Accessibility and stability of the coalition structure core

Abstract

This article shows that, for any transferable utility game in coalitional form with a nonempty coalition structure core, the number of steps required to switch from a payoff configuration out of the coalition structure core to a payoff configuration in the coalition structure core is less than or equal to $(n^2+4n)/4$, where $n$ is the cardinality of the player set. This number improves the upper bounds found so far. We also provide a sufficient condition for the stability of the coalition structure core, i.e. a condition which ensures the accessibility of the coalition structure core in one step. On the class of simple games, this sufficient condition is also necessary and has a meaningful interpretation.

Appendices

Appendix 1: Proof of Lemma 1

Point 1 Pick any $\mathcal P^*$-inefficient payoff configuration $(x,\mathcal{P})$. On the one hand, we have $\sum _{C^{\prime }\in \mathcal{P}}v(C^{\prime })=\sum _{C^{\prime }\in \mathcal{P}}x(C^{\prime })=x(N)=\sum _{C^{\prime }\in \mathcal{P^*}}x(C^{\prime })$. On the other hand, since $(x^*,\mathcal{P}^*)\in \text{ CS }(v)$ we have $\sum _{C^{\prime }\in \mathcal{P}}v(C^{\prime })\le \sum _{C^{\prime }\in \mathcal{P}^*}v(C^{\prime })=v^*$. Thus, we obtain $\sum _{C^{\prime }\in \mathcal{P^*}}x(C^{\prime })\le \sum _{C^{\prime }\in \mathcal{P^*}}v(C^{\prime })$, i.e.

$$\begin{aligned} \sum _{C^{\prime }\in \mathcal{P^*}}e(x,C^{\prime })\ge 0. \end{aligned}$$

Since $(x,\mathcal{P})$ is $\mathcal P^*$-inefficient, there is some $C^{\prime }\in \mathcal{P^*}$ such that $e(x,C^{\prime })\ne 0$. This means that there exists $C\in \mathcal{P^*}$ (possibly equal to coalition $C^{\prime }$) such that $e(x,C)>0$.

Point 2 Pick any $\mathcal P^*$-efficient payoff configuration $(x, \mathcal{P})$ and any coalition $C\in \Omega$ such that $e(x,C)>0$. We have to show that there exists a player in $P(C,\mathcal{P})\setminus C$ who is overpaid in $(x,\mathcal{P})$.

We first show that $P(C,\mathcal{P})\setminus C\ne \emptyset$. The result is obvious if $C$ is not the union of elements of $\mathcal P$. Furthermore, $e(x,C)>0$ implies that $C\not \in \mathcal{P}$. So, by way of contradiction, assume that $C$ is the union of at least two elements of $\mathcal P$ and denote by $\{C^1,\dots , C^k\}$ the set of these $k\in \{2,\dots ,|\mathcal{P}|\}$ elements. Next, consider the partition $\mathcal{P}^{\prime }=(\mathcal{P}\setminus \{C^1,\dots , C^k\})\cup \{C\}$. Then, $\sum _{C\in \mathcal{P}^{\prime }}v(C^{\prime })>\sum _{C^{\prime }\in \mathcal{P}}x(C^{\prime })=\sum _{C^{\prime }\in \mathcal{P}}v(C^{\prime })=v^*$, which contradicts the definition of $v^*$.

Second, assume for the sake of contradiction that none of the players in $P(C,\mathcal{P})\setminus C$ is overpaid in $(x,\mathcal{P})$. On the one hand, since $x(C)<v(C)\le x^*(C)$, we directly obtain:

$$\begin{aligned} x(P(C,\mathcal{P})\setminus C)< x^*(P(C,\mathcal{P})\setminus C). \end{aligned}$$

(2)

On the other hand, by definition of a payoff configuration, $(x^*,\mathcal{P^*})\in \text{ CS }(v)$ and $\mathcal{P^*}$-efficiency of $(x,\mathcal P)$, we have:

$$\begin{aligned} \forall C^{\prime } \in \mathcal{P}, \quad x(C^{\prime })=v(C^{\prime })\le x^*(C^{\prime }) \quad \text{ and }\quad x(N) = x^*(N), \end{aligned}$$

which forces the equality $x(C^{\prime })=x^*(C^{\prime })$ for each $C^{\prime }\in \mathcal{P}$. Since $P(C,\mathcal{P})$ is the union of disjoint elements of $\mathcal{P}$, we immediately get $x(P( C,\mathcal{P}))=x^*(P(C,\mathcal{P}))$, which contradicts (2).

