Axiomatization and implementation of a class of solidarity values for TU-games

Abstract

A new class of values combining marginalistic and egalitarian principles is introduced for cooperative TU-games. It includes some modes of solidarity among the players by taking the collective contribution of some coalitions to the grand coalition into account. Relationships with other class of values such as the Egalitarian Shapley values and the Procedural values are discussed. We propose a strategic implementation of our class of values in subgame perfect Nash equilibrium. Two axiomatic characterizations are provided: one of the whole class of values, and one of each of its extreme points.

Appendix

In this Appendix, we employ the following extra definition. For any non-empty coalition $T\subseteq N$, the Dirac TU-game $\mathbf {1}_T\in V_N$ is defined as: $\mathbf {1}_T(T)=1$, and $\mathbf {1}_T(S)=0$ for each other S.

Proof

(Proposition 6)

Point 1 follows from (6).

Point 2. Consider the vector of constants $B^p$ as defined in point 2 of Proposition 6. From the definition of $\hbox {Sh}(B^p v)$, for each $i\in N$, we have:

$$\begin{aligned} \hbox {Sh}_i (B^p v)= & {} \sum _{S \in 2^N: S \ni i} \frac{(n - s)! (s -1) !}{n !} \bigl ( b_sv (S) - b_{s -1} v(S {\setminus } i ) \bigr ) \\= & {} \sum _{\begin{array}{c} S \in 2^N: S \ni i, \\ s \le p \end{array}} \frac{(n - s) ! (s -1) !}{n !} \bigl (b_sv (S) - b_{s -1}v(S {\setminus } i) \bigr ) \\&+ \sum _{\begin{array}{c} S \in 2^N: S \ni i, \\ s > p \end{array}} \frac{(n - s) ! (s -1) !}{n !} \bigl (b_{s} v (S) - b_{s -1}v(S {\setminus } i) \bigr ) \\= & {} \sum _{\begin{array}{c} S \in 2^N: S \ni i, \\ s \le p \end{array}} \frac{(n - s) ! (s -1) !}{n !} \bigl (v (S) - v(S {\setminus } i) \bigr ) \\&+\sum _{\begin{array}{c} S \in 2^N: S \ni i, \\ s = p+1 \end{array}} \frac{(n - s) ! (s -1) !}{n !} \bigl ( - v(S {\setminus } i) \bigr ) + \frac{1}{n} v(N) \\= & {} \sum _{\begin{array}{c} S \in 2^N: S \ni i, \\ s \le p \end{array}} \frac{(n - s) ! (s -1) !}{n !} \bigl (v (S) - v(S {\setminus } i) \bigr ) \\&+\sum _{\begin{array}{c} S \in 2^N: S \ni i, \\ s = p+1 \end{array}} \frac{(n - s) ! (s -1) !}{n !} \bigl ( - v(S {\setminus } i) \bigr ) \\&+ \sum _{\begin{array}{c} S \in 2^N: S \ni i, \\ s = p+1 \end{array}} \frac{(n - s) ! (s -1) !}{n !} v(N) \\= & {} \sum _{\begin{array}{c} S \in 2^N: S \ni i, \\ s \le p \end{array}} \frac{(n - s) ! (s -1) !}{n !} \bigl (v (S) - v(S {\setminus } i) \bigr ) \\&+\sum _{\begin{array}{c} S \in 2^N: S \ni i, \\ s = p+1 \end{array}} \frac{(n - s ) ! (s - 1) ! }{n !}\bigl (v (N) - v(S {\setminus } i) \bigr ) \\= & {} \hbox {Sol}_i^p(v), \end{aligned}$$

where the last equality follows from (7).

Point 3. Consider the procedure $l^p$ as defined in point 3 of Proposition 6. First, for any permutation $\sigma \in \Sigma _N$, the contribution vector $r^{\sigma , l^p}$ given by (1) writes:

$$\begin{aligned}&\forall i\in N,\quad r_i^{\sigma ,l^p}(v)\\&=\left\{ \begin{array}{ll} v(P_i^{\sigma })-v(P_i^{\sigma }{\setminus } i) &{} \quad \text {if }\sigma (j)\le p\text { and }\sigma (i)=\sigma (j), \\ \sum _{j\in (N{\setminus } P_i^\sigma )\cup i}v(P_j^{\sigma })-v(P_j^{\sigma }{\setminus } j) &{} \quad \text {if }\sigma (j)\ge p+1\text { and }\sigma (i)=p+1, \\ 0 &{}\quad \text {otherwise.} \end{array} \right. \end{aligned}$$

or equivalently,

$$\begin{aligned} \forall i\in N,\quad r_i^{\sigma ,l^p}(v)=\left\{ \begin{array}{ll} v(P_i^\sigma )-v(P_i^\sigma {\setminus } i) &{} \quad \text {if }\sigma (i)=\sigma (j)\le p, \\ v(N)-v(P_{\sigma ^{-1}(p)}^\sigma ) &{} \quad \text {if }\sigma (i)=p+1\le \sigma (j), \\ 0 &{}\quad \text {otherwise.} \end{array} \right. \end{aligned}$$

