The price of ‘one person, one vote’

Abstract

I study the design of binary voting rules in environments where agents are heterogenous either in the stakes that they have in the decision or in the quality of information they possess regarding the correct course of action. The price of ‘one person, one vote’ is defined as the reduction in welfare resulting from constraining the voting rule to treat the agents symmetrically. I analyze how this price depends on the parameters of the environment, particularly on the level of heterogeneity in the society. The results shed light on the tradeoff between equality and efficiency in the context of constitutional design.

Appendices

Appendix A: Proofs for Section 2

Proof of Lemma 3

From Lemma 1,

$$\begin{aligned} W^*(a,p) = \sum _{S\subseteq [n]} p^{|S|}(1-p)^{n-|S|} \max \left\{ a(S),a(S^c)\right\} . \end{aligned}$$

Now, for a fixed coalition S, the mapping $a\longmapsto \max \left\{ a(S),a(S^c)\right\}$ is convex as the maximum of two linear functions. Since a convex combination of convex functions is also convex, it follows that $W^*$ is a convex function of a. In addition, it is clear that $W^*$ is symmetric in a, i.e., that a permutation of the coordinates of a does not change the value of $W^*$. It is well-known [see, e.g., Marshal et al. (2011, Proposition C.2, p. 97)] that any function that is both convex and symmetric is monotonically non-decreasing with respect to the majorization partial order, which proves the assertion regarding $W^*$.

Second, if a majorizes $a'$ then $a[n]=a'[n]$, so it follows immediately from Lemma 2 that $W^{**}(a,p)=W^{**}(a',p)$. $\square$

Proof of Proposition 1

The proposition follows directly from Lemma 3. Indeed, if a majorizes $a'$ then

$$\begin{aligned} \tau (a,p) = W^*(a,p)-W^{**}(a,p) \ge W^*(a',p)-W^{**}(a',p) = \tau (a',p), \end{aligned}$$

and

$$\begin{aligned} \rho (a,p) = 1-\frac{W^{**}(a,p)}{W^*(a,p)} \ge 1-\frac{W^{**}(a',p)}{W^*(a',p)} = \rho (a',p). \end{aligned}$$

Proof of Proposition 2

1. Given $\delta >0$, let $c^{\delta +} = \frac{a[n]}{a[n]+\delta } a^{\delta +}$. Recall that I assume in the proposition that a is ordered so that $a_1\ge a_2\ge \cdots \ge a_n$, and notice that in moving from a to $c^{\delta +}$ this order has not changed. For any $k=1,\ldots ,n$,

$$\begin{aligned} \sum _{i=1}^k c_i^{\delta +} = \frac{a[n]}{a[n]+\delta } \sum _{i=1}^k a_i^{\delta +} = \frac{a[n]}{a[n]+\delta } \left( \delta +\sum _{i=1}^k a_i\right) \ge \sum _{i=1}^k a_i, \end{aligned}$$

with equality when $k=n$. In other words, $c^{\delta +}$ majorizes a, so it follows from Proposition 1 that $\rho (c^{\delta +},p)\ge \rho (a,p)$. But recall that multiplying the utility functions of all agents by a constant does not affect $\rho$, since both $W^*$ and $W^{**}$ are multiplied by that constant. Hence, $\rho (c^{\delta +},p)= \rho (a^{\delta +},p)$. Combining these one gets

$$\begin{aligned} \rho (a^{\delta +},p) = \rho (c^{\delta +},p) \ge \rho (a,p), \end{aligned}$$

(3)

which is the required inequality for $\rho$.

For $\tau$, notice first that inequality (3) above can be rewritten as

$$\begin{aligned} W^*(a,p)W^{**}(a^{\delta +},p) \le W^{**}(a,p)W^*(a^{\delta +},p). \end{aligned}$$

Since clearly $W^*(a^{\delta +},p)>W^*(a,p)>0$, I can rewrite this inequality as

$$\begin{aligned} \frac{W^{**}(a^{\delta +},p) - W^{**}(a,p)}{W^*(a^{\delta +},p) - W^*(a,p)} \le \frac{W^{**}(a,p)}{W^*(a,p)} \le 1, \end{aligned}$$

where the last inequality follows from the definitions of $W^*$ and $W^{**}$. Rearranging gives

$$\begin{aligned} W^*(a^{\delta +},p) - W^{**}(a^{\delta +},p) \ge W^*(a,p) - W^{**}(a,p), \end{aligned}$$

or

$$\begin{aligned} \tau (a^{\delta +},p)\ge \tau (a,p), \end{aligned}$$

as needed.

2. The proof for $\rho$ is similar to part 1 above. Define $c^{\delta -} = \frac{a[n]}{a[n]-\delta } a^{\delta -}$. It is easy to check that $c^{\delta -}$ majorizes a, and that $\rho (c^{\delta -},p)= \rho (a^{\delta -},p)$. It then follows from Proposition 1 that

$$\begin{aligned} \rho (a^{\delta -},p) = \rho (c^{\delta -},p) \ge \rho (a,p). \end{aligned}$$

However, I cannot conclude in a similar way that $\tau (a^{\delta -},p)\ge \tau (a,p)$, since here $W^*(a^{\delta -},p)<W^*(a,p)$. Instead, I present a direct proof that does not rely on Proposition 1.

By Lemma 2,

$$\begin{aligned} W^{**}(a,p) - W^{**}(a^{\delta -},p) = \frac{\delta }{n}\sum _{k=0}^n {n \atopwithdelims ()k} p^k(1-p)^{n-k} \max \left\{ k, n-k\right\} . \end{aligned}$$

Consider now the difference

$$\begin{aligned}&W^*(a,p) - W^*(a^{\delta -},p)\\&\quad = \sum _{k=0}^n p^{k}(1-p)^{n-k}\sum _{\{S:|S|=k\}} \left[ \max \left\{ a(S),a(S^c)\right\} - \max \left\{ a^{\delta -}(S),a^{\delta -}(S^c)\right\} \right] . \end{aligned}$$

The difference between the two maxima is positive only in the two cases where (1) $n\in S$ and $a(S)>a(S^c)$; or (2) $n\in S^c$ and $a(S^c)>a(S)$. In both cases the difference is at most $\delta$. Therefore,

$$\begin{aligned}&W^*(a,p) - W^*(a^{\delta -},p) \le \delta \sum _{k=0}^n p^{k}(1-p)^{n-k} \nonumber \\&\quad \times \,\left[ \left| \left\{ S:~ |S|=k,~n\in S,~ a(S)>a(S^c)\right\} \right| \right. \nonumber \\&\left. \quad + \left| \left\{ S:~ |S|=k,~n\in S^c,~ a(S)<a(S^c)\right\} \right| \right] . \end{aligned}$$

