Binary mechanism for the allocation problem with single-dipped preferences

Abstract

In this study, we consider the problem of fairly allocating a fixed amount of a perfectly divisible resource among agents with single-dipped preferences. It is known that any efficient and strategy-proof rule violates several fairness requirements. We alternatively propose a simple and natural mechanism, in which each agent announces only whether he or she demands a resource and the resource is divided equally among the agents who demand it. We show that any Nash equilibrium allocation of our mechanism belongs to the equal-division core. In addition, we show that our mechanism is Cournot stable. In other words, from any message profile, any path of better-replies converges to a Nash equilibrium.

Appendix

1.1 Proof of Theorem 1

Claim 1

For each $R\in {\mathcal {R}}^{N},$ and each $S\subseteq N$ , if $m^{S}\in NE(\Gamma _{B},R)$ , then $S\subseteq N_{\left| S\right| }^{^{\prime }}(R).$

Proof

We distinguish three cases.

Case 1: $\left| S\right| =0.$ Since $S=\phi$, $S=\phi \subseteq N_{\left| S\right| }^{^{\prime }}(R)$.

Case 2: $\left| S\right| =1.$ Suppose there is $i\in S$ such that $i\notin N_{1}^{^{\prime }}(R).$ Since $i\notin N_{1}^{^{\prime }}(R)=\left\{ i\in N\text { }|\text { }1\,R_{i}\,\frac{1}{n}\right\}$,

$$\begin{aligned} g_{i}\left( 0,m_{-i}^{S}\right) =\frac{1}{n}\,P_{i}\,1=g_{i}\left( m^{S}\right) . \end{aligned}$$

Hence, $m^{S}\notin NE(\Gamma ,R).$

Case 3: $\left| S\right| =k\ge 2.$ Suppose there is $i\in S$ such that $i\notin N_{k}^{^{\prime }}(R).$ Since $i\notin N_{k}^{^{\prime }}(R)=\left\{ i\in N\text { }|\text { }\frac{1}{k} \,R_{i}\,0\right\}$,

$$\begin{aligned} g_{i}\left( 0,m_{-i}^{S}\right) =0\,P_{i}\,\frac{1}{k}=g_{i}\left( m^{S}\right) . \end{aligned}$$

Hence, $m^{S}\notin NE(\Gamma _{B},R).\blacksquare$

Claim 2

For each $R\in {\mathcal {R}}^{N},$ and each $S\subseteq N$, if $m^{S}\in NE(\Gamma _{B},R)$, then $N_{\left| S\right| +1}\subseteq S.$

Proof

We distinguish three cases.

Case 1: $\left| S\right| =0.$ Suppose there is $i\in N_{1}(R)$ such that $i\notin S$. Since $i\in N_{1}(R)=\left\{ i\in N\text { }| \text { }1\,P_{i}\,\frac{1}{n}\right\}$

$$\begin{aligned} g_{i}\left( 1,m_{-i}^{S}\right) =1\,P_{i}\,\frac{1}{n}=g_{i}\left( m^{S}\right) . \end{aligned}$$

Hence, $m^{S}\notin NE(\Gamma _{B},R).$

Case 2: $\left| S\right| =k\in \left\{ 1,\ldots ,n-1\right\} .$ Suppose there is $i\in N_{k+1}(R)$ such that $i\notin S$. Then, since $i\in N_{k+1}(R)=\left\{ i\in N\text { }|\text { }\frac{1}{k+1} \,P_{i}\,0\right\}$

$$\begin{aligned} g_{i}\left( 1,m_{-i}^{S}\right) =\frac{1}{k+1}\,P_{i}\,0=g_{i}\left( m^{S}\right) . \end{aligned}$$

Hence, $m^{S}\notin NE(\Gamma _{B},R).$

Case 3: $\left| S\right| =n$. Since $N_{n+1}(R)=\phi$, $N_{n+1}=\phi \subseteq S$.■

Claim 3

For each $R\in {\mathcal {R}}^{N},$ and each $S\subseteq N$, if $N_{\left| S\right| +1}(R)\subseteq S\subseteq N_{\left| S\right| }^{^{\prime }}(R)$, then $m^{S}\in NE(\Gamma _{B},R).$

Proof

We distinguish four cases. Case 1: $\left| S\right| =0.$ For each $j\in N\backslash S=N$, since $j\notin N_{1}(R)=\left\{ i\in N\text { }|\text { }1\,P_{i}\, \frac{1}{n}\right\} ,$

$$\begin{aligned} g_{j}\left( m^{S}\right) =\frac{1}{n}\,R_{j}\,1=g_{j}\left( 1,m_{-j}^{S}\right) . \end{aligned}$$

Hence, $m^{S}\in NE(\Gamma _{B},R).$

Case 2: $\left| S\right| =1.$ For each $i\in S,$ since $\ i\in N_{1}^{^{\prime }}(R)=\left\{ i\in N\text { }|\text { }1\,R_{i}\,\frac{1}{ n}\right\} ,$

$$\begin{aligned} g_{i}\left( m^{S}\right) =1\,R_{i}\,\frac{1}{n}=g_{i}\left( 0,m_{-i}^{S}\right) . \end{aligned}$$

For each $j\in N\backslash S$, since $j\notin N_{2}(R)=\left\{ i\in N\text { } |\text { }\frac{1}{2}\,P_{i}\,0\right\} ,$

$$\begin{aligned} g_{j}\left( m^{S}\right) =0\,R_{j}\,\frac{1}{2}=g_{j}\left( 1,m_{-j}^{S}\right) . \end{aligned}$$

Hence, $m^{S}\in NE(\Gamma _{B},R).$

Case 3: $\left| S\right| =k\in \left\{ 2,\ldots ,n-1\right\} .$ For each $i\in S,$ since $\ i\in N_{k}^{^{\prime }}(R)=\left\{ i\in N\text { }|\text { }\frac{1}{k}\,R_{i}\,0\right\} ,$

$$\begin{aligned} g_{i}\left( m^{S}\right) =\frac{1}{k}\,R_{i}\,0=g_{i}\left( 0,m_{-i}^{S}\right) . \end{aligned}$$

For each $j\in N\backslash S$, since $j\notin N_{k+1}(R)=\left\{ i\in N\text { }|\text { }\frac{1}{k+1}\,P_{i}\,0\right\} ,$

$$\begin{aligned} g_{j}\left( m^{S}\right) =0\,R_{j}\,\frac{1}{k+1}=g_{j}\left( 1,m_{-j}^{S}\right) . \end{aligned}$$

Hence, $m^{S}\in NE(\Gamma _{B},R).$

Case 4: $\left| S\right| =n.$ For each $i\in S=N,$ since $\ i\in N_{n}^{^{\prime }}(R)=\left\{ i\in N\text { }|\text { }\frac{1}{n} \,R_{i}\,0\right\} ,$

$$\begin{aligned} g_{i}\left( m^{S}\right) =\frac{1}{n}\,R_{i}\,0=g_{i}\left( 0,m_{-i}^{S}\right) . \end{aligned}$$

Hence, $m^{S}\in NE(\Gamma _{B},R).$■

1.2 Proof of Theorem 2

Claim 4

Suppose $\left| N\right| \ge 3$ . Then, for each $R\in {\mathcal {R}}^{N},$ and each $S\subseteq N,$ if $m^{S}\in NE(\Gamma _{B},R)$ , then $x^{S}\in WP(R).$

Proof

We distinguish three cases.

