Sharing the surplus in games with externalities within and across issues

Abstract

We consider issue-externality games in which agents can cooperate on multiple issues and externalities are present both within and across issues, that is, the amount a coalition receives in one issue depends on how the players are organized on all the issues. Examples of such games are several firms competing in multiple markets, and countries negotiating both a trade agreement (through, e.g., WTO) and an environmental agreement (e.g., Kyoto Protocol). We propose a way to extend (Shapley) values for partition function games to issue-externality games. We characterize our proposal through axioms that extend the Shapley axioms to our more general environment. The solution concept that we propose can be applied to many interesting games, including inter-temporal situations where players meet sequentially.

Appendix

Proof of Proposition 1

(i) Consider two games $(N,A,v)$ and $(N,A,v^{\prime })$. Since $\phi ^{*}$ satisfies linearity, we have

$$\begin{aligned} \phi _{k}^{*}(N(A),\hat{v}+\widehat{v^{\prime }})=\phi _{k}^{*}(N(A), \hat{v})+\phi _{k}^{*}(N(A),\widehat{v^{\prime }})\quad \text { for every }k\in N(A). \end{aligned}$$

Also, following (1), it is easy to check that $\widehat{ v+v^{\prime }}=\hat{v}+\widehat{v^{\prime }}$. Hence,

$$\begin{aligned}&\Phi _{i}^{*}(N,A,v+v^{\prime })=\underset{a\in A}{\sum }\phi _{i(a)}^{*}(N(A),\widehat{v+v^{\prime }})=\underset{a\in A}{\sum }\phi _{i(a)}^{*}(N(A),\hat{v}+\widehat{v^{\prime }}) \\&\quad =\underset{a\in A}{\sum }\phi _{i(a)}^{*}(N(A),\hat{v})+\underset{a\in A}{ \sum }\phi _{i(a)}^{*}(N(A),\widehat{v^{\prime }})=\Phi _{i}^{*}(N,A,v)+\Phi _{i}^{*}(N,A,v^{\prime }) \end{aligned}$$

for all $i\in N,$ and $\Phi ^{*}$ satisfies part $1.1$ of the linearity axiom. Similarly, for the multiplication by a scalar $\lambda$, it is the case that $\lambda \phi _{k}^{*}(N(A),\hat{v})=\phi _{k}^{*}(N(A),\lambda \hat{v})$ for every $k\in N(A)$ and $\widehat{\lambda v} =\lambda \hat{v}.$ Hence,

$$\begin{aligned}&\Phi _{i}^{*}(N,A,\lambda v)=\underset{a\in A}{\sum }\phi _{i(a)}^{*}(N(A),\widehat{\lambda v})=\underset{a\in A}{\sum }\phi _{i(a)}^{*}(N(A),\lambda \hat{v})\\&\quad =\underset{a\in A}{\sum }\lambda \phi _{i(a)}^{*}(N(A),\hat{v})=\lambda \Phi _{i}^{*}(N,A,v) \end{aligned}$$

for all $i\in N,$ and $\Phi ^{*}$ satisfies part $1.2$ of the linearity axiom.

(ii) The player anonymity axiom of $\phi ^{*}$ implies that $\phi _{\sigma k}^{*}(N(A),\sigma \hat{v})=\phi _{k}^{*}(N(A),\hat{v})$ for any $k\in N(A)$ and for any permutation $\sigma$ on the set $N(A).$ Take now a permutation $\sigma _{N}$ on the set $N$ and denote $\sigma _{N(A)}$ the permutation on the set $N(A)$ that associates player $i(a)$ with $(\sigma _{N}\left( i\right) )(a),$ for every $i\in N, a\in A$. Consider the game $(N,A,v)$. Then,

$$\begin{aligned}&\widehat{\sigma _{N}v}(T,Q)=\underset{a\in A}{\sum }\sigma _{N}v\left( \widetilde{T}(a);a;(\widetilde{Q}(b))_{b\in A}\right) =\underset{a\in A}{ \sum }v(\sigma _{N}\widetilde{T}(a);a;\left( \sigma _{N}\widetilde{Q} (b))_{b\in A}\right) \\&\quad = \underset{a\in A}{\sum }v((\widetilde{\sigma _{N(A)}T})(a);a;( ( \widetilde{\sigma _{N(A)}Q}) (b))_{b\in A}) =\hat{v}(\sigma _{N(A)}T,\sigma _{N(A)}Q)\\&\quad =\left( \sigma _{N(A)}\hat{v}\right) (T,Q). \end{aligned}$$

Consequently,

$$\begin{aligned}&\Phi _{i}^{*}(N,A,\sigma _{N}v)=\underset{a\in A}{\sum }\phi _{i(a)}^{*}(N(A),\widehat{\sigma _{N}v})=\underset{a\in A}{\sum }\phi _{i(a)}^{*}(N(A),\sigma _{N(A)}\hat{v})\\&\quad = \underset{a\in A}{\sum }\phi _{\sigma _{N(A)}\left( i(a)\right) }^{*}(N(A),\hat{v})=\underset{a\in A}{\sum }\phi _{\left( \sigma _{N}\left( i\right) \right) (a)}^{*}(N(A),\hat{v})=\Phi _{\sigma _{N}\left( i\right) }^{*}(N,A,v) \end{aligned}$$

for each $i\in N$. Hence, $\Phi ^{*}$ satisfies the player anonymity axiom.

