Dominance in spatial voting with imprecise ideals

Abstract

We introduce a dominance relationship in spatial voting with Euclidean preferences, by treating voter ideal points as balls of radius $\delta$. Values $\delta >0$ model imprecision or ambiguity as to voter preferences from the perspective of a social planner. The winning coalitions may be any consistent monotonic collection of voter subsets. We characterize the minimum value of $\delta$ for which the $\delta$-core, the set of undominated points, is nonempty. In the case of simple majority voting, the core is the yolk center and $\delta$ is the yolk radius.

Appendices

Appendix

1.1 A- Proofs

Throughout this section, $\left( N,W,\{q^{i}\}_{i\in N},\delta \right) \equiv \left( V,\delta \right)$ is a $m-$dimensional spatial voting game where $B(q^{i},\delta )$ is the imprecise ideals locale of the voter $i\in N$.

1.1.1 Proof of Proposition 1

Suppose y is $\delta$-dominated by some point z with respect to a coalition $T\in W$. Let $\mathcal {H}=\{x:||x-z||=||x-y||\}$ be the hyperplane orthogonal to and bisecting segment $\left[ y,z\right]$, Fig. 5a. Let $\mathcal {H}^{z}$ (respectively $\mathcal {H}^{y}$) be the open halfspace defined by $\mathcal {H}$ that contains z (respectively y). By definition of $\delta$-dominance, $\forall i\in T,$ $B(q^{i},\delta )\subset \mathcal {H}^{z}$. Hence the set $\mathcal {S}\equiv \bigcup _{i\in T} B(q^{i},\delta )\subset \mathcal {H}^{z}$. Since $\mathcal {H}^{z}$ is convex, the convex hull $Conv(T,\delta )\subset \mathcal {H}^{z}$ as well. Then $y\in \mathcal {H}^{y}\Rightarrow y\not \in \mathcal {H}^{z}\Rightarrow y\not \in Conv(T,\delta )$.

Conversely, suppose $y\not \in Conv(T,\delta )$. The set $\mathcal {S}$ is a finite union of compact sets and therefore is compact. It is known from Caratheodory’s Theorem (see e.g. Theorem  11.1.8.6 in (Berger 1978)) that in $\mathbb {R} ^{m}$ the convex hull of any compact set is compact. Hence the convex hull $Conv(T,\delta )$ is compact. By (Rockafellar 1970, Corollary 11.4.2 p. 99), there exists a hyperplane $\Pi =\{x:\pi \cdot x=\pi _{0}\}$ strictly separating $\left\{ y\right\}$ from $Conv(T,\delta )$, such that $\pi \cdot y<\pi _{0}$ and $\pi \cdot x>\pi _{0},\ \forall x\in Conv(T,\delta )$, Fig. 5-(b).

Let p be the projection of y onto $\Pi$, so $\pi \cdot p=\pi _{0}$ and $p=y+\alpha \pi$ for some scalar $\alpha >0$.

Fig. 5

$\delta$-dominated positions

Geometrically one can see that p is closer than y to every point in $Conv(T,\delta )$ because $p\in \Pi$ and y is on the other side of $\Pi$ from $Conv(T,\delta )$. Algebraically, for all $x\in Conv(T,\delta )$,

$$\begin{aligned} ||p-x||^{2}&=(p-x)\cdot (y+\alpha \pi -x)\\&=\alpha \pi \cdot p-\alpha \pi \cdot x+p\cdot y+||x||^{2}-x\cdot p-x\cdot y\\&<\alpha \pi _{0}-\alpha \pi _{0}+p\cdot y+||x||^{2}-x\cdot p-x\cdot y\\&<\alpha \pi \cdot x-\alpha \pi \cdot y+p\cdot y+||x||^{2}-x\cdot p-x\cdot y\\&=(x-y)\cdot (x+\alpha \pi -p)=||y-x||^{2}. \end{aligned}$$

Therefore, p $\delta$-dominates y with respect to coalition T, which completes the proof. $\square$

1.1.2 Proof of Corollary 1

The proof of the Corollary follows immediately from Proposition 1 and the definition of $\delta$-core. $\square$

1.1.3 Proof of Corollary 2

Consider the function $\rho$ defined from $\mathbb {R} ^{m}$ to $\mathbb {R}$ by: $\forall z\in \mathbb {R} ^{m}$, $\rho (z)=\underset{S\in W}{\max }$ $d\left( z,Conv(S)\right)$. Finding a minimum for the function $\rho$ may be restricted to $Conv\left( N \right)$, the convex hull of $\left\{ q^{i}\right\} _{i\in N}$. Indeed, if $z\notin Conv\left( N\right)$, then there is a point l such that for all $S\in W,$ $d\left( l,Conv(S)\right) <d\left( z,Conv(S)\right)$. However for all $S\in W$, $d\left( l,Conv(S)\right) <d\left( z,Conv(S)\right)$ implies that $\underset{S\in W}{\max }$ $d\left( l,Conv(S)\right) <\underset{S\in W}{\max }$ $d\left( z,Conv(S)\right) ,$ i.e. $\rho (l)<\rho (z).$

