Buck-passing dumping in a garbage-dumping game

Abstract

We study stable strategy profiles in a pure exchange game of bads, where each player dumps his or her bads such as garbage onto someone else. Hirai et al. (Mathematical Social Sciences 51(2):162–170, 2006) show that cycle dumping, in which each player follows an ordering and dumps his or her bads onto the next player, is a strong Nash equilibrium and that self-disposal is $\alpha$-stable for some initial distributions of bads. In this paper, we show that a strategy profile of bullying, in which all players dump their bads onto a single player, becomes $\alpha$-stable for every exchange game of bads. We also provide a necessary and sufficient condition for a strategy profile to be $\alpha$-stable in an exchange game of bads. In addition, we show that repeating an exchange after the first exchange makes self-disposal stationary.

Appendix

1.1 Proof of Proposition 2

Proposition 2. Let $b\in B^N$. A strategy profile $x\in X^N_b$ is $\alpha$-stable if and only if for any $S\subseteq N$ there is $i\in S$, such that $\sum _{j\in N\setminus S}b^j\ge r^i_x$.

Proof

Let $x\in X^N_b$ be $\alpha$-stable. By the definition of $\alpha$-stability, for any $S\subseteq N$ and any $y^S\in X^S_b$, there is $z^{N\setminus S}\in X^{N\setminus S}_b$, such that

$$\begin{aligned} v^i(y^S,z^{N\setminus S})\le v^i(x) \text { for some}{i\in S}. \end{aligned}$$

Since $v^i$ is represented by $u^i$, the inequality is equivalent to $u^i(r^i_{(y^S,z^{N\setminus S})})\le u^i(r^i_x)$. Moreover, since $u^i$ is a strictly decreasing function, this is equivalent to $r^i_{(y^S,z^{N\setminus S})}\ge r^i_x$. Hence, x is $\alpha$-stable if and only if for any $S\subseteq N$ and any $y^S\in X^S_b$, there is $z^{N\setminus S}\in X^{N\setminus S}_b$, such that

$$\begin{aligned} r^i_{(y^S,z^{N\setminus S})}\ge r^i_x \text { for some}{i\in S}. \end{aligned}$$

Below, fixing $S\subseteq N$, we show that the following two statements are equivalent:

• (i) For any $y^S\in X^S_b$, there is $z^{N\setminus S}\in X^{N\setminus S}_b$, such that $r^i_{(y^S,z^{N\setminus S})}\ge r^i_x$ for some $i\in S$.

• (ii) There is $i\in S$, such that $\sum _{j\in N\setminus S}b^j\ge r^i_x$.

We first show (ii) $\Rightarrow$ (i). Let i be the player satisfying the condition of (ii). Set ${\bar{z}}^{ji}=b^j$ for all $j\in N\setminus S$. It readily follows that $\sum _{j\in N\setminus S}{\bar{z}}^{ji}= \sum _{j\in N\setminus S}b^j$. Hence, for any $y^S\in X^S_b$,

$$\begin{aligned} r^i_{(y^S,z^{N\setminus S})}=\sum _{j\in S}y^{ji}+\sum _{j\in N\setminus S}{\bar{z}}^{ji}\ge \sum _{j\in N\setminus S}b^j \overset{\text {(ii)}}{\ge } r^i_x \end{aligned}$$

We now show (i) $\Rightarrow$ (ii). Assume that for any $i\in S$,

$$\begin{aligned} \sum _{j\in N\setminus S}b^j< r^i_x. \end{aligned}$$

(1)

Let ${\bar{y}}^S$ be a profile satisfying ${\bar{y}}^{ji}=0$ for all $j,i \in S$. It holds that for any $z^{N\setminus S}\in X^{N\setminus S}_b$ and any $i\in S$,

$$\begin{aligned} r^i_{({\bar{y}}^S,z^{N\setminus S})}=\sum _{j\in S}{\bar{y}}^{ji}+\sum _{j\in N\setminus S}z^{ji} \le \sum _{j\in N\setminus S}b^j\overset{{(1)}}{<}r^i_x. \end{aligned}$$

This contradicts (i). $\square$

1.2 Proof of Lemma 1

Lemma 1. Let $n\ge 4$. For any $b\in B^N$ and any $\sigma \in \Psi ^N$, no strategy profile of the following $2n-1$ profiles can be defined as a nonnegative convex combination of the other $2n-2$ profiles:

$$\begin{aligned} x^\sigma (b),\ {\hat{x}}^i(b) \text { for all}{i\in N\setminus \{1\}},\ x^{\sigma i}(b) \text { for all}{i\in N\setminus \{\lambda (1)\}}. \end{aligned}$$

If $n=3$, then the same holds for the following four profiles:

$$\begin{aligned} x^\sigma (b),\ {\hat{x}}^2 (b),\ {\hat{x}}^3 (b),\ x^{\sigma 1}(b). \end{aligned}$$

Proof

We prove the statement for $n\ge 4$. Let $b\in B^N$ and $\sigma \in \Psi ^N$.

