The informational basis of scoring rules

Abstract

We consider voting wherein voters assign a certain score to each of the many available alternatives. We study the normative properties of procedures that aggregate the scores collected in the ballot box. A vast class of ballot aggregators, including procedures based on the pairwise comparison of alternatives, satisfy May’s famous conditions in our framework. We prove that, within such a plethora of procedures, scoring rules are singled out by a property related to their informational basis: in order to determine the winner, they do not take into account the specific distribution of scores chosen by each voter. The result is shown to hold regardless of the introduction of asymmetry among the alternatives.

Appendices

Appendix 1: Ballot aggregators based on the pairwise comparisons of alternatives

In this section we provide a definition of some ballot aggregators based on the pairwise comparisons of alternatives, the Condorcet rules. Usually such a comparison is based on the whole ranking over alternatives that is asked to each voter. We adapt the procedures to our framework in the sense that the pairwise comparison is based on the information that is made available by the balloting procedure. The definitions are adaptations of those one can find in Fishburn (1977) and Laslier (1997).

For each $m \in N$ and $\alpha \in A$, let $s_m(\alpha ,v)$ denote the score that voter m has assigned to alternative $\alpha$ at the vote profile $v \in V_B$. In our framework this amounts to count the number of ballots that this voter has cast in favor of alternative $\alpha$. For all $v \in V_B$, and for any pair of alternatives $\alpha , \beta \in A$, let $p(\alpha , \beta ,v)$ be the number of voters which assign a higher score to alternative $\alpha$ than to alternative $\beta$, i.e.

$$\begin{aligned} p(\alpha ,\beta ,v)=\left. \#\left\{ m\in N\right| s_m(\alpha ,v) \ge s_m(\beta ,v)\right\} . \end{aligned}$$

The pairwise comparison of the alternatives $\alpha ,\beta \in A$ is based on such scores. We say that $\alpha$ is not defeated by $\beta$ at the profile $v \in V_B$, noted $\alpha D_{v}\beta$, if and only if $p(\alpha ,\beta ,v) \ge p (\beta ,\alpha ,v)$ (alternatives $\alpha$ and $\beta$ are tied at the pairwise comparison if $p(\alpha ,\beta ,v)=p(\beta ,\alpha ,v)$).

Moreover, for all $v \in V_B$ and $\alpha \in A$, we denote:

• $b(\alpha ,v)=\sum \nolimits _{\beta \in A}p(\alpha ,\beta ,v);$

• $c(\alpha ,v)=\#\{\beta \in A: \alpha D_{v}\beta \}-\#\{\beta \in A: \beta D_{v}\alpha \};$

• $d(\alpha , v) =$ number of voters in the largest subset from v for which $\alpha$ ties or beats every other alternatives on the basis of simple majority with respect to this subset. If there is no such subset then $d(\alpha ,v) = 0$;

• $e(\alpha ,v)=\min _{\beta \in A{\setminus }\{\alpha \}}p(\alpha ,\beta ,v).$

1.1 Black’s rule

For all $v \in V_B$, $W_{B}(v) = \{\alpha \in A: \beta D_{v} \alpha \, \text {for no}\,\beta \in A\}$ if the latter set is nonempty. Otherwise W is determined by Borda count over the vote profile:

$$\begin{aligned} W_{B}(v) =\left\{ \alpha \in A:b(\alpha ,v)\ge b(\beta ,v)\quad \text {for all}\quad \beta \in A\right\} . \end{aligned}$$

1.2 Copeland’s rule

For all $v \in V_B$, $W_{Cop}(v) =\{\alpha \in A: c(\alpha ,v) \ge c(\beta ,v)$ for all $\beta \in A\}.$

1.3 Young’s rule

For all $v \in V_B$ and $d(\alpha ,v)$, let $d^{*}(\alpha ,v) =$ $\lim _{n\rightarrow \infty }d(\alpha ,nv)/n$. Then

$$\begin{aligned} W_{You}(v)=\left\{ \alpha \in A: d^{*}(\alpha ,v)\ge d^{*}(\beta ,v)\quad \text {for all}\quad \beta \in A\right\} . \end{aligned}$$

Note that $nV_B$, for any integer n, stands for the fact the vote profile v being replicated n times.

