Weighted values and the core in NTU games

Abstract

Monderer et al. (Int J Game Theory 21(1):27–39, 1992) proved that the core is included in the set of the weighted Shapley values in TU games. The purpose of this paper is to extend this result to NTU games. We first show that the core is included in the closure of the positively weighted egalitarian solutions introduced by Kalai and Samet (Econometrica 53(2):307–327, 1985). Next, we show that the weighted version of the Shapley NTU value by Shapley (La Decision, aggregation et dynamique des ordres de preference, Editions du Centre National de la Recherche Scientifique, Paris, pp 251–263, 1969) does not always include the core. These results indicate that, in view of the relationship to the core, the egalitarian solution is a more desirable extension of the weighted Shapley value to NTU games. As a byproduct of our approach, we also clarify the relationship between the core and marginal contributions in NTU games. We show that, if the attainable payoff for the grand coalition is represented as a closed-half space, then any element of the core is attainable as the expected value of marginal contributions.

Appendix

Proof of Claim 1

Let $V\in \mathcal {G}$, $S\subseteq N$, $|S|\ge 2$, $j\in S$ and $z\in \mathbb {R}^{S\backslash \{j\}}$. We define $Y\subseteq \mathbb {R}$ by

$$\begin{aligned} Y=\left\{ r\in \mathbb {R}: \left( (z_k)_{k\in S\backslash \{j\}}, r\right) \in V(S)\right\} \end{aligned}$$

By N2 (closed), Y is closed. Thus, to prove Claim 1, it suffices to prove that Y is bounded from above and non-empty.

Let $x\in \partial V(S)$. By N2 (convex), there exists a normal vector $\lambda$ such that

$$\begin{aligned} V(S)\subseteq \{y\in \mathbb {R}^S: y\cdot \lambda \le x\cdot \lambda \}. \end{aligned}$$

Moreover, by N3, there exists $\delta >0$ such that $\lambda _j\ge \delta$. It follows that Y is bounded from above.

We prove $Y\ne \emptyset$, i.e., there exists $r \in \mathbb {R}$ such that $\left( (z_k)_{k\in S\backslash \{j\}}, r\right) \in V(S)$. Assume, by way of contradiction, that for any $r\in \mathbb {R}$, $\left( (z_k)_{k\in S\backslash \{j\}}, r\right) \notin V(S)$. We define $Z\subseteq \mathbb {R}^S$ by

$$\begin{aligned} Z=\left\{ \left( (z_k)_{k\in S\backslash \{j\}}, r\right) : r\in \mathbb {R}\right\} . \end{aligned}$$

The two sets V(S) and Z are convex, and the intersection between the two sets is empty. By the separation theorem, there exists $p\in \mathbb {R}^S$, $p\ne \mathbf {0}$, such that

$$\begin{aligned} p\cdot x \le p\cdot y \quad \text {for all } x\in V(S), y\in Z. \end{aligned}$$

(16)

Suppose that $p_j>0$. Then, by taking an arbitrary sequence $\{y^k\}_{k=1}^\infty \subseteq Z$ such that $y_j^k \rightarrow -\infty$, we have $p\cdot y^k\rightarrow -\infty$, which contradicts Eq. (16). Similarly, if we suppose that $p_j<0$, we have a contradiction with Eq. (16). As a result, $p_j=0$.

Suppose to the contrary that $p_i<0$ for some $i\in S\backslash \{j\}$. Consider the sequence $\{x^l\}_{l=1}^\infty \subseteq V(S)$ such that $x_i^l\rightarrow -\infty$ and $x_h^l=x_h^{l+1}$ for all $h\in S$, $h\ne i$, $l=1,2,\ldots$. By N2 (comprehensive), such a sequence always exists. Then, $p\cdot x^l\rightarrow +\infty$, which contradicts Eq. (16). It follows that $p\ge \mathbf {0}$. Since $p\ne \mathbf {0}$, there exists a player $i'\in S\backslash \{j\}$ such that $p_{i'}>0$.

Let $x\in \partial V(S)$ be arbitrarily given. For any $m \in \mathbb {N}$, let $\tilde{x}^m$ denote the following vector:

$$\begin{aligned} \tilde{x}^m_j=x_j-m, \tilde{x}^m_i=x_i \quad \text {for all } i\in S\backslash \{j\}. \end{aligned}$$

By N2 (comprehensive), $\tilde{x}^m \in V(S)$. Then, for any $m\in \mathbb {N}$, there exists $x^m$ such that⁶

$$\begin{aligned} x^m_j&=\tilde{x}_j^m, \\ x^m_h&=x_h \quad \text {for all } h\in S, h\ne i', h\ne j, \\ x^m&\in \partial V(S). \end{aligned}$$

Consider the sequence $\{x^m\}_{m=1}^\infty \subseteq \partial V(S)$. Since $x^m_j\rightarrow -\infty$, by N4, we have $x^m_{i'} \rightarrow +\infty$. Since $p_j=0$ and $p_{i'}>0$, we have $p\cdot x^m\rightarrow +\infty$, which contradicts Eq. (16). $\square$

Proof of Claim 2

The proof of $\Omega (v)\subseteq cl \mathcal {W}(v)$ is given by Theorem 4 of Kalai and Samet (1987). We prove the converse set-inclusion. Let $x\in cl \mathcal {W}(v)$. Then, there exists a sequence $\{w^k\}_{k=1}^\infty \subseteq \varDelta ^N_{++}$ such that

$$\begin{aligned} x=\lim _{k\rightarrow \infty }\phi ^{w^k}(v). \end{aligned}$$

Since $\{w^k\}_{k=1}^\infty \subseteq \varDelta ^N_{++}$ is a bounded sequence, there exists a convergent subsequence. Assume, for notational simplicity, that $\{w^k\}_{k=1}^\infty$ itself converges. Let $w^*$ denote the limit point.