We begin with $x^\sigma (b)$. We assume that there is a collection of non-negative real numbers that sum to 1, ${\hat{c}}^i$ with $i\in N\setminus \{1\}$ and $c^{\sigma i}$ with $i\in N\setminus \{\lambda (1)\}$, such that

$$\begin{aligned} x^\sigma (b) = \sum _{i\in N\setminus \{1\}} {\hat{c}}^i {\hat{x}}^i(b) + \sum _{i\in N\setminus \{\lambda (1)\}} c^{\sigma i} x^{\sigma i}(b). \end{aligned}$$

If there is $i\in N\setminus \{1\}$, such that ${\hat{c}}^{i}>0$, then for any player $j\in N\setminus \{1\}$, we have $x^\sigma (b)^{j1} \ge {\hat{c}}^{i} {\hat{x}}^{i}(b)^{j1} = {\hat{c}}^{i} b^j > 0$. However, in the $\sigma$-cycle profile $x^\sigma (b)$, there is a player $j^*$ in $N\setminus \{1\}$ who dumps no bads to player 1, namely, $x^\sigma (b)^{j^*1}=0$. This is a contradiction. If there is $i\in N\setminus \{\lambda (1)\}$ such that $c^{\sigma i}>0$, then $x^\sigma (b)^{i1} \ge c^{\sigma i} x^{\sigma i}(b)^{i1} = c^{\sigma i} b^{i} > 0$. However, since $\lambda (1)$ is the only player who dumps bads to player 1 in $x^\sigma (b)$, we have $x^\sigma (b)^{i'1}=0$ for every $i'\in N\setminus \{\lambda (1)\}$. This is a contradiction. Hence all the coefficients above are zero, while $x^\sigma (b)$ contains positive entries, which implies that $x^\sigma (b)$ can not be written as a non-negative convex combination of the other profiles.

Now, let $i^*\in N\setminus \{1\}$. We suppose that

$$\begin{aligned} {\hat{x}}^{i^*}(b) = c^\sigma x^\sigma (b) + \sum _{i\in N\setminus \{1,i^*\}} {\hat{c}}^i {\hat{x}}^i(b) + \sum _{i\in N\setminus \{\lambda (1)\}} c^{\sigma i} x^{\sigma i}(b) \end{aligned}$$

holds for some collection of non-negative coefficients that sum to 1. Assume $c^\sigma >0$. For any $i\in N\setminus \{1,\lambda (1)\}$, ${\hat{x}}^{i^*}(b)^{i \eta (i)}=0$. However, $x^\sigma (b)^{h \eta (h)}=b^{h}>0$ for some $h\in N\setminus \{1,\lambda (1)\}$. This is a contradiction. If there is $i\in N\setminus \{1,i^*\}$ such that ${\hat{c}}^{i}>0$, then ${\hat{x}}^{i^*}(b)^{1i}=0< b^1= {\hat{x}}^{i}(b)^{1i}$, which is a contradiction. If there is $i\in N\setminus \{\lambda (1)\}$, such that $c^{\sigma i}>0$, then since $n\ge 4$ we obtain a contradiction in the same manner as the case $c^\sigma >0$ by replacing $x^\sigma (b)$ with $x^{\sigma i}(b)$. Thus, ${\hat{x}}^{i^*}(b)$ can not be written as a non-negative convex combination of the other profiles.

