Effectiveness, Decisiveness, and Success in Weighted Voting Systems: Collective Behavior and Voting Measures

Abstract

Efficiency, decisiveness, and success in a voting system depend not only on the voting rules but also on the collective behavior of the voters. The voting behavior is modeled by a voting measure which describes the interdependence (or independence) of the voters. In this paper, we define and investigate a large class of voting measures. This class can be characterized as those voting measures which are invariant under permuting the voters and which allow a natural extension to an arbitrary number of voters. The class includes the Penrose–Banzhaf measure (independent, impartial behavior), the Shapley–Shubik measure (impartial anonymous behavior). We analyze the efficiency and the success for these voting measures in weighted voting systems.

A Mathematical Appendix

1.1 A.1 Hoeffding’s Inequality

For the reader’s convenience, in this appendix, we present a few mathematical results needed in the main text. In particular, we formulate Hoeffding’s inequality.

Theorem 1

(Hoeffding’s Inequality)

Suppose $X_{i}, i=1,\ldots ,N$ are independent random variables such that $X_{i}\in [a_{i},b_{i}]$ almost surely.

Set $\sigma ^2=\sum _{i=1}^{N} (b_{i}-a_{i})^2$. Then

$$\begin{aligned}&\mathbb {P}\Big (\sum _{i=1}^{N} X_{i}~\ge ~\sum _{i=1}^{N}\mathbb {E}(X_{i}) +\lambda \Big )~ \le ~\,e^{-2\frac{\lambda ^{2}}{\sigma ^2}}\qquad \textrm{and}\end{aligned}$$

(A.1)

$$\begin{aligned}&\mathbb {P}\Big (\sum _{i=1}^{N} X_{i}~\le ~\sum _{i=1}^{N}\mathbb {E}(X_{i}) -\lambda \Big )~ \le ~\,e^{-2\frac{\lambda ^{2}}{\sigma ^2}} \end{aligned}$$

(A.2)

For a proof of Theorem 1, see e.g., (Pollard, 1984).

An immediate consequence of Hoeffding’s inequality is the following proposition. As before $P_{p}$ with $0\le p\le 1$ denotes the probability measure on $\{0,1\}^{N}$ given by:

$$\begin{aligned} P_{p}\big (x_{1},x_{2},\ldots ,x_{N}\big )~=~p^{\,\sum x_{i}}\,\big (1-p\big )^{N-\sum x_{i}} \end{aligned}$$

(A.3)

and $E_{p}$ denotes expectation with respect to $P_{p}$.

Proposition 1

Let $X_{i}, i=1,\ldots ,N$ be random variables with distribution $P_{p}$ and $w_{1},\ldots , w_{N}\in [0,\infty )$, then for $\lambda \ge 0$

$$\begin{aligned} P_{p}\Big (\big |\sum _{i=1}^{N}w_{i} X_{i}-p \sum _{i=1}^{N}w_{i}\big |\ge \alpha \sum _{i=1}^{N} w_{i}\Big ) \le 2\,e^{{-2\alpha ^{2}\, \frac{(\sum w_{i})^{2}}{\sum w_{i}^{2}}}} \end{aligned}$$

(A.4)

Proof

The random variables $Y_{i}=w_{i}X_{i}$ are independent (under $P_{p}$) and take values in $[0,w_{i}]]$. Moreover, $E_{p}(Y_{i})=p w_{i}$. Thus, (A.4) follows from Theorem 1.    $\square$

Corollary 1

Suppose the Laakso–Taagepera index $\textrm{LT}_{N}=\frac{\sum w_{i}^{2}}{(\sum w_{i})^{2}}$ goes to infinity as $N\rightarrow \infty$ then

$$\begin{aligned} \textrm{If } \alpha >p\quad&P_{p}\Big (\sum _{i=1}^{N}w_{i} X_{i}~\ge \alpha \,\sum _{i=1}^{N}w_{i}\Big )~\rightarrow ~0\qquad \textrm{as }~N\rightarrow \infty \end{aligned}$$

(A.5)

$$\begin{aligned} \textrm{If } \alpha <p\quad&P_{p}\Big (\sum _{i=1}^{N}w_{i} X_{i}~\le \alpha \,\sum _{i=1}^{N}w_{i}\Big )~\rightarrow ~0\qquad \textrm{as }~N\rightarrow \infty \end{aligned}$$

(A.6)

1.2 A.2 Urn Models

We start by the following observation:

Proposition 2

Suppose $c\ge 0$. For $(x_{1},x_{2},\ldots ,x_{N})\in \{ 0,1 \}^{N}$ and $1\le k < N$ we set $n_{k}=\sum _{j=1}^{k}x_{j}$. Then

$$\begin{aligned}&\mathbb {P}_{A,B,c}\left( x_{1},x_{2},\ldots ,x_{N}\right) \nonumber \\ ~=~&\mathbb {P}_{A,B,c}\left( x_{1},x_{2},\ldots ,x_{k}\right) \, \mathbb {P}_{A+n_{k}c,B+(k-n_{k})c,c}\left( x_{k+1},x_{k+2},\ldots ,x_{N}\right) \end{aligned}$$

