A note on limit results for the Penrose–Banzhaf index

Abstract

It is well known that the Penrose–Banzhaf index of a weighted game can differ starkly from corresponding weights. Limit results are quite the opposite, i.e., under certain conditions the power distribution approaches the weight distribution. Here we provide parametric examples that give necessary conditions for the existence of limit results for the Penrose–Banzhaf index.

Appendix

To prove Propositions 2 and 3 we need a small numerical estimate and a tightening of the general bound $\Vert x\Vert _\infty \le \Vert x\Vert _1$ in our setting.

Lemma 3

For $n\ge 11$ we have $2n^3/2.6^n\le \tfrac{1}{n}$.

Lemma 4

For $w,w'\in \mathbb {R}^n_{\ge 0}$ with $\Vert w\Vert _1=\Vert w'\Vert _1=1$, we have $\Vert w-w'\Vert _{\infty }\le \frac{1}{2}\Vert w-w'\Vert _1$.

Proof

With $S:=\{1\le i\le n\mid w_i\le w'_i\}$ and $A:=\sum _{i\in S} \left( w'_i-w_i\right)$, $B:=\sum _{i\in N\backslash S} \left( w_i-w'_i\right)$, where $N=\{1,\dots ,n\}$, we have $A-B=0$ since $\Vert w\Vert _1=\Vert w'\Vert _1$ and $w,w'\in \mathbb {R}_{\ge 0}^n$. Thus, $\Vert w-w'\Vert _1=2A$ and $\Vert w-w'\Vert _{\infty }\le \max \{A,B\}=A$. $\square$

Proof of Proposition 2

We easily check $\Vert w\Vert _1=1$ and $v_n=[\tfrac{1}{2};w]$. For $v=[q;k,1,\dots ,1]$ with m times weight 1 we have $\eta _1(v)=\sum _{i=1}^{k}{m \atopwithdelims (){q-i}}$ and $\eta _2(v)={{m-1}\atopwithdelims (){q-1}}+{{m-1}\atopwithdelims (){q-k-1}}$ so that

$$\begin{aligned} \eta _1(v_n) = \sum _{i=1}^{2n^2} {{2n^3}\atopwithdelims (){n^3+n^2-i}}\ge {{2n^3}\atopwithdelims (){n^3}} \end{aligned}$$

(2)

and

$$\begin{aligned} \eta _2(v_n)= & {} {{2n^3-1}\atopwithdelims (){n^3+n^2-1}}+{{2n^3-1}\atopwithdelims (){n^3-n^2-1}} = {{2n^3-1}\atopwithdelims (){n^3+n^2-1}}+{{2n^3-1}\atopwithdelims (){n^3+n^2}} \nonumber \\= & {} {{2n^3}\atopwithdelims (){n^3+n^2}} \end{aligned}$$

(3)

using $q=n^3+n^2$, $k=2n^2$, and $m=2n^3$.

Since $(1+\tfrac{1}{n})^n$ is monotonically increasing we have $(1+\tfrac{1}{n})^n\ge 2.6$ for $n\ge 11$, so that

$$\begin{aligned} \frac{\eta _1(v_n)}{\eta _2(v_n)}\ge & {} \frac{\left( n^3+n^2\right) !\left( n^3-n^2\right) !}{\left( n^3\right) !\left( n^3\right) !}\\= & {} \frac{\prod \nolimits _{i=1}^{n^2}n^3+i}{\prod \nolimits _{i=1}^{n^2}n^3+i-n^2}\ge \left( 1+\frac{n^2}{n^3}\right) ^{n^2} =\left( \left( 1+\tfrac{1}{n}\right) ^n\right) ^n\ge 2.6^n. \end{aligned}$$

From

$$\begin{aligned} \mathrm{PBnI}_1(v_n)= & {} \frac{\eta _1(v_n)}{\eta _1(v_n)+m\cdot \eta _2(v_n)}=1-\frac{m\cdot \eta _2(v_n)}{\eta _1(v_n)+m\cdot \eta _2(v_n)} \\\ge & {} 1-m\cdot \frac{\eta _2(v_n)}{\eta _1(v_n)}\ge 1-\frac{2n^3}{2.6^n}, \end{aligned}$$

$w_1=\tfrac{1}{n+1}\le \tfrac{1}{n}$, and $2n^3/2.6^n\le \tfrac{1}{n}$ for $n\ge 11$, see Lemma 3, we deduce

$$\begin{aligned} \Vert \mathrm{PBnI}(v_n)-w\Vert _\infty \ge \left| \mathrm{PBnI}_1(v_n)-w_1\right| \ge 1-\tfrac{2}{n}. \end{aligned}$$

From Lemma 4, we then conclude $\Vert \mathrm{PBnI}(v_n)-w\Vert _1\ge 2-\tfrac{4}{n}$. $\square$

