The SD-prenucleolus for TU games

Abstract

We introduce and characterize a new solution concept for TU games: The Surplus Distributor Prenucleolus. The new solution is a lexicographic value although it is not a weighted prenucleolus. The SD-prenucleolus satisfies core stability, strong aggregate monotonicity, null player out property in the class of balanced games and coalitional monotonicity in the class of monotonic games with veto players. We characterize the solution in terms of balanced collections of sets and we provide a simple formula for computing it in the class of monotonic games with veto players.

Appendix

Proof of Lemma 3.

Proof

Assume that the lemma is not true. Consider the minimal number $k$ such that there exist coalitions $S,T$ with $S\in {\mathcal {H}}_{k}, T\in {\mathcal {H}}_{k+1}\setminus {\mathcal {H}}_{k}$, and $F(T)\le F(S)$. Consider the stage of the algorithm where the collection ${\mathcal {H}}_{k-1}$ is fixed. Since $S\in {\mathcal {H}}_{k}$,

$$\begin{aligned} F_{{\mathcal {H}}_{k-1}}(S)=\min \limits _{U\not \in {\mathcal {H}}_{k-1}}F_{{\mathcal {H}}_{k-1}}(U). \end{aligned}$$

Note that $F(T)=F_{{\mathcal {H}}_{k}}(T)$ since $T\in {\mathcal {H}} _{k+1}\setminus {\mathcal {H}}_{k}$. The minimality of $k$ implies that:

• 1. For every $i\in \sigma _{{\mathcal {H}}_{k-1}}(T)$ it holds that

  $$\begin{aligned} f_{{\mathcal {H}}_{k},F}(i,T)=f_{{\mathcal {H}}_{k-1},F}(i,T)<F(S). \end{aligned}$$

• 2. For every $i\in \sigma _{{\mathcal {H}}_{k}}(T)\setminus \sigma _{{\mathcal {H}}_{k-1}}(T)$ it holds that

  $$\begin{aligned} f_{{\mathcal {H}}_{k},F}(i,T)=F(S). \end{aligned}$$

We consider three cases:

• 1. $\sigma _{{\mathcal {H}}_{k}}(T)\ne T$. Since $F_{{\mathcal {H}}_{k}}(T)=F(T)\le F(S)$, it holds that

  $$\begin{aligned} F(S)&\ge F_{{\mathcal {H}}_{k}}(T)=\frac{x(T)-v(T)-\sum \nolimits _{i\in \sigma _{ {\mathcal {H}}_{k}}(T)}f_{{\mathcal {H}}_{k},F}(i,T)}{|T|-\left| \sigma _{{\mathcal {H}}_{k}}(T)\right| }\\&= \frac{x(T)-v(T)\!-\!\sum \nolimits _{i\in \sigma _{{\mathcal {H}}_{k-1}}(T)}f_{ {\mathcal {H}}_{k-1},F}(i,T)\!-\!F(S)\left( \left| \sigma _{{\mathcal {H}}_{k}}(T)\right| \!-\!\left| \sigma _{ {\mathcal {H}}_{k-1}}(T)\right| \right) }{|T|\!-\!\left| \sigma _{{\mathcal {H}}_{k}}(T)\right| } \end{aligned}$$

  This implies that

  $$\begin{aligned} x(T)-v(T)-\sum \limits _{i\in \sigma _{{\mathcal {H}}_{k-1}}(T)}f_{{\mathcal {H}}_{k-1},F}(i,T)\le F(S)\left( \left| T\right| -\left| \sigma _{{\mathcal {H}}_{k-1}}(T)\right| \right) . \end{aligned}$$

  Therefore

  $$\begin{aligned} \frac{x(T)-v(T)-\sum \nolimits _{i\in \sigma _{{\mathcal {H}}_{k-1}}(T)}f_{{\mathcal {H}}_{k-1},F}(i,T)}{\left| T\right| -\left| \sigma _{{\mathcal {H}}_{k-1}}(T)\right| }\le F(S) \end{aligned}$$

  and consequently $F_{{\mathcal {H}}_{k-1}}(T)\le F(S)$ contradicting the fact that

  $$\begin{aligned} F(S)=F_{{\mathcal {H}}_{k-1}}(S)=\min \limits _{U\not \in {\mathcal {H}}_{k-1}}F_{{\mathcal {H}}_{k-1}}(U). \end{aligned}$$

  Therefore $F_{{\mathcal {H}}_{k-1}}(T)=\min \limits _{U\not \in {\mathcal {H}}_{k-1}}F_{{\mathcal {H}}_{k-1}}(U)$ and from Algorithm 1 it holds that $T\in {\mathcal {H}}_{k}$ contradicting that $T\in {\mathcal {H}}_{k+1}\setminus {\mathcal {H}}_{k}$.