Because the expression does not depend on $\sigma (j)$, we have:

$$\begin{aligned} \forall i\in N,\quad r_i^{\sigma ,l^p}(v)=\left\{ \begin{array}{ll} v(P_i^\sigma )-v(P_i^\sigma {\setminus } i) &{} \quad \text {if }\sigma (i)\le p, \\ v(N)-v(P_{\sigma ^{-1}(p)}^\sigma ) &{} \quad \text {if }\sigma (i)=p+1, \\ 0 &{} \quad \text {otherwise.} \end{array} \right. \end{aligned}$$

Second, the previous observation implies that the Procedural value induced by $l^p$ assigns to a player $i\in N$ in a game $v\in V_N$, the payoff

$$\begin{aligned} \begin{array}{rl} \hbox {Pv}_i^{l^p}(v) &{} \displaystyle =\frac{1}{n!}\left( \sum _{\sigma \in \Sigma _N:\sigma (i)\le p}\bigl (v(P_i^\sigma )-v(P_i^\sigma {\setminus } i)\bigr )+\sum _{\sigma \in \Sigma _N:\sigma (i)=p+1}\bigl (v(N)-v(P_{\sigma ^{-1}(p)}^\sigma )\bigr )\right) \\ &{} \displaystyle =\frac{1}{n!}\left( \sum _{\sigma \in \Sigma _N:\sigma (i)\le p}\bigl (v(P_i^\sigma )-v(P_i^\sigma {\setminus } i)\bigr )+\sum _{\sigma \in \Sigma _N:\sigma (i)>p}\frac{v(N)-v(P_{\sigma ^{-1}(p)}^\sigma )}{n-p}\right) \\ &{} \displaystyle =\frac{1}{n!}\sum _{\sigma \in \Sigma _N}c_i^{\sigma ,p}(v) \\ &{} =\hbox {Sol}^p_i(v), \end{array} \end{aligned}$$

where the second equality follows from the fact that the number of permutations in which $\sigma (i)$ is greater than p is $(n-p)$ times larger than the number of permutations in which $\sigma (i)$ is equal to $p+1$. $\square$

Proof

(Proposition 8) Assume that $\Phi \in \mathbf{Sol}_N$. Then there is a probability distribution $(\alpha _0,\ldots ,\alpha _{n-1})$ such that:

$$\begin{aligned} \Phi =\sum _{p=0}^{n-1}\alpha _p\hbox {Sol}^p. \end{aligned}$$

By point 2 of Proposition 6, we have:

$$\begin{aligned} \forall v\in V_N,\quad \Phi (v)=\sum _{p=0}^{n-1}\alpha _p\hbox {Sh}(B^p v). \end{aligned}$$

Define the vector of constants $B^{\Phi }$ as:

$$\begin{aligned} B^{\Phi } = \sum _{ p = 0 }^{n -1}\alpha _p B^p. \end{aligned}$$

By Linearity of the Shapley value and Remark 3, we get:

$$\begin{aligned} \Phi (v)=\sum _{p=0}^{n-1}\alpha _p\hbox {Sh}(B^pv)=\hbox {Sh}\biggl (\sum _{p=0}^{n-1}\alpha _p(B^pv)\biggr )=\hbox {Sh}\biggl (\biggl (\sum _{p=0}^{n-1}\alpha _pB^p\biggr )v\biggr )=\hbox {Sh}(B^{\Phi }v) \end{aligned}$$

Using the definition of $B^p$, we obtain:

$$\begin{aligned} b_0^{\Phi }=0,\,\,b_n^{\Phi }=1,\,\,\hbox {and, for each }\,\,s\in \{1,\dots ,n-1\},\,\,b_s^{\Phi }=\sum _{s=p}^{n-1}\alpha _p. \end{aligned}$$

Because, for each $p\in \{0,\dots ,n-1\}$, $\alpha _p\in [0,1]$, we conclude that

$$\begin{aligned} 1=b_n^{\Phi }\ge b_1^{\Phi }\ge b_2^{\Phi }\ge \cdots \ge b_{n-1}^\Phi \ge b_0^{\Phi }=0, \end{aligned}$$

as desired.

Reciprocally, consider a value $\Phi$ on $V_N$ such that $\Phi (v)=\hbox {Sh}(B^{\Phi }v)$, where the vector of constants $B^{\Phi }$ is such that:

$$\begin{aligned} 1=b_n^{\Phi }\ge b_1^{\Phi }\ge b_2^{\Phi }\ge \cdots \ge b_{n-1}^{\Phi }\ge b_0^{\Phi }=0. \end{aligned}$$

(8)

Define the collection of real numbers $(\alpha _p:p\in \{0,\dots ,n-1\})$ as follows:

$$\begin{aligned} \forall p\in \{1,\dots ,n-2\},\,\,\alpha _p=b_p^{\Phi }-b_{p+1}^{\Phi },\,\,\alpha _{n-1}=b_{n -1}^{\Phi }\,\,\hbox { and }\,\,\alpha _0=1-b_{1}^{\Phi }. \end{aligned}$$

