Comparable axiomatizations of the average tree solution and the Myerson value

Abstract

Combination of a TU-game and an undirected graph representing cooperation restrictions among the players is called a TU-game with communication structure. For TU-games with communication structure, the average tree solution is defined as the average of the marginal contribution vectors corresponding to all spanning trees of the undirected graph. In this paper, we provide new characterizations for the average tree solution and the Myerson value on the class of TU-games with connected cycle-free communication structure. On this class of games, we show that the average tree solution is the unique solution satisfying linearity, efficiency, satellite symmetry, and satellite marginality. Together with linearity and efficiency, by using network symmetry and network marginality we also characterize the Myerson value and provide a comparison with the average tree solution.

Appendix

In this Appendix, we provide the proofs of the propositions and lemmata which are used for the characterizations of the average tree solution and the Myerson value.

Proof of Proposition 5.1

First, take any $(N,u_N,L)\in {{\mathcal {G}}}_N^{L}$ and $i\in N$. Note that $m_i^{T(i)}(N,u_N,L)=1$ and $m_i^{T(j)}(N,u_N,L)=0$ holds for all $j\in N$, $j\ne i$. Since the average tree solution is the average of the marginal contribution vectors corresponding to all spanning trees, this implies $AT_i(N,u_N,L)=\frac{1}{n}$ for all $i\in N$. Now, take any $(N,\mathbf{0 },L)\in {{\mathcal {G}}}_N^{L}$ and $i\in N$. Note that $m_i^{T(j)}(N,\mathbf{0 },L)=0$ holds for all $j\in N$. Since the average tree solution is the average of all such marginal contribution vectors, this implies $AT_i(N,\mathbf{0 },L)=0$ for all $i\in N$. $\square$

Proof of Proposition 5.2

Take any $(N,u_S,L) \in {{\mathcal {G}}}_N^{L}$ where $S\in C^L(N)$ and $|E^L(S)|=1$. First, consider any $i\in N{\setminus } S$. Note that $m_i^{T(j)}(N,u_S,L)=0$ holds for all $j\in N$ which implies $AT_i(N,u_S,L)=0$. Next, consider any $i\in S{\setminus } E^L(S)$. Note that $m_i^{T(j)}(N,u_S,L)=0$ holds for all $j\ne i$ and $m_i^{T(i)}(N,u_S,L)=1$. Since the average tree solution is the average of the marginal contribution vectors corresponding to all spanning trees, this implies $AT_i(N,u_S,L)=\frac{1}{n}$. Finally, consider $i\in E^L_S$. Note that $m_i^{T(j)}(N,u_S,L)=0$ holds for all $j\in S$, $j\ne i$, $m_i^{T(j)}(N,u_S,L)=1$ holds for all $j\in N{\setminus } S$, and $m_i^{T(i)}(N,u_S,L)=1$. Since the average tree solution is the average of the marginal contribution vectors corresponding to all spanning trees, this implies $AT_i(N,u_S,L)=\frac{n-|S|+1}{n}$ for $i\in E^L(S)$. $\square$

Proof of Lemma 5.3

Take any connected cycle-free undirected graph (N, L), $S\in C^L(N)$ and let $j\in E^L(S)$. Define $S'=N{\setminus } M$ where $M=\{i\in N\mid i\in Q$ for some $Q\in {\widehat{C}}^L(N{\setminus } \{j\})$ with $Q\cap S =\emptyset \}$. Since $M\cap S=\emptyset$, we have $S'\supseteq S$. Also, by construction $S'\in C^L(N)$ and since (N, L) is a cycle-free undirected graph, it holds that $E^L(S')=\{j\}~\text{and}~E^L(Q)\neq\{j\}~\text{for any}~Q\in C^L(N)~\text{with}~S'\supset Q\supseteq S$. $\square$

Proof of Proposition 5.4

Take any $(N,u_S,L) \in {{\mathcal {G}}}_N^{L}$ with $S\in C^L(N)$ and $|E^L(S)|>1$. First consider any $i\in N{\setminus } S$. Note that $m_i^{T(j)}(N,u_S,L)=0$ holds for all $j\in N$ which implies $AT_i(N,u_S,L)=0$. Next, consider any $i\in S{\setminus } E^L(S)$. Note that $m_i^{T(j)}(N,u_S,L)=0$ holds for all $j\in N$, $j\ne i$ and $m_i^{T(i)}(N,u_S,L)=1$. Since the average tree solution is the average of the marginal contribution vectors corresponding to all spanning trees, this implies $AT_i(N,u_S,L)=\frac{1}{n}$. Finally, consider any $i\in E^L(S)$ and let $S'={K^L_i(S)}$. Note that $m_i^{T(j)}(N,u_S,L)=m_i^{T(j)}(N,u_{S'},L)$ holds for all $j\in N$ which implies $AT_i(N,u_{S},L)=AT_i(N,u_{S'},L)$. Since $S'\in C^L(N)$ and $E^L(S')=\{i\}$, by Proposition 5.2 we have $AT_i(N,u_{S'},L)=\frac{n-|S'|+1}{n}$ which means $AT_i(N,u_{S},L)=\frac{n-|K^L_i(S)|+1}{n}$. $\square$