Case 1: $\left| S\right| =0.$ Suppose that $m^{S}\in NE(\Gamma _{B},R).$ By Theorem 1, since for each $i\in N\backslash S=N,$ $i\notin N_{1}(R)=\left\{ i\in N\text { }|\text { }1\,P_{i}\,\frac{1}{n} \right\} ,$ we obtain $x_{i}^{S}=\frac{1}{n}\,R_{i}\,1$. Hence, by single-dippedness of $R_{i}$, for each $i\in N,$ if $y_{i}\,P_{i}\,x_{i}^{S}$ then $y_{i}<\frac{1}{n}$. Suppose there is $y\in A$ such that for each $i\in N,$ $y_{i}\,P_{i}\,x_{i}^{S}$. Then, since for each $i\in N,\ y_{i}<\frac{1}{ n},$

$$\begin{aligned} \sum \limits _{i\in N}y_{i}<\frac{\left| N\right| }{n}=1, \end{aligned}$$

which contradicts $y\in A=\left\{ (x_{1},...,x_{n})\in {\mathbb {R}} _{+}^{n}\text { }|\text { }\sum \nolimits _{i\in N}x_{i}=1\right\} .$ Therefore, $x^{S}\in WP(R).$

Case 2: $\left| S\right| =1.$ Let $S=\left\{ j\right\} .$ Suppose that $m^{S}\in NE(\Gamma _{B},R).$ By Theorem 1, since for each $i\in N\backslash \left\{ j\right\} ,$ $i\notin N_{2}(R)=\left\{ i\in N\text { }|\text { }\frac{1}{2}\,P_{i}\,0\right\} ,$ $x_{i}^{S}=0\,R_{i}\,\frac{1}{2}$ . Hence, by single-dippedness of $R_{i}$, for each $i\in N\backslash \left\{ j\right\} ,$ if $y_{i}\,P_{i}\,x_{i}^{S}$ then $y_{i}>\frac{1}{2}$. Suppose that there is $y\in A$ such that for each $i\in N,$ $y_{i}\,P_{i}\,x_{i}^{S}$ . Then, since for each $i\in N\backslash \left\{ j\right\} ,\ y_{i}>\frac{1}{ 2},$

$$\begin{aligned} \sum \limits _{i\in N}y_{i}\ge \sum \limits _{i\in N\backslash \left\{ j\right\} }y_{i}>\frac{\left| N\right| -1}{2}\ge 1, \end{aligned}$$

which contradicts $y\in A.$ Therefore, $x^{S}\in WP(R).$

Case 3: $\left| S\right| \ge 2.$ Let $k=\left| S\right| \ge 2.$ Suppose that $m^{S}\in NE(\Gamma _{B},R^{N})$. By Theorem 1, since for each $i\in S,$ $i\in N_{k}^{\prime }(R)=\left\{ i\in N \text { }|\text { }\frac{1}{k}\,P_{i}\,0\right\} ,$ $x_{i}^{S}=\frac{1}{k} \,R_{i}\,0$. Hence, by single-dippedness of $R_{i}$, for each $i\in S$, if $y_{i}\,P_{i}\,x_{i}^{S}$, then, $y_{i}>\frac{1}{k}$. Suppose there is $y\in A$ such that for each $i\in N,$ $y_{i}\,P_{i}\,x_{i}^{S}$. Then, since for each $i\in S,\ y_{i}>\frac{1}{k},$

$$\begin{aligned} \sum \limits _{i\in N}y_{i}\ge \sum \limits _{i\in S}y_{i}>\frac{\left| S\right| }{k}=1, \end{aligned}$$

which contradicts $y\in A.$ Therefore, $x^{S}\in WP(R).$■

Claim 5

Suppose $\left| N\right| \ge 4$ . Then, for each $R\in {\mathcal {R}}^{N}$ , and each $S\subseteq N,$ if $m^{S}\in NE(\Gamma _{B},R)$ , then $x^{S}\in EC(R).$

Proof

We distinguish four cases.

Case 1: $\left| S\right| =0.$ Suppose that $m^{S}\in NE(\Gamma _{B},R)$ and that there are $T\subseteq N$ and $y\in A$ such that for each $i\in T,$ $y_{i}\,P_{i}\,x_{i}^{S}$ and $\ \sum \nolimits _{i\in T}y_{i}=\frac{\left| T\right| }{n}$. We know from the proof of Claim 4 that for each $i\in N$, if $\ y_{i}\,P_{i}\,x_{i}^{S}$, then $y_{i}< \frac{1}{n}$. Hence,

$$\begin{aligned} \sum \limits _{i\in T}y_{i}<\frac{\left| T\right| }{n}, \end{aligned}$$

which is a contradiction. Therefore, $x^{S}\in EC(R).$

Case 2: $\left| S\right| =1.$ Suppose that $m^{S}\in NE(\Gamma _{B},R)$ and that there are $T\subseteq N$ and $y\in A$ such that for each $i\in T,$ $y_{i}\,P_{i}\,x_{i}^{S}$ and $\ \sum \nolimits _{i\in T}y_{i}=\frac{\left| T\right| }{n}$. If $T=\left\{ j\right\} ,$ then since $j\in \left\{ i\in N\text { }|\text { }1\,R_{i}\,\frac{1}{n}\right\} ,$ we have $\ x_{i}^{S}=1\,R_{i}\,\frac{1}{n}=\frac{\left| T\right| }{n}$ , which is a contradiction. Hence, $T\backslash \left\{ j\right\} \ne \phi$ . We know from the proof of Claim 4 that for each $i\in N\backslash \left\{ j\right\}$, if $\ y_{i}\,P_{i}\,x_{i}^{S}$ then $y_{i}>\frac{1}{2}$. Hence,

$$\begin{aligned} \frac{\left| T\right| }{n}=\sum \limits _{i\in T}y_{i}\ge \sum \limits _{i\in T\backslash \left\{ j\right\} }y_{i}>\frac{\left| T\backslash \left\{ j\right\} \right| }{2}\ge \frac{1}{2}. \end{aligned}$$

