Positional Voting Systems Generated by Cumulative Standings Functions

Abstract

Positional voting systems are a class of voting systems where voters rank order the candidates from best to worst and a set of winners is selected using the positions of the candidates in the voters’ preference orders. Although scoring rules are the best known positional voting systems, this class includes other voting systems proposed in the literature as, for example, the Majoritarian Compromise or the $q$-Approval Fallback Bargaining. In this paper we show that some of these positional voting systems can be integrated in a model based on cumulative standings functions. The proposed model allows us to establish a general framework for the analysis of these voting systems, to extend to them some results in the literature for the particular case of the scoring rules, and also facilitates the study of the social choice properties considered in the paper: monotonicity, Pareto-optimality, immunity to the absolute winner paradox, Condorcet consistency, immunity to the absolute loser paradox and immunity to the Condorcet loser paradox.

Appendix

Proof of Theorem 2

Let $F$ be a CSF, let $\varvec{p}$ be a profile and let $A_{i}$ be a winning candidate for this profile. Suppose that some voters raise $A_{i}$ in their rankings without changing the orders of the remaining candidates. Let $\varvec{p}^{\prime }$ be this new profile. It is straightforward to check that $\varvec{V}^{\prime i}\ge \varvec{V}^i$ and $\varvec{V}^{\prime j}\le \varvec{V}^j$ for all $j\ne i$. Since $A_{i}$ is a winning candidate and $F$ is monotonic we have

$$\begin{aligned} F(\varvec{V}^{\prime i})\ge F(\varvec{V}^i)\ge F(\varvec{V}^j)\ge F(\varvec{V}^{\prime j}) \end{aligned}$$

for all $j\ne i$. Therefore, $A_{i}$ continues to be a winning candidate for the new profile $\varvec{p}^{\prime }$. $\square$

Before proving Theorem 3, we give the following lemma.

Lemma 1

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 $k \in \{2,\dots ,m\}$.

Proof

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 ranked $A_j$ higher. Therefore, $V_{k}^{i} \le V_{k-1}^{j}$ for all $k \in \{2,\dots ,m\}$. $\square$

Proof of Theorem 3

In all cases we are going to prove that, given a profile $\varvec{p}$, if a candidate $A_i$ is not Pareto-optimal, then $A_i$ cannot be a winning candidate.

• 1. Lexicographic Rule If $A_i$ is not Pareto-optimal, then, by Lemma 1, $V_{1}^{i}=0$ and, consequently, $A_i$ is not a winning candidate.

• 2. Condorcet’s Practical Method Consider $m=3$. If $A_i$ is not Pareto-optimal, then he/she is dominated by a candidate $A_{j}$. We distinguish two cases:

  • (a) If $V_{1}^{j} > \lfloor n/2\rfloor$, then $A_{j}$ is the only winner.

  • (b) If $V_{1}^{j} \le \lfloor n/2\rfloor$, then, by Lemma 1, $V_{1}^{i} = 0$, $V_{2}^{j} \ge V_{3}^{i} = n$ and $V_{2}^{i} \le V_{1}^{j}< n$. So, $F_{C}(\varvec{V}^{j}) = 1$ and $F_{C}(\varvec{V}^{i}) < 1$. Therefore, $A_i$ cannot be a winning candidate.

• 3. Obata and Ishii’s Method If $A_i$ is not Pareto-optimal, then, by Lemma 1, $V_{1}^{i}=0$ and there exists another candidate $A_j$ with $V_{k}^{i} \le V_{k-1}^{j}$ for all $k \in \{2,\dots ,m\}$. Let $l\in \{2,\dots ,m\}$ such that

  $$\begin{aligned} F_{OI}(\varvec{V}^{i}) = \max \left\{ V_{1}^{i}, \frac{V_{2}^{i}}{2}, \dots , \frac{V_{m-1}^{i}}{m-1}, \frac{V_{m}^{i}}{m}\right\} = \frac{V_{l}^{i}}{l}. \end{aligned}$$

  Since $V_{l}^{i} > 0$ and $V_{l}^{i} \le V_{l-1}^{j}$, we have

  $$\begin{aligned} F_{OI}(\varvec{V}^{i}) = \frac{V_{l}^{i}}{l} < \frac{V_{l-1}^{j}}{l-1} \le F_{OI}(\varvec{V}^{j}). \end{aligned}$$

  Therefore, $A_i$ is not a winning candidate.