(4)

To bound this last expression, notice that since $a_n\le a_i$ for any other agent $i\in [n]$,

$$\begin{aligned} \Big |\big \{S:~ |S|=k,~n\in S,~ a(S)>a(S^c)\big \} \Big | \le \Big |\big \{S:~ |S|=k,~i\in S,~ a(S)>a(S^c)\big \} \Big |. \end{aligned}$$

(5)

Summing up (5) over all $i\in [n]$ and dividing by n gives

$$\begin{aligned} \Big |\big \{S:~ |S|= & {} k,~n\in S,~ a(S)>a(S^c)\big \} \Big | \nonumber \\\le & {} \frac{1}{n}\sum _{i\in [n]} \Big |\big \{S:~ |S|=k,~i\in S,~ a(S)>a(S^c)\big \} \Big |\nonumber \\= & {} \frac{k}{n} \Big |\big \{S:~ |S|=k,~ a(S)>a(S^c)\big \}\Big |, \end{aligned}$$

(6)

where the last equality is due to the fact that each winning coalition S of size k is counted k times in the sum. Repeating the argument, one obtains the inequality

$$\begin{aligned} \Big |\big \{S:~ |S|= & {} k,~n\in S^c,~ a(S)<a(S^c)\big \} \Big |\nonumber \\\le & {} \frac{1}{n}\sum _{i\in [n]} \Big |\big \{S:~ |S|=k,~i\in S,~ a(S)<a(S^c)\big \}\Big |\nonumber \\= & {} \frac{n-k}{n} \Big |\big \{S:~ |S|=k,~ a(S)<a(S^c)\big \}\Big |. \end{aligned}$$

(7)

Plugging (6) and (7) into (4) yields

$$\begin{aligned}&W^*(a,p) - W^*(a^{\delta -},p) \le \delta \sum _{k=0}^n p^{k}(1-p)^{n-k} \\&\qquad \times \,\left[ \frac{k}{n} \Big |\big \{S:~ |S|=k,~ a(S)>a(S^c)\big \}\Big | + \frac{n-k}{n} \Big |\big \{S:~ |S|=k,~ a(S)<a(S^c)\big \}\Big | \right] \\&\quad \le \frac{\delta }{n} \sum _{k=0}^n p^{k}(1-p)^{n-k} \max \big \{k,n-k\big \} \\&\qquad \times \, \left[ \Big |\big \{S:~ |S|=k,~ a(S)>a(S^c)\big \}\Big | + \Big |\big \{S:~ |S|=k,~ a(S)<a(S^c)\big \}\Big | \right] \\&\quad \le \frac{\delta }{n} \sum _{k=0}^n {n \atopwithdelims ()k} p^{k}(1-p)^{n-k} \max \big \{k,n-k\big \}. \end{aligned}$$

Therefore, $W^*(a,p) - W^*(a^{\delta -},p) \le W^{**}(a,p) - W^{**}(a^{\delta -},p)$, or $\tau (a,p) \le \tau (a^{\delta -},p)$ as needed. $\square$

Proof of Proposition 3

From Lemma 2, if (a, b) sum-majorizes $(a',b')$ then $W^{**}(a,b,p)=W^{**}(a',b',p)$. To prove the proposition I therefore need to show that $W^*(a,b,p)\ge W^*(a',b',p)$. I do this by showing that a PD transfer from an agent with high stakes to an agent with lower stakes (as measured by the sum $a_i+b_i$) results in lower value of $W^*$. As explained in the text, sum-majorization is characterized by finite sequences of such transfers, so this is sufficient to conclude that $W^*(a,b,p)\ge W^*(a',b',p)$.

Fix the vectors a, b and suppose that $a_i+b_i>a_j+b_j$. Let $0<\delta < (a_i+b_i)-(a_j+b_j)$ and denote by $a^\delta$ the vector obtained from a by replacing $a_i$ by $a_i-\delta$, replacing $a_j$ by $a_j+\delta$, and keeping all other coordinates unchanged. From Lemma 1,

$$\begin{aligned}&W^*(a,b,p)-W^*(a^\delta ,b,p)\\&\quad = \sum _{k=0}^n p^k(1-p)^{n-k} \sum _{\{S:|S|=k\}} \left[ \max \left\{ a(S),b(S^c)\right\} -\max \left\{ a^\delta (S),b(S^c)\right\} \right] . \end{aligned}$$

For any coalition S that contains both i and j, or that contains neither of them, $a^\delta (S)=a(S)$ holds. For coalitions S that contain i but not j one has $a^\delta (S)=a(S)-\delta$. And for coalitions S that contain j but not i one has $a^\delta (S)=a(S)-\delta$. Therefore, for each $k=0,1,\ldots ,n$,

$$\begin{aligned}&\sum _{\left\{ S:|S|=k\right\} } \left[ \max \left\{ a(S),b(S^c)\right\} - \max \left\{ a^\delta (S),b(S^c)\right\} \right] \nonumber \\&\quad = \sum _{\{S:|S|=k,~i\in S,~j\notin S\}} \left[ \max \left\{ a(S),b(S^c)\right\} - \max \left\{ a(S)-\delta ,b(S^c)\right\} \right] \nonumber \\&\qquad -\, \sum _{\left\{ S:|S|=k,~j\in S,~i\notin S\right\} } \max \left\{ a(S)+\delta ,b(S^c)\right\} - \left[ \max \left\{ a(S),b(S^c)\right\} \right] . \end{aligned}$$

(8)

For every coalition S in the last sum (i.e., $|S|=k$, $j\in S,~ i\notin S$), let ${\bar{S}}$ be defined by ${\bar{S}} = (S{\setminus } \{j\}) \cup \{i\}$. Note that ${\bar{S}}$ is in the first sum (i.e., $|{\bar{S}}|=k$, $i\in {\bar{S}},~ j\notin {\bar{S}}$), and that the mapping $S\longmapsto {\bar{S}}$ is one-to-one. Also, notice that $a({\bar{S}}) = a(S)-a_j+a_i$ and $b({\bar{S}}^c) = b(S^c)-b_i+b_j$, so

$$\begin{aligned} a({\bar{S}})-b({\bar{S}}^c) = a(S)-b(S^c)+(a_i+b_i)-(a_j+b_j) \ge a(S)-b(S^c)+\delta , \end{aligned}$$

(9)

where the inequality is by the bound on $\delta$.