Let $i^*\in N\setminus \{\lambda (1)\}$. We assume that

$$\begin{aligned} x^{\sigma i^*}(b) = c^\sigma x^\sigma (b) + \sum _{i\in N\setminus \{1\}} {\hat{c}}^i {\hat{x}}^i(b) + \sum _{i\in N\setminus \{\lambda (1),i^*\}} c^{\sigma i} x^{\sigma i}(b) \end{aligned}$$

holds for some collection of non-negative coefficients that sum to 1. If $c^\sigma >0$ then we have $x^{\sigma i^*}(b)^{i^* \eta (i^*)}=0<b^{i^*}=x^\sigma (b)^{i^* \eta (i^*)}$, which is a contradiction. If there is $i\in N\setminus \{1\}$, such that ${\hat{c}}^{i}>0$, then at least one player $j^*\in N\setminus \{1,i^*,\lambda (1)\}$ satisfies $x^{\sigma i^*}(b)^{j^*1}=0$, while ${\hat{x}}^i(b)^{j1}=1$ for every $j\in N\setminus \{1\}$. This is a contradiction. If there is $i\in N\setminus \{\lambda (1),i^*\}$ such that $c^{\sigma i}>0$, then $x^{\sigma i^*}(b)^{i^* \eta (i^*)}=0<b^{i^*}=x^{\sigma i}(b)^{i^* \eta (i^*)}$, which is a contradiction. Thus, $x^{\sigma i^*}(b)$ can not be written as a non-negative convex combination of the other profiles. $\square$

1.3 Proof of Proposition 3

Proposition 3. For any $n\ge 3$, any $b\in B^N$, and any $\sigma \in \Psi ^N$, all nonnegative convex combinations of $E(n,b,\sigma )$ are $\alpha$-stable.

Proof

In view of Lemma 1, the statement for $n=3$ is straightforward to show, since the number of the profiles is four. Below, we prove the statement for $n\ge 4$. Let $b\in B^N$ and $\sigma \in \Psi ^N$. Let y(b) be a non-negative convex combination of the profiles in $E(n,b,\sigma )$:

$$\begin{aligned} y(b) =c^\sigma x^\sigma (b) + \sum _{i\in N\setminus \{1\}} {\hat{c}}^i {\hat{x}}^i(b) + \sum _{i\in N\setminus \{\lambda (1)\}} c^{\sigma i} x^{\sigma i}(b). \end{aligned}$$

In view of Proposition 2, we show that for every coalition $S\subseteq N$, there is $i\in N$, such that $\sum _{j\in N\setminus S}b^j\ge r^i_{y}$.

We have

$$\begin{aligned} r_y =c^\sigma r_{x^\sigma } + \sum _{i\in N\setminus \{1\}} {\hat{c}}^i r_{{\hat{x}}^i} + \sum _{i\in N\setminus \{\lambda (1)\}} c^{\sigma i} r_{x^{\sigma i}}. \end{aligned}$$

Let $h\in N\setminus \{1\}$. We have

$$\begin{aligned} r^{h}_y= & {} c^\sigma r^{h}_{x^\sigma } + \sum _{i\in N\setminus \{1\}} {\hat{c}}^i r^{h}_{{\hat{x}}^i} + \sum _{i\in N\setminus \{\lambda (1)\}} c^{\sigma i} r^{h}_{x^{\sigma i}}\nonumber \\= & {} c^\sigma r^{h}_{x^\sigma } + \sum _{i\in N\setminus \{1\}} {\hat{c}}^i r^{h}_{{\hat{x}}^i} + \sum _{i\in N\setminus \{\lambda (1),\lambda (h)\}} c^{\sigma i} r^{h}_{x^{\sigma i}} + c^{\sigma \lambda (h)} r^{h}_{x^{\sigma \lambda (h)}}\nonumber \\= & {} c^\sigma b^{\lambda (h)} + \sum _{i\in N\setminus \{1\}} {\hat{c}}^i r^{h}_{{\hat{x}}^i} + \sum _{i\in N\setminus \{\lambda (1),\lambda (h)\}} c^{\sigma i} b^{\lambda (h)} + c^{\sigma \lambda (h)} 0\nonumber \\= & {} \left( c^\sigma + \sum _{i\in N\setminus \{\lambda (1),\lambda (h)\}} c^{\sigma i} \right) b^{\lambda (h)} + \sum _{i\in N\setminus \{1\}} {\hat{c}}^i r^{h}_{{\hat{x}}^i}. \end{aligned}$$

(2)