By (8):

$$\begin{aligned} \forall p\in \{0,1\dots ,n-1\},\,\,\alpha _p\in [0,1],\,\,\hbox { and }\,\,\sum _{p=0}^{n-1}\alpha _p=1, \end{aligned}$$

which means that $(\alpha _p:p\in \{0,\dots ,n-1\})$ can be viewed as a probability distribution over the positions $p\in \{0,\dots ,n-1\}$. Furthermore, it holds that:

$$\begin{aligned} \forall s\in \{1,\ldots ,n-1\},\quad \sum _{p=s}^{n-1}\alpha _p=b_s^{\Phi }. \end{aligned}$$

From this, it follows that:

$$\begin{aligned} \sum _{p=0}^{n-1}\alpha _p B^p=\biggl (0,\sum _{p=1}^{n-1}\alpha _p,\sum _{p=2}^{n-1}\alpha _p,\ldots ,\sum _{p=n-2}^{n-1}\alpha _p,\alpha _{n-1},\sum _{p=0}^{n-1}\alpha _p\biggr )=B^{\Phi }. \end{aligned}$$

Therefore, using Remark 3, we obtain:

$$\begin{aligned} \Phi (v)= & {} \hbox {Sh}(B^{\Phi } v)=\hbox {Sh}\biggl (\biggl (\sum _{p=0}^{n-1}\alpha _pB^p \biggr )v\biggr )=\hbox {Sh}\biggl (\sum _{p=0}^{n-1}\alpha _p(B^pv)\biggr )\\= & {} \sum _{p=0}^{n-1}\alpha _p\hbox {Sh}(B^pv)=\hbox {Sol}^{\alpha }(v), \end{aligned}$$

as desired. Finally, the second statement of Proposition 8 immediately follows from the previous steps. $\square$

Proof

(Proposition 9) First, assume that Mechanism (A) possesses at least one SPNE. We show successively that points 1, 2, 3 and 4 of Proposition 9 hold.

Point 1. By definition of $\hat{H}_a$, $a \in A$, and the constraint placed on bids, we have:

$$\begin{aligned} \sum _{a \in A} \alpha _{a} \hat{H}_a = \sum _{a \in A} \sum _{i \in N} \alpha _a \hat{h}_a^i = \sum _{i \in N} \sum _{a \in A} \alpha _a \hat{h}_a^i = \sum _{i \in N} 0 = 0. \end{aligned}$$

Therefore, it is sufficient to prove that, on each SPNE of Mechanism (A), $\hat{H}_a = \hat{H}_{a'}$ for each pair of distinct objects a and $a'$ of A. Consider any SPNE of Mechanism (A). Assume, for the sake of contradiction, that the assertion is false. This implies that there is $a' \in A {\setminus } \Omega _A$. We show that any player has an incentive to change unilaterally his/her bids. So, consider any $i \in N$ and any positive real number $\varepsilon >0$, and define the alternative list of bids $(\tilde{h}_{a}^{i})_{a \in A}$ for player i:

• (i) For the selected object $a' \in A {\setminus } \Omega _A$, $\tilde{h}_{a'}^{i} = \hat{h}_{a'}^{i} + \sum _{a \in \Omega _A} \alpha _a \, \varepsilon$;

• (ii) For each $a \in \Omega _A$, $\tilde{h}_{a}^{i} = \hat{h}_{a}^{i} - \alpha _{a'} \varepsilon$;

• (iii) For each $a \in A {\setminus } (\Omega _A \cup \{a'\})$, $\tilde{h}_{a}^{i} = \hat{h}_{a}^{i}$.

Note that the bids $(\tilde{h}_{a}^{i})_{a \in A}$ satisfy the constraint on bids since:

$$\begin{aligned} \sum _{a \in A} \alpha _a \tilde{h}_{a}^{i}= & {} \sum _{a \in \Omega _A} \alpha _a \bigl (\hat{h}_{a}^{i} - \alpha _{a'} \varepsilon \bigr ) + \alpha _{a'} \bigl (\hat{h}_{a'}^{i} +\sum _{a \in \Omega _A} \alpha _a \, \varepsilon \bigr ) + \sum _{a \in A {\setminus } (\Omega _A \cup \{a'\})} \alpha _a \, \hat{h}_{a}^{i} . \nonumber \\= & {} \sum _{a \in A} \alpha _a \hat{h}_{a}^{i} - \sum _{a \in \Omega _A} \alpha _a \alpha _{a'} \varepsilon + \alpha _{a'} \sum _{a \in \Omega _A} \alpha _a \, \varepsilon \nonumber \\= & {} \sum _{a \in A} \alpha _a \hat{h}_{a}^{i} \nonumber \\= & {} 0. \end{aligned}$$

(9)

where the last equality comes from the fact that the equilibrium bids $(\hat{h}_{a}^{i})_{a \in A}$ satisfy the constraint on bids by definition. From this, we obtain the following aggregate bids:

• (iv) For the selected object $a' \in A {\setminus } \Omega _A$, $\tilde{H}_{a'} = \hat{H}_{a'} + \sum _{a \in \Omega _A} \alpha _a \varepsilon$;

• (v) For each $a \in \Omega _A$, $\tilde{H}_{a} = \hat{H}_{a} - \alpha _{a'} \varepsilon$;

• (vi) For each $a \in A {\setminus } (\Omega _A \cup \{a'\})$ $\tilde{H}_{a} = \hat{H}_{a}$.