I claim that for every such S

$$\begin{aligned}&\max \left\{ a({\bar{S}}),b({\bar{S}}^c)\right\} - \max \left\{ a({\bar{S}})-\delta ,b({\bar{S}}^c)\right\} \nonumber \\&\quad \ge \max \left\{ a(S)+\delta ,b(S^c)\right\} - \max \left\{ a(S),b(S^c)\right\} , \end{aligned}$$

(10)

and I prove it by considering three possible cases:

Case 1: [$a(S)\ge b(S^c)$]. Here the right-hand side of (10) is equal to $\delta$. From (9), $a({\bar{S}})-b({\bar{S}}^c) \ge \delta$, which implies that the left-hand side of (10) is equal to $\delta$ as well.

Case 2: [$a(S) < b(S^c) \le a(S)+\delta$]. In this case the right-hand side of (10) is equal to $a(S)+\delta -b(S^c)$. As for the left-hand side, it follows again from (9) that $a({\bar{S}})\ge b({\bar{S}}^c)$, so the left-hand side is equal to $a({\bar{S}})-\max \{a({\bar{S}})-\delta ,b({\bar{S}}^c)\} = \min \{\delta , a({\bar{S}})-b({\bar{S}}^c)\} \ge \min \{\delta , a(S)-b(S^c)+\delta \} = a(S)-b(S^c)+\delta$. It follows that (10) holds in this case as well.

Case 3: [$b(S^c)> a(S)+\delta$]. In this last case the right-hand side of (10) is equal to 0. Since the left-hand side of (10) is clearly non-negative the proof is complete.

To conclude the proof of the proposition, note that (10) together with the fact that the mapping from S to ${\bar{S}}$ is one-to-one implies that (8) is non-negative for every k. This implies in turn that $W^*(a,b,p)\ge W^*(a^\delta ,b,p)$ as needed. $\square$

Proof of Proposition 4

Fix a. To simplify notation I write $g(p)=W^*(a,p)$ and $h(p)=W^{**}(a,p)$, and view g, h as functions defined on the interval (0, 1) (suppressing the dependence on the fixed vector a). For every $0\le k\le n$ let

$$\begin{aligned} g_k=\sum _{\left\{ S: |S|=k\right\} } \max \left\{ a(S),a(S^c)\right\} \quad \text { and }\quad h_k=\frac{a[n]}{n} {n \atopwithdelims ()k} \max \left\{ k, n-k\right\} . \end{aligned}$$

Then it follows from Lemmas 1 and 2 that

$$\begin{aligned} g(p) = \sum _{k=0}^n p^k(1-p)^{n-k} g_k\quad \text { and } \quad h(p) = \sum _{k=0}^n p^k(1-p)^{n-k} h_k. \end{aligned}$$

Since $g_k=g_{n-k}$ and $h_k=h_{n-k}$ for all k, it follows that g and h are symmetric around $p=0.5$, i.e. $g(p)=g(1-p)$ and $h(p)=h(1-p)$ for every $p\in (0,1)$. To prove the proposition one therefore needs to show that both $g-h$ and $1-\frac{h}{g}$ increase for $p\in (0,0.5)$.

Both g and h are polynomials and hence differentiable on (0, 1). In addition, g is bounded above zero on this interval, so the ratio h / g is differentiable. I show that $h'(p)\le 0$ and $h'(p)\le g'(p)$ for $p\in (0,0.5)$, which is sufficient to prove the result. Indeed, from $h'\le g'$ immediately follows that $g-h$ is increasing. Also, since $g\ge h>0$ it follows that $h'g-g'h \le h'g-h'h = h'(g-h)\le 0$, which shows that $\frac{h}{g}$ is decreasing.

To compute the derivatives $g'$ and $h'$, notice that for any function of the form $f(p)=\sum _{k=0}^n p^k(1-p)^{n-k}f_k$ with $f_k=f_{n-k}$ for all k the derivative is given by

$$\begin{aligned} f'(p)= & {} \sum _{k=0}^{n} f_k \left( kp^{k-1}(1-p)^{n-k}-(n-k)p^k(1-p)^{n-k-1}\right) \\= & {} \sum _{k=0}^{n-1} ((k+1) f_{k+1}-(n-k)f_{k})p^{k}(1-p)^{n-k-1}\\= & {} \sum _{k=0}^{\left\lfloor \frac{n-2}{2} \right\rfloor } ((k+1) f_{k+1}-(n-k)f_{k}) p^k(1-p)^k \left[ (1-p)^{n-2k-1}-p^{n-2k-1}\right] , \end{aligned}$$

where in the last equality $f_k=f_{n-k}$ was used. Thus, to prove that $h'\le 0$ and $h' \le g'$ on (0, 0.5) it is sufficient to show that

$$\begin{aligned} (k+1)h_{k+1}-(n-k)h_k\le 0 \end{aligned}$$

(11)

and

$$\begin{aligned} (k+1)h_{k+1}-(n-k)h_k\le (k+1)g_{k+1}-(n-k)g_k \end{aligned}$$

(12)

for every $k=0,\ldots ,\lfloor (n-2)/2\rfloor$. For every such k,

$$\begin{aligned}&(k+1)h_{k+1} - (n-k)h_k \\&= (k+1)\frac{a[n]}{n} {n \atopwithdelims ()k+1} \max \left\{ k+1, n-k-1\right\} - (n-k)\frac{a[n]}{n} {n \atopwithdelims ()k} \max \left\{ k, n-k\right\} \\&= a[n]{n-1 \atopwithdelims ()k}\left[ (n-k-1)-(n-k)\right] = -a[n]{n-1 \atopwithdelims ()k}, \end{aligned}$$

which proves (11).