Proof of Proposition 5.5

Take any $(N,u_S,L) \in {{\mathcal {G}}}_N^{L}$ where $S\notin C^L(N)$. Let $S'=M^L_S$ and note that $m_i^{T(j)}(N,u_S,L)=m_i^{T(j)}(N,u_{S'},L)$ for all $i,j\in N$. This means $AT_i(N,u_{S},L)=AT_i(N,u_{S'},L)$ for all $i\in N$. Since $M^L_S\in C^L(N)$, by Proposition 5.4 we have $AT_i(N,u_{S'},L)=0$ for all $i\in N{\setminus } S'$, $AT_i(N,u_{S'},L)=\frac{1}{n}$ for all $i\in S'{\setminus } E^L(S')$, and $AT_i(N,u_{S'},L)=\frac{n-|K^L_i(S')|+1}{n}$ for all $i\in E^L(S')$ which completes the proof. $\square$

Proof of Proposition 5.7

Take any $(N,\mathbf{0 },L) \in {{\mathcal {G}}}_N^{L}$ and $\pi \in \Pi _N$. Note that, $\mathbf{0 }^L(S)=0$ holds for all $S\in 2^N$ which means $m_i^{\pi }(N,\mathbf{0 },L)=0$ for all $i\in N$. Since the Myerson value is the average of the marginal contribution vectors corresponding to all permutations, this implies $MV_i=(N,\mathbf{0 },L)=0$ for all $i\in N$. $\square$

Proof of Proposition 5.8

Take any $(N,u_S,L) \in {{\mathcal {G}}}_N^{L}$ with $S\in C^L(N)$. First consider any $i\in N{\setminus } S$. Note that, $m_i^{\pi }(N,u_S,L)=0$ holds for all $\pi \in \Pi _N$ which implies $MV_i(N,u_S,L)=0$. Next, consider any $i\in S$ and note that $m_i^{\pi }(N,u_S,L)=1$ if ${\bar{\pi }}^i\supseteq S$ and $m_i^{\pi }(N,u_S,L)=0$ if ${\bar{\pi }}^i\nsupseteq S$. Moreover, the number of permutations for which ${\bar{\pi }}^i\supseteq S$ holds is equal to n!/|S| i.e., $|\{\pi \in \Pi _N:{\bar{\pi }}^i\supseteq S\}|=n!/|S|$. Since the total number of permutations on N is equal to n! and the Myerson value is the average of the marginal contributions corresponding to all permutations, this implies $MV_i=1/|S|$. $\square$

Proof of Proposition 5.9

Take any $(N,u_S,L) \in {{\mathcal {G}}}_N^{L}$ with $S\notin C^L(N)$. Let $S'=M^L_S$. Since $S'\in C^L(N)$, by Proposition 5.8 we have $MV_i(N,u_{S'},L)=0$ if $i\in N{\setminus } S'$ and $MV_i(N,u_{S'},L)=1/|S'|$ if $i\in S'$. Note that, $m^{\pi }(N,u_S,L)=m^{\pi }(N,u_{S'},L)$ holds for all $\pi \in \Pi _N$ which means $MV(N,u_S,L)=MV(N,u_{S'},L)$. This implies $MV_i(N,u_{S},L)=0$ if $i\in N{\setminus } M^L_S$ and $MV_i(N,u_{S},L)=1/|M^L_S|$ if $i\in M^L_S$ $\square$

