Condorcet winners on median spaces

Abstract

We characterize the outcome of majority voting for single-peaked preferences on median spaces. This large class of preferences covers a variety of multi-dimensional policy spaces including products of lines (e.g. grids), trees, and hypercubes. Our main result is the following: If a Condorcet winner (i.e. a winner in pairwise majority voting) exists, then it coincides with the appropriately defined median (“the median voter”). This result generalizes previous findings which are either restricted to a one-dimensional policy space or to the assumption that any two voters with the same preference peak must have identical preferences. The result applies to models of spatial competition between two political candidates. A bridge to the graph-theoretic literature is built.

Appendix: Proof of Proposition 3

Proposition 3

(Existence) A graph $(X,E)$ guarantees existence of a Condorcet winner for any profile of single-peaked preferences if and only if it is a tree.

Proof

We have to prove two directions.

• IF Suppose $\succeq$ is a profile of single-peaked preferences on a tree graph $(X,E)$. Let $q$ be a Weber point, i.e. $q\in \arg \min _{x\in X}\sum _{i\in N}{d(x,x_i^*)}$.¹⁶ Consider any alternative $x\ne q$. We will show that x cannot be preferred over q by a majority of voters. Since $(X,E)$ is a tree graph, there is a unique path between q and x. Let $N(q):=\{y\in X \mid qy\in E \}$, i.e. the set of graphic neighbors of node q. Let y be the neighbor of q that lies graphicly between q and x, i.e. $y\in N(q)$ such that $(x,y,q)\in T_E$. Since trees are connected without circles, we have for any node z, either $d(z,y)<d(z,q)$ or $d(z,q)<d(z,y)$. A well-known result on tree graphs is that for a profile of peaks $(x^*_i)_{i\in N}$ a Weber point q satisfies $\forall y\in N(q), \,\# \{i\in N \; | \; d(x_i^*,q)<d(x_i^*,y) \}\ge \frac{n}{2}$ (cf., e.g. Bandelt and Barthélémy 1984). Thus, for at least $\frac{n}{2}$ voters we have $d(x_i^*,q)<d(x_i^*,y)$, which means that $(x_i^*,q,x)\in T_E$ such that single-peakedness implies $q\succ _i x$.

• ONLY IF For any graph that is not a tree, we will provide a single-peaked preference profile of three voters such that there is no Condorcet winner. Trees are connected and cyclic. Recall that a circle (or cycle) $\mathcal K$ is a sequence of $K(\ge 3)$ distinct nodes $(x_{1},\ldots ,x_{K})$ such that $x_{k}x_{k+1}\in {E}$ for all $k\in \{1,\ldots ,K-1\}$ and $x_Kx_1 \in E$. We first consider unconnected graphs (case 1), then connected graphs where the smallest circle is of odd length (case 2), and finally connected graphs where the smallest circle is of even length (case 3).

• Case 1 $(X,E)$ is unconnected. Then there are two nodes, say $a$ and $b$, which are not connected by any path. Let c be some other node. Note that none of the three nodes $a,\, b,\, c$ is graphicly between the two others. Denote $R:=X\setminus \{a,b,c \}$. Consider three voters N with the following strict preference orderings: $(a,b,c,\sigma _1(R) ),\, (b,c,a, \sigma _2(R) )$, and $(c,a,b,\sigma _3(R) )$, where $\sigma _i$ are orders on R which do not violate single-peakedness. This profile of preferences is single-peaked on $(X,E)$. In majority voting any alternative in R is strictly defeated, while among the three alternatives $a,\, b, \,c$, we have the classic Condorcet cycle.