• 4. Contreras, Hinojosa and Mármol’s Method If $A_i$ is not Pareto-optimal, then, by Lemma 1, $V_{1}^{i}=0$ and there exists another candidate $A_j$ with $V_{k}^{i} \le V_{k-1}^{j}$ for all $k \in \{2,\dots ,m\}$. Let $l\in \{2,\dots ,m\}$ such that

  $$\begin{aligned} F_{CHM}(\varvec{V}^{i}) = \max \left\{ V_{1}^{i}, \frac{V_{1}^{i} + V_{2}^{i}}{3}, \dots , \frac{2}{m(m+1)}\sum _{k=1}^{m} V_{k}^{i}\right\} = \frac{2}{l(l+1)}\sum _{k=1}^{l} V_{k}^{i}. \end{aligned}$$

  Since $\sum _{k=1}^{l} V_{k}^{i} > 0\,$ and $\sum _{k=1}^{l} V_{k}^{i} \le \sum _{k=1}^{l-1} V_{k}^{j}$, we have

  $$\begin{aligned} F_{CHM}(\varvec{V}^{i}) = \frac{2}{l(l+1)}\sum _{k=1}^{l} V_{k}^{i} < \frac{2}{(l-1)l}\sum _{k=1}^{l-1} V_{k}^{j} \le F_{CHM}(\varvec{V}^{j}). \end{aligned}$$

  Therefore, $A_i$ is not a winning candidate.

• 5. Geometric rule If $A_i$ is not Pareto-optimal, then, by Lemma 1, $V_{1}^{i}=0$ and, consequently, $F_{G}(\varvec{V}^{i})=0$. So, $A_{i}$ is not a winning candidate. $\square$

Proof of Theorem 4

• 1. Obata and Ishii’s Method Let $A_i$ be the absolute winner of a profile $\varvec{p}$. In this case, $V_{1}^{i}\ge \lfloor n/2 \rfloor + 1$; so, $F_{OI}(\varvec{V}^i)\ge \lfloor n/2 \rfloor + 1$. Now, consider $j\ne i$. Then $V_{k}^{j}/k \le n/2$ for all $k \in \{1,\dots ,m\}$. Therefore, $F_{OI}(\varvec{V}^j)\le n/2 < \lfloor n/2 \rfloor + 1 \le F_{OI}(\varvec{V}^i)$, for all $j\ne i$. Consequently, $A_i$ is the only winner.

• 2. Contreras, Hinojosa and Mármol’s Method Let $A_i$ be the absolute winner of a profile $\varvec{p}$. In this case, $V_{1}^{i}\ge \lfloor n/2 \rfloor + 1$; so, $F_{CHM}(\varvec{V}^i)\ge \lfloor n/2 \rfloor + 1$. Now, consider $j\ne i$. Note that

  $$\begin{aligned} F_{CHM}(\varvec{V}^j) = \max _{l=1,\dots ,m}\left\{ \frac{2}{l(l+1)}\sum _{k=1}^{l} V_{k}^{j} \right\} \end{aligned}$$

  and that $V_{1}^{j} \le n-\big (\lfloor n/2 \rfloor + 1 \big )$. Then, for all $l \in \{1,\dots ,m\}$, we have

  $$\begin{aligned} \frac{2}{l(l+1)}\sum _{k=1}^{l} V_{k}^{j}&\le \frac{2}{l(l+1)}\Big (n - \big (\lfloor n/2 \rfloor + 1 \big ) + (l-1)n\Big ) \\&= \frac{2}{l(l+1)}\Big (ln - \big (\lfloor n/2 \rfloor + 1 \big )\Big ) = 2\left( \frac{n}{l+1}-\frac{\lfloor n/2 \rfloor + 1}{l(l+1)}\right) . \end{aligned}$$

  We are going to prove that the maximum value of the last expression is achieved when $l=2$. We distinguish two cases:

  • (a) If $l=1$, then

    $$\begin{aligned} \frac{n}{2}-\frac{\lfloor n/2 \rfloor + 1}{2}&\le \frac{n}{3}-\frac{\lfloor n/2 \rfloor + 1}{6} \\&\Leftrightarrow \frac{n}{6} \le \frac{\lfloor n/2 \rfloor + 1}{3}\\&\Leftrightarrow n \le 2\big (\lfloor n/2 \rfloor + 1\big ), \end{aligned}$$

    which is true.

