Joint games and compatibility

Abstract

We introduce the concepts of joint games and compatibility. In a joint game, members of the grand coalition have the option to split and participate in different underlying games, thereby maximizing their total worths. In order to determine whether the grand coalition will remain intact, we introduce the notion of compatibility of these games. A set of games is compatible if the core of the joint game is non-empty. We find a necessary and sufficient condition for compatibility.

Appendix: Proofs

Proof of Theorem 1

We prove that the games $v_1,\ldots ,v_K$ are compatible if and only if

$$\begin{aligned} {\mathop {\mathop {\max }\limits _{f_1+\cdots +f_K=1}}\limits _{f_i:N\rightarrow {\mathbb {R}}_+}} \sum \limits _{1\le i\le K}\int ^\mathrm{cav} f_i \mathrm{d}v_i\le v_1\bullet \cdots \bullet v_K(N) \end{aligned}$$

(6.1)

in six steps.

Step 1 Suppose that $u(X)=\inf \nolimits _{P\in {\mathcal {Q}}}\int X \mathrm{d}P$, where ${\mathcal {Q}}$ is a comprehensive,⁴ convex and closed set of additive games. We show that $u(X)=-\infty$ whenever one of the coordinates of $X$ is negative. Indeed, let $X=(x_1,\ldots ,x_n)\in {\mathbb {R}}^n$ such that $x_{r_0}<0$, for some $1\le r_0\le n$, and let $P=(p_1,\ldots ,p_n)\in {\mathcal {Q}}$. For any $\lambda >0$, define an additive game $P^\lambda =(p^\lambda _{1},\ldots ,p^\lambda _{n})$ as $p^\lambda _{r}=p_r$ if $r\not =r_0$, and $p^\lambda _{r_0}= p_{r_0}+\lambda$. Since ${\mathcal {Q}}$ is comprehensive, $P^\lambda \in {\mathcal {Q}}$. Therefore,

$$\begin{aligned} u(X)=\min \limits _{P\in {\mathcal {Q}}}\int X\mathrm{d}P\le \int X\mathrm{d}P^\lambda =\int X\mathrm{d}P+\lambda x_{r_0}, \end{aligned}$$

which tends to $-\infty$ when $\lambda \rightarrow \infty$.

Step 2 Suppose that $u_i(X)=\inf \nolimits _{P_i\in {\mathcal {Q}}_i}\int X \mathrm{d}P_i$, for $i=1,\ldots ,K$, where ${\mathcal {Q}}_i$ is a comprehensive set of additive games. The convolution of $u_1,\ldots ,u_K$, denoted $u_1\star \cdots \star u_K$, is defined as

$$\begin{aligned} u_1\star \cdots \star u_K(X):=\sup \limits _{\left\{ (X_l)_{l=1}^K\in ({\mathbb {R}}^n)^K\vert X_1+\cdots +X_K=X\right\} }u_1(X_1)+\cdots +u_K(X_K). \end{aligned}$$

We claim that

$$\begin{aligned} u_1\star \cdots \star u_K(X)=\inf \limits _{P\in \cap _{i}{\mathcal {Q}}_i}\int X\mathrm{d}P. \end{aligned}$$

In order to prove this claim, we use convex analysis. Consider a set of convex functions $f_1,\ldots ,f_K$ from ${\mathbb {R}}^n$ to ${\mathbb {R}}\cup \{+\infty \},$ whose conjugates are defined as

$$\begin{aligned} f_i^*(P)=\sup \limits _{X\in {\mathbb {R}}^n}\left\{ \int X \mathrm{d}P-f_i(X)\right\} ,\quad i=1,\ldots ,K. \end{aligned}$$

The infimal convolution of these $K$ functions is

$$\begin{aligned} h(X)=\inf \limits _{\left\{ (X_l)_{l=1}^K\in ({\mathbb {R}}^n)^K\vert X_1+\cdots +X_K=X\right\} }\{f(X_1)+\cdots +f(X_K)\}. \end{aligned}$$

We state the following version of Theorem 20, d) in Rockafellar (1973) to be used later. $\square$

Theorem 4

Suppose that $\exists \bar{X}\in {\mathbb {R}}^n$ and $M\in {\mathbb {R}}$ such that the set

$$\begin{aligned} {\mathcal {K}}:=\left\{ (X_1,\ldots ,X_K)\in ({\mathbb {R}}^n)^K\Bigg \vert \begin{array}{ll}X_1+\cdots +X_K=\bar{X}, \\ f_1(X_1)+\cdots +f_K(X_K)\le M\end{array}\right\} \end{aligned}$$

(6.2)

is non-empty and bounded. Then, $h^*=f^*_1+\cdots +f^*_K$.