(A.7)

Proof

For $k=1$, (A.7) is just the definition of the urn process. The general case follows by iterating the first step.    $\square$

To prove 1.2 we start with a special case

Proposition 3

$$\begin{aligned}&\mathbb {P}_{A,B,c}\left( x_{1},\dots ,x_{k-1},x_{k},x_{k+1},x_{k+2},\ldots ,x_{N}\right) ~\nonumber \\ =~&\mathbb {P}_{A,B,c}\left( x_{1},\dots ,x_{k-1},x_{k+1},x_{k},x_{k+2},\ldots ,x_{N}\right) \end{aligned}$$

(A.8)

Proof

If $x_{k}=x_{k+1}$, the assertion is trivial. So, we assume $x_{k}\not =x_{k+1}$ We treat the case $x_{k}=0, x_{k+1}=1$, the other one being similar. We apply Proposition 2 three times to obtain with $\ell =\sum _{j}^{k-1} x_{j}$

$$\begin{aligned}&\mathbb {P}_{A,B,c}\left( x_{1},\dots ,x_{k-1},x_{k},x_{k+1},x_{k+2},\ldots ,x_{N}\right) ~\nonumber \\ =~&\mathbb {P}_{A,B,c}\left( x_{1},\dots ,x_{k-1}\right) ~ \mathbb {P}_{A+\ell c,B+(k-1-\ell )c,c}(0) ~\mathbb {P}_{A+\ell c,B+(k-\ell )c,c}(1)~\times \nonumber \\ \times ~&\mathbb {P}_{A+(\ell +1)c,B+(k-\ell ),c}\left( x_{k+2},\ldots ,x_{N}\right) ~\nonumber \\ =~&\mathbb {P}_{A,B,c}\left( x_{1},\dots ,x_{k-1}\right) ~ \frac{B+(k-1-\ell )c}{A+B+(k-1)c} ~\frac{A+\ell c}{A+B+kc}~\times \nonumber \\ \times ~&\mathbb {P}_{A+(\ell +1)c,B+(k-\ell ),c}\left( x_{k+2},\ldots ,x_{N}\right) ~\nonumber \\ =~&\mathbb {P}_{A,B,c}\left( x_{1},\dots ,x_{k-1}\right) ~ \frac{A+\ell c}{A+B+(k-1)c} ~\frac{B+\left( k-\left( \ell +1\right) \right) c}{A+B+kc}~\times \nonumber \\ \times ~&\mathbb {P}_{A+(\ell +1)c,B+(k-\ell ),c}\left( x_{k+2},\ldots ,x_{N}\right) ~\nonumber \\ =~&\mathbb {P}_{A,B,c}\left( x_{1},\dots ,x_{k-1},x_{k+1},x_{k},x_{k+2},\ldots ,x_{N}\right) \end{aligned}$$

(A.9)

Iterating Propositions 2 and 3 proves Proposition 1. To prove Theorem 1, we first note:

Remark 1

For $a,b>0$ we have

$$\begin{aligned} B(a,b)~=~\frac{\Gamma (a)\Gamma (b)}{\Gamma (a+b)}\,, \end{aligned}$$

(A.10)

where $\Gamma (x)=\int \limits _{0}^{\infty } t^{x-1}\,e^{-t}\,dt$ is the Gamma function.

For a proof as well as for details about the Gamma function, see e.g., (Georgii, 2008).

We now prove (42) by proving that all moments of the measures on the left-hand side converge to the moments of the right-hand side.

From (Kirsch, 2019), Theorem 5, we learn that the $k^{th}$ moment under $\mathbb {P}_{A,B,c}$ of $M_{N}$ converges to

$$\begin{aligned} m_{k}~:=~\mathbb {E}_{A,B,c}\left( X_{1}\cdot X_{2}\cdot \ldots \cdot X_{k}\right) \end{aligned}$$

(A.11)

but $m_{k}$ can be computed by (34) giving

$$\begin{aligned} m_{k}~=~\frac{a^{\,(1,k)}}{(a+b)^{(1,k)}} \end{aligned}$$

(A.12)

The moments of $\beta _{a,b}$ are given by $\bar{m}_{k}=\frac{B(a+k,b)}{B(a,b)}$. By Proposition 1

$$\begin{aligned} B(a+k,b)~&=~\frac{\Gamma (a+k)\Gamma (b)}{\Gamma (a+b+k)}\nonumber \\&=~\frac{a^{(k,1)}}{(a+b)^{(k,1)}}\,\frac{\Gamma (a)\Gamma (b)}{\Gamma (a+b)}\,. \end{aligned}$$

(A.13)

Above we used that $\Gamma (x+k)=x^{(k,1)}\Gamma (x)$ which follows from the well-known equality $\Gamma (x)=(x-1)\Gamma (x-1)$. This proves 1.