Since $n\ge 4$ and $\frac{\left| T\right| }{n}>\frac{1}{2}$, we have $\left| T\right| \ge 3$. Hence, $\left| T\backslash \left\{ j\right\} \right| \ge 2$, so that

$$\begin{aligned} \sum \limits _{i\in T}y_{i}\ge \sum \limits _{i\in T\backslash \left\{ j\right\} }y_{i}>\frac{\left| T\backslash \left\{ j\right\} \right| }{2}\ge 1\ge \frac{\left| T\right| }{n}, \end{aligned}$$

which is a contradiction. Therefore, $x^{S}\in EC(R).$

Case 3: $\left| S\right| \in \left\{ 2,\ldots ,n-1\right\} .$ Let $k=\left| S\right| \ge 2.$ Suppose that $m^{S}\in NE(\Gamma _{B},R^{N})$ and that there are $T\subseteq N$ and $y\in A$ such that for each $i\in T,$ $y_{i}\,P_{i}\,x_{i}^{S}$ and$\ \sum \nolimits _{i\in T}y_{i}= \frac{\left| T\right| }{n}$. We know from the proof of Claim 4 that for each $i\in S,$ if $y_{i}\,P_{i}\,x_{i}^{S}$, then $y_{i}>\frac{1}{k}$. By Theorem 1, since for each $i\in N\backslash S,$ $i\notin N_{k+1}(R)=\left\{ i\in N\text { }|\text { }\frac{1}{k+1}\,P_{i}\,0\right\} ,$ we obtain $x_{i}^{S}=0\,R_{i}\,\frac{1}{k+1}$. By single-dippedness of $R_{i}$, for each $i\in N\backslash S,$ if $y_{i}\,P_{i}\,x_{i}^{S}$, then $y_{i}>\frac{1}{k+1}$. Hence,

$$\begin{aligned} \sum \limits _{i\in T}y_{i}=\sum \limits _{i\in T\cap S}y_{i}+\sum \limits _{i\in T\backslash S}y_{i}>\frac{\left| T\cap S\right| }{k}+\frac{\left| T\backslash S\right| }{k+1}\ge \frac{ \left| T\cap S\right| }{n}+\frac{\left| T\backslash S\right| }{n}=\frac{\left| T\right| }{n}, \end{aligned}$$

which is a contradiction. Therefore, $x^{S}\in EC(R).$

Case 4: $\left| S\right| =n.$ Let $m^{S}\in NE(\Gamma _{B},R).$ Suppose there are $T\subseteq N$ and $y\in A$ such that for each $i\in T,$ $y_{i}\,P_{i}\,x_{i}^{S}$ and $\ \sum \nolimits _{i\in T}y_{i}=\frac{ \left| T\right| }{n}$. We know from the proof of Claim 4 that for each $i\in N$, if $\ y_{i}\,P_{i}\,x_{i}^{S}$, then $y_{i}>\frac{1}{n}$. Hence,

$$\begin{aligned} \sum \limits _{i\in T}y_{i}>\frac{\left| T\right| }{n}, \end{aligned}$$

which is a contradiction. Therefore, $x^{S}\in EC(R).$■

Claim 6

For each $R\in {\mathcal {R}}^{N}$ , and each $S\subseteq N,$ if $m^{S}\in NE(\Gamma _{B},R)$ , then $x^{S}\in ELB(R).$

Proof

Obvious.■

Claim 7

$F_{B}$ satisfies anonymity.

Proof

Obvious.■

1.3 Proof of Theorem 3

Let us recall some notation. For each $R_{i}\in {\mathcal {R}}$, define $\delta (R_{i})\in \left[ 0,1\right]$ as follows. If $1\,R_{i}\,0,$ then let $\ \delta (R_{i})\in \left[ 0,\frac{1}{2}\right]$ be such that $2\delta (R_{i})\,I_{i}\,0$. If $0\,P_{i}\,1$, then let $\delta (R_{i})\in \left( \frac{1}{2},1\right]$ be such that $(2\delta (R_{i})-1)\,I_{i}\,1$. For each $R\in {\mathcal {R}}^{N},$ let $P:\prod \limits _{i\in N}\left\{ 0,1\right\} \rightarrow {\mathbb {R}}$ be such that for each $m\in \prod \limits _{i\in N}\left\{ 0,1\right\}$,

$$\begin{aligned} P(m)=\left\{ \begin{array}{cc} \sum \limits _{i\in N}\frac{m_{i}^{2}}{\delta (R_{i})}-2\sum \limits _{i\in N}m_{i}-2\sum \limits _{i\in N}\sum \limits _{j\in N\backslash \left\{ i\right\} }m_{i}m_{j} &{} \text {if there is }i\in N,\text { }m_{i}=1 \\ \frac{2n}{1+n}-2 &{} \text {otherwise} \end{array} \right. \end{aligned}$$

We show that for each $R\in {\mathcal {R}}^{N},$ P is a potential function of $G(R,\Gamma _{B}).$

We distinguish two cases.

Case 1 There is $j\in N\backslash \left\{ i\right\}$ such that $m_{j}=1.$

In this case, $g_{i}(0,m_{-i})=0$ and $g_{i}(1,m_{-i})=\frac{1}{ 1+\sum \nolimits _{j\in N\backslash \left\{ i\right\} }m_{j}}.$ By the definition of $\delta (R_{i}),$ $g_{i}(1,x_{-i})\,R_{i}\,g_{i}(0,x_{-i})$ if and only if $2\delta (R_{i})\le \frac{1}{1+\sum \nolimits _{j\in N\backslash \left\{ i\right\} }m_{j}}.$

The potential function can be written as

$$\begin{aligned} P(m_{i},m_{-i})= & {} \frac{m_{i}^{2}}{\delta (R_{i})}-2m_{i}\left( 1+\sum \limits _{j \in N\backslash \left\{ i\right\} }m_{j}\right) +Q_{1}\left( m_{-i}\right) \\= & {} \frac{1}{\delta (R_{i})}\left( m_{i}-\delta (R_{i})\left( 1+\sum \limits _{j\in N\backslash \left\{ i\right\} }m_{j}\right) \right) ^{2}+Q_{2}\left( m_{-i}\right) \end{aligned}$$

where $Q_{1}$ and $Q_{2}$ represent the collective terms that do not depend on $m_{i}$. We know from this equation that $P(1,m_{-i})\ge P(0,m_{-i})$ if and only if

$$\begin{aligned} \delta (R_{i})\left( 1+\sum \limits _{j\in N\backslash \left\{ i\right\} }m_{j}\right) \le \frac{1}{2}. \end{aligned}$$

This inequality can be transformed into

$$\begin{aligned} 2\delta (R_{i})\le \frac{1}{1+\sum \nolimits _{j\in N\backslash \left\{ i\right\} }m_{j}}. \end{aligned}$$

Therefore, $g_{i}(1,m_{-i})\,R_{i}\,\,g_{i}(0,m_{-i})$ if and only if $P(1,m_{-i})\ge P(0,m_{-i}).$