Let $f_i=-u_i$. We verify that the condition of Theorem 4 is satisfied. Take $\bar{X}=\overrightarrow{0}$ and $M=0$. First, for every $i=1,\ldots ,K$ ${\mathcal {Q}}_i$ is comprehensive and therefore, $f_i$ is non-positive on ${\mathbb {R}}^n_+$. In particular, $f_i(\bar{X})\le 0$, implying that $\bar{X}^K:=(\underbrace{\bar{X},\ldots ,\bar{X}}_{K ~\mathrm{times}})\in {\mathcal {K}}$. Second, according to Step 1, ${\mathcal {K}}\subseteq ({\mathbb {R}}^n_+)^K$. This implies that $\bar{X}^K$ is the only point in ${\mathcal {K}}$, meaning that ${\mathcal {K}}$ is non-empty and bounded.

Having verified the condition of Theorem 4, we proceed to apply it to $f_i$. By the definition of the conjugate function, $f^*_i(Q)=\sup \nolimits _{X\in {\mathbb {R}}^n}\left\{ \int XdQ+u_i(X)\right\}$. Since for all $\lambda >0, u_i(\lambda X)=\lambda u_i( X)$, we have

$$\begin{aligned} \begin{array}{llllll} f^*_i(Q)&{}=\sup \limits _{X\in {\mathbb {R}}^n}\left\{ \int XdQ+u_i(X)\right\} \\ &{}=\sup \limits _{\lambda X\in {\mathbb {R}}^n}\left\{ \int \lambda XdQ+u_i(\lambda X)\right\} \\ &{}=\lambda \sup \limits _{X\in {\mathbb {R}}^n}\left\{ \int XdQ+u_i(X)\right\} \\ &{}=\lambda f^*_i(Q). \end{array} \end{aligned}$$

This implies that either $f^*_i(Q)=0$ or $f^*_i(Q)=+\infty$.

We now elaborate on $(f^*_1+\cdots +f^*_K)(P)$. If $P$ is such that for any $i, -P\in {\mathcal {Q}}_i$, then by $\int X\mathrm{d}P+u_i(X)\le \int X\mathrm{d}P-\int X\mathrm{d}P=0$, we have $f^*_i(P)=0$. On the other hand, if there exists $i~{\mathrm{such~that}} -P\not \in {\mathcal {Q}}_i$, then due to the assumption that ${\mathcal {Q}}_i$ is convex and closed, there is a unique $R\in {\mathcal {Q}}_i$ such that $R$ is the closest point (in the Euclidean norm) to $-P$ in ${\mathcal {Q}}_i$. Let, $X:=-P-R\not =0$. The integral $\int X\mathrm{d}(-P)- \int X\mathrm{d}R'$ is bounded away from 0 for every $R'\in {\mathcal {Q}}_i$. Thus, $f^*_i(-P)=+\infty$. This implies that $f^*_1+\cdots +f^*_K$ is zero iff $-P\in \cap _{i=1}^K{\mathcal {Q}}_i$ and $+\infty$ otherwise. That is,

$$\begin{aligned} (f^*_1+\cdots +f^*_K)(P)=\left\{ \begin{array}{ll} 0, &{}\quad \text {if}\,\,\, {-}P\in \bigcap \nolimits _{i=1}^K{\mathcal {Q}}_i\\ +\infty ,&{}\quad \text {otherwise} \end{array}\right. . \end{aligned}$$

(6.3)

Now define, $g(X)=-\inf \nolimits _{P\in -\bigcap \nolimits _{i=1}^K{\mathcal {Q}}_i}\int X\mathrm{d}P$. Since $\bigcap \nolimits _{i=1}^K{\mathcal {Q}}_i$ is a comprehensive, convex and closed set, the same argument just used implies that $g^*$ is equal to (6.3). Using Theorem 4, we conclude that $h^*=f^*_1+\cdots +f^*_K=g^*$. Since $h$ and $g$ are both lower semicontinuous functions, it implies that they are equal (see Rockafellar 1973). Thus, $h(X)= -\inf \nolimits _{-P\in \bigcap \nolimits _{i=1}^K{\mathcal {Q}}_i}\int X\mathrm{d}P$, which implies,

$$\begin{aligned} u_1\star \cdots \star u_K(X)= & {} \sup \limits _{\left\{ (X_l)_{l=1}^K\in ({\mathbb {R}}^n)^K\vert X_1+\cdots +X_K=X\right\} }u_1(X_1)+\cdots +u_K(X_K) \\= & {} \inf \limits _{\left\{ (X_l)_{l=1}^K\in ({\mathbb {R}}^n)^K\vert X_1+\cdots +X_K=X\right\} }\{f(X_1)+\cdots +f(X_K)\} \\= & {} h(X)= -\inf \limits _{-P\in \bigcap \limits _{i=1}^K{\mathcal {Q}}_i}\int X\mathrm{d}P=\inf \limits _{P\in \bigcap \limits _{i=1}^K{\mathcal {Q}}_i}\int X\mathrm{d}P , \end{aligned}$$

which proves our claim.

