A Class of Social-Shapley Values of Cooperative Games with Graph Structure

Abstract

This paper is devoted to a class of Social-Shapley values for cooperative games with graph structure. The Social-Shapley value compromises the utilitarianism of the Shapley value and the egalitarianism of the Solidarity value, in which the sociality is reflected by the Solidarity value. Through defining the corresponding properties in graph-restricted games, the paper axiomatically characterizes the Social-Shapley value when the coefficient is given exogenously. Moreover, we axiomatize the class of all possible Social-Shapley values in the graph-restricted games.

Appendix

1.1 Proof of Lemma 3

Given $i\in N$, let $\langle v,L\rangle =\mathcal {G}^0_N(i)\bigcap \mathcal {G}^1_N(i)$, then for each coalition $S\in C^L_{N \backslash \{i\}}$, we get $v^L(S\cup \{i\})=v^(S)$ and $\sum _{k\in S\cup \{i\}}[v^{L}(S\cup \{i\})-v^L(S\cup \{i\}\backslash \{k\})]=0$. Thus, the following equation holds,

$$\begin{aligned} \sum _{k\in S}[v^L(S\cup \{i\})-v^L(S\cup \{i\}\backslash \{k\})]=0. \end{aligned}$$

In this equation, let $S=\{j\}$, where $j\ne i$, we get $v^L(\{i,j\})=v^L(\{i\})$. By $\langle v,L\rangle \in \mathcal {G}^0_N(i)$, we deduce $v^L(\{i\})=0$ and $v^L(\{i,j\})=v^L(\{j\})$. Therefore, $v^L(\{j\})=v^L(\{i,j\})=v^L(\{i\})=0$.

Using mathematical induction on the cardinality of S, we can verify that $v^L(S)=v^L(S\cup \{i\})=0$ exists for each $S\in C^L_{N\backslash \{i\}}$, which means $\langle v,L\rangle$ is the null game, hence $\mathcal {G}^0_N(i)\bigcap \mathcal {G}^1_N(i)=\emptyset$.

Note that the dimension of the space $\mathcal {G}_N(i)$ equals $2^n-2$. According to $\mathcal {G}^0_N(i)\subset \mathcal {G}_N(i)$, $\mathcal {G}^1_N(i)\subset \mathcal {G}_N(i)$, it remains to prove that the dimensions of the linear space $\mathcal {G}^0_N(i)$ and $\mathcal {G}^1_N(i)$ are at least $2^{n-1}-1$, respectively. Define two classes of games as

$$\begin{aligned} \mathcal {B}^0_N(i)=\{\langle u_T,L\rangle |i\notin T, T\ne \emptyset \}, \ \ \ \mathcal {B}^1_N(i)=\{\langle w_T,L\rangle |i\notin T, T\ne \emptyset \}, \end{aligned}$$

where $\langle w_T,L\rangle$ is defined as in Sect. 3 and

$$\begin{aligned} u^L_T(S)=\left\{ \begin{array}{ll} 1, &{} S=T \text { or } S=T\cup \{i\}; \\ 0, &{} \text {otherwise.} \\ \end{array}\right. \end{aligned}$$

Clearly, both $\mathcal {B}^0_N(i)$ and $\mathcal {B}^1_N(i)$ are composed of a set of linearly independent games.

Besides, since $\mathcal {B}^0_N(i)\subset \mathcal {G}^0_N(i)$, $\mathcal {B}^1_N(i)\subset \mathcal {G}^1_N(i)$ and the cardinalities of $\mathcal {B}^0_N(i)$ and $\mathcal {B}^1_N(i)$ are respectively $2^{n-1}-1$, we can deduce the cardinalities of $\mathcal {G}^0_N(i)$ and $\mathcal {G}^1_N(i)$ are respectively $2^{n-1}-1$.    $\square$

1.2 Proof of Theorem 2

It is easy to check that $\varPhi ^\lambda (v,L)$ satisfies component efficiency, additivity, symmetry, rationality, proportionality of Shapley value and proportionality of Solidarity value, the uniqueness will be proved as follow:

Step 1. We will prove for each $i\in N$, there exists $\lambda _i\in \mathbb {R}$, so that the following equality establish, $\langle v,L\rangle \in \mathcal {G}^L_N$,

$$\begin{aligned} \varPhi _i(v,L)=\lambda _i Sh_i(v,L)+(1-\lambda _i)Sol_i(v,L). \end{aligned}$$

(3)

