The equal-surplus Shapley value for chance-constrained games on finite sample spaces

Abstract

Many interactions from linear production problems, financial markets, or sequencing problems are modeled by cooperative games where payoffs to a coalition of players is a random variable. For this class of cooperative games, we introduce a two-stage value as an ex-ante agreement among players. Players are first promised their prior Shapley shares which are exactly their respective shares by the Shapley value of the expectation game. The final payoff vector is obtained by equally re-allocating the surplus when a realization of the random payoff of the grand coalition is observed. In support of the tractability of the newly introduced value called equal-surplus Shapley value, we provide a simple and compact formula. Depending on which probability distributions over the sample spaces are admissible, we present several characterization results of the equal-surplus Shapley value. This is achieved by using some classical axioms together with some other appealing axioms such as the independence of local duplication which simply requires that individual shares in a game remain unchanged when only certain events are duplicated in the sample space of a coalition without altering the probability of observing the others.

Appendices

Appendices

In the main text, we have presented some allocation rules and assert that each of them meets some properties. We show here that each such rule effectively satisfies the announced properties.

1.1 Proof of Proposition 7

Proposition 7

Assume that $\varpi$ is uniform on $\varOmega$ and that $|\varOmega _S|=|\varOmega _T|$ for all $S,T\in {\mathcal {C}}_N{\setminus } \{N\}$.

Axioms $\left( E\right)$, $\left( A\right)$, $\left( NP^{*}\right)$ and $\left( SSym\right)$ are independent on ${\mathcal {CC}}(N,\varOmega ,\varpi ).$

Proof

Assume that $\varpi$ is uniform on $\varOmega$ and that $|\varOmega _S|=|\varOmega _T|$ for all coalitions $S,T\in {\mathcal {C}}_N{\setminus } \{N\}.$

• 1. $F_{i}^{1}\left( v,k\right) =\frac{1}{n}v\left( N,k\right)$ for all $v\in {\mathcal {CC}}(N,\varOmega ,\varpi )$, for all coalitions S, for all $k\in \varOmega _N$ and for all $i\in N$. Then $F^{1}$ is $\left( E\right)$, $\left( A\right)$ and $\left( SSym\right)$; but not $\left( NP^*\right)$. Proving this is straightforward and is omitted.

• 2. $F^{2}\left( v\right) =2\varPsi \left( v\right)$ for all $v\in {\mathcal {CC}}(N,\varOmega ,\varpi )$. Then $F^{2}$ is $\left( NP^*\right)$, $\left( A\right)$ and $\left( SSym\right)$; but not $\left( E\right)$. Proving this is straightforward and is omitted.

• 3. Given $v\in {\mathcal {CC}}(N,\varOmega ,\varpi )$, denote by $N^*(v)$ the set of all null players in the expectation game $E_v$ and for all $S\in {\mathcal {C}}_N$, let $V(S)=\{v(S,l):l\in \varOmega _S\}$ be the set of all possible worths of coalition S. Define the value $F^{3}$ for all $v\in {\mathcal {CC}}(N,\varOmega ,\varpi )$ and for all $i\in N$ by

  $$\begin{aligned} F^3_i(v,k)=0 if i\in N^*(v); \text {and } \displaystyle F^3_i(v,k)=\frac{v(N,k)}{|N\backslash N^*(v)|} \text {otherwise.} \end{aligned}$$

  • (a) It is clear that $F^3$ satisfies (E) and $(NP^*)$.

  • (b) Suppose that i and j are two stochastically symmetric players in a c.c. game $v\in {\mathcal {CC}}(N,\varOmega ,\varpi )$. To see that i and j are symmetric players in the expectation game, consider $S\subseteq N\backslash \{i,j\}$.

    $$\begin{aligned} E_v(S\cup \{i\})=&\sum _{l\in \varOmega _{S\cup \{i\}}}\varpi ((S\cup \{i\},l)v((S\cup \{i\},l) \\ =&\sum _{x\in V(S\cup \{i\})}x\sum _{l\in \varOmega _{S\cup \{i\}}:v(S\cup \{i\},l)=x}\varpi (S\cup \{i\},l)\\ =&\sum _{x\in V(S\cup \{j\})}x\sum _{l\in \varOmega _{S\cup \{j\}}:v(S\cup \{j\},l)=x}\varpi (S\cup \{j\},l) \\&\text { since } i \text { and} j \text { are stochastically symmetric in } v\\ =&\sum _{l\in \varOmega _{S\cup \{j\}}}\varpi ((S\cup \{j\},l)v((S\cup \{j\},l)\\ =&E_v(S\cup \{j\}) \end{aligned}$$