Case 2 For each $j\in N\backslash \left\{ i\right\}$, $m_{j}=0.$

In this case, $g_{i}(0,m_{-i})=\frac{1}{n}$ and $g_{i}(1,m_{-i})=1.$ By the definition of $\delta (R_{i}),$ $g_{i}(1,m_{-i})\,R_{i}\,\,g_{i}(0,m_{-i})$ if and only if $2\delta (R_{i})-1\le \frac{1}{n}.$

Since $\ P(0,m_{-i})=\frac{2n}{1+n}-2$ and $P(1,m_{-i})=\frac{1}{\delta (R_{i})}-2$, $P(1,m_{-i})\ge P(0,m_{-i})$ if and only if

$$\begin{aligned} \frac{1}{\delta (R_{i})}-2\ge \frac{2n}{1+n}-2. \end{aligned}$$

This inequality can be transformed into

$$\begin{aligned} 2\delta (R_{i})-1\le \frac{1}{n}. \end{aligned}$$

Therefore, $g_{i}(1,m_{-i})\,R_{i}\,\,g_{i}(0,m_{-i})$ if and only if $P(1,m_{-i})\ge P(0,m_{-i}).\blacksquare$■

1.4 Logical relationships among axioms

Here we investigate several relationships among axioms and check whether $F_{B}$ satisfies several well-known axioms.

Example 1

As mentioned in Remark 1, strong Pareto efficiency is incompatible with Maskin monotonicity. To confirm this, suppose $\left| N\right| \ge 2$ and let $R\in {\mathcal {R}}^{N}$ be such that for each $i\in N,$ $d(R_{i})=\frac{1}{2}$ and $1\,P_{i}\,0.$ We can easily check that

$$\begin{aligned} SP(R)=\left\{ x^{\left\{ i\right\} }\in A\text { }|i\in N\right\} =\left\{ x\in A\text { }|\exists i\in N,\text { }x_{i}=1\right\} . \end{aligned}$$

Suppose there is a solution $F:{\mathcal {R}}\longrightarrow 2^{A}\setminus \left\{ \emptyset \right\}$ satisfying both strong Pareto efficiency and Maskin monotonicity. Then, there is $i\in N$ such that $\ x^{\left\{ i\right\} }\in F(R).$ Let ${\overline{R}}\in {\mathcal {R}}^{N}$ be such that (1) for $i\in N,\ d({\overline{R}}_{i})=\frac{1}{2}$ and $1\,{\overline{I}} _{i}\,0$, and (2) for each $j\in N\backslash \left\{ i\right\} ,$ $\overline{ R}_{j}=R_{j}.$ Since for each $j\in N\backslash \left\{ i\right\} ,$ $L( {\overline{R}}_{j},0)=L(R_{j},0),$ and for $i\in N,$

$$\begin{aligned} L({\overline{R}}_{i},1)= & {} \left[ 0,1\right] \\\supseteq & {} L(R_{i},1), \end{aligned}$$

by Maskin monotonicity of F, we have $x^{\left\{ i\right\} }\in F( {\overline{R}}).$ However, since $SP({\overline{R}})=\left\{ x^{\left\{ j\right\} }\in A\text { }|j\in N\backslash \left\{ i\right\} \right\}$, $x^{\left\{ i\right\} }\notin SP({\overline{R}})$. Therefore, $F({\overline{R}} )\nsubseteq SP({\overline{R}}).$

Example 2

As mentioned in Remark 2, envy-freeness is incompatible with weak Pareto efficiency. To confirm this, let $R\in {\mathcal {R}}^{N}$ be such that for each $i\in N,$ $d(R_{i})=\frac{1}{2}$ and $1\,P_{i}\,0.$ We first show that if $x\in WP(R),$ then there is $i\in N$ such that $x_{i}=0.$ If for each $i\in N,$ $x_{i}>0$, then there is $j\in N$ such that $1>x_{j}>0$ and for each $i\ne j,$ $0<x_{i}\le \frac{1}{2}.$ By single-dippedness of $R_{i},$ for $j\in N,\ 1\,P_{j}\,x_{j}$ and for each $i\ne j,$ $0\,P_{i}\,x_{i}$. Hence, $x\notin WP(R).$ Therefore,

$$\begin{aligned} WP(R)\subseteq \left\{ x\in A\text { }|\exists i\in N,\text { }x_{i}=0\right\} . \end{aligned}$$

Let $x^{\prime }\in \left\{ x\in A\text { }|\exists i\in N,\text { } x_{i}=0\right\}$ be such that there are distinct $i,j,k\in N$ such that $x_{i}^{\prime }=0$, $x_{j}^{\prime }>0$, and $x_{k}^{\prime }>0$. If $x^{\prime }\in EF(R),$ then $x_{j}^{\prime }\,R_{j}\,x_{i}^{\prime }=0$ and $x_{k}^{\prime }\,R_{k}\,x_{i}^{\prime }=0$. By single-dippedness of $R_{j}$ and $R_{k}$, we have $x_{j}^{\prime }>\frac{1}{2}$ and $x_{k}^{\prime }> \frac{1}{2}.$ Hence,

$$\begin{aligned} \sum \limits _{i\in N}x_{i}^{\prime }\ge x_{j}^{\prime }+x_{k}^{\prime }> \frac{1}{2}+\frac{1}{2}=1, \end{aligned}$$

which contradicts $x^{\prime }\in A.$ Therefore, whenever $x\in WP(R)\cap EF(R)$, there is $i\in N$ such that $x_{i}=1,$ so that

$$\begin{aligned} WP(R)\cap EF(R)\subseteq \left\{ x^{\left\{ i\right\} }\in A\text { }|i\in N\right\} . \end{aligned}$$

For each $x^{\left\{ i\right\} }\in A,$ since for each $j\ne i,\ 1=x_{i}\,P_{j}\,x_{j}=0$, we have $x^{\left\{ i\right\} }\notin EF(R).$ Therefore,

$$\begin{aligned} WP(R)\cap EF(R)=\phi . \end{aligned}$$

Remark 7

A solution F satisfies welfare-domination under preference replacement if for each $R\in {\mathcal {R}}^{N},$ each $x\in F(R),$ each $i\in N,$ each $R_{i}^{\prime }\in {\mathcal {R}}$, and each $x^{\prime }\in F(R_{i}^{\prime },R_{-i}),$ either $x_{j}\,R_{j}\,x_{j}^{ \prime }$, for each $j\in N\backslash \left\{ i\right\}$ or $x_{j}^{\prime }\,R_{j}\,x_{j}$, for each $j\in N\backslash \left\{ i\right\}$ (Thomson 1993; Klaus et al. 1997). Note that if F satisfies weak Pareto efficiency, anonymity, and Maskin monotonicity, then $\ F$ does not satisfy welfare-domination under preference replacement. Hence, $F_{B}$ also violates welfare-domination under preference replacement.

