Ranking Candidates Through Convex Sequences of Variable Weights

Abstract

Scoring rules are a well-known class of positional voting systems where fixed scores are assigned to the different ranks. Nevertheless, since the winners may change according to the scores used, the choice of the scoring vector is not obvious. For this reason several methods have been suggested so that each candidate may be evaluated with the most favorable scoring vector for him/her. In this paper we propose a new model that allows to use different scoring vector for each candidate and avoid some shortcomings of other methods suggested in the literature. Moreover we give a closed expression for the score obtained by each candidate and, in this way, it is possible to rank the candidates without solving the proposed model.

Appendix

Proof of Lemma 1

Given Model (3), we consider the following change of variables:

$$\begin{aligned} \left\{ \begin{aligned}&s_{j}=w_{j}-w_{j+1}, \quad \text {for all } j\in \{1,\dots , k-1\},\\&s_{k}=w_{k}. \end{aligned} \right. \end{aligned}$$

It is easy to check that $w_{j}=\sum _{p=j}^{k} s_{p}\,$ for all $j\in \{1,\dots , k\}$ and that the set of constraints

$$\begin{aligned} \left\{ \begin{aligned}&w_j-w_{j+1} \ge w_{j+1}-w_{j+2},&j=1,\dots ,k-2,\\&w_{k-1}-w_{k} \ge w_{k}, \\&w_k \ge \varepsilon . \end{aligned} \right. \end{aligned}$$

is equivalent to the set

$$\begin{aligned} \left\{ \begin{aligned}&s_{j} \ge s_{j+1},&j=1,\dots ,k-1,\\&s_{k} \ge \varepsilon . \end{aligned} \right. \end{aligned}$$

Let us make yet another change of variables:

$$\begin{aligned} \left\{ \begin{aligned}&W_{j}=s_{j}-s_{j+1}, \quad \text {for all } j\in \{1,\dots , k-1\},\\&W_{k}=s_{k}-\varepsilon . \end{aligned} \right. \end{aligned}$$

From the previous relationships, variables $W_{j}$ can be expressed as functions of variables $w_{j}$ in the following way:

$$\begin{aligned} \left\{ \begin{aligned}&W_{j} = s_{j}-s_{j+1} = w_{j}-2w_{j+1}+w_{j+2}, \quad \text {for all }\quad j\in \{1,\dots , k-2\},\\&W_{k-1} = s_{k-1}-s_{k} = w_{k-1}-2w_{k},\\&W_{k} = w_{k}-\varepsilon . \end{aligned} \right. \end{aligned}$$

Furthermore, it is easy to see that $s_{j}=\sum _{l=j}^{k} W_{l}+\varepsilon \,$ for all $j\in \{1,\dots , k\}$. Therefore,

$$\begin{aligned} w_{j}&= \sum _{p=j}^{k} s_{p} = \sum _{p=j}^{k} \left( \sum _{l=p}^{k} W_{l}+\varepsilon \right) = \mathop {\mathop {\sum }\limits _{{j\le p \le k}}}\limits _{{p\le l\le k}} W_{l} + \sum _{p=j}^{k} \varepsilon = \sum _{j\le p \le l\le k} W_{l} + (k+1-j)\varepsilon \\&= \sum _{l=j}^{k} \sum _{p=j}^{l} W_{l} + (k+1-j)\varepsilon = \sum _{l=j}^{k} (l+1-j)W_{l} + (k+1-j)\varepsilon . \end{aligned}$$

Next we write the remaining expressions of Model (3) as functions of the variables $W_{j}$:

$$\begin{aligned} \sum _{j=1}^k v_{oj} w_j&= \sum _{j=1}^k v_{oj}\left( \sum _{l=j}^{k} (l+1-j)W_l + (k+1-j)\varepsilon \right) \\&= \mathop {\mathop {\sum }\limits _{{1\le j \le k}}}\limits _{{j\le l\le k}} (l+1-j) v_{oj} W_{l} + \varepsilon \sum _{j=1}^k (k+1-j) v_{oj}\\&= \sum _{1\le j \le l\le k} (l+1-j) v_{oj} W_{l} + \varepsilon \sum _{j=1}^k (k+1-j) v_{oj}\\&= \mathop {\mathop {\sum }\limits _{{1\le l \le k}}}\limits _{{1\le j\le l}} (l+1-j) v_{oj} W_{l} + \varepsilon \sum _{j=1}^k (k+1-j) v_{oj}\\&= \sum _{l=1}^k W_{l} \left( \sum _{j=1}^{l} (l+1-j) v_{oj}\right) + \varepsilon \sum _{j=1}^k (k+1-j) v_{oj}\\&= \sum _{j=1}^k W_{j} \left( \sum _{l=1}^{j} (j+1-l) v_{ol}\right) + \varepsilon \sum _{l=1}^k (k+1-l) v_{ol}, \end{aligned}$$

where the last equality is obtained by changing the role of j and l. If we denote $\sum _{l=1}^{j} (j+1-l) v_{ol}$ by $V_{oj}$ for all $j\in \{1,\dots ,k\}$, then we have

$$\begin{aligned} \sum _{j=1}^k v_{oj} w_j = \sum _{j=1}^k V_{oj} W_j + \varepsilon V_{ok}. \end{aligned}$$

