Coalition-weighted Shapley values

Abstract

We introduce a new class of values for coalitional games: the coalition-weighted Shapley values. Weights can be assigned to coalitions, not just to players, and zero-weights are admissible. The Shapley value belongs to this class. Coalition-weighted Shapley values recommend for each game the allocation defined by the Shapley value of a weighted game obtained as a linear convex combination of the associated marginal games. Coalition-weighted Shapley values are random order values and Harsanyi values. Positively weighted Shapley values and weighted Shapley values can be seen as the limit of a sequence of iterated coalition-weighted Shapley values. We provide axiomatic characterizations of coalition-weighted Shapley values through properties that do not involve the weights. Finally, we discuss how to extend our model to include exogenous coalition structures as in the hierarchical and Owen values.

Appendix

1.1 Proof of Theorem 3.5

Let $R\in \mathcal {N}$, $\delta =\bigl \{\delta _S \bigr \}_{S\in \mathcal {N}}\in \Delta$, and $i\in R$. Then, by linearity of the Shapley value and equality (4),

$$\begin{aligned} \Phi ^{\delta }_i(u_R)&={{\,\textrm{Sh}\,}}_i (u_R^{\delta }) = \left( \sum _{S\in \mathcal {N}(N\backslash R)} \delta _S \right) {{\,\textrm{Sh}\,}}_i(u_R) + \sum _{S\not \in \mathcal {N}(N\backslash R)}\delta _S {{\,\textrm{Sh}\,}}_i(u_{R\cap S})\\&= \sum _{S\in \mathcal {N}(N\backslash R)} \delta _S \frac{1}{|R|} +\sum _{\genfrac{}{}{0.0pt}1{S\not \in \mathcal {N}(N\backslash R)}{i\in S}} \delta _S\frac{1}{|R\cap S|}. \end{aligned}$$

Then $\Phi ^{\delta }_i(u_R)=\sum_{S\in \mathcal {N}} a^R_i(S) \delta _S$ where $a^R_i(S)=\tfrac{1}{|R|}$ if $S \cap R=\emptyset$, $a^R_i(S)=\tfrac{1}{|S\cap R|}$ if $i\in S\cap R$, and $a^R_i(S)=0$ otherwise. In particular, if $\lambda \in \Lambda$ then $\Phi ^{\lambda }_i(u_R)=\sum_{k\not \in R}\ \lambda _k \tfrac{1}{|R|}+\lambda _i=\lambda _i+\tfrac{\lambda (N\backslash R)}{|R|}=\lambda _i+\tfrac{1-\lambda (R)}{|R|}$.

1.2 Proof of Proposition 3.9

For each $R\in \mathcal {N}$ and $\delta =\bigl \{\delta _S \bigr \}_{S\in \mathcal {N}}\in \Delta$, we have, from equality (4), that $u_R^{\delta }(S)=1$ if $R\subset S$ and $u_R^{\delta }(S)=\sum_{T\in \mathcal {N}(R\cap S)} \delta _T$ otherwise. Note that if $R\cap S=\emptyset$ then $u_R^{\delta }(S)=0$. Now, if $v=\sum_{R \in \mathcal {N}} \alpha _R^v u_R\in G^N$ and $S\in 2^N$ then, by Proposition 3.2, $v^{\delta }(S)= \sum_{R\subset S} \alpha _R^v +\sum_{R\not \subset S}\ \alpha _R^v u^{\delta }_R(S)=v(S)+\sum_{R\not \subset S}\ \alpha _R^v \Bigl ( \sum_{T\in \mathcal {N}(R\cap S)} \delta _T\Bigr )$. The result follows immediately because ${{\,\textrm{Sh}\,}}$ is linear.