    Thus, i and j are symmetric players in $E_v$. By the definition of $F^3$, $F^3_i(v,k)=F^3_j(v,k)$ for all $k\in \varOmega _N$. Therefore $F^3$ satisfies (SSym).

  • (c) Consider $\{i,j\}\in {\mathcal {C}}_N$. Pose $u=\left( {\widetilde{\gamma }}_{\{i\}}\right) _{\varpi }$ and $v=\left( {\widetilde{\gamma }}_{\{i,j\}}\right) _{\varpi }$. Let $k\in \varOmega _S$. We have $N^*(u)=N\backslash \{i\}$ and $N^*(u+v)=N\backslash \{i,j\}$. By the definition of $F^3$,

    $$\begin{aligned} F^3_i(u,k)+F^3_i(v,k)=1+\frac{1}{2}=\frac{3}{2}\text { and } F^3_i(u+v,k)=1. \end{aligned}$$

    Therefore $F^3(v+u)\ne F^3(u)+F^3(v)$ since $F^3_i(u+v,k)\ne F_i^3(u,k)+F_i^3(v,k)$. This proves that $F^3$ does not satisfy (A).

• 4. Given two distinct players i and j in N, denote by a the n-tuple defined by $a_i=1$, $a_j=-1$ and $a_h=0$ for all $h\in N\backslash \{i,j\}$. Let the value $F^{4}$ be defined for all $v\in {\mathcal {CC}}(N,\varOmega ,\varpi )$ and for all $k\in \varOmega _N$ by

  $$\begin{aligned} F^4(v,k)= & {} \varPsi (v,k) +\left[ \sum _{l\in \varOmega _{\{i,j\}}}\varpi (\{i,j\},l)v(\{i,j\},l)-\sum _{l\in \varOmega _{\{i\}}}\varpi (\{i\},l)v(\{i\},l)\right. \\&\left. \quad -\sum _{l\in \varOmega _{\{j\}}}\varpi (\{j\},l)v(\{j\},l)\right] a. \end{aligned}$$

  • (a) Since the terms of $\varPsi (v,k)$ sum to v(N, k) and the terms of a sum to zero, it follows that the terms of $F^4(v,k)$ sum to v(N, k). Therefore $F^{4}$ satisfies $\left( E\right)$.

  • (b) Suppose that u is a TU-game on N and $h\in N$ is a null player in u. If $h\in N\backslash \{i,j\}$, then by the definition of a, $a_h=0$ and $F_h^4\left( {\widetilde{u}}_{\varpi },k\right) =\varPsi _h({\widetilde{u}}_{\varpi },k)=0$ since $\varPsi$ is $(NP^*)$. Now, without lost generality, suppose that $h=i$. Since $\varPsi$ is $\left( NP^*\right)$, $\varPsi _i({\widetilde{u}}_{\varpi },k)=0$. Moreover, ${\widetilde{u}}_{\varpi }(S,l)=u(S)$ for all coalitions S and for all $l\in \varOmega _S$. Thus, by the definition of $F^4$, we have:

    $$\begin{aligned} F_{i}^{4}\left( {\widetilde{u}}_{\varpi },k\right)= & {} \left[ \sum _{l\in \varOmega _{\{i,j\}}}\varpi (\{i,j\},l)u(\{i,j\})-\sum _{l\in \varOmega _{\{i\}}}\varpi (\{i\},l)u(\{i\})\right. \\&\left. -\sum _{l\in \varOmega _{\{j\}}}\varpi (\{j\},l)u(\{j\})\right] a_i \\= & {} \left[ \underbrace{u(\{i,j\})-u(\{j\})}_0-\underbrace{u(\{i\})}_0\right] a_i \text { since }\\&\varpi \text { is a probability distribution function}=0. \end{aligned}$$

    We conclude that $F^4$ satisfies $(NP^*)$.