On the one hand, by choosing a sufficiently small $\varepsilon >0$, the set $\Omega _A$ is not altered by this unilateral deviation. On the other hand, the new expected payoff $\tilde{m}_i$ received by i is given by:

$$\begin{aligned} \tilde{m}_i = \hat{m}_i + \frac{n -1}{n} \alpha _{a'} \varepsilon > \hat{m}_i, \end{aligned}$$

which is in contradiction with the assumption that the players play a SPNE. We conclude that there is no $a' \in A {\setminus } \Omega _A$, and so $\Omega _A = A$.

Point 2. By point 1, we know if the players play a SPNE, then $\hat{H}_a = 0$ for each $a \in A$ and $\Omega _A = A$. It follows that the expected payoffs are given by:

$$\begin{aligned} \forall i \in N, \quad \hat{m}_i = \frac{ \sum _{a\in A} \hat{z}^i_a }{|A|} = \frac{ \sum _{a\in A}\bigl (\hat{g}_a^{i} - \hat{h}_{a}^{i} \bigr ) }{|A|}. \end{aligned}$$

To prove point 2, we proceed again by contradiction. Thus, assume that there $i \in N$ and two distinct objects $a, a' \in A$ such that $\hat{z}_{a}^{i} > \hat{z}_{a'}^{i}$. Then there exist two distinct objects in A, say for simplicity, a and $a'$ such that $\hat{z}_{a}^{i}> \hat{m}_i >\hat{z}_{a'}^{i}$. We show that player i can strictly improve his/her expected payoff by unilaterally changing his/her list of bids. Indeed, consider the following list of bids for i: for any real number $\varepsilon >0$,

• (i) $\tilde{h}_{a}^{i} = \hat{h}^{i}_{a} +\alpha _{a'} \varepsilon$;

• (ii) $\tilde{h}_{a'}^{i} = \hat{h}^{i}_{a'} - \alpha _{a} \varepsilon$;

• (iii) For each $o \in A {\setminus } \{a, a'\}$, $\tilde{h}_{o}^{i} = \hat{h}^{i}_{o}$.

We easily verify that this new list of bids satisfies the constraint on bids:

$$\begin{aligned} \sum _{o \in A} \alpha _o \tilde{h}_{o}^{i} = \sum _{o \in A} \alpha _o \hat{h}_{o}^{i} +\alpha _{a} \alpha _{a'} \ \varepsilon - \alpha _{a'} \alpha _{a} \ \varepsilon = \sum _{o \in A} \alpha _o \hat{h}_{o}^{i}= 0. \end{aligned}$$

From this, we obtain:

• (i) $\tilde{H}_{a} = \hat{H}_{a} + \alpha _{a'} \varepsilon$;

• (ii) $\tilde{H}_{a'} = \hat{H}_{a'} - \alpha _{a} \varepsilon$;

• (iii) For each $o \in A {\setminus } \{a, a'\}$, $\tilde{H}_{o} = \hat{H}_{o}$.

It follows that $\Omega _A = \{a\}$, and so:

$$\begin{aligned} \tilde{m}_i = \tilde{z}_{a}^{i} = \hat{z}^{i}_{a} - \alpha _{a'} \varepsilon \frac{n-1}{n}. \end{aligned}$$

By assumption $\hat{z}_{a}^{i} > \hat{m}_i$. Therefore, there is $\varepsilon >0$ such that $\tilde{m}_i > \hat{m}_{i}$, which contradicts the fact that $\hat{m}_{i}$ is i’s equilibrium expected payoff. Conclude that the initial assumption is false, and so point 2 holds.

Point 3. Pick any $i \in N$. By point 2, we have:

$$\begin{aligned} \forall a \in A, \quad \hat{m}_i = \hat{z}_{a}^{i} = \hat{g}_a^{i} - \hat{h}_{a}^{i}. \end{aligned}$$

Since, by assumption, $\sum \nolimits _{a \in A} \alpha _a= 1$ and $\sum \nolimits _{a \in A}\alpha _a \hat{h}_{a}^{i} = 0$, we obtain:

$$\begin{aligned} \hat{m}_i = \sum _{a \in A} \alpha _a \hat{m}_i = \sum _{a \in A} \alpha _a \bigl ( \hat{g}_a^{i} - \hat{h}_{a}^{i}\bigr ) = \sum _{a\in A} \alpha _a \hat{g}_a^{i} -\sum _{a \in A} \alpha _a \hat{h}_{a}^{i} = \sum _{a\in A} \alpha _a \hat{g}_a^{i}, \end{aligned}$$

which is the desired result.