Step 3 For a set of games $v_i, i=1,\ldots ,K, \, \text {and any additive game}~P$, it is easy to see that

$$\begin{aligned} P(S)\ge v_1\bullet \cdots \bullet v_K(S),~ \forall S\subseteq N~\mathbf{iff}~P(S)\ge v_i(S),\quad \forall S\subseteq N,\quad \forall i=1,\ldots ,K. \end{aligned}$$

Step 4 Consider the games $v_1,\ldots ,v_K$ and the functions $u_i(X)=\inf \nolimits _{P\in {\mathcal {Q}}_i}\int X \mathrm{d}P$, where ${\mathcal {Q}}_i$ is the set of all additive games $P$ such that $P\ge v_i$. ${\mathcal {Q}}_i$ is comprehensive, convex and closed, and therefore, Step 1 applies. By the definition of the $\star$ operator and Step 1 ($u_i(X_i)=-\infty$ for every $X_i\not \in {\mathbb {R}}^n_+$), for $X\ge 0$, we therefore have,

$$\begin{aligned} \left( \int ^\mathrm{cav}\cdot \ \mathrm{d}v_1\star \cdots \star \int ^\mathrm{cav} \cdot \ \mathrm{d}v_K\right) (X)= & {} {\mathop {\mathop {\max }\limits _{X_1+\cdots +X_K=X}}\limits _{\forall i, X_i\ge 0}}\int ^\mathrm{cav}X_1\mathrm{d}v_1\\&\quad +\,\cdots + \int ^\mathrm{cav}X_K\mathrm{d}v_K. \end{aligned}$$

Note that the maximum is replacing the supremum because all concave functions, by Remark 3, are continuous and the set $\left\{ (X_i)_{i=1}^K\vert X_1+\cdots +X_K=X, \forall i, X_i\ge 0\right\}$ is compact.

Step 5 For every $S\subseteq N$, the following equality holds,

$$\begin{aligned} B_{v_1\bullet \cdots \bullet v_K}(S)={\mathop {\mathop {\max }\limits _{f_1+\cdots +f_K=\mathbf{1 }_S}}\limits _{\forall i, f_i\ge 0}} \sum \limits _{1\le i\le K}\int ^\mathrm{cav} f_i \mathrm{d}v_i. \end{aligned}$$

Indeed,

$$\begin{aligned} \int ^\mathrm{cav}Xd(v_1\bullet \cdots \bullet v_K)&=\min \limits _{P\ge v_1\bullet \cdots \bullet v_K}\int ^\mathrm{cav}X\mathrm{d}P\\&=\min \limits _{\cap \{P\ge v_i\}}\int X\mathrm{d}P\\&=\left( \int ^\mathrm{cav} \cdot ~ \mathrm{d}v_1\star \cdots \star \int ^\mathrm{cav} \cdot ~ \mathrm{d}v_K\right) (X)\\&= {\mathop {\mathop {\max }\limits _{X_1+\cdots +X_K=X}}\limits _{\forall i, X_i\ge 0}}\int ^\mathrm{cav}X_1\mathrm{d}v_1+\cdots + \int ^\mathrm{cav}X_K\mathrm{d}v_K. \end{aligned}$$

In the first equality, we use the definition of the concave integral; in the second equality, we use Step 3; in the third, we use Step 2; and in the fourth equality, we use Step 4. Using $X=\mathbf{1 }_S$, we get the result.

Step 6 Given Step 5 and (4.4) we have,

$$\begin{aligned} C(v_1\bullet \cdots \bullet v_K)\not =\emptyset&\Leftrightarrow B_{v_1\bullet \cdots \bullet v_K}(N)\le v_1\bullet \cdots \bullet v_K(N)\\&\Leftrightarrow {\mathop {\mathop {\max }\limits _{f_1+\cdots +f_K=1}}\limits _{\forall i, f_i\ge 0}} \sum \limits _{1\le i\le K}\int ^\mathrm{cav} f_i \mathrm{d}v_i\le v_1\bullet \cdots \bullet v_K(N). \end{aligned}$$

$\square$

Proof of Theorem 2

Let $(v_1, N_1),\ldots , (v_K, N_K)$ be $K$ games and recall that we have defined $N=N_1\cup \cdots \cup N_K$. We show first that if $(v_1, N_1),\ldots , (v_K, N_K)$ are compatible then,

$$\begin{aligned} {\mathop {\mathop {\max }\limits _{f_1+\cdots +f_K=1}}\limits _{f_i\in {\mathcal {F}}_i,~ i=1,\ldots ,K}} \sum \limits _{1\le i\le K}\int ^\mathrm{cav} f_{i|N_i} \mathrm{d}v_i\le v_1\bullet \cdots \bullet v_K(N). \end{aligned}$$

Suppose that $f_1+\cdots +f_K=1$, where $f_i\in {\mathcal {F}}_i,~ i=1,\ldots ,K$ and that $\int ^\mathrm{cav} f_{i|N_i} \mathrm{d}v_i=\sum \nolimits _{j=1}^{r_i}v_i(S_i^j)$, where $S_i^j\subseteq N_i$ and $\sum \nolimits _{j=1}^{r_i}\mathbf{1 }_{S_i^j}=f_i$. Then, $\sum \nolimits _{1\le i\le K}\int ^\mathrm{cav} f_{i|N_i} \mathrm{d}v_i=\sum \nolimits _{1\le i\le K}\sum \nolimits _{j=1}^{r_i}v_i(S_i^j)\le v_1\bullet \cdots \bullet v_K(N)$. The inequality holds because $\sum \nolimits _{1\le i\le K}\sum \nolimits _{j=1}^{r_i}\mathbf{1 }_{S_i^j}=f_1+\cdots +f_K=\mathbf{1 }_N$.