(iii) We first prove that if $j\in N$ is a dummy player in the game $(N,A,v),$ then all the replicas $j(a),$ for all $a\in A,$ are dummy players in $(N(A),\hat{v})$. Consider any $(T,Q)\in ECL (N(A))$ and any $(T^{\prime },Q^{\prime })$ obtained from $(T,Q)$ by changing the affiliation of player $j(a).$ For any such $(T^{\prime },Q^{\prime }),$ it is always the case that $Q^{\prime }(b)=Q(b)$ for any $b\ne a,$ since we are changing the affiliation of a player that belongs to $N(a).$ There are two possibilities:

a) It can be the case that $Q^{\prime }(a)=Q(a)$. Then,

$$\begin{aligned} \hat{v}(T^{\prime },Q^{\prime })=\underset{b\in A}{\sum }v(\widetilde{T^{\prime }}(b);b;(\widetilde{Q^{\prime }}(c))_{c\in A})=\underset{b\in A}{ \sum }v(\widetilde{T}(b);b;(\widetilde{Q}(c))_{c\in A})=\hat{v}(T,Q). \end{aligned}$$

b) Or it can be the case that $Q^{\prime }(a)\ne Q(a)$ when $j(a)$ changes affiliation. In this case,

$$\begin{aligned} v(\widetilde{T^{\prime }}(b);b;(\widetilde{Q^{\prime }}(c))_{c\in A})=v( \widetilde{T}(b);b;(\widetilde{Q}(c))_{c\in A}) \end{aligned}$$

for any embedded coalition $(\widetilde{T}(b);b;(\widetilde{Q}(c))_{c\in A})$ and for all $b\in A$ because $(\widetilde{T^{\prime }}(b);b;(\widetilde{ Q^{\prime }}(c))_{c\in A})$ can be deduced from $(\widetilde{T}(b);b;( \widetilde{Q}(c))_{c\in A})$ by changing the affiliation of the dummy player $j$ within issue $a$ in $(N,A,v).$ Hence again, $\hat{v}(T^{\prime },Q^{\prime })=\hat{v}(T,Q).$

This ends the proof that all the replicas $j(a),$ for all $a\in A,$ are dummy players in $(N(A),\hat{v}).$

If $\phi ^{*}$ satisfies the dummy player axiom, then $\phi _{j(a)}^{*}(N(A),\hat{v})=0$ for all $a\in A$ since $j(a)$ is a dummy player in $(N(A),\hat{v}).$ Therefore,

$$\begin{aligned} \Phi _{j}^{*}(N,A,v)=\underset{a\in A}{\sum }\phi _{j(a)}^{*}(N(A), \hat{v})=0 \end{aligned}$$

and $\Phi ^{*}$ satisfies the dummy player axiom.

(iv) Consider a dummy player $j\in N$ in the game $(N,A,v)$ and a particular issue $a\in A$. First, since $\phi ^{*}$ satisfies the strong dummy player property and $j(a)$ is a dummy player in $(N(A),\hat{v}),$

$$\begin{aligned} \phi _{k}^{*}(N(A)\backslash j(a),\hat{v}_{-j(a)})=\phi _{k}^{*}(N(A),\hat{v}) \end{aligned}$$

for all $k\in N(A)\backslash j(a)$. Second, player $j(b)$, for $b\ne a$, is also a dummy player in the game $(N(A)\backslash j(a),\hat{v}_{-j(a)})$. (If we have two dummy players in any PFG, the second dummy player is still dummy in the game where we have eliminated the first one). Applying this procedure sequentially to all the issues in $A$, and denoting $j(A)=\underset{a\in A}{\cup }j(a)$, we have that

$$\begin{aligned} \phi _{k}^{*}(N(A)\backslash j(A),\hat{v})=\phi _{k}^{*}(N(A)\backslash j(A),\hat{v}_{-j(A)}) \end{aligned}$$

for all $k\in N(A)\backslash j(A)$. Therefore,

$$\begin{aligned}&\Phi _{i}^{*}(N\backslash j,A,v_{-j})=\underset{a\in A}{\sum }\phi _{i(a)}^{*}(\left( N\backslash j\right) (A),\widehat{v_{-j}})=\underset{ a\in A}{\sum }\phi _{i(a)}^{*}(N(A)\backslash j(A),\hat{v}_{-j(A)})\\&\quad =\underset{a\in A}{\sum }\phi _{i(a)}^{*}(N(A),\hat{v})=\Phi _{i}^{*}(N,A,v) \end{aligned}$$

for all $i\in N\backslash j$ and $\Phi ^{*}$ satisfies the strong dummy player axiom.$\square$

Proof of Proposition 2

(i) This property is trivially satisfied.