Proof of Proposition 3

We easily check $\Vert w\Vert _1=1$ and $v_n=[\tfrac{1}{2};w]$. For players $1\le i\le 2n+1$ examples of swing coalitions are given by n other players of weight $2n^2$ and $n^3$ players of weight 1, so that

$$\begin{aligned} \eta _i(v_n)\ge {{2n}\atopwithdelims (){n}}\cdot {{2n^3}\atopwithdelims (){n^3}}. \end{aligned}$$

For players of weight 1, i.e., $2n+2\le i\le 2n+1+2n^3$, we have

$$\begin{aligned} \eta _i(v_n)= & {} \sum _{j=0}^{2n+1} {{2n+1}\atopwithdelims ()j}\cdot {{2n^3-1}\atopwithdelims (){3n^3+n^2 -j\cdot 2n^2-1}}\\\le & {} (n+1)\cdot {{2n+1}\atopwithdelims ()n} \cdot {{2n^3-1}\atopwithdelims (){n^3-n^2-1}}\\&+\, (n+1)\cdot {{2n+1}\atopwithdelims ()n} \cdot {{2n^3-1}\atopwithdelims (){n^3+n^2-1}}\\= & {} (n+1)\cdot {{2n+1}\atopwithdelims ()n} \cdot {{2n^3}\atopwithdelims (){n^3+n^2}}\\= & {} (2n+1) \cdot {{2n}\atopwithdelims ()n} \cdot {{2n^3}\atopwithdelims (){n^3+n^2}}. \end{aligned}$$

Similar as in the proof of Proposition 2, we conclude

$$\begin{aligned} \frac{\mathrm{PBnI}_1(v_n)}{\mathrm{PBnI}_{2n+2}(v_n)}=\frac{\eta _1(v_n)}{\eta _{2n+2}(v_n)} \ge \frac{2.6^n}{2n+1}, \end{aligned}$$

(4)

noting that $\eta _1(v_n)=\eta _i(v_n)$ for all $1\le i\le 2n+1$ and $\eta _{2n+2}(v_n)=\eta _i(v_n)$ for all $2n+2\le i\le 2n+1+2n^3$ due to symmetry. With this we compute

$$\begin{aligned} \mathrm{PBnI}_1(v_n)-w_1= & {} \frac{\eta _1(v_n)}{(2n+1)\cdot \eta _1(v_n)+2n^3\cdot \eta _{2n+2}(v_n)}-\frac{1}{3n+1}\\= & {} \frac{1}{2n+1}\cdot \left( 1-\frac{2n^3\cdot \eta _{2n+2}(v_n)}{(2n+1)\cdot \eta _1(v_n)+2n^3\cdot \eta _{2n+2}(v_n)}\right) -\frac{1}{3n+1}\\&\overset{n\ge 3}{\ge } \frac{1}{2n+1}\cdot \left( \frac{3}{10}-\frac{2n^3}{2n+1}\cdot \frac{\eta _{2n+2}(v_n)}{\eta _{1}(v_n)}\right) \\&\overset{\text {(}4\text {)}}{\ge } \frac{1}{2n+1}\cdot \left( \frac{3}{10}-\frac{2n^3}{2.6^n}\right) \\&\overset{\text {Lemma}~3}{\ge } \frac{1}{2n+1}\cdot \left( \frac{3}{10}-\frac{1}{n}\right) \overset{n\ge 10}{\ge } \frac{1}{2n+1}\cdot \frac{1}{5} \end{aligned}$$

for $n\ge 11$ (using Inequality (4) and Lemma 3). Thus,

$$\begin{aligned} \Vert \mathrm{PBnI}(v_n)-w\Vert _1\ge \sum _{i=1}^{2n+1}\left| \mathrm{PBnI}_i(v_n)-w_i\right| = (2n+1)\cdot \left| \mathrm{PBnI}_1(v_n)-w_1\right| \ge \frac{1}{5}. \end{aligned}$$

$\square$

The details for our briefly sketched last example from Sect. 3 are given by

Proposition 4

For $n\in \mathbb {N}$ and $q\in [0,1]$, let

$$\begin{aligned} v_{n,q}=[q\cdot 3n;\overset{n}{\overbrace{2,\dots ,2}},\overset{n}{\overbrace{1,\dots ,1}}] \end{aligned}$$

with n times weight 2 and n times weight 1. Relative weights for a relative quota of q are given by $w_{n,q}=\left( 2,\dots ,2,1,\dots ,1\right) /(3n)$, i.e., $\Vert w_{n,q}\Vert _1=1$ and $v_{n,q}=[q;w_{n,q}]$. Then the function $f(q):=\lim _{n\rightarrow \infty } \Vert \mathrm{PBnI}(v_{n,q})-w_{n,q} \Vert _1$ satisfies $f(q)=f(1-q)\in [0,\tfrac{1}{3}]$ for all $q\in [0,1]$ and is strictly monotonically increasing in $[\tfrac{1}{2},1]$.