For inequality (12), note that

$$\begin{aligned}&(k+1)g_{k+1} - (n-k)g_k \\&= (k+1)\sum _{\{S: |S|=k+1\}} \max \left\{ a(S),a(S^c)\right\} - (n-k)\sum _{\{S: |S|=k\}} \max \left\{ a(S),a(S^c)\right\} \\&=\sum _{\{S: |S|=k\}}\sum _{i\in S^c} \max \left\{ a(S\cup \{i\}),a((S\cup \{i\})^c)\right\} - (n-k)\sum _{\{S: |S|=k\}} \max \left\{ a(S),a(S^c)\right\} \\&=\sum _{\{S: |S|=k\}}\sum _{i\in S^c} \left[ \max \left\{ a(S\cup \{i\}),a((S\cup \{i\})^c)\right\} - \max \left\{ a(S),a(S^c)\right\} \right] . \end{aligned}$$

For each coalition S (with $|S|=k$) and for each $i\in S^c$ the above difference is bounded below by $-a_i$. I therefore get that

$$\begin{aligned} (k+1)g_{k+1} - (n-k)g_k\ge & {} -\sum _{\{S: |S|=k\}}\sum _{i\in S^c} a_i\\= & {} - {n-1 \atopwithdelims ()k}a[n] = (k+1)h_{k+1}-(n-k)h_k, \end{aligned}$$

which concludes the proof. $\square$

Proof of Proposition 5

First, since $\rho$ is invariant to scalings of the utility functions, I may assume that $a[n]=1$. Second, for any fixed p, $W^*$ and $W^{**}$ can be extended continuously from $a\in \mathbb {R}^n_{++}$ to $a\in \mathbb {R}^n_+$, and $W^*(a,p)>0$ for any $a\in \mathbb {R}^n_+$ except the origin. It follows that $\rho$ can be extended continuously to the simplex $\{a\in \mathbb {R}^n_+ :~ a[n]=1\}$. Now, Proposition 1 implies that for any given p the maximum of $\rho$ with respect to a is attained at a maximal element of the majorization ordering, i.e. in one of the corners of the simplex. Proposition 4 then implies that $\rho$ is maximal at $p=0.5$. Plugging these values into the expressions for $W^*$ and $W^{**}$ yields

$$\begin{aligned} W^*(a,p)=1, \end{aligned}$$

and

$$\begin{aligned} W^{**}(a,p) = \frac{1}{n}\left( \frac{1}{2}\right) ^n\sum _{k=0}^n {n \atopwithdelims ()k} \max \left\{ k, n-k \right\} . \end{aligned}$$

Simple manipulation gives

$$\begin{aligned} W^{**}(a,p) = \frac{1}{2} + \left( \frac{1}{2}\right) ^n {n-1 \atopwithdelims ()\lceil \frac{n-1}{2}\rceil }. \end{aligned}$$

Thus, $\bar{\rho } (n) = \frac{1}{2} - \left( \frac{1}{2}\right) ^n {n-1 \atopwithdelims ()\lceil \frac{n-1}{2}\rceil }$ as needed.

Finally, the expression $\left( \frac{1}{2}\right) ^{n-1} {n-1 \atopwithdelims ()\lceil \frac{n-1}{2}\rceil }$ is the probability that out of $n-1$ fair coin flips exactly $\lceil \frac{n-1}{2}\rceil$ are heads. It is well-known and easy to see that this probability is decreasing in n and converges to 0. Therefore $\bar{\rho } (n)$ is increasing in n and converges to $\frac{1}{2}$. $\square$

Appendix B: Proofs for Section 3

For the proofs of this section it is convenient to introduce some auxiliary random variables and to express the relevant probabilities using these variables. Given a society $\alpha$, let $X_1,\ldots ,X_n$ be independent random variables with $\mathbb {P}_\alpha (X_i=+1)=1-\mathbb {P}_\alpha (X_i=-1)=\alpha _i$. Let ${\bar{X}} = \sum _{i=1}^n X_i$. Also, if $i\in [n]$ then I denote ${\bar{X}}_{-i} = \sum _{j\ne i} X_j$, and for $i,j\in [n]$ I denote ${\bar{X}}_{-ij} = \sum _{l\ne i,j} X_l$.

Recall that $\beta _i=\log \frac{\alpha _i}{1-\alpha _i}$. Let $Y_1,\ldots ,Y_n$ be independent random variables with $\mathbb {P}_\alpha (Y_i=\beta _i)=1-\mathbb {P}_\alpha (Y_i=-\beta _i)=\alpha _i$. Define ${\bar{Y}}$, ${\bar{Y}}_{-i}$ and ${\bar{Y}}_{-ij}$ in an analogous way.

Notice that the probabilities of a correct decision under the optimal and optimal anonymous rules are simply given by $W^*(\alpha )=\mathbb {P}_\alpha ({\bar{Y}}> 0)+\frac{1}{2}\mathbb {P}_\alpha ({\bar{Y}}= 0)$ and $W^{**}(\alpha )=\mathbb {P}_\alpha ({\bar{X}}> 0)$, respectively.

Proof of Lemma 6

I first prove monotonicity of $W^{**}$. Notice that this function is continuously differentiable on $(0.5,1)^n$. Thus, by Marshal et al. (2011, Theorem A.4, p. 84), $W^{**}$ is monotonic with respect to the majorization ordering if and only if it satisfies the condition

$$\begin{aligned} (\alpha _i-\alpha _j)\left( \frac{\partial W^{**}}{\partial \alpha _i} - \frac{\partial W^{**}}{\partial \alpha _j} \right) \ge 0 \end{aligned}$$

(13)

at any point $\alpha$ and for any pair of agents i, j. I compute the partial derivatives of $W^{**}$ and show that (13) holds.

Given $\alpha$ and i,

$$\begin{aligned} W^{**}(\alpha )=\mathbb {P}_\alpha ({\bar{X}}> 0) = \alpha _i\mathbb {P}_\alpha ({\bar{X}}_{-i}> -1) + (1-\alpha _i)\mathbb {P}_\alpha ({\bar{X}}_{-i} > 1). \end{aligned}$$

Hence,

$$\begin{aligned} \frac{\partial W^{**}}{\partial \alpha _i} = \mathbb {P}_\alpha ({\bar{X}}_{-i}> -1) - \mathbb {P}_\alpha ({\bar{X}}_{-i} > 1) = \mathbb {P}_\alpha ({\bar{X}}_{-i} =0), \end{aligned}$$

where the last equality follows from the assumption that n is odd. In words, the marginal increase in the probability of a correct decision with respect to $\alpha _i$ is the probability that i is pivotal, i.e., the probability that exactly half of the other agents get an h signal. Now, given two agents i and j, the difference between the corresponding partial derivatives is

$$\begin{aligned} \frac{\partial W^{**}}{\partial \alpha _i} - \frac{\partial W^{**}}{\partial \alpha _j}= & {} \mathbb {P}_\alpha ({\bar{X}}_{-i} =0) - \mathbb {P}_\alpha ({\bar{X}}_{-j} =0) \\= & {} \left[ \alpha _j \mathbb {P}_\alpha ({\bar{X}}_{-ij} =-1)+(1-\alpha _j)\mathbb {P}_\alpha ({\bar{X}}_{-ij} =1)\right] \\&- \left[ \alpha _i \mathbb {P}_\alpha ({\bar{X}}_{-ij} =-1)+(1-\alpha _i)\mathbb {P}_\alpha ({\bar{X}}_{-ij} =1)\right] \\= & {} (\alpha _i-\alpha _j)\left[ \mathbb {P}_\alpha ({\bar{X}}_{-ij} =1)-\mathbb {P}_\alpha ({\bar{X}}_{-ij} =-1)\right] . \end{aligned}$$