1.4 Condorcet’s rule (or Minimax function or Simpson–Kramer rule)

For all $v \in V_B$,

$$\begin{aligned} W_{Con}(v)= \left\{ \alpha \in A: e(\alpha ,v)\ge e(\beta ,v)\quad \text {for all}\quad \beta \in A\right\} . \end{aligned}$$

1.5 Kemeny’s rule

We first need to define the Kemeny distance between the vote of some agent $i \in N$ at some profile $v \in V$ (i.e., $v^i$) and some other profile $u \in V$. The Kemeny distance, denoted by $\delta (v^i,u)$, is the number of pairs $(\alpha ,\beta )$ whose corresponding relative score at $v^i$ and u differ:

$$\begin{aligned} \delta (v^i,u)=\#\left\{ (\alpha ,\beta )\in A\times A|s_{\alpha }\left( v^i\right) >s_{\beta }\left( v^i\right) \quad \text {and}\quad s_{\alpha }(u)<s_{\beta }(u)\right\} . \end{aligned}$$

For any vote profile v,

Remark 1

Under approval, Borda and plurality balloting, all the rules listed here satisfy Universal Domain, Anonymity, Neutrality and Positive Responsiveness. Under approval and Borda balloting, they all fail to satisfy Ballot anonymity.

Appendix 2: Proof of Theorem 2

In proving Theorem 2 we use the following equivalent definition of qualified scoring rules. The ballot aggregator W is the qualified scoring rule on B if and only if, for all $\alpha ,\beta \in A$, there exists a strictly positive real number $\widehat{r^{\alpha \beta }}$, with $\widehat{r^{\beta \alpha }}=\frac{1}{\widehat{r^{\alpha \beta }}}$, such that for all $v \in V_B$,

$$\begin{aligned} \alpha \in W(v) \Longleftrightarrow s_{\alpha }(v)\ge \widehat{r^{\alpha \beta }}s_{\beta }(v)\quad \text {for all}\quad \beta \ne \alpha , \end{aligned}$$

with $\widehat{r^{\alpha \beta }}=\widehat{r^{\alpha \gamma }}\widehat{r^{\gamma \beta }}$ for any $\gamma \ne \alpha , \beta$. Let a ballot aggregator W satisfy all the listed conditions. The following lemmata, together with lemma 1 will be useful for the proof. We define for any strictly positive integer p, $p/0=\infty$. The balloting procedure B is assumed to be neutral and rich.

Fix some alternative $\gamma \in A$. For some $u \in V_B$ with $s_{\gamma }(u)>0$, and any $\alpha \in A$, let $r_{\alpha }(u)=\frac{s_{\alpha }(u)}{s_{\gamma }(u)}$. Two vote profiles $u,u^{\prime } \in V_B$ have the same score ratios if $r(u)=r(u^{\prime })=r \in {\mathbb {Q}}_+^{|A|}$ with $r(u)=\{r_{\alpha }(u)\}_{\alpha \in A}$. For a fixed $r \in {\mathbb {Q}}_+^{|A|}$ the family of such vote profiles is denoted by $V_{\gamma }^r$ with $V_{\gamma }^r=\{u \in V_B | r(u)=r\}$.

The proof is divided in three steps.

• Step A shows that any two vote profiles whose score profiles are proportional must lead to the same winning set.

• Step B shows how the winning set changes when the score ratio between alternatives changes.

• Step C concludes the proof.

1.1 Step A

Lemma 3

Consider some $r=\{r_{\alpha }\}_{\alpha \in A}$ with $0 \le r_{\alpha }<\infty$ for each $\alpha \in A$. There exists at least some $v^{\prime } \in V_{\gamma }^{r}$ such that, for any $v \in V_{\gamma }^r$ we have $s(v)=m s(v^{\prime })$ for some $m \in {\mathbb {N}}$.