  • (c) $F^4$ verifies (A) since $\varPsi$ verifies (A) and the coefficient of vector a in the definition of $F^4$ is linear.

  • (d) Pose $v=\varUpsilon ^{k,\{i,j\}}$ for some $k\in \varOmega _{\{i,j\}}$. Players i and j are stochastically symmetric in v. Moreover,

    $$\begin{aligned} F_{i}^{4}\left( v,k\right) =\varPsi _{i}(v,k)+1 \text { and } F_{j}^{4}\left( v,k\right) =\varPsi _{j}(v,k)-1. \end{aligned}$$

    Since $\varPsi _{i}(v,k)=\varPsi _{j}(v,k)$ by symmetry of $\varPsi$, we deduce that $F^4_i((v,k)\ne F_j^4(v,k)$. Therefore $F^{4}$ does not satisfy $\left( SSym\right)$.

In summary, the four axioms are independent. $\square$

1.2 Proof of Proposition 8

Proposition 8

The axioms $\left( E\right)$, $\left( A\right)$, $\left( NP^{*}\right)$, $\left( SSym\right)$ and $\left( ILD\right)$ are independent on ${\mathcal {CC}}^{r}\left( N\right) .$

Proof

Each of the four values presented in Proposition 7 fails to satisfy exactly one axiom among $\left( E\right)$, $\left( A\right)$, $\left( NP^{*}\right)$ and $\left( SSym\right)$. Each of those four values obviously satisfies (ILD). Therefore, we only have to prove that (ILD) can not be deduced from the other four axioms in consideration on ${\mathcal {CC}}^{r}\left( N\right) .$ To prove this, consider a pair $\{a,b\}$ of integers and the collection of sample spaces $\varOmega ^0$ such that $\varOmega ^0_S=\{a,b\}$ for all $S\in {\mathcal {C}}_N$. Denote by $\varpi ^0$ the uniform probability distribution function on $\varOmega ^0$ and let $\varpi _p$ be the probability distribution function defined for all coalitions S by $\varpi _p(S,a)=\frac{2}{p}$ and $\varpi (S,b)=1-\frac{2}{p}$ where p is a prime number such that $p\ge 3$. Now, define the value $F^5$ on ${\mathcal {CC}}^{r}\left( N\right)$ by

$$\begin{aligned} F^5\left( v\right) =\left\{ \begin{array}{ll} \varPsi \left( {{\widehat{v}}} \right) &{}\quad \text {if } v\in {\mathcal {CC}}(N,\varOmega ^{0},\varpi _p) \\ \varPsi \left( v\right) &{}\quad \text {otherwise} \end{array} \right. \end{aligned}$$

where the game ${{\widehat{v}}}$ is obtained from v by substituting to $\varpi _p$ the uniform probability distribution function $\varpi ^0$ on $\varOmega ^{0}.$

• 1. The value $F^{5}$ satisfies $\left( E\right)$ since $\varPsi$ verifies (E).

(a):

To prove that $F^5$ satisfies $(NP^*)$, suppose that u is a TU-game on N and $i\in N$ is a null player in u. We have to prove that for all $k\in \varOmega _N$, $F^5_i\left( v,k\right) =0$ where $v={\widetilde{u}}_{\varpi }$ for an arbitrary probability distribution function $\varpi$ on a collection $\varOmega$ of sample spaces. First suppose that $v\in {\mathcal {CC}}(N,\varOmega ^{0},\varpi _p)$; that is $\varOmega =\varOmega ^0$ and $\varpi =\varpi _p$. Then ${\widehat{v}}={\widetilde{u}}_{\varpi ^0}$ and $F^5_i\left( v,k\right) =\varPsi _i\left( {\widetilde{u}}_{\varpi ^0},k\right) =0$ since $\varPsi$ satisfies $(NP^*)$. Now, suppose that $v\notin {\mathcal {CC}}(N,\varOmega ^{0},\varpi _p)$. Then $F^5_i\left( v,k\right) =\varPsi _i\left( {\widetilde{u}}_{\varpi },k\right) =0$ since $\varPsi$ satisfies $(NP^*)$. Thus, we conclude that $F^5$ satisfies $(NP^*)$.