As for the inverse direction, for any positive number, $M$, and $i=1,\ldots ,K$, define the game $(v_i^M,N)$ as follows.

$$\begin{aligned} v_i^M(S)= v_i^M(S\cap N_i)-|S\setminus N_i|\cdot M. \end{aligned}$$

The worth $v_i^M(S)$ coincides with $v_i(S)$ as long as $S\subseteq N_i$. Any player out of $N_i$ is worth $-M$, which is also her contribution to any coalition in $N_i$. When $M$ is sufficiently large, in order to obtain $v_1^M\bullet \cdots \bullet v_K^M(S)$, it is optimal to split $S$ as $S=S_1\cup \cdots \cup S_K$, where for any $i, S_i\subseteq N_i$. Since there are finitely many coalitions in $N$, for sufficiently large $M, v_1^M\bullet \cdots \bullet v_K^M(S)= v_1\bullet \cdots \bullet v_K(S)$ for every $S\subseteq N$. It implies that $(v_1, N_1),\ldots , (v_K, N_K)$ are compatible whenever $(v_1^M, N_1),\ldots , (v_K^M, N_K)$ are.

We now use Theorem 1 and apply it to $(v_1^M, N_1),\ldots , (v_K^M, N_K)$. Theorem 1 states that these games are compatible if and only if,

$$\begin{aligned} {\mathop {\mathop {\max }\limits _{f_1+\cdots +f_K=1}}\limits _{f_i:N\rightarrow {\mathbb {R}}_+}} \sum \limits _{1\le i\le K}\int ^\mathrm{cav} f_i \mathrm{d}v_i^M\le v_1^M\bullet \cdots \bullet v_K^M(N). \end{aligned}$$

(6.4)

We claim that for $M$ large enough,

$$\begin{aligned} {\mathop {\mathop {\max }\limits _{f_1+\cdots +f_K=1}}\limits _{\forall i, f_i:N\rightarrow {\mathbb {R}}_+}} \sum \limits _{1\le i\le K}\int ^\mathrm{cav} f_i \mathrm{d}v_i^M = {\mathop {\mathop {\max }\limits _{f_1+\cdots +f_K=1}}\limits _{\forall i, f_i\in {\mathcal {F}}_i}} \sum \limits _{1\le i\le K}\int ^\mathrm{cav} f_i \mathrm{d}v_i^M. \end{aligned}$$

(6.5)

We show first that for every $f_i:N\rightarrow {\mathbb {R}}_+, i=1,\ldots ,K$ that satisfy $f_1+\cdots +f_K=1$, there are $f'_i\in {\mathcal {F}}_i, i=1,\ldots ,K$ that satisfy $f_1+\cdots +f_K=1$ and $M$ sufficient large such that,

$$\begin{aligned} \sum \limits _{1\le i\le K}\int ^\mathrm{cav} f'_i \mathrm{d}v_i^M\ge \sum \limits _{1\le i\le K}\int ^\mathrm{cav} f_i \mathrm{d}v_i^M. \end{aligned}$$

(6.6)

Suppose that $f_i:N\rightarrow {\mathbb {R}}_+, i=1,\ldots ,K$ satisfy $f_1+\cdots +f_K=1$ and that there are $j$ and $\ell \not \in N_j$ with $f_j(\ell )>0$. Denote by $f_i1_{N_i}$ the function that coincides with $f_i$ on $N_i$ and is equal to 0 out of $N_i$. Note that $\int ^\mathrm{cav} f_i \mathrm{d}v_i^M=\int ^\mathrm{cav} f_i1_{ N_i} \mathrm{d}v_i-M\sum \nolimits _{\ell \not \in N_i}f_i(\ell )$. For every $\ell \in N$, there is an index $i(\ell )$ such that $\ell \in N_{i(\ell )}$. For every $i$ define,

$$\begin{aligned} f_i'=f_i1_{N_i} +\sum _{\ell ;~ i(\ell )=i} {\mathop {\mathop {\max }\limits _{j=1,\ldots ,K}}\limits _{\mathrm{s.t.}~\ell \not \in N_j}} f_j(\ell ). \end{aligned}$$

The function $f_i'$ is 0 on $\ell \not \in N_i$. Therefore, $f_i'\in {\mathcal {F}}_i$. Moreover, on $\ell \in N_i, f_i'(\ell )\ge f_i(\ell )$ with strict inequality when there is $j$ such that $f_j(\ell )>0, \ell \not \in N_j$ and $i(\ell )=i$. In the latter case, $f_i'(\ell )$ is getting the total value of all $f_j(\ell )$ for which $\ell \not \in N_j$. It is clear that $f'_1+\cdots +f'_K=1$.