Proof

Let us pick some $r=\{r_{\alpha }\}_{\alpha \in A}$ with $0 \le r_{\alpha }<\infty$ for each $\alpha \in A$ and consider the set $V_{\gamma }^r$. Choose $v^{\prime }\in V_{\gamma }^r$ with the smallest value of $s_{\gamma }$. If $V_{\gamma }^r$ is non-empty, $v^{\prime }$ exists as $s_{\gamma }$ is integer valued. By construction $s(v^{\prime })=r s_{\gamma }(v^{\prime })$ so that, for all $\alpha \in A$, $r_{\alpha } s_{\gamma }(v^{\prime })$ is an integer. Moreover for all $m \in {\mathbb {N}}_+$ and for all $\alpha \in A$, $m r_{\alpha } s_{\gamma }(vv)$ is still an integer. Hence, for all $v \in V_B$ such that $s(v)=ms(v^{\prime })$, $v \in V_{\gamma }^r$. Assume, by contradiction, that there exists $u \in V_{\gamma }^r$ and some strictly positive ${\widehat{m}}\in \mathbb {R}_+{\setminus }{\mathbb {N}}_+$ such that $s(u)={\widehat{m}}s(v^{\prime })$. Let $\lfloor {\widehat{m}} \rfloor$ be the greatest integer such that $\lfloor {\widehat{m}} \rfloor <{\widehat{m}}$. This implies that $0<{\widehat{m}}-\lfloor {\widehat{m}} \rfloor <1$. Take $u^{\prime } \in V_B$ with $s(u^{\prime })=\lfloor {\widehat{m}} \rfloor s(v^{\prime })$. One should notice that, for all $\alpha \in A$, $s_\alpha (u^{\prime })< s_\alpha (u)$, therefore $u^{\prime }$ exists as u exists by assumption and the domain is rich. Moreover by construction, $u^{\prime }\in V_{\gamma }^r$. Take also $u^{\prime \prime } \in V_B$ such that $s(u^{\prime \prime })=s(u)-s(u^{\prime })$. The vote profile $u^{\prime \prime }$ exists because the balloting procedure is rich. Again, it suffices to observe that by construction, for all $\alpha \in A$, $s_\alpha (u^{\prime \prime }) < s_\alpha (u)$. For each $\alpha \in A$,

$$\begin{aligned} s_{\alpha }\left( u^{\prime \prime }\right) =s_{\alpha }(u)-s_{\alpha }\left( u^{\prime }\right) ={\widehat{m}}s_{\alpha }\left( v^{\prime }\right) -\lfloor {\widehat{m}} \rfloor s_{\alpha }\left( v^{\prime }\right) =\left( {\widehat{m}}-\lfloor {\widehat{m}} \rfloor \right) s_{\alpha }(v^{\prime })<s_{\alpha }\left( v^{\prime }\right) \end{aligned}$$

(2)

and

$$\begin{aligned} \frac{s_{\alpha }\left( u^{\prime \prime }\right) }{s_{\gamma }\left( u^{\prime \prime }\right) }=\frac{\left( {\widehat{m}}-\lfloor {\widehat{m}} \rfloor \right) \left( s_{\alpha }\left( v^{\prime }\right) \right) }{\left( {\widehat{m}}-\lfloor {\widehat{m}}\rfloor \right) \left( s_{\gamma }\left( v^{\prime }\right) \right) }=r_{\alpha }. \end{aligned}$$

(3)

From Eq. (3), $u^{\prime \prime } \in V_{\gamma }^r$. However, in light of Eq. (2), $s_{\alpha }(u^{\prime \prime })<s_{\alpha }(v^{\prime })$ which contradicts the definition of $v^{\prime }$. $\square$

Lemma 4

Let a ballot aggregator $W:V_B \rightrightarrows A$ satisfy $\textit{BA}$ and $\textit{CP}$. For any $r \in {\mathbb {Q}}^{|A|}_+$ and $\gamma \in A$ and for each $u,v \in V_{\gamma }^r$, $W(u)=W(v)$.