Appendix 2: Proof of Lemma 2

Point 1 Assume that $(x^{\prime },\mathcal{P^{\prime }})$ is the $C$-transformation of $(x,\mathcal{P})$ with respect to $(x^*,\mathcal{P}^*)\in \text{ CS }(v)$. Consider any player $i\in N$ such that $i\in N\setminus O(x)$. If $i\in U(x, C)$ then either $x^{\prime }_i=x^*_i$ if the gap $x^*_i-x_i$ is filled up by the excess $e(x,C)$ or $x^{\prime }_i<x^*_i$. In both cases, $i\in N\setminus O(x^{\prime })$. If $i\in (N\setminus P(C,\mathcal{P}))\cup (C\setminus U(x,C))$, then $x^{\prime }_i=x_i\le x^*_i$ and thus $i\in N\setminus O(x^{\prime })$. Finally, if $i\in P(C,\mathcal{P})\setminus C$ then $x^{\prime }_i=0 \le x^*_i$, which means once again that $i\in N\setminus O(x^{\prime })$. The claim follows.

Point 2 Fix any $\mathcal P^*$-inefficient payoff configuration, which we denote by $(x^0,\mathcal{P}^0)$. Let us construct a sequence of $m+1$ payoff configurations $((x^0,\mathcal{P}^0),(x^1,\mathcal{P}^1),\dots , (x^m,\mathcal{P}^m))$ and a sequence of $m$ coalitions $(C^1,\dots , C^m)$ as follows. For each $k\in \{1,\dots ,m\}$ such that $(x^{k-1},\mathcal{P}^{k-1})$ is a $\mathcal P^*$-inefficient payoff configuration, we can choose $C^k\in \mathcal{P^*}$ such that $x^{k-1}(C^k)<v(C^k)$ by point 1 of Lemma 1. Then, define $(x^k,\mathcal{P}^k)$ as the $C^k$-transformation of $(x^{k-1},\mathcal{P}^{k-1})$ with respect to $(x^*,\mathcal{P}^*)$. By definition of the $C^k$-transformation, $k\in \{1,\dots ,m\}$, the payoff configuration $(x^k,\mathcal{P}^k)$ o.i.-dominates $(x^{k-1},\mathcal{P}^{k-1})$. It follows that $(x^m,\mathcal{P}^m)$ is trans-accessible from $(x^0,\mathcal{P}^0)$. It remains to prove that $(x^m,\mathcal{P}^m)$ is $\mathcal P^*$-efficient after $m\le p^*$ steps. Recall that for each $k\in \{1,\dots ,m\}$, the coalition $C^k$ is chosen such that $C^k\in \mathcal{P^*}$ and $x^{k-1}(C^k)<v(C^k)$. In addition, $x^k(C^k)=v(C^k)$ and the payoff of any player in $C^k$ will not be altered in the subsequent steps since for each $C^h\in \mathcal{P^*}$, $k< h \le m$, $C^k\cap P(C^h,\mathcal{P}^h)=\emptyset$. This implies that once $C^k$ has been chosen, it cannot be chosen in a subsequent step, i.e. the sequence $(C^1,\dots ,C^m)$ contains $m$ distinct coalitions belonging to $\mathcal P^*$. Because $\mathcal{P^*}$ contains $p^*$ coalitions, the sequence of coalitions $(C^1,\dots ,C^m)$ contains at most $p^*$ elements. Therefore, there exists some integer $m\le p^*$ such that we cannot choose $C^{m+1}\in \mathcal{P^*}$ such that $e(x^m, C^{m+1})>0$. This means that $x^m(C^{\prime })\ge v(C^{\prime })$ for each $C^{\prime }\in \mathcal{P^*}$ and thus that $\sum _{C^{\prime }\in \mathcal{P^*}}x^m(C^{\prime })\ge \sum _{C^{\prime }\in \mathcal{P^*}}v(C^{\prime })=v^*$. By definition of $v^*$, we necessarily have $x^m(C^{\prime })=v(C^{\prime })$ for each $C^{\prime } \in \mathcal{P^*}$. We conclude that $(x^m,\mathcal{P}^m)$ is a $\mathcal P^*$-efficient payoff configuration, as claimed.