To confirm this, suppose $\left| N\right| \ge 3$ and F satisfies weak Pareto efficiency, anonymity, and Maskin monotonicity. Let $R\in {\mathcal {R}}^{N}$ be such that for each $i\in N,$ $d(R_{i})=\frac{1}{2}$ and $1\,P_{i}\,0\,I_{i}\,\frac{9}{10}.$ As seen in Example 2, since

$$\begin{aligned} WP(R)\subseteq \left\{ x\in A\text { }|\exists i\in N,\text { }x_{i}=0\right\} , \end{aligned}$$

there is $i\in N$ such that $x_{i}=0.$ Let $x\in \left\{ x\in A\text { } |\exists i\in N,\text { }x_{i}=0\right\}$ be such that $x_{i}=0$ and for each $j\in N\backslash \left\{ i\right\} ,$ $x_{j}\in \left( 0,\frac{9}{10} \right) .$ Then, for $x^{\left\{ i\right\} }\in A,$ we have $x_{i}^{\left\{ i\right\} }=1\,P_{i}\,0=x_{i}$ and for each $j\in N\backslash \left\{ i\right\} ,$ $x_{j}^{\left\{ i\right\} }=0\,P_{j}\ x_{j},$ so that $x\notin WP(R).$ Hence, for each $x\in F(R)\subseteq WP(R),$ either (1) there are distinct $i,j\in N$ such that $x_{i}=0$ and $\ x_{j}>\frac{9}{10},$ (2) there are distinct $i,j,k\in N$ such that $x_{i}=0,$ $x_{j}=\frac{9}{10},$ and $x_{k}\in \left( 0,\frac{1}{10}\right]$, or (3) there are distinct $i,j,k\in N$ such that $x_{i}=x_{j}=0$ and $x_{k}\in \left( 0,\frac{9}{10} \right)$ holds. Let $x\in F(R).$ We distinguish three cases.

Case 1: There are distinct $i,j\in N$ such that $x_{i}=0,$ and $\ x_{j}>\frac{9}{10}.$ Let $x^{\prime }\in WP(R)$ be such that $x_{i}^{\prime }=x_{j},$ $x_{j}^{\prime }=x_{i},$ and for each $k\in N\backslash \left\{ i,j\right\} ,$ $x_{k}^{\prime }=x_{k}.$ Since F satisfies anonymity, $x^{\prime }\in F(R).$ For $k\in N\backslash \left\{ i,j\right\} ,$ let $R_{k}^{\prime }\in {\mathcal {R}}$ be such that $\frac{9}{10}\,P_{k}^{\prime }\,0$ and $L(R_{k},x_{k})=L(R_{k}^{\prime },x_{k}).$ Then, by Maskin monotonicity of F, $x,x^{\prime }\in F(R_{k}^{\prime },R_{-k}).$ For $x\in F(R)$ and $x^{\prime }\in F(R_{k}^{\prime },R_{-k}),$ we must have $x_{i}^{\prime }\,P_{i}\,x_{i},$ and $x_{j}\,P_{j}\,x_{j}^{\prime }.$ Therefore, F does not satisfy welfare-domination under preference replacement.

Case 2: There are distinct $i,j,k\in N$ such that $x_{i}=0,$ $x_{j}= \frac{9}{10},$ and $x_{k}\in \left( 0,\frac{1}{10}\right] .$ Let $x^{\prime }\in WP(R)$ be such that $x_{j}^{\prime }=x_{k},$ $x_{k}^{\prime }=x_{j},$ and for each $i\in N\backslash \left\{ j,k\right\} ,$ $x_{i}^{\prime }=x_{i}.$ Since F satisfies anonymity, $x^{\prime }\in F(R).$ Let $R_{i}^{\prime }\in {\mathcal {R}}$ be such that $d(R_{i}^{\prime })=1.$ Since $L(R_{i},0)\subseteq L(R_{i}^{\prime },0)=\left[ 0,1\right] ,$ by Maskin monotonicity of F, $x,x^{\prime }\in F(R_{i}^{\prime },R_{-i}).$ For $x\in F(R)$ and $x^{\prime }\in F(R_{i}^{\prime },R_{-i}),$ we must have $x_{j}\,P_{j}\,x_{j}^{\prime },$ and $x_{k}^{\prime }\,P_{k}\,x_{k}.$ Therefore, F does not satisfy welfare-domination under preference replacement.

Case 3: There are distinct $i,j,k\in N$ such that $x_{i}=x_{j}=0$ and $x_{k}\in \left( 0,\frac{9}{10}\right)$. Let $x^{\prime }\in WP(R)$ be such that $x_{j}^{\prime }=x_{k},$ $x_{k}^{\prime }=x_{j},$ and for each $i\in N\backslash \left\{ j,k\right\} ,$ $x_{i}^{\prime }=x_{i}.$ Since F satisfies anonymity, $x^{\prime }\in F(R).$ Let $R_{i}^{\prime }\in \mathcal { R}$ be such that $d(R_{i}^{\prime })=1.$ Since $L(R_{i},0)\subseteq L(R_{i}^{\prime },0)=\left[ 0,1\right] ,$ by Maskin monotonicity of F, $x,x^{\prime }\in F(R_{i}^{\prime },R_{-i}).$ For $x\in F(R)$ and $x^{\prime }\in F(R_{i}^{\prime },R_{-i}),$ we must have $x_{j}\,P_{j}\,x_{j}^{\prime },$ and $x_{k}^{\prime }\,P_{k}\,x_{k}.$ Therefore, F does not satisfy welfare-domination under preference replacement.

Remark 8

A solution F satisfies weak non-bossiness if for each $R\in {\mathcal {R}}^{N},$ each $x\in F(R),$ each $i\in N,$ each $R_{i}^{\prime }\in {\mathcal {R}}$, and each $x^{\prime }\in F(R_{i}^{\prime },R_{-i}),$ if $x_{i}=x_{i}^{\prime }$, then $x=x^{\prime }$ (Klaus 2001; Thomson 2016). Note that if F satisfies weak Pareto efficiency, anonymity, and Maskin monotonicity, then $\ F$ does not satisfy weak non-bossiness. Hence, $F_{B}$ also violates weak non-bossiness.

To confirm this, suppose $\left| N\right| \ge 3$ and F satisfies weak Pareto efficiency, anonymity, and Maskin monotonicity. Let $R\in {\mathcal {R}}^{N}$ be such that for each $i\in N,$ $d(R_{i})=\frac{1}{2}$ and $1\,P_{i}\,0\,I_{i}\,\frac{9}{10}.$ As seen in Example 3, if $x\in F(R)\subseteq WP(R),$ then either (1) there are distinct $i,j\in N$ such that $x_{i}=0,$ and $\ x_{j}>\frac{9}{10},$ (2) there are distinct $i,j,k\in N$ such that $x_{i}=0,$ $x_{j}=\frac{9}{10},$ and $x_{k}\in \left( 0,\frac{1 }{10}\right]$, or (3) there are distinct $i,j,k\in N$ such that $x_{i}=x_{j}=0$ and $x_{k}\in \left( 0,\frac{9}{10}\right)$ holds. Let $x\in F(R).$ We distinguish three cases.