Proof

In order to prove our statement, we show that for each $v \in V_{\gamma }^r$, $W(v)=W(v^{\prime })$. Note that due to Lemma 3 and $\textit{BA}$, given $r \in {\mathbb {Q}}^{|A|}_+$ and $\gamma \in A$, there exists some $v^{\prime } \in V_{\gamma }^r$ such that any $v \in V_{\gamma }^r$ and some integer m such that $s(v)=m s(v^{\prime })$. Hence, one can decompose v in m identical complementary vote profiles $v^{\prime }$ (up to a permutation of the ballots) with identical winning set due to $\textit{BA}$. $\textit{CP}$ then directly implies that $W(v)=W(v^{\prime })$ as required. $\square$

1.2 Step B

Lemma 5

Let a ballot aggregator $W:V_B \rightrightarrows A$ satisfy $\textit{UD}$, SPR and $\textit{GA}$. For any $v \in V_B {\setminus } \overline{v}_{0}$ and any $\alpha \in A$, if $s_{\alpha }(v)=0$, then $\alpha \notin W(v)$.

Proof

Fix some alternative $\beta \in A$. For the sake of contradiction consider a vote profile $v \in V_B {\setminus } \overline{v}_{0}$ such that $s_{\beta }(v)=0$, $s_{\gamma }(v)>0$ for any $\gamma \ne \beta$ and $\beta \in W(v)$. Consider now a vote profile $v^{\{1\}}\in V_B$ with $s_{\delta }(v^{\{1\}})=0$ for some $\delta \ne \beta$ and $s_{\gamma }(v^{\{1\}})=s_{\gamma }(v)$ for any $\gamma \ne \delta$. It follows that $s_{\beta }(v^{\{1\}})=s_{\delta }(v^{\{1\}})=0$ and $s_{\gamma }(v^{\{1\}})>0$ for any $\gamma \ne \delta ,\beta$. Moreover, $v^{\{1\}}\succ _{\beta \delta }v$ so by SPR one has $\beta \in W(v^{\{1\}})$. Using this logic it is possible to build a sequence of vote profiles $v^{\{1\}},v^{\{2\}},\ldots ,v^{\{|A|-2\}} \in V_B$, each time setting to zero the score of some alternative $\delta \ne \beta$. So, by SPR, for each $i=1,\ldots ,|A|-2$, $\beta \in W(v^{\{i\}})$ as $v^{\{i\}}\succ _{\beta \delta }v^{\{i+1\}}$. In particular, $\beta \in W(v^{\{|A|-2\}})$. By construction, at $v^{\{|A|-2\}}$, for some $\alpha \ne \beta$, $s_{\alpha }(v^{\{|A|-2\}})>0$ while for any $\gamma \ne \alpha$, $s_{\gamma }(v^{\{|A|-2\}})=0$. Finally, as by $\textit{GA}$, $W(\overline{v}_0)=A$ and since $v^{\{|A|-2\}}\succ _{\alpha \beta }\overline{v}_0$ then SPR implies that $\beta \not \in W(v^{\{|A|-2\}})$, the desired contradiction. $\square$

Let

$$\begin{aligned} R^{\alpha }=\left\{ r \in {\mathbb {Q}}_+ | \exists v \in V_B \ \text {s.t.} \frac{s_{\alpha }(v)}{s_{\beta }(v)}=r\quad \text {and}\quad \alpha \in W(v)\right\} . \end{aligned}$$

and

$$\begin{aligned} R^{\beta }=\left\{ r \in {\mathbb {Q}}_+|\exists v \in V_B \ \text {s.t.} \frac{s_{\alpha }(v)}{s_{\beta }(v)}=r\quad \text {and}\quad \beta \in W(v)\right\} , \end{aligned}$$

Notice that by Lemma 5 both sets are not empty. Let $r^{\alpha \beta }=\inf R^{\alpha }$ and $r^{\beta \alpha }=\sup R^{\beta }$. Since N is finite, both $r^{\alpha \beta }$ and $r^{\beta \alpha }$ are well defined and, by Lemma 5, $r^{\beta \alpha }< \infty$ and $r^{\alpha \beta }>0$.