The function $\rho$ is continuous as it is the maximum of finitely many continuous functions. By Caratheodory’s Theorem $Conv\left( N\right)$ is compact. It follows that $\rho$ attains its minimum at a point $z^{*}$. Let $\delta ^{*}=\rho (z^{*})=\underset{Z}{\min }\left( \underset{S}{\max }\text { }d\left( z,Conv(S)\right) \right)$. By definition, $\delta ^{*}$ coincides with r the radius of a g-yolk.

$$\begin{aligned} \begin{array}{lll} Moreover, \delta ^{*}=\rho (z^{*})=\underset{S\in W}{\max } d\left( z^{*},Conv(S)\right) &{} \iff &{} \forall S\in W, d\left( z^{*},Conv(S)\right) \le \delta ^{*}\\ &{} \iff &{} \forall S\in W, z^{*}\in Conv(S,\delta ^{*})\\ &{} \iff &{} z^{*}\in \underset{S\in W}{\cap }Conv(S,\delta ^{*} )=C(V,\delta ^{*}) \end{array} \end{aligned}$$

(1)

Therefore, $C(V,\delta ^{*})\ne \varnothing$. Furthermore, if $\delta \ge \delta ^{*}$ then $C(V,\delta ^{*})\subseteq C\left( V,\delta \right)$ i.e. $C\left( V,\delta \right) \ne \varnothing$.

Now, for a $\delta \ge 0,$ assume that $C\left( V,\delta \right) \ne \varnothing .$ Then there exists $x\in C\left( V,\delta \right) =\underset{S\in W}{\cap }Conv(S,\delta )$, such that for all $S\in W,$ $d\left( x,Conv(S)\right) \le \delta .$ It follows that $\rho \left( x\right) =\underset{S\in W}{\max }$ $d\left( x,Conv(S)\right) \le \delta$. We already know that $\delta ^{*}\le \rho \left( x\right)$. Then $\delta ^{*} \le \delta$ i.e. $C\left( V,\delta \right) \ne \varnothing \Rightarrow \delta \ge \delta ^{*}$. $\blacksquare$

1.1.4 Proof of Corollary 3

According to the proof of Corollary 2, $z\in C\left( V,\delta ^{*}\right) =C\left( V,r\right)$, where r is the radius of a g-yolk if and only if for all $S\in W,$ $B\left( z,r\right) \cap Conv(S)\ne \varnothing$. The radius r of all g-yolks is unique since it is minimal. If there is a unique center c such that B(c, r) is a g-yolk, then by definition the $\delta -core$ is the singleton c. $\blacksquare$

B- an example

In this example, we find the center C and radius r of the g-yolk depend on the set of winning coalitions. For $q^{1}=(1,3);$ $q^{2}=(2,0);$ $q^{3}=(0,2);$ $q^{4}=(2,2)$ and $q^{5}=O=(0,0),$ we show that : $\left( i\right)$ for simple majority voting $c=(1,\frac{1+\sqrt{5}}{2})$ and $r=\frac{\sqrt{10}-\sqrt{2}}{4}$; $\left( ii\right)$ for weighted voting game $V=\left[ 51;6,40,77,40\right]$, $c=(1,\frac{\sqrt{10}-1}{3})$ and $r=\frac{\sqrt{10}-1}{3}$ and $\left( iii\right)$ for weighted voting game $V=\left[ 54;6,40,77,40\right]$, $c=(1,r)$ and $r=\sqrt{2}-1$. Figures 4(i), (ii), iii) provided earlier in Sect. 3.3 relate to this example.

