Positionalist voting rules: a general definition and axiomatic characterizations

Abstract

The paper proposes a general definition of positionalist voting rules. Unlike the commonly-employed scoring rules, our notion of positionalism allows for non-linear criteria to be included in the requisite class. We define a voting rule as positionalist if, for any profile of strict individual orderings, any two alternatives are compared collectively solely on the basis of their positional scores according to the individual rankings. In contrast to the class of scoring rules, however, we do not restrict attention to linear aggregation procedures. Two plausible unanimity properties are examined in the context of our new class of positionalist rules and, moreover, we characterize the lexicographic extensions that refine the plurality rule and its inverse counterpart.

Appendix: Scoring rules and representable rules

In this Appendix, we prove that all scoring rules are representable in the sense of Gärdenfors (1973). Moreover, although the reverse implication is not valid in general, the two notions are equivalent in the presence of neutrality. We first provide the formal definition of representability.

Definition 11

A voting rule F is representable if and only if there exists a function $g :\mathcal{P} \times \mathbf{A} \rightarrow \mathbb {R}$ such that, for all $\mathbf{R} \in \mathcal{P}^N$ and for all $a,b \in \mathbf{A}$,

$$\begin{aligned} a F(\mathbf{R}) b \; \Leftrightarrow \; \sum _{i=1}^N g(R_i,a) \ge \sum _{i=1}^N g(R_i,b). \end{aligned}$$

The proof of the result under neutrality employs Theorem 5.1 of Gärdenfors (1973, pp. 13–14), and we require the following definition for this purpose.

Two strict preference orderings $R,R' \in \mathcal{P}$have the same frame if there exists a one-to-one mapping $\phi :\mathbf{A} \rightarrow \mathbf{A}$ such that aRb if and only if $\phi (a)R'\phi (b)$ for all $a,b \in \mathbf{A}$; see Gärdenfors (1973, p. 12) for this definition applied to the set of all (not necessarily strict) orderings. Note that any two strict preference orderings have the same frame. This is the case because, for any two $R,R' \in \mathcal{P}$, we can define $\phi$ to be the one-to-one mapping that assigns, to each alternative $a \in \mathbf{A}$, the unique alternative $b \in \mathbf{A}$ such that the rank of a in R is the same as the rank of b in $R'$.

Theorem 9

• (a) A scoring rule must be representable. The reverse implication is not valid.

• (b) A neutral and representable voting rule must be a scoring rule.

Proof

(a) Suppose that F is a scoring rule. Thus, there exists an A-tuple of weights $w = (w_1,\ldots ,w_A) \in \mathbb {R}^A$ such that, for all $\mathbf{R} \in \mathcal{P}^N$ and for all $a,b \in \mathbf{A}$,

$$\begin{aligned} a F(\mathbf{R}) b \; \Leftrightarrow \; \sum _{j=1}^A w_j \cdot s_j(\mathbf{R},a) \ge \sum _{j=1}^A w_j \cdot s_j(\mathbf{R},b). \end{aligned}$$

(26)

Recall that $s_j(\mathbf{R},a)$ is the number of individuals i who place a in position j according to their ordering $R_i$. That is, for each of these individuals, $p(R_i,a)$ is equal to j. It follows that the term

$$\begin{aligned} w_j \cdot s_j(\mathbf{R},a) \end{aligned}$$

is equal to the sum

$$\begin{aligned} \sum _{i \in \mathbf{N} \mid p(R_i,a) = j} w_j \end{aligned}$$

for each possible position $j \in \{1,\ldots ,A\}$ and, therefore, the sum

$$\begin{aligned} \sum _{j=1}^A w_j \cdot s_j(\mathbf{R},a) \end{aligned}$$

is equal to the sum

$$\begin{aligned} \sum _{j=1}^A \sum _{i \in \mathbf{N} \mid p(R_i,a) = j} w_j. \end{aligned}$$

Equivalently, this sum can be expressed as

$$\begin{aligned} \sum _{i=1}^N w_{p(R_i,a)} \end{aligned}$$

and, defining

$$\begin{aligned} g(R_i,a) = w_{p(R_i,a)} \end{aligned}$$

for all $i \in \mathbf{N}$, for all $R_i \in \mathcal{P}$ and for all $a \in \mathbf{A}$, we can use (26) to obtain

$$\begin{aligned} a F(\mathbf{R}) b\Leftrightarrow & {} \sum _{j=1}^A w_j \cdot s_j(\mathbf{R},a) \ge \sum _{j=1}^A w_j \cdot s_j(\mathbf{R},b) \\\Leftrightarrow & {} \sum _{i=1}^N g(R_i,a) \ge \sum _{i=1}^N g(R_i,b) \end{aligned}$$

for all $\mathbf{R} \in \mathcal{P}^N$ and for all $a,b \in \mathbf{A}$. Thus, F is representable.