(b):

By noting that ${\widehat{u+v}}={\widehat{u}}+{{\widehat{v}}}$ for all $u,v\in {\mathcal {CC}}(N,\varOmega ^{0},\varpi _p)$, it follows that $F^5$ verifies (A) since $\varPsi$ verifies (A).

(c):

Suppose that i and j are two stochastically symmetric players in a c.c. game $v\in {\mathcal {CC}}(N,\varOmega ,\varpi )$. Let $k\in \varOmega _N$. First suppose that $v\notin {\mathcal {CC}}(N,\varOmega ^{0},\varpi _p)$. By the definition of $F^5$, $F^5(v)=\varPsi (v)$. Since $\varPsi$ verifies (SSym), it follows that $F^5_i\left( v,k\right) =F_j^5\left( v,k\right)$ for all $k\in \varOmega _N$. Now, suppose that $v\in {\mathcal {CC}}(N,\varOmega ^{0},\varpi _p)$. By the definition of $\varOmega ^0$, i and j are such that $v(S\cup \{i\},l)=v(S\cup \{j\},l)$ for all $S\subseteq N\backslash \{i,j\}$ and for all $l\in \varOmega _{S\cup \{i\})}=\{a,b\}=\varOmega _{S\cup \{j\})}$. From v to ${\widehat{v}}$, only the probability distribution function changes. Therefore, ${\widehat{v}}(S\cup \{i\},l)={\widehat{v}}(S\cup \{j\},l)$ for all $S\subseteq N\backslash \{i,j\}$ and for all $l=\{a,b\}$. This proves that i and j are stochastically symmetric in ${\widehat{v}}$ and that $F^5(v)=\varPsi ({\widehat{v}})$. Since $\varPsi$ verifies (SSym), it follows that $F^5_i\left( v,k\right) =F_j^5\left( v,k\right)$ for all $k\in \varOmega _N$. This prove that $F^5$ satisfies (SSym).

(d):

Consider $i\in N$. Pose $S=N\backslash \{i\}$, $v=\varUpsilon ^{a,S}\in {\mathcal {CC}}(N,\varOmega ^{0},\varpi _p)$ and $u=v^{S,b,b'}$ where u is obtained from v by only duplicating, in $\varOmega _S$, b into b and $b'$. Note that $v\in {\mathcal {CC}}(N,\varOmega ^{0},\varpi _p)$, ${\widehat{v}}\in {\mathcal {CC}}(N,\varOmega ^{0},\varpi ^0)$ and $u\notin {\mathcal {CC}}(N,\varOmega ^{0},\varpi _p)$. Also note that,

$$\begin{aligned} E_u=E_v=\frac{2}{p}\gamma _S+\left( 1-\frac{2}{p}\right) \gamma _N \text { and } E_{{\widehat{v}}}=\frac{1}{p}\gamma _S+\left( 1-\frac{1}{p}\right) \gamma _N. \end{aligned}$$

(14)

Therefore

$$\begin{aligned} F^5_i(v,a)=\varPsi _i({\widehat{v}},a)=Shap_i\left( \frac{1}{p}\gamma _S+\left( 1-\frac{1}{p}\right) \gamma _N\right) =\frac{1}{n}-\frac{1}{np} \end{aligned}$$

and

$$\begin{aligned} F^5_i(u,a)=\varPsi _i(u,a)=Shap_i\left( \frac{2}{p}\gamma _S+\left( 1-\frac{2}{p}\right) \gamma _N\right) =\frac{1}{n}-\frac{2}{np} \end{aligned}$$

It follows that $F^5_i(u,a)\ne F^5_i(v,a)$. Since u is obtained from v by a duplication of b in $\varOmega _S$, we conclude that $F^{5}$ does not satisfy $\left( ILD\right)$. The proof is thus completed. $\square$

1.3 Proof of Proposition 10

Proposition 10

Axioms $\left( E\right)$, $\left( A\right)$, $\left( NP^{*}\right)$, $\left( SSym\right)$ and $\left( MCI\right)$ are independent on ${\mathcal {CC}}\left( N\right) .$

