Representation-compatible power indices

Abstract

This paper studies power indices based on average representations of a weighted game. If restricted to account for the lack of power of null voters, average representations become coherent measures of voting power, with power distributions being proportional to the distribution of weights in the average representation. This makes these indices representation-compatible, a property not fulfilled by classical power indices. In this paper we introduce two computationally cheaper alternatives to the existing representation-compatible power indices, and study the properties of a family that now comprises four measures.

Appendix

1.1 Example $\left[ 3;2,1,1\right]$

The sets of minimal winning and maximal losing coalitions for the game $\left[ 3;2,1,1\right]$ are, respectively, $\{\{1,2\},\{1,3\},\{2,3\}\}$ and $\{\{1\},\{2,3\}\}$. Since voters 2 and 3 are equivalent, there are two equivalence classes in this game. There are no nulls.

Using Lemma 2, we obtain the following constraints:

$$\begin{aligned} w_1+w_2>w_1\Longleftrightarrow & {} w_2>0;\\ w_1+w_3>w_1\Longleftrightarrow & {} w_3>0;\\ w_1+w_2>w_2+w_3\Longleftrightarrow & {} w_1>w_3;\\ w_1+w_3>w_2+w_3\Longleftrightarrow & {} w_1>w_2. \end{aligned}$$

In addition, $w_1,w_2,w_3\ge 0$ and $w_1+w_2+w_3=1$. Eliminating $w_3$ and removing the redundant constraints yields the following inequalities: $w_2>0$, $w_2<1-w_1$, $w_2>1-2w_1$, $w_2<w_1$. Since $1-2w_1<w_1$ and $1-w_1>0$, we have $w_1\in \big (\frac{1}{3},1\big )$. For $w_1\in \big (\frac{1}{3},\frac{1}{2}\big )$, we have $w_2\in \big (1-2w_1,w_1\big )$. For $w_1\in \big [\frac{1}{2},1\big )$, we have $w_2\in \big (0,1-w_1\big )$. The polytope is thus given by

$$\begin{aligned} \mathsf {V}^d(v)=\left\{ (w_1,w_2)\in \mathbb {R}^2_{\ge 0}\mid w_2\ge 0, w_2\le 1-w_1, w_2\ge 1-2w_1, w_2\le w_1\right\} . \end{aligned}$$

Since there are no nulls in this game, the null-revealing polytope $\mathsf {V}^d(v)$ coincides with its non-revealing counterpart $\mathsf {V}(v)$.

For voter 1, we have

$$\begin{aligned} \iint \limits _{\mathsf {V}^d} w_1{\text {d}}w_1{\text {d}}w_2= & {} \int _{\frac{1}{3}}^{\frac{1}{2}}\int _{1-2w_1}^{w_1} w_1{\text {d}}w_2{\text {d}}w_1+\int _{\frac{1}{2}}^{1}\int _{0}^{1-w_1} w_1{\text {d}}w_2{\text {d}}w_1\\= & {} \frac{1}{54}+\frac{1}{12}=\frac{11}{108}. \end{aligned}$$

For voter 2, we obtain

$$\begin{aligned} \iint \limits _{\mathsf {V}^d} w_2{\text {d}}w_1{\text {d}}w_2= & {} \int _{\frac{1}{3}}^{\frac{1}{2}}\int _{1-2w_1}^{w_1} w_2{\text {d}}w_1{\text {d}}w_2+\int _{\frac{1}{2}}^{1}{\text {d}}w_1\int _{0}^{1-w_1} w_2{\text {d}}w_2{\text {d}}w_1\\= & {} \frac{1}{48}+\frac{5}{432}=\frac{7}{216}. \end{aligned}$$

The total volume of the polytope is given by

$$\begin{aligned} \iint \limits _{\mathsf {V}^d}{\text {d}}w_1{\text {d}}w_2=\int _{\frac{1}{3}}^{\frac{1}{2}}\int _{1-2w_1}^{w_1}{\text {d}}w_2{\text {d}}w_1+\int _{\frac{1}{2}}^{1}\int _{0}^{1-w_1}{\text {d}}w_2{\text {d}}w_1=\frac{1}{8}+\frac{1}{24}=\frac{1}{6}, \end{aligned}$$

This yields the following vector of average (normalized) feasible weights $\left( \frac{11}{18},\frac{7}{36},\frac{7}{36}\right)$.

The polytope for the average representation defined by Lemma 3 is given by

$$\begin{aligned} \mathsf {R}^d(v)=\left\{ (q,w_1,w_2)\in \mathbb {R}^3_{\ge 0}\mid w_1+w_2\ge q,w_1\le q,1-w_1\le q,1-w_2\ge q\right\} , \end{aligned}$$

since need to take into account the inequalities from the minimal winning and the maximal losing coalitions only, due to $w_1,w_2\ge 0$ and $w_3=1-w_1-w_2\ge 1-w_2-q\ge 0$. From $w_1\le q$ and $1-w_1\le q$ follows $\frac{1}{2}\le q\le 1$. We can rewrite the four inequalities to $1-q\le w_1\le q$, with $1-q\le q$ for all $q\ge \frac{1}{2}$, and $q-w_1\le w_2\le 1-q$, with $q-w_1\le 1-q$ iff $w_1\ge 2q-1$. Since $2q-1\ge 1-q$ iff $q\ge \frac{2}{3}$ and $2q-1\le q$ for all $q\le 1$, we have,

