The uniqueness of the pivotal mechanisms without strategy-proofness

Abstract

Moulin (J Public Econ 31:53–78, 1986; Theorem 4) characterizes the pivotal mechanisms without imposing strategy-proofness under the assumption of the full domain. In this paper, we provide a domain property that is necessary and sufficient for Moulin’s characterization without strategy-proofness to hold. We also provide examples of domains that do (not) satisfy our domain property.

Appendix

1.1 A. Proof of Theorem 1

(i) $\varvec{\Rightarrow }$ (ii). Let us show that if a domain $\prod _{j=1}^n D_j$ violates Property 1, then there exists a mechanism that satisfies the five axioms and is not Pareto-dominated by a pivotal mechanism. Suppose that $\prod _{j=1}^n D_j$ violates Property 1. Let $(d^*,t^*_1, \ldots , t^*_n) \in {\mathscr {M}}(\prod _{j=1}^n D_j)$ be a pivotal mechanism. Since $\prod _{j=1}^n D_j$ violates Property 1, there exist $i \in N, {\bar{v}} \in \prod _{j=1}^n D_j$, and $\varepsilon >0$ such that for each $v' \in \prod _{j=1}^n D_j$ with ${\bar{v}} \ \begin{array}{c} \rightarrow \\ i \end{array} \ v'$,

$$\begin{aligned} \left\{ \max _{y \in X} \sum _{j=1}^n v'_j(y)-\sum _{j \ne i} \min _{y \in E(\sum _{k \ne j}v'_k)} v'_j(y)\right\} -\left\{ u_i\left( d^*(\bar{v}), t^*_i({\bar{v}}); {\bar{v}}_i \right) \right\} \ge \varepsilon . \end{aligned}$$

(9)

Let $(d, t_1, \ldots , t_n) \in {\mathscr {M}}(\prod _{j=1}^n D_j)$ be such that for each $v \in \prod _{j=1}^n D_j$,

$$\begin{aligned}&\text {(i)} \ d(v) \in E\left( \sum _{j=1}^n v_j \right) , \nonumber \\&\text {(ii)} \ t_i(v) =-v_i\left( d(v)\right) \nonumber \\&\quad +\inf _{v \ \begin{array}{c} \rightarrow \\ i \end{array} \ v'} \left\{ \max _{y \in X} \sum _{j=1}^n v'_j(y)-\sum _{j \ne i} \min _{y \in E(\sum _{k \ne j}v'_k)} v'_j(y)\right\} \nonumber \\&\text {(iii)} \ t_j(v)= -v_j\left( d(v)\right) +\min _{y \in E(\sum _{k \ne j}v_k)} v_j(y) \ \ \text { for all }j \ne i. \end{aligned}$$

(10)

Note that by the same argument as Eq. (6) in Section 3, for each $v \in \prod _{j=1}^n D_j$,

$$\begin{aligned} u_i\left( d(v), t_i(v); v_i \right) =v_i\left( d(v)\right) +t_i(v)&=\inf _{v \ \begin{array}{c} \rightarrow \\ i \end{array} \ v'} \left\{ \max _{y \in X} \sum _{j=1}^n v'_j(y)-\sum _{j \ne i} \min _{y \in E(\sum _{k \ne j}v'_k)} v'_j(y)\right\} \nonumber \\&\ge u_i\left( d^*(v), t^*_i(v); v_i \right) . \end{aligned}$$

(11)

Moreover, by definition of $(d, t_1, \ldots , t_n)$ and Inequality (9), we have

$$\begin{aligned} u_i\left( d({\bar{v}}), t_i({\bar{v}}); {\bar{v}}_i \right) ={\bar{v}}_i\left( d(\bar{v})\right) +t_i({\bar{v}})&\ge \inf _{{\bar{v}} \ \begin{array}{c} \rightarrow \\ i \end{array} \ v'} \left\{ \max _{y \in X} \sum _{j=1}^n v'_j(y)-\sum _{j \ne i} \min _{y \in E(\sum _{k \ne j}v'_k)} v'_j(y)\right\} \\&\ge u_i\left( d^*({\bar{v}}), t^*_i({\bar{v}}); {\bar{v}}_i \right) +\varepsilon > u_i\left( d^*({\bar{v}}), t^*_i({\bar{v}}); {\bar{v}}_i \right) . \end{aligned}$$

Therefore, $(d, t_1, \ldots , t_n)$ is not Pareto dominated by a pivotal mechanism. In addition, obviously $(d, t_1, \ldots , t_n)$ satisfies efficiency. Let us show that $(d, t_1, \ldots , t_n)$ satisfies feasibility, no free ride, no disposal of utility, and distribution.