Thus, to prove that (13) holds it is enough to show that $\mathbb {P}_\alpha ({\bar{X}}_{-ij} =1)\ge \mathbb {P}_\alpha ({\bar{X}}_{-ij} =-1)$ for any pair of agents i and j. By Samuels (1965, pp. 1272–1274), the distribution of ${\bar{X}}_{-ij}$ is unimodal, first increasing, then decreasing, and the mode can be either a single odd integer or a pair of adjacent odd integers. Furthermore, since $\alpha _l>0.5$ for all agents, it follows from Theorem 2 in Samuels (1965) that any mode must be positive. Hence, $\mathbb {P}_\alpha ({\bar{X}}_{-ij} =1)\ge \mathbb {P}_\alpha ({\bar{X}}_{-ij} =-1)$ which concludes the proof for $W^{**}$.

Moving on to $W^*$, I cannot immediately use condition (13) since $W^*$ is not everywhere differentiable. Specifically, $W^*$ is continuously differentiable only on the set $D=\{\alpha ~:~ \mathbb {P}_\alpha ({\bar{Y}}=0)=0\}$, as can be seen from Lemma 4. I claim however that if $(\alpha _i -\alpha _j) \left( \frac{\partial W^*}{\partial \alpha _i}-\frac{\partial W^*}{\partial \alpha _j}\right) \ge 0$ everywhere in D then the proposition will be proved. Indeed, given two agents i and j, let $d\in \mathbb {R}^n$ be defined by $d_i=1$, $d_j=-1$ and $d_l=0$ for any other agent l. By Marshal et al. (2011, Lemma A.2, p. 81), if $\alpha _i\ge \alpha _j$ implies that the mapping $t\longmapsto W^*(\alpha +t d)$ is non-decreasing for $t\ge 0$ (where it’s defined), then the proof is complete. But it is immediate to check that for every $\alpha$ the interval $\{\alpha +t d : t\ge 0\}$ intersects $D^c$ only finitely many times. Thus, the function $t\longmapsto W^*(\alpha +t d)$ is continuously differentiable at all but finitely many points on its domain (and continuous everywhere), so to show that it is non-decreasing it is enough to show that its derivative is non-negative wherever the derivative exists. But this is equivalent to showing that $\frac{\partial W^*}{\partial \alpha _i} \ge \frac{\partial W^*}{\partial \alpha _j}$ in D.

Now, fix $\alpha \in D$ and notice that

$$\begin{aligned} W^*(\alpha )=\mathbb {P}_\alpha ({\bar{Y}}> 0) = \alpha _i \mathbb {P}_\alpha ({\bar{Y}}_{-i}> -\beta _i) + (1-\alpha _i)\mathbb {P}_\alpha ({\bar{Y}}_{-i} > \beta _i). \end{aligned}$$

Since $\alpha \in D$, $\mathbb {P}_\alpha (-\beta _i-\epsilon< {\bar{Y}}_{-i}< -\beta _i+\epsilon ) = \mathbb {P}_\alpha (\beta _i-\epsilon< {\bar{Y}}_{-i} < \beta _i+\epsilon )=0$ for small enough $\epsilon >0$, and hence $\frac{\partial }{\partial \alpha _i}\mathbb {P}_\alpha ({\bar{Y}}_{-i}> -\beta _i) = \frac{\partial }{\partial \alpha _i}\mathbb {P}_\alpha ({\bar{Y}}_{-i} > \beta _i)=0$. Thus,

$$\begin{aligned} \frac{\partial W^*}{\partial \alpha _i} = \mathbb {P}_\alpha ({\bar{Y}}_{-i}> -\beta _i) - \mathbb {P}_\alpha ({\bar{Y}}_{-i} > \beta _i) = \mathbb {P}_\alpha (|{\bar{Y}}_{-i}| < \beta _i), \end{aligned}$$

which, as in the anonymous case above, is the probability that agent i is pivotal.

Notice that $\alpha _i\ge \alpha _j$ implies that $\beta _i\ge \beta _j$, and therefore that $-\beta _i+\beta _j\le 0 \le \beta _i-\beta _j$. Hence, $\alpha _i\ge \alpha _j$ implies that

$$\begin{aligned} \frac{\partial W^*}{\partial \alpha _i} - \frac{\partial W^*}{\partial \alpha _j}= & {} \mathbb {P}_\alpha (|{\bar{Y}}_{-i}|< \beta _i) -\mathbb {P}_\alpha (|{\bar{Y}}_{-j}|< \beta _j) \\= & {} \left[ \alpha _j \mathbb {P}_\alpha (-\beta _i-\beta _j< {\bar{Y}}_{-ij}< \beta _i-\beta _j )\right. \\&\left. +\, (1-\alpha _j)\mathbb {P}_\alpha (-\beta _i+\beta _j< {\bar{Y}}_{-ij}< \beta _i+\beta _j )\right] \\&- \left[ \alpha _i \mathbb {P}_\alpha (-\beta _i-\beta _j< {\bar{Y}}_{-ij}<-\beta _i+\beta _j ) \right. \\&\left. +\, (1-\alpha _i)\mathbb {P}_\alpha ( \beta _i-\beta _j< {\bar{Y}}_{-ij}< \beta _i+\beta _j )\right] \\\ge & {} (\alpha _i-\alpha _j)\left[ \mathbb {P}_\alpha (\beta _i-\beta _j< {\bar{Y}}_{-ij}< \beta _i+\beta _j ) \right. \\&\left. -\, \mathbb {P}_\alpha (-\beta _i-\beta _j< {\bar{Y}}_{-ij} <-\beta _i+\beta _j ) \right] , \end{aligned}$$

so it is sufficient to show that

$$\begin{aligned} \mathbb {P}_\alpha (\beta _i-\beta _j< {\bar{Y}}_{-ij}< \beta _i+\beta _j ) \ge \mathbb {P}_\alpha (-\beta _i-\beta _j< {\bar{Y}}_{-ij} <-\beta _i+\beta _j ). \end{aligned}$$