Suppose that $w^*_i>0$ for all $i\in N$. Then, by letting $\omega =(\{N\}, w^*)$, we have

$$\begin{aligned} \lim _{k\rightarrow \infty }\phi ^{w^k}(v)=\phi ^\omega (v), \end{aligned}$$

which implies that $x\in \Omega (v)$.

Suppose that $w^*_i=0$ for some $i\in N$. We define the ordered partition $\{T^r, T^{r-1}, \ldots , T^1\}$ of N inductively as follows:⁷

$$\begin{aligned} T^1=\{i\in N: \lim _{k\rightarrow \infty }w^k_i>0\}, \end{aligned}$$

and, for each $q=2, \ldots , r$, given that $T^s$, $1\le s\le q-1$, is determined, we define

$$\begin{aligned} T^q=\left\{ i\in N\backslash \cup _{s=1}^{q-1} T^s: \lim _{k\rightarrow \infty } \frac{w_i^k}{\sum _{j\in N\backslash \cup _{s=1}^{q-1} T^s}w_j^k}>0\right\} . \end{aligned}$$

We remark that, for each $q=2, \ldots , r$, and $i\in N\backslash \cup _{s=1}^{q-1} T^s$, the sequence

$$\begin{aligned} \frac{w_i^k}{\sum _{j\in N\backslash \cup _{s=1}^{q-1} T^s}w_j^k}, \text { } k=1, 2, \ldots , \end{aligned}$$

is bounded. So, if the above sequence does not converge, we can choose a convergent subsequence. For simplicity we assume that the above sequence always converges. Since N and r are finite, the above inductive definition is well-defined. For each $q=1, \ldots , r$, we define $w^{T^q}\in \varDelta ^{T^q}_{++}$ by

$$\begin{aligned} w^{T^q}_i=\lim _{k\rightarrow \infty } \frac{w_i^k}{\sum _{j\in N\backslash \cup _{s=1}^{q-1} T^s}w_j^k} \quad \text {for all } i\in T^q. \end{aligned}$$

We define the generalized weight vector $\omega$ by

$$\begin{aligned} \omega =(T^r, \ldots , T^1, w^{T^r}, \ldots , w^{T^1}). \end{aligned}$$

We prove that, for any $i\in T^q$, $q=1, \ldots , r$,

$$\begin{aligned} \lim _{k\rightarrow \infty }\phi ^{w^k}_i(v)=\phi ^\omega _i(v). \end{aligned}$$

Choose an arbitrary $q\in \{1, \ldots , r\}$ and $i\in T^q$. Let $S\subseteq N$, $S\ne \emptyset$. If $i\notin S$, then $\lim _{k\rightarrow \infty }\phi ^{w^k}_i(u_S)=0$. If $i\in S$, set

$$\begin{aligned} q'=\min \left\{ s\in \{1, \ldots , q\}: T^s\cap S\ne \emptyset \right\} . \end{aligned}$$

If $q'<q$, then

$$\begin{aligned} \lim _{k\rightarrow \infty }\phi ^{w^k}_i(u_S)=\lim _{k\rightarrow \infty } \frac{\frac{w^k_i}{\sum _{j\in \cup _{s=q'}^{r} T^s}w^k_j}}{\frac{\sum _{j\in S}w^k_j}{\sum _{j\in \cup _{s=q'}^{r} T^s}w^k_j}}=0. \end{aligned}$$

On the other hand, if $q'=q$, then

$$\begin{aligned} \lim _{k\rightarrow \infty }\phi ^{w^k}_i(u_S)=\lim _{k\rightarrow \infty } \frac{\frac{w^k_i}{\sum _{j\in \cup _{s=q'}^{r} T^s}w^k_j}}{\frac{\sum _{j\in S}w^k_j}{\sum _{j\in \cup _{s=q'}^{r} T^s}w^k_j}}=\frac{w^{T^q}_i}{\sum _{j\in S\cap T^q}w^{T^q}_j}. \end{aligned}$$

Since S is arbitrarily chosen, together with the facts that the set of the unanimity games forms a basis of the set of TU games and $\phi ^w$, $\phi ^\omega$ are linear, we obtain

$$\begin{aligned} \lim _{k\rightarrow \infty }\phi ^{w^k}_i(v)=\phi ^\omega _i(v). \end{aligned}$$

Since q and $i\in T^q$ are arbitrarily chosen, the proof completes. $\square$