On the costly voting model: the mean rule

Abstract

In this paper, we study a model in which a policy is chosen by a group of people through the formation of a committee. Attending the committee is costly, and each person decides whether to take part in it or not. Our work complements Osborne, Rosenthal and Turner (2000) since we allow various types of costs. We work with the mean compromise function, and we do not restrict the distribution of the favourite policy of the group members. Under these assumptions, we establish the existence of pure Nash equilibrium and show that, in comparison to the case of the median compromise function, the outcome of the committee’s work is less random and is not likely to be extreme. In the case of constant costs of participation, we show that when the costs increase, the size of the equilibrium committee decreases and the spread of the outcomes increases.

Appendix

1.1 1 Proof of Proposition 1

The proof of Proposition 1 relies on the following simple lemma about means of numbers.

Lemma 4

Let $X\subset \{x_1, \ldots ,x_n\}$ be a subset of $X_n$ and suppose $x_i,x_j\notin X$. Suppose $|m(X)-x_i|\le |m(X)-x_j|$ then

• (1) $m(X\cup x_i)$ belongs to the interval bounded by m(X) and $x_i$, namely

  $$\begin{aligned} m(X\cup x_i)-x_i=\frac{|X|}{|X|+1}(m(X)-x_i). \end{aligned}$$

  (5)

• (2) $|m(X\cup x_i)-x_i|\le |m(X\cup x_j)-x_j|$.

• (3) $|m(X\cup x_i\cup x_j)-x_i|\le |m(X\cup x_i \cup x_j)-x_j|$.

Proof

• (1) By the definition of the mean we have

  $$\begin{aligned} m(X\cup x_i)=\frac{1}{|X+1|}(x_i+\sum _{x_k\in X}x_k)=\frac{x_i+|X|\cdot m(X)}{|X+1|}. \end{aligned}$$

  Subtracting $x_i$ from the left- and the right-hand side expressions, we obtain Equality (5).

• Statement (2) follows from Equality (5).

• (3) First note that the statement does not change if we subtract the value m(X) from $x_i, x_j$ and each number in X. As a result we will have $m(X)=0$, and the condition $|m(X)-x_i|\le |m(X)-x_j|$ becomes $|x_i|\le |x_j|$. The inequality that we need to prove is the following one:

  $$\begin{aligned} \left| \frac{|X|\cdot m(X)+x_i+x_j}{|X|+2}-x_i\right| \le \left| \frac{|X|\cdot m(X)+x_i+x_j}{|X|+2}-x_j\right| . \end{aligned}$$

  Since we assume now that $m(X)=0$, we need to prove that $|x_j-(|X|+1)x_i|<|x_i-(|X|+1)x_j|$. This clearly follows from the inequality $|x_i|\le |x_j|$. $\square$

$\square$

Proof of Proposition 1

Both statements of Proposition 1 (a) and (b) are proved by induction. The statement (a) follows easily from the definition of $x_{\sigma _k}$, so we will only prove statement (b).

For $k=1$, the statement (b) is obvious since $m(X_1)=x_1$ and $x_{\sigma _1}=x_1$. Suppose by induction that $x_{\sigma _k}$ is the closest element to $m(X_k)$ among all elements of $X_k$, and let us deduce that $x_{\sigma _{k+1}}$ is the closest to $m(X_{k+1})$.

Note that $x_{\sigma _{k+1}}$ is not further from $m(X_{k-1})$ than $x_{\sigma _k}$. Since $X_{k+1}=X_{k-1}\cup x_{\sigma _k}\cup x_{\sigma _{k+1}}$, by statement (3) of Lemma 4

$$\begin{aligned} |x_{\sigma _{k+1}}-m(X_{k+1})|\le |x_{\sigma _k}-m(X_{k+1})|. \end{aligned}$$

(6)

Now consider the following two possibilities separately:

• (i) $m(X_k)$ lies in the segment bounded by $x_{\sigma _k}$ and $x_{\sigma _{k+1}}$. In this case, $m(X_{k+1})$ lies also between $x_{\sigma _k}$ and $x_{\sigma _{k+1}}$. By property a) no element of $X_{k-1}$ lies between $x_{\sigma _k}$ and $x_{\sigma _{k+1}}$. Hence elements of $X_{k-1}$ are further from $m(X_{k+1})$ than either $x_{\sigma _k}$ or $x_{\sigma _{k+1}}$ (or both). I.e., by (6) they are all further from $m(X_{k+1})$ than $x_{\sigma _{k+1}}$, and the proposition is proven.

