Competitive outcomes and the inner core of NTU market games

Abstract

We consider the inner core as a solution concept for cooperative games with non-transferable utility (NTU) and its relationship to payoffs of competitive equilibria of markets that are induced by NTU games. An NTU game is an NTU market game if there exists a market such that the set of utility allocations a coalition can achieve in the market coincides with the set of utility allocations the coalition can achieve in the game. In this paper, we introduce a new construction of a market based on a closed subset of the inner core which satisfies a strict positive separability. We show that the constructed market represents the NTU game and, further, has the given closed set as the set of payoff vectors of competitive equilibria. It turns out that this market is not uniquely determined, and thus, we obtain a class of markets. Our results generalize those relating to competitive outcomes of NTU market games in the literature.

A Appendix

1.1 A.1 Proof Lemma 1

Proof

• As $V(S)=C^S-\mathbb {R}^S_+$ for all $S \in \mathcal {N}$ it is enough to show that for all $S \in \mathcal {N}$ the payoff vectors in the set $C^S$ can be achieved by coalition $S$ in the market $\mathcal {E}_{V,A,\varepsilon }$. Consider an arbitrary $S \in \mathcal {N}$. Let $z\in C^S$. We show, that there exists a feasible $S$-allocation $\left( x^i\right) _{i\in S}$ with $\left( y^i\right) _{i\in S}$ such that $u^i\left( x^i\right) =z_i$ for all $i\in S$. Define for $i \in S$ the consumption plan

  $$\begin{aligned} x^i=\left( z^{\left\{ i\right\} },0,z^{\left\{ i\right\} },0,0\right) \end{aligned}$$

  and let

  $$\begin{aligned} y^i=\frac{1}{|S|}\left( z,-e^S,z,-e^S,-e^S\right) \end{aligned}$$

  be the (same) production plan for all $i \in S$. We observe that by the definition of the consumption sets $x^i \in X^i$ for all $i \in S$. With regard to the production sets for $S\ne N$, we have immediately $y^i \in Y^i$ for all $i\in S$. For $S=N$ note that $z \in V(N) \subseteq \tilde{V}(N)$ and thus $P_A(z)=z$. Hence, we have $y^i \in Y^i$ for all $i\in N$. Observe that

  $$\begin{aligned} \sum \limits _{i\in S}\left( x^i-\omega ^i\right) =\sum \limits _{i\in S}y^i. \end{aligned}$$

  Hence, $\left( x^i\right) _{i\in S}$ is a feasible $S$-allocation and

  $$\begin{aligned} u^i\left( x^i\right) =z_i \quad \text{ for } \text{ all } i\in S. \end{aligned}$$

• Let

  $$\begin{aligned} \left( \bar{x}^{(1)i},0,\bar{x}^{(3)i},0,0\right) _{i\in S} \end{aligned}$$

  be a feasible $S$-allocation with

  $$\begin{aligned} \left( \bar{y}^{(1)i},\bar{y}^{(2)i},\bar{y}^{(3)i},\bar{y}^{(4)i},\bar{y}^{(5)i}\right) _{i\in S} \end{aligned}$$

  in the market $\mathcal {E}_{V,A,\varepsilon }$. The feasibility implies

  $$\begin{aligned}&\left( \sum \limits _{i\in S}\bar{x}^{(1)i}, -e^S,\sum \limits _{i\in S}\bar{x}^{(3)i},-e^S,-e^S\right) \nonumber \\&\quad =\sum \limits _{i\in S}\left( \bar{y}^{(1)i},\bar{y}^{(2)i},\bar{y}^{(3)i},\bar{y}^{(4)i},\bar{y}^{(5)i}\right) . \end{aligned}$$

  (5)

  Each production set is a convex cone of a union of convex sets. Hence, an arbitrary production plan can be written in the following way: Choose one suitable element from each of the convex sets and build a linear combination (with non-negative coefficients) of these elements. For the 1st and the 2nd block of $n$ commodities we obtain, that there exist $\alpha ^i_R\in \mathbb {R}_+$ for all $R\in \mathcal {N},\, z^i_R\in C^R$ for all $R\in \mathcal {N}\setminus \{N\}$ and $\tilde{z}^i_N\in \tilde{C}^N$, such that

  $$\begin{aligned} \left( \bar{y}^{(1)i},\bar{y}^{(2)i}\right) = \sum \limits _{R\in \mathcal {N}\setminus \{N\}} \alpha ^i_R \left( z^i_R, - e^R\right) + \alpha ^i_N \left( P_A\left( \tilde{z}^i_N\right) , - e^N\right) . \end{aligned}$$