The third equality holds, because (i) in the profiles $x^\sigma$ and $x^{\sigma i}$, the bads player h receives are equal to the initial endowment of player $\lambda (h)$, and (ii) in the profile $x^{\sigma \lambda (h)}$, player h receives no bads as player $\lambda (h)$ dumps his bads all to player 1. Moreover, we have

$$\begin{aligned} (2)= & {} \left( c^\sigma + \sum _{i\in N\setminus \{\lambda (1),\lambda (h)\}} c^{\sigma i} \right) b^{\lambda (h)} + {\hat{c}}^h b^1\nonumber \\\le & {} \left( c^\sigma + \sum _{i\in N\setminus \{\lambda (1),\lambda (h)\}} c^{\sigma i} \right) b^{\lambda (h)} + {\hat{c}}^h b^{\lambda (h)}\nonumber \\\le & {} b^{\lambda (h)}. \end{aligned}$$

(3)

The first equality holds, because $r^{h}_{{\hat{x}}^i}=b^1$ holds only for $i=h$. The second inequality holds as $b^1\le ... \le b^n$. Since the coefficients sum to 1, we obtain the third inequality. Thus, (3) shows that $r^{h}_y \le b^{\lambda (h)}$ for every $h\in N\setminus \{1\}$, which implies that if a coalition S contains a player $h\in N\setminus \{1\}$ but not $\lambda (h)$, then, in view of Proposition 2, the coalition does not $\alpha$-improve. Therefore, below, we focus on a coalition defined as $\{\sigma (1),\ldots ,\sigma (s)\}$ for some $1\le s\le n-1$.

For any $s\ge 2$, coalition $S:=\{\sigma (1),\ldots ,\sigma (s)\}$ contains player $\sigma (2)$. From (3), it follows that $r^{\sigma (2)}_y\le b^{\sigma (1)}=b^{1}= \min _{\emptyset \ne T\subseteq N} \sum _{j\in T}b^j \le \sum _{j\in N\setminus S}b^j$. Hence, for any $s\ge 2$, coalition $S=\{\sigma (1),\ldots ,\sigma (s)\}$ does not $\alpha$-improve. We now consider the coalition $S=\{1\}$. We have

$$\begin{aligned} r^{1}_y= & {} c^\sigma r^{1}_{x^\sigma } + \sum _{i\in N\setminus \{1\}} {\hat{c}}^i r^{1}_{{\hat{x}}^i} +\sum _{i\in N\setminus \{\lambda (1)\}} c^{\sigma i} r^{1}_{x^{\sigma i}}\nonumber \\= & {} c^\sigma b^{\lambda (1)} + \sum _{i\in N\setminus \{1\}} {\hat{c}}^i \left( \sum _{j\in N\setminus \{1\}}b^j \right) + \sum _{i\in N\setminus \{\lambda (1)\}}c^{\sigma i} \left( b^{\lambda (1)}+b^{i} \right) \nonumber \\= & {} c^\sigma b^{\lambda (1)} + \sum _{j\in N\setminus \{1\}}b^j\left( \sum _{i\in N\setminus \{1\}} {\hat{c}}^i \right) + \sum _{i\in N\setminus \{\lambda (1)\}}c^{\sigma i} b^{\lambda (1)}+\sum _{i\in N\setminus \{\lambda (1)\}}c^{\sigma i} b^{i}. \end{aligned}$$

(4)

For simplicity, let $\Sigma {\hat{c}}:= \sum _{i\in N\setminus \{1\}} {\hat{c}}^i$. We have