Lemma 6

Let a ballot aggregator $W: V_B \rightrightarrows A$ satisfy $\textit{UD}$, $\textit{BA}$, SPR, $\textit{GA}$ and $\textit{CP}$. For all $u,v \in V_B \backslash \overline{v}_{0}$ such that, for some pair of alternatives $\alpha ,\beta \in A$, $\frac{s_{\alpha }(u)}{s_{\beta }(u)}>\frac{s_{\alpha }(v)}{s_{\beta }(v)}$,

$$\begin{aligned} \alpha \in W(v)\Longrightarrow \beta \not \in W(u). \end{aligned}$$

Proof

Let $u,v \in V_B \backslash \overline{v}_{0}$ and $\alpha ,\beta \in A$ be such that $\frac{s_{\alpha }(u)}{s_{\beta }(u)}>\frac{s_{\alpha }(v)}{s_{\beta }(v)}$. Assume, for the sake of contradiction, that $\alpha \in W(v)$ and $\beta \in W(u)$. It follows then that $r^{\beta \alpha } > r^{\alpha \beta }$. Let also $E^{+}=\{v \in V_B | \frac{s_{\alpha }(v)}{s_{\beta }(v)}=r^{\beta \alpha }$ and $\beta \in W(v) \}$ and $E^{-}=\{v \in V_B | \frac{s_{\alpha }(v)}{s_{\beta }(v)}=r^{\alpha \beta } \ \text {and} \ \alpha \in W(v)\}$. Take $u^{\prime } \in E^{+}$ and $v^{\prime }\in E^{-}$ such that $s_{\gamma }(u^{\prime })=s_{\gamma }(v^{\prime })=0$ for all $\gamma \ne \alpha ,\beta$. These profiles exist as the balloting procedure is rich and W satisfies SPR.¹⁰ We can now consider the following cases:

• 1. $s_{\alpha }(u^{\prime }) > s_{\alpha }(v^{\prime })$ and $s_{\beta }(u^{\prime }) \le s_{\beta }(v^{\prime })$ [or $s_{\alpha }(u^{\prime }) \ge s_{\alpha }(v^{\prime })$ and $s_{\beta }(u^{\prime }) < s_{\beta }(v^{\prime })$]. This yields a contradiction with SPR as by definition $u^{\prime }\succ _{\alpha \beta }v^{\prime }$ implying that $\beta \not \in W(u^{\prime })$.

• 2. $s_{\alpha }(u^{\prime }) \le s_{\alpha }(v^{\prime })$ and $s_{\beta }(u^{\prime }) \ge s_{\beta }(v^{\prime })$. This contradicts $r^{\beta \alpha } > r^{\alpha \beta }$.

• 3. $s_{\alpha }(u^{\prime }) < s_{\alpha }(v^{\prime })$ and $s_{\beta }(u^{\prime }) < s_{\beta }(v^{\prime })$. In this case, the vote profile $u^{\prime \prime }$ such that $v^{\prime }=u^{\prime } \mid _{u^{\prime \prime }}$ can be defined. Assume first that $W(u^{\prime })\ne W(v^{\prime })$ then by $\textit{CP}$, $\beta \notin W(u^{\prime \prime })$ so that one must have $W(u^{\prime \prime })=\{\alpha \}$. Assume then $W(u^{\prime }) = W(v^{\prime })$, necessarily one must have $W(u^{\prime \prime })=\{\alpha ,\beta \}$ otherwise one would have a contradiction with $\textit{CP}$. In any case $\alpha \in W(u^{\prime \prime })$. However, $\frac{s_{\alpha }(u^{\prime \prime })}{s_{\beta }(u^{\prime \prime })}<r^{\alpha \beta }$ by construction which entails a contradiction with the definition of $r^{\alpha \beta }$.