Case 1: There are distinct $i,j\in N$ such that $x_{i}=0,$ and $\ x_{j}>\frac{9}{10}.$ Let $x^{\prime }\in WP(R)$ be such that $x_{i}^{\prime }=x_{j},$ $x_{j}^{\prime }=x_{i},$ and for each $k\in N\backslash \left\{ i,j\right\} ,$ $x_{k}^{\prime }=x_{k}.$ Since F satisfies anonymity, $x^{\prime }\in F(R).$ For $k\in N\backslash \left\{ i,j\right\} ,$ let $R_{k}^{\prime }\in {\mathcal {R}}$ be such that $\frac{9}{10}\,P_{k}^{\prime }\,0$ and $L(R_{k},x_{k})=L(R_{k}^{\prime },x_{k}).$ Then, by Maskin monotonicity of F, $x,x^{\prime }\in F(R_{k}^{\prime },R_{-k}).$ For $x\in F(R)$ and $x^{\prime }\in F(R_{k}^{\prime },R_{-k}),$ we must have $x_{k}=x_{k}^{\prime },$ and $x_{i}\ne x_{i}^{\prime }.$ Therefore, F does not satisfy weak non-bossiness.

Case 2: There are distinct $i,j,k\in N$ such that $x_{i}=0,$ $x_{j}= \frac{9}{10},$ and $x_{k}\in \left( 0,\frac{1}{10}\right] .$ Let $x^{\prime }\in WP(R)$ be such that $x_{j}^{\prime }=x_{k},$ $x_{k}^{\prime }=x_{j},$ and for each $i\in N\backslash \left\{ j,k\right\} ,$ $x_{i}^{\prime }=x_{i}.$ Since F satisfies anonymity, $x^{\prime }\in F(R).$ Let $R_{i}^{\prime }\in {\mathcal {R}}$ be such that $d(R_{i}^{\prime })=1.$ Since $L(R_{i},0)\subseteq L(R_{i}^{\prime },0)=\left[ 0,1\right] ,$ by Maskin monotonicity of F, $x,x^{\prime }\in F(R_{i}^{\prime },R_{-i}).$ For $x\in F(R)$ and $x^{\prime }\in F(R_{i}^{\prime },R_{-i}),$ we must have $x_{i}=x_{i}^{\prime },$ and $x_{j}\ne x_{j}^{\prime }.$ Therefore, F does not satisfy weak non-bossiness.

Case 3: There are distinct $i,j,k\in N$ such that $x_{i}=x_{j}=0$ and $x_{k}\in \left( 0,\frac{9}{10}\right)$. Let $x^{\prime }\in WP(R)$ be such that $x_{j}^{\prime }=x_{k},$ $x_{k}^{\prime }=x_{j},$ and for each $i\in N\backslash \left\{ j,k\right\} ,$ $x_{i}^{\prime }=x_{i}.$ Since F satisfies anonymity, $x^{\prime }\in F(R).$ Let $R_{i}^{\prime }\in \mathcal { R}$ be such that $d(R_{i}^{\prime })=1.$ Since $L(R_{i},0)\subseteq L(R_{i}^{\prime },0)=\left[ 0,1\right] ,$ by Maskin monotonicity of F, $x,x^{\prime }\in F(R_{i}^{\prime },R_{-i}).$ For $x\in F(R)$ and $x^{\prime }\in F(R_{i}^{\prime },R_{-i}),$ we must have $x_{i}=x_{i}^{\prime },$ and $x_{j}\ne x_{j}^{\prime }.$ Therefore, F does not satisfy weak non-bossiness.

Finally, we extend the model so that changes in the set of agents and the amount to be allocated are allowed. We introduce some notation. Let ${\mathbb {N}}$ be the set of potential agents indexed by natural numbers. Let ${\mathcal {N}}$ be the class of non-empty and finite subsets of ${\mathbb {N}}$. Each agent $i\in {\mathbb {N}}$ has a single-dipped preference $R_{i}$ over ${\mathbb {R}} _{+}.$ Let ${\mathcal {R}}_{+}$ denote the set of single-dipped preferences over ${\mathbb {R}} _{+}.$ Let $\Omega \in {\mathbb {R}} _{+}$ denote the amount of the resource.

An economy is defined by $e\equiv \left( N,\left( R_{i}\right) _{i\in N},\Omega \right)$, in which $N\in {\mathcal {N}}$ is the set of agents, $\ R_{i}\in {\mathcal {R}}_{+}$ is i’s preference over ${\mathbb {R}} _{+}$, and $\Omega \in {\mathbb {R}} _{+}$ is the amount of the resource to be allocated among N. For each $N\in {\mathcal {N}},$ let ${\mathcal {E}}^{N}$ denote the set of economies in which the set of agents is denoted by N. For each $e\in {\mathcal {E}}^{N}$, an allocation $x\in {\mathbb {R}} _{+}^{N}$ is feasible for e if $\sum \nolimits _{i\in N}x_{i}=\Omega .$ A solution is a mapping F which associates with each economy $e\in {\mathcal {E}}^{N}$, a non-empty subset of feasible allocations for e.

For each $e\in {\mathcal {E}}^{N},$ the binary mechanism $\Gamma _{B}$ is defined such that for each $i\in N,$ $M_{i}=\left\{ 0,1\right\}$, and for each $m\in \underset{i\in N}{\prod }M_{i}$,

$$\begin{aligned} g_{i}(e,m)=\left\{ \begin{array}{cc} \frac{\Omega }{\left| \left\{ i\in N\text { }|m_{i}=1\right\} \right| } &{} \text {if }m_{i}=1 \\ 0 &{} \text {if }m_{i}=0\text { and }\left\{ i\in N\text { }|m_{i}=1\right\} \ne \emptyset \\ \frac{\Omega }{\left| N\right| } &{} \text {if }m_{i}=0\text { and } \left\{ i\in N\text { }|m_{i}=1\right\} =\emptyset . \end{array} \right. \end{aligned}$$

Let $F_{B}$ denote the solution implemented by the binary mechanism.

Remark 9

A solution F satisfies consistency if for each $N\in {\mathcal {N}}$, each $e=\left( N,\left( R_{i}\right) _{i\in N},\Omega \right) \in {\mathcal {E}}^{N}$, each $x\in F(e)$, each $N^{\prime }\subset N,$ and each $e^{\prime }=\left( N,\left( R_{i}\right) _{i\in N^{\prime }},\Omega -\sum \nolimits _{i\in N\backslash N^{\prime }}x_{i}\right) \in {\mathcal {E}}^{N^{\prime }},$ we have $\left( x_{i}\right) _{i\in N^{\prime }}\in F(e^{\prime })$ (Thomson 1994). Here we show that $F_{B}$ satisfies consistency.