$$\begin{aligned} \iint \limits _{\mathsf {R}^d} w_1{\text {d}}w_1{\text {d}}w_2{\text {d}}q= & {} \int _{\frac{1}{2}}^{\frac{2}{3}}{\text {d}}q\int _{1-q}^{q}w_1{\text {d}}w_1\int _{q-w_1}^{1-q}{\text {d}}w_2\\&+\,\int _{\frac{2}{3}}^{1}{\text {d}}q\int _{2q-1}^{q}w_1{\text {d}}w_1\int _{q-w_1}^{1-q}{\text {d}}w_2 = \frac{31}{7776}+\frac{1}{243}=\frac{7}{864}\\ \iint \limits _{\mathsf {R}^d} w_2{\text {d}}w_1{\text {d}}w_2{\text {d}}q= & {} \int _{\frac{1}{2}}^{\frac{2}{3}}{\text {d}}q\int _{1-q}^{q}{\text {d}}w_1\int _{q-w_1}^{1-q}w_2{\text {d}}w_2\\&+\,\int _{\frac{2}{3}}^{1}{\text {d}}q\int _{2q-1}^{q}{\text {d}}w_1\int _{q-w_1}^{1-q}w_2{\text {d}}w_2 = \frac{29}{15{,}552}+\frac{1}{972}=\frac{5}{1728},\\ \iint \limits _{\mathsf {R}^d}{\text {d}}w_1{\text {d}}w_2{\text {d}}q= & {} \int _{\frac{1}{2}}^{\frac{2}{3}}{\text {d}}q\int _{1-q}^{q}{\text {d}}w_1\int _{q-w_1}^{1-q}{\text {d}}w_2+\int _{\frac{2}{3}}^{1}{\text {d}}q\int _{2q-1}^{q}{\text {d}}w_1\int _{q-w_1}^{1-q}{\text {d}}w_2\\= & {} \frac{5}{648}+\frac{1}{162}=\frac{1}{72}, \end{aligned}$$

so that the average representation is given by $\left( \frac{7}{12},\frac{5}{24},\frac{5}{24}\right)$.

We now move from individual voting weights to weights aggregated by equivalence classes, as if voters belonging to the same class form a voting bloc with weight being equal to the sum of weights of its members.

The game has two classes: class A comprises voter 1, whereas voters 2 and 3 form class B. Let $w_a$ be the voting weight of class A, which equals the weight of the first voter $w_a=w_1$. The AWTI polytope degenerates to an interval $\mathsf {V}^t(v)=\left\{ w_a\in \mathbb {R}_{\ge 0}\mid 3w_a\ge 1, w_a\le 1\right\} .$ We have,

$$\begin{aligned} \int \limits _{\mathsf {V}^t} w_a{\text {d}}w_a=\int _{\frac{1}{3}}^1 w_a{\text {d}}w_a=\frac{4}{9}\quad \text{ and }\quad \int \limits _{\mathsf {V}^t} {\text {d}}w_a=\int _{\frac{1}{3}}^1 {\text {d}}w_a=\frac{2}{3}. \end{aligned}$$

The voting power of class A according to AWTI equals $\frac{2}{3}$, which is the power of the first voter. The power of class B equals $\frac{1}{3}$. Since all voters comprising a class share its power equally, the AWTI power vector for the voters reads $\left( \frac{2}{3},\frac{1}{6},\frac{1}{6}\right)$.

We now turn to the final index. The ARTI polytope is given by

$$\begin{aligned} \mathsf {R}^t(v)=\left\{ (q,w_a)\in \mathbb {R}_{\ge 0}^2\mid 3w_a\ge 1, 2q\le 1+w_a, q\ge 1-w_a, q\ge w_a\right\} , \end{aligned}$$

where $w_a$ is the voting weight of class A. We have,

$$\begin{aligned} \iint \limits _{\mathsf {R}^t} w_a{\text {d}}w_a{\text {d}}q= & {} \int _{\frac{1}{3}}^{\frac{1}{2}}w_a{\text {d}}w_a\int _{1-w_a}^{\frac{1+w_a}{2}}{\text {d}}q+\int _{\frac{1}{2}}^{1}w_a{\text {d}}w_a\int _{w_a}^{\frac{1+w_a}{2}}{\text {d}}q=\frac{1}{108}+\frac{1}{24}=\frac{11}{216},\\ \iint \limits _{\mathsf {R}^t} {\text {d}}w_a{\text {d}}q= & {} \int _{\frac{1}{3}}^{\frac{1}{2}}{\text {d}}w_a\int _{1-w_a}^{\frac{1+w_a}{2}}{\text {d}}q+\int _{\frac{1}{2}}^{1}{\text {d}}w_a\int _{w_a}^{\frac{1+w_a}{2}}{\text {d}}q=\frac{1}{48}+\frac{1}{16}=\frac{1}{12}. \end{aligned}$$

The power distribution according to the ARTI is $\left( \frac{11}{18},\frac{7}{36},\frac{7}{36}\right)$ (Fig. 1; Tables 9, 10, 11, 12).

Fig. 1

The distribution of Euclidean distances between the indices

Table 9 Representation-compatible power indices for $n\le 5$

Table 10 Representation-compatible power indices for $n\le 5$ (cont.)

Table 11 Representation-compatible power indices for $n\le 5$ (cont.)

Table 12 Representation-compatible power indices for $n\le 5$ (cont.)