(ii) If $d$ is a dummy issue in the game $(N,A,v),$ then all the replicas $i(d),$ for any $i\in N,$ are dummy players in $(N(A),\hat{v}),$ because, by the definition of dummy issue, $\hat{v}(T,Q)=\hat{v}(T^{\prime },Q^{\prime })$ for all $(T^{\prime },Q^{\prime })$ obtained from $(T,Q)$ by changing the affiliation of player $i(d),$ for any $i\in N$.

Given that $\phi ^{*}$ satisfies the strong dummy player axiom, then if the $n$ dummy players $i(d)$ are dropped off $N(A)$

$$\begin{aligned} \phi _{k}^{*}(N(A)\backslash (i(d))_{i\in N},\hat{v}_{-(i(d))_{i\in N}})=\phi _{k}^{*}(N(A),\hat{v}), \end{aligned}$$

which implies that for all $i\in N$

$$\begin{aligned}&\Phi _{i}^{*}(N,A\backslash d,v_{-d})=\underset{a\in A}{\sum }\phi _{i(a)}^{*}(N(A\backslash d),\widehat{v_{-d}})\\&\quad =\underset{a\in A}{\sum } \phi _{i(a)}^{*}(N(A)\backslash \{i(d)\}_{i\in N},\hat{v}_{-(i(d))_{i\in N}}) =\underset{a\in A}{\sum }\phi _{i(a)}^{*}(N(A),\hat{v})=\Phi _{i}^{*}(N,A,v) \end{aligned}$$

and $\Phi ^{*}$ satisfies the dummy issue axiom.$\square$

Proof of Proposition 3

(i) Consider the game $\left( N,A,v\right)$ and, for any $a\in A$, define $\left( N,A,v_{a}\right)$ as

$$\begin{aligned} v_{a}(S;a;P^{A})\equiv & {} v(S;a;P^{A})\quad \text { for all }(S;a;P^{A})\in ECL (N,A) \\ v_{a}(S;b;P^{A})\equiv & {} 0\quad \text { for all }b\in A\backslash a\text {, } (S;b;P^{A})\in ECL (N,A). \end{aligned}$$

It is immediate that $v=\sum _{a\in A}v_{a}$. The linearity of $\phi ^{*}$ implies the linearity of $\Phi ^{*}$ (Proposition 1); hence,

$$\begin{aligned} \Phi ^{*}(N,A,v)=\sum _{a\in A}\Phi ^{*}(N,A,v_{a})\text {.} \end{aligned}$$

Similarly, consider the game $\left( N,A,v_{\sigma _{N}}\right)$, where $\sigma _{N}$ is a permutation of the set of players $N$. Remember that the function $v_{\sigma _{N}}$ is defined as $v_{\sigma _{N}}(S;a;P^{A})\equiv v\left( S;a;\left( P^{a},O^{A\backslash a}\right) \right)$ for all $(S;a;P^{A})\in ECL (N,A),$ where $O^{b}=\sigma _{N}P^{b}$ or $O^{b}=P^{b}$ for all $b\in A\backslash a$. Let $B\subset A\backslash a$ be the subset of issues where $\sigma _{N}$ applies, i.e., $O^{b}=\sigma _{N}P^{b}$ for all $b\in B$ and $O^{b}=P^{b}$ for all $b\in A\backslash B\backslash a$. For any particular $a\in A$, we define $\left( N,A,\left( v_{\sigma _{N}}\right) _{a}\right)$ as

$$\begin{aligned} \left( v_{\sigma _{N}}\right) _{a}(S;a;P^{A})\equiv & {} v_{\sigma _{N}}(S;a;P^{A})\quad \text { for all }(S;a;P^{A})\in ECL (N,A) \\ \left( v_{\sigma _{N}}\right) _{a}(S;b;P^{A})\equiv & {} 0\quad \text { for all }b\in A\backslash a\text {, }(S;b;P^{A})\in ECL (N,A). \end{aligned}$$

Given that $v_{\sigma _{N}}=\sum _{a\in A}\left( v_{\sigma _{N}}\right) _{a}$, the linearity of $\Phi ^{*}$ implies

$$\begin{aligned} \Phi ^{*}(N,A,v_{\sigma _{N}})=\sum _{a\in A}\Phi ^{*}\left( N,A,\left( v_{\sigma _{N}}\right) _{a}\right) \text {.} \end{aligned}$$

We now prove that $\Phi _{i}^{*}(N,A,\left( v_{\sigma _{N}}\right) _{a})=\Phi _{i}^{*}(N,A,v_{a})$ for all $i\in N$ for whom $\sigma _{N}(i)=i$, which will prove part (i) of the proposition.