(14)

Notice that ${\bar{Y}}_{-ij}=\beta (S)-\beta (S^c)$, where $S=\{l\ne i,j : Y_l=\beta _l\}$ and the complement is relative to the set $[n]{\setminus } \{i,j\}$. Also, recall that $\beta (S)<\beta (S^c)$ is equivalent to $\alpha (S){\bar{\alpha }}(S^c) < {\bar{\alpha }}(S)\alpha (S^c)$. Therefore, any realized S at which ${\bar{Y}}_{-ij}=\beta (S) -\beta (S^c)<0$ has smaller probability than the realization $S^c$ (at which ${\bar{Y}}_{-ij} =\beta (S^c)-\beta (S)>0$). It follows that for every $0<c<d$, $\mathbb {P}_\alpha (c< {\bar{Y}}_{-ij}< d ) \ge \mathbb {P}_\alpha (-d< {\bar{Y}}_{-ij} < -c )$, which proves (14) $\square$

Proof of Proposition 6

1. Fix $\alpha _1\ge \cdots \ge \alpha _{n-1}$. I prove that $\tau (\alpha ) = W^*(\alpha ) - W^{**}(\alpha )$ is (weakly) decreasing in $\alpha _n\in (0.5, \alpha _{n-1})$. Note that $\tau$ is a continuous and piece-wise linear function of $\alpha _n$, and that it’s derivative exists at any point in which $\mathbb {P}_\alpha ({\bar{Y}}=0)=0$. Thus, it is sufficient to show that $\frac{\partial W^*}{\partial \alpha _n} \le \frac{\partial W^{**}}{\partial \alpha _n}$ at any such point. From the proof of the previous lemma one has that $\frac{\partial W^*}{\partial \alpha _n} = \mathbb {P}_\alpha (|{\bar{Y}}_{-n}| < \beta _n)$ and that $\frac{\partial W^{**}}{\partial \alpha _n} = \mathbb {P}_\alpha ({\bar{X}}_{-n} =0)$. Thus, one needs to prove that $\mathbb {P}_\alpha (|{\bar{Y}}_{-n}| < \beta _n) \le \mathbb {P}_\alpha ({\bar{X}}_{-n} =0)$, i.e., that the probability that agent n is pivotal under simple majority is at least as large as the probability that he is pivotal under the optimal rule.

The argument is based on the following two claims, whose proofs follow the proof of the proposition.

Claim 1

Let M be a set with an even number m of elements and let $\Gamma$ be a collection of subsets of M such that no two different sets in $\Gamma$ contain each other. Then there exists a one-to-one mapping f from $\Gamma$ to $\Gamma ':= \{T\subseteq M ~:~ |T|=m/2\}$ such that $S\subseteq f(S)$ or $f(S)\subseteq S$ for every $S\in \Gamma$.

Claim 2

Let $S\subseteq [n]{\setminus }\{n\}$ satisfy $|\beta (S)-\beta (S^c)|<\beta _n$ (complements are relative to the set $[n]{\setminus }\{n\}$). Then for every coalition $T\subseteq [n]{\setminus }\{n\}$ of size $|T|=\frac{n-1}{2}$ satisfying $T\supseteq S$ or $T\subseteq S$, $\alpha (S){\bar{\alpha }}(S^c)+\alpha (S^c){\bar{\alpha }}(S) \le \alpha (T){\bar{\alpha }}(T^c)+\alpha (T^c){\bar{\alpha }}(T)$.

Let $\Gamma =\{S \subseteq [n]{\setminus }\{n\} ~:~ |\beta (S)-\beta (S^c)|< \beta _n\}$ and $\Gamma '=\{T \subseteq [n]{\setminus }\{n\} ~:~ |T| = \frac{n-1}{2}\}$. Note that if $S\in \Gamma$ and $S\subsetneqq {\tilde{S}}$ then

$$\begin{aligned} |\beta ({\tilde{S}})-\beta ({\tilde{S}}^c)|= & {} |\beta (S)-\beta (S^c) + \beta ({\tilde{S}}{\setminus } S)+\beta (S^c{\setminus } {\tilde{S}}^c)| \\\ge & {} -\beta _n + \beta ({\tilde{S}}{\setminus } S)+\beta (S^c{\setminus } {\tilde{S}}^c) \ge \beta _n, \end{aligned}$$

where the first inequality follows from $S\in \Gamma$ and the last inequality from $\beta _n \le \beta _i$ for all $i\in [n]{\setminus }\{n\}$. It follows that $\Gamma$ satisfies the condition of Claim 1, so there is a one-to-one mapping $f:\Gamma \rightarrow \Gamma '$ such that $S\subseteq f(S)$ or $f(S)\subseteq S$ for every $S\in \Gamma$. Therefore,

$$\begin{aligned} \mathbb {P}_\alpha (|{\bar{Y}}_{-n}| < \beta _n) = \sum _{S \in \Gamma } \alpha (S){\bar{\alpha }}(S^c)= & {} \frac{1}{2}\sum _{S \in \Gamma } [\alpha (S){\bar{\alpha }}(S^c) + \alpha (S^c){\bar{\alpha }}(S)] \\\le & {} \frac{1}{2}\sum _{S \in \Gamma } [\alpha (f(S)){\bar{\alpha }}(f(S)^c) + \alpha (f(S)^c){\bar{\alpha }}(f(S))]\\\le & {} \frac{1}{2}\sum _{T \in \Gamma '} [\alpha (T){\bar{\alpha }}(T^c) + \alpha (T^c){\bar{\alpha }}(T)]\\= & {} \sum _{T \in \Gamma '} \alpha (T){\bar{\alpha }}(T^c) = \mathbb {P}_\alpha ({\bar{X}}_{-n} =0), \end{aligned}$$

where the first equality follows from the definition of $\Gamma$, the second equality from the fact that $\Gamma$ is closed under complements, the first inequality from Claim 2 and the fact that either $S\subseteq f(S)$ or $f(S)\subseteq S$ for every $S\in \Gamma$, the second inequality from the fact that f is one-to-one, the next equality from the fact that $\Gamma '$ is closed under complements, and the final equality from the definition of $\Gamma '$. This completes the proof that $\tau$ is non-increasing in $\alpha _n$ over $(0.5, \alpha _{n-1})$. Since $\rho (\alpha ) = \frac{\tau (\alpha )}{W^*(\alpha )}$, and since $W^*$ is weakly increasing in each of its arguments, it follows that $\rho$ is also non-increasing in $\alpha _n$ over $(0.5, \alpha _{n-1})$.

