Generalized consistent ranking and the formation of self-enforcing coalitions

Abstract

Agents endowed with powers compete for a divisible prize by forming coalitions. When a coalition wins, all non-members are eliminated. The winning coalition then divides the prize among its members according to a given sharing rule. We investigate the case where the sharing rule satisfies a property we call consistent ranking. Consistent ranking ensures that agents’ rankings of competing coalitions coincide. Sharing rules such as equal and proportional sharing satisfy this property. We also examine a larger class of sharing rules that satisfy a property we call generalized consistent ranking where agents can rank coalitions even though the sharing rule does not satisfy consistent ranking. For instance, a convex combination of equal and proportional sharing, which we call combination sharing, violates consistent ranking but satisfies generalized consistent ranking under certain conditions. For these different sharing rules, we characterize rules on choosing coalitions (called transition correspondence) that satisfy two main axioms: self-enforcement, which requires that no further deviation happens after a coalition has formed, and rationality, which requires that agents pick the coalition that gives them their highest payoff. We find that a transition correspondence that satisfies self-enforcement and rationality always exists for sharing rules that satisfy generalized consistent ranking (and hence, consistent ranking). In the case of combination sharing, one way to satisfy generalized consistent ranking is to restrict the configuration of powers in society to satisfy size-power monotonicity, where larger coalitions have higher powers.

A. Appendix: Proofs

Proof of Lemma  1

Proof

Consider the sets

$$\begin{aligned} A^u=\{S | S\in \phi (S, \pi ), |S|\le u\} \end{aligned}$$

and

$$\begin{aligned} B^u=\{T | T\in {\tilde{\phi }}(T, \pi ), |T|\le u\}. \end{aligned}$$

We will prove by induction on the size of u that $A^u=B^u.$

This is clearly true if $u=1,$ because any singleton coalition is self-enforcing.

For the induction hypothesis, assume that $A^{u-1}=B^{u-1}.$

Consider $S\in A^u.$ Then $S\in \phi (S, \pi )$. Therefore, since $\phi$ is minimalistic, there is no $Q\subset S$ such that $Q\in W_{(S, \pi )}$ and $Q\in \phi (Q, \pi _{Q}).$

Therefore, since $A^{u-1}=B^{u-1},$ there is no $Q\subset S$ such that $Q\in W_{(S, \pi )}$ and $Q\in {\tilde{\phi }} (Q, \pi _{Q}).$

Hence, $S\in {\tilde{\phi }}(S, \pi )$ and $S\in B^u.$ Thus $A^u\subseteq B^u$.

We can similarly prove that $B^u\subseteq A^u.$ $\square$

Proof of Proposition 1

Proof

Step 1. $\phi ^*$ is SE and minimalistic.

Proof. To show SE, take any $X \in \phi ^*(S, \pi )$. There are two cases: either $X=S$ or $X \in Q(S,\pi )$. If $X=S$, then $X \in \phi ^*(S, \pi )=\phi ^*(X, \pi _{X})$. If $X \in Q(S,\pi )$, then $X \in \phi ^*(X, \pi _{X})$ by definition of the set $Q(S,\pi )$.

On the other hand, $\phi ^*$ is minimalistic because $\xi$ is cross-monotonic. That is, at the coalition formation game $(S, \pi )$, the set S is chosen only if $Q(S,\pi )=\emptyset$.

Step 2. $\phi ^*$ satisfies RAT.

Proof. Take $T\in \phi ^*(S, \pi )$ and consider a coalition Z such that $Z\in W_{(S, \pi )}$ such that $Z\in \phi ^*(Z, \pi _{Z})$.

($\Rightarrow$) First assume that $Z\not \in \phi ^*(S, \pi ).$ Since $T\in \phi ^*(S, \pi )$ we have that

$$\begin{aligned} T \in \displaystyle {\text {*}}{arg\,max}_{M \in Q(S, \pi )\cup \{S\}} R^\xi (M,\pi _M) \end{aligned}$$

Notice that Z is winning and self-enforcing within S, therefore $Z\in Q(S, \pi )\cup \{S\}$. Moreover, since $Z\not \in \phi ^*(S, \pi )$, we have that $Z \not \in \displaystyle {\text {*}}{arg\,max}_{M \in Q(S, \pi )\cup \{S\}} R^\xi (M,\pi _M)$. Hence, $R^\xi (T,\pi _T)>R^\xi (Z,\pi _Z)$.

