Reduced games and egalitarian solutions

Abstract

For a class of reduced games satisfying a monotonicity property, we introduce a family of set-valued solution concepts based on egalitarian considerations and consistency principles, and study its relation with the core. Regardless of the reduction operation we consider, the intersection between both sets is either empty or a singleton containing the lexmax solution Arin et al. (Math Soc Sci 46:327–345, 2003). This result induces a procedure for computing the lexmax solution for a class of games that contains games with large core Sharkey (Int J Game Theory 11:175–182, 1982).

Appendix

Proof of Proposition 3

Let (N, v) be a game, $\alpha \in \mathcal {A}_t$ and $x\in \phi ^{\alpha }(N,v)$ generated by $\pi =(T_1,\ldots , T_t)$, with $t>1$. For $k\in \{1,\ldots ,t-1\}$, let us denote $N_k=N{\setminus } T_1\cup \cdots \cup T_k$. Let $N_0=N$ and $v=r^{N_0}_{\alpha ,x}(v).$ For $k \le t-1,$ $i\in T_k$ and $j\in T_{k+1}$, we have

$$\begin{aligned} \begin{array}{c} x_i=\frac{r^{N_{k-1}}_{\alpha ,x}(v)(T_k)}{|T_k|}\quad \text {and}\quad x_j=\frac{r^{N_k}_{\alpha ,x}(v)(T_{k+1})}{|T_{k+1}|}= \frac{ r^{N_k}_{\alpha ,x_{|N_{k-1}}}\left( r^{N_{k-1}}_{\alpha ,x}(v) \right) (T_{k+1})}{|T_{k+1}|}. \end{array} \end{aligned}$$

(12)

We distinguish two cases:

• Case 1: $T_{k+1}=N_k$. In this situation, for $j\in T_{k+1}$,

  $$\begin{aligned} x_j=\frac{ r^{N_{k-1}}_{\alpha ,x}(v)(N_{k-1})- r^{N_{k-1}}_{\alpha ,x}(v)(T_k)}{|N_k|}. \end{aligned}$$

  (13)

  Suppose $x_j > x_i$, for $i\in T_k$ and $j\in T_{k+1}$. Then, combining (12) and (13) we obtain

  $$\begin{aligned} r^{N_{k-1}}_{\alpha ,x}(v)(N_{k-1})> \frac{ r^{N_{k-1}}_{\alpha ,x}(v)(T_k)}{|T_k|}(|N_k|+|T_k|) \end{aligned}$$

  or, equivalently,

  $$\begin{aligned} \frac{r^{N_{k-1}}_{\alpha ,x}(v)(N_{k-1})}{|N_{k-1}|}> \frac{ r^{N_{k-1}}_{\alpha ,x}(v)(T_k)}{|T_k|}, \end{aligned}$$

  in contradiction with the fact that $\displaystyle T_k\in \arg \max _{\emptyset \not = T\subseteq N_{k-1}}\left\{ \frac{r^{N_{k-1}}_{\alpha ,x}(v)(T)}{|T|}\right\} .$

• Case 2: $T_{k+1}\subset N_k$. In this case, there is $Q^*\in \alpha (T_k)$ such that, for all $j\in T_{k+1},$

  $$\begin{aligned} x_j=\frac{ r^{N_{k-1}}_{\alpha ,x}(v)(T_{k+1}\cup Q^*)-|Q^*|\frac{ r^{N_{k-1}}_{\alpha ,x}(v)(T_k)}{|T_k|}}{|T_{k+1}|}. \end{aligned}$$

  (14)

  If $x_j >x_i$ for $i\in T_k,$ then combining (12) and (14) we have

  $$\begin{aligned} r^{N_{k-1}}_{\alpha ,x}(v)(T_{k+1}\cup Q^*)>\frac{ r^{N_{k-1}}_{\alpha ,x}(v)(T_k)}{|T_k|} (|T_{k+1}|+|Q^*|) \end{aligned}$$

  or, equivalently,

  $$\begin{aligned} \frac{r^{N_{k-1}}_{\alpha ,x}(v)(T_{k+1}\cup Q^*)}{|T_{k+1}\cup Q^*|}>\frac{ r^{N_{k-1}}_{\alpha ,x}(v)(T_k)}{|T_k|}, \end{aligned}$$

  in contradiction with the fact that $\displaystyle T_k\in \arg \max _{\emptyset \not = T\subseteq N_{k-1}}\left\{ \frac{r^{N_{k-1}}_{\alpha ,x}(v)(T)}{|T|}\right\} .$

Hence, for any $k\in \{1,\ldots , t-1\}$, $x_j\le x_i$ for all $i\in T_k$ and all $j\in T_{k+1},$ which concludes the proof. $\square$