Analogously,

$$\begin{aligned} \sum _{j=1}^k (n-v_{oj}) w_j = n \sum _{j=1}^k w_{j} - \sum _{j=1}^k v_{oj}w_{j} = n \sum _{j=1}^k w_{j} - \sum _{j=1}^k V_{oj} W_j - \varepsilon V_{ok}. \end{aligned}$$

Given that

$$\begin{aligned} \sum _{j=1}^k w_j&= \sum _{j=1}^k \left( \sum _{l=j}^{k} (l+1-j)W_l + (k+1-j) \varepsilon \right) \\&= \mathop {\mathop {\sum }\limits _{{1\le j \le k}}}\limits _{{j\le l\le k}} (l+1-j)W_{l} + \varepsilon \sum _{j=1}^k (k+1-j) = \sum _{1\le j \le l\le k} (l+1-j)W_{l} + \varepsilon \frac{k(k+1)}{2}\\&= \mathop {\mathop {\sum }\limits _{{1\le l \le k}}}\limits _{{1\le j\le l}} (l+1-j)W_{l} + \varepsilon \frac{k(k+1)}{2} = \sum _{l=1}^k W_{l} \left( \sum _{j=1}^{l} (l+1-j)\right) + \varepsilon \frac{k(k+1)}{2}\\&= \sum _{l=1}^k \frac{l(l+1)}{2} W_{l} + \varepsilon \frac{k(k+1)}{2}, \end{aligned}$$

we have

$$\begin{aligned} \sum _{j=1}^k (n-v_{oj}) w_j&= \sum _{j=1}^k \frac{nj(j+1)}{2} W_{j} + \varepsilon \frac{nk(k+1)}{2}- \sum _{j=1}^k V_{oj} W_j - \varepsilon V_{ok}\\&= \sum _{j=1}^k \left( \frac{nj(j+1)}{2}-V_{oj}\right) W_j +\left( \frac{nk(k+1)}{2}- V_{ok}\right) \varepsilon . \end{aligned}$$

Therefore, the constraint

$$\begin{aligned} \sum _{j=1}^k (n-v_{oj}) w_j\le m-1 \end{aligned}$$

can be written as

$$\begin{aligned} \sum _{j=1}^k \left( \frac{nj(j+1)}{2}-V_{oj}\right) W_j +\left( \frac{nk(k+1)}{2}- V_{ok}\right) \varepsilon \le m-1. \end{aligned}$$

If we consider

$$\begin{aligned} \delta _{o}(\varepsilon ) = (m-1)- \left( \frac{nk(k+1)}{2}- V_{ok}\right) \varepsilon , \end{aligned}$$

then Model (3) can be expressed as

$$\begin{aligned} \widehat{Z}_o^{*}(\varepsilon ) = \max&\,\, \sum _{j=1}^k V_{oj} W_j + \varepsilon V_{ok},\\ {\mathrm {s.t.}}&\,\, \sum _{j=1}^k \left( \frac{nj(j+1)}{2}-V_{oj}\right) W_j \le \delta _{o}(\varepsilon ),\\&\,\, W_j\ge 0,\quad j=1,\dots ,k. \end{aligned}$$

$\square$

Proof of Theorem 1

Model (4) is equivalent to the following one:

$$\begin{aligned} \widehat{Z}'_o(\varepsilon ) = \max&\,\, \sum _{j=1}^k V_{oj} W_j,\\ {\mathrm {s.t.}}&\,\, \sum _{j=1}^k \left( \frac{nj(j+1)}{2}-V_{oj}\right) W_j \le \delta _{o}(\varepsilon ),\\&\,\, W_j\ge 0,\quad j=1,\dots ,k. \end{aligned}$$

Moreover $\widehat{Z}_o^{*}(\varepsilon )=\widehat{Z}'_o(\varepsilon ) + \varepsilon V_{ok}$. It is well known that if a linear program has an optimal solution, then its dual also has an optimal solution and the optimal values for both problems are equal. Therefore, it is sufficient to solve the dual of the previous problem, that is,

$$\begin{aligned} \min&\,\, \delta _{o}(\varepsilon ) X,\\ {\mathrm {s.t.}}&\,\, \left( \frac{nj(j+1)}{2}-V_{oj}\right) X\ge V_{oj},\quad j=1,\dots ,k,\\&\,\,\, X\ge 0. \end{aligned}$$

It is easy to check that the optimal solution is

$$\begin{aligned} X^{*}=V_{o}^{*} =\displaystyle \max _{j=1,\dots ,k}\frac{V_{oj}}{\displaystyle \frac{nj(j+1)}{2}-V_{oj}}. \end{aligned}$$

Therefore, $\widehat{Z}'_o(\varepsilon ) = \delta _{o}(\varepsilon ) V_{o}^{*}$ and $\widehat{Z}_o^{*}(\varepsilon ) = \delta _{o}(\varepsilon ) V_{o}^{*} + \varepsilon V_{ok}$. $\square$