Choose any game $\langle w,L\rangle \in \mathcal {B}^1_N(i)\subset \mathcal {G}^1_N(i)$ such that $Sh_i(w,L)>0$, let $\lambda _i=\frac{\varPhi _i(w,L)}{Sh_i(w,L)}$, by proportionality of Shapley value, for all $\langle v,L\rangle \in \mathcal {G}^1_N(i)$, we have

$$\begin{aligned} \varPhi _i(v,L)=\lambda _i Sh_i(v,L). \end{aligned}$$

Similarly, choose any game $\langle u,L\rangle \in \mathcal {B}^0_N(i)\subset \mathcal {G}^0_N(i)$ such that $Sol_i(u,L)>0$, let $\gamma _i=\frac{\varPhi _i(u,L)}{Sol_i(u,L)}$, combining proportionality of Solidarity value, for each $\langle v,L\rangle \in \mathcal {G}^0_N(i)$, the following equality holds:

$$\begin{aligned} \varPhi _i(v,L)=\gamma _i Sol_i(v,L). \end{aligned}$$

According to the definitions of $\mathcal {G}^0_N(i)$ and $\mathcal {G}^1_N(i)$, given $\langle v,L\rangle \in \mathcal {G}^0_N(i)$, we have

$$\begin{aligned} \varPhi _i(v,L)=\lambda _i Sh_i(v,L)+\gamma _i Sol_i(v,L). \end{aligned}$$

(4)

Considering game $\langle w,L\rangle \in \mathcal {G}^L_N$, where

$$\begin{aligned} w^L(S)=\left\{ \begin{array}{ll} n, &{} S=N; \\ 0, &{} \text {otherwise,} \\ \end{array}\right. \end{aligned}$$

by component efficiency and symmetry, it follows $\varPhi _i(w,L)=1$, $i\in N$. Similarly, it follows that

$$\begin{aligned} Sh_i(w,L)=Sol_i(w,L)=1, \ i\in N. \end{aligned}$$

In the Eq. (4), let $\langle w,L\rangle =\langle v,L\rangle$, we get $\gamma _i=1-\lambda _i$. Consequently, the Eq. (3) holds for any $\langle v,L\rangle \in \mathcal {G}_N(i)$.

Now, defining a game $\langle u,L\rangle \in \mathcal {G}^L_N$ as follow:

$$\begin{aligned} u^L(S)=\left\{ \begin{array}{ll} 0, &{} S=\emptyset ; \\ 1, &{} S\ne \emptyset . \\ \end{array} \right. \end{aligned}$$

Obviously $\langle u,L\rangle \notin \mathcal {G}_N(i)$.

Note that as linear space, the dimensions of $\mathcal {G}^L_N$ and $\mathcal {G}_N(i)$ are $2^n-1$ and $2^n-2$ respectively, hence, any $\langle v,L\rangle \in \mathcal {G}^L_N$ can be represented as follow

$$\begin{aligned} v=v^0+cu, \end{aligned}$$

where $\langle v^0,L\rangle \in \mathcal {G}_N(i)$ and c is a constant coefficient. According to component efficiency, additivity, symmetry, it follows that

$$\begin{aligned} \varPhi _i(v,L)=\varPhi _i(v^0,L)+\varPhi _i(cu,L)=\lambda _i Sh_i(v^0,L)+(1-\lambda _i)Sol_i(v^0,L)+\frac{c}{n}. \end{aligned}$$

In addition, we have

$$\begin{aligned} \lambda _i Sh_i(v,L)+(1-\lambda _i)Sol_i(v,L)=\lambda _i Sh_i(v^0,L)+(1-\lambda _i)Sol_i(v^0,L)+\frac{c}{n}. \end{aligned}$$

Consequently, Eq. (3) holds for any $\langle v,L\rangle \in \mathcal {G}^L_N$.

Step 2. We will prove that coefficient $\lambda _i$ in Eq. (3) is independent of i. Assuming that $n\ge 2$, for $\langle v,L\rangle \in \mathcal {G}^L_N$, component efficiency implies:

$$\begin{aligned} \sum _{i\in N}[\lambda _i Sh_i(v,L)+(1-\lambda _i)Sol_i(v,L)]=v^L(N). \end{aligned}$$

By component efficiency of Solidarity value, this equation is equivalent to

$$\begin{aligned} \sum _{i\in N}\lambda _i[Sh_i(v,L)-Sol_i(v,L)]=0. \end{aligned}$$

By component efficiency of the Shapley and Solidarity value, we have

$$\begin{aligned}&\sum _{i=2}^n(\lambda _i-\lambda _1)[Sh_i(v,L)-Sol_i(v,L)] \\= & {} \sum _{i\in N}\lambda _i[Sh_i(v,L)-Sol_i(v,L)]-\sum _{i\in N}\lambda _1[Sh_i(v,L)-Sol_i(v,L)] \\= & {} 0. \end{aligned}$$

This means for $\langle v,L\rangle \in \mathcal {G}^L_N$,

$$\begin{aligned} \sum _{i=2}^n\gamma _i[Sh_i(v,L)-Sol_i(v,L)]=0, \end{aligned}$$

(5)

where $\gamma _i=\lambda _i-\lambda _1$, $i=2,3,\ldots ,n$.