Feasibility Take any $v \in \prod _{j=1}^n D_j$. Note that

$$\begin{aligned} u_i\left( d(v), t_i(v); v_i \right)&=\inf _{v \ \begin{array}{c} \rightarrow \\ i \end{array} \ v'} \left\{ \max _{y \in X} \sum _{j=1}^n v'_j(y)-\sum _{j \ne i} \min _{y \in E(\sum _{k \ne j}v'_k)} v'_j(y)\right\} , \\ u_j\left( d(v), t_j(v); v_j \right)&= \min _{y \in E(\sum _{k \ne j}v_k)} v_j(y) \ \ \text { for all }j \ne i. \end{aligned}$$

Therefore, by $v \ \begin{array}{c} \rightarrow \\ i \end{array} \ v$,

$$\begin{aligned} \sum _{j=1}^n t_j(v)&=\left\{ \max _{y \in X} \sum _{j=1}^n v_j(y)+\sum _{j=1}^n t_j(v)\right\} -\max _{y \in X} \sum _{j=1}^n v_j(y)\\&=\sum _{j=1}^n u_j\left( d(v), t_j(v); v_j \right) - \max _{y \in X} \sum _{j=1}^n v_j(y) \\&=\inf _{v \ \begin{array}{c} \rightarrow \\ i \end{array} \ v'} \left\{ \max _{y \in X} \sum _{j=1}^n v'_j(y)-\sum _{j \ne i} \min _{y \in E(\sum _{k \ne j}v'_k)} v'_j(y)\right\} \\&\quad -\left\{ \max _{y \in X} \sum _{j=1}^n v_j(y)-\sum _{j \ne i} \min _{y \in E(\sum _{k \ne j}v_k)} v_j(y)\right\} \\&\le 0. \end{aligned}$$

Hence $\sum _{j=1}^n t_j(v) \le 0$.

No free ride Take any $j \in N$ and any $v \in \prod _{j=1}^n D_j$. If $j \ne i$, then clearly $u_j\left( d(v), t_j(v); v_j \right) =v_j\left( d(v)\right) +t_j(v)=\min _{y \in E(\sum _{k \ne j}v_k)} v_j(y)$. Suppose that $j=i$. Note that since the pivotal mechanism $(d^*,t^*_1, \ldots , t^*_n)$ satisfies no free ride, it follows that $u_i\left( d^*(v), t^*_i(v); v_i \right) \ge \min _{y \in E(\sum _{j \ne i} v_j)} v_i(y)$. Then, by Eq. (11), we have

$$\begin{aligned} u_i\left( d(v), t_i(v); v_i \right) \ge u_i\left( d^*(v), t^*_i(v); v_i \right) \ge \min _{y \in E(\sum _{j \ne i} v_j)} v_i(y). \end{aligned}$$

Therefore, $(d,t_1, \ldots , t_n)$ satisfies no free ride.

No disposal of utility Take any $j \ne i$, any $v \in \prod _{j=1}^n D_j$, and any $v'_j \in D_j$ with $v'_j \ge v_j$. Then,

$$\begin{aligned} u_j\left( d(v'_j, v_{-j}), t_j(v'_j, v_{-j}); v_j \right)= & {} \min _{y \in E(\sum _{k \ne j} v_k)} v'_j(y) \ge \min _{y \in E(\sum _{k \ne j}v_k)} v_j(y)\\= & {} u_j\left( d(v), t_j(v); v_j \right) . \end{aligned}$$

Next, take any $v \in \prod _{j=1}^n D_j$ and any $v'_i \in D_i$ with $v'_i \ge v_i$. Then, since $(v_i, v_{-i}) \ \begin{array}{c} \rightarrow \\ i \end{array} \ (v'_i, v_{-i})$ and the relation $\begin{array}{c} \rightarrow \\ i \end{array}$ is transitive,

$$\begin{aligned}&\inf _{(v'_i, v_{-i}) \ \begin{array}{c} \rightarrow \\ i \end{array} \ v'} \left\{ \max _{y \in X} \sum _{j=1}^n v'_j(y)-\sum _{j \ne i} \min _{y \in E(\sum _{k \ne j}v'_k)} v'_j(y)\right\} \\&\quad \ge \inf _{(v_i, v_{-i}) \ \begin{array}{c} \rightarrow \\ i \end{array} \ v'} \left\{ \max _{y \in X} \sum _{j=1}^n v'_j(y)-\sum _{j \ne i} \min _{y \in E(\sum _{k \ne j}v'_k)} v'_j(y)\right\} . \end{aligned}$$