1.3 Proof of Proposition 4.1

Let $v \in G^N$ and $\delta =\bigl \{\delta _S \bigr \}_{S\in \mathcal {N}}\in \Delta$. Then,

$$\begin{aligned} \sum _{\sigma \in \Sigma } \left( \sum _{r=1}^n \frac{\delta _{T_r^{\sigma }}}{r!(n-r)!}\right) m^{\sigma }(v)&= \sum _{R \in \mathcal {N}} \frac{1}{|R|!(n-|R|)!} \left( \sum _{\sigma =(\sigma _{N \backslash R},\sigma _R)} \delta _{R}m^{\sigma }(v)\right) \\&= \sum _{R \in \mathcal {N}} \delta _{R} \left( \sum _{\sigma =(\sigma _{N \backslash R},\sigma _R)}\left( {\begin{array}{c}n\\ |R|\end{array}}\right) \frac{1}{n!} m^{\sigma }(v)\right) \\&= \sum _{R \in \mathcal {N}} \delta _{R} \left( \sum _{\sigma =(\sigma _{N \backslash R},\sigma _R)}\left( {\begin{array}{c}n\\ |R|\end{array}}\right) \frac{1}{n!} m^{\sigma }(v^{R})\right) \\&= \sum _{R \in \mathcal {N}} \delta _{R} {{\,\textrm{Sh}\,}}(v^{R}) = \Phi ^{\delta }(v), \end{aligned}$$

where the second and third equalities hold by (2) and (6) respectively.

1.4 Proof of Theorem 5.1

First, note that since $\Phi ^{\lambda }(u_N)=\lambda$ for all $\lambda \in \Lambda$ then $\Phi ^{\lambda }\not = \Phi ^{\lambda '}$ if $\lambda ,\lambda '\in \Lambda$ and $\lambda \not =\lambda '$. Fixed $\lambda \in \Lambda$. We know that $\Phi ^{\lambda }$ satisfies (E), (N), (A), and (Pos). Clearly $\lambda _i=\Phi ^{\lambda }_i(u_N)$ for all $i\in N$. Moreover, by Theorem 3.5, if $R\in \mathcal {N}$ and $i,j\in R$ then $\Phi ^{\lambda }_i(u_R)-\Phi ^{\lambda }_j(u_R)=\lambda _i - \lambda _j$. Therefore $\Phi ^{\lambda }$ also satisfies (DI).

Assume that $f:G^N \rightarrow \mathbb {R}^N$ is a value that satisfies (E), (N), (A), (Pos), and (DI). For each $j\in N$ let $\lambda _j=f_j(u_N)$ and $\lambda =\bigl (\lambda _j\bigr )_{j\in N}$. Now, we have that $\lambda \in \Lambda$ because, by (Pos), $\lambda _j\ge 0$ for all $j\in N$, and by (E), $\lambda (N)=\sum_{j\in N}\ f_j(u_N)=1$. We claim that $f=\Phi ^{\lambda }$. First, observe that f is linear, since f satisfies (A) and (Pos) and therefore (H). So, it suffices to prove that $f(u_R)=\Phi ^{\lambda }(u_R)$ for all $R\in \mathcal {N}$. Given $R\in \mathcal {N}$ and $j\not \in R$ then j is a null player in $u_R$. Thus, by (N), we have $f_j(u_R)=\Phi ^{\lambda }_j(u_R)=0$ and hence, by (E), $f_{i}(u_{\{i\}}) = \Phi ^{\lambda }_i(u_{\{i\}})$. If $R=\{i_1, \dots , i_r\}$, with $r>1$, then, by (DI), $f_{i_1}(u_R)=f_{i_k}(u_R)+ f_{i_1}(u_N)-f_{i_k}(u_N)=f_{i_k}(u_R)+\lambda _{i_1}-\lambda _{i_k}$ for all $k=2, \dots , r$. Adding these equations, by (E), we have that

$$\begin{aligned} (r-1)f_{i_1}(u_R)&=\sum _{k=2}^{r} f_{i_k}(u_R) +(r-1)\lambda _{i_1}-\sum _{k=2}^{r} \lambda _{i_k}\\&=1-f_{i_1}(u_R)+(r-1)\lambda _{i_1}-\bigl (1-\lambda _{i_1}-\lambda (N\backslash R)\bigr ). \end{aligned}$$

Therefore, $rf_{i_1}(u_R)=r\lambda _{i_1}+\lambda (N\backslash R)$ which, in turn, implies that $f_{i_1}(u_R)=\lambda _{i_1}+\tfrac{1}{r}\lambda (N\backslash R)=\Phi ^{\lambda }_{i_1}(u_R)$. Similarly, we can prove that $f_{i_k}(u_R)=\Phi ^{\lambda }_{i_k}(u_R)$ for all $k=2, \dots , r$.