$$\begin{aligned}&(4) =c^\sigma b^{\lambda (1)} + \sum _{j\in N\setminus \{1\}}b^j \Sigma {\hat{c}} + \sum _{i\in N\setminus \{\lambda (1)\}}c^{\sigma i} b^{\lambda (1)}+\sum _{i\in N\setminus \{\lambda (1)\}}c^{\sigma i} b^{i}\\&=c^\sigma b^{\lambda (1)} + \left[ b^{\lambda (1)} \Sigma {\hat{c}} +\sum _{j\in N\setminus \{1,\lambda (1)\}}b^j \Sigma {\hat{c}} \right] \\&\quad + \sum _{i\in N\setminus \{\lambda (1)\}}c^{\sigma i} b^{\lambda (1)}+ \left[ c^{\sigma 1} b^{1} + \sum _{i\in N\setminus \{1,\lambda (1)\}}c^{\sigma i} b^{i} \right] \\&= \left[ \sum _{i\in N\setminus \{1,\lambda (1)\}} (c^{\sigma i} + \Sigma {\hat{c}}) b^{i} \right] + \left[ c^\sigma + \Sigma {\hat{c}} + \sum _{i\in N\setminus \{\lambda (1)\}} c^{\sigma i} \right] b^{\lambda (1)} + c^{\sigma 1} b^{1}\\&= \left[ \sum _{i\in N\setminus \{1,\lambda (1)\}} (c^{\sigma i} + \Sigma {\hat{c}}) b^{i} \right] + b^{\lambda (1)} + c^{\sigma 1} b^{1}\\&\le \left[ \sum _{i\in N\setminus \{1,\lambda (1)\}} (c^{\sigma i} + \Sigma {\hat{c}}) b^{i} \right] + b^{\lambda (1)} + \sum _{i\in N\setminus \{1,\lambda (1)\}} c^{\sigma 1}b^{i} \\&= \left[ \sum _{i\in N\setminus \{1,\lambda (1)\}} (c^{\sigma 1}+c^{\sigma i} + \Sigma {\hat{c}}) b^{i} \right] + b^{\lambda (1)}\\&\le \sum _{i\in N\setminus \{1,\lambda (1)\}} b^{i} + b^{\lambda (1)}\\&= \sum _{i\in N\setminus \{1\}} b^{i}. \end{aligned}$$

Hence, coalition $\{1\}$ does not $\alpha$-improve. Thus, no coalition $\alpha$-improves, and y(b) is $\alpha$-stable. $\square$

1.4 Proof of Lemma 2

Lemma 2. Let $\mathcal {P}\in \Pi ^*(N)$, $\sigma _\mathcal {P}\in \Psi ^{\mathcal {P}}$, and $b\in B^N$. If for any $\mathcal {Q}\subsetneq \mathcal {P}$ with $\mathcal {P}(n)\in \mathcal {Q}$ there is $j\in \cup _{T\in \mathcal {Q}} T$, such that $b^j\le \sum _{i\in N\setminus (\cup _{T\in \mathcal {Q}} T)}b^i$, then strategy profile $x^{\sigma _\mathcal {P}}(b)$ is $\alpha$-stable.

Proof

We consider coalition S that satisfies $\emptyset \ne S\subsetneq N$. If there is a coalition $T\in \mathcal {P}$, such that $T\cap S\ne \emptyset$ and $T\setminus S \ne \emptyset$, then for some $h\in S$, $\lambda (h)\in T\setminus S$ and $h\in T\cap S$. From the definition of $x^{\sigma _\mathcal {P}}$, we have

$$\begin{aligned} r^h_{x^{\sigma _\mathcal {P}}}=b^{\lambda (h)} \overset{\lambda (h)\not \in S}{\le } \sum _{j\in N\setminus S}b^j, \end{aligned}$$

and S does not $\alpha$-improve. Therefore, for coalition S to $\alpha$-improve, S needs to satisfy $S=\cup _{T\in \mathcal {Q}}T$ for some $\mathcal {Q}\subsetneq \mathcal {P}$. If $\mathcal {P}(n)\not \in \mathcal {Q}$, then $n\in N\setminus S$. Hence, for some $h\in S$, $r^h_{x^{\sigma _\mathcal {P}}}=b^{\lambda (h)}\le b^n \le \sum _{j\in N\setminus S}b^j$, which means that S does not $\alpha$-improve. If $\mathcal {P}(n)\in \mathcal {Q}$, then in view of the assumption of the claim, there is $j\in \cup _{T\in \mathcal {Q}} T$, such that $b^j\le \sum _{i\in N\setminus (\cup _{T\in \mathcal {Q}} T)}b^i$. We fix this player j and consider $\eta (j)$. It holds that $r^{\eta (j)}_{x^{\sigma _\mathcal {P}}}=b^{j}\le \sum _{i\in N\setminus S}b^i$, which means that S does not $\alpha$-improve. Therefore, no coalition $\alpha$-improves $x^{\sigma _\mathcal {P}}(b)$. $\square$

1.5 Proof of Lemma 3

Lemma 3. Let $\mathcal {P}\in \Pi ^*(N)$ with $\mathcal {P}(n)=\mathcal {P}(1)$. For any $b\in B^N$ and any $\sigma _\mathcal {P}\in \Psi ^\mathcal {P}$, strategy profile $x^{\sigma _\mathcal {P}}(b)$ is $\alpha$-stable.