Without loss of generality let $\left| N\right| \ge 3,$ $e=\left( N,\left( R_{i}\right) _{i\in N},1\right) \in {\mathcal {E}}^{N},$ and $x^{S}\in F_{B}(e).$ It suffices to show that for each $i\in N,$ and each $e^{\prime }=\left( N\backslash \left\{ i\right\} ,\left( R_{j}\right) _{i\in N\backslash \left\{ i\right\} },1-x_{i}^{S}\right) \in {\mathcal {E}} ^{N\backslash \left\{ i\right\} },$ we have $\left( x_{j}^{S}\right) _{j\in N\backslash \left\{ i\right\} }\in F_{B}(e^{\prime }).$ We distinguish six cases.

Case 1: $\left| S\right| =0.$ Since $m^{S}$ is a Nash equilibrium of $\ \Gamma _{B}$ for $e\in {\mathcal {E}}^{N}$, for each $j\in N\backslash \left\{ i\right\} ,$

$$\begin{aligned} g_{j}\left( e,0,m_{N\backslash \left\{ j\right\} }^{S}\right) =\frac{1}{\left| N\right| }\,R_{j}\,1=g_{j}\left( e,1,m_{N\backslash \left\{ j\right\} }^{S}\right) , \end{aligned}$$

so that by single-dippeness of $R_{j},$ $\frac{1}{\left| N\right| } \,R_{j}\,\frac{\left| N\right| -1}{\left| N\right| }.$ Since $x_{i}^{S}=\frac{1}{\left| N\right| },$ for each $j\in N\backslash \left\{ i\right\} ,$

$$\begin{aligned} g_{j}\left( e^{\prime },0,m_{N\backslash \left\{ i,j\right\} }^{S}\right) =\frac{1}{ \left| N\right| }\,R_{j}\,\frac{\left| N\right| -1}{ \left| N\right| }=g_{j}\left( e^{\prime },1,m_{N\backslash \left\{ i,j\right\} }^{S}\right) , \end{aligned}$$

so that $\left( m_{j}^{S}\right) _{j\in N\backslash \left\{ i\right\} }$ is a Nash equilibrium of $\ \Gamma _{B}$ for $e^{\prime }\in {\mathcal {E}} ^{N\backslash \left\{ i\right\} }$. Hence, $\left( x_{j}^{S}\right) _{j\in N\backslash \left\{ i\right\} }\in F_{B}(e^{\prime }).$

Case 2: $\left| S\right| =1$ and $i\in S.$ Since $x_{i}^{S}=1,$ the set is feasible allocations for $e^{\prime }=\left( N\backslash \left\{ i\right\} ,\left( R_{j}\right) _{i\in N\backslash \left\{ i\right\} },1-x_{i}^{S}\right)$ is a singleton. Hence, by non-emptiness of $F_{B},$ $\left( x_{j}^{S}\right) _{j\in N\backslash \left\{ i\right\} }\in F_{B}(e^{\prime }).$

Case 3: $\left| S\right| =1$ and $i\notin S.$ Let $\left| S\right| =\left\{ j\right\} .$ Since $m^{S}$ is a Nash equilibrium of $\ \Gamma _{B}$ for $e\in {\mathcal {E}}^{N}$ , for $j\in N\backslash \left\{ i\right\} ,$ we have

$$\begin{aligned} g_{j}\left( e,1,m_{N\backslash \left\{ j\right\} }^{S}\right) =1\,R_{j}\,\frac{1}{ \left| N\right| }=g_{j}\left( e,0,m_{N\backslash \left\{ j\right\} }^{S}\right) , \end{aligned}$$

so that by single-dippeness of $R_{j},$ $1\,P_{j}\,\frac{1}{\left| N\right| -1}.$ Since $x_{i}^{S}=0,$ for $j\in N\backslash \left\{ i\right\} ,$

$$\begin{aligned} g_{j}\left( e^{\prime },1,m_{N\backslash \left\{ i,j\right\} }^{S}\right) =1\,P_{j}\, \frac{1}{\left| N\right| -1}=g_{j}\left( e^{\prime },0,m_{N\backslash \left\{ i,j\right\} }^{S}\right) . \end{aligned}$$

Since $m^{S}$ is a Nash equilibrium of $\ \Gamma _{B}$ for $e\in {\mathcal {E}} ^{N}$ , for each $k\in N\backslash \left\{ i,j\right\} ,$ we have

$$\begin{aligned} g_{k}\left( e,0,m_{N\backslash \left\{ k\right\} }^{S}\right) =0\,R_{k}\,\frac{1}{2} =g_{k}\left( e,1,m_{N\backslash \left\{ k\right\} }^{S}\right) . \end{aligned}$$

Since $x_{i}^{S}=0,$ for each $k\in N\backslash \left\{ i,j\right\} ,$

$$\begin{aligned} g_{k}\left( e^{\prime },0,m_{N\backslash \left\{ i,k\right\} }^{S}\right) =0\,R_{k}\, \frac{1}{2}=g_{k}\left( e^{\prime },1,m_{N\backslash \left\{ i,k\right\} }^{S}\right) , \end{aligned}$$

so that $\left( m_{j}^{S}\right) _{j\in N\backslash \left\{ i\right\} }$ is a Nash equilibrium of $\ \Gamma _{B}$ for $e^{\prime }\in {\mathcal {E}} ^{N\backslash \left\{ i\right\} }$. Hence, $\left( x_{j}^{S}\right) _{j\in N\backslash \left\{ i\right\} }\in F_{B}(e^{\prime }).$

Case 4: $\left| S\right| =2$ and $i\in S.$ Let $\left| S\right| =\left\{ i,j\right\} .$ Since $m^{S}$ is a Nash equilibrium of $\ \Gamma _{B}$ for $e\in {\mathcal {E}}^{N}$ , for $j\in S\backslash \left\{ i\right\} ,$ we have

$$\begin{aligned} g_{j}\left( e,1,m_{N\backslash \left\{ j\right\} }^{S}\right) =\frac{1}{2} \,R_{j}\,0=g_{j}\left( e,0,m_{N\backslash \left\{ j\right\} }^{S}\right) , \end{aligned}$$

so that by single-dippeness of $R_{j},$ $\frac{1}{2}\,R_{j}\,\frac{1}{ \left| N\right| -1}.$ Since $x_{i}^{S}=\frac{1}{2},$ for $j\in S\backslash \left\{ i\right\} ,$

$$\begin{aligned} g_{j}\left( e^{\prime },1,m_{N\backslash \left\{ i,j\right\} }^{S}\right) =\frac{1}{2} \,R_{j}\,\frac{1}{\left| N\right| -1}=g_{j}\left( e^{\prime },0,m_{N\backslash \left\{ i,j\right\} }^{S}\right) . \end{aligned}$$