• (ii) $m(X_k)$ does not lie between $x_{\sigma _k}$ and $x_{\sigma _{k+1}}$. We will assume that $m(X_k)\le x_{\sigma _{k+1}}\le x_{\sigma _k}$ (the case $m(X_k)\ge x_{\sigma _{k+1}}\ge x_{\sigma _k}$ is treated similarly). Using these inequalities and the induction assumption for any $x_i\in X_k$ we have

  $$\begin{aligned} |x_i-m(X_k)|\ge |x_{\sigma _k}-m(X_k)|\ge |x_{\sigma _{k+1}}-m(X_k)|. \end{aligned}$$

  At the same time, by Lemma 4, $m(X_{k+1})$ belongs to the interval bounded by $x_{\sigma _{k+1}}$ and $m(X_k)$. So one can deduce from the above inequality that $|x_i-m(X_{k+1})|\ge |x_{\sigma _{k+1}}-m(X_{k+1})|$. Hence the proposition holds.

$\square$

1.2 2 Proof of Proposition 2

The proof of Proposition 2 will rely on the following lemma.

Lemma 5

Let v be a concave single-peaked symmetric function defined on $[-1,1]$. Suppose $0<|y_1|<|y_2|<1$, $0<|z_1|<|z_2|<1$, $|y_1|\le |z_1|$, $|y_2-y_1|\le |z_2-z_1|$, and $y_1\cdot y_2>0$, $z_1\cdot z_2>0$. Then

$$\begin{aligned} v(z_1)-v(z_2)\ge v(y_1)-v(y_2). \end{aligned}$$

Proof

Since v is symmetric, we can change the signs of $y_1, y_2$ and $z_1,z_2$, to make them all positive, in particular $0<y_1\le z_1$.

Note also, that for any concave function v, and any x and a positive t, the function $v_t(x)=v(x)-v(x+t)$ is increasing. Indeed, if $x_2>x_1$ then from the definition of concavity it follows that $v(x_1)+v(x_2+t)\le v(x_2)+v(x_1+t)$ for all positive t. Applying this fact to our situation, we get

$$\begin{aligned} v(z_1)-v(z_2)\ge v(y_1)-v(y_1+z_2-z_1). \end{aligned}$$

Since v(x) is single-peaked, it is decreasing for positive x, and so it follows from our assumptions that $v(y_1+z_2-z_1)\le v(y_2)$. By plugging the latter inequality into the previous one we finish the proof of the lemma. $\square$

Proof of Proposition 2

Now we give the proof of Proposition 2 and split it into two parts.

(1). First, we will prove that if $\mathcal{A}(X_k)$ is not profitable for a player i not from the committee $X_k$ then it is not profitable for the player $\sigma _{k+1}$, i.e., the player corresponding to the unique element in $X_{k+1}\setminus X_{k}$.

According to Eqs. (1) and (2), the profit that player i gets from changing his strategy is equal to

$$\begin{aligned} v(m(X_k\cup x_i)-x_i)-v(m(X_k)-x_i)-c(k+1). \end{aligned}$$

(7)

The profit that player $\sigma _{k+1}$ gets from changing his strategy is equal to

$$\begin{aligned} v(m(X_k\cup x_{\sigma _{k+1}})-x_{\sigma _{k+1}})-v(m(X_k)-x_{\sigma _{k+1}})-c(k+1). \end{aligned}$$

(8)

By our assumption the first expression is non-negative, so to prove Claim 1) it is enough to show that the second expression is larger than the first. Here we use Lemma 5. In order to apply this lemma in our situation, set

$$\begin{aligned} y_1&=m(X_k\cup x_i)-x_i,\ y_2=m(X_k)-x_i\\ z_1&=m(X_k\cup x_{\sigma _{k+1}})-x_{\sigma _{k+1}}, \ z_2=m(X_k)-x_{\sigma _{k+1}} \end{aligned}$$

Note that $y_1,y_2,z_1,z_2$ defined this way satisfy the inequalities of Lemma 5. Indeed, by statement (1) of Lemma 4 we have $y_1=\frac{k}{k+1}y_2$, and $z_1=\frac{k}{k+1}z_2$, and the definition of the element $x_{\sigma _{k+1}}$ gives us the following inequality

$$\begin{aligned} |y_2|=|m(X_k)-x_i|\le |m(X_k)-x_{\sigma _{k+1}}|=|z_2|. \end{aligned}$$

So, applying Lemma 5 to $(y_1,y_2,z_1,z_2)$, we see that the profit (8) of player $\sigma _{k+1}$ to change his strategy is larger than the profit (7) of player i to change his strategy. Hence if the action profile $\mathcal{A}(X_k)$ is not profitable for player i, it is not profitable for player $\sigma _{k+1}$ as well. Claim (1) is proved.