  (6)

  As $P_A\left( \tilde{C}^N\right) =C^N$ there exists $z^i_N\in C^N$ such that $P_A\left( \tilde{z}^i_N\right) =z^i_N$ and hence (6) simplifies to

  $$\begin{aligned} \left( \bar{y}^{(1)i},\bar{y}^{(2)i}\right) =\sum \limits _{R\in \mathcal {N}} \alpha ^i_R \left( z^i_R, - e^R\right) . \end{aligned}$$

  (7)

  As feasibility implies $\left( \sum \nolimits _{i\in S}\bar{x}^{(1)i},-e^S\right) =\sum \nolimits _{i\in S}\left( \bar{y}^{(1)i},\bar{y}^{(2)i}\right)$, for the 2nd block of $n$ coordinates we have from combining the respective coordinates in (5) and (7)

  $$\begin{aligned} e^S = \sum \limits _{i\in S}\sum \limits _{R\in \mathcal {N}}\alpha ^i_R e^R= \sum \limits _{R\in \mathcal {N}}\left( \sum \limits _{i\in S} \alpha _R^i\right) e^R. \end{aligned}$$

  (8)

  Thus, $\alpha ^i_R>0$ implies $R\in \mathcal {S}=\{R' \subseteq S| R' \ne \emptyset \}$ and if we define $\alpha \left( R\right) =\sum \nolimits _{i\in S}\alpha ^i_R$, then (8) implies that $\left( \alpha \left( R\right) \right) _{R\in \mathcal {S}}$ is a balanced family for the coalition $S$. Looking at the 1st block of $n$ coordinates, we have from combining the respective coordinates in (5) and (7)

  $$\begin{aligned} \sum \limits _{i\in S} \bar{x}^{(1)i}&=\sum \limits _{R\in \mathcal {S}} \sum \limits _{i\in S} \alpha ^i_R z^i_R =\sum \limits _{\{R\in \mathcal {S}|\alpha \left( R\right) >0\}} \alpha (R)\left( \frac{1}{\alpha \left( R\right) } \sum \limits _{i\in S} \alpha ^i_R z^i_R\right) . \end{aligned}$$

  (9)

  Since for all $R \in \mathcal {S}$ the set $C^R$ is convex, we have

  $$\begin{aligned} \frac{1}{\alpha \left( R\right) } \sum \limits _{i\in S} \alpha ^i_R z^i_R \in C^R \end{aligned}$$

  (10)

  and hence, using total balancedness in (9) together with (10), we have $\sum \nolimits _{i\in S} \bar{x}^{(1)i} \in V(S)$. From the definition of the utility function, we obtain

  $$\begin{aligned} u^i\left( \bar{x}^{(1)i},0,\bar{x}^{(3)i},0,0\right) \le \bar{x}^{(1)i}_i. \end{aligned}$$

  Since $\left( \bar{x}^{(1)i}_i\right) _{i \in S} \le \sum \nolimits _{i\in S} \bar{x}^{(1)i} \in V(S)$ we have by the $S$-comprehensiveness of $V\left( S\right)$ that $\left( u^i\left( \bar{x}^{(1)i},0,\bar{x}^{(3)i},0,0\right) \right) _{i\in S}\in V\left( S\right)$.

$\square$

1.2 A.2 Proof of the Claim used in the Proof of Proposition 1

Proof

Suppose there exists a production plan $y^i \in Y^i$ such that $\hat{p} \cdot y^i> 0$. We first distinguish two cases:

• the production plan is of the form $\left( c^S, -e^S, c^S, -e^S,-e^S\right)$ for $S \in \mathcal {N} \setminus \{N\}$ and $c^S\in C^S$,

• or of the form $\left( P_A \left( \tilde{c}^N \right) , -e^N ,\tilde{c}^N, -e^N,-e^N \right)$ for $c^N\in C^N$.