Proof

If there is a coalition $T\in \mathcal {P}$, such that $T\cap S\ne \emptyset$ and $T\setminus S \ne \emptyset$, then S does not $\alpha$-improve in the same manner as Lemma 2. Hence, to $\alpha$-improve, S must satisfy $S=\cup _{T\in \mathcal {Q}}T$ for some $\mathcal {Q}\subsetneq \mathcal {P}$. If $\mathcal {P}(n)\not \in \mathcal {Q}$, then player n is in $N\setminus S$, and for some $h\in S$, $b^{h}\le b^n \le \sum _{j\in N\setminus S}b^j$. If $\mathcal {P}(n)\in \mathcal {Q}$, then player 1 is in S, and for some $h\in N\setminus S$, $b^1\le b^h \le \sum _{j\in N\setminus S}b^j$. $\square$

1.6 Proof of Proposition 4

Proposition 4. Let $\mathcal {P}\in \Pi ^*(N)$ with $\mathcal {P}(n)=\mathcal {P}(1)$, $\sigma _\mathcal {P}\in \Psi ^\mathcal {P}$, $t\in \mathbb {R}_+$, and $i\in N\setminus T_\mathcal {P}$. For any $b\in B^N$, the following two statements are equivalent:

• i. $x^{\sigma _\mathcal {P}}_{ti}(b)$ is $\alpha$-stable,

• ii. $t\le \sum _{j\in N}b^j - (b^i + b^{\lambda (i)})$.

Proof

In this proof, we fix $\mathcal {P}$, $\sigma _\mathcal {P}$, t, i, and b as mentioned in the proposition and write $r_{**}:=r_{x^{\sigma _\mathcal {P}}_{ti}}$ and $r_{*}:=r_{x^{\sigma _\mathcal {P}}}$ for simplicity. We begin with [(ii)$\Rightarrow$(i)]. Below, we consider that a coalition S satisfies $\emptyset \ne S\subsetneq N$. Claim 1 is used for Claims 2–5. Claims 2–5 are used for the $\alpha$-improvement of coalition S. $\square$

Claim 1

For any $j\in N\setminus \{i\}$, $r^j_{**}\le r^j_{*}$. For the player i, $r^i_{**}\ge r^i_{*}$.

Proof of Claim 1. For every player $j\in N\setminus \{i\}$, $r^j_{**} = b^{\lambda (j)}- x^{\sigma _\mathcal {P}}_{ti}(b)^{\lambda (j) i} \le b^{\lambda (j)} = r^j_{*}$. For the fixed player i, $r^i_{**} = b^{\lambda (i)}+ t \ge b^{\lambda (i)} = r^i_{*}$. //

Claim 2

For any S with $\emptyset \ne S\subsetneq N$, if $i\in N\setminus S$, then S does not $\alpha$-improve.

Proof of Claim 2. In view of lemma 3, $x^{\sigma _\mathcal {P}}$ is $\alpha$-stable. Hence, for the coalition S, there is $h\in S$, such that $r^h_{*}\le \sum _{j\in N\setminus S}b^j$. Since $h\ne i$, it follows from Claim 1 that

$$\begin{aligned} r^h_{**} \overset{{Claim 1}}{\le } r^h_* \le \sum _{j\in N\setminus S}b^j, \end{aligned}$$

which means that S does not $\alpha$-improve. //

Claim 3

For any S with $\emptyset \ne S\subsetneq N$, if $T_\mathcal {P}\subseteq S$, then S does not $\alpha$-improve.

Proof of Claim 3. Since $\{1,n\}\subseteq T_\mathcal {P}$, $\sigma _{T_\mathcal {P}}(1)=1$. We write $h:=\sigma _{T_\mathcal {P}}(2)$. Since $i\in N\setminus T_\mathcal {P}$, $h\ne i$. We have

$$\begin{aligned} r^h_{**} \overset{{Claim 1}}{\le } r^h_* = b^{\lambda (h)} =b^1 \le \min _{j\in N}b^j. \end{aligned}$$

Since S is a proper subset of N, for some $h'\in N\setminus S$, $\min _{j\in N}b^j \le b^{h'} \le \sum _{j\in N\setminus S}b^j$. Hence, S does not $\alpha$-improve. //

Claim 4

For any S with $\emptyset \ne S\subsetneq N$, if S satisfies the following condition, then S does not $\alpha$-improve:

$$\begin{aligned} \text {there is }T\in \mathcal {P}\setminus \{\mathcal {P}(i)\} \text { such that }T\cap S\ne \emptyset \text { and }T\setminus S\ne \emptyset . \end{aligned}$$