Defining games $\langle v_k,L\rangle$, $k=2,3,\ldots ,n$, as follow

$$\begin{aligned} v^L_k(S)=\left\{ \begin{array}{ll} 1, &{} S=N\backslash \{k\}; \\ 0, &{} \text { otherwise.} \\ \end{array} \right. \end{aligned}$$

According to Eq. (5),

$$\begin{aligned} \sum _{i=2}^n\gamma _i[Sh_i(v_k,L)-Sol_i(v_k,L)]=0, \ k=2,\ldots ,n \end{aligned}$$

and

$$\begin{aligned} Sh_i(v_k,L)=\left\{ \begin{array}{ll} -\frac{1}{n}, &{} \ i=k; \\ \frac{1}{n(n-1)}, &{} \ i\ne k, \\ \end{array} \right. , \ \ Sol_i(v_k,L)=\left\{ \begin{array}{ll} -\frac{1}{n^2}, &{} \ i=k; \\ \frac{1}{n^2(n-1)}, &{} \ i\ne k, \\ \end{array} \right. \end{aligned}$$

we get the linear system

$$\begin{aligned} (n-1)\gamma _k-\sum _{i\ne k}\gamma _i-0, \ k=2,3, \ldots , n. \end{aligned}$$

The only solution of the above linear system is $\gamma _2=\gamma _3=\cdots ,\gamma _n=0$.

Then for every $i\in N$, it implies $\lambda _i-\lambda _1=0$, $i=2,3,\ldots , n$. Now we have proved that coefficient $\lambda _i$ is independent of i and $\lambda _i=\lambda _1=\lambda$, $i=2,3,\ldots , n$.

Step 3. The coefficient $\lambda$ will be proved belonging to interval [0, 1] in this step.

Assuming that $n\ge 2$, for any player $i\in N$, considering the following unanimity games:

$$\begin{aligned} u^L_{N\backslash \{i\}}(S)=\left\{ \begin{array}{ll} 1, &{} \ S=N \text { or } S=N\backslash \{i\}; \\ 0, &{} \ \text {otherwise.} \\ \end{array} \right. \end{aligned}$$

By applying the property of rationality to $\langle u_{N\backslash \{i\}},L\rangle$, we obtain $\varPhi _i(u_{N\backslash \{i\}},L)\ge 0$. In addition, combining the following equations:

$$\begin{aligned} Sh_i(u_{N\backslash \{i\}},L)=0, \ \ Sol_i(u_{N\backslash \{i\}},L)=\frac{n-1}{n^2}, \ \lambda _i=\lambda , \end{aligned}$$

we have

$$\begin{aligned} 0\le \varPhi _i(u_{N\backslash \{i\}},L)=\lambda Sh_i(u_{N\backslash \{i\}},L)+(1-\lambda )Sol_i(u_{N\backslash \{i\}},L)=\frac{(1-\lambda )(n-1)}{n^2}. \end{aligned}$$

Evidently, we deduce that $\lambda \le 1$.

Considering the game $\langle \bar{w},L\rangle$, which is defined as follow,

$$\begin{aligned} \bar{w}^L(S)=\left\{ \begin{array}{ll} -1, &{} \ S=N; \\ -n, &{} \ S=N\backslash \{i\}; \\ 0, &{} \ \text {otherwise.} \\ \end{array} \right. \end{aligned}$$

Applying the property of rationality to $\langle \bar{w},L\rangle$, we get $\varPhi _i(\bar{w},L)\ge 0$. By the conditions:

$$\begin{aligned} Sh_i(\bar{w},L)=\frac{n-1}{n}, \ \ Sol_i(\bar{w},L)=0, \end{aligned}$$

we have

$$\begin{aligned} 0\le \varPhi _i(\bar{w},L)=\lambda Sh_i(\bar{w},L)+(1-\lambda )Sol_i(\bar{w},L)=\frac{\lambda (n-1)}{n}, \end{aligned}$$

which implies $\lambda \ge 0$. Then for $n\ge 2$, we have proven the conclusion. When $n=1$, the theorem is obviously established.    $\square$