd, we distinguish two cases:

  • (a) If $p=m/2$, then

    $$\begin{aligned} \sum _{k=1}^{\frac{m}{2}-1} (m+k) S_k + \sum _{k=\frac{m}{2}+1}^{m-1} k S_k&= \sum _{k=1}^{\frac{m}{2}-1} (m+k) S_k + \sum _{k=\frac{m}{2}+1}^{m-1} k S_{m-k}\\&= \sum _{k=1}^{\frac{m}{2}-1} (m+k) S_k + \sum _{k=1}^{\frac{m}{2}-1} (m-k) S_k\\&= 2m \sum _{k=1}^{\frac{m}{2}-1} S_k = m \sum _{k=1}^{m-1} S_k = m. \end{aligned}$$

  • (b) If $p<m/2$, then

    $$\begin{aligned}&\sum _{k=1}^{p-1} (m+k) S_k + \sum _{k=p+1}^{m-1} k S_k \\&\quad = \sum _{k=1}^{p-1} (m+k) S_k + \sum _{k=p+1}^{\frac{m}{2}-1} k S_k + \frac{m}{2} S_{\frac{m}{2}} + \sum _{k=\frac{m}{2}+1}^{m-p-1} k S_{m-k} + \sum _{k=m-p+1}^{m-1} k S_{m-k}\\&\quad = \sum _{k=1}^{p-1} (m+k) S_k\! +\! \sum _{k=p+1}^{\frac{m}{2}-1} k S_k\! +\! \frac{m}{2} S_{\frac{m}{2}} + \sum _{k=p+1}^{\frac{m}{2}-1} (m-k) S_{k} + \sum _{k=1}^{p-1} (m-k) S_{k}\\&\quad = 2m \sum _{k=1}^{p-1} S_k + m \sum _{k=p+1}^{\frac{m}{2}-1} S_{k} + \frac{m}{2} S_{\frac{m}{2}} \le m \left( 2\sum _{k=1}^{\frac{m}{2}-1} S_k + S_{\frac{m}{2}}\right) = m. \end{aligned}$$

    $\square$

Proof of Theorem 5

If the scoring rule associated with $\varvec{S}$ satisfies reversal symmetry, then, by Theorem 3, $S_{k}=S_{m-k}$ for all $k\in \{1,\dots ,m-1\}$. We distinguish two cases:

• 1. If $m-1$ is even, then

  $$\begin{aligned} \sum _{k=1}^{m-1} k S_{k} \!=\! \sum _{k=1}^{\frac{m-1}{2}} k S_{k} + \sum _{k=\frac{m+1}{2}}^{m-1} k S_{m-k} \!=\! \sum _{k=1}^{\frac{m-1}{2}} k S_{k} + \sum _{k=1}^{\frac{m-1}{2}} (m-k) S_{k} = m \sum _{k=1}^{\frac{m-1}{2}} S_{k} \!=\!\frac{m}{2}. \end{aligned}$$

• 2. If $m-1$ is odd, then

  $$\begin{aligned} \sum _{k=1}^{m-1} k S_{k}&=\! \sum _{k=1}^{\frac{m}{2}-1} k S_{k} \!+\! \frac{m}{2} S_{\frac{m}{2}} \!+\!\! \sum _{k=\frac{m}{2}+1}^{m-1} k S_{m-k} \!\!=\!\! \sum _{k=1}^{\frac{m}{2}-1} k S_{k} \!+\! \frac{m}{2} S_{\frac{m}{2}}\! \!+\! \sum _{k=1}^{\frac{m}{2}-1} (m-k) S_{k}\\&= m \sum _{k=1}^{\frac{m}{2}-1} S_{k} + \frac{m}{2} S_{\frac{m}{2}} = \frac{m}{2} \left( 2\sum _{k=1}^{\frac{m}{2}-1} S_{k} + S_{\frac{m}{2}}\right) = \frac{m}{2} \sum _{k=1}^{m-1} S_{k} = \frac{m}{2}. \end{aligned}$$

Therefore, in both cases, $\sum _{k=1}^{m-1} k S_{k} = m/2$ and, by Theorem 2, the scoring rule associated with $\varvec{S}$ is immune to the absolute loser paradox.$\square$

Proof of Corollary 6

It is obvious from Theorems 1, 3, 4 and 5.$\square$