Therefore,

$$\begin{aligned} u_i\left( d(v'_i, v_{-i}), t_i(v'_i, v_{-i}); v'_i \right) \ge u_i\left( d(v), t_i(v); v_i \right) . \end{aligned}$$

Distribution It suffices to show that for each distinct $k, \ell \in N$, each $v \in \prod _{j=1}^n D_j$ with $E(\sum _{j=1}^n v_{j})=\{z\}$, and each $v'_{\ell } \in D_{\ell }$ such that

$$\begin{aligned} v'_{\ell }(z)>v_{\ell }(z) \ \ \text { and } \ \ v'_{\ell }(x)=v_{\ell }(x) \ \ \text { for all }x \in X \setminus \{z\}, \end{aligned}$$

we have

$$\begin{aligned} u_{k}\left( d(v'_{\ell }, v_{-{\ell }}), t_k(v'_{\ell }, v_{-{\ell }}); v_{k} \right) \ge u_{k}\left( d(v_{\ell }, v_{-{\ell }}), t_k(v_{\ell }, v_{-{\ell }}); v_{k} \right) . \end{aligned}$$

Take any distinct $k, \ell \in N$.

Let us consider the case with $k \ne i$. Take any $v \in \prod _{j=1}^n D_j$ such that for some $z \in X, E(\sum _{j=1}^n v_{j})=\{z\}$. We first show that

$$\begin{aligned} v_{k}(z) \ge \min _{y \in E(\sum _{j \ne k}v_{j})} v_{k}(y) . \end{aligned}$$

(12)

Suppose, by contradiction, that $\min _{y \in E(\sum _{j \ne k}v_{j})} v_{k}(y) > v_{k}(z)$. Then,

$$\begin{aligned} \max _{y \in X} \sum _{j=1}^n v_{j}(y)\ge & {} \min _{y \in E(\sum _{j \ne k}v_{j})} v_{k}(y) + \max _{y \in X} \sum _{ j \ne k}v_{j}(y) >v_k(z)\\&\quad +\,\max _{y \in X} \sum _{ j \ne k}v_{j}(y) \ge \sum _{j=1}^n v_{j}(z), \end{aligned}$$

a contradiction to $E(\sum _{j=1}^n v_{j})=\{z\}$. Therefore, Eq. (12) holds.

Now, take any $v'_{\ell } \in D_{\ell }$ such that $v'_{\ell }(z)>v_{\ell }(z)$ and $v'_{\ell }(x)=v_{\ell }(x)$ for all $x \in X \setminus \{z\}$. Then,

$$\begin{aligned} E\left( v'_{\ell }+\sum _{j \ne k, \ell } v_{j} \right) \subset E\left( \sum _{j \ne k} v_{j}\right) \cup \{z\}. \end{aligned}$$

(13)

Therefore, by $k \ne i$ and Eqs. (12) and (13),

$$\begin{aligned} u_{k}\left( d(v'_{\ell }, v_{-{\ell }}), t_k(v'_{\ell }, v_{-{\ell }}); v_{k} \right)&=\min _{y \in E( v'_{\ell }+\sum _{j \ne k, \ell } v_{j})} v_{k}(y) \ge \min _{y \in E(\sum _{j \ne k} v_{j}) \cup \{z\}} v_{k}(y) \\&= \min _{y \in E(\sum _{j \ne k} v_{j}) } v_{k}(y) = u_{k}\left( d(v_{\ell }, v_{-{\ell }}), t_k(v_{\ell }, v_{-{\ell }}); v_{k} \right) . \end{aligned}$$