Proof of Proposition 4

Let (N, v) be a game and $x\in \phi ^{\alpha _D}(N,v)$ generated by $\pi =(T_1,\ldots , T_t)$, with $t>1$. For $k\in \{1,\ldots ,t-1\}$ let us denote $N_k=N{\setminus } T_1\cup \cdots \cup T_k$. Let $N_0=N$ and $v=r^{N_0}_{\alpha _D,x}(v).$ For $k \le t-1,$ $i\in T_k$ and $j\in T_{k+1}$ we have

$$\begin{aligned} \begin{array}{c} x_i=\frac{r^{N_{k-1}}_{\alpha _D,x}(v)(T_k)}{|T_k|}\quad \text {and}\quad x_j=\frac{r^{N_k}_{\alpha _D,x}(v)(T_{k+1})}{|T_{k+1}|}. \end{array} \end{aligned}$$

(15)

If $k < t-1$, we distinguish two cases:

• Case 1: $x_j=\frac{v\left( T_{k+1}\right) }{|T_{k+1}|}$.

  In this situation,

  $$\begin{aligned} x_j=\frac{v\left( T_{k+1}\right) }{|T_{k+1}|} \le \frac{ r^{N_{k-1}}_{\alpha _D,x}(v)(T_{k+1})}{|T_{k+1}|} \le \frac{ r^{N_{k-1}}_{\alpha _D,x}(v)(T_{k})}{|T_{k}|}=x_i, \end{aligned}$$

  (16)

  where the first inequality follows from the definition of $\alpha _D$ and the second one from the fact that $\displaystyle T_k\in \arg \max _{\emptyset \not = T\subseteq N_{k-1}}\left\{ \frac{r^{N_{k-1}}_{\alpha _D,x}(v)(T)}{|T|}\right\} .$

• Case 2: $x_j=\frac{v\left( T_1\cup \cdots \cup T_k\cup T_{k+1}\right) -x\left( T_1\cup \cdots ,\cup T_k\right) }{|T_{k+1}|}.$

  Notice first that $x(T_k)=r^{N_{k-1}}_{\alpha _D,x}(v)(T_k).$ Then,

  $$\begin{aligned} x_j= & {} \frac{v\left( T_1\cup \cdots \cup T_k\cup T_{k+1}\right) -x\left( T_1\cup \cdots ,\cup T_{k-1}\right) -x(T_k)}{|T_{k+1}|}\\\le & {} \frac{r^{N_{k-1}}_{\alpha _D,x}(v)(T_k\cup T_{k+1})-x(T_k)}{|T_{k+1}|} \\ {}= & {} \frac{r^{N_{k-1}}_{\alpha _D,x}(v)(T_k\cup T_{k+1})-r^{N_{k-1}}_{\alpha _D,x}(v)(T_k)}{|T_{k+1}|} \\\le & {} \frac{r^{N_{k-1}}_{\alpha _D,x}(v)(T_k)}{|T_k|}=x_i, \end{aligned}$$

  where the first inequality follows from the definition of $\alpha _D$ and the second one from the fact that $\displaystyle T_k\in \arg \max _{\emptyset \not = T\subseteq N_{k-1}}\left\{ \frac{r^{N_{k-1}}_{\alpha _D,x}(v)(T)}{|T|}\right\} .$

If $i\in T_{t-1}$ and $j\in T_t$, then $x_j=\frac{v\left( T_1\cup \cdots \cup T_{t-1}\cup T_{t}\right) -x\left( T_1\cup \cdots ,\cup T_{t-1}\right) }{|T_{t}|}.$ Thus, as in the above Case 2 it can be shown that $x_j\le x_i$.

To see that $\alpha _D\not \in \mathcal {A}_t$, consider the game (N, v) with set of players $N=\{1,2,3,4,5\}$ and characteristic function as follows:

$$\begin{aligned} v(\{4\})= & {} 0.95, v(\{1,4\})=v(\{1,3,4\})=1.9, v(\{2,3\})=v(\{1,2,3\})=1.05, v(\{3,4\})=1,\\ v(\{2,3,4\})= & {} 2.8, v(\{1,2,3,4\})=2, v(\{1,2,3,4,5\})=3.8\quad \text {and}\quad v(S)=0,\quad \text {otherwise}. \end{aligned}$$

Take $x=\left( 0.95, 0.6\hat{3}, 0.6\hat{3}, 0.95, 0.6\hat{3}\right) .$ Routine verification shows that

$$\begin{aligned} r^{\{2,3,5\}}_{\alpha _D, x_{|\{1,2,3,5\}}}\left( r^{\{1,2,3,5\}}_{\alpha _D, x}(v)\right) (\{2,3\})=1.85 > r^{\{2,3,5\}}_{\alpha _D, x}(v)(\{2,3\})=1.05. \end{aligned}$$

$\square$

Proof of Lemma 1

Let (N, v) be a game and $x\in \phi ^{\alpha _{DM}}(N,v)$ generated by $\pi _x=(T_1,\ldots , T_t)$. Let $T_1\cup \cdots \cup T_{q^*}=\{i\in N\,|\,x_i\ge x_j\,\,\text {for all}\,\, j\in N\}.$ Notice first that $q^*< t$ since, otherwise, $x=\left( \frac{v(N)}{|N|},\ldots , \frac{v(N)}{|N|}\right)$ which implies $\displaystyle N\in \arg \max _{\emptyset \not = T\subseteq N}\left\{ \frac{v(T)}{|T|}\right\}$, in contradiction with $N_1\not = N$.