2. For an example in which $\rho (\alpha ^{\delta +})>\rho (\alpha )$ and $\tau (\alpha ^{\delta +})>\tau (\alpha )$ simply take $\alpha$ to be a constant vector. Then $\rho (\alpha )=\tau (\alpha )=0$, but after increasing $\alpha _1$ sufficiently both measures become positive.

I now give an example in which $\rho (\alpha ^{\delta +})<\rho (\alpha )$ and $\tau (\alpha ^{\delta +}) <\tau (\alpha )$. Let $n=7$ with $\alpha _1=\alpha _2=\alpha _3=\alpha _4=1-\epsilon$ and $\alpha _5=\alpha _6=\alpha _7=0.5+\epsilon$ for small $\epsilon >0$. I show that slightly increasing $\alpha _1$ decreases both $\rho$ and $\tau$.

Consider first the partial derivative of $W^*$ with respect to $\alpha _1$ (which exists since $\mathbb {P}_\alpha ({\bar{Y}}=0)=0$). As has already been shown, this derivative is equal to the probability $\mathbb {P}_\alpha (|{\bar{Y}}_{-1}| < \beta _1)$. But notice that if agents 2, 3 and 4 all got an h signal (i.e., $Y_2 =Y_3=Y_4>0$) then ${\bar{Y}}_{-1} > \beta _1$ regardless of the signals of the other agents, so agent 1 is not pivotal. Thus,

$$\begin{aligned} \frac{\partial W^*}{\partial \alpha _1} = \mathbb {P}_\alpha (|{\bar{Y}}_{-1}| < \beta _1) \le 1-\mathbb {P}_\alpha (Y_2=Y_3=Y_4>0) = 1- (1-\epsilon )^3. \end{aligned}$$

The partial derivative of $W^{**}$ with respect to $\alpha _1$ is equal to $\mathbb {P}_\alpha ({\bar{X}}_{-1}=0)$. One possible realization yielding ${\bar{X}}_{-1}=0$ is $X_2=X_3=X_4=1$ and $X_5=X_6=X_7=-1$. Therefore,

$$\begin{aligned} \frac{\partial W^{**}}{\partial \alpha _1} = \mathbb {P}_\alpha ({\bar{X}}_{-1}=0) \ge (1-\epsilon )^3(0.5-\epsilon )^3. \end{aligned}$$

It follows that $\frac{\partial W^*}{\partial \alpha _1} < \frac{\partial W^{**}}{\partial \alpha _1}$ whenever $\epsilon >0$ is sufficiently small. This implies that $\tau (\alpha ^{\delta +}) < \tau (\alpha )$ for $\delta >0$ sufficiently small. In addition, for every such $\delta$, $\rho (\alpha ^{\delta +}) = \frac{\tau (\alpha ^{\delta +})}{W^*(\alpha ^{\delta +})} < \frac{\tau (\alpha )}{W^*(\alpha )} = \rho (\alpha )$. $\square$

Proof of Claim 1

The proof of the claim is based on the following fact. For any $1\le k< m/2$ denote $\Lambda _k=\{S\subseteq M ~:~ |S|=k\}$. Then for every such k there exists a one-to-one mapping $g_k:\Lambda _k \rightarrow \Lambda _{k+1}$ such that $S\subseteq g_k(S)$ for all $S\in \Lambda _k$. The existence of such mappings can be proved by induction on m.

Now, consider a collection $\Gamma$ which satisfies the condition in the claim. As a first step I argue that I may assume that $\Gamma$ only contains sets of cardinality less than m / 2. Indeed, suppose that the claimis proved under this additional requirement. Then given a general collection $\Gamma$ one can partition $\Gamma$ into the three sets $\Gamma _1=\{S\in \Gamma ~:~ |S|<m/2\}$, $\Gamma _2=\{S\in \Gamma ~:~ |S|=m/2\}$ and $\Gamma _3=\{S\in \Gamma ~:~ |S|>m/2\}$. Let $f_1$ be the required function for $\Gamma _1$. Let $f_2$ be defined by $f(S)=S$ for any $S\in \Gamma _2$. Let $f_3$ be the function corresponding to $\{S^c ~:~ S\in \Gamma _3\}$. Then it is easy to check that the function f defined by $f(S)=f_1(S)$ for $S\in \Gamma _1$, $f(S)=f_2(S)$ for $S\in \Gamma _2$, and $f(S)=f_3(S^c)$ for $S\in \Gamma _3$ satisfies the requirement for the entire collection $\Gamma$.

Thus, suppose that $\Gamma$ contains only sets of cardinality less than m / 2 and satisfies the condition in the claim. For each $S\in \Gamma$ define $f(S) = g_{\frac{m}{2}-1}( \ldots ( g_{|S|+1} ( g_{|S|} (S))))$. Then f maps $\Gamma$ into $\Gamma '$ and clearly $S\subseteq f(S)$ for all $S\in \Gamma$. Also, it follows from the properties of $\{g_k\}_k$ and from the assumption that no two sets in $\Gamma$ contain one another that f is one-to-one. This concludes the proof. $\square$

Proof of Claim 2

Let S and T be as in the claim and suppose that $S\subseteq T$ (the case $T\subseteq S$ is analogous). If $S=T$ then the claim is trivially true, so suppose that S is a strict subset of T. Then simple algebra gives

$$\begin{aligned}&\left[ \alpha (S){\bar{\alpha }}(S^c)+\alpha (S^c){\bar{\alpha }}(S) \right] - \left[ \alpha (T){\bar{\alpha }}(T^c) + \alpha (T^c){\bar{\alpha }}(T) \right] \\&\quad =\left[ {\bar{\alpha }} (T{\setminus } S) - \alpha (T{\setminus } S) \right] \times \left[ \alpha (S) {\bar{\alpha }}(T^c) - {\bar{\alpha }}(S) \alpha (T^c) \right] . \end{aligned}$$