• 2. $\sigma _{{\mathcal {H}}_{k-1}}(T)\ne T, \sigma _{{\mathcal {H}}_{k}}(T)=T$. Since $\sigma _{{\mathcal {H}}_{k-1}}(T)\ne T$, it holds that

  $$\begin{aligned} F_{{\mathcal {H}}_{k}}(T)&= x(T)-v(T)-\sum \limits _{i\in T}f_{{\mathcal {H}} _{k},F}(i,T)+\max \limits _{i\in T}f_{{\mathcal {H}}_{k},F}(i,T)\\&= x(T)-v(T)-\sum \limits _{i\in T}f_{{\mathcal {H}}_{k},F}(i,T)+F(S)\\&= x(T)-v(T)-\sum \limits _{i\in \sigma _{{\mathcal {H}}_{k-1}}(T)}f_{{\mathcal {H}} _{k-1},F}(i,T)-F(S)\left( \left| T\right| -\left| \sigma _{{\mathcal {H}}_{k-1}}(T)\right| \right) \\&\quad +F(S). \end{aligned}$$

  Since $F_{{\mathcal {H}}_{k}}(T)=F(T)\le F(S)$, it can be obtained that

  $$\begin{aligned} x(T)-v(T)-\sum \limits _{i\in \sigma _{{\mathcal {H}}_{k-1}}(T)}f_{{\mathcal {H}}_{k-1},F}(i,T)\le F(S)\left( \left| T\right| -\left| \sigma _{{\mathcal {H}}_{k-1}}(T)\right| \right) . \end{aligned}$$

  This implies that

  $$\begin{aligned} \frac{x(T)-v(T)-\sum \nolimits _{i\in \sigma _{{\mathcal {H}}_{k-1}}(T)}f_{{\mathcal {H}}_{k-1},F}(i,T)}{\left| T\right| -\left| \sigma _{{\mathcal {H}}_{k-1}}(T)\right| }\le F(S). \end{aligned}$$

  Therefore, $F_{{\mathcal {H}}_{k-1}}(T)\le F(S)$ and as in the previous case a contradiction is obtained.

• 3. $\sigma _{{\mathcal {H}}_{k-1}}(T)=\sigma _{{\mathcal {H}}_{k}}(T)=T$. Then

  $$\begin{aligned} F_{{\mathcal {H}}_{k-1}}(T)=F_{{\mathcal {H}}_{k}}(T) \end{aligned}$$

  Since $T\not \in {\mathcal {H}}_{k}$, it can be concluded that

  $$\begin{aligned} F_{{\mathcal {H}}_{k}}(T)=F_{{\mathcal {H}}_{k-1}}(T)>\min \limits _{U\not \in {\mathcal {H}}_{k-1}}F_{{\mathcal {H}}_{k-1}}(U)=F_{{\mathcal {H}}_{k-1}}(S). \end{aligned}$$

$\square$

Proof of Proposition 6.

Proof

Consider games $(N,v)$ and $(N,v_{A})$ where $v_{A}(N)=v(N)+A|N|$ and $v_{A}(S)=v(S)$ for $S\ne N$. Let $x\in X(N,v)$ and $y\in X(N,v_{A})$ such that $y_{i}=x_{i}+A$ for each $i\in N$.