For any $i\in N$,

$$\begin{aligned} \Phi _{i}^{*}(N,A,v_{a})=\underset{b\in A}{\sum }\phi _{i(b)}^{*}(N(A),\widehat{v_{a}}) \end{aligned}$$

where

$$\begin{aligned} \widehat{v_{a}}(T,Q)=\underset{b\in A}{\sum }v_{a}(\widetilde{T}(b);b;\left( \widetilde{Q}(c))_{c\in A}\right) =v_{a}(\widetilde{T}(a);a;\left( \widetilde{Q}(c))_{c\in A}\right) \end{aligned}$$

for any $(T,Q)\in ECL (N(A))$, since the other terms in the sum are zero by construction of the function $v_{a}$. Also,

$$\begin{aligned} \Phi _{i}^{*}\left( N,A,\left( v_{\sigma _{N}}\right) _{a}\right) =\underset{b\in A}{ \sum }\phi _{i(b)}^{*}\left( N(A),\widehat{\left( v_{\sigma _{N}}\right) _{a}}\right) \end{aligned}$$

where

$$\begin{aligned} \widehat{\left( v_{\sigma _{N}}\right) _{a}}(T,Q)=\underset{b\in A}{\sum } \left( v_{\sigma _{N}}\right) _{a}(\widetilde{T}(b);b;\left( \widetilde{Q} (c))_{c\in A}\right) =\left( v_{\sigma _{N}}\right) _{a}(\widetilde{T} (a);a;\left( \widetilde{Q}(c))_{c\in A}\right) \end{aligned}$$

for any $(T,Q)\in ECL (N(A))$. We notice that, by definition of $v_{\sigma _{N}}, \left( v_{\sigma _{N}}\right) _{a}(S;a;P^{A})=v(S;a;\left( P^{a},O^{A\backslash a}\right) )$ for all $(S;a;P^{A})\in ECL (N,A),$ where $O^{b}=\sigma _{N}P^{b}$ or $O^{b}=P^{b}$ for all $b\in A\backslash a$ (and $\left( v_{\sigma _{N}}\right) _{a}(S;b;P^{A})=0$ for all $b\in A\backslash a, (S;b;P^{A})\in ECL (N,A)$). Since $\left( v_{\sigma _{N}}\right) _{a}$ only permutes the roles of the players involved in a subset of issues $B\subset A\backslash a, \widehat{\left( v_{\sigma _{N}}\right) _{a}}$ only permutes the roles of the players in each $N(b)$, for all $b\in B$. In fact, $\widehat{\left( v_{\sigma _{N}}\right) _{a}}=\sigma _{N(A)}\widehat{ v_{a}}$, where the permutation $\sigma _{N(A)}$ is as follows:

$$\begin{aligned} \sigma _{N(A)}(i(c))= & {} i(c)\quad \text { for all }i\in N\text { and all }c\in A\backslash B. \\ \sigma _{N(A)}(i(b))= & {} \left( \sigma _{N}\left( i\right) \right) (b)\quad \text { for all }i\in N\text { and all }b\in B\text {.} \end{aligned}$$

Given that $\phi ^{*}$ satisfies player anonymity,

$$\begin{aligned} \phi _{i(c)}^{*}\left( N(A),\widehat{\left( v_{\sigma _{N}}\right) _{a}}\right) =\phi _{i(c)}^{*}(N(A),\sigma _{N(A)}\widehat{v_{a}})=\phi _{i(c)}^{*}(N(A),\widehat{v_{a}}) \end{aligned}$$

for all $i\in N$ and $c\in A\backslash B$ and

$$\begin{aligned} \phi _{i(b)}^{*}(N(A),\widehat{\left( v_{\sigma _{N}}\right) _{a}})=\phi _{\left( \sigma _{N}\left( i\right) \right) (b)}^{*}(N(A),\sigma _{N(A)} \widehat{v_{a}})=\phi _{\left( \sigma _{N}\left( i\right) \right) (b)}^{*}(N(A),\widehat{v_{a}}) \end{aligned}$$

for all $i\in N$ and all $b\in B$. In particular, $\phi _{i(b)}^{*}(N(A), \widehat{\left( v_{\sigma _{N}}\right) _{a}})=\phi _{i(b)}^{*}(N(A), \widehat{v_{a}})$ for all $i\in N$ for whom $\sigma _{N}(i)=i$. This implies that $\Phi _{i}^{*}(N,A,\left( v_{\sigma _{N}}\right) _{a})=\Phi _{i}^{*}(N,A,v_{a})$ for any $i\in N$ for whom $\sigma _{N}(i)=i$, and the result holds.