We refrain from giving a rigorous proof. Symmetry around $q=\tfrac{1}{2}$, i.e., $f(q)=f(1-q)$ follows by considering the dual game. For a relative quota q near 0 or near 1, all players are equivalent so that $\mathrm{PBnI}(v_{n,q})=\tfrac{1}{2n}\cdot (1,\dots ,1)$, which gives $f(0)=f(1)=\tfrac{1}{3}$. From Lindner and Machover (2004, Theorem 3.6), we conclude $f(\tfrac{1}{2})=0$. To check the existence of the limit and monotonicity numerically, we state

$$\begin{aligned} \eta _{1}(v_{n,q})= & {} \sum _{i=0}^{n-1} {{n-1}\atopwithdelims (){i}}\cdot \left( {{n}\atopwithdelims (){\lceil q\cdot 3n\rceil -2i-2}} + {{n}\atopwithdelims (){\lceil q\cdot 3n\rceil -2i-1}}\right) \\= & {} \sum _{i=0}^{n-1} {{n-1}\atopwithdelims (){i}}\cdot {{n+1}\atopwithdelims (){\lceil q\cdot 3n\rceil -2i-1}}\\ \eta _{n+1}(v_{n,q})= & {} \sum _{i=0}^n {{n}\atopwithdelims (){i}}\cdot {{n-1}\atopwithdelims (){\lceil q\cdot 3n\rceil -2i-1}} \end{aligned}$$

noting that convergence is rather slow and requires high-precision computations. We remark that error bounds in general local limit theorems for lattice distributions like, e.g. (Petrov 1975, Theorem 2 in Chapter VII) are (inevitably) too weak to determine f(q) analytically. While the summands of $\eta _1(v_{n,q})$ and $\eta _{n+1}(v_{n,q})$ are unimodal and quickly sloping outside a small neighborhood around the almost coinciding peaks, the intuitive idea to bound the sums in terms of their maximal summands is not too easy to pursue. The maximal summand is not attained at $i\approx q\cdot n$, as one could expect. Even approximating $\log _2 {n\atopwithdelims ()k}$ by $n\cdot H(k/n)$, where $H(p)=-p\log _2(p)-(1-p)\log _2(1-p)$ is the binary entropy of p, gives that the maximum summand is attained for $i\approx n\cdot g(q)$, where

$$\begin{aligned} g(q)=\frac{\tilde{g}(q)^{\tfrac{1}{3}}}{12}-\frac{-3q^2+3q+\tfrac{1}{2}}{\tilde{g}(q)^{\tfrac{1}{3}}}+q \end{aligned}$$

and

$$\begin{aligned} \tilde{g}(q)=-216q^3+324q^2-108q+6\sqrt{972q^4-1944q^3+864q^2+108q+6}. \end{aligned}$$

Numerically, we can check that this fancy function g satisfies $q\le g(q)\le 1.07\cdot g(q)$ for all $\tfrac{1}{2}\le q\le 1$ and is, of course, symmetric to $q=\tfrac{1}{2}$.

Given the numerical results for f(q), we can state that $\frac{8}{3}\cdot \left| q-\tfrac{1}{2}\right| ^3$ and $\tfrac{1}{3}-H(q)/3\log _2(2)$ correspond to curves that look similar to f(q) and have a rather small absolute error.

For $w\in \mathbb {R}_{\ge 0}^n$ with $w\ne 0$ the Laakso–Taagepera index is given by

$$\begin{aligned} L(w) =\left( \sum \limits _{i=1}^{n}w_i\right) ^2 / \sum \limits _{i=1}^{n} w_i^2. \end{aligned}$$

In general, we have $1\le L(w)\le n$. If the weight vector w is normalized, then the formula simplifies to $L(w)=1/\sum _{i=1}^n w_i^2$. Under the name “effective number of parties” the index is widely used in political science to measure party fragmentation, see, e.g., (Laakso and Taagapera 1979). We observe the following relations between the maximum relative weight $\Delta =\Delta (w)$ and the Laakso–Taagepera index L(w):

Lemma 5

(Kurz 2018, Lemma 3) For $w\in \mathbb {R}_{\ge 0}^n$ with $\Vert w\Vert _1=1$, we have

$$\begin{aligned} \frac{1}{\Delta }\le \frac{1}{\Delta \left( 1-\alpha (1-\alpha )\Delta \right) }\le L(w)\le \frac{1}{\Delta ^2+\frac{(1-\Delta )^2}{n-1}}\le \frac{1}{\Delta ^2} \end{aligned}$$

for $n\ge 2$, where $\alpha :=\frac{1}{\Delta }-\left\lfloor \frac{1}{\Delta }\right\rfloor \in [0,1)$. If $n=1$, then $\Delta =L(w)=1$.