Since $\alpha _i>1-\alpha _i$ for every i it follows that ${\bar{\alpha }} (T{\setminus } S) < \alpha (T{\setminus } S)$, so to complete the proof I need to show that $\alpha (S) {\bar{\alpha }}(T^c) \ge {\bar{\alpha }}(S) \alpha (T^c)$. Rearranging and taking $\log$ of both sides shows that this is equivalent to $\beta (S) \ge \beta (T^c)$. This latter inequality follows from

$$\begin{aligned} \beta (S)-\beta (T^c) = \beta (S)-(\beta (S^c)-\beta (T{\setminus } S)) \ge -\beta _n +\beta (T{\setminus } S) \ge 0, \end{aligned}$$

where the first inequality follows from the assumption of the claim and the last inequality from the assumption that S is a strict subset of T and that $\beta _n\le \beta _i$ for every agent i. $\square$

Proof of Proposition 7

The functions $W^*(\alpha )$ and $W^{**}(\alpha )$ are continuous on $(0.5,1)^n$ and can be continuously extended to the compact set $[0.5,1]^n$. I prove that the maximum of $\tau$ on this set is attained at $(1,0.5,\ldots ,0.5)$.

Fix an agent i and $\alpha _{-i}$. Then $W^{**}(\cdot ;\alpha _{-i})$ is a linear function of $\alpha _i$ with slope $\mathbb {P}_\alpha ({\bar{X}}_{-i} =0)$, and $W^*(\cdot ;\alpha _{-i})$ is piece-wise linear in $\alpha _i$ with slope $\mathbb {P}_\alpha (|{\bar{Y}}_{-i}| < \beta _i)$ at any point $\alpha _i$ in which it is differentiable (see the proof of Lemma 6). Since $\beta _i$ is monotonically increasing in $\alpha _i$ it follows that $W^*(\cdot ;\alpha _{-i})$ is convex. Therefore, the difference $\tau (\cdot ;\alpha _{-i}) = W^*(\cdot ;\alpha _{-i}) - W^{**}(\cdot ;\alpha _{-i})$ is a convex function of $\alpha _i$. This implies that the maximum of this function is attained at one of the boundaries, either at $\alpha _i=0.5$ or at $\alpha _i=1$.

It follows that $\tau (\alpha )$ is maximized at a point in which each agent is either perfectly informed or completely uninformed. At $\alpha =(0.5,\ldots ,0.5)$ one has $\tau (\alpha )=0$, so this cannot be a maximizer. At any other corner of the cube at least one agent is perfectly informed, so that $W^*(\alpha )=1$. Since $W^{**}$ is increasing in each of its arguments, it follows that it is minimal when only one agent, say agent 1, is perfectly informed and everyone else is uninformed. At this point $\alpha$,

$$\begin{aligned} W^{**}(\alpha )= & {} \sum _{\left\{ S:~ |S|\ge \frac{n+1}{2}\right\} } \alpha (S){\bar{\alpha }}(S^c) = \sum _{\left\{ S:~ |S|\ge \frac{n+1}{2},~ 1\in S\right\} } \left( \frac{1}{2}\right) ^{|S|-1} \left( \frac{1}{2}\right) ^{n-|S|} \\= & {} \left( \frac{1}{2}\right) ^{n-1} \cdot \#\left\{ S:~ |S|\ge \frac{n+1}{2},~ 1\in S\right\} \\= & {} \left( \frac{1}{2}\right) ^{n-1} \cdot \frac{2^{n-1} - {n-1 \atopwithdelims ()\frac{n-1}{2}}}{2} = \frac{1}{2} + \left( \frac{1}{2}\right) ^n {n-1 \atopwithdelims ()\frac{n-1}{2} }. \end{aligned}$$

Thus, ${\bar{\tau }} (n) = \frac{1}{2} - \left( \frac{1}{2}\right) ^n {n-1 \atopwithdelims ()\frac{n-1}{2}}$ as needed. The rest of the proof is the same as in the proof of Proposition 5. $\square$

Appendix C: Heterogeneous probabilities example

Consider a society of $n=4$ agents with $a_1=4$, $a_2=a_3=3$, $a_4=1$, and $b_i=a_i$ for all i. Let the vector of probabilities be $(p_1,p_2,p_3,p_4)=(1-\epsilon ,0.5,0.5,\epsilon )$, where $\epsilon >0$ is small. I compare $\tau$ and $\rho$ between this society and the society with $a^{\delta +}$, i.e., the society in which $a_1=4+\delta$ and everything else is unchanged.

Note first that whenever at least three of the agents support the same alternative (either A or B) then it is optimal to choose that alternative. It follows that the optimal rule and the optimal anonymous rule coincide at any such realization of types. So the rules may differ only if two agents support A and the other two support B. Now, since agent 1 is much more likely to prefer A and agent 4 is much more likely to prefer B, it follows that conditional on a tie (two agents support each alternative) the optimal anonymous rule chooses A. On the other hand, the optimal rule chooses A when the pair that prefers A is either $\{1,2\}$ or $\{1,3\}$ or $\{2,3\}$, and chooses B otherwise. Therefore,

$$\begin{aligned} \tau (a,p)= & {} W^*(a,p)-W^{**}(a,p) \\= & {} p_1(1-p_2)(1-p_3)p_4(6-5)+(1-p_1)p_2(1-p_3)p_4(7-4)\\&+\,(1-p_1)(1-p_2)p_3p_4(7-4)\\= & {} \frac{1}{4}\left[ \epsilon (1-\epsilon )+6\epsilon ^2\right] . \end{aligned}$$

When moving to $a^{\delta +}$, as long as $\delta <1$ the optimal and optimal anonymous rules do not change. A similar computation then gives

$$\begin{aligned} \tau (a^{\delta +},p)= & {} W^*(a^{\delta +},p)-W^{**}(a^{\delta +},p) \\= & {} \frac{1}{4}\left[ (1-\delta )\epsilon (1-\epsilon )+(6+2\delta )\epsilon ^2\right] \\= & {} \tau (a,p)-\frac{\delta }{4}[\epsilon (1-\epsilon ) - 2\epsilon ^2] < \tau (a,p). \end{aligned}$$

This also implies that

$$\begin{aligned} \rho (a^{\delta +},p) = \frac{\tau (a^{\delta +},p)}{W^*(a^{\delta +},p)} <\frac{\tau (a,p)}{W^*(a,p)} = \rho (a,p). \end{aligned}$$

Hence, increasing the inequality of stakes in the society resulted in a decrease of both measures of the price of OPOV, contrary to the conclusions of Sect. 2.3. $\square$