The allocation of marginal surplus for cooperative games with transferable utility

Abstract

Marginal contribution is a significant index to measure every player’s ability to cooperate in cooperative games. Several solutions for cooperative games are defined in terms of marginal contribution, including the Shapley value and the Solidarity value. In this paper, we introduce marginal surplus as an alternative index to describe the contribution level of every player. We define a new solution for cooperative games, namely the average-surplus value, which is determined by an underlying procedure of sharing marginal surplus. Then we characterize the average-surplus value by introducing the A-null surplus player property and the revised balanced contributions property. We also propose the AS-potential function to implement the average-surplus value. Finally, we provide a non-cooperative game, the outcome of which coincides with the average-surplus value in subgame perfect equilibria.

Appendix

Proof of Theorem 3

It is sufficient to prove that $A{\tilde{P}}$ satisfies efficiency, symmetry, additivity and the A-null surplus player property.

• Efficiency: By Eq. (5.1) and Eq.(5.6), we have

  $$\begin{aligned} \sum _{i\in N}A_i{\tilde{P}}(N,v)&= \sum _{i\in N}\left[ D_i{\tilde{P}}(N,v)+v(i)+\frac{1}{n}(v(N\backslash i)-\sum _{j\in N\backslash i}v(j))\right] \\&= v(N)-\frac{1}{n}\sum _{i\in N}(v(N\backslash i)+v(i))+\sum _{i\in N}v(i)\\&\qquad +\frac{1}{n}\sum _{i\in N}(v(N\backslash i)-\sum _{j\in N\backslash i}v(j))\\&= v(N). \end{aligned}$$

• Additivity: It is easy to check that the AS-potential function ${\tilde{P}}$ satisfies additivity by Eq. (5.5) , that is, ${\tilde{P}}(N,v+w)={\tilde{P}}(N,v)+{\tilde{P}}(N,w)$ for all $\langle N,v\rangle , \langle N,w\rangle \in {\mathcal {G}}^N$. Thus, $D{\tilde{P}}(N,v)$ and $A{\tilde{P}}(N,v)$ also satisfy additivity by Eqs. (5.2) and (5.6).

• Symmetry: Let i and j be symmetric players. Firstly, we prove that ${\tilde{P}}(S\backslash i, v|_{S\backslash i})={\tilde{P}}(S\backslash j, v|_{S\backslash j})$ for all $S\subseteq N, S\ni i,j$ by induction. It is obvious that ${\tilde{P}}(i, v|_i)=0={\tilde{P}}(j, v|_j)$ for $S=\{i,j\}$. Suppose that the conclusion holds for $2\le |S|< n$. Then, for $|S|=n$, we have

  $$\begin{aligned}&{\tilde{P}}(N\backslash i, v|_{N\backslash i})-{\tilde{P}}(N\backslash j, v|_{N\backslash j}) \\&\quad = \frac{1}{n-1}\left[ \sum _{k\in N\backslash i}{\tilde{P}}(N \backslash \{i,k\}, v|_{N \backslash \{i,k\}})+v(N\backslash i) -\frac{1}{n-1}\sum _{k\in N\backslash i}(v(N \backslash \{i,k\})+v(k))\right] \\&\qquad -\frac{1}{n-1}\left[ \sum _{k\in N\backslash j}{\tilde{P}}(N \backslash \{j,k\}, v|_{N \backslash \{j,k\}})+v(N\backslash j) -\frac{1}{n-1}\sum _{k\in N\backslash j}(v(N \backslash \{j,k\})+v(k))\right] \\&\quad = \frac{1}{n-1}\left[ \sum _{k\in N\backslash i}{\tilde{P}}(N \backslash \{i,k\}, v|_{N \backslash \{i,k\}}) - \sum _{k\in N\backslash j}{\tilde{P}}(N \backslash \{j,k\}, v|_{N \backslash \{j,k\}})\right] \\&\quad = 0. \end{aligned}$$

  Since $v(N\backslash i)=v(N\backslash j)$ and $v(i)=v(j)$, then symmetry of $A{\tilde{P}}$ is verified as follows,

  $$\begin{aligned} A_i{\tilde{P}}(N,v)-A_j{\tilde{P}}(N,v)&= D_i{\tilde{P}}(N,v)+v(i)+\frac{1}{n}(v(N\backslash i)-\sum _{k\in N\backslash i}v(k))\\&\quad -\left[ D_j{\tilde{P}}(N,v)+v(j)+\frac{1}{n}(v(N\backslash j)-\sum _{k\in N\backslash j}v(k))\right] \\&= {\tilde{P}}(N\backslash i, v|_{N\backslash i})-{\tilde{P}}(N\backslash j, v|_{N\backslash j})\\&= 0. \end{aligned}$$