• 4. $s_{\alpha }(u^{\prime }) > s_{\alpha }(v^{\prime })$ and $s_{\beta }(u^{\prime }) > s_{\beta }(v^{\prime })$. One can use an argument similar to that used in case 3 in order to obtain a contradiction.$\square$

1.3 Step C

Lemma 6 implies that for any $\alpha ,\beta \in A$, $r^{\alpha \beta } \ge r^{\beta \alpha }$.

If $r^{\alpha \beta }=r^{\beta \alpha }$, we let $r^{\alpha \beta }=\widehat{r^{\alpha \beta }}=r^{\beta \alpha }$.

If $r^{\alpha \beta }>r^{\beta \alpha }$, we then let $r^{\alpha \beta } > \widehat{r^{\alpha \beta }} > r^{\beta \alpha }$.

The following lemma concludes the proof of Theorem 2.

Lemma 7

Let a ballot aggregator $W: V_B \rightrightarrows A$ satisfy $\textit{UD}$, $\textit{BA}$, SPR, $\textit{GA}$ and $\textit{CP}$. For any $v \in V_B \backslash \overline{v}_0$ and $\alpha \in A$ ,

$$\begin{aligned} \alpha \in W(v) \Longleftrightarrow s_{\alpha }(v) \ge \widehat{r^{\alpha \beta }}s_{\beta }(v)\quad \text {for}\,\text {any}\quad \beta \ne \alpha . \end{aligned}$$

Proof

As far as the $\ if$ part is concerned, note that Lemma 6 implies that for any $\alpha ,\beta \in A$, $r^{\alpha \beta } \ge r^{\beta \alpha }$. Hence, for any $\alpha \in A$ and $v \in V_B \backslash \overline{v}_0$,

$$\begin{aligned} \alpha \in W(v) \Longrightarrow s_{\alpha }(v) \ge r^{\alpha \beta }s_{\beta }(v)\quad \text {for}\,\text {any}\quad \beta \ne \alpha . \end{aligned}$$

Since by definition, $r^{\alpha \beta }\ge \widehat{r^{\alpha \beta }}$, it follows that

$$\begin{aligned} \alpha \in W(v) \Longrightarrow s_{\alpha }(v) \ge \widehat{r^{\alpha \beta }}s_{\beta }(v) \text {for}\,\text {any } \beta \ne \alpha . \end{aligned}$$

Now, the only if part of the statement remains to be proved. Suppose, for the sake of contradiction, that there exists $\alpha \in A$ and $v^{\prime } \in V_B \backslash \overline{v}_0$ such that $s_{\alpha }(v^{\prime }) \ge \widehat{r^{\alpha \beta }}s_{\beta }(v^{\prime })$ for any $\beta \ne \alpha$ and $\alpha \not \in W(v^{\prime })$. The winner at $v^{\prime }$ must then be some $\beta \ne \alpha$. For any $\beta \ne \alpha$ such that $\frac{s_{\alpha }(v^{\prime })}{s_{\beta }(v^{\prime })}>\widehat{r^{\alpha \beta }}$ one also has $\frac{s_{\alpha }(v^{\prime })}{s_{\beta }(v^{\prime })}>r^{\beta \alpha }$ since $\widehat{r^{\alpha \beta }}\ge r^{\beta \alpha }$. Therefore $\beta \not \in W(v^{\prime })$ by definition of $r^{\beta \alpha }$. Let us then consider, as a possible winner at $v^{\prime }$, some $\beta \ne \alpha$ such that $\frac{s_{\alpha }(v^{\prime })}{s_{\beta }(v^{\prime })} = \widehat{r^{\alpha \beta }}$. It follows that $r^{\alpha \beta }=\widehat{r^{\alpha \beta }}=r^{\beta \alpha }$ since otherwise there is a contradiction with $\beta \in W(v^{\prime })$. By definition of $r^{\alpha \beta }$ there must exist some $u^{\prime } \in V_B \backslash \overline{v}_0$ such that $\alpha \in W(u^{\prime })$ and $\frac{s_{\alpha }(u^{\prime })}{s_{\beta }(u^{\prime })} = r^{\alpha \beta }$. Take the vectors $u^{+}$ and $v^{+}$ sucht that $s_{\alpha }(u^{+})=s_{\alpha }(u^{\prime })$, $s_{\beta }(u^{+})=s_{\beta }(u^{\prime })$, $s_{\alpha }(v^+)=s_{\alpha }(v^{\prime })$, $s_{\beta }(v^{+})=s_{\beta }(v^{\prime })$ and $s_{\delta }(u^{+})=s_{\delta }(v^{+})=0$ for any $\beta \ne \alpha ,\delta$ (these profiles exist since the balloting procedure is rich). SPR implies that $\alpha \notin W(v^+)$ and that $\alpha \in W(u^+)$. This yields a contradiction with Lemma 4 ($u^+$ and $v^+$ have the same score ratios but a different winning set), concluding the proof. $\square$