  • (b) If $l \in \{3,\dots ,m\}$, then

    $$\begin{aligned} \frac{n}{l+1}-\frac{\lfloor n/2 \rfloor + 1}{l(l+1)}&\le \frac{n}{3}-\frac{\lfloor n/2 \rfloor + 1}{6} \\&\Leftrightarrow \; \big (\lfloor n/2 \rfloor + 1\big ) \frac{l(l+1)-6}{6l(l+1)} \le n \frac{l-2}{3(l+1)}\\&\Leftrightarrow \; \big (\lfloor n/2 \rfloor + 1\big ) \frac{(l+3)(l-2)}{l} \le 2n(l-2)\\&\Leftrightarrow \; \big (\lfloor n/2 \rfloor + 1\big )(l+3) \le 2ln, \end{aligned}$$

    which is true.

  Therefore,

  $$\begin{aligned} F_{CHM}(\varvec{V}^j)&\le 2\left( \frac{n}{3}-\frac{\lfloor n/2 \rfloor + 1}{6}\right) \\&< 2\left( \frac{2\big (\lfloor n/2 \rfloor + 1\big )}{3}-\frac{\lfloor n/2 \rfloor + 1}{6}\right) = \lfloor n/2 \rfloor + 1 \le F_{CHM}(\varvec{V}^i), \end{aligned}$$

  for all $j\ne i$. Consequently, $A_i$ is the only winner. $\square$

Proof of Theorem 5

First, we are going to prove that the Lexicographic rule, the Condorcet’s practical method (with $m > 4$), Fallback Bargaining, the Obata and Ishii’s method, the Contreras, Hinojosa and Mármol’s method and the Geometric rule are vulnerable to the absolute loser paradox. Consider twice the forward cyclic list of orders generated by $A_{2}\,A_{3}\cdots A_{m}$. 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 $8m-9$ voters where each order is considered $4$ times but the first one, which is considered $3$ times:

• $3$ voters: $A_{1}\,A_{2}\,A_{3}\cdots A_{m-1}\,A_{m}$

• $4$ voters: $A_{1}\,A_{3}\,A_{4}\cdots A_{m}\,A_{2}$

• $4$ voters: $A_{1}\,A_{4}\,A_{5}\cdots A_{2}\,A_{3}$

• $\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

• $4$ voters: $A_{1}\,A_{m}\,A_{2}\cdots A_{m-2}\,A_{m-1}$

• $4$ voters: $A_{2}\,A_{3}\cdots A_{m-1}\,A_{m}\,A_{1}$

• $4$ voters: $A_{3}\,A_{4}\cdots A_{m}\,A_{2}\,A_{1}$

• $4$ voters: $A_{4}\,A_{5}\cdots A_{2}\,A_{3}\,A_{1}$

• $\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$

• $4$ voters: $A_{m}\,A_{2}\cdots A_{m-2}\,A_{m-1}\,A_{1}.$

Notice that $A_{1}$ is the absolute loser in the profile $\varvec{p}$. In Table 8 we show the cumulative standings of the candidates according to this profile. It is worth noting that $V_{k}^{m} = 4(2k-1)$ for all $k\in \{1,\dots ,m-1\}$.

Table 8 Cumulative standings of profile $\varvec{p}$ (Absolute loser paradox)

We are going to show that $A_{1}$ is a winning candidate for the PVSs listed in the statement of Theorem 5 and, consequently, these PVSs are vulnerable to the absolute loser paradox. Given that $\varvec{V}^{m}\ge \varvec{V}^{i}$ for all $i\in \{2,\dots ,m-1\}$, for any CSF $F$ we have $F(\varvec{V}^{m}) \ge F(\varvec{V}^{i})$ (and, therefore, $A_{m}\succcurlyeq A_{i}$) for all $i\in \{2,\dots ,m-1\}$. Let us see that $A_{1}\succcurlyeq A_{m}$ for the PVSs considered:

• 1. Lexicographic Rule Since $4m-5 > 4$ for all $m\ge 3$, we have that $A_{1}\succ A_{m}$.