Regarding the left-hand side (LHS) of (6.6), note that since $f_i'\in {\mathcal {F}}_i, \int ^\mathrm{cav} f'_i \mathrm{d}v_i^M=\int ^\mathrm{cav} f'_{i|N_i} \mathrm{d}v_i$, meaning that it does not depend on $M$, and therefore, the LHS does not depend on $M$ as well.

On the right-hand side (RHS), note that

$$\begin{aligned} \sum \limits _{1\le i\le K}\int ^\mathrm{cav} f_i \mathrm{d}v_i^M \le \sum \limits _{i\not = j}\int ^\mathrm{cav} f_i1_{N_i} \mathrm{d}v_i+ \int ^\mathrm{cav} f_j1_{N_j} \mathrm{d}v_j-f_j(\ell )M, \end{aligned}$$

which tends to $-\infty$ as $M\rightarrow \infty$. We obtain that for $M$ large enough, the LHS is strictly greater than the RHS. Note that the $M$ we found depends on the functions $f_1,\ldots ,f_K$ under discussion. However, due to the fact that the convex integral is continuous and that the set $\{(f_1,\ldots ,f_K); ~ \forall i, f_i, N\rightarrow {\mathbb {R}}_+, f_1+\cdots +f_K=1\}$ is compact, we conclude that there is one $M$ such that Eq. (6.5) holds, as desired.

We obtain that there is $M$ large enough such that [by Eqs. (6.4 and (6.5)], $v_1,\ldots ,v_K$ are compatible if and only if,

$$\begin{aligned} {\mathop {\mathop {\max }\limits _{f_1+\cdots +f_K=1}}\limits _{\forall i, f_i\in {\mathcal {F}}_i}} \sum \limits _{1\le i\le K}\int ^\mathrm{cav} f_i \mathrm{d}v_i^M \le v_1^M\bullet \cdots \bullet v_K^M(N). \end{aligned}$$

(6.7)

However, for $M>0$ when $f_i\in {\mathcal {F}}_i, \int ^\mathrm{cav} f_i \mathrm{d}v_i^M =\int ^\mathrm{cav} f_i \mathrm{d}v_i$. Moreover, when $M$ is sufficiently large $v_1^M\bullet \cdots \bullet v_K^M(N)=v_1\bullet \cdots \bullet v_K(N)$. This implies Eq. (4.6), and the proof of Theorem 2 is complete. $\square$

Proof of Theorem 3

Let $(v_1,N_1),(v_2,N_2)$ be convex games.

Step 1: We first assume that $N_1=N_2$. Denote by $C(v_i)$ the core of $v_i$. Since $v_i$ is convex, by Remark 6,

$$\begin{aligned} \int ^\mathrm{cav}X\mathrm{d}v_i=\min \limits _{P\in C(v_i)}\int ^\mathrm{cav} X\mathrm{d}P,\quad \,X\ge 0. \end{aligned}$$

We therefore have,

$$\begin{aligned} {\mathop {\mathop {\max }\limits _{f_1+f_2=1}}\limits _{f_1,f_2\ge 0}}\sum \limits _i\int ^\mathrm{cav}f_i\mathrm{d}v_i= & {} {\mathop {\mathop {\max }\limits _{f_1+f_2=1}}\limits _{f_1,f_2\ge 0}}\sum \limits _i\min \limits _{P_i\in C(v_i)}\int ^\mathrm{cav} f_i\mathrm{d}P_i\nonumber \\= & {} {\mathop {\mathop {\max }\limits _{f_1+f_2=1}}\limits _{f_1,f_2\ge 0}}\min \limits _{(P_i)_i\in C(v_1) \times C(v_2)}\sum \limits _i\int ^\mathrm{cav} f_i\mathrm{d}P_i\nonumber \\= & {} \min \limits _{(P_i)_i\in C(v_1) \times C(v_2)}{\mathop {\mathop {\max }\limits _{f_1+f_2=1}}\limits _{f_1,f_2\ge 0}}\sum \limits _i\int ^\mathrm{cav} f_i\mathrm{d}P_i, \end{aligned}$$

(6.8)

where the last equality is obtained by using the minimax theorem, as explained in what follows. Consider the zero-sum game in which player 1’s set of strategies is $\{f_1+f_2=1,f_1,f_2\ge 0\}$ and $C(v_1) \times C(v_2)$ is player 2’s set of strategies. Both sets are compact. Finally, the payoff function $((P_i)_i,(f_i)_i)\mapsto \sum \nolimits _i\int f_i\mathrm{d}P_i$ is bilinear. The minimax theorem allows us now to change the order of the minimum and the maximum in order to obtain the last equality.

Let $(P^*_1,P^*_2)$ and $(f^*_1,f^*_2)$ be the optimal strategies of the players. In particular, $(f^*_1,f^*_2)$ solves the maximization problem in the LHS of Eq. (6.8).