It is sufficient to show that if the following holds for some $k\ge 0$

• 1. $F(S,y)=F(S,x)+A$ for each $S\in {\mathcal {H}}_{k}$

• 2. The collections ${\mathcal {H}}_{k}(y)$ and ${\mathcal {H}}_{k}(x)$ coincide

then the two statements hold also for $k+1$ (for $k=0$ it is evident). Note that for every $T\subsetneq N, i\in \sigma _{{\mathcal {H}}_{k}}(T)$ it holds that

$$\begin{aligned} f_{{\mathcal {H}}_{k},F_{v_{A}}}(i,T)=\min \limits _{U:U\in {\mathcal {H}}_{k},i\in U\subset T}(F_{v}(U)+A)=f_{{\mathcal {H}}_{k},F_{v}}(i,T)+A. \end{aligned}$$

Consider a coalition $T\subsetneq N$. We distinguish two cases:

• 1. $\sigma _{{\mathcal {H}}_{k}}(T)\ne T$. Then

  $$\begin{aligned} F_{{\mathcal {H}}_{k}}(T,y)&= \frac{y(T)-v_{A}(T)-\sum \nolimits _{i\in \sigma _{ {\mathcal {H}}_{k}}(T)}f_{{\mathcal {H}}_{k},F}^{v_{A}}(i,T)}{\left| T\right| -\left| \sigma _{{\mathcal {H}}_{k}}(T)\right| }\\&= \frac{x(T)+A|T|-v(T)-\sum \nolimits _{i\in \sigma _{{\mathcal {H}}_{k}}(T)}f_{ {\mathcal {H}}_{k},F}^{v}(i,T)-A\left| \sigma _{{\mathcal {H}}_{k}}(T)\right| }{|T|-\left| \sigma _{ {\mathcal {H}}_{k}}(T)\right| }\\&= \frac{x(T)-v(T)-\sum \nolimits _{i\in \sigma _{{\mathcal {H}}_{k}}(T)}f_{{\mathcal {H}}_{k},F}^{v}(i,T)}{|T|-\left| \sigma _{{\mathcal {H}}_{k}}(T)\right| }+A=F_{{\mathcal {H}}_{k}}^{v}(T,x)+A \end{aligned}$$

• 2. $\sigma _{{\mathcal {H}}_{k}}(T)=T$ Then

  $$\begin{aligned} F_{{\mathcal {H}}_{k}}^{v_{A}}(T,y)&= y(T)-v_{A}(T)-\sum \limits _{i\in T}f_{{\mathcal {H}}_{k},F}^{v_{A}}(i,T)+\max \limits _{i\in T}f_{{\mathcal {H}}_{k},F}^{v_{A}}(i,T)\\&= x(T)+A|T|-v(T)-\sum \limits _{i\in T}f_{{\mathcal {H}}_{k},F}^{v_{A}}(i,T)\\&\quad -A|T|+ \max \limits _{i\in T}f_{{\mathcal {H}}_{k},F}^{v_{A}}(i,T)+A\\&= x(T)-v(T)-\sum \limits _{i\in T}f_{{\mathcal {H}}_{k},F}^{v_{A}}(i,T)+\max \limits _{i\in T}f_{{\mathcal {H}}_{k},F}^{v_{A}}(i,T)\\&\quad +A=F_{{\mathcal {H}}_{k}}^{v}(T,x)+A. \end{aligned}$$

  In this way, for every coalition $T\subsetneq N$ it holds that

  $$\begin{aligned} F_{{\mathcal {H}}_{k}}^{v_{A}}(T,y)=F_{{\mathcal {H}}_{k}}^{v}(T,x)+A \end{aligned}$$

  and therefore collections ${\mathcal {H}}_{k+1}$ in both games ($(N,v_{A})$ and $(N,v))$ coincide.

$\square$

Proof of Proposition 8.

Proof

Consider a balanced game $(N,v)$ where $i\in N$ is a null player. Let $x\in C(N,v)$ be a core allocation. To prove the NPO property of the SD-prenucleolus it is sufficient to show that for every $S\subset N\setminus \{i\}$

$$\begin{aligned} F(S,x)=F(S\cup \{i\},x). \end{aligned}$$

It is immediately apparent that $x_{i}=0$ and coalition $\{i\}$ has the minimal satisfaction, which is 0. Therefore if $P\in \underset{S\subset N\backslash \{i\}}{\arg \min } \frac{x(S)-v(S)}{\left| S\right| }$ then $F(P,x)=F\left( P\cup \{i\},x\right)$.