($\Leftarrow$) Now, assume that $R^\xi (T,\pi _T)>R^\xi (Z,\pi _Z)$. Then, $Z \not \in \displaystyle {\text {*}}{arg\,max}_{M \in Q(S, \pi )\cup \{S\}} R^\xi (M,\pi _M)$. Hence, $Z \not \in \phi ^*(S, \pi )$

Step 3. Consider any cross-monotonic sharing rule and transition correspondences $\phi$ and ${\tilde{\phi }}$ that are self-enforcing and minimalistic. Then, the sets of coalitions that are self-enforcing coincide. That is,

$$\begin{aligned} \{S | S\in \phi (S, \pi )\}=\{T | T\in {\tilde{\phi }}(T, \pi )\}. \end{aligned}$$

Proof. The proof of this result is a straightforward consequence of Lemma 1.

Step 4. There exists a unique transition correspondence that meets SE and RAT.

Proof. Consider a transition correspondence $\phi$ that is SE and RAT. Then, $\phi$ is minimalistic because the sharing rule is cross-monotonic. We will show that $\phi =\phi ^*.$

Since $\phi$ and $\phi ^*$ are SE and RAT, by step 3, we have,

$$\begin{aligned} \{T | T\in \phi (T, \pi _T)\}=\{T | T\in \phi ^*(T, \pi _T)\} \end{aligned}$$

(2)

Suppose $X\in \phi (X, \pi )$. Then, by equation 2, $X\in \phi ^*(X, \pi )$. By RAT, for any $S\subset X$, $S\ne X$, then $S\not \in \phi (X,\pi )$. Hence, $\phi (X, \pi )=\phi ^*(X, \pi ).$

On the other hand, suppose $S\in \phi (X, \pi )$, where $S\not =X$. Then, by RAT, $\xi _i(S,\pi _S)\ge \xi _i(T,\pi _T)$ for $i\in S\cap T$ for any coalition T such that $T\in \phi (T, \pi _T)$ and $T\in W_{(X,\pi )}$. Therefore, by consistent ranking, $R^\xi (S,\pi _S)\ge R^\xi (T,\pi _T)$ for any coalition T such that $T\in \phi (T, \pi _T)$ and $T\in W_{(X,\pi )}$. Hence, $S\in \phi ^*(X, \pi )$ and $\phi (X, \pi )\subset \phi ^*(X, \pi )$. We could similarly show that $\phi ^*(X, \pi )\subset \phi (X, \pi ).$ Therefore, $\phi (X, \pi )=\phi ^*(X, \pi ).$

Step 5. $\phi ^{*}$ is superior to any transition correspondence that is SE and minimalistic.

Proof. We prove this step by contradiction. Suppose $\phi ^{*}$ is not superior to the SE and minimalistic transition correspondence ${\hat{\phi }}$. Then, there exists a game $(N, \pi )$ such that $S, T \subset N$ where $S\in \phi ^{*}(N,\pi )$ and $T\in {\hat{\phi }}(N,\pi )$ such that $\xi _i(T,\pi _T)\ge \xi _{i}(S,\pi _S)$ for some $i\in T\cap S$.

By step 3, since T is self-enforcing for ${{\hat{\phi }}}$, it follows that it is also self-enforcing for $\phi ^*.$ Therefore, $T\in Q(N, \pi )$.

Since $T\not \in \phi ^{*}(N,\pi )$, it follows that $T \not \in \displaystyle {\text {*}}{arg\,max}_{M \in Q(N, \pi )\cup \{N\}} R^\xi (M,\pi _M)$. Since $S \in \phi ^{*}(N,\pi )$, it follows that $S \in \displaystyle {\text {*}}{arg\,max}_{M \in Q(N, \pi )\cup \{N\}} R^\xi (M,\pi _M)$. Since S and T are winning within N (by the definition of a transition correspondence), we have $T\cap S\ne \emptyset$. Therefore, $R^\xi (S,\pi _{S})>R^\xi (T,\pi _T)$, which implies that $\xi _{i}(S, \pi _S)>\xi _{i}(T, \pi _T)$ for $i\in T\cap S$. This is a contradiction.

$\square$

Proof of Proposition 2

Proof

We prove this result in three steps.

Step 1. Suppose $\phi$ and ${\tilde{\phi }}$ are transition correspondences that satisfy SE and RAT under the sharing rule $\xi$. Then, $\phi ={\tilde{\phi }}$.

Proof. Suppose there is a game $(N,\pi )$ such that $\phi (N,\pi )\not ={\tilde{\phi }}(N,\pi )$. Without loss of generality, let $T\in {\tilde{\phi }}(N,\pi )$ and $T\not \in \phi (N,\pi )$. Since RAT implies MIN, by Lemma 1, the set of self-enforcing coalitions generated under $\phi$ and ${\tilde{\phi }}$ coincide. Since T was not chosen in $\phi$, by rationality, there exists $S\in SEC(N,\pi )$ such that $\xi _i(S,\pi _S)> \xi _i(T,\pi _T)$ for some $i\in T\cap S$. By rationality of ${\tilde{\phi }}$, $\xi _i(T,\pi _T)\ge \xi _i(S,\pi _S)$ for all $i\in T\cap S$. This is a contradiction.