For every $j\in N$, denote⁵ $M(j)=\{i; ~ P^*_i(j) \ge P^*_{-i}(j)\}={\mathrm{argmax}_{i=1,2}} P^*_i(j)$. When $i\in M(j), P^*_i(j)$ is greater than or equal to $P^*_{-i}(j)$. Since $(f^*_1,f^*_2)$ solves

$$\begin{aligned} {\mathop {\mathop {\max }\limits _{f_1+f_2=1}}\limits _{f_1,f_2\ge 0}}\sum \limits _i\int ^\mathrm{cav} f_i\mathrm{d}P^*_i, \end{aligned}$$

(6.9)

we infer that if $f_i^*(j)>0$, then $i\in M(j)$. Indeed, if $i\not \in M(j)$, then $P^*_{-i}(j)> P^*_{i}(j)$. Define now, $f_i'(j')=f_i^*(j'), f_{-i}'(j')=f_{-i}^*(j')$ for every $j'\not = j, f_i'(j)=0$ and $f_{-i}'(j)=f_{-i}^*(j)+ f_{i}^*(j)$. One obtains that $f_1'$ and $f_2'$ are nonnegative, $f_1'+f_2'=1$ and,

$$\begin{aligned} \sum \limits _i\int ^\mathrm{cav} f_i^*\mathrm{d}P^*_i<\sum \limits _i\int ^\mathrm{cav} f_i'\mathrm{d}P^*_i, \end{aligned}$$

a contradiction. Furthermore, any two nonnegative functions, $f_1'$ and $f_2'$, such that $f_1'+f_2'=1$ and that satisfy the implication $[f_i'(j)>0 \Rightarrow i\in M(j)]$ solve Eq. (6.9).

Claim There are two disjoint sets $S_1$ and $S_2$ such that $S_1\cup S_2=N$ (i.e., $\mathbf{1 }_{S_1}+ \mathbf{1 }_{S_2}=1$) and

• (a) $j \in S_i$ implies $i\in M(j)$ for every $j\in N$ and $i=1,2$;

• (b) $\int \mathbf{1 }_{S_i}\mathrm{d}P^*_i=P^*_i(S_i)= v_i(S_i)$.

Before we prove the claim, we argue that (a) and (b) together would complete the proof. Indeed, by the previous paragraph, (a) would imply that $(\mathbf{1 }_{S_1},\mathbf{1 }_{S_2})$ solves Eq. (6.9). (b) would imply that $\sum \nolimits _i \int \mathbf{1 }_{S_i}\mathrm{d}P^*_i(j)\!=\! \sum \nolimits _i v_i(S_i)$. Together we would obtain

$$\begin{aligned} {\mathop {\mathop {\max }\limits _{f_1+f_2=1}}\limits _{f_1,f_2\ge 0}}\sum \limits _i\int ^\mathrm{cav}f_i\mathrm{d}v_i&=\sum \limits _i\int ^\mathrm{cav}f_i^*\mathrm{d}v_i =\sum \limits _i\int ^\mathrm{cav} f_i^*\mathrm{d}P^*_i\\&=\sum \limits _i\int ^\mathrm{cav} \mathbf{1 }_{S_i}\mathrm{d}P^*_i=\sum _i v_i(S_i)\le v_1 \bullet v_2(N). \end{aligned}$$

Theorem 1 implies that $v_1$ and $v_2$ are compatible. $\square$

Proof of the claim

The game $v_i$ is convex, and therefore, the concave integral takes a special form. Let $\pi _i$ be a permutation over $N$ such that $f^*_i(\pi _i(1))\le f^*_i(\pi _i(2))\le \cdots \le f^*_i(\pi _i(n))$. Define $A_i(j)=\{j'\in N; ~ f^*_i(j')\ge f^*_i(\pi _i(j))\}$. It is clear that $A_i(n)\subseteq A_i(n-1)\subseteq \cdots \subseteq A_i(1)$. Note that $f^*_i$ attains its maximum on the set $A_i(n)$ and $A_i(1)=N$. By Lovász (1983),⁶

$$\begin{aligned} \int ^\mathrm{cav}f^*_i\mathrm{d}v_i= \sum _{j=1}^n \Big (f^*_i(\pi _i(j))- f^*_i(\pi _i(j-1))\Big ) v_i(A_i(j)), \end{aligned}$$

where $f^*_i(\pi _i(0))=0$.

Recall that $f^*_1+f^*_2=1$.

• Case 1: $\max f^*_1<1$. Then, $\min f^*_2>0$ implies $2\in M(j), \forall j$, and therefore, $A_{2}(1)=N$. In this case, set $S_1=\emptyset$ and $S_2=N$. Properties (a) and (b) are satisfied, and the proof of the claim is complete.

• Case 2: $\max f^*_2<1$. This is similar to the previous case.

• Case 3: $\max f^*_1=\max f^*_2=1$. Denote by $j^*$ the smallest index such that $f_2^*$ is strictly positive on $A_{2}(j^*)$ (it exists because otherwise, $f_2^*=0$ contradicting $\max f^*_2=1$). Note that the complement of $A_{2}(j^*)$ is the set where $f^*_2=0$, which is precisely where $f^*_1=1$, namely $A_1(n)$. Define, $S_1=A_1(n)$ and $S_2= A_{2}(j^*)$. The sets $S_1$ and $S_2$ are complements of one another. Moreover, (a) is satisfied (because $j\in S_i$ implies that $f^*_i(j)$ is positive and therefore in $M(i)$).