(ii) Consider the game $\left( N,A,v\right)$, a set $M$ of $a$-externality players, and $b\ne a.$ We will show that if $\phi ^{*}$ satisfies linearity and player anonymity in PFG, then $\Phi _{i}^{*}\left( N,A,v_{M,ab}\right) =\Phi _{i}^{*}\left( N,A,v\right)$ for all $i\in N$. Notice that $\Phi _{i}^{*}(N,A,v)=\underset{c\in A}{\sum }\phi _{i(c)}^{*}(N(A),\widehat{v}),$ where $\widehat{v}(T,Q)=\underset{c\in A}{\sum }v(\widetilde{T}(c);c;\left( \widetilde{Q}(d))_{d\in A}\right)$, and $\Phi _{i}^{*}(N,A, v_{M,ab})=\underset{c\in A}{\sum }\phi _{i(c)}^{*}(N(A),\widehat{v_{M,ab}}),$ where $\widehat{v_{M,ab}}(T,Q)=\underset{c\in A}{\sum }v_{M,ab}(\widetilde{T}(c);c;\left( \widetilde{Q}(d))_{d\in A}\right)$. We consider the following permutation $\sigma _{N(A)}$ on the set $N(A):\ \sigma _{N(A)}(i(a))=i(b)$ and $\sigma _{N(A)}(i(b))=i(a)$ for all $i\in M$ and $\sigma _{N(A)}(k)=k$ otherwise. Applying the permutation $\sigma _{N(A)}$ to the value function $\widehat{v}$ has the same effect as going from $v$ to $v_{M,ab}$: It moves the roles of players in $M\$from issue $a$ to issue $b$. Hence, $\sigma _{N(A)}\widehat{v}=\widehat{v_{M,ab}}$.

Given that the value $\phi ^{*}$ satisfies anonymity, it is the case that

$$\begin{aligned} \phi _{i(c)}^{*}(N(A),\widehat{v_{M,ab}})=\phi _{i(c)}^{*}(N(A),\sigma _{N(A)}\widehat{v})=\phi _{\sigma _{N(A)}(i(c))}^{*}(N(A), \widehat{v}). \end{aligned}$$

Given that $\sigma _{N(A)}$ only permutes replicas of the same players (those in $M$), it is the case that

$$\begin{aligned} \underset{c\in A}{\sum }\phi _{i(c)}^{*}(N(A),\widehat{v_{M,ab}})= \underset{c\in A}{\sum }\phi _{\sigma _{N(A)}(i(c))}^{*}(N(A),\widehat{v} )=\underset{c\in A}{\sum }\phi _{i(c)}^{*}(N(A),\widehat{v}) \end{aligned}$$

(since $\phi _{\sigma _{N(A)}(i(a))}^{*}(N(A),\widehat{v})+\phi _{\sigma _{N(A)}(i(b))}^{*}(N(A),\widehat{v})\!=\!\phi _{i(b)}^{*}(N(A),\widehat{v })+\phi _{i(a)}^{*}(N(A),\widehat{v})$ for $i\in M$). Therefore, $\Phi _{i}^{*}\left( N,A,v_{M,ab}\right) =\Phi _{i}^{*}\left( N,A,v\right)$ as we wanted to prove.$\square$

Proof of Theorem 1

The sufficiency part of the theorem is a corollary of Propositions 1, 2, and 3. We prove the necessity part through a series of steps. Take any game $(N,A,v)$ in ${\mathcal {G}}$.

Step 1. For any $a\in A$, we define the following game $(N,A,v_{a})$:

$$\begin{aligned} v_{a}(S;a;P^{A})\equiv & {} v(S;a;P^{A})\quad \text {for all }(S;a;P^{A})\in ECL (N,A) \\ v_{a}(S;b;P^{A})\equiv & {} 0\quad \text {for all }b\in A\backslash a\text {, } (S;b;P^{A})\in ECL (N,A). \end{aligned}$$

That is, the worth of a coalition on issue $a$ in the game $v_{a}$ is the same as that in $v$; however, the worth of a coalition on any other issue is zero in game $v_{a}$. Note that the organization of the players on issues other than $a$ influences the worth of coalitions in issue $a$ in the game $v_{a}$ in the same way as it does in $v$.