Next, consider the case with $k = i$. Take any $v \in \prod _{j=1}^n D_j$ with $E(\sum _{j=1}^n v_j)=\{z\}$ and any $v'_{\ell } \in D_{\ell }$ such that $v'_{\ell }(z)>v_{\ell }(z)$ and $v'_{\ell }(x)=v_{\ell }(x)$ for all $x \in X \setminus \{z\}$. Then, since $(v_{\ell }, v_{-{\ell }}) \ \begin{array}{c} \rightarrow \\ i \end{array} \ (v'_{\ell }, v_{-{\ell }})$, we have

$$\begin{aligned}&\inf _{(v'_{\ell }, v_{-{\ell }}) \ \begin{array}{c} \rightarrow \\ i \end{array} \ v''} \left\{ \max _{y \in X} \sum _{j=1}^n v''_j(y)-\sum _{j \ne i} \min _{y \in E(\sum _{h \ne j} v''_{h})} v''_j(y)\right\} \\&\quad \ge \inf _{(v_{\ell }, v_{-{\ell }}) \ \begin{array}{c} \rightarrow \\ i \end{array} \ v''} \left\{ \max _{y \in X} \sum _{j=1}^n v''_j(y)-\sum _{j \ne i} \min _{y \in E(\sum _{h \ne j} v''_{h})} v''_{j}(y)\right\} . \end{aligned}$$

Therefore, $u_{i}\left( d(v'_{\ell }, v_{-{\ell }}), t_i(v'_{\ell }, v_{-{\ell }}); v_{i} \right) \ge u_{i}\left( d(v_{\ell }, v_{-{\ell }}), t_i(v_{\ell }, v_{-{\ell }}); v_{i} \right)$.

(ii) $\varvec{\Rightarrow }$ (i). Suppose that a domain $\prod _{j=1}^n D_j$ satisfies Property 1. Take any mechanism $(d, t_1, \ldots , t_n) \in {\mathscr {M}}(\prod _{j=1}^n D_j)$ that satisfies efficiency, feasibility, no free ride, no disposal of utility, and distribution. Let $(d^*,t^*_1, \ldots , t^*_n) \in {\mathscr {M}}(\prod _{j=1}^n D_j)$ be a pivotal mechanism. Let us show that for each $i \in N$ and each $v \in \prod _{j=1}^n D_j, u_i\left( d^*(v), t^*_i(v); v_i\right) \ge u_i\left( d(v), t_i(v); v_i\right)$. Take any $i \in N$, any $v \in \prod _{j=1}^n D_j$, and any $\varepsilon >0$. Since $\prod _{j=1}^n D_j$ satisfies Property 1, there exists $v' \in \prod _{j=1}^n D_j$ with $v \ \begin{array}{c} \rightarrow \\ i \end{array} \ v'$ such that

$$\begin{aligned} \left\{ \max _{y \in X} \sum _{j=1}^n v'_j(y)-\sum _{j \ne i} \min _{y \in E(\sum _{k \ne j} v'_k)} v'_j(y)\right\} - \left\{ u_i\left( d^*(v), t^*_i(v); v_i\right) \right\} <\varepsilon . \end{aligned}$$

(14)

By the same argument as Eq. (4) in Sect. 3, it follows that

$$\begin{aligned} u_i\left( d(v'), t_i(v'); v'_i\right) \le \max _{y \in X} \sum _{j=1}^n v'_j(y)-\sum _{j \ne i} \min _{y \in E(\sum _{k \ne j} v'_k)} v'_j(y). \end{aligned}$$

(15)

Combining Eqs. (14) and (15), we have

$$\begin{aligned} u_i\left( d(v'), t_i(v'); v'_i\right) \le \max _{y \in X} \sum _{j=1}^n v'_j(y)-\sum _{j \ne i} \min _{y \in E(\sum _{k \ne j} v'_k)} v'_j(y)< u_i\left( d^*(v), t^*_i(v); v_i\right) +\varepsilon . \end{aligned}$$

In addition, by $v \ \begin{array}{c} \rightarrow \\ i \end{array} \ v'$ and Lemma 1,

$$\begin{aligned} u_i\left( d(v), t_i(v); v_i\right)\le & {} u_i\left( d(v'), t_i(v'); v'_i\right) \\< & {} u_i\left( d^*(v), t^*_i(v); v_i\right) +\varepsilon . \end{aligned}$$

Since $\varepsilon$ was chosen arbitrarily, $u_i\left( d(v), t_i(v); v_i\right) \le u_i\left( d^*(v), t^*_i(v); v_i\right)$. $\square$

