On the characterizations of viable proposals

Abstract

Sengupta and Sengupta (Int Econ Rev 35:347–359, 1994) consider a payoff configuration of a TU game as a viable proposal if it challenges each legitimate contender. Lauwers (Int Econ Rev 43:1369–1371, 2002) prove that the set of viable proposals is nonempty for every game. In the present paper, we prove that the set of viable proposals coincides with the coalition structure core if there exists an undominated proposal; otherwise, it coincides with the set of accessible proposals. This characterization result implies that a proposal is a viable proposal if and only if it is undominated or accessible. Moreover, we prove that the set of viable proposals includes the minimal dominant set, which is another nonempty extension of the coalition structure core introduced by Kóczy and Lauwers (Games Econ Behav 61:277–298, 2007). In particular, we prove that the set of viable proposals of a cohesive game coincides with the minimal dominant set.

Appendix: Proof of Proposition 1

Proof of Proposition 1

(a) Let $(x,\mathcal {P})\in \Omega (N,v)$ be any given proposal. Since $C(N,v)=\emptyset$, by the Bondareva-Shapley Theorem, it follows that

$$\begin{aligned} \underset{S\in \mathcal {B}}{\sum }\lambda _{S} v(S) > \underset{S\in \mathcal {P}}{\sum }v(S) =x(N) =\underset{S\in \mathcal {B}}{\sum }\lambda _{S}x(S). \end{aligned}$$

Hence, we have $\sum _{S\in \mathcal {B}}\lambda _S[v(S)-x(S)]>0$. Combining with the fact that $\lambda _S>0$ for all $S\in \mathcal {B}$, we have that there exists a coalition $T\in \mathcal {B}$ such that $v(T) >x(T)$. Moreover, since $x(T)\ge \sum _{i\in T}v(\{i\})$, we have $|T|\ge 2$.

(b) Let $T\in \mathcal {B}$ be a coalition with $|T|\ge 2$. It suffices to prove that the core of the subgame $(T, v_{T})$ is nonempty. Suppose, to the contrary, that the core of $(T,v_T)$ is empty. Then there exists a T-balanced collection $\mathcal {B}'$ with balancing weights $\{\lambda '_{S}\} _{S\in \mathcal {B}'}$ satisfying $\sum \nolimits _{S\in \mathcal {B}'}\lambda '_{S}v_{T}(S)>v_{T}(T).$ This implies

$$\begin{aligned} \sum \limits _{S\in \mathcal {B}'}\lambda _{T} \lambda ' _{S} v(S) +\sum \limits _{S\in \mathcal {B}\backslash \{T\} }\lambda _{S} v(S)>\sum \limits _{S\in \mathcal {B}}\lambda _{S}v( S). \end{aligned}$$

However, this is impossible since $\mathcal {B}'\cup (\mathcal {B}\backslash \{T\})$ is an N-balanced collection with balancing weights $\{ \lambda _T \lambda '_{S}\}_{S\in \mathcal {B}'}\cup \{\lambda _{S}\}_{S\in \mathcal {B}\backslash \{ T\}}$. Hence, the core of $(T, v_{T})$ is nonempty.

(c) Let $(x_0,\mathcal {P}_0)\in \Omega (N,v)$ be any given proposal. To prove that $(x,\mathcal {R}(T))$ is accessible from $(x_0,\mathcal {P}_0)$, we inductively construct a sequence of proposals

$$\begin{aligned} (x_0,\mathcal {P}_0)\rightarrow _{T_1} (x_1,\mathcal {P}_1)\rightarrow _{T_2} \cdots \rightarrow _{T_{j}} (x_j,\mathcal {P}_j)\rightarrow _{T_{j+1}} \cdots , \end{aligned}$$

(4)

where $T_k\in \mathcal {B}$ for $k=1,2,\ldots$, and prove that sequence (4) does terminate at $(x,\mathcal {R}(T))$.

Suppose that the sequence of proposals $\{(x_k,\mathcal {P}_k)\}_{k=0}^j$ has been constructed to satisfy $(x_{k-1},\mathcal {P}_{k-1})\rightarrow _{T_k} (x_k,\mathcal {P}_k)$ with $T_k\in \mathcal {B}$ for $k=1,\ldots ,j$. Let $A_k=\{i\in T:x_k^i>v(\{i\})\}$ for $k=0,\ldots ,j$. We first assume that $A_j=\emptyset$, i.e., $x_j^i=v(\{i\})$ for $i\in T$. Since $(T,v_T)$ is cohesive, it follows that $v(T)>\sum _{i\in T}v(\{i\})=x_j(T)$, and hence $(x,\mathcal {R}(T))$ dominates $(x_j,\mathcal {P}_j)$ by T. This implies that $(x,\mathcal {R}(T))$ is accessible from $(x_0,\mathcal {P}_0)$ and we are done.