• Case 2 $(X,E)$ is connected and the smallest circle is of odd length K. If $K=3$, denote a smallest circle by $\mathcal K =(a,b,c )$ and let $R:=X\setminus \{a,b,c \}$. Consider three voters with preferences on $a,\, b,\, c$ exactly as in case 1 and the preferences on the rest R are added in some single-peaked order $\sigma _i$. This profile of preferences is single-peaked on $(X,E)$. There is no Condorcet winner such as in case 1. Now, let $K\ge 5$. Let us consider one smallest circle $\mathcal K$ and label it as follows: $\mathcal K =(a,b_1,b_2,\ldots , b_{\frac{K-1}{2}},d, c_{\frac{K-1}{2}-1}, c_{\frac{K-1}{2}-2}, \ldots ,c_1)$ such as illustrated in Fig. 3 panel (a). Let Z denote the set of nodes on the circle. Now, consider three voters with the following strict preference orderings:

  $$\begin{aligned} \succeq _1&= (a,b_1,c_1,b_2,c_2,\ldots ,b_{\frac{K-1}{2}-1},c_{\frac{K-1}{2}-1},b_{\frac{K-1}{2}},d,\sigma _1(X\setminus Z)) \\ \succeq _2&= (b_1,b_2,\ldots ,b_{\frac{K-1}{2}},d,a,c_1,c_2,\ldots ,c_{\frac{K-1}{2}-1}, \sigma _2(X\setminus Z))\\ \succeq _3&= (c_1,c_2,\ldots ,c_{\frac{K-1}{2}-1},d,a,b_1,b_2,\ldots ,b_{\frac{K-1}{2}}, \sigma _3(X\setminus Z)), \end{aligned}$$

  where the alternatives in $X\setminus Z$ are listed in some single-peaked order $\sigma _i$. Since $\mathcal K$ is a smallest circle, there are no paths between members of the circle which are shorter than the shortest path on the circle. Thus, there is no node $y\in X\setminus Z$ such that $(x,y,z)\in T_E$ for some $x,z \in Z$. Therefore, the profile $\succeq$ is single-peaked on $(X,H)$. In majority voting alternative a defeats any alternative except d. $b_1$ defeats d and, finally, d defeats a.

• Case 3 $(X,E)$ is connected and the smallest circle is of even length K. Let us consider one smallest circle $\mathcal K$ and label it as follows: $\mathcal K =(a,b_1,b_2,\ldots , b_{\frac{K}{2}-1},d, c_{\frac{K}{2}-1}, c_{\frac{K}{2}-2}, \ldots ,c_1)$ such as illustrated in Fig. 3 panel (b). Let Z denote the set of nodes on the circle. Any pair $x,y\in Z$ has a shortest path using only nodes of the circle. However, there might be additional shortest paths for pairs at the opposite sides of the circle, i.e. when $d(x,y)=\frac{K}{2}$.¹⁷ Let Y denote the set of all nodes that belong to such a path but not to the circle, i.e. $Y:=\{y\in X\setminus Z \; | \; \exists x,z\in Z \text{ with } (x,y,z)\in T_E \}$, and let $R:=X\setminus (Z\cup Y)$. Now, consider three voters with the following strict preference orderings:

  $$\begin{aligned} \succeq _1&= (a,b_1,c_1,b_2,c_2,\ldots ,b_{\frac{K}{2}-1},c_{\frac{K}{2}-1},\sigma _1(Y), d, \tilde{\sigma }_1(R)) \\ \succeq _2&= (b_1,b_2,\ldots ,b_{\frac{K}{2}-1},d,a,c_1,c_2,\ldots ,c_{\frac{K}{2}-2},\sigma _2(Y),c_{\frac{K}{2}-1}, \tilde{\sigma }_2(R))\\ \succeq _3&= (c_1,c_2,\ldots ,c_{\frac{K}{2}-1},d,a,b_1,b_2,\ldots ,b_{\frac{K-1}{2}-2}, \sigma _3(Y),b_{\frac{K-1}{2}-1}, \tilde{\sigma }_3(R)), \end{aligned}$$

  where $\sigma _i$ (resp. $\tilde{\sigma }_i$) puts the alternatives in Y (resp. R) in some single-peaked order. As it can be checked, this profile $\succeq$ is single-peaked on $(X,H)$.¹⁸ In majority voting alternative a defeats any alternative except d. $b_{1}$ defeats d and, finally, d defeats a.

$\square$

Fig. 3

A circle of length K, where K is odd in panel (a) and K is even in panel (b)