Assume that for coalitions that obtain their satisfaction before step $k$ it holds that $F(S,x)=F(S\cup \{i\},x)$. We now prove that for step $k$ and any coalition $S\in {\mathcal {H}}_{k}, S\subset N\backslash \{i\}$ it also holds that

$$\begin{aligned} F_{{\mathcal {H}}_{k}}(S,x)=F_{{\mathcal {H}}_{k}}\left( S\cup \{i\},x\right) . \end{aligned}$$

(1)

Note that

$$\begin{aligned} \sigma _{{\mathcal {H}}_{k}}\left( S\cup \{i\}\right) =S\cup \{i\}\Leftrightarrow \sigma _{{\mathcal {H}}_{k}}(S)=S. \end{aligned}$$

We consider two cases (relevant and completed coalitions):

• 1. $\sigma _{{\mathcal {H}}_{k}}\left( S\cup \{i\}\right) \ne S\cup \{i\}$. Then

  $$\begin{aligned} F_{{\mathcal {H}}_{k}}(S\cup \{i\})&= \frac{x(S\cup \{i\})-v(S\cup \{i\})-\sum \nolimits _{j\in \sigma _{{\mathcal {H}}_{k}}\left( S\cup \{i\}\right) }f_{{{\mathcal {H}}_{k}},F}(j,S)}{|S|+1-\left| \sigma _{{\mathcal {H}}_{k}}(S\cup \{i\})\right| }\\&= \frac{x(S)-v(S)-\sum \limits _{j\in \sigma _{{\mathcal {H}}_{k}}(S)}f_{{{\mathcal {H}}_{k}},F}(j,S)}{|S|+1-\left| \sigma _{{\mathcal {H}}_{k}}(S)\right| -1}=F_{{\mathcal {H}}_{k}}(S). \end{aligned}$$

• 2. $\sigma _{{\mathcal {H}}_{k}}(S\cup \{i\})=S\cup \{i\}$. Then

  $$\begin{aligned} F_{{\mathcal {H}}_{k}}\left( S\cup \{i\}\right)&= x\left( S\cup \{i\}\right) -v\left( S\cup \{i\}\right) \\&\quad -\sum \limits _{j\in S\cup \{i\}}f_{{{\mathcal {H}}_{k}},F}(j,S)+\max \limits _{j\in S\cup \{i\}}f_{{{\mathcal {H}}_{k}},F}(j,S\cup \{i\})\\&= x(S)-v(S)-\sum \limits _{j\in S}f_{{{\mathcal {H}}_{k}},F}(j,S)+\max \limits _{j \in S}f_{{{\mathcal {H}}_{k}},F}(j,S\cup \{i\}) \end{aligned}$$

In order to show that $F_{{\mathcal {H}}_{k}}(S\cup \{i\})=F_{{\mathcal {H}}_{k}}(S)$ it suffices to check that

$$\begin{aligned} \max \limits _{j\in S}f_{{{\mathcal {H}}_{k}},F}(j,S\cup \{i\})=\max \limits _{j\in S}f_{{{\mathcal {H}}_{k}},F}(j,S). \end{aligned}$$

But

$$\begin{aligned} f_{{{\mathcal {H}}_{k}},F}(j,S\cup \{i\})=\min \limits _{T\in {{\mathcal {H}}_{k}} ,j\in T\subset S\cup \{i\}}F_{{\mathcal {H}}_{k}}(T). \end{aligned}$$

Since the fact (1) holds for $k^{\prime }<k$, it can be concluded that $F_{{\mathcal {H}}_{k}}(T)=F_{{\mathcal {H}}_{k}}(T\cup \{i\})$ for every $T\in {{\mathcal {H}}_{k}}$. Therefore

$$\begin{aligned} \min \limits _{T\in {{\mathcal {H}}_{k}},j\in T\subset S\cup \{i\}}F_{{\mathcal {H}}_{k}}(T)=\min \limits _{T\in {{\mathcal {H}}_{k}},j\in T\subset S}F_{{\mathcal {H}}_{k}}(T). \end{aligned}$$

This implies that

$$\begin{aligned} f_{{{\mathcal {H}}_{k}},F}(j,S\cup \{i\})=f_{{{\mathcal {H}}_{k}},F}(j,S) \end{aligned}$$

and the proposition has been proved.$\square$