• The A-null surplus player property: Let i be an A-null surplus player in $\langle N,v\rangle$, then $v(S)-\frac{1}{s}\sum _{j\in S}(v(S\backslash j)+v(j))=0$ for all $S\subseteq N$ and $S\ni i$. Obviously, i is also an A-null surplus player in all subgames $\langle S, v|_S\rangle$. Next, we prove that $A_i{\tilde{P}}(S,v|_S)=v(i)$ for all $\langle S, v|_S\rangle$ by induction, in particular, $A_i{\tilde{P}}(N,v)=v(i)$. It is trivial that $A_i{\tilde{P}}(i,v|_i)=v(i)$ with $S=\{i\}$. Suppose that $A_i{\tilde{P}}(S,v|_S)=v(i)$ for all subgames with $|S|\le n-1$, that is, for all $j\in N\backslash i$,

  $$\begin{aligned} D_i{\tilde{P}}(N\backslash j,v|_{N\backslash j})+v(i)+\frac{1}{n-1} (v(N\backslash \{i,j\})-\sum _{k\in N\backslash \{i,j\}}v(k))=v(i). \end{aligned}$$

  Thus, by Eq. (5.4), we have

  $$\begin{aligned}&nA_i{\tilde{P}}(N,v)\\&\quad = n{\tilde{P}}(N,v)-n{\tilde{P}}(N\backslash i, v|_{N\backslash i})+nv(i)+ v(N\backslash i)-\sum _{j\in N\backslash i}v(j) \\&\quad = \left[ \sum _{j\in N}{\tilde{P}}(N\backslash j, v|_{N\backslash j}) +v(N)-\frac{1}{n}\sum _{j\in N}(v(N\backslash j)+v(j))\right] \\&\qquad - \left[ \sum _{j\in N\backslash i}{\tilde{P}}(N\backslash \{i,j\}, v|_{N\backslash \{i,j\}}) +v(N\backslash i)-\frac{1}{n-1}\sum _{j\in N\backslash i} (v(N\backslash \{i,j\})+v(j))\right] \\&\qquad -{\tilde{P}}(N\backslash i, v|_{N\backslash i})+nv(i)+ v(N\backslash i)-\sum _{j\in N\backslash i}v(j)\\&\quad = \sum _{j\in N\backslash i}D_i{\tilde{P}}(N\backslash j,v|_{N\backslash j})+ \frac{1}{n-1}\sum _{j\in N\backslash i}v(N\backslash \{i,j\}) +nv(i)-\frac{n-2}{n-1}\sum _{j\in N\backslash i}v(j) \\&\quad = \sum _{j\in N\backslash i}\left[ D_i{\tilde{P}}(N\backslash j,v|_{N\backslash j})+v(i)+\frac{1}{n-1} (v(N\backslash \{i,j\})-\sum _{k\in N\backslash \{i,j\}}v(k))\right] +v(i)\\&\quad = nv(i). \end{aligned}$$

$\square$

Proof of Theorem 4

The theorem is proved by induction on the number k of players. Obviously, the result holds for $k=1$, since $AS_i(\{i\},v)=v(i)$ for every one-person TU-game $\langle \{i\},v\rangle$. Suppose that the result holds for all $k<n$, then we show that it also holds for $k=n$.

Let $N=\{1,2,\ldots ,n\}$. We now show that the average-surplus value is indeed a SPE outcome by considering the following strategies.

• At stage 1, each player, $i\in N$, makes bids $b_j^i = AS_j(N,v)- AS_j(N\backslash i,v|_{N\backslash i})-\frac{1}{n}(v(N)-v(N\backslash i)-v(i))$ for every player $j\in N\backslash i$.

• At stage 2, The proposer $\alpha$ makes offers $x_j^{\alpha }= AS_j(N\backslash \alpha ,v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$ to every player $j\in N\backslash \alpha$.

• At stage 3, each player $j\in N\backslash \alpha$ will accept the offer if $x_j^{\alpha }\ge AS_j(N\backslash \alpha ,v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$, otherwise the offer is rejected.