Since $m^{S}$ is a Nash equilibrium of $\ \Gamma _{B}$ for $e\in {\mathcal {E}} ^{N}$ , for each $k\in N\backslash \left\{ i,j\right\} ,$ we have

$$\begin{aligned} g_{k}\left( e,0,m_{N\backslash \left\{ k\right\} }^{S}\right) =0\,R_{k}\,\frac{1}{3} =g_{k}\left( e,1,m_{N\backslash \left\{ k\right\} }^{S}\right) , \end{aligned}$$

so that by single-dippedness of $R_{k},$ $0\,P_{k}\,\frac{1}{4}.$ Since $x_{i}^{S}=\frac{1}{2},$ for each $k\in N\backslash \left\{ i,j\right\} ,$

$$\begin{aligned} g_{k}\left( e^{\prime },0,m_{N\backslash \left\{ i,k\right\} }^{S}\right) =0\,P_{k}\, \frac{1}{4}=g_{k}\left( e^{\prime },1,m_{N\backslash \left\{ i,k\right\} }^{S}\right) , \end{aligned}$$

so that $\left( m_{j}^{S}\right) _{j\in N\backslash \left\{ i\right\} }$ is a Nash equilibrium of $\ \Gamma _{B}$ for $e^{\prime }\in {\mathcal {E}} ^{N\backslash \left\{ i\right\} }$. Hence, $\left( x_{j}^{S}\right) _{j\in N\backslash \left\{ i\right\} }\in F_{B}(e^{\prime }).$

Case 5: $\left| S\right| \ge 2$ and $i\notin S.$ Since $m^{S}$ is a Nash equilibrium of $\Gamma _{B}$ for $e\in {\mathcal {E}}^{N}$ , for each $j\in S,$ we have

$$\begin{aligned} g_{j}\left( e,1,m_{N\backslash \left\{ j\right\} }^{S}\right) =\frac{1}{\left| S\right| }\,R_{j}\,0=g_{j}\left( e,0,m_{N\backslash \left\{ j\right\} }^{S}\right) . \end{aligned}$$

Since $x_{i}^{S}=0,$ for each $j\in S,$

$$\begin{aligned} g_{j}\left( e^{\prime },1,m_{N\backslash \left\{ i,j\right\} }^{S}\right) =\frac{1}{ \left| S\right| }\,R_{j}\,0=g_{j}\left( e^{\prime },0,m_{N\backslash \left\{ i,j\right\} }^{S}\right) . \end{aligned}$$

Since $m^{S}$ is a Nash equilibrium of $\ \Gamma _{B}$ for $e\in {\mathcal {E}} ^{N}$ , for each $k\in N\backslash S,$ $k\ne i$, we have

$$\begin{aligned} g_{k}\left( e,0,m_{N\backslash \left\{ k\right\} }^{S}\right) =0\,R_{k}\,\frac{1}{ \left| S\right| +1}=g_{k}\left( e,1,m_{N\backslash \left\{ k\right\} }^{S}\right) , \end{aligned}$$

Since $x_{i}^{S}=0,$ for each for each $k\in N\backslash S,$ $k\ne i$, we have, 

$$\begin{aligned} g_{k}\left( e^{\prime },0,m_{N\backslash \left\{ i,k\right\} }^{S}\right) =0\,R_{k}\, \frac{1}{\left| S\right| +1}=g_{k}\left( e^{\prime },1,m_{N\backslash \left\{ i,k\right\} }^{S}\right) , \end{aligned}$$

so that $\left( m_{j}^{S}\right) _{j\in N\backslash \left\{ i\right\} }$ is a Nash equilibrium of $\Gamma _{B}$ for $e^{\prime }\in {\mathcal {E}} ^{N\backslash \left\{ i\right\} }$. Hence, $\left( x_{j}^{S}\right) _{j\in N\backslash \left\{ i\right\} }\in F_{B}(e^{\prime }).$

Case 6: $\left| S\right| \ge 3$ and $i\in S.$ Since $m^{S}$ is a Nash equilibrium of $\Gamma _{B}$ for $e\in {\mathcal {E}}^{N}$ , for each $j\in S\backslash \left\{ i\right\} ,$ we have

$$\begin{aligned} g_{j}\left( e,1,m_{N\backslash \left\{ j\right\} }^{S}\right) =\frac{1}{\left| S\right| }\,R_{j}\,0=g_{j}\left( e,0,m_{N\backslash \left\{ j\right\} }^{S}\right) , \end{aligned}$$

Since $x_{i}^{S}=0,$ for each $j\in S\backslash \left\{ i\right\} ,$

$$\begin{aligned} g_{j}\left( e^{\prime },1,m_{N\backslash \left\{ i,j\right\} }^{S}\right) =\frac{1}{ \left| S\right| }\,R_{j}\,0=g_{j}\left( e^{\prime },0,m_{N\backslash \left\{ i,j\right\} }^{S}\right) . \end{aligned}$$

Since $m^{S}$ is a Nash equilibrium of $\Gamma _{B}$ for $e\in {\mathcal {E}} ^{N}$ , for each $k\in N\backslash S,$ $k\ne i$, we have

$$\begin{aligned} g_{k}\left( e,0,m_{N\backslash \left\{ k\right\} }^{S}\right) =0\,R_{k}\,\frac{1}{ \left| S\right| +1}=g_{k}\left( e,1,m_{N\backslash \left\{ k\right\} }^{S}\right) , \end{aligned}$$

so that by single-dippedness of $R_{k},$ $0\,P_{k}\,\frac{\left| S\right| -1}{\left| S\right| }\times \frac{1}{\left| S\right| }.$¹² Since $x_{i}^{S}= \frac{1}{\left| S\right| },$ for each for each $k\in N\backslash S,$ $k\ne i$, we have, 

$$\begin{aligned} g_{k}\left( e^{\prime },0,m_{N\backslash \left\{ i,k\right\} }^{S}\right) =0\,P_{k}\, \frac{\left| S\right| -1}{\left| S\right| }\times \frac{1}{ \left| S\right| }=g_{k}\left( e^{\prime },1,m_{N\backslash \left\{ i,k\right\} }^{S}\right) , \end{aligned}$$

so that $\left( m_{j}^{S}\right) _{j\in N\backslash \left\{ i\right\} }$ is a Nash equilibrium of $\Gamma _{B}$ for $e^{\prime }\in {\mathcal {E}} ^{N\backslash \left\{ i\right\} }$. Hence, $\left( x_{j}^{S}\right) _{j\in N\backslash \left\{ i\right\} }\in F_{B}(e^{\prime }).$