Proof of Claim 4. Since $T\cap S\ne \emptyset$ and $T\setminus S\ne \emptyset$, there is a player $h\in T\cap S$, such that $\lambda (h)\in T\setminus S$. Considering that $h\in T\ne \mathcal {P}(i)$ implies $h\ne i$, we have

$$\begin{aligned} r^h_{**} \overset{{Claim 1}}{\le } r^h_* = b^{\lambda (h)} \overset{\lambda (h)\not \in S}{\le } \sum _{j\in N\setminus S}b^j. \end{aligned}$$

Hence, S does not $\alpha$-improve. //

Claim 5

For any S with $\emptyset \ne S\subsetneq N$ and $S\cap T_\mathcal {P}=\emptyset$, if S satisfies the following condition, then S does not $\alpha$-improve:

$$\begin{aligned} \text {there is }T\in \mathcal {P}\setminus \{\mathcal {P}(i)\} \text { such that }T\cap S\ne \emptyset . \end{aligned}$$

Proof of Claim 5. Since $T\cap S\ne \emptyset$, we consider a player $h\in T\cap S$. Moreover, since $h\in T\ne \mathcal {P}(i)$, we obtain $h\ne i$. We have

$$\begin{aligned} r^h_{**} \overset{{Claim 1}}{\le } r^h_* = b^{\lambda (h)} \le b^{n} \end{aligned}$$

Since $n\in T_\mathcal {P}$ and $S\cap T_\mathcal {P}=\emptyset$, we have $n\in N\setminus S$, which implies

$$\begin{aligned} b^{n} \le \sum _{j\in N\setminus S}b^j. \end{aligned}$$

Hence, S does not $\alpha$-improve. //

In view of Claim 4, for a coalition S to $\alpha$-improve, it must obey the following condition: for some partition $\mathcal {Q}\subseteq \mathcal {P}\setminus \{\mathcal {P}(i)\}$, $S= (\cup _{T\in \mathcal {Q}}T) \cup (S\cap \mathcal {P}(i))$, where $\mathcal {Q}$ may be $\emptyset$. Moreover, by Claim 3 and the fact that $T_\mathcal {P}$ is an element of $\mathcal {P}$, the previous condition must become the following form: for some partition $\mathcal {Q}\subseteq \mathcal {P}\setminus \{\mathcal {P}(i), T_\mathcal {P}\}$, $S= (\cup _{T\in \mathcal {Q}}T) \cup (S\cap \mathcal {P}(i))$. Hence, $S\cap T_\mathcal {P}=\emptyset$. By Claim 5, $(\cup _{T\in \mathcal {Q}}T)$ must be empty. Hence, $S=S\cap \mathcal {P}(i)$, which is equivalent to the condition $S\subseteq \mathcal {P}(i)$. By Claim 2, S satisfies $i\in S\subseteq \mathcal {P}(i)$.

If there is $h\in S$, such that $h\ne i$, then we have $r^h_{**} \overset{{Claim 1}}{\le } r^h_* = b^{\lambda (h)} \le b^{n} \overset{n\in N\setminus S}{\le } \sum _{j\in N\setminus S}b^j$. Hence, S does not $\alpha$-improve. Therefore, it suffices to show $S=\{i\}$ does not $\alpha$-improve. By the condition of t, we obtain $r^i_{**} = b^{\lambda (i)}+t \le b^{\lambda (i)} + \sum _{j\in N\setminus \{i, \lambda (i) \}}b^j = \sum _{j\in N\setminus \{i\}}b^j$, which means that S does not $\alpha$-improve. Thus, for every $S\subseteq N$, S does not $\alpha$-improve $x^{\sigma _\mathcal {P}}_{ti}(b)$: $x^{\sigma _\mathcal {P}}_{ti}(b)$ is $\alpha$-stable.

Now, we assume that the opposite inequality holds for the condition of t, $t> \sum _{j\in N\setminus \{i, \lambda (i) \}}b^j$. Then, in the same manner as above, $r^i_{**} > \sum _{j\in N\setminus \{i\}}b^j$. Given Proposition 2 and the fact that i is the only member of S, $S=\{i\}$ $\alpha$-improves. Hence, $x^{\sigma _\mathcal {P}}_{ti}(b)$ is not $\alpha$-stable. This establishes [not (ii) $\Rightarrow$ not (i)].