Point 4. This point follows from point 2 and point 3. Indeed, we have:

$$\begin{aligned} \forall i \in N, \forall a \in A, \quad \hat{g}_{a}^{i} - \hat{h}_{a}^{i} = \alpha _{a} \hat{g}_{a}^{i} + \sum _{o \in A {\setminus } a} \alpha _{o} \hat{g}_{o}^{i}, \end{aligned}$$

which is the desired result. Note that the list of bids $(\hat{h}_{a}^{i})_{a \in A}$ is feasible since:

$$\begin{aligned} \sum _{a \in A} \alpha _a \hat{h}_{a}^{i} = \sum _{a \in A} \alpha _a \hat{g}_{a}^{i} - \sum _{a \in A} \alpha _a \biggl ( \sum _{o \in A} \alpha _{o} \hat{g}_{o}^{i}\biggr ) = 0. \end{aligned}$$

It remains to shows that Mechanism (A) has a SPNE if and only if the total payoff $\sum \nolimits _{i \in N} \hat{g}_a^i$ does not depend on $a \in A$.

To show that the existence of a SPNE implies that the total payoff $\sum \nolimits _{i \in N} \hat{g}_a^i$ does not depend on $a \in A$, we proceed as follows. By point 4,

$$\begin{aligned} \forall i \in N, \forall a \in A, \quad \hat{h}_{a}^{i} = (1 - \alpha _{a}) \hat{g}_{a}^{i} - \sum _{o \in A {\setminus } a} \alpha _{o} \hat{g}_{o}^{i}, \end{aligned}$$

It follows that:

$$\begin{aligned} \hat{H}_a = \sum _{i \in N} \hat{h}_{a}^{i} = \sum _{i \in N} \hat{g}_{a}^{i} - \sum _{i \in N} \sum _{o \in A} \alpha _{o} \hat{g}_{o}^{i}. \end{aligned}$$

By point 1, for each $a \in A$, $\hat{H}_a = 0$. Therefore,

$$\begin{aligned} \bigl [\forall a \in A, \quad \hat{H}_a = 0 \bigr ] \Longleftrightarrow \biggl [\forall a \in A, \quad \sum _{i \in N} \hat{g}_{a}^{i} = \sum _{o \in A} \alpha _{o} \sum _{i \in N} \hat{g}_{o}^{i}\biggr ], \end{aligned}$$

which amounts to saying that $\sum \nolimits _{i \in N} \hat{g}_{a}^{i}$ does not depend on $a \in A$. Conclude that if Mechanism (A) possesses a SPNE, then $\sum \nolimits _{i \in N} \hat{g}_{a}^{i}$ does not depend on $a \in A$.

Reciprocally, assume that $\sum \nolimits _{i \in N} \hat{g}_{a}^{i}$ does not depend on $a \in A$. To show that Mechanism (A) possesses at least one SPNE, we proceed as follows. Assume that a SPNE is played in each subgame $G_a$, $a \in A$, of Mechanism (A). Such equilibria exist by assumption and the associated payoffs are given by $(\hat{g}_a^i)_{i \in N}$ for each object $a \in A$. Assume that, at Stage 1, each player $i \in N$ announces the list of bids $(\hat{h}_a^i)_{a \in A}$ given in point 4. It remains to show that no player has an incentive to unilaterally change his or her list of bids. Assume that there is $i \in N$ who plans to announce the alternative list of bids $(h_a^i)_{a \in A}$. By assumption on the set of feasible lists of bids, we have $\sum \nolimits _{a \in A} \alpha _a h_a^i = 0$. Because $\sum \nolimits _{a \in A} \alpha _a \hat{h}_a^i = 0$ as well (see the proof of point 4.), there necessarily exists an object $a \in A$ such that $h^i_a > \hat{h}_a^i$. Let $A_0 \subseteq A$ be the subset of objects that maximize the value $h^i_a - \hat{h}_a^i$, and denote by $\ell >0$ the maximal value obtained by choosing any object in $A_0$. By point 1, $\hat{H}_a = 0$ for each $a \in A$. Thus, we have $H_a > 0$ for each $a \in A_0$. It follows that $\Omega _A \subseteq A_0$. On each subgame $G_a, a \in \Omega _A$, i’s reward is given by:

$$\begin{aligned} z_a^i = \hat{z}_a^i - \frac{n - 1}{n}\ell , \end{aligned}$$

which implies that player i has no interest to unilaterally deviate from $(\hat{h}_a^i)_{a \in A}$ to announce $(h_a^i)_{a \in A}$. Conclude that Mechanism (A) possesses at least one SPNE. $\square$

Proof

(Proposition 10) Consider any zero-monotonic TU-game $v \in V_N$ and any probability distribution $\alpha =(\alpha _p)_{p=0}^{n-1}$. We first solve the game at Stage 3, that is the sequential bargaining game $G_{p, \sigma }$, where $p\in A$ and $\sigma \in \Sigma _N$, where A is the support of $\alpha$. We show that $G_{p, \sigma }$ has a unique SPNE with payoffs:

$$\begin{aligned} \forall i \in N, \quad \hat{g}_{\sigma ,p}^i(v) =\left\{ \begin{array}{cl} v (P_i^{\sigma })- v(P_i^{\sigma }{\setminus } i) &{} \quad \hbox {if } \sigma (i)\le p, \\ \displaystyle v(N)-v(P_{\sigma ^{-1}(p)}^{\sigma })&{} \quad \hbox {if } \sigma (i) = p +1, \\ 0 &{} \quad \hbox {if }\sigma (i) \ge p +2. \end{array} \right. \end{aligned}$$

Recall that by convention $P^{\sigma }_{\sigma ^{-1}(0)}=\emptyset$.