First we show that $T_1\cup \cdots \cup T_{q^*}\subseteq N_1$. Let $i\in T_1\cup \cdots \cup T_{q^*}$. If $i\in T_1$, clearly $i\in N_1$. If $i\in T_h$ for some $h\in \{2,\ldots ,q^*\},$ then there is $Q^*\subseteq T_1\cup \cdots \cup T_{h-1}$ such that

$$\begin{aligned} x_i=\frac{v(T_1)}{|T_1|}=\frac{r^{N{\setminus } T_1\cup \cdots \cup T_{h-1}}_{\alpha _{DM},x}(v)(T_h)}{|T_h|}=\frac{v(T_h\cup Q^*)-|Q^*|\frac{v(T_1)}{|T_1|}}{|T_h|}. \end{aligned}$$

(17)

Reordering terms in (17), we have that $\frac{v(T_1)}{|T_1|}=\frac{v(T_h\cup Q^*)}{|T_h\cup Q^*|},$ which implies $\displaystyle T_h\cup Q^* \in \arg \max _{\emptyset \not = T\subseteq N}\left\{ \frac{v(T)}{|T|}\right\}$, and thus $i\in N_1.$

To show the reverse inclusion, take $i\in N_1$ and suppose $i\not \in T_1\cup \cdots \cup T_{q^*}$. Then, there is $R^*\in M_1$ such that $i\in R^*$. Next we show that $R^*{\setminus } T_1\cup \cdots \cup T_{q^*}\not = T_{q^*+1}\cup \cdots \cup T_{t}$. Indeed, if $R^*{\setminus } T_1\cup \cdots \cup T_{q^*} = T_{q^*+1}\cup \cdots \cup T_{t}$, then $N=T_1\cup \cdots \cup T_{q^*} \cup R^*$. As we have seen before, $T_1\cup \cdots \cup T_{q^*}\subseteq N_1$. This inclusion, together with $\displaystyle R^*\in \arg \max _{\emptyset \not = T\subseteq N}\left\{ \frac{v(T)}{|T|}\right\}$, imply $N_1=N$, a contradiction. Hence,

$$\begin{aligned} \begin{array}{lll} \frac{v(T_1)}{|T_1|}&{}>&{} \frac{r^{N{\setminus } T_1 \cup \cdots \cup T_{q^*}}_{\alpha _{DM},x}(v)(T_{q^*+1})}{|T_{q^*+1}|} \ge \frac{r^{N{\setminus } T_1 \cup \cdots \cup T_{q^*}}_{\alpha _{DM},x}(v)(R^*{\setminus } T_1\cup \cdots \cup T_{q^*})}{|R^*{\setminus } T_1\cup \cdots \cup T_{q^*}|} \\ \\ &{}\ge &{}\frac{v(R^*)-x(R^*\cap \{T_1\cup \cdots \cup T_{q^*}\}}{|R^*{\setminus } T_1\cup \cdots \cup T_{q^*}|} = \frac{v(R^*)-|R^*\cap \{T_1\cup \cdots \cup T_{q^*}\}|\frac{v(T_1)}{|T_1|}}{|R^*{\setminus } T_1\cup \cdots \cup T_{q^*}|}, \end{array} \end{aligned}$$

(18)

where the first inequality follows from the definition of $T_1\cup \cdots \cup T_{q^*}$, the second one from $\displaystyle T_{q^*+1}\in \arg \max _{\emptyset \not = T\subseteq N{\setminus } T_1\cup \cdots \cup T_{q^*}}\left\{ \frac{r^{N{\setminus } T_1\cup \cdots \cup T_{q^*}}_{\alpha _{DM},x}(v)(T)}{|T|}\right\}$, and the last one from the definition of the $\alpha _{DM}$-max reduced game and the fact that $R^*{\setminus } T_1\cup \cdots \cup T_{q^*}\not = T_{q^*+1}\cup \cdots \cup T_{t}$. From (18) it follows $\frac{v(T_1)}{|T_1|}> \frac{v(R^*)}{|R^*|}$, in contradiction with $\displaystyle R^*\in \arg \max _{\emptyset \not = T\subseteq N}\left\{ \frac{v(T)}{|T|}\right\}$. Hence, $i\in T_1 \cup \cdots \cup T_{q^*}$ and $N_1=T_1\cup \cdots \cup T_{q^*}$. $\square$