1.7 Proof of Proposition 5

Proposition 5. For any $b\in B^N$, the self-disposal profile $x^*(b)$ is the only m-stable profile in $X^N_b$.

Proof

Let $b\in b^N$. We first show that if profile $x\in X^N_b$ is not the self-disposal profile $x^*(b)$, then x is not m-stable. Assume that there is a player $i\in N$, such that $x^{ih}>0$ for some $h\in N\setminus \{i\}$. We define $y^i\in X_b^i$ as follows: $y^{ii}=b^i$ and $y^{ij}=0$ for every $j\in N\setminus \{i\}$. We have

$$\begin{aligned} \sum _{j\in N\setminus \{i\}}r^j_x - \sum _{j\in N\setminus \{i\}}r^j_{y^i,x^{-i}}= & {} \sum _{j\in N\setminus \{i\}} \sum _{l\in N} x^{lj} - \sum _{j\in N\setminus \{i\}}\left( y^{ij} + \sum _{l\in N\setminus \{i\}} x^{lj} \right) \\= & {} \sum _{j\in N\setminus \{i\}} ( x^{ij} - y^{ij} )\\= & {} \sum _{j\in N\setminus \{i\}} x^{ij}\\> & {} 0. \end{aligned}$$

Hence, $\{i\}$ m-deviates from x, and profile x is not m-stable.

Now, we show that $x^*(b)$ is m-stable. We write $x^*:=x^*(b)$ for simplicity. Assume that a coalition S m-deviates from $x^*$. By the definition of m-deviation, there is $y^S\in X_{b}^S$, such that for every $i\in S$:

$$\begin{aligned} \sum _{j\in N\setminus \{i\}}r^j_{y^S,x^{* N\setminus S}} < \sum _{j\in N\setminus \{i\}}r^j_{x^*}. \end{aligned}$$

(5)

Let $T:=\{i\in S | y^{ij}>0 \text { for some }j\in N\setminus \{i\}\}$. Note that T is nonempty, because if $y^{ij}=0$ for every $j\in N\setminus \{i\}$ and every $i\in S$, then $y^S=x^{*S}$, which contradicts (5). For every $i\in T$, define

$$\begin{aligned} I^i(y^S):= \sum _{h\in S\setminus \{i\}}y^{hi} \text { and }O^i(y^S):= \sum _{h\in N\setminus \{i\}}y^{ih}. \end{aligned}$$

For every $i\in T$, we have

$$\begin{aligned} \sum _{j\in N\setminus \{i\}}r^j_{y^S,x^{* N\setminus S}} = \left( \sum _{j\in N\setminus \{i\}}b^j \right) + O^i(y^S) - I^i(y^S). \end{aligned}$$

(6)

Since $\sum _{j\in N\setminus \{i\}}r^j_{x^*} = \sum _{j\in N\setminus \{i\}}b^j$, (5) implies

$$\begin{aligned} \sum _{j\in N\setminus \{i\}}r^j_{y^S,x^{* N\setminus S}} < \sum _{j\in N\setminus \{i\}}b^j. \end{aligned}$$

(7)

It follows from (6) and (7) that for every $i\in T$,

$$\begin{aligned} O^i(y^S) < I^i(y^S). \end{aligned}$$

(8)

In view of the definition of T, $y^{ii}=b^i$ for every $i\in S\setminus T$. Hence, we have $I^i(y^S) \overset{\text {def}}{=} \sum _{h\in S\setminus \{i\}}y^{hi}=\sum _{h\in T\setminus \{i\}}y^{hi}$ and $O^i(y^S) \overset{\text {def}}{=} \sum _{h\in N\setminus \{i\}}y^{ih}=\sum _{h\in T\setminus \{i\}}y^{ih}+\sum _{h\in N\setminus T}y^{ih}$. We obtain

$$\begin{aligned} \sum _{i\in T}I^i(y^S)= & {} \sum _{i\in T} \sum _{h\in T\setminus \{i\}}y^{hi}\\= & {} \sum _{i\in T} \sum _{h\in T\setminus \{i\}}y^{ih}\\\le & {} \sum _{i\in T} \left( \sum _{h\in T\setminus \{i\}}y^{ih} + \sum _{h\in N\setminus T}y^{ih} \right) \\= & {} \sum _{i\in T}O^i(y^S). \end{aligned}$$

This contradicts (8). $\square$