1.5 Proof of Theorem 5.3

Let $\delta =\bigl \{\delta _S \bigr \}_{S\in \mathcal {N}}\in \Delta$ be a weight family and consider the set $\mathcal {I}=\{ (R,i)\in \mathcal {N}\times N :|R|>1,\ i\in R\}$ with cardinality $|\mathcal {I}|=\sum_{r=2}^{{n}} r\left( {\begin{array}{c}n\\ r\end{array}}\right)$. We introduce some notations:

• Let $\Phi _{\mathcal {I}} \in \mathbb {R}^{|\mathcal {I}|}$ be the column vector whose coordinates are $\Phi ^{\delta }_i(u_R)$ for $(R,i)\in \mathcal {I}$.

• Given $(R,i)\in \mathcal {I}$ denote by $a_i^R\in \mathbb {R}^{2^n-1}$ the row vector whose coordinates are $a_i^R(S)$ for all $S\in \mathcal {N}$.

• Let $A=\Bigl ( a_i^R(S)\Bigr )_{ \underset{S\in \mathcal {N}}{(R,i)\in \mathcal {I}} }$ be the $|\mathcal {I}|\times (2^n-1)$-matrix whose rows are the vectors $a_i^R$ for $(R,i)\in \mathcal {I}$.

From Theorem 3.5, we have that $A \delta = \Phi _{\mathcal {I}}$.¹⁵

Given $i,j\in N$ let $R\in \mathcal {N}$, $|R|\ge 3$, such that $\{i,j\} \subset R$. Then, for each $S\in \mathcal {N}$,

$$\begin{aligned} \begin{array} {ll} a^{R\backslash \{j\}}_i(S)=a^{R\backslash \{i\}}_j(S)=0 &{}\quad \text {if } \{i,j\}\cap S=\emptyset , \ S\cap R\not =\emptyset \\ a^{R\backslash \{j\}}_i(S)=a^{R\backslash \{i\}}_j(S)=\tfrac{1}{|R|-1} &{}\quad \text {if } \{i,j\}\cap S=\emptyset , \ S\cap R=\emptyset \\ a^{R\backslash \{j\}}_i(S)=a^{R\backslash \{i\}}_j(S)=\tfrac{1}{|S\cap R\backslash \{i\}|} &{}\quad \text {if } \{i,j\}\subset S\\ a^{R\backslash \{j\}}_i(L\cup \{i\})=a^{R\backslash \{i\}}_j(L\cup \{j\})=\tfrac{1}{| (L\cup \{i\})\cap R\backslash \{i\}|} &{}\quad \text {if } L\subset N\backslash \{i,j\} \\ \end{array} \end{aligned}$$

(7)

Let us prove the necessity. We already know that, for each $\delta \in \Delta$, $\Phi ^{\delta }$ satisfies (E), (N), (A), and (Pos). Let us prove that $\Phi ^{\delta }$ satisfies (DC). Theorem 3.5 and relationships (7) easily imply that if $R\in \mathcal {N}$, $|R|\ge 3$, and $\sigma \in \Pi (R)$ with $\sigma (i)\not = i$ for all $i\in R$, then

$$\begin{aligned} \Phi ^{\delta }_i(u_{R\backslash \{\sigma (i)\}})=\sum_{S\in \mathcal {N}}a^{R\backslash \{\sigma (i)\}}_i(S) \delta _S = \sum_{S\in \mathcal {N}}{ } a^{R\backslash \{i\}}_{\sigma (i)}(S) \delta _S = \Phi ^{\delta }_{\sigma (i)}(u_{R\backslash \{i\}}). \end{aligned}$$

Therefore, $\sum_{i\in R} \Phi ^{\delta }_{i} (u_{R\backslash \{\sigma (i)\}})=\sum_{i\in R} \ \Phi ^{\delta }_{\sigma (i)} (u_{R\backslash \{i\}})$.