(2). Suppose that the action profile $\mathcal{A}(X_{k}\cup x_{\sigma _{k+1}})=\mathcal{A}(X_{k+1})$ is profitable for $\sigma _{k+1}$. Then it is profitable as well for all players from $X_{k+1}$.

Here the reasoning is identical to the reasoning for proving (1), the only difference is that we should use the fact that $x_{\sigma _{k+1}}$ is the closest to $m(X_{k+1})$ among all the elements of $X_{k+1}$. This is exactly the statement of Proposition 1. Hence the theorem is proven. $\square$

1.3 3 Proof of Theorem 2

The proof of Theorem 2 relies on several lemmas and propositions, and we start by introducing the following useful notations. Write $X'\preceq X+c$ if there is a one-to-one correspondence $\varphi : X\rightarrow X'$ such that for any $x_i\in X$ we have $\varphi (x_i)\le x_i+ c$. For any committee X by r(X) we denote the largest $x_i \in X$, and by l(X) we denote the smallest $x_i \in X$.

The following lemma is straightforward.

Lemma 6

Suppose that $|X|=|X'|$ and c is a constant, then the following holds.

• (1) If $X'\preceq X+c$, then $m(X')\le m(X)+c$.

• (2) $X'\preceq X+c$ if and only if for any y we have

  $$\begin{aligned} |X\cap [0,y]|\le |X'\cap [0,y+c]|. \end{aligned}$$

The first statement follows from the definition of the mean. To prove the second statement, consider the unique map $\varphi : X\rightarrow X'$, which preserves the order of policies (i.e., if for $x_i,x_j\in X$ we have $x_i<x_j$, then $\varphi (x_i)<\varphi (x_j)$).

To state the next proposition, let us give a formal definition of a committee with one gap.

Definition 3

We say that a committee X is a committee with one gap if there exist i and j satisfying $2\le i+1<j\le n$ such that $X=\{x_1,\ldots , x_i\}\cup \{x_j,\ldots , x_n\}$. Furthermore, for such a committee X with one gap, we denote $x_i$ by l(X) and $x_j$ by r(X).

Proposition 5

For any equilibrium committee X with less than n members, there is a committee $X'$ of the same size, with one gap, and satisfying the following conditions.

• (1) $X'-2C\preceq X \preceq X'+2C$.

• (2) For any $x_i\in X'$ we have $|x_i-m(X')|\ge (|X'|-3)C$.

• (3) For any $x_j\notin X'$ we have $|x_j-m(X')|\le (|X'|+3)C$.

• (4) $m(X')$ lies inside the gap of $X'$.

• (5) $2(|X|-1)C<r(X')-l(X')<2(|X|+1)C$.

Proof

• (1) There is a unique committee $X'$ with one gap that has the following properties:

  $$\begin{aligned} |X\cap [0, m(X)]|=|X'\cap [0,m(X)]|,\;\;\; |X\cap [ m(X),1]|=|X'\cap [m(X),1]|. \end{aligned}$$

  To obtain $X'$ from X we take all the members of X to the left from m(X) and push them further to the left to fill in all empty places (if there are any). The members of X to the right from m(X) are similarly pushed to the right. From Corollary 2 it follows that the sets X and $X'$ coincide outside of the interval $[m(X)-(|X|+1)C, m(X)+(|X|+1)C]$, and are empty inside the interval $[m(X)-(|X|-1)C, m(X)+(|X|-1)C]$. This implies that the conditions of statement (2) of Lemma 6 are satisfied for $c=2C$ and for both pairs $(X, X')$ and $(X', X)$. Hence, applying this lemma we obtain statement (1).

• (2) Applying statement (2) of Lemma 6 and statement (1) of Proposition 5 and Claim 1), we get $m(X)-2C\le m(X') \le m(X)+2C$. At the same time, by its construction, $X'$ is empty inside the interval $[m(X)-(|X|-1)C, m(X)+(|X|-1)C]$. The combination of these statements proves the claim.

• (3) By construction, $X'$ contains all the points $x_j$ that lie outside the interval $[m(X)-(|X|+1)C, m(X)+(|X|+1)C]$. Hence, again the claim follows from the inequalities $m(X)-2C\le m(X') \le m(X)+2C$.