Assume, without loss of generality, that $\sigma \in \Sigma _N$ is such that $\sigma (i) = i$ for each $i \in N$. Assume further that $G_{p, \sigma }$ reaches a decision node of player 2: player 2 at position $\sigma (2) = 2$ receives the opportunity to make an offer $x_{1}^{2} \in \mathbb {R}$ to player 1 at position $\sigma (1) = 1$. By the rules of $G_{p, \sigma }$ if player 1 refuses the offer $x_{1}^{2}$, he or she can guarantee v(1) at the next round. So, player 2 will offer $x_{1}^{2}= v(1)$. Player 2’s payoff is equal to $v(P_{2} ^{\sigma })- v(1)$, where $P_{2}^{\sigma } = \{1, 2\}$. Assume now that player 3 at position $\sigma (3) = 3$ has the opportunity to make an offer $x_{2}^{3} \in \mathbb {R}$ to player 2 and an offer $x_{1}^{3} \in \mathbb {R}$ to player 1. By the previous step, proposer 3 knows that player 2 at position 2 can guarantee $v(P_{2}^{\sigma }) - v(P_{2}^{\sigma } {\setminus } 2)$; he also knows that player 1 can guarantee $v(P_{2} ^{\sigma } {\setminus } 2)$. Therefore, player 3 will offer $x_{2}^{3} = v(P_{2}^{\sigma }) - v(P_{2} ^{\sigma } {\setminus } 2)$ and $x_{1}^{3} = v(P_{2}^{\sigma } {\setminus } 2)$. Players 1 and 2 have no incentive to refuse the proposal made by player 3. It follows that player 3’s payoff is equal to:

$$\begin{aligned} v(P_{3} ^{\sigma }) - x_{2}^{3} - x_{1}^{3} = v(P_{3}^{\sigma }) - v(P_{2}^{\sigma }) = v(P_{3}^{\sigma }) - v(P_{3}^{\sigma }{\setminus } 3). \end{aligned}$$

Continuing in this fashion for each agent i at position $\sigma (i)\le p$, we obtain that each of them can guarantee the payoff $v (P_{i}^{\sigma })- v(P_{i}^{\sigma }{\setminus } i)$. Now, consider player $p+1$ at position $\sigma (p+1) = p+1$. He/she is the first proposer in $G_{p, \sigma }$. Given the payoff that each player at position $i \le p$ can guarantee, he/she will offer to each of them $x_{i}^{p+1} = v (P_{i}^{\sigma })- v(P_{i}^{\sigma }{\setminus } i)$. Each other agent in position $i \ge p+ 2$ obtain a null payoff if he/she refuses the offer $x_{i}^{p} \in \mathbb {R}$ made by player $p+1$. Knowing that, player $p+1$ will offer $x_{i}^{p+1} = 0$ to each of them. No player has an incentive to refuse the offer made by $p+1$. So, his/her proposal can be unanimously accepted by all the agents. In such a case, $p+1$ will obtain the following payoff:

$$\begin{aligned} v(N) - \sum _{i \le p} \biggl (v (P_{i}^{\sigma })- v(P_{i}^{\sigma }{\setminus } i) \biggr ) + \sum _{i \ge p+2} 0 = v(N) - v (P_{p}^{\sigma }), \end{aligned}$$

which is non-negative since v is zero-monotonic. This means that it is in the interest of player $p+1$ that this proposal be accepted. Conclude that these strategies constitute the only SPNE of $G_{p, \sigma }$.

At Stage 2 players bid over the set of permutations $\Sigma _N$. Note that in the sequential bargaining game $G_{p, \sigma }$, the total equilibrium payoffs $\sum \nolimits _{i \in N} g_{\sigma ,p}^i(v) = v(N)$ whatever $\sigma \in \Sigma _N$. Thus, for each fixed position $p \in A$, we can apply Proposition 9 to the Mechanism $M_p$ formed by Stage 2 and Stage 3 of Mechanism B, where, for each $\sigma \in \Sigma _N$, the rewards are given by:

$$\begin{aligned} \forall i \in N, \quad z_{p, \sigma } ^i = g_{p, \sigma }^i - h^i_{\sigma } + \frac{H_{\sigma }}{n}. \end{aligned}$$

where $g_{p, \sigma }^i$ is the payoff function of player i in $G_{p, \sigma }$. By Proposition 9, each SPNE of Mechanism $M_p$ induces, for each player $i \in N$, the same expected payoff given by:

$$\begin{aligned} \hat{m}_i = \frac{1}{n !} \hat{g}_{\sigma ,p}^i(v) = \hbox {Sol}_i^p(v), \end{aligned}$$

where second equality comes from point 3. of Proposition 6.