Let us prove the sufficiency. Assume that $f:G^N \rightarrow \mathbb {R}^N$ is a value that satisfies (E), (N), (A), (Pos), and (DC). We have to show that there is $\delta \in \Delta$ such that $f=\Phi ^{\delta }$. First, observe that f is linear, since f satisfies (A) and (Pos) and then (H). So, it suffices to prove that there is $\delta \in \Delta$ such that $f(u_R)=\Phi ^{\delta }(u_R)$ for all $R\in \mathcal {N}$. Let $f_{\mathcal {I}}\in \mathbb {R}^{|\mathcal {I}|}$ be the column vector whose coordinates are $f_i(u_R)$ for $(R,i)\in \mathcal {I}$. Therefore, we have to prove that the linear system $A\delta =f_{\mathcal {I}}$ has a non-negative solution $\delta \in \Delta$. Now, since the solution set for the homogenous system $A\delta =0$ has dimension $n-1$, we know that ${{\,\textrm{rank}\,}}(A)=2^n-n$. Denote by $F_i^R=\bigl (a_i^R| f_i(u_R) \bigr )$ the row of the augmented matrix $(A|f_{\mathcal {I}})$ corresponding to $(R,i)\in \mathcal {I}$. By efficiency, for each $R,R'\in \mathcal {N}$ we have that $\sum_{i\in R} \ F_i^R = \sum_{j\in R'}\ F_j^{R'}$. Since f satisfies (DC), for each $R\in \mathcal {N}$, $|R|\ge 3$, and each derangement $\sigma \in \Pi (R)$ we have $\sum_{i\in R} \ F_{i}^{R\backslash \{\sigma (i)\}}=\sum_{i\in R} \ F_{\sigma (i)}^{R\backslash \{i\}}$. Then, there are rows of matrix $(A|f_{\mathcal {I}})$ that are linearly dependent. After some cumbersome calculations one can show that ${{\,\textrm{rank}\,}}(A|f_{\mathcal {I}})={{\,\textrm{rank}\,}}(A)=2^n-n$.¹⁶ As a consequence, the system $A\delta =f_{\mathcal {I}}$ of $|\mathcal {I}|$ simultaneous linear equations in $2^n-1$ unknowns has infinitely many solutions. We select a set of $(2^n-n)$ linearly independent rows of $(A|f_{\mathcal {I}})$ that contains the rows $(a_i^N|f_{i}(u_N))$ for $i\in N$, that we denote $(\tilde{A}|\tilde{b})$. For convenience, let us assume that the first n rows of $(\tilde{A}|\tilde{b})$ are precisely $(a_i^N|f_{i}(u_N))$, $i=\{1,\dots , n\}$. Our problem is to find a vector $\delta \in \mathbb {R}^{2^n-1}$ such that:

$$\begin{aligned} \left. \begin{aligned} \tilde{A}\delta&=\tilde{b}\\ \delta&\ge 0 \end{aligned} \right\} \end{aligned}$$

(S)

Clearly, finding $\delta \in \mathbb {R}^{2^n-1}$ satisfying the system of constraints (S) is equivalent to solving the linear programming problem:

$$\begin{aligned} \begin{aligned} \text {minimize }\ {}&\sum _{S\in \mathcal {N}} \delta _S \\ \text {subject to }\ {}&\tilde{A}\delta =\tilde{b}, \ \delta \ge 0\\ \end{aligned} \end{aligned}$$

(P)

The dual problem associated with (P) is:

$$\begin{aligned} \begin{aligned} \text {maximize }\ {}&\langle \tilde{b}, y \rangle \\ \text {subject to }\ {}&\tilde{A}^t y \le 1\\ \end{aligned} \end{aligned}$$

(D)

Consider the vector $y^*$ such that $y_j^*=1$ if $j\in \{1,\dots , n\}$ and $y_j^*=0$ otherwise. Clearly, $y^*$ is feasible for (D), in fact $\tilde{A}^ty^*=1$ so it is the unique basic feasible solution. Since, f satisfies (Pos) we know that $\tilde{b}\ge 0$, so $y^*$ is optimal for (D) and the corresponding value of the dual objective function is $\langle \tilde{b}, y^* \rangle = 1$. Applying the duality theorem of linear programming we conclude that (P) has an optimal solution $\bar{\delta } \in \mathbb {R}^{2^n-1}$. Obviously, $\bar{\delta } \in \Delta$ and $f=\Phi ^{\bar{\delta }}$.