It is immediate that

$$\begin{aligned} v=\sum _{a\in A}v_{a}. \end{aligned}$$

Therefore, if $\Phi$ satisfies the axiom of linearity then,

$$\begin{aligned} \Phi (N,A,v)=\sum _{a\in A}\Phi (N,A,v_{a}). \end{aligned}$$

Step 2. For each $(N,A,v_{a}),$ we now define a related game $(N(A),A,w_{a})$, which is similar to $(N,A,v_{a})$ except that we add $(\left| A\right| -1)n$ dummy players. More precisely, for each $b\in A\setminus a,$ let $N(b)=\left\{ i(b)\mid i\in N\right\}$ be the $b$-replica of $N$ and for convenience, let $N(a)\equiv N$ (i.e., $N(a)$ is the original set of players). Then, the set of players in the new game is $N(A)=\cup _{b\in A}N(b)$ with $N(A)\setminus N(a)$ being dummy players. Therefore, for every $a\in A, (N(A),A,w_{a})$ is defined as follows:²⁰

$$\begin{aligned} w_{a}(T;a;Q^{A})\equiv v_{a}(\widetilde{T}(a);a;(\widetilde{Q^{b}}(a))_{b\in A}) \end{aligned}$$

for all $(T;a;Q^{A})\in ECL (N(A),A)$ (i.e., for all vector $Q^{A}$ of $\left| A\right|$ partitions of $N(A)$ and any $T\in Q^{a}$), and

$$\begin{aligned} w_{a}(T;b;Q^{A})\equiv v_{a}(\widetilde{T}(a);b;(\widetilde{Q^{b}}(a))_{b\in A})=0 \end{aligned}$$

for all $b\in A\backslash a$ and all $(T;b;Q^{A})\in ECL (N(A),A)$.

Given that $\Phi$ satisfies the axioms of strong dummy player and player anonymity (2’), we have

$$\begin{aligned} \Phi _{i}(N(A),A,w_{a})= & {} \Phi _{i}(N,A,v_{a})\quad \text { for all }i\in N(a)=N \\ \Phi _{i}(N(A),A,w_{a})= & {} 0\quad \text { for all }i\in N(A)\backslash N(a)\text {.} \end{aligned}$$

Step 3. Next, for each $a\in A,$ we define another game $(N(A),A,z_{a})$ that is related to $(N(A),A,w_{a})$ in the following sense. First, as in $(N(A),A,w_{a})$, a coalition of players obtains worth only on issue $a$. Second, only players in $N(a)$ create worth. Third, the inter-issue externalities in $(N(A),A,z_{a})$ are “similar” to those in $(N(A),A,w_{a})$; however, there is one important difference: In game $(N(A),A,z_{a}),$ the externalities originating from each issue $b\in A\backslash a$ are exerted by players in $N(b),$ rather than by players in $N(a)$ as in game $(N(A),A,w_{a})$. That is, the game $(N(A),A,z_{a})$ is defined as follows:

$$\begin{aligned} z_{a}(T;a;Q^{A})\equiv w_{a}(T;a;R^{A}) \end{aligned}$$

for all $(T;a;Q^{A})\in ECL (N(A),A)$, where $R^{A}$ is a vector of $\left| A\right|$ partitions of $N(A)$ such that $R^{a}=Q^{a}$ and for every $b\in A\backslash a, R^{b}$ is obtained from $Q^{b}$ by exchanging the memberships of $i(a)$ and $i(b)$ for each $i\in N,$ ²¹ and

$$\begin{aligned} z_{a}(T;b;Q^{A})\equiv 0 \end{aligned}$$

for all $b\in A\backslash a$ and all $(T;b;Q^{A})\in ECL (N(A),A)$.

Note that $z_{a}(T;a;Q^{A})=v_{a}\left( \widetilde{T}(a);a;\left( \widetilde{ Q^{b}}(b)\right) _{b\in A}\right)$ for all $(T;a;Q^{A})\in ECL (N(A),A)$.²²

We claim that, by issue-externality anonymity axiom,

$$\begin{aligned} \sum _{b\in A}\Phi _{i(b)}(N(A),A,z_{a})=\Phi _{i}(N,A,w_{a})\quad \text { for all } i\in N=N(a). \end{aligned}$$

(4)

We prove this claim by decomposing the change from $(N(A),A,w_{a})$ to $(N(A),A,z_{a})$ in $\left| N\right| \left( \left| A\right| -1\right)$ stages. In each stage, we switch the membership of some $i(a)\in N(a)$ with that of $i(b)\in N(b)$ in the partition $P^{b}$ on some issue $b\in A\backslash a$. In doing so, $i(b)$ takes the role of $i(a)$ in generating externalities from issue $b.$ Note that the identities of the players who create worth (always on issue $a)$ remain the same. Then, by the issue-externality anonymity axiom, the value of every player different from $i(a)$ and $i(b)$ should not change; hence, the sum of the values for players $i(a)$ and $i(b)$ should not change either. Repeating this argument cross-issues implies that after $\left| A\right| -1$ stages of switching the membership of $i(a)\in N(a)$ with $i(b)\in N(b)$ for every issue $b\in A\backslash a,$ the sum of the values for all replicas of player $i$ remains unchanged, while the value of each of the remaining players stays the same throughout these stages. By repeating the above stages for all $i(a)\in N(a),$ we complete our transformation from $(N(A),A,w_{a})$ to $(N(A),A,z_{a})$ and obtain Eq. (4).