Appendix 3: Independence of the axioms

We now present some examples [some of which have been adapted from Houy (2007)] that prove that the axioms used in the characterization are independent. For the sake of simplicity we consider three alternatives, namely $\alpha , \beta \in A$ or $\alpha , \beta , \gamma \in A$. For each $v \in V_B$, the total score vector is denoted by $s(v)=(s_{\alpha }(v),s_{\beta }(v), s_{\gamma }(v))$.

1. Coalition permanency Take any two alternatives $\alpha ,\beta \in A$. For each $v \in V$,

• (a) If $s_{\alpha }(v)=s_{\beta }(v)=0$, then $W(v)=\{\alpha ,\beta \}$

• (b) If $s_{\alpha }(v)=s_{\beta }(v)\ge 1$ and $s_{\alpha }(v)$ odd, then $W(v)=\{\alpha \}$.

• (c) If $s_{\alpha }(v)=s_{\beta }(v)\ge 1$ and $s_{\alpha }(v)$ even, then $W(v)=\{\beta \}$.

• (d) If $s_{\alpha }(v)>s_{\beta }(v)$, then $W(v)=\{\alpha \}$.

• (e) If $s_{\alpha }(v)<s_{\beta }(v)$, then $W(v)=\{\beta \}$.

Such a ballot aggregator satisfies $\textit{UD}$, $\textit{BA}$, SPR and $\textit{GA}$. However, it does not satisfy $\textit{CP}$. To see why, take some vote profile u with $s_{\alpha }(u)=s_{\beta }(u)=1$ so that, by definition, $W(u)=\{\alpha \}$. By $\textit{CP}$, any vote profile v with $s_{\alpha }(u)=s_{\beta }(u)=2$ must satisfy $W(v)=\{\alpha \}$, which is in contradiction with $W(v)=\{\beta \}$ which holds by definition.

2. Strong positive responsiveness See Example 2.

3. General abstention

For each $v \in V_B$, $W(v)=\{\alpha \}$. Such a ballot aggregator satisfies $\textit{UD}$, $\textit{BA}$, $\textit{CP}$ and SPR. However, it does not satisfy $\textit{GA}$.

4. Ballot anonymity

For each $v\in V_B$, let $\hat{S}(v)= \sum \nolimits _{i=1}k^{i}v^{i}$, that is a weighted sum of the ballots.

• (a) If $\hat{S}_{\alpha }(v)=\hat{S}_{\beta }(v)$, then $W(v)=\{\alpha ,\beta \}$

• (b) If $\hat{S}_{\alpha }(v)>\hat{S}_{\beta }(v)$, then $W(v)=\{\alpha \}$.

• (c) If $\hat{S}_{\alpha }(v)<\hat{S}_{\beta }(v)$, then $W(v)=\{\beta \}$.

Such a ballot aggregator satisfies $\textit{UD}$, $\textit{GA}$, CP and SPR. However, it does not satisfy BA.