1.6 Proof of Proposition 5.4

First of all, consider again the matrix $A=\Bigl ( a_i^R(S)\Bigr )_{ \genfrac{}{}{0.0pt}1{(R,i)\in \mathcal {I}}{S\in \mathcal {N}}}$ whose rows are the vectors $a_i^R$ for $(R,i)\in \mathcal {I}$. Let ${{\,\textrm{Sh}\,}}_{\mathcal {I}}\in \mathbb {R}^{|\mathcal {I}|}$ be the column vector whose coordinates are ${{\,\textrm{Sh}\,}}_i(u_R)=a_i^R(N)=\tfrac{1}{|R|}$ for $(R,i)\in \mathcal {I}$. For each $t\in \{1, \dots , n\}$, let $\mathcal {N}_t=\{T\in \mathcal {N}:|T|=t\}$ be the family of coalitions of size t. If $(R,i)\in \mathcal {I}$ and $t\in \{1, \dots , n\}$ it is easy to check that:

$$\begin{aligned} \sum _{T\in \mathcal {N}_t} a_i^R(T)=\left( {\begin{array}{c}n\\ t\end{array}}\right) \frac{1}{|R|}. \end{aligned}$$

(8)

Therefore, for each $t\in \{1, \dots , n-1\}$ the vector $\delta ^t=\bigl \{\delta ^t_S \bigr \}_{S\in \mathcal {N}}\in \mathbb {R}^{2^n-1}$, where $\delta ^t_N=-\left( {\begin{array}{c}n\\ t\end{array}}\right)$, $\delta ^t_T=1$ if $T\in \mathcal {N}_t$, and $\delta ^t_S=0$ otherwise, satisfies $A\delta ^t=0$. Then, the set of vectors $\{\delta ^t:t=1, \dots , n-1\}$ is linearly independent and any linear combination of them is a solution for the homogenous system $A\delta =0$. In other words, given $h=(h_1, \dots , h_{n-1}) \in \mathbb {R}^{n-1}$ the vector $\delta ^h=\sum _{t=1}^{ {n-1}\ } h_t\delta ^t \in \mathbb {R}^{2^n-1}$ satisfies $A\delta ^h=0$. Naturally, $\delta ^h_T=h_t$ if $T\in \mathcal {N}_t$, $t\in \{1,\dots , n-1\}$, and $\delta ^h_N=-\sum_{t=1}^{{n-1}{ }} \left( {\begin{array}{c}n\\ t\end{array}}\right) h_t$.

Now, we claim that the solution set for the homogenous system $A\delta =0$ is the $(n-1)$-dimensional linear subspace $\{\delta ^h:h \in \mathbb {R}^{n-1}\}$. First observe that if a vector $\delta \in \mathbb {R}^{2^n -1}$ has all its coordinates zero except for two corresponding to coalitions of the same size then $\delta$ is not a solution of the homogeneous system $A\delta =0$. Indeed, let $\delta = \bigl \{\delta _S \bigr \}_{S\in \mathcal {N}} \in \mathbb {R}^{2^n-1}$ such that $\delta _T\not =\delta _{T'}$, $\delta _T\not =0$, for some $T,T'\in \mathcal {P}$, with $|T|=|T'|$, and $\delta _S=0$ otherwise. Then $A\delta$ is the column vector with coordinates $a_i^R(T)\delta _T+a_i^R(T')\delta _{T'}$, $(R,i)\in \mathcal {I}$. In particular, take $(R,j)\in \mathcal {I}$ such that $j\in T\cap R$ and $T'\cap R\not =\emptyset$. Clearly, $a_j^R(T)\not =0$ and $a_j^R(T')=0$, so $a_j^R(T)\delta _T+a_j^R(T')\delta _{T'}=a_j^R(T)\delta _T\not =0$ and $A\delta \not =0$. Now, we proceed by contradiction. So assume that $\delta = \bigl \{\delta _S \bigr \}_{S\in \mathcal {N}} \in \mathbb {R}^{2^n-1}$ satisfies $\delta _T\not =\delta _{T'}$ for some $T,T'\in \mathcal {P}$, with $|T|=|T'|$, and $A\delta =0$. Consider the vector $\hat{\delta }\in \mathbb {R}^{2^n -1}$ obtained from $\delta$ by simply exchanging the values $\delta _T$ and $\delta _{T'}$. Now, the T-column of matrix A is the vector with coordinates $a_i^R(T)$, $(R,i)\in \mathcal {I}$, while the $T'$-column of matrix A is the vector with coordinates $a_i^R(T')$, $(R,i)\in \mathcal {I}$. Since $|T|=|T'|$, the $T'$-column of A is just a rearrangement of the elements of the T-column of A. Therefore, $A\hat{\delta }=A\delta =0$, so $A(\delta -\hat{\delta })=0$. But, $\delta -\hat{\delta }$ is a vector with null coordinates except for two corresponding to coalitions T and $T'$ and then, as we have shown above, it cannot be a solution of the homogeneous system. This contradiction allows us to assert that, in fact, any solution for the homogenous system $A\delta =0$ belongs to $\{\delta ^h:h \in \mathbb {R}^{n-1}\}$.