That the reverse implication is not true in general is established by means of the following example. Let $a^0,b^0 \in \mathbf{A}$ be two distinct fixed alternatives. Furthermore, let $R^0 \in \mathcal{P}$ be a fixed strict preference ordering such that $a^0 R^0 b^0 R^0 c$ for all $c \in \mathbf{A} {\setminus } \{a^0,b^0\}$. Define the function $g :\mathcal{P} \times \mathbf{A} \rightarrow \mathbb {R}$ by

$$\begin{aligned} g(R,a) = A - p(R,a) \end{aligned}$$

for all $R \in \mathcal{P} {\setminus } \{R^0\}$ and for all $a \in \mathbf{A}$;

$$\begin{aligned} g(R^0,a) = A - p(R^0,a) \end{aligned}$$

for all $a \in \mathbf{A} {\setminus } \{a^0\}$; and

$$\begin{aligned} g(R^0,a^0) = 2 \cdot (A-1). \end{aligned}$$

Thus, the values of g correspond to the Borda numbers with the exception of $g(R^0,a^0)$, which assigns twice the Borda number to the greatest element $a^0$ in $\mathbf{A}$ according to $R^0$. Define F by letting, for all $\mathbf{R} \in \mathcal{P}^N$ and for all $a,b \in \mathbf{A}$,

$$\begin{aligned} a F(\mathbf{R}) b \; \Leftrightarrow \; \sum _{i=1}^N g(R_i,a) \ge \sum _{i=1}^N g(R_i,b). \end{aligned}$$

It is immediate that F is representable. To prove that F is not a scoring rule, suppose that $\mathbf{A} = \{a^0,b^0,c\}$ and $\mathbf{N} = \{1,2\}$. Let $b^0 R a^0 R c$ and consider the profile $\mathbf{R} = (R^0,R) \in \mathcal{P}^2$. It follows that

$$\begin{aligned} g(R^0,a^0) + g(R,a^0) = 4 + 1 = 5 > 3 = 1 + 2 = g(R^0,b^0) + g(R,b^0) \end{aligned}$$

and hence $a^0 P_{(R^0,R)} b^0$. But we also have

$$\begin{aligned} \mathbf{s}(\mathbf{R},a^0) = (1,1,0) = \mathbf{s}(\mathbf{R},b^0) \end{aligned}$$

and, therefore, any scoring rule has to declare $a^0$ and $b^0$ indifferent. Thus, F cannot be a scoring rule.

(b) Suppose that F is a neutral and representable voting rule. Thus, there exists a function $g :\mathcal{P} \times \mathbf{A} \rightarrow \mathbb {R}$ such that, for all $\mathbf{R} \in \mathcal{P}^N$ and for all $a,b \in \mathbf{A}$,

$$\begin{aligned} a F(\mathbf{R}) b \; \Leftrightarrow \; \sum _{i=1}^N g(R_i,a) \ge \sum _{i=1}^N g(R_i,b). \end{aligned}$$

By Theorem 5.1 of Gärdenfors (1973, pp. 13–14), neutrality is equivalent to the property that, for any two preference orderings R and $R'$ that have the same frame (with the mapping $\phi$), we must have

$$\begin{aligned} g(R,a) - g(R,b) = g(R',\phi (a)) - g(R',\phi (b)) \end{aligned}$$

(27)

for all $a,b \in \mathbf{A}$. As mentioned in the paragraph preceding the theorem statement, any two strict preference orderings have the same frame. Now fix a strict preference ordering $R^0 \in \mathcal{P}$. It follows that, for any profile $\mathbf{R} \in \mathcal{P}^N$ and for any two alternatives $a,b \in \mathbf{A}$, the relative ranking of a and b according to $F(\mathbf{R})$ can be determined by means of the given ordering $R^0$. To see that this is indeed the case, let $\mathbf{R} \in \mathcal{P}^N$ and, for all $i \in \mathbf{N}$, define the function $\phi _i$ by letting

$$\begin{aligned} \phi _i(a) = c \; \Leftrightarrow \; p(R_i,a) = p(R^0,c) \end{aligned}$$

for all $a \in \mathbf{A}$. To simplify the exposition, suppose that, without loss of generality, the alternatives in $\mathbf{A}$ are numbered so that $a_1 R^0 \cdots R^0 a_A$ and define, for all $j \in \{1,\ldots ,A\}$,

$$\begin{aligned} w_j = g(R^0,a_j). \end{aligned}$$

(28)

Theorem 5.1 of Gärdenfors (1973) allows us to invoke (27) to conclude that

$$\begin{aligned} a F(\mathbf{R}) b \; \Leftrightarrow \; \sum _{i=1}^N g(R^0,\phi _i(a)) \ge \sum _{i=1}^N g(R^0,\phi _i(b)) \end{aligned}$$

for all $a,b \in \mathbf{A}$ and, substituting (28), it follows that

$$\begin{aligned} a F(\mathbf{R}) b \; \Leftrightarrow \; \sum _{j=1}^A w_j \cdot s_j(\mathbf{R},a) \ge \sum _{j=1}^A w_j \cdot s_j(\mathbf{R},b) \end{aligned}$$

for all $a,b \in \mathbf{A}$ so that F is a scoring rule. $\square$