# Anonymous and neutral majority rules

## Abstract

In the standard arrovian framework and under the assumption that individual preferences and social outcomes are linear orders on the set of alternatives, we provide necessary and sufficient conditions for the existence of anonymous and neutral rules and for the existence of anonymous and neutral majority rules. We also determine a general method for constructing and counting these rules and we explicitly apply it to some simple cases.

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1. 1.

When alternatives are two, Campbell and Kelly (2011, 2013) prove that social choice functions satisfying monotonicity and suitable weak versions of anonymity and neutrality are consistent with the majority principle, whether the number of individuals is odd or even.

2. 2.

For the sake of completeness, following Can and Storken (2013, Example 4), we prove again that fact in Sect. 5.

3. 3.

Theorem A is just a rephrasing of Theorem 5, while Theorem B of Theorem 12.

4. 4.

The use of group theory in social choice theory is not a novelty. Kelly (1991), for instance, discusses the role of symmetry in the arrovian framework through suitable subgroups of the symmetric group. Many results proved within the topological approach to social choices developed by Chichilnisky (1980) require the use of algebraic concepts and in particular that of symmetric group. The geometric approach and the symmetry arguments introduced by Donald Saari to understand paradoxes in voting, have been recently cast in a fully algebraic framework by Daugherty et al. (2009), inaugurating what is now called algebraic voting theory.

5. 5.

When alternatives are two and individual and social preferences can express indifference between them, the problem to count the anonymous and neutral rules was solved by Perry and Powers (2008).

6. 6.

A partition of a nonempty set $$X$$ is a set of nonempty pairwise disjoint subsets of $$X$$ whose union is $$X$$.

7. 7.

Actually, Eğecioğlu (2009) claims that the formula for $$R$$ was previously proved by Giritligil and Doğan (An Impossibility Result on Anonymous and Neutral Social Choice Functions, preprint). However, we could not find that reference.

## References

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## Acknowledgments

We wish to thank Domenico Menicucci for reading and commenting a preliminary draft. We are also grateful to two anonymous referees for providing useful suggestions allowing to improve the readability of the paper. In particular, one of the referees contributed to simplify the structure of the proof of Proposition 2 and proposed a more direct approach to the proof of Proposition 18. Daniela Bubboloni was supported by the MIUR project “Teoria dei gruppi ed applicazioni (2009).”

## Author information

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### Corresponding author

Correspondence to Michele Gori.

## Appendix

### Proof of Proposition 17

Since $$p\in \mathcal {P}(b)$$ we know that there are a partition $$B=\{B_k\}_{k=1}^{r(b)}$$ of $$H$$ with $$|B_k|=b_k$$ and distinct $$q_1,\ldots q_{r(b)}\in \mathcal {L}(N)$$ such that, for every $$i\in H$$ and $$k\in \{1\ldots ,r(b)\}$$,

\begin{aligned} p_i=q_k\quad \hbox { if and only if} \quad i\in B_k. \end{aligned}
(15)

Consider the subgroup $$U$$ of $$S_h$$ which leaves fixed each $$B_k$$, that is,

\begin{aligned} U=\left\{ \varphi \in S_h: \forall k\in \{1,\ldots ,r(b)\}, \varphi ( B_k)=B_k\right\} . \end{aligned}

Clearly $$U$$ is isomorph to $$\times _{k=1 }^{r(b)}\mathrm {Sym}(B_k)$$ and then also to $$\times _{j=1 }^h S_j^{a_j(b)}.$$

We show that $$\mathrm {Stab}_G(p)=U\times \{id\}$$. Let $$(\varphi ,id)\in U \times \{id\}$$ and see that $$p^{(\varphi ,id)}=p$$, that is, for every $$i\in H, p_{\varphi (i)}=p_i$$. Since $$B$$ is a partition of $$H,$$ we can verify that, for every $$k\in \{1,\ldots ,r(b)\}$$ and $$i\in B_k, p_{\varphi (i)}=p_i.$$ But if $$i\in B_k$$ also $$\varphi (i)\in B_k,$$ by definition of $$U$$, and thus $$p_i=q_k=p_{\varphi (i)},$$ by (15). Next let $$g\in \mathrm {Stab}_G(p)$$. Then, by Proposition 2, we have that $$g=(\varphi ,id)$$ for some $$\varphi \in S_h$$ and thus $$p_{\varphi (i)}=p_i$$ for all $$i\in H.$$ We need to see that $$\varphi \in U$$, that is, $$\varphi (B_k)\subseteq B_k$$ for all $$k\in \{1,\ldots ,r(b)\}.$$ But if $$i\in B_k$$ we have $$p_i=q_k$$ and thus also $$p_{\varphi (i)}=q_k,$$ which by (15) gives $$\varphi (i)\in B_k.$$