• 2. Condorcet’s Practical Method Since neither $A_{1}$ nor $A_{m}$ have a majority of first ranks and $4m-5 > 12$ for all $m\ge 5$, we have that $A_{1}\succ A_{m}$.

• 3. Fallback Bargaining Since $V_{m-1}^{i} < 8m-9 = n$ when $i\in \{1,m\}$, we have that $F_{n}(\varvec{V}^{1}) = F_{n}(\varvec{V}^{m}) = 0$ and, consequently, $A_{1}\succcurlyeq A_{m}$.

• 4. Obata and Ishii’s Method Since $V_{k}^{m} = 4(2k-1)$ for all $k\in \{1,\dots ,m-1\}$, the scores of candidates $A_{1}$ and $A_{m}$ are

  $$\begin{aligned} F_{OI}(\varvec{V}^{1})&= \max \left\{ 4m-5,\frac{4m-5}{2}, \dots ,\frac{4m-5}{m-1},\frac{8m-9}{m}\right\} ,\\ F_{OI}(\varvec{V}^{m})&= \max \left\{ 4, 6,\dots , \frac{4(2m-3)}{m-1},\frac{8m-9}{m}\right\} . \end{aligned}$$

  It is straightforward to check the following inequalities:

  $$\begin{aligned} 4m-5&> \frac{8m-9}{m} \quad (\text {for all } m\ge 3),\\ \frac{4(2m-3)}{m-1}&> \frac{8m-9}{m} \quad (\text {for all } m\ge 3), \end{aligned}$$

  and

  $$\begin{aligned} \frac{V_{k+1}^{m}}{k+1} = \frac{4(2(k+1)-1)}{k+1}&> \frac{4(2k-1)}{k} = \frac{V_{k}^{m}}{k} \quad (\text {for all } k\in \{1,\dots ,m-2\}). \end{aligned}$$

  Therefore, we get

  $$\begin{aligned} F_{OI}(\varvec{V}^{1}) = 4m-5,\qquad F_{OI}(\varvec{V}^{m}) = \frac{4(2m-3)}{m-1}. \end{aligned}$$

  Given that

  $$\begin{aligned} 4m-5 > \frac{4(2m-3)}{m-1} \quad (\text {for all } m\ge 3), \end{aligned}$$

  we have $A_{1}\succ A_{m}$.

• 5. Contreras, Hinojosa and Mármol’s Method The score obtained by candidates $A_{1}$ is

  $$\begin{aligned} F_{CHM}(\varvec{V}^{1}) =&\max \Bigg \{ 4m-5, \frac{2}{3}(4m-5), \dots ,\frac{2}{m}(4m-5),\\&\qquad \quad \frac{2}{m(m+1)}\big ((m-1)(4m-5)+(8m-9)\big )\Bigg \}. \end{aligned}$$

  It is straightforward to check that the inequality

  $$\begin{aligned} 4m-5 > \frac{2}{m(m+1)}\big ((m-1)(4m-5)+(8m-9)\big ) \end{aligned}$$

  is satisfied for all $m\ge 3$. Therefore, $F_{CHM}(\varvec{V}^{1}) = 4m-5$. To calculate the score of candidate $A_{m}$ we take into account that

  $$\begin{aligned} \sum _{l=1}^{k} V_{l}^m = 4 \sum _{l=1}^{k} (2l-1) = 4 k^2, \end{aligned}$$

  for all $k\in \{1,\dots ,m-1\}$. Therefore

  $$\begin{aligned} \frac{2}{k(k+1)}\sum _{l=1}^{k} V_{l}^m&= \frac{8k}{k+1} \quad (\text {for all } k\in \{1,\dots ,m-1\}),\\ \frac{2}{m(m+1)}\sum _{l=1}^{m} V_{l}^m&= \frac{2}{m(m+1)} \left( 4(m-1)^2 + (8m-9)\right) . \end{aligned}$$