Step 4. For each $(N(A),A,z_{a}),$ we now define a related game $(N(A),A,r_{a})$ such that all externalities are generated from issue $a.$ Recall that in $(N(A),A,z_{a}),$ for any $(T;a;Q^{A})\in ECL (N(A),A)$, the worth of $T$ depends only on $\left( Q^{b}(b\right) )_{b\in A}$; moreover, only a coalition of players in $N(a)$ can create worth and it does so only on issue $a.$ In fact, for each $b\in A\backslash a, N(b)$ is a set of $b$-externality players in $(N(A),A,z_{a}).$ We define the game $(N(A),A,r_{a})$ by encoding the externalities exerted by $N(b)$ for all $b\in A\backslash b$ in $z_{a}$:

$$\begin{aligned} r_{a}(T;a;Q^{A})\equiv z_{a}(T;a;R^{A}) \end{aligned}$$

for all $(T;a;Q^{A})\in ECL (N(A),A)$, where $R^{A}$ is a vector of $\left| A\right|$ partitions of $N(A)$ such that $R^{a}=Q^{a}$ and for every $b\in A\backslash a, R^{b}$ is such that $R^{b}\cap N(b)=Q^{a}\cap N(b).$ Thus, $r_{a}$ can be obtained from $z^{a}$ from $(\left| A\right| -1)$ steps of transformation, each involving moving the externalities induced by $N(b)$, for a particular $b\in A\backslash a$, from issue $b$ to issue $a$.

Note that $r_{a}(T;a;Q^{A})=v_{a}\left( \widetilde{T}(a);a;\left( \widetilde{ Q^{a}}(b)\right) _{b\in A}\right)$ for all $(T;a;Q^{A})\in ECL (N(A),A)$.

By the issue-externality symmetry axiom,

$$\begin{aligned} \Phi _{k}(N(A),A,r_{a})=\Phi _{k}(N,A,w_{a})\quad \text { for all}\quad k\in N(A). \end{aligned}$$

We also note that all issues in $A\backslash a$ are dummy issues in $(N(A),A,r_{a})$.

Step 5. Finally, we define game $(N(A),a,s_{a})$ by eliminating the set of dummy issues $A\backslash a$ in $(N(A),A,r_{a}),$ that is,

$$\begin{aligned} s_{a}(T;a;Q)\equiv r_{a}(T;a;Q^{A}) \end{aligned}$$

for any $(T;a;Q)\in ECL (N(A),a)$ and any vector $Q^{A}$ of $\left| A\right|$ partitions of $N(A)$ that satisfies $Q^{a}=Q$. By the dummy issue axiom, we have

$$\begin{aligned} \Phi _{k}(N(A),a,s_{a})=\Phi _{k}(N(A),A,r_{a})\quad \text { for all}\quad k\in N(A). \end{aligned}$$

Note that $(N(A),a,s_{a})$ is a game with a single issue ($a$ in this case). Therefore, we can consider $(N(A),a,s_{a})$ as a PFG, which we denote $(N(A),\widetilde{s_{a}})$. Moreover, when it is applied to games with only one issue, the issue symmetry axiom implies that the value $\Phi$ depends only on the function that gives the worth of each embedded coalition, not on the identity of the issue itself. Thus, $\Phi$ also defines a value for $PFG .$ Let $\phi$ be this value. Hence,

$$\begin{aligned} \phi _{k}(N(A),\widetilde{s_{a}})=\Phi _{k}(N(A),a,s_{a})\quad \text { for all}\quad k\in N(A). \end{aligned}$$

Therefore, steps 1–5 allow us to obtain the following series of equalities for every $i\in N$:

$$\begin{aligned}&\Phi _{i}(N,A,v)=\sum _{a\in A}\Phi _{i}(N,A,v_{a})=\sum _{a\in A}\Phi _{i(a)}(N(A),A,w_{a}) \\&\quad =\sum _{a\in A}\sum _{b\in A}\Phi _{i(b)}(N(A),A,z_{a})=\sum _{a\in A}\sum _{b\in A}\Phi _{i(b)}(N(A),A,r_{a})\\&\quad =\sum _{a\in A}\sum _{b\in A}\Phi _{i(b)}(N(A),a,s_{a}) \\&\quad =\sum _{a\in A}\sum _{b\in A}\phi _{i(b)}(N(A),\widetilde{s_{a}})=\sum _{b\in A}\sum _{a\in A}\phi _{i(b)}(N(A),\widetilde{s_{a}}). \end{aligned}$$