In the first case, we obtain using the definition of $\hat{p}$

$$\begin{aligned} \hat{p} \cdot y^i&= \lambda ^a \cdot c^S -\frac{2}{3}\left( \lambda ^a\circ \, a\right) \cdot e^S +\lambda ^a\cdot c^S -\frac{2}{3}\left( \lambda ^a\circ \, a\right) \cdot e^S -\frac{2}{3}\left( \lambda ^a\circ \, a\right) \cdot e^S \\&=2 \left( \lambda ^a \cdot c^S - \left( \lambda ^a\circ \, a\right) \cdot e^S \right) \\&= 2 \left( \sum \limits _{i \in S}\left( \lambda ^a_ic^S_i - \lambda ^a_i a_i\right) \right) \end{aligned}$$

If the last expression was strictly positive, then coalition $S$ could improve upon $a$ using the vector of utility weights $\lambda ^a$ with $c^S$. This in contradiction to the fact that $a$ is chosen to be in the inner core.

In the second case, we have

$$\begin{aligned} \hat{p} \cdot y^i&= \lambda ^a \cdot P_A \left( \tilde{c}^N \right) \!-\!\frac{2}{3}\left( \lambda ^a\circ \, a\right) \cdot e^N \!+\!\lambda ^a\cdot \tilde{c}^N -\frac{2}{3}\left( \lambda ^a\circ \, a\right) e^N -\frac{2}{3}\left( \lambda ^a\circ \, a\right) e^N\\&= \left( \lambda ^a \cdot c^N - \left( \lambda ^a\circ \, a\right) \cdot e^N \right) +\left( \lambda ^a \cdot \tilde{c}^N - \left( \lambda ^a\circ \, a\right) \cdot e^N \right) \\&= \left( \sum \limits _{i \in N}\left( \lambda ^a_ic^N_i - \lambda ^a_i a_i\right) \right) +\left( \sum \limits _{i \in N}\left( \lambda ^a_i\tilde{c}^N_i - \lambda ^a_i a_i\right) \right) \end{aligned}$$

From the argumentation from the first case, we know that the first summand is non-positive. Moreover, the definition of $\tilde{V}(N)$ (on page 8) implies that the second summand is non-positive as well. Therefore, in both cases, $\hat{p} \cdot y^i \le 0$. Repeating this argument for arbitrary $y^i\in Y^i$ being in the convex cone of sets derived from elements of this form shows the claim. $\square$

1.3 A.3 Proof of Lemma 3

For the proof of Lemma 3, we follow the idea of Clippel and Minelli (2005).

Proof

Let $\left( \left( \hat{x}^i\right) _{i\in N},\left( \hat{y}^i\right) _{i\in N},\hat{p}\right)$ be a competitive equilibrium with $\hat{p}\in \mathbb {R}^\ell _+\setminus \left\{ 0\right\}$. For each individual $i\in N$ define the set

$$\begin{aligned} C^i=\left\{ \left( u,m\right) \in \mathbb {R}^2|\exists z^i\in X^i:u\le u^i\left( z^i\right) -u^i\left( \hat{x}^i\right) ,m\le \hat{p} \cdot \left( \omega ^i+\hat{y}^i-z^i\right) \right\} . \end{aligned}$$

By the concavity of $u^i$ and by the linearity of the scalar product, the set $C^i$ is convex for each $i\in N$. On the other hand, $C^i\cap \mathbb {R}^2_{++}=\emptyset$, as $\hat{x}^i$ is optimal for individual $i$ in his budget set.

• Suppose $(u,m) \in C^i$ and $(u,m) \gg 0$, then there exists $z^i \in X^i$ with $u^i(\hat{x}^i)<u^i(z^i)$ and $\hat{p} \cdot z^i< \hat{p} \cdot (\omega ^i + \hat{y}^i)$ which means $z^i$ gives individual $i$ a higher utility as $\hat{x}^i$ and is affordable under the price system $\hat{p}$. This is in contradiction to the optimality of $\hat{x}^i$.

Therefore, $(0,0)$ is not in the interior of the convex set $C^i$. Hence, by the supporting hyperplane theorem¹⁴ there exists a nonzero, non-negative vector $\left( \alpha ^i,\beta ^i\right) \in \mathbb {R}^2_+$ such that we can separate $(0,0)$ from $C^i$ and obtain

$$\begin{aligned} 0\ge \alpha ^i u+\beta ^i m \end{aligned}$$

(11)

for all $\left( u,m\right) \in C^i.$ Note, that for any $z^i \in X^i$ we have that

$$\begin{aligned} \left( u^i\left( z^i\right) -u^i\left( \hat{x}^i\right) ,\hat{p}\cdot \left( \omega ^i+\hat{y}^i-z^i \right) \right) \in C^i. \end{aligned}$$

By (11) we obtain

$$\begin{aligned} \alpha ^i u^i \left( \hat{x}^i\right) \ge \alpha ^i u^i\left( z^i\right) -\beta ^i \hat{p}\cdot \left( z^i-\omega ^i-\hat{y}^i\right) \end{aligned}$$

for all $z^i\in X^i$.