We now show that (b) is satisfied. As $(P^*_1,P^*_2)$ and $(f^*_1,f^*_2)$ are the optimal strategies, we have

$$\begin{aligned} \sum \limits _{i=1}^2\int ^\mathrm{cav}f^*_i\mathrm{d}v_i&= \sum \limits _{i=1}^2\int f_i^*\mathrm{d}P^*_i\\&= \sum \limits _{i=1}^2\sum _{j=1}^n \Big (f^*_i(\pi _i(j))- f^*_i(\pi _i(j-1))\Big )P^*_i(A_i(j))\\&\ge \sum \limits _{i=1}^2\sum _{j=1}^n \Big (f^*_i(\pi _i(j))- f^*_i(\pi _i(j-1))\Big )v_i(A_i(j))\\&=\sum \limits _{i=1}^2 \int ^\mathrm{cav}f^*_i\mathrm{d}v_i, \end{aligned}$$

where the inequality is due to the fact that $P^*_i$ is in the core of $v_i$. We obtain that this inequality is actually an equality, and therefore, $P^*_i(A_i(j))=v_i(A_i(j))$ for every $j$. In particular, it holds for $S_i, i=1,2$, and hence (b). This shows the claim and the proof of Theorem 3 in case of identical grand coalitions.

Step 2 We now lift the restriction that $N_1\!=\!N_2$ and allow different grand coalitions, $N_1$ and $N_2$. Denote $N\!=\!N_1\cup N_2$ and use the same technique we employed in the proof of Theorem 2: For $i\!=\!1,2$ and a positive $M$, define $v_i^M$ over $N$. Recall that in $v_i^M$ the worth of $\ell \notin N_i$ is $-M$, which is also her contribution to any coalition she does not belong to. Thus, $v_i^M$ is also convex. Furthermore, from the proof of Theorem 2, we know that when $M$ is sufficiently large, $v_1^M$ and $v_2^M$ are compatible if and only if $v_1$ and $v_2$.

Using Step 1, $v_1^M$ and $v_2^M$ are compatible as convex games. Thus, when $M$ is large enough, $v_1$ and $v_2$ are compatible, as desired. $\square$

Proof of Proposition 1

Denote $v=v_1\bullet \cdots \bullet v_K$. Suppose first that $v(N)=v_1(N)$. We show that $C(v)=C(v_1)\cap \cdots \cap C(v_K)$. Assume that $P\in C(v_1)\cap \cdots \cap C(v_K)$. Let $S\subseteq N$ and let $S_1,\ldots , S_K$ be a partition of $S$ to pairwise disjoint sets such that $v(S)=\sum \nolimits _i v_i(S_i)$. One obtains, $P(S)=\sum \nolimits _iP(S_i)\ge \sum \nolimits _i v_i(S_i)=v(S).$ Furthermore, $P(N)=v_1(N)=v(N)$ and therefore, $P\in C(v)$. Now assume that $P\in C(v)$. For every coalition $S, v(S)\ge v_i(S)$, implying $P(S)\ge v(S)\ge v_i(S)$ for every $i=1,\ldots ,K$. Since $P(N)=v(N)=v_i(N), P\in C(v_i)$ for every $i=1,\ldots ,K$.

As for the inverse direction, $C(v)=C(v_1)\cap \cdots \cap C(v_K)$ readily implies that $v(N)=v_1(N)$, which completes the proof. $\square$

Proof of Proposition 2

We prove the theorem by induction on the number of games. First assume that $K=2$. We have