We now prove that $\hat{v}=\sum _{a\in A}\widetilde{s_{a}}.$ Consider any partition $Q$ of $N(A)$ and any coalition $T\in Q$. By construction,

$$\begin{aligned} \widetilde{s_{a}}(T;Q)=s_{a}(T;a;Q)=r_{a}(T;a;Q^{A}), \end{aligned}$$

where $Q^{A}$ is any vector of $\left| A\right|$ partitions of $N(A)$ that satisfies $Q^{a}=Q$. Also,

$$\begin{aligned}&r_{a}(T;a;Q^{A})=v_{a}\left( \widetilde{T}(a);a;\left( \widetilde{Q^{a}} (b)\right) _{b\in A}\right) =v_{a}(\widetilde{T}(a);a;\left( \widetilde{Q} (b)\right) _{b\in A})\\&\quad =v(\widetilde{T}(a);a;\left( \widetilde{Q}(b)\right) _{b\in A}). \end{aligned}$$

Hence,

$$\begin{aligned} \sum _{a\in A}\widetilde{s_{a}}(T;Q)=\sum _{a\in A}v(\widetilde{T}(a);a;\left( \widetilde{Q}(b)\right) _{b\in A})=\hat{v}(T,Q). \end{aligned}$$

Finally, linearity of $\Phi$ implies that the value $\phi$ is also linear and $\phi _{k}(N(A),\hat{v})=\sum _{a\in A}\phi _{k}(N(A),\widetilde{s_{a}})$ for all $k\in N(A).$ Therefore,

$$\begin{aligned} \Phi _{i}(N,A,v)=\sum _{b\in A}\phi _{i(b)}(N(A),\hat{v}) \end{aligned}$$

which completes the proof of Theorem 1.$\square$

Proof of Proposition 4

Take two games $(N,A,v)$ and $(N,B,w),$ with $A\cap B=\varnothing$, and consider a value $\Phi$ that satisfies the dummy issue axiom. We add to the first game $\left| B\right|$ dummy issues, obtaining the game $(N,A\cup B,v^{\prime })$ where $v^{\prime }$ is a characteristic function such that

$$\begin{aligned} v^{\prime }(S;a;P^{A\cup B})= & {} v(S;a;P^{A})\quad \text { for all }a\in A,S\in P^{a},P^{a}\in P^{A} \\ v^{\prime }(S;b;P^{A\cup B})= & {} 0\quad \text { for all }b\in B,S\in P^{b},\text { and }P^{b}\in P^{B}\text {.} \end{aligned}$$

By the dummy issue property, $\Phi$ assigns the same payoff in both games to any player $i\in N,$ i.e.,

$$\begin{aligned} \Phi _{i}(N,A,v)=\Phi _{i}(N,A\cup B,v^{\prime }). \end{aligned}$$

Similarly, if we add to the game $(N,B,w)$ a set of $\left| A\right|$ dummy issues, we obtain the game $(N,A\cup B,w^{\prime })$ where $w^{\prime }$ is a characteristic function such that

$$\begin{aligned} w^{\prime }(S;a;P^{A\cup B})= & {} 0\quad \text { for all }a\in A,S\in P^{a},P^{a}\in P^{A} \\ w^{\prime }(S;b;P^{A\cup B})= & {} w(S;b;P^{A})\quad \text { for all }b\in B,S\in P^{b},\text { and }P^{b}\in P^{B}. \end{aligned}$$

Again, by the dummy issue axiom, we have

$$\begin{aligned} \Phi _{i}(N,B,w)=\Phi _{i}(N,A\cup B,w^{\prime }),\quad \text { for all }i\in N. \end{aligned}$$

Since $\Phi$ satisfies linearity,

$$\begin{aligned} \Phi _{i}(N,A\cup B,v^{\prime })+\Phi _{i}(N,A\cup B,w^{\prime })=\Phi _{i}(N,A\cup B,v^{\prime }+w^{\prime }). \end{aligned}$$

Finally, we notice that the game $(N,A\cup B,v^{\prime }+w^{\prime })$ is equivalent to $(N,A\cup B,v^{\prime }\cup w^{\prime });$ hence,

$$\begin{aligned} \Phi _{i}(N,A,v)+\Phi _{i}(N,B,w)=\Phi _{i}(N,A\cup B,v^{\prime }\cup w^{\prime }) \end{aligned}$$

and the independence axiom is satisfied.$\square$