As $\hat{p}\cdot \omega ^i >0$, it follows from the above inequality that we have $\alpha ^i>0$.

• To see this suppose $\alpha ^i=0$ ($\beta ^i>0$). Then, as in equilibrium $\hat{p} \cdot \hat{y}^i=0$, we obtain from the above inequality

  $$\begin{aligned} 0 \le \hat{p} \cdot \left( z^i-\omega ^i-\hat{y}^i\right) \quad \text{ for } \text{ all } \, z^i\in X^i, \end{aligned}$$

  which is not true, as $0\in X^i$ and $\hat{p}\cdot \hat{y}^i=0$. Thus, $\alpha ^i>0$.

Moreover, monotonicity and locally non-satiation of the utility function imply that $\beta ^i>0$.

Let $\hat{\lambda }^i=\frac{\alpha ^i}{\beta ^i}>0$. Summing up over all $i\in S$ we obtain

$$\begin{aligned} \sum \limits _{i\in S} \lambda ^i u^i\left( \hat{x}^i\right) \ge \sum \limits _{i\in S}\lambda ^i u^i\left( z^i\right) -\hat{p}\cdot \sum \limits _{i\in S}\left( z^i-\omega ^i-\hat{y}^i\right) \end{aligned}$$

for all $S\subseteq N$ and for all $z^i\in \mathbb {R}^{\ell }_+$ with $i\in S$.

Suppose $\left( u^i\left( \hat{x}^i\right) \right) _{i\in N}$ is not in the inner core. Then, for each $\lambda \gg 0$, there exists $S\subseteq N$ and a feasible $S$-allocation $\left( \tilde{x}^i \right) _{i \in S}$ satisfying

$$\begin{aligned} \sum \limits _{i\in S} \lambda ^i u^i\left( \tilde{x}^i \right) > \sum \limits _{i\in S}\lambda ^i u^i\left( \hat{x}^i \right) . \end{aligned}$$

In particular, for the case $\lambda =\hat{\lambda }$, there exists $S\subseteq N$ and a feasible $S$-allocation $\left( \bar{x}^i\right) _{i \in S}$ (feasible with the production plan $\left( \bar{y}^i\right) _{i \in S}$) such that

$$\begin{aligned} \sum \limits _{i\in S} \hat{\lambda }^i u^i\left( \bar{x}^i \right) > \sum \limits _{i\in S}\hat{\lambda }^i u^i\left( \hat{x}^i \right) . \end{aligned}$$

Due to the feasibility we have,

$$\begin{aligned} \sum \limits _{i\in S} \left( \bar{x}^i-\omega ^i-\bar{y}^i\right) \le 0. \end{aligned}$$

We obtain

$$\begin{aligned} \sum \limits _{i\in S} \hat{\lambda }^i u^i\left( \bar{x}^i\right)&> \sum \limits _{i\in S} \hat{\lambda }^i u^i\left( \hat{x}^i\right) \\&\ge \sum \limits _{i\in S}\hat{\lambda }^i u^i\left( \bar{x}^i\right) -\hat{p}\cdot \sum \limits _{i\in S}\left( \bar{x}^i-\omega ^i-\hat{y}^i\right) \\&\ge \sum \limits _{i\in S}\hat{\lambda }^i u^i\left( \bar{x}^i\right) -\hat{p}\cdot \sum \limits _{i\in S}\left( \bar{x}^i-\omega ^i\right) \\&\ge \sum \limits _{i\in S}\hat{\lambda }^i u^i\left( \bar{x}^i\right) -\hat{p}\cdot \sum \limits _{i\in S}\bar{y}^i. \end{aligned}$$

Therefore, $0 < \hat{p}\cdot \sum \limits _{i\in S}\bar{y}^i$. This is a contradiction. Hence, $\left( u^i\left( \hat{x}^i\right) \right) _{i\in N}$ is in the inner core. $\square$