Finally, construct a new mechanism formed by Stage 1 of Mechanism B and the Mechanisms ${\mathbf M}_{\mathbf p}$, $p \in A$, with equilibrium payoffs $\hat{g}_p^{i} = \hbox {Sol}_i^p(v)$. This mechanism coincides with Mechanism B. Remark that in Mechanism $M_p$ the total payoff $\sum _{i \in N} \hbox {Sol}_i^p(v) = v(N)$ does not depend on p. So, we can apply Proposition 9 to Mechanism B. We obtain that each SPNE of Mechanism B induces, for each player $i \in N$, the same expected payoff given by:

$$\begin{aligned} \sum _{p = 0}^{n -1} \alpha _p \hat{g}_p^{i} = \sum _{p = 0}^{n -1} \alpha _p \hbox {Sol}_i^p(v) = \hbox {Sol}^{\alpha }_i (v), \end{aligned}$$

which proves that Mechanism B implements $\hbox {Sol}^{\alpha } (v)$. $\square$

Proof

(Proposition 11) By Propositions 5 and 6, $\hbox {Sol}^p$ satisfies Efficiency, Linearity, Anonymity, and so it also satisfies Additivity and Equal treatment of equals. From the definition of $c^{p,\sigma }$ given in (5), we easily conclude that $\hbox {Sol}^p$ satisfies the p-null player axiom.

To prove that there exists a unique value that satisfies Efficiency, Equal treatment of equals, Additivity, and the p-null player axiom for some $p\in \{1,\ldots ,n-1\}$, consider any such value $\Phi$ on $V_N$. We have already underlined that the combination of these axioms implies that $\Phi$ also satisfies Linearity and Anonymity. By Proposition 5, there exists a unique vector of constants $B^{\Phi }=(b_s^{\Phi }:s\in \{0,1,\ldots ,n\})$ such that $b_0^{\Phi }=0$, $b_n^{\Phi }=1$, and $\Phi (v)=\hbox {Sh}(B^{\Phi }v)$. By Proposition 6, it remains to show that $B^{\Phi }=B^p$.

Fix any player $i\in N$. For any size $s\in \{1,\ldots ,n-1\}$, any coalition $S\subseteq N$ of size s such that $S \ni i$, consider the TU-game $\mathbf {1}_S + \mathbf {1}_{S {\setminus } i}.$ For each $s \not = p+1$, player i is p-null. By the p-null player axiom, we have:

$$\begin{aligned} 0= & {} \Phi _i (\mathbf {1}_S + \mathbf {1}_{S {\setminus } i}) = \hbox {Sh}_i (B^{\Phi }(\mathbf {1}_S + \mathbf {1}_{S {\setminus } i})) \\= & {} \frac{(n -s)! (s -1) !}{n !} (b_s^{\Phi } - b_{s-1}^{\Phi }), \,\, \text{ and } \text{ so } \,\, b_s^{\Phi } = b_{s -1}^{\Phi }. \end{aligned}$$

We conclude that:

$$\begin{aligned} b_1^{\Phi } = \cdots = b_{p}^{\Phi } \,\, \hbox { and } \,\, b_{p+1}^{\Phi } = \cdots = b_{n -1}^{\Phi }. \end{aligned}$$

(10)

Next, define the TU-game $v_{p, i}$ as follows:

$$\begin{aligned} \forall S \in 2^N , \,\, v_{p, i}(S) = \left\{ \begin{array}{ll} 1 &{} \quad \text{ if } s \ge p+1, \\ 1 &{} \quad \text{ if } s = p \text{ and } S \not \ni i, \\ 0 &{} \quad \text{ if } s = p \text{ and } S \ni i, \\ 0 &{} \quad \text{ if } s < p. \end{array} \right. \end{aligned}$$

Player i is p-null in $v_{p, i}$. By the p-null player axiom, we have:

$$\begin{aligned} 0= & {} \Phi _i (v_{p, i}) \\= & {} \hbox {Sh}_i (B^{\Phi } v_{p, i}) \\= & {} \sum _{S \in 2^N: S \ni i} \frac{(n -s)! (s -1) !}{n !} (b_s^{\Phi } - b_{s-1}^{\Phi }) \\= & {} \sum _{S \in 2^N: S \ni i, s \ge p+1} \frac{(n -s)! (s -1) !}{n !} (b_s^{\Phi } - b_{s-1}^{\Phi }) \\= & {} \sum _{s =p+1}^{n} \left( {\begin{array}{c}n -1\\ s -1\end{array}}\right) \frac{(n -s)! (s -1) !}{n !} (b_s^{\Phi } - b_{s-1}^{\Phi }) \\= & {} \frac{1}{n} \sum _{s =p+1}^{n} (b_s^{\Phi } - b_{s-1}^{\Phi }) \\= & {} \frac{1}{n} (b_n^{\Phi } - b_{p}^{\Phi }). \end{aligned}$$