Proof

The key idea is to optimize $\sum \nolimits _{i=1}^n w_i^2$ with respect to the constraints $w\in \mathbb {R}^n$, $\Vert w\Vert _1=1$, and $\Delta (w)=\Delta$.

For $n=1$, we have $w_1=1$, $\Delta (w)=1$, $\alpha =0$, and $L(w)=1$, so that we assume $n\ge 2$ in the remaining part of the proof. For $w_i\ge w_j$ consider $a:=\frac{w_i+w_j}{2}$ and $x:=w_i-a$, so that $w_i=a+x$ and $w_j=a-x$. With this we have $w_i^2+w_j^2=2a^2+2x^2$ and $(w_i+y)^2+(w_j-y)^2=2a^2+2(x+y)^2$. Let us assume that $w^\star$ minimizes $\sum _{i=1}^n w_i^2$ under the conditions $w\in \mathbb {R}_{\ge 0}$, $\Vert w\Vert _1=1$, and $\Delta (w)=\Delta$. (Since the target function is continuous and the feasible set is compact and non-empty, a global minimum indeed exists.) W.l.o.g. we assume $w_1^\star =\Delta$. If there are indices $2\le i,j\le n$ with $w_i^\star >w_j^\star$, i.e., $x>0$ in the above parameterization, then we may choose $y=-x$. Setting $w_i':=w_i^\star +y=a=\frac{w_i^\star +w_j^\star }{2}$, $w_j':=w_j^\star -y=a =\frac{w_i^\star +w_j^\star }{2}$, and $w_h':=w_h^\star$ for all $1\le h\le n$ with $h\notin \{i,j\}$, we have $w'\in \mathbb {R}_{\ge 0}^n$, $\Vert w'\Vert _1=1$, $\Delta (w')=\Delta$, and $\sum _{h=1}^n \left( w_h'\right) ^2=\sum _{h=1}^n \left( w_h^\star \right) ^2\,-\,x^2$. Since this contradicts the minimality of $w^\star$, we have $w_i^\star =w_j^\star$ for all $2\le i,j\le n$, so that we conclude $w_i^\star =\frac{1-\Delta }{n-1}$ for all $2\le i\le n$ from $1=\Vert w^\star \Vert _1=\sum \nolimits _{h=1}^n w_h^\star$. Thus, $L(w)\le 1/\left( \Delta ^2+\frac{(1-\Delta )^2}{n-1}\right)$, which is tight. Since $\Delta \le 1$ and $n\ge 2$, we have $1/\left( \Delta ^2+\frac{(1-\Delta )^2}{n-1}\right) \le \frac{1}{\Delta ^2}$, which is tight if and only if $\Delta =1$, i.e., $n-1$ of the weights have to be equal to zero.

Now, let us assume that w maximizes $\sum _{i=1}^n w_i^2$ under the conditions $w\in \mathbb {R}_{\ge 0}$, $\Vert w\Vert _1=1$, and $\Delta (w)=\Delta$. (Due to the same reason a global maximum indeed exists.) Due to $1=\Vert w\Vert _1\le n\Delta$ we have $0<\Delta \le 1/n$, where $\Delta =1/n$ implies $w_i=\Delta$ for all $1\le i\le n$. In that case we have $L(w)=n$ and $\alpha =0$, so that the stated lower bounds for L(w) are valid. In the remaining cases, we assume $\Delta >1/n$. If there would exist two indices $1\le i,j\le n$ with $w_i\ge w_j$, $w_i<\Delta$, and $w_j>0$, we may strictly increase the target function by moving weight from $w_j$ to $w_i$ (this corresponds to choosing $y>0$), by an amount small enough to still satisfy the constraints $w_i\le \Delta$ and $w_j\ge 0$. Since $\Delta >0$, we can set $a:=\lfloor 1/\Delta \rfloor \ge 0$ with $a\le n-1$ due to $\Delta >1/n$. Thus, for a maximum solution, we have exactly a weights that are equal to $\Delta$, one weight that is equal to $1-a\Delta \ge 0$ (which may indeed be equal to zero), and $n-a-1$ weights that are equal to zero. With this and $a\Delta =1-\alpha \Delta$ we have $\sum _{i=1}^n w_i^2=a\Delta ^2 (1-a\Delta )^2=\Delta -\alpha \Delta ^2+\alpha ^2\Delta ^2=\Delta (1-\alpha \Delta +\alpha ^2\Delta ) =\Delta \left( 1-\alpha (1-\alpha )\Delta \right) \le \Delta$. Here, the latter inequality is tight if and only if $\alpha =0$, i.e., $1/\Delta \in \mathbb {N}$. $\square$