1.2 B. Proof of Lemma 2

Suppose that a domain $\prod _{j=1}^n D_j$ satisfies Property 1*. Take any $i \in N$, any $v \in \prod _{j=1}^n D_j$, and any $\varepsilon >0$. Then, there exists $v' \in \prod _{j=1}^n D_j$ with $v \ \, \begin{array}{c} \rightarrow \\ i \end{array} \ \, v'$ such that for some $z \in E(\sum _{j \ne i} v_j)$ the following two conditions are satisfied:

• (i) $\left\{ v'_i(z)+ \sum _{j \ne i} v_j(z)\right\} -\max _{y \in X} \sum _{j=1}^n v_j(y) < \varepsilon$,

• (ii) $E(\sum _{k \ne j} v'_k)=\{z\}$ for all $j \in N$.

Note that by Condition (ii), $E(\sum _{j =1}^n v'_j)=\{z\}$. Therefore, by Conditions (i) and (ii),

$$\begin{aligned}&\max _{y \in X} \sum _{j=1}^n v'_j(y)-\sum _{j \ne i} \min _{y \in E(\sum _{k \ne j} v'_k)} v'_j(y) =\sum _{j=1}^n v'_j(z)-\sum _{j \ne i} v'_j(z)\\&\quad =v'_i(z) <\max _{y \in X} \sum _{j=1}^n v_j(y)- \sum _{j \ne i} v_j(z)+\varepsilon =\max _{y \in X} \sum _{j=1}^n v_j(y)-\max _{y \in X} \sum _{j \ne i} v_j(y)+\varepsilon . \end{aligned}$$

Thus,

$$\begin{aligned} \left\{ \max _{y \in X} \sum _{j=1}^n v'_j(y)-\sum _{j \ne i} \min _{y \in E(\sum _{k \ne j} v'_k)} v'_j(y)\right\} -\left\{ \max _{y \in X} \sum _{j=1}^n v_j(y)-\max _{y \in X} \sum _{j \ne i} v_j(y)\right\} <\varepsilon . \end{aligned}$$

Therefore, the domain $\prod _{j=1}^n D_j$ satisfies Property 1. $\square$

1.3 C. Proof of Lemma 3

Take any $i \in N$, any $v \in \prod _{j=1}^n D_j$, and any $\varepsilon >0$. Fix some $z \in E(\sum _{j \ne i} v_j)$. Note that $\max _{y \in X} \sum _{j=1}^n v_j(y)- \sum _{j \ne i} v_j(z) \ge v_i(z)$. Then, by Property 2*, there exists $v'_i \in D_i$ such that:

• (i) $\max _{y \in X} \sum _{j=1}^n v_j(y)- \sum _{j \ne i} v_j(z)< v'_i(z) <\max _{y \in X} \sum _{j=1}^n v_j(y)- \sum _{j \ne i} v_j(z) +\varepsilon$,

• (ii) $v'_i(y)= v_i(y)$ for all $y \in X \setminus \{z\}$.

First, Condition (i) implies that $\left\{ v'_i(z)+ \sum _{j \ne i} v_j(z)\right\} -\max _{y \in X} \sum _{j=1}^n v_j(y) < \varepsilon$. Second, Conditions (i) and (ii) together imply that for each $y \ne z$,

$$\begin{aligned} v'_i(y) +\sum _{j \ne i} v_j(y) =v_i(y) +\sum _{j \ne i} v_j(y)< v'_i(z) +\sum _{j \ne i} v_j(z), \end{aligned}$$

and hence, $E(v'_i+\sum _{j \ne i} v_j)=\{z\}$.

On the other hand, again by Property 2*, there exists $v'_{-i} \in \prod _{j \ne i} D_j$ such that for each $j \ne i$,

$$\begin{aligned} v'_j(y)= {\left\{ \begin{array}{ll} v_j(z) + \delta \ &{}\text { if }y=z,\\ v_j(y) \ &{}\text {otherwise}, \end{array}\right. } \end{aligned}$$

where $\delta >0$ is a sufficiently large number such that $E(\sum _{k \ne j'} v'_k)=\{z\}$ for all $j' \in N$. Then, obviously $v \ \, \begin{array}{c} \rightarrow \\ i \end{array} \ \, v'$ holds. Therefore, the domain $\prod _{j=1}^n D_j$ satisfies Property 1*. This and Lemma 2 in turn imply that $\prod _{j=1}^n D_j$ satisfies Property 1. $\square$