It is obvious that the outcome of the strategies is the average-surplus value. We will verify that the strategies constitute a SPE. At stage 3, each player $j\in N\backslash \alpha$ can obtain $AS_j(N\backslash \alpha ,v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$ by the induction hypothesis, and the proposer $\alpha$ just receives $v(\alpha )-\sum _{j\in N\backslash \alpha } \frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$ in the case of rejection. Thus, the strategies are best responses at stage 3 and stage 2 as long as $v(N)-v(N\backslash \alpha )\ge v(\alpha )$ for all superadditive TU-games, since

$$\begin{aligned}&v(N)-\sum _{j\in N\backslash \alpha }\left[ AS_j(N\backslash \alpha ,v|_{N\backslash \alpha }) +\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))\right] \\&\quad \ge v(\alpha )-\sum _{j\in N\backslash \alpha } \frac{1}{n}(v(N)-v(N\backslash \alpha ) -v(\alpha )) \\&\qquad \Leftrightarrow \ \ v(N)-v(N\backslash \alpha )\ge v(\alpha ). \end{aligned}$$

At stage 1, by the revised balanced contributions property, for all $i\in N$, we have

$$\begin{aligned} B^i&= \sum _{j\in N\backslash i}(b_j^i-b_i^j) \\&= \sum _{j\in N\backslash i}[AS_j(N,v)-AS_j(N\backslash i,v|_{N\backslash i}) -\frac{1}{n}(v(N)-v(N\backslash i)-v(i))\\&\quad -(AS_i(N,v)-AS_i(N\backslash j,v|_{N\backslash j}) -\frac{1}{n}(v(N)-v(N\backslash j)-v(j)))]\\&= 0. \end{aligned}$$

Therefore, if a player, $i\in N$, increases his total bid $\sum _{j\in N\backslash i} b_j^i$, he will become the proposer, but his payoff will decrease. If a player, $i\in N$, decreases his total bid $\sum _{j\in N\backslash i} b_j^i$, then his payoff is invariable since other player will be chosen as the proposer. Thus, the strategy is a best response at stage 1. Hence, the above strategies constitute a SPE.

We now prove that any SPE yields the average-surplus value by a series of claims.

Claim (a):

In any SPE, at stage 3, every player $j\in N\backslash \alpha$ will accept the offer if $x_j^\alpha > AS_j(N\backslash \alpha ,v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$. The offer is rejected if there exists at least one player $j \in N\backslash \alpha$ such that $x_j^\alpha < AS_j(N\backslash \alpha ,v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$.

Note that in the case of rejection at stage 3, the payoff of each player, $j\in N\backslash \alpha$, is $AS_j (N\backslash \alpha , v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$ by the induction hypothesis. Thus, a player, $j\in N\backslash \alpha$, will accept the offer if $x_j^\alpha > AS_j (N\backslash \alpha , v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$ since he can improve his payoff, and he will reject the offer if $x_j^\alpha < AS_j (N\backslash \alpha , v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$. Therefore, claim (a) is proved by using the same argument for all $j\in N\backslash \alpha$.

Claim (b):

If $v(N)> v(N\backslash \alpha ) +v(\alpha )$, the SPE strategies starting from stage 2 are as follows. At stage 2, the proposer $\alpha$ offers $x_j^\alpha =AS_j(N\backslash \alpha ,v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$ to each player $j\in N\backslash \alpha$; at stage 3, each player, $j\in N \backslash \alpha$, rejects any offer $x_j^\alpha < AS_j(N\backslash \alpha ,v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$ and accepts the offer otherwise. If $v(N)= v(N\backslash \alpha ) +v(\alpha )$, there exist SPE strategies besides the previous SPE strategies. At stage 2, the proposer $\alpha$ offers $x_j^\alpha \le AS_j(N\backslash \alpha ,v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$ to each player $j\in N\backslash \alpha$; at stage 3, each player, $j\in N \backslash \alpha$, rejects any offer $x_j^\alpha \le AS_j(N\backslash \alpha ,v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$ and accepts the offer otherwise.

We verify that the previous strategies constitute a SPE. Suppose that $v(N)> v(N\backslash \alpha ) +v(\alpha )$. In that case, the offer made by the proposer $\alpha$ is rejected and then the proposer $\alpha$ obtains $v(\alpha )-\sum _{j\in N\backslash \alpha }\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$, which cannot be part of a SPE, since the proposer $\alpha$ can improve his payoff by offering $AS_j(N\backslash \alpha ,v|_{N\backslash \alpha }) +\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))+\varepsilon / (n-1)$ to each player $j\in N\backslash \alpha$ with $0<\varepsilon < v(N)- v(N\backslash \alpha ) -v(\alpha )$ so that the offer is accepted by claim (a). Therefore, it implies that $x_j^\alpha \ge AS_j(N\backslash \alpha ,v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$ for all $j\in N\backslash \alpha$ in any SPE. However, an offer with $x_j^\alpha > AS_j(N\backslash \alpha ,v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$ for some $j\in N\backslash \alpha$ cannot be part of a SPE. The reason is that the proposer $\alpha$ can improve his payoff by offering $AS_j(N\backslash \alpha ,v|_{N\backslash \alpha }) +\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))+\varepsilon / (n-1)$ to each player $j\in N\backslash \alpha$ with $\varepsilon < x_j^\alpha - AS_j(N\backslash \alpha ,v|_{N\backslash \alpha })-\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$ and $\varepsilon > 0$. Hence, $x_j^\alpha =AS_j(N\backslash \alpha ,v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$ for all $j\in N\backslash \alpha$, and acceptation of the offer implies that each player $j\in N \backslash \alpha$ accepts any offer $x_j^\alpha \ge AS_j(N\backslash \alpha ,v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$.