Proof of Proposition 1

We distinguish two cases:

• 1. If a candidate $A_{o}$ gets all the first ranks, then he/she is the winner according with Model (5). On the other hand, since $v_{o1}=n$, we have $V_{oj}=j v_{o1} + (j-1) v_{o2} + \cdots + v_{oj}= jn$. Therefore,

  $$\begin{aligned} Z_{o} = \max _{j=1,\dots ,k} \frac{V_{oj}}{\displaystyle \frac{j(j+1)}{2}} = \max _{j=1,\dots ,k} \frac{2n}{j+1}=n. \end{aligned}$$

  Let $A_i$ be a different candidate. Then $v_{i1}=0$, $V_{ij}=(j-1) v_{i2} + \cdots + v_{ij} \le (j-1) n$, and

  $$\begin{aligned} Z_{i} = \max _{j=1,\dots ,k} \frac{V_{ij}}{\displaystyle \frac{j(j+1)}{2}} \le \max _{j=1,\dots ,k} \frac{2(j-1)n}{j(j+1)} < n. \end{aligned}$$

  Therefore $Z_{o} > Z_{i}$.

• 2. If no candidate obtains all the first ranks, let $A_{o}$ and $A_i$ be two candidates such that $V_{o1}<n$ and $V_{i1}<n$. Given that

  $$\begin{aligned}&\,\, \frac{V_{oj}}{\displaystyle \frac{nj(j+1)}{2}-V_{oj}} > \frac{V_{il}}{\displaystyle \frac{nl(l+1)}{2}-V_{il}}\\&\Leftrightarrow \,\, \frac{nl(l+1)}{2}V_{oj} - V_{il}V_{oj} > \frac{nj(j+1)}{2}V_{il} - V_{oj}V_{il}\\&\Leftrightarrow \,\, \frac{l(l+1)}{2}V_{oj} > \frac{j(j+1)}{2}V_{il} \,\, \Leftrightarrow \,\, \frac{V_{oj}}{\displaystyle \frac{j(j+1)}{2}} > \frac{V_{il}}{\displaystyle \frac{l(l+1)}{2}} \end{aligned}$$

  for all $j,l\in \{1,\dots ,k\}$, we have

  $$\begin{aligned}&\,\, \max _{j=1,\dots ,k}\frac{V_{oj}}{\displaystyle \frac{nj(j+1)}{2}-V_{oj}} > \max _{j=1,\dots ,k}\frac{V_{ij}}{\displaystyle \frac{nj(j+1)}{2}-V_{ij}} \\&\Leftrightarrow \,\, \max _{j=1,\dots ,k} \frac{V_{oj}}{\displaystyle \frac{j(j+1)}{2}} > \max _{j=1,\dots ,k} \frac{V_{il}}{\displaystyle \frac{l(l+1)}{2}}; \end{aligned}$$

  that is,

  $$\begin{aligned} \widehat{Z}_o^{*} > \widehat{Z}_i^{*} \;\Leftrightarrow \; Z_{o} > Z_{i}. \end{aligned}$$

$\square$

Proof of Theorem 2

$$\begin{aligned} \overline{Z}_o&= \frac{1}{\varepsilon _{o}^{*}} \int _{0}^{\varepsilon _{o}^{*}}\widehat{Z}_o^{*}(\varepsilon )\, d\varepsilon \\&= \frac{1}{\varepsilon _{o}^{*}}\int _{0}^{\varepsilon _{o}^{*}} \left( (m-1)V_{o}^{*} + \varepsilon \left( V_{ok}-V_{o}^{*}\left( \frac{nk(k+1)}{2}-V_{ok}\right) \right) \right) \, d\varepsilon \\&= (m-1)V_{o}^{*} + \left( V_{ok}-V_{o}^{*}\left( \frac{nk(k+1)}{2}-V_{ok}\right) \right) \frac{1}{\varepsilon _{o}^{*}} \int _{0}^{\varepsilon _{o}^{*}}\varepsilon \, d\varepsilon \\&= (m-1)V_{o}^{*} + \left( V_{ok}-V_{o}^{*}\left( \frac{nk(k+1)}{2}-V_{ok}\right) \right) \frac{\varepsilon _{o}^{*}}{2}\\&= (m-1)V_{o}^{*} + \left( V_{ok}-V_{o}^{*}\left( \frac{nk(k+1)}{2}-V_{ok}\right) \right) \frac{m-1}{2\left( \displaystyle \frac{nk(k+1)}{2}-V_{ok}\right) }\\&= (m-1)V_{o}^{*}+ \frac{m-1}{2}\frac{V_{ok}}{2\left( \displaystyle \frac{nk(k+1)}{2}-V_{ok}\right) } - \frac{m-1}{2}V_{o}^{*}\\&= \frac{m-1}{2}\left( V_{o}^{*}+\frac{V_{ok}}{\displaystyle \frac{nk(k+1)}{2}-V_{ok}}\right) = (m-1)\frac{V_{o}^{*} + V_{o}^{k}}{2}. \end{aligned}$$

$\square$