It follows that

\begin{aligned} \left| \mathrm {Stab}_G(p)\right| =\left| U\right| =\left| \times _{j=1 }^h S_j^{a_j(b)}\right| =\prod _{j=1}^h\,j!^{\,a_j(b)} \end{aligned}

which, by (8), gives immediately (12). $$\square$$

### Proof of Proposition 18

For every $$p\in \mathcal {P}(b)$$ and $$j\in \{1,\ldots ,h\}$$, define

\begin{aligned} Q_j(p)=\big \{q_0\in \mathcal {L}(N): |\{i\in H: p_i=q_0\}|=j \big \}. \end{aligned}

It is immediately observed that, for every $$p\in \mathcal {P}(b), j\in \{1,\ldots ,h\}$$ and $$(\varphi ,\psi )\in G$$,

\begin{aligned} Q_j(p^{(\varphi ,\psi )})=\psi Q_j(p). \end{aligned}
(16)

Moreover, for every $$p\in \mathcal {P}(b)$$, define $$Q(p)=\left( Q_j(p)\right) _{j=1}^h$$ and note that it belongs to the set

\begin{aligned}&\mathcal {Q}(b)=\left\{ \left( Q_j\right) _{j=1}^h : \forall j,k\in \{1,\ldots ,h\} \text{ with } j\ne k, Q_j\subseteq \mathcal {L}(N), \right. \\&\left. |Q_j|=a_j(b), Q_j\cap Q_k=\varnothing \right\} . \end{aligned}

Since $$\sum _{j=1}^ha_j(b)=r(b)$$, we have that

\begin{aligned} |\mathcal {Q}(b)|=\left( {\begin{array}{c}n!\\ a_1(b)\end{array}}\right) \left( {\begin{array}{c}n!-a_1(b)\\ a_2(b)\end{array}}\right) \ldots \left( {\begin{array}{c}n!-r(b)+a_h(b)\\ a_h(b)\end{array}}\right) =\frac{\left( {\begin{array}{c}n!\\ r(b)\end{array}}\right) r(b)!}{\prod _{j=1}^ha_j(b)!}. \end{aligned}

For every $$Q=\left( Q_j\right) _{j=1}^h\in \mathcal {Q}(b)$$, let $$\mathcal {P}(Q)=\{p\in \mathcal {P}:Q(p)=Q\}$$. Note that the set $$\mathcal {P}(Q)$$ is well defined and nonempty because $$\sum _{j=1}^h ja_j(b)=h.$$ Since the profiles in $$\mathcal {P}(Q)$$ are obtained one from the other simply acting with a permutation of the individuals, $$\mathcal {P}(Q)$$ is contained in a unique orbit. Consider then the function $$f:\mathcal {Q}(b)\rightarrow \{p^G: p\in \mathcal {P}(b)\}$$ defined, for every $$Q\in \mathcal {Q}(b)$$, as the unique orbit containing $$\mathcal {P}(Q)$$ and prove that, for every $$p\in \mathcal {P}(b)$$,

\begin{aligned} f^{-1}(p^G)=\left\{ \left( Q_j(p^{(id,\psi )})\right) _{j=1}^h \in \mathcal {Q}(b): \psi \in S_n \right\} . \end{aligned}

Let $$Q=\left( Q_j(p^{(id,\psi )})\right) _{j=1}^h$$ where $$\psi \in S_n$$. Then $$p^{(id,\psi )}\in \mathcal {P}(Q)$$ and therefore $$p^G=f(Q).$$ This shows that $$\left\{ \left( Q_j(p^{(id,\psi )})\right) _{j=1}^h \in \mathcal {Q}(b): \psi \in S_n \right\} \subseteq f^{-1}(p^G)$$. To get the other inclusion let $$Q\in f^{-1}(p^G)$$: then $$f(Q)=p^G$$, which implies $$\mathcal {P}(Q)\subseteq p^G$$. Picking $$q\in \mathcal {P}(Q)$$, we have that there exists $$(\varphi , \psi )\in G$$ such that $$q=p^{(\varphi , \psi )}$$. Then, by (16), it follows that, for every $$j\in \{1,\ldots ,h\}$$,

\begin{aligned} Q_j=Q_j(q)=Q_j(p^{(\varphi , \psi )})=\psi Q_j(p)=Q_j(p^{(id,\psi )}), \end{aligned}

which says $$Q=\left( Q_j(p^{(id,\psi )})\right) _{j=1}^h$$.