• (4) The statement is obviously correct if $|X|\le 2$. Let us prove it for $|X|\ge 3$. Recall from the construction of $X'$ that m(X) lies inside the gap of $X'$ and the minimal distance from m(X) to members of $X'$ is no smaller than the minimal distance to the members of X. By Corollary 2, the latter distance is at least $(|X|-1)C$. Now, from (1) we know that $|m(X)-m(X')|\le 2C$. And since $|X|-1\ge 2$, it follows that $m(X')$ lies inside of the gap of $X'$ as well.

• (5) From the construction of $X'$ and Corollary 2 it follows that $m(X)-l(X')\in [(|X|-1)C,(|X|+1)C]$ and $r(X')-m(X)\in [(|X|-1)C,(|X|+1)C]$. This clearly implies the claim.

$\square$

Proposition 6

Suppose that X and $X'$ are committees with one gap containing the mean inside their gaps. If $l(X)\le l(X'), r(X)\le r(X')$, then $m(X)>m(X')$.

Proof

Let us consider the set $X''$ that is obtained from X by throwing out all elements $x_i$ that do not belong to $X'$ and such that $x_i\ge r(X)$ from X. Since $m(X)<r(X)$, we clearly have $m(X'')<m(X)$.

Now note that $X''$ is obtained from $X'$ by throwing out all the elements that lie in the interval $(l(x), m(X'))$ from $X'$. Hence by the reasoning as above $m(X'')>m(X')$. This concludes the proof. $\square$

Now we are ready to give the proof of Theorem 2.

Proof of Theorem 2

Let $X'$ be the committee with one gap obtained from X using Proposition 5. Let us consider four cases:

• (1) $l(X')\ge l(X_k)$, $r(X')\le r(X_k)$. In this case

  $$\begin{aligned} 2(k-1)C\le 2(|X|-1)C\le r(X')-l(X')\le r(X_k)-l(X_k)\le 2(k+1)C. \end{aligned}$$

  Indeed, the first and third inequalities hold since $X_k\subset X'$. The second inequality follows from statement (1) of Corollary 2 and the inequality $k\le |X|$. The fourth inequality follows from statement (2) of Corollary 2. We conclude $k\le |X'|\le k+2$. Assume now by contradiction $m(X)-m(X_k)>6C$. Then from statement (2) of Proposition 5 it follows that $m(X')-m(X_k)>4C$. We deduce finally

  $$\begin{aligned} r(X')-m(X')<r(X_k)-m(X_k)-4C< (k-3)C\le (|X'|-3)C. \end{aligned}$$

  This contradicts statement (2) of Proposition 5. By identical reasoning, we prove that $m(X_k)-m(X)<6C$. This finishes the proof of the first claim.

• (2) $l(X')\ge l(X_k)$, $r(X')\le r(X_k)$. By reasoning similar to the above, we get $k\ge |X'|\ge k-2$ and deduce $|m(X_k)-m(X)|<6C$.

• (3) $l(X')> l(X_k)$, $r(X')> r(X_k)$. In this case, by Proposition 6 we have $m(X')\le m(X_k)$. Now we can apply statement (2) and (3) of Proposition 5, to get the following inequalities

  $$\begin{aligned} (|X'|-3)C&\le m(X')-l(X')< m(X_k)-l(X_k)\le (k+1)C.\\ (|X'|+3)C&\ge r(X')-m(X')> r(X_k)-m(X_k)\ge (k-1)C. \end{aligned}$$

  Taking these two inequalities together, we deduce that $||X|-k|\le 3$. Since we know that $m(X')\le m(X_k)$, it follows that $m(X)\le m(X_k)+2C$. Hence we only need to show that $m(X_k)-m(X)\le 6C$. Assuming by contradiction that the opposite is true, we get $m(X_k)-m(X')>4C$. Now consider two situations: first, $|X|\ge k$. In this case

  $$\begin{aligned} m(X')-l(X')<m(X_k)-l(X_k)-4C<(k-3)C\le ||X|-3|C, \end{aligned}$$

  which contradicts statement (2) of Proposition 5. Secondly, suppose $|X|\le k$. In this case

  $$\begin{aligned} r(X')-m(X')>r(X)-m(X)+4C> (k+3)C\ge (|X|+3)C, \end{aligned}$$

  which contradicts statement (3) of Proposition 5.

• (4) $l(X')< l(X_k)$, $r(X')< r(X_k)$. This case is similar to case (3).

$\square$