Proof of Lemma 5.10

Take any $(N,\mathbf{0 },L)\in {{\mathcal {G}}}_N^{L}$ and consider any $i,j\in N$ with $\{i,j\}\in L$. Note that, $\mathbf{0 }(S)=0$ for all $S\in 2^N$ which means that $\mu _i^{(N,\mathbf{0 },L)}(N{\setminus } S)=\mu _j^{(N,\mathbf{0 },L)}(N{\setminus } Q)=0$ for all $S\in \Psi _i^L$ and for all $Q\in \Psi _j^L$. Hence, i and j are satellite symmetric players and by satellite symmetry, we have $\xi _i(N,\mathbf{0 },L)=\xi _j(N,\mathbf{0 },L)$. Since (N, L) is a connected undirected graph, there exists a path between any two players. This means that all players in $(N,\mathbf{0 },L)$ are satellite symmetric. Since $\mathbf{0 }(N)=0$ and $\xi$ : $\mathcal{G}_N^{L}\rightarrow {\mathbb {R}}^n$ satisfies efficiency, this implies $\xi _i(N,\mathbf{0 },L)=0$ for all $i\in N$. Now, take any $(N,u_N,L)\in {{\mathcal {G}}}_N^{L}$ and consider any $i,j\in N$ with $\{i,j\}\in L$. Note that, $u_N(S)=u_N(N{\setminus } S)=0$ for all $S\in \Psi _i^L$ and $u_N(Q)=u_N(N{\setminus } Q)=0$ for all $Q\in \Psi _j^L$ which implies $\mu _i^{(N,u_N,L)}(N{\setminus } S)=\mu _j^{(N,u_N,L)}(N{\setminus } Q)=0$ for all $S\in \Psi _i^L$ and for all $Q\in \Psi _j^L$. This means i and j are satellite symmetric players and by satellite symmetry we have $\xi _i(N,u_N,L)=\xi _j(N,u_N,L)$. Since (N, L) is a connected undirected graph, there exists a path between any two players which means that all players in $(N,u_N,L)$ are satellite symmetric. Hence, together with the fact that $u_N(N)=1$, efficiency implies $\xi _i(N,u_N,L)=1/n$ for all $i\in N$. $\square$

Proof of Lemma 5.11

Take any $(N,u_S,L)\in {{\mathcal {G}}}_N^{L}$ with $S\in C^L(N)$ and let $E^L(S)=\{j\}$ for some $j\in N$. For all players $i\in N{\setminus } S$, it holds that $\mu _i^{(N, u_S,L)}(Q)=\mu _i^{(N,\mathbf{0 },L)}(Q)$ for $Q=N$ and $Q=N{\setminus } S$ for some $S\in \Psi _i^L$. From satellite marginality and Lemma 5.10, it follows that $\xi _i(N,u_S,L)=\xi _i(N,\mathbf{0 },L)=0$ for all $i\in N{\setminus } S$. Regarding any $i\in S{\setminus }\{j\}$, it holds that $\mu _i^{(N, u_S,L)}(Q)=\mu _i^{(N,u_N,L)}(Q)$ for $Q=N$ and $Q=N{\setminus } S$ for some $S\in \Psi _i^L$. By satellite marginality and Lemma 5.10, we have $\xi _i(N, u_S,L)=\xi _i(N,u_N,L)=1/n$ for all $i\in S{\setminus } \{j\}$. Since $u_S(N)=1$, efficiency implies $\xi _j(N,u_S,L)=(n-|S|+1)/n$. $\square$

Proof of Lemma 5.12

Take any $(N,u_S,L) \in {{\mathcal {G}}}_N^{L}$ with $S\in C^L(N)$ and $|E^L(S)|>1$. First consider any $i\in N{\setminus } S$ and note that $\mu _i^{(N, u_S,L)}(Q)=\mu _i^{(N,\mathbf{0 },L)}(Q)$ for $Q=N$ and $Q=N{\setminus } S$ for some $S\in \Psi _i^L$. From satellite marginality and Lemma 5.10, it follows that $\xi _i(N,u_S,L)=\xi _i(N,\mathbf{0 },L)=0$ for all $i\in N{\setminus } S$. Next, consider any $i\in S{\setminus } E^L(S)$. Note that $\mu _i^{(N, u_S,L)}(Q)=\mu _i^{(N,u_N,L)}(Q)$ for $Q=N$ and $Q=N{\setminus } S$ for some $S\in \Psi _i^L$. By satellite marginality and Lemma 5.10, this implies $\xi _i(N, u_S,L)=\xi _i(N,u_N,L)=1/n$ for all $i\in S{\setminus } E^L(S)$. Finally, consider any $i\in E^L(S)$ and let $S'={K^L_i(S)}$. Note that $\mu _i^{(N, u_S,L)}(Q)=\mu _i^{(N,u_{S'},L)}(Q)$ for $Q=N$ and $Q=N{\setminus } S$ for some $S\in \Psi _i^L$. By satellite marginality, this implies $\xi _i(N,u_{S},L)=\xi _i(N,u_{S'},L)$. By Lemma 5.11, we know that $\xi _i(N,u_{S'},L)=\frac{n-|S'|+1}{n}$ which means $\xi _i(N,u_{S},L)=\frac{n-|K^L_i(S)|+1}{n}$. $\square$