Let us show now that, for every $$p\in \mathcal {P}(b)$$,

\begin{aligned} \left| \left\{ \left( Q_j(p^{(id,\psi )})\right) _{j=1}^h \in \mathcal {Q}(b): \psi \in S_n \right\} \right| =n! \end{aligned}

by proving that, for every $$\psi _1,\psi _2\in S_n$$,

\begin{aligned} \left( Q_j(p^{(id,\psi _1)})\right) _{j=1}^h=\left( Q_j(p^{(id,\psi _2)})\right) _{j=1}^h \quad \Rightarrow \quad \psi _1=\psi _2. \end{aligned}

Assume that, for every $$j\in \{1,\ldots ,h\}$$, $$Q_j(p^{(id,\psi _1)})=Q_j(p^{(id,\psi _2)})$$. Then there exists $$\varphi \in S_h$$ such that $$p^{(id,\psi _1)}=[p^{(id,\psi _2)}]^{(\varphi ,id)}$$, which by (7), gives $$p^{(\varphi , \psi _1^{-1}\psi _2)}=p$$. Thus, $$(\varphi , \psi _1^{-1}\psi _2)\in \mathrm {Stab}_G(p)$$ and by Proposition 2, it follows that $$\psi _1^{-1}\psi _2=id$$, that is, $$\psi _1=\psi _2.$$

As a consequence

\begin{aligned} |\{p^G: p\in \mathcal {P}(b)\}|=\frac{1}{n!}|\mathcal {Q}(b)|=\frac{\left( {\begin{array}{c}n!\\ r(b)\end{array}}\right) r(b)!}{n!\prod _{j=1}^ha_j(b)!}, \end{aligned}

and the proof is complete. $$\square$$

### Proof of equality (13)

It is well known (see, for instance, Feller (1957)) that the number $$W(m,k)$$ of ways of distributing $$m\in \mathbb {N}$$ indistinguishable balls into $$k\in \mathbb {N}$$ distinguishable boxes is given by

\begin{aligned} W(m,k)=\left( {\begin{array}{c}m+k-1\\ k-1\end{array}}\right) . \end{aligned}

We also show that

\begin{aligned} W(m,k)=\sum _{b \in \Pi _k(m)}\frac{\left( {\begin{array}{c}k\\ r(b)\end{array}}\right) r(b)!}{\prod _{j=1}^ma_j(b)\,!} \end{aligned}
(17)

Let $$r\in \mathbb {N}$$ with $$r\le k$$ be fixed. We associate with each partition $$b\in \Pi (m)$$ such that $$r(b)=r$$ some distributions of the $$m$$ balls into the $$k$$ boxes: form bunches of $$b_1$$ balls, $$b_2$$ balls, up to $$b_r$$ balls exhausting the $$m$$ available balls. Note that there is only a way to do that because the balls are indistinguishable.

Now choose $$r$$ boxes among the $$k$$ available to distribute the $$r$$ bunches of balls each in a different box: since the boxes are distinguishable you have $$k$$ choices for the first box in which you put the $$b_1$$ balls, $$k-1$$ choices for the second box in which you put the $$b_2$$ balls, up to $$k-r+1$$ choices for the box in which you put the last bunch of $$b_r$$ balls. Note that $$a_j(b)\in \mathbb {N}\cup \{0\}$$ counts the number of bunches of order $$j$$ for $$j\in \{1,\dots ,m\}:$$ since the balls are indistinguishable the $$a_j(b)$$ bunches of order $$j$$ may be interchanged among themselves producing the same final result. Thus we reach $$\frac{\left( {\begin{array}{c}k\\ r\end{array}}\right) r!}{\prod _{j=1}^ma_j(b)\,!}$$ different distributions associated with $$b\in \Pi (m)$$ with $$r(b)=r.$$ It is clear that, with $$r$$ fixed, different choices of the partition $$b$$ produce different distributions. It is also clear that the varying of $$r$$ leads to different distributions, because $$r$$ is the number of boxes involved in the distribution of the balls. In other words, we have in total reached

\begin{aligned} \sum _{r=1}^k\;\sum _{b\in \Pi (m),r(b)=r}\frac{\left( {\begin{array}{c}k\\ r\end{array}}\right) r!}{\prod _{j=1}^ma_j(b)\,!}=\sum _{b\in \Pi _k(m)}\frac{\left( {\begin{array}{c}k\\ r(b)\end{array}}\right) r(b)!}{\prod _{j=1}^ma_j(b)\,!} \end{aligned}

ways of distributing the $$m$$ indistinguishable balls into the $$k$$ distinguishable boxes. On the other hand, if any distribution is given, we may think that it arises from our procedure. Namely, look into each box, count the balls inside and extract them. Then, consider the number $$r$$ of boxes containing at least one ball and arrange the number of balls found in the boxes in a non-decreasing order reaching a partition $$b\in \Pi (m)$$, with $$r(b)=r$$. Now among the possibilities of our procedure starting from $$b\in \Pi (m)$$, there is the one consisting in putting the bunches in the boxes where they originally were. Thus, we have shown (17). $$\square$$

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