In what follows, we assume that $A_j\ne \emptyset$. By (a), there exists a proposal $T_{j+1}\in \mathcal {B}$ such that $v(T_{j+1})>x_j(T_{j+1})$. Let $\mathcal {P}_{j+1}=\mathcal {R}(T_{j+1})$. We consider three cases.

Case I. $T_{j+1}\cap A_j \ne \emptyset$. Let $(x_{j+1},\mathcal {P}_{j+1})\in \Omega (N,v)$ be the proposal defined by

$$\begin{aligned} x_{j+1}^i= {\left\{ \begin{array}{ll} x_j^i+[v(T_{j+1})-x_j(T_{j+1})]/|T_{j+1}\cap A_j|,&{} \text {if } i\in T_{j+1}\cap A_j,\\ x_j^i, &{}\quad \text {if } i\in T_{j+1}\backslash A_j,\\ v(\{i\}),&{}\quad \text {otherwise.} \end{array}\right. } \end{aligned}$$

Clearly, $(x_{j+1},\mathcal {P}_{j+1})$ dominates $(x_j,\mathcal {P}_j)$ by $T_{j+1}$. Let $A_{j+1}=\{i\in T:x_{j+1}^i>v(\{i\})\}$. Then we have $A_{j}\supseteq T_{j+1}\cap A_j=A_{j+1}\ne \emptyset$.

Case II. $T_{j+1}\cap A_j =\emptyset$ and $T_{j+1}\backslash T \ne \emptyset$. Let $(x_{j+1},\mathcal {P}_{j+1})\in \Omega (N,v)$ be the proposal defined by

$$\begin{aligned} x_{j+1}^i= {\left\{ \begin{array}{ll} x_j^i+[v(T_{j+1})-x_j(T_{j+1})]/|T_{j+1}\backslash T|,&{}\quad \text {if } i\in T_{j+1}\backslash T,\\ x_j^i=v(\{i\}), &{}\quad \quad \text {if } i\in T_{j+1}\cap T,\\ v(\{i\}),&{}\quad \text {otherwise.} \end{array}\right. } \end{aligned}$$

Again, one can easily check that $(x_{j+1},\mathcal {P}_{j+1})$ dominates $(x_j,\mathcal {P}_j)$ by $T_{j+1}$. Let $A_{j+1}=\{i\in T:x_{j+1}^i>v(\{i\})\}$. Then we have $A_{j+1}=\emptyset$. This implies that $(x,\mathcal {R}(T))$ dominates $(x_{j+1},\mathcal {P}_{j+1})$ by T and we are done.

Case III. $T_{j+1}\cap A_j =\emptyset$ and $T_{j+1}\subseteq T$. Let $\lambda =\frac{v(T_{j+1})-x_j(T_{j+1})}{x(T_{j+1})-x_j(T_{j+1})}$. Since $x(T_{j+1})\ge v(T_{j+1})>x_j(T_{j+1})$, it follows that $1\ge \lambda >0$. Let $(x_{j+1},\mathcal {P}_{j+1})\in \Omega (N,v)$ be the proposal defined by

$$\begin{aligned} x_{j+1}^i= {\left\{ \begin{array}{ll} x_j^i+\lambda (x^i-x_j^i),&{}\quad \text {if } i\in T_{j+1},\\ v(\{i\}),&{}\quad \text {otherwise}, \end{array}\right. } \end{aligned}$$

where $x^i_j=v(\{i\})$ for each $i\in T_{j+1}$. Clearly, $(x_{j+1},\mathcal {P}_{j+1})$ dominates $(x_j,\mathcal {P}_j)$ by $T_{j+1}$ and $x^i\ge x_{j+1}^i$ for each $i\in T_{j+1}$. Moreover, since $A_j\ne \emptyset$ and $T_{j+1}\cap A_j=\emptyset$, it follows that $T_{j+1}\ne T$. By the fact that $(T,v_T)$ is cohesive, we have $v(T)>v(T_{j+1})+\sum _{i\in T\backslash T_{j+1}}v(\{i\})$. This implies that $(x,\mathcal {R}(T))$ dominates $(x_{j+1},\mathcal {P}_{j+1})$ by T and we are done.