  It is easy to check the following inequalities:

  $$\begin{aligned}&\frac{8k}{k+1} < \frac{8(k+1)}{k+2} \quad (\text {for all } k\in \{1,\dots ,m-2\}),\\&\frac{2}{m(m+1)} \left( 4(m-1)^2 + (8m-9)\right) < \frac{8(m-1)}{m}. \end{aligned}$$

  Therefore

  $$\begin{aligned} F_{CHM}(\varvec{V}^{m})=\frac{8(m-1)}{m}. \end{aligned}$$

  Given that

  $$\begin{aligned} 4m-5 > \frac{8(m-1)}{m}\quad (\text {for all } m\ge 3), \end{aligned}$$

  we have $A_{1}\succ A_{m}$.

• 6. Geometric Rule The scores of candidates $A_{1}$ and $A_{m}$ are

  $$\begin{aligned} F_{G}(\varvec{V}^{1})&= (4m-5)^{m-1},\\ F_{G}(\varvec{V}^{m})&= 4^{m-1}\prod _{k=1}^{m-1} (2k-1) = \frac{4^{m-1}(2m-2)!}{2^{m-1}(m-1)!} = 2^{m-1}\frac{(2m-2)!}{(m-1)!}. \end{aligned}$$

  We are going to prove that, for all $m\ge 3$,

  $$\begin{aligned} 2^{m-1}\frac{(2m-2)!}{(m-1)!} < (4m-5)^{m-1}. \end{aligned}$$

  The proof is by induction on $m$. If $m=3$, then $48 < 49$. Suppose now the result is true for $m=k$, and let us see that the claim holds for $m=k+1$. By hypothesis of induction we have

  $$\begin{aligned} 2^{k}\frac{(2k)!}{k!} = 2\frac{(2k-1)2k}{k}\cdot 2^{k-1}\frac{(2k-2)!}{(k-1)!} < 4(2k-1)(4k-5)^{k-1}. \end{aligned}$$

  Since

  $$\begin{aligned} (4k-1)^k&= (4k-1)\Big ((4k-5)+4\Big )^{k-1}\\&> (4k-2)\left( (4k-5)^{k-1}+(k-1)4(4k-5)^{k-2}\right) \\&> (4k-2)\left( (4k-5)^{k-1}+(4k-5)^{k-1}\right) \\&= 4(2k-1)(4k-5)^{k-1}, \end{aligned}$$

  we get

  $$\begin{aligned} 2^{k}\frac{(2k)!}{k!} < (4k-1)^k. \end{aligned}$$

  Therefore $A_{1}\succ A_{m}$.

Next we are going to prove that the Condorcet’s practical method is immune to the absolute loser paradox when $m \le 4$. Let $A_i$ be the absolute loser of a profile $\varvec{p}$. Then $V_{1}^{i} < \lfloor n/2 \rfloor + 1$. We are going to prove that there exists $j\ne i$ such that $V_{2}^{j}\ge V_{2}^{i}$ and, consequently, $A_{i}$ cannot be a winning candidate.

Given that $V_{2}^{i}\le n - (\lfloor n/2 \rfloor + 1) = \lfloor (n-1)/2 \rfloor$ and $\displaystyle \sum _{j=1}^{m} V_{2}^{j} = 2n$, then we have $\displaystyle \sum _{j\ne i} V_{2}^{j} \ge 2n - \lfloor (n-1)/2 \rfloor$. Therefore, there exists $j\ne i$ such that

$$\begin{aligned} V_{2}^{j} \ge \frac{2n - \lfloor (n-1)/2 \rfloor }{m-1}. \end{aligned}$$

Since

$$\begin{aligned} \frac{2n - \lfloor (n-1)/2 \rfloor }{m-1} > \lfloor (n-1)/2 \rfloor&\Leftrightarrow 2n > m \lfloor (n-1)/2 \rfloor \Leftrightarrow \frac{2}{m} n > \lfloor (n-1)/2 \rfloor , \end{aligned}$$

which is true when $m\le 4$, we get

$$\begin{aligned} V_{2}^{j} \ge \frac{2n - \lfloor (n-1)/2 \rfloor }{m-1} > \lfloor (n-1)/2 \rfloor \ge V_{2}^{i}. \end{aligned}$$

$\square$