Proof of Lemma 5.13

Take any $(N,u_S,L) \in {{\mathcal {G}}}_N^{L}$ where $S\notin C^L(N)$. Let $S'=M^L_S$ and note that for any $i\in N$, it holds that $\mu _i^{(N, u_S,L)}(Q)=\mu _i^{(N,u_{S'},L)}(Q)$ for $Q=N$ and $Q=N{\setminus } S$ for some $S\in \Psi _i^L$. By satellite marginality, this implies $\xi _i(N,u_{S},L)=\xi _i(N,u_{S'},L)$ for all $i\in N$. Since $M^L_S\in C^L(N)$, by Lemma 5.12 we have $\xi _i(N,u_{S'},L)=0$ for all $i\in N{\setminus } S'$, $\xi _i(N,u_{S'},L)=\frac{1}{n}$ for all $i\in S'{\setminus } E^L(S')$, and $\xi _i(N,u_{S'},L)=\frac{n-|K^L_i(S')|+1}{n}$ for all $i\in E^L(S')$ which completes the proof. $\square$

Proof of Lemma 5.15

Take any $(N,\mathbf{0 },L)\in {{\mathcal {G}}}_N^{L}$ and consider any $i,j\in N$ with $\{i,j\}\in L$. Note that, $\mathbf{0 }(S)=0$ for all $S\in 2^N$ which means that $\mu _i^{(N,\mathbf{0 },L)}(S)=\mu _j^{(N,\mathbf{0 },L)}(Q)=0$ for all $S\in C^L(N{\setminus } \{j\})$ and for all $Q\in C^L(N{\setminus } \{i\})$. Hence, i and j are network symmetric players and by network symmetry, we have $\xi _i(N,\mathbf{0 },L)=\xi _j(N,\mathbf{0 },L)$. Since (N, L) is a connected undirected graph, all players in $(N,\mathbf{0 },L)$ are network symmetric players. Since $\mathbf{0 }(N)=0$ and $\xi$ : ${{\mathcal {G}}}_N^{L}\rightarrow {\mathbb {R}}^n$ satisfies efficiency, this implies $\xi _i(N,\mathbf{0 },L)=0$ for all $i\in N$. $\square$

Proof of Lemma 5.16

Take any $(N,u_S,L)\in {{\mathcal {G}}}_N^{L}$ with $S\in C^L(N)$. For all players $i\in N{\setminus } S$, it holds that $\mu _i^{(N, u_S,L)}(S)=\mu _i^{(N,\mathbf{0 },L)}(S)$ for all $S\in C^L(N)$. From network marginality and Lemma 5.15, it follows that $\xi _i(N,u_S,L)=\xi _i(N,\mathbf{0 },L)=0$ for all $i\in N{\setminus } S$. Now, consider any $i,j\in S$ such that $\{i,j\}\in L$. Note that $\mu _i^{(N,u_S,L)}(S)=\mu _j^{(N,u_S,L)}(Q)=0$ for all $S\in C^L(N{\setminus } \{j\})$ and for all $Q\in C^L(N{\setminus } \{i\})$ which means i and j are network symmetric players. By network symmetry, we have $\xi _i(N,u_S,L)=\xi _j(N,u_S,L)$. Since $S\in C^L(N)$, all players in S are satellite symmetric players. Since $u_S(N)=1$ and $\xi$ : ${{\mathcal {G}}}_N^{L}\rightarrow {\mathbb {R}}^n$ satisfies efficiency, this implies $\xi _i(N,u_S,L)=1/|S|$ for all $i\in S$. $\square$

Proof of Lemma 5.17

Take any $(N,u_S,L) \in {{\mathcal {G}}}_N^{L}$ where $S\notin C^L(N)$. Let $S'=M^L_S$ and note that for any $i\in N$, it holds that $\mu _i^{(N, u_S,L)}(Q)=\mu _i^{(N,u_{S'},L)}(Q)$ for all $Q\in C^L(N)$. By network marginality, this implies $\xi _i(N,u_{S},L)=\xi _i(N,u_{S'},L)$ for all $i\in N$. Since $M^L_S\in C^L(N)$, by Lemma 5.16 we have $\xi _i(N,u_{S'},L)=0$ for all $i\in N{\setminus } S'$ and $\xi _i(N,u_{S'},L)=\frac{1}{|S'|}$ for all $i\in S'$ which completes the proof. $\square$