Consider the system $A\delta ={{\,\textrm{Sh}\,}}_{\mathcal {I}}$ of $|\mathcal {I}|$ simultaneous linear equations in $2^n-1$ unknowns. Since $\bar{\delta }=\bigl \{\bar{\delta }_S \bigr \}_{S\in \mathcal {N}}\in \mathbb {R}^{2^n-1}$, where $\bar{\delta }_N=1$ and $\bar{\delta }_S=0$ otherwise, is a specific solution to $A\delta ={{\,\textrm{Sh}\,}}_{\mathcal {I}}$, the solution set for $A\delta ={{\,\textrm{Sh}\,}}_{\mathcal {I}}$ can be described as $\{ \bar{\delta }+\delta ^h:\delta ^h \text { is a solution to } A\delta =0\}$. Therefore, a vector $\delta \in \mathbb {R}^{2^n-1}$ is a solution to $A\delta ={{\,\textrm{Sh}\,}}_{\mathcal {I}}$ if and only if there exists $h=(h_1, \dots , h_{n-1})\in \mathbb {R}^{n-1}$ such that $\delta _T=h_t$ if $|T|=t<n$ and $\delta _N= 1- \sum _{t=1}^{n-1} \left( {\begin{array}{c}n\\ t\end{array}}\right) h_t$. Or, equivalently, $\delta \in \mathbb {R}^{2^n-1}$ is a solution to $A\delta ={{\,\textrm{Sh}\,}}_{\mathcal {I}}$ if and only if $\delta _T=\delta _{T'}$ for all $T,T'\in \mathcal {N}_t$ with $t<n$ and $\delta _N=1-\sum_{S\in \mathcal {P}} \delta _S$.

We know that $\Phi ^{\delta }(u_{\{i\}})={{\,\textrm{Sh}\,}}(u_{\{i\}})=e_i$ for all $\delta \in \Delta$ and $i\in N$, and that $\Phi ^{\delta }_k(u_R)={{\,\textrm{Sh}\,}}_k (u_R)=0$ for all $\delta \in \Delta$, $R\in \mathcal {N}$ and $k\in N$ such that $k\not \in R$. Therefore, by linearity, there is $\delta \in \Delta$ such that ${{\,\textrm{Sh}\,}}=\Phi ^{\delta }$ if and only if there is $\delta \in \Delta$ such that ${{\,\textrm{Sh}\,}}_i(u_R)=\Phi ^{\delta }_i(u_R)$ for all $(R,i)\in \mathcal {I}$, or equivalently if $\delta \in \Delta$ is a (non-negative) solution of the system $A\delta ={{\,\textrm{Sh}\,}}_{\mathcal {I}}$. And Proposition 5.4 follows.¹⁷