$\left( i\right)$ Let $F=(0.5,1.5)=\left( q^{1}q^{5}\right) \cap \left( q^{2}q^{3}\right)$. The radius of the inscribed circle of triangle $q^{1}q^{2}F$ is the ratio of its area to semiperimeter. A unitary orientation vector of the line $\left( q^{1}q^{2}\right)$ is $\frac{\overrightarrow{q^{1}q^{2}}}{\left\| \overrightarrow{q^{1}q^{2}}\right\| }=\left( \frac{1}{\sqrt{10}},-\frac{3}{\sqrt{10}}\right)$. A normal vector of $\left( q^{1}q^{2}\right)$ is $\overrightarrow{u}=\left( \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right)$. Line $\left( q^{1}q^{2}\right)$ is the set of points $p=(x_{1},x_{2})$ that check the normalized equation $\overrightarrow{u}\cdot \overrightarrow{q^{2}p}=0$ i.e. $\overrightarrow{u}\cdot \overrightarrow{Op}=\overrightarrow{u}\cdot \overrightarrow{Oq^{2}}$. It follows that $\left( \frac{3}{\sqrt{10}},\frac{1}{\sqrt{10}}\right) \cdot (x_{1},x_{2})=\frac{6}{\sqrt{10}}$. The distance from F to $\left( q^{1}q^{2}\right)$ is therefore $\frac{6}{\sqrt{10}}-\overrightarrow{u} \cdot \overrightarrow{OF}=(\frac{1}{\sqrt{10}})\left( 6-(1.5+1.5)\right) =\frac{3}{\sqrt{10}}$. Therefore the area of the triangle $q^{1}q^{2}F$ is $\frac{1}{2}(\frac{3}{\sqrt{10}})\overline{q^{1}q^{2}}=\frac{3}{2}$. Its semiperimeter is $\frac{1}{2}(\sqrt{10}+\frac{3}{2}\sqrt{2}+\frac{1}{2} \sqrt{10})$ $=\frac{3}{4}(\sqrt{10}+\sqrt{2})$. The radius is $r=\frac{2}{\sqrt{10}+\sqrt{2}}=\frac{1}{4}(\sqrt{10}-\sqrt{2})$.

The normalized equation for $\left( q^{2}q^{3}\right)$ is $\frac{1}{\sqrt{2}}(x_{1}+x_{2})=\sqrt{2}$. The $c_{1}$ coordinate of the yolk center is 1, by symmetry. Let the yolk center have coordinates $(1,c_{2})$. It must satisfy $\frac{1}{\sqrt{2}}(1+c_{2})=\sqrt{2}+r$. Hence $\frac{\sqrt{2}}{2}+c_{2}\frac{\sqrt{2}}{2}=\sqrt{2}+\frac{\sqrt{10}}{4}-\frac{\sqrt{2}}{4}$ $\Rightarrow c_{2}=\sqrt{2}\left( \frac{\sqrt{2}}{4}+\frac{\sqrt{10}}{4}\right) =\frac{1+\sqrt{5}}{2}$.

To summarize, the yolk’s center is $c=(1,\frac{1+\sqrt{5}}{2})$ and its radius is $r=\frac{\sqrt{10}-\sqrt{2}}{4}$. It also must be verified that this circle intersects the median line defined by the segment $\left[ q^{3},q^{4}\right]$. We require $\frac{1+\sqrt{5}}{2}+r\ge 2$. The calculation is $\frac{1}{4}(2+2\sqrt{5}+\sqrt{10}-\sqrt{2})\approx \frac{8.21}{4}>2$.

$\left( ii\right)$ For the weighted voting case with threshold 51 the minimal coalitions are $\left\{ 2,5\right\}$, $\left\{ 1,3,5\right\}$, $\left\{ 1,2,4\right\}$, and also $\left\{ 3,4,5\right\}$ and $\left\{ 2,3,4\right\}$. The first three are met by the inscribed circle of triangle $q^{1}q^{2}q^{5}$. The triangle’s area is $\frac{1}{2}(2)(3)=3$. Its semiperimeter is $\frac{1}{2}\left( \sqrt{10}+\sqrt{10}+2\right) =\sqrt{10}+1$. The radius is therefore $r=\frac{3}{\sqrt{10}+1}=\frac{\sqrt{10} -1}{3}$. The center has coordinates $(1,\frac{\sqrt{10}-1}{3})$ because the circle is tangent to the convex combinations of the winning coalition $\left\{ 2,5\right\}$.

$\left( iii\right)$ For the weighted voting case with threshold 54, voter 1 is not a member of any minimal winning coalition. Those coalitions are $\left\{ 2,5\right\}$, $\left\{ 3,4,5\right\}$, and $\left\{ 2,3,4\right\}$. The yolk is the inscribed circle of the triangle with vertices $q^{2}$, $q^{5}$, and $\left( q^{2}q^{3}\right) \cap \left( q^{4}q^{5}\right) =(1,1)$. The triangle area is 1 since it is $\frac{1}{4}$ of $2^{2}$ by symmetry of the square $q^{3}q^{4}q^{2}q^{5}$. The semiperimeter is $\frac{1}{2}(2+2\sqrt{2})=\sqrt{2}+1$. The radius is therefore $r=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$. As in the previous case, the circle must be tangent to $\left( q^{2}q^{5}\right)$, so its center is $(1,\sqrt{2}-1)$. $\blacksquare$