Since $b_n^{\Phi } = 1$, it follows that $b_{p}^{\Phi } =1$. By (10), we obtain:

$$\begin{aligned} b_1^{\Phi } = \cdots = b_{p}^{\Phi } = 1. \end{aligned}$$

(11)

This gives the result for $p = n -1$. To complete to proof for each other $p \le n -2$, note that player $i \in N$ is p-null for each $p \le n -2$ in the TU-game $\mathbf {1}_{N {\setminus } i}$. By the p-null player axiom, we have:

$$\begin{aligned} 0 = \Phi _i (\mathbf {1}_{N {\setminus } i}) = \hbox {Sh}_i (B^{\Phi } \mathbf {1}_{N {\setminus } i}) = -\frac{ b_{n-1}^{\Phi } }{n}, \,\, \hbox { and so } \,\, b_{n-1}^{\Phi } = 0. \end{aligned}$$

By (10), we obtain:

$$\begin{aligned} b_{p+1}^{\Phi } = \cdots = b_{n -1}^{\Phi } = 0. \end{aligned}$$

(12)

By (11) and (12), we conclude that $B^{\Phi } = B^p$, as desired. $\square$

Proof

(Proposition 12) We first prove that each $\Phi \in \mathbf{Sol}_N$ satisfies all the axioms of the statement of the Proposition 12. By Corollary 1, we have $\mathbf{Pv}_N \supseteq \mathbf{Sol}_N$. Therefore, by Proposition 4, each $\Phi \in \mathbf{Sol}_N$ satisfies Efficiency, Additivity, Desirability, Monotonicity. It remains to show that any $\Phi \in \mathbf{Sol}_N$ satisfies Null player in a null environment for positive games. Pick any positive $v\in V_N$ such that $v(N)=0$, and any null player $i\in N$ in v. For any $\sigma \in \Sigma _N$ and any $p\in \{0,\dots ,n-1\}$, the fact that i is null in v and $v(N)=0$ implies that (5) can be rewritten as follows:

$$\begin{aligned} c_i^{\sigma ,p}(v)= \left\{ \begin{array}{ll} 0 &{} \quad \text{ if } \sigma (i)\le p, \\ - \frac{\displaystyle v(P_{\sigma ^{-1}(p)}^{\sigma })}{\displaystyle n-p} &{} \quad \hbox { if } \sigma (i)>p. \end{array} \right. \end{aligned}$$

By positivity of v, $c_i^{\sigma ,p}(v)\le 0$, which in turn implies that $\hbox {Sol}^p_i(v)\le 0$ for each $p\in \{0,\dots ,n-1\}$, and consequently $\Phi _i(v)\le 0$, as desired.

Reciprocally, pick any $\Phi$ which satisfies Efficiency, Additivity, Desirability, Monotonicity, and Null player in a null environment for positive games. By Proposition 4, $\Phi \in \mathbf{Pv}_N$, and by Corollary, $\Phi \in \mathbf{EAL}_N.$ By Proposition 5, there is a unique vector of constants $B^{\Phi }= (b_s^{\Phi }: s \in \{0, 1, \ldots , n\})$ such that $b_0^{\Phi } = 0$, $b^{\Phi }_n = 1$ and:

$$\begin{aligned} \forall v \in V_N, \quad \Phi (v) = \hbox {Sh}(B^{\Phi } v). \end{aligned}$$

By Theorem 2 by Radzik and Driessen (2013), we also know that each real number $b_s^{\Phi } \in [0, 1]$. By Proposition 8, it remains to show that, for each $s \in \{1, \ldots , n -2\}$, $b_s^{\Phi }\ge b_{s+1}^{\Phi }$. To that end, pick any $i\in N$ and any non-empty coalition $S\subseteq N{\setminus } i$, $S \not = N {\setminus } i$, and consider the TU-game $\mathbf {1}_{S\cup i}+\mathbf {1}_S$. This TU-game is positive, the environment is null $(\mathbf {1}_{S\cup i}+\mathbf {1}_S)(N)=0$ since $S \not = N{\setminus } i$, and i is a null player in $\mathbf {1}_{S\cup i}+\mathbf {1}_S$. Therefore, Null player in a null environment for positive games yields:

$$\begin{aligned} \Phi _i(\mathbf {1}_{S\cup i}+\mathbf {1}_S)\le 0. \end{aligned}$$

(13)

On the other hand, we have:

$$\begin{aligned} \Phi _i(\mathbf {1}_{S\cup i}+\mathbf {1}_S)= \hbox {Sh}_i(B^{\Phi }(\mathbf {1}_{S\cup i}+\mathbf {1}_S)) =\frac{s!(n-s-1)!}{n!}(b_{s+1}^{\Phi }-b_s^{\Phi }). \end{aligned}$$

(14)

Combining (13) and (14) yields $b_s^{\Phi } \ge b_{s+1}^{\Phi }$ for each $s\in \{1,\dots ,n-2\}$, as desired.

$\square$