1.7 Proof of Proposition 5.5

Let $\delta =\bigl \{\delta _S \bigr \}_{S\in \mathcal {N}}\in \Delta$ and $b_{\mathcal {I}}\in \mathbb {R}^{|\mathcal {I}|}$ be the column vector whose coordinates are $\Phi ^{\delta }_i(u_R)$ for $(R,i)\in \mathcal {I}$. Therefore, $\delta$ is a non-negative solution of the linear system $A\delta =b_{\mathcal {I}}$. For each $t\in \{1, \dots , n-1\}$ let $T_t \in \mathcal {N}_t$ be a coalition of size t such that $\delta _{T_t}=\min \{ \delta _T:T\in \mathcal {N}_t\}$. Take $\bar{\delta } \in \mathbb {R}^{2^n-1}$ such that $\bar{\delta }_T=\delta _{T_t}$ for all $T\in \mathcal {N}_t$ with $t<n$ and $\bar{\delta }_N=1-\sum_{S\in \mathcal {P}} \bar{\delta }_{S}$. Clearly $\bar{\delta } \in \Delta$ and, by Proposition 5.4, $A\bar{\delta }=0$. Now, take $\delta '=\delta -\bar{\delta }$. It is straightforward to check that $\delta '\in \Delta$ and $A\delta '=b_{\mathcal {I}}$ or, equivalently, $\Phi ^{\delta }_i(u_R)=\Phi ^{\delta '}_i(u_R)$ for $(R,i)\in \mathcal {I}$.

1.8 Proof of Theorem 6.1

Given $R \in \mathcal {N}$ and $\lambda \in \Lambda$, define the (additive) game $\lambda _{| R} \in G^N$ as $\lambda _{| R} (S)= \lambda (R\cap S)$ for $S\in 2^N$. Then, we have $u_R^{\lambda }=\sum_{i\in N}{ }\ \lambda _i u_R^{\{i\}}=\sum_{i\in R} \ \lambda _i u_{\{i\}} + \sum_{j\not \in R}\ \lambda _j u_R=\lambda _{| R}+\lambda (N\backslash R) u_R$, so,¹⁸

$$\begin{aligned} u_R^{\lambda }= \bigl (1- \lambda (R) \bigr ) u_R + \lambda _{| R}. \end{aligned}$$

(9)

Next, let us show that, for each $m \in \mathbb {N}$, the game $(u_R^{\lambda })^{m}$ can be written as:

$$\begin{aligned} (u_R^{\lambda })^{m}= {\left\{ \begin{array}{ll} \lambda _{|N} &{} \text {if } R=N\\ (1-\lambda (R))^m u_R + \left( \displaystyle \sum _{s=0}^{m-1} (1-\lambda (R))^s \right) \lambda _{|R} &{} \text {if } R\in \mathcal {P} \end{array}\right. }. \end{aligned}$$

(10)

Since $u_N^{\lambda }=\lambda _{|N}\in G^N$ is an additive game, we have that $(u_R^{\lambda })^{m}=\lambda _{|N}$ for all $m\in \mathbb {N}$. Assume that $R\in \mathcal {P}$. We proceed by induction. The case when $m=1$ is given by equality (9). Let $m\ge 2$ and, by the induction hypothesis, assume that the result holds for $m-1$, that is,

$$\begin{aligned} (u_R^{\lambda })^{m-1}=(1- \lambda (R))^{m-1} u_R + \left( \sum _{s=0}^{m-2} (1-\lambda (R))^s \right) \lambda _{|R}. \end{aligned}$$

Then,

$$\begin{aligned} (u_R^{\lambda })^m&=\bigl ( (u_R^{\lambda })^{m-1} \bigr )^{\lambda }=\left( (1- \lambda (R))^{m-1} u_R + \left( \sum _{s=0}^{m-2} (1-\lambda (R))^s \right) \lambda _{|R} \right) ^\lambda \\&= (1- \lambda (R))^{m-1} u_R^{\lambda } + \left( \sum _{s=0}^{m-2} (1-\lambda (R))^s \right) \lambda _{|R} \\&= (1- \lambda (R))^{m-1}\bigl ( \bigl (1- \lambda (R) \bigr ) u_R + \lambda _{| R} \bigr ) + \left( \sum _{s=0}^{m-2} (1-\lambda (R))^s \right) \lambda _{|R} \\&= (1-\lambda (R))^m u_R + \left( \sum _{s=0}^{m-1} (1-\lambda (R))^m \right) \lambda _{|R}. \end{aligned}$$

The third equality holds because the $\lambda$ operator is linear and $\lambda _{| R} \in G^N$ is an additive game.