If $v(N) = v(N\backslash \alpha ) +v(\alpha )$, the proposer $\alpha$ offers at least $AS_j(N\backslash \alpha ,v|_{N\backslash \alpha })+\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$ to each player $j\in N\backslash \alpha$ so that the offer can be accepted by the same argument in the previous case. The proposer gets $v(\alpha )-\sum _{j\in N\backslash \alpha }\frac{1}{n}(v(N)-v(N\backslash \alpha )-v(\alpha ))$ in the case of rejection, which is identity to the payoff in the case of acceptation. Therefore, any offer that leads to a rejection also is a SPE.

Claim (c):

In any SPE, the net bid $B^i=0$ for all $i\in N$.

Let $\Lambda =\{i\in N|B^i=\max _{j\in N}B^j\}$. If $\Lambda = N$, the net bid $B^i=0$ for all $i\in N$ due to $\sum _{i\in N}B^i=0$. Otherwise, for all $j\in \Lambda$, he can improve his expected payoff by slightly changing his bids without altering the set $\Lambda$. Let $j\notin \Lambda$ and $i\in \Lambda$. Player i changes his strategy by making bids $b^{'i}_k=b_k^i+\delta$ for all $k\in \Lambda \backslash i$, $b_j^{'i}=b^i_j-|\Lambda |\delta$, and $b^{'i}_l=b^i_l$ for all $l\notin \Lambda$ and $l\ne j$. Then the net bids are $B^{'k}=B^k-\delta$ for all $k\in \Lambda$; $B^{'j}=B^j+|\Lambda |\delta$; $B^{'l}=B^l$ for all $l\notin \Lambda$ and $l\ne j$. Because $B^l<B^i$ for all $l\notin \Lambda$, there must exist $\delta >0$ such that $B^j+|\Lambda |\delta <B^i-\delta$ and $B^{'l}<B^{'i}=B^{'k}$ for all $k\in \Lambda$. Therefore, $\Lambda$ remains unchanged, but player i’s expected payoff increases.

Claim (d):

In any SPE, the payoff of every player is invariable whoever is chosen as the proposer.

The net bids of all players are the same by claim (c). Then every player would not strictly prefer to be the proposer. Otherwise, he has to enhance his bids, which will result in a decrease of his payoff. On the other hand, if a player prefers that the proposer is one of the other players, he needs to decrease his bids, which makes no difference to his payoff.

Claim (e):

In any SPE, the final payoff of every player coincides with the average-surplus value.

Firstly, if a player, $i\in N$, is the proposer, his final payoff is $y_i^i=v(N)-v(N\backslash i)- \sum _{j\in N\backslash i}(b_j^i+\frac{1}{n}(v(N)-v(N\backslash i)-v(i)))$. Then, if a player, $j\in N\backslash i$, is the proposer, player i’s final payoff is $y_i^j=AS_i(N\backslash j, v|_{N\backslash j})+\frac{1}{n}(v(N)-v(N\backslash j)-v(j))+b_i^j$. Therefore, the sum of player i’s payoff over all possible choices is as follows,

$$\begin{aligned} \sum _{j\in N}y_i^j&= v(N)-v(N\backslash i)- \sum _{j\in N\backslash i}\left( b_j^i+\frac{1}{n}(v(N)-v(N\backslash i)-v(i))\right) \\&\quad +\sum _{j\in N \backslash i}(AS_i(N\backslash j, v|_{N\backslash j}) +\frac{1}{n}(v(N)-v(N\backslash j)-v(j))+b_i^j) \\&= v(N)-\frac{1}{n}\sum _{j\in N}(v(N\backslash j)+v(j))+v(i) +\sum _{j\in N\backslash i}AS_i(N\backslash j, v|_{N\backslash j})\\&= nAS_i(N,v), \end{aligned}$$

where the last equality holds by Corollary 2. By claim (d), we have $y_i^j=y_i^k$ for all $j,k\in N$, and then we conclude that $y_i^j=AS_i(N,v)$ for all $j \in N$. Hence, the final payoff of every player coincides with the average-surplus value. $\square$