Step 2. Show that $UD^{\xi }(N,\pi )\not =\emptyset$.

Since $\phi$ is a transition correspondence, $\phi (N,\pi )\not =\emptyset$. Let $T\in \phi (N,\pi )$. Since $\phi$ satisfies SE, $T\in SEC(N,\pi )$. By RAT, for all $S\in SEC(N,\pi )$, $\xi _i(T,\pi _T)\ge \xi _i(S,\pi _S)$ for all $i\in S\cap T$. Therefore, $T\in UD^{\xi }(N,\pi )$.

Step 3. We show that $\phi (N,\pi )= UD^{\xi }(N,\pi )$ satisfies SE and RAT.

If $T\in {\phi }(N,\pi )= UD^{\xi }(N,\pi )$, by Definition 6, $T\in SEC(N,\pi )$. Thus, $\phi$ satisfies SE.

Let coalition $T\in \phi (N,\pi )=UD^{\xi }(N,\pi )$. By definition of $UD^{\xi }(N,\pi )$, we have that $\xi _i(T,\pi _T)\ge \xi _i(S,\pi _S)$ for all $S\in SEC(N,\pi )$ and for all $i\in S\cap T$. Thus, $\phi$ satisfies RAT.

$\square$

Proof of Proposition 3

Proof

To prove part (i), let the game be $(N,\pi )\in {\bar{G}}$ and assume combination sharing and fix a $\lambda$.

For any game $(S,\pi _{S})\subset (N,\pi )$, agent i’s share in a coalition S is:

$$\begin{aligned} \xi _{i}(S,\pi _{S})=\lambda \cdot \frac{1}{|S|} + (1-\lambda )\cdot \frac{\pi _{i}}{\pi (S)} \end{aligned}$$

Since $(N,\pi )\in {\bar{G}}$, if there exists another game $(T,\pi _{T})\subset (N,\pi )$ such that $|S|<|T|$, it must also be true that $\pi (S)<\pi (T)$. Therefore,

$$\begin{aligned} \xi _{i}(S,\pi _{S})>\xi _{i}(T,\pi _{T}) \end{aligned}$$

for all agents in $i\in S\cap T$.

Hence, we can construct a ranking function

$$\begin{aligned} R^{CS^{\lambda }}(S,\pi )=\lambda \cdot \frac{1}{|S|} + (1-\lambda )\cdot \frac{\pi _{i}}{\pi (S)} \end{aligned}$$

where coalitions with smaller size (and therefore lesser power by size-power monotonicity) will yield a higher value of $R^{CS^{\lambda }}(S,\pi )$. Hence, the sharing rule satisfies consistent ranking, and by Proposition 1, the transition correspondence $\phi ^{**}$ is the unique transition correspondence satisfying SE and RAT over the domain of SPM games.

To prove part (ii), assume a transition correspondence $\phi$ that satisfies SE and RAT. Suppose we have two arbitrary coalitions S and T, such that $(S,\pi _{S})\in {\bar{G}}$ and $(T,\pi _{T})\in {\bar{G}}$. With combination sharing, agents in the intersection of S and T will have a higher share in S whenever

$$\begin{aligned} \xi _{i}(S,\pi _{S})=\lambda \frac{1}{|S|}+\left( 1-\lambda \right) \frac{\pi _{i}}{\pi (S)}>\lambda \frac{1}{|T|}+\left( 1-\lambda \right) \frac{\pi _{i}}{\pi (T)}=\xi _{i}(T,\pi _{T}) \end{aligned}$$

As $\lambda \rightarrow 1$, $\xi _{i}(S,\pi _{S})\rightarrow \frac{1}{|S|}$ and $\xi _{i}(T,\pi _{T})\rightarrow \frac{1}{|T|}$. Hence to maintain the inequality, it must be true that $|S|<|T|$. As $\lambda \rightarrow 0$, $\xi _{i}(S,\pi _{S})\rightarrow \frac{\pi _{i}}{\pi (S)}$ and $\xi _{i}(T,\pi _{T})\rightarrow \frac{\pi _{i}}{\pi (T)}$. Hence to maintain the inequality, it must be true that $\pi (S)<\pi (T)$. Therefore, for a ranking function $R^{CS^{\lambda }}$ to exist for all values of $\lambda$ and for all games in ${\bar{G}}$, we should have that for any two coalitions S and T, such that $(S,\pi _{S})\in {\bar{G}}$ and $(T,\pi _{T})\in {\bar{G}}$, if $|S|<|T|$ then $\pi (S)<\pi (T)$.

The proof that $\phi =\phi ^{**}$ is similar to Step 4 in the proof of Proposition 1, restricted to the domain of SPM games.

$\square$