Now, suppose that $\lambda \in \Lambda ^+$ and let us show that $\lim_{m \rightarrow \infty} (\Phi ^{\lambda })^m(u_R)={{\,\textrm{Sh}\,}}^{\lambda } (u_R)$ for each $R \in \mathcal {N}$. First, assume that $R\in \mathcal {P}$. Since $0< 1-\lambda (R)<1$ we have that $\lim_{m \rightarrow \infty }(1-\lambda (R))^m=0$ and $\displaystyle \sum _{s=0}^{\infty } (1-\lambda (R))^s =\frac{1}{\lambda (R)}$. Now, from (10), for each $R \in \mathcal {N}$ we have that $\lim_{m \rightarrow \infty }\ (u_R^{\lambda })^m=\frac{1}{\lambda (R)} \lambda _{|R}$ and $\lim_{m \rightarrow \infty }\ (\Phi ^{\lambda })^m (u_R)={{\,\textrm{Sh}\,}}\bigl (\frac{1}{\lambda (R)} \lambda _{|R} \bigr )={{\,\textrm{Sh}\,}}^{\lambda } (u_R)$.

1.9 Proof of Proposition 6.2

Let $R\in \mathcal {N}$ and $i\in N$. If $i\not \in R$ then ${{\,\textrm{Sh}\,}}^{\lambda }_i(u_R)={{\,\textrm{Sh}\,}}^{\lambda }_i((u_R)^{\lambda })=0$. If $i\in R$ then, by (9), we have that $u_R^{\lambda }= \bigl (1- \lambda (R) \bigr ) u_R + \lambda _{| R}$. Therefore,

$$\begin{aligned} {{\,\textrm{Sh}\,}}^{\lambda }_i((u_R)^{\lambda })&={{\,\textrm{Sh}\,}}_i \bigl ( (1- \lambda (R)) u_R + \lambda _{| R} \bigr ) =(1-\lambda (R)) {{\,\textrm{Sh}\,}}^{\lambda }_i(u_R)+{{\,\textrm{Sh}\,}}^{\lambda }_i(\lambda _{| R})\\&= (1-\lambda (R)) \frac{\lambda _i}{\lambda (R)}+ \lambda _i= \frac{\lambda _i}{\lambda (R)} = {{\,\textrm{Sh}\,}}^{\lambda }_i(u_R). \end{aligned}$$

Since the positively weighted Shapley value is linear, we have ${{\,\textrm{Sh}\,}}^{\lambda }(v)={{\,\textrm{Sh}\,}}^{\lambda }(v^{\lambda })$ for all $v\in G^N$.

Let $\omega =(\lambda ,\mathcal {S})\in \Omega$ where $\lambda =\bigl (\lambda _i \bigr )_{i\in N}\in \Lambda ^+$ and $\mathcal {S}=(S_1, \dots , S_m)$. For each $R\in \mathcal {N}$, applying equality (3), we have that the hierarchical game $(u_R)_{\mathcal {S}}\in G^N$ is the unanimity game $u_{R \cap S_k} \in G^N$ where $k=\max \{ i\in N:S_i\cap R\not = \emptyset \}$. Therefore, ${{\,\textrm{Sh}\,}}^{\lambda } ((u_R)_{\mathcal {S}})={{\,\textrm{Sh}\,}}^{\omega } (u_R)$. If $v=\sum_{R \in \mathcal {N}} \alpha _R^v u_R \in G^N$, since positively weighted Shapley values and weighted Shapley values are linear and $v_{\mathcal {S}}=\sum_{R \in \mathcal {N}} \alpha _T^v (u_R)_{\mathcal {S}}$, we conclude that ${{\,\textrm{Sh}\,}}^{\lambda } (v_{\mathcal {S}})={{\,\textrm{Sh}\,}}^{\omega }(v)$.

1.10 Proof of Corollary 6.3

Let $v\in G^N$, $\omega =(\lambda ,\mathcal {S})\in \Omega$, and $v_{\mathcal {S}}\in G^N$ be the hierarchical game associated with the ordered partition $\mathcal {S}$. From Theorem 6.1 and Proposition 6.2, we have that $\lim _{m \rightarrow \infty } (\Phi ^{\lambda })^m(v_{\mathcal {S}})={{\,\textrm{Sh}\,}}^{\lambda } (v_{\mathcal {S}})={{\,\textrm{Sh}\,}}^{\omega }(v)$.