Proof

Each of the four values presented in Proposition 7 fails to satisfy exactly one axiom among $\left( E\right)$, $\left( A\right)$, $\left( NP^{*}\right)$ and $\left( SSym\right)$. Each of those four values obviously satisfies (MCI). Now, we have proved that the value $F^5$ in the proof of Proposition 8 satisfies $\left( E\right)$, $\left( A\right)$, $\left( NP^{*}\right)$ and $\left( SSym\right)$. To prove that $F^5$ fails to meet (MCI), consider $i\in N$. Pose $S=N\backslash \{i\}$, $v=\varUpsilon ^{a,S}\in {\mathcal {CC}}(N,\varOmega ^{0},\varpi _p)$ and $u=v^{S,a\sim b}$ where u is obtained from v by merging, in $\varOmega _S$, a and b into a. Note that $v\in {\mathcal {CC}}(N,\varOmega ^{0},\varpi _p)$, ${\widehat{v}}\in {\mathcal {CC}}(N,\varOmega ^{0},\varpi ^0)$ and $u\notin {\mathcal {CC}}(N,\varOmega ^{0},\varpi _p)$. Since the expectation game does not change by applying an MC-merging operation, (14) still holds. Therefore,

$$\begin{aligned} F^5_i(v,a)=\varPsi _i({\widehat{v}},a)=Shap_i\left( \frac{1}{p}\gamma _S+\left( 1-\frac{1}{p}\right) \gamma _N\right) =\frac{1}{n}-\frac{1}{np} \end{aligned}$$

and

$$\begin{aligned} F^5_i(u,a)=\varPsi _i(u,a)=Shap_i\left( \frac{2}{p}\gamma _S+\left( 1-\frac{2}{p}\right) \gamma _N\right) =\frac{1}{n}-\frac{2}{np} \end{aligned}$$

It follows that $F^5_i(u,a)\ne F^5_i(v,a)$. Since u is obtained from v by an MC-merging operation, we conclude that $F^{5}$ does not satisfy $\left( MCI\right)$. The proof is thus completed.

$\square$

1.4 Proof of Proposition 11

Proposition 11

Axioms $\left( E\right)$, $\left( A\right)$, $\left( NP^{*}\right)$, $\left( SSym\right)$ and $\left( MKI\right)$ are independent on ${\mathcal {CC}}\left( N\right) .$

Proof

Each of the four values presented in Proposition 7 fails to satisfy exactly one axiom among $\left( E\right)$, $\left( A\right)$, $\left( NP^{*}\right)$ and $\left( SSym\right)$. Each of those four values obviously satisfies (MKI). Now, we have proved that the value $F^5$ in the proof of Proposition 8 satisfies $\left( E\right)$, $\left( A\right)$, $\left( NP^{*}\right)$ and $\left( SSym\right)$. To prove that $F^5$ fails to meet (MKI), consider $i\in N$. Pose $S=N\backslash \{i\}$, $v=\varUpsilon ^{a,S}\in {\mathcal {CC}}(N,\varOmega ^{0},\varpi _p)$ and $u=v^{S,a\wedge b}$ where u is obtained from v by an MK-merging operation, in $\varOmega _S$. Note that $v\in {\mathcal {CC}}(N,\varOmega ^{0},\varpi _p)$, ${\widehat{v}}\in {\mathcal {CC}}(N,\varOmega ^{0},\varpi ^0)$ and $u\in {\mathcal {CC}}(N,\varOmega ^{0},\varpi _p)$. Since the expectation game does not change by applying an MK-merging operation, (14) still holds. Therefore,

$$\begin{aligned} F^5_i(v,a)=\varPsi _i({\widehat{v}},a)=Shap_i\left( \frac{1}{p}\gamma _S+\left( 1-\frac{1}{p}\right) \gamma _N\right) =\frac{1}{n}-\frac{1}{np} \end{aligned}$$

and

$$\begin{aligned} F^5_i(u,a)=\varPsi _i((\widehat{u},a)=Shap_i\left( \frac{2}{p}\gamma _S+\left( 1-\frac{2}{p}\right) \gamma _N\right) =\frac{1}{n}-\frac{2}{np} \end{aligned}$$

It follows that $F^5_i(u,a)\ne F^5_i(v,a)$. Since u is obtained from v by an MK-merging operation, we conclude that $F^{5}$ does not satisfy $\left( MKI\right)$. The proof is thus completed.

$\square$