Suppose, to the contrary, that the construction of sequence (4) does not come to an end. Let $a_k=x_k(A_k)$ for $k=0,1,2,\ldots$. Since the corresponding sequence

$$\begin{aligned} A_0\supseteq T_1\cap A_0=A_1\supseteq \cdots A_j\supseteq T_{j+1}\cap A_j=A_{j+1}\supseteq \cdots \end{aligned}$$

is weakly decreasing, there exists a non-negative integer r such that $A_r\ne \emptyset$ and $A_r=A_{r+t}$ for $t=1,2,\ldots$. This implies that $A_r\subseteq T_{r+t}$ for $t=1,2,\ldots$, and the sequence $a_r<a_{r+1}<a_{r+2}<\cdots$ is strictly increasing. Moreover, since the number of coalitions in $\mathcal {B}$ is finite, there exists a positive integer $t^*$ such that

$$\begin{aligned} \bigcap _{t=1}^{\infty } T_{r+t}=\bigcap _{t=1}^{t^*} T_{r+t}\supseteq A_r \end{aligned}$$

(5)

and there exist two distinct positive integers $k_1<k_2$ such that $T_{r+t^*+k_1}=T_{r+t^*+k_2}$. By (5), we have that

$$\begin{aligned} x^i_{r+t^*+k_1}=x^i_{r+t^*+k_2}= {\left\{ \begin{array}{ll} x_{r+t^*}^i,&{}\quad \text {if } i\in \left( \bigcap _{t=1}^{t^*} T_{r+t}\right) \backslash A_r,\\ v(\{i\}),&{} \quad \text {if } i\in N\backslash \left( \bigcap _{t=1}^{t^*} T_{r+t}\right) . \end{array}\right. } \end{aligned}$$

This implies

$$\begin{aligned} a_{r+t^*+k_1}&=x_{r+t^*+k_1}(A_{r+t^*+k_1})=x_{r+t^*+k_1}(A_r)=v(T_{r+t^*+k_1})-x_{r+t^*+k_1}(T_{r+t^*+k_1}\backslash A_r)\\&=v(T_{r+t^*+k_2})-x_{r+t^*+k_2}(T_{r+t^*+k_2}\backslash A_r)=x_{r+t^*+k_2}(A_r)=a_{r+t^*+k_2}, \end{aligned}$$

contradicting the fact $a_{r+t^*+k_1}<a_{r+t^*+k_2}$. $\square$

To better illustrate the iterative process for generating a dominating chain that terminates at an accessible proposal in the proof of Proposition 1 (c), it can be useful to consider the following example.

Example 3

Consider the three-player game (N, v) given by

$$\begin{aligned} v(\{1,2,3\})&=v(\{1,2\})=v(\{1,3\})=v(\{2,3\})=3,\\ v(\{1\})&=v(\{2\})=v(\{3\})=0. \end{aligned}$$

Let $T_1=\{1,2\},T_2=\{1,3\},T_3=\{2,3\}$. Clearly, $\mathcal {B}=\{T_1,T_2,T_3\}$ is an N-balanced collection with balancing weights $\lambda _1=\lambda _2=\lambda _3=1/2$ that maximizes the amount $\sum _{S\in \mathcal {B}}\lambda _Sv(S)$ and $(T_j,v_{T_j})$ is cohesive for $j=1,2,3$. Since

$$\begin{aligned} v^*=\max _{\mathcal {P}\in \Pi }\sum _{S\in \mathcal {P}}v(S)=v(\{1,2\})+v(\{3\})=3<9/2=\sum _{j=1}^3\lambda _jv(T_j), \end{aligned}$$

by the Bondareva–Shapley Theorem, it follows that $C(N,v)=\emptyset$. One can verify that the iterative process in the proof of Proposition 1 (c) generates a dominating chain from $(x_0,\mathcal {P}_0)=(1,1,1,N)$ to $(x,\mathcal {R}(T_3))=(0,1,2,\{2,3\},\{1\})$ with the associated decreasing sequence $A_0=\{2,3\}\supseteq T_1\cap A_0=A_1=\{2\}\supseteq T_2\cap A_1=\emptyset$ as follows:

$$\begin{aligned} (1,1,1,N)\rightarrow _{T_1} (1,2,0,\{1,2\},\{3\})\rightarrow _{T_2}(3,0,0,\{1,3\},\{2\})\rightarrow _{T_3}(0,1,2,\{2,3\},\{1\}). \end{aligned}$$