$$\begin{aligned} \tilde{v}_1\bullet \tilde{v}_2(S)&={\mathop {\mathop {\max }\limits _{S_1\cap S_2=\emptyset }}\limits _{S_1\cup S_2=S}}\left[ \tilde{v}_1\left( S_1\right) +\tilde{v}_2\left( S_2\right) \right] \\&={\mathop {\mathop {\max }\limits _{S_1\cap S_2=\emptyset }}\limits _{S_1\cup S_2=S}}\left[ {\mathop {\mathop {\max }\limits _{S^i_1\cap S^j_1=\emptyset }}\limits _{S^1_1\cup \cdots \cup S^n_1=S_1}}\left\{ v_1\left( S^1_1\right) +\cdots +v_1\left( S^n_1\right) \right\} \right. \\&\quad \left. +\,{\mathop {\mathop {\max }\limits _{S^i_2\cap S^j_2=\emptyset }}\limits _{S^1_2\cup \cdots \cup S^n_2=S_2}}\left\{ v_2\left( S^1_2\right) +\cdots +v_2\left( S^n_2\right) \right\} \right] \\&={\mathop {\mathop {\max }\limits _{S_1\cap S_2=\emptyset }}\limits _{S_1\cup S_2=S}}{\mathop {\mathop {\max }\limits _{S^i_1\cap S^j_1=\emptyset }}\limits _{S^1_1\cup \cdots \cup S^n_1=S_1}}{\mathop {\mathop {\max }\limits _{S^i_2\cap S^j_2=\emptyset }}\limits _{S^1_2\cup \cdots \cup S^n_2=S_2}}\left[ v_1\left( S^1_1\right) +\cdots +v_1\left( S^n_1\right) \right. \\&\left. \quad +\,v_2\left( S^1_2\right) +\cdots +v_2\left( S^n_2\right) \right] \\&={\mathop {\mathop {\max }\limits _{T_i\cap T_j=\emptyset }}\limits _{T_1\cup \cdots \cup T_n=S}}\left[ {\mathop {\mathop {\max }\limits _{S^1_1\cap S^1_2=\emptyset }}\limits _{S^1_1\cup S^1_2=:T_1}}\left\{ v_1\left( S^1_1\right) +v_2\left( S^1_2\right) \right\} \right. \\&\left. \quad +\,\cdots +{\mathop {\mathop {\max }\limits _{S^n_1\cap S^n_2=\emptyset }}\limits _{S^n_1\cup S^n_2=:T_n}}\left\{ v_1\left( S^n_1\right) +v_2\left( S^n_2\right) \right\} \right] \\&={\mathop {\mathop {\max }\limits _{T_i\cap T_j=\emptyset }}\limits _{T_1\cup \cdots \cup T_n=S}}[v_1\bullet v_2(T_1)+\cdots +v_1\bullet v_2(T_n)]\\&=\widetilde{v_1\bullet v_2}(S). \end{aligned}$$

Now assume the induction hypothesis: The statement of the theorem holds for $2,\ldots ,K-1$ games. Let $v:=v_1\bullet \cdots \bullet v_{K-1}$. By using the induction hypothesis and by the semigroup structure of $({\mathcal {G}}(N),\bullet )$, we have,

$$\begin{aligned} \widetilde{v_1\bullet \cdots \bullet v_K}&=\widetilde{v\bullet v_K}\\&=\tilde{v}\bullet \tilde{v}_K\\&=\left( \tilde{v}_1\bullet \cdots \bullet \tilde{v}_{K-1}\right) \bullet \tilde{v}_K\\&=\tilde{v}_1\bullet \cdots \bullet \tilde{v}_K. \end{aligned}$$

$\square$

Proof of Proposition 3

We need the following three lemmas.

Lemma 1

For any game $v$,

$$\begin{aligned} \{P\mathrm{linear }| P\ge v\}=\{P\mathrm{linear }| P\ge \tilde{v}\}=\{P{\mathrm{linear }}| P\ge v^c\}. \end{aligned}$$

Proof

This is an immediate result of the linearity of $P$ and the definition of $\tilde{v}$ and $v^c$.

$\square$

An immediate consequence of this lemma is the following one,

Lemma 2

For any game $v$,

$$\begin{aligned} B_v=B_{\tilde{v}}=B_{v^c}. \end{aligned}$$

The third lemma is,

Lemma 3

For any game $v$,

$$\begin{aligned} C^c(v)\not =\emptyset \quad \mathrm{iff } \quad C(\tilde{v})\not =\emptyset . \end{aligned}$$

Proof

By (4.4), $C^c(v)\not =\emptyset$ is equivalent to $B_{v^c}(N)\le v^c(N)$. Since $v^c(N)=\tilde{v}(N)$, by Lemma 2, $B_{v^c}(N)\le v^c(N)$ is equivalent to $B_{\tilde{v}}(N)\le \tilde{v}(N)$, which is, by (4.4), equivalent to $C(\tilde{v})\not =\emptyset$. $\square$

Now we complete the proof of the proposition. We have,

$$\begin{aligned} v_1,\ldots ,v_K \text { are c-compatible }&\Leftrightarrow C^c(v_1\bullet \cdots \bullet v_K)\not =\emptyset \\&\Leftrightarrow C(\widetilde{v_1\bullet \cdots \bullet v_K})\not =\emptyset \\&\Leftrightarrow B_{\widetilde{v_1\bullet \cdots \bullet v_K}}(N)\le \widetilde{v_1\bullet \cdots \bullet v_K}(N)\\&\Leftrightarrow B_{\tilde{v}_1\bullet \cdots \bullet \tilde{v}_K}(N)\le \widetilde{v_1\bullet \cdots \bullet v_K}(N)\\&\Leftrightarrow B_{\tilde{v}_1\bullet \cdots \bullet \tilde{v}_K}(N)\le \tilde{v}_1,\ldots ,\tilde{v}_K(N)\\&\Leftrightarrow C(\tilde{v}_1\bullet \cdots \bullet \tilde{v}_K)\not =\emptyset \\&\Leftrightarrow \tilde{v}_1,\ldots ,\tilde{v}_K\text { are compatible}. \end{aligned}$$

The first equivalence is the definition of compatibility. The second one is due to Lemma 3. The third equivalence is by (4.4), while the fourth and the fifth are due to Proposition 2. The last two equivalences result from (4.4) and the definition of compatibility, respectively.$\square$