On the equal treatment imputations subset in the bargaining set for smooth vector-measure games with a mixed measure space of players

Abstract

This paper studies vector-measure games in which the set of players is represented by a mixed measure space with atoms and with an atom-less part. The bargaining set introduced in Mas-Colell (J Math Econ 18(2):129–139, 1989) for a continuum exchange economy is adapted and analyzed within this (TU) market game. It is shown that the equal treatment imputations in the bargaining set have some interesting properties. In particular, we prove an extension of the budgetary exploitation property of core allocations in mixed markets to equal-treatment imputations in the Mas-Colell bargaining set.

Appendices

Appendix 1

Claim 2

The$\hat{C}$imputations (in the monopoly case) with two types of players are$x = \left( {x_{a} ,x_{0} } \right)$such that$f\left( {\omega_{a} + \alpha \omega_{0} } \right) \le x_{a} + \alpha x_{0}$for every number$\alpha \in \left( {0,1} \right)$which is equivalent to the existence of some super-differential price vector$p \in \partial f\left( {\mu \left( T \right)} \right)$with$x_{0} \le p.\omega_{0}$.

Proof

Since $\left( {\mu \left( T \right) - \frac{1}{n}\omega_{0} } \right) \in R_{ + + }^{M}$ for $n = 2,3, \ldots ,$ by concavity of $f$ there exist super-differentials $p_{n} \in \partial f\left( {\mu \left( T \right) - \frac{1}{n}\omega_{0} } \right)$.

$f\left( {\omega_{a} + \alpha \omega_{0} } \right) \le x_{a} + \alpha x_{0}$ for $\alpha \in [0,1]$, it follows:

$$ f\left( {\omega_{a} + \left( {1 - \frac{1}{n}} \right)\omega_{0} } \right) \le x_{a} + \left( {1 - \frac{1}{n}} \right)x_{0} $$

(1)

By efficiency, $x_{a} + x_{0} = f\left( {\omega_{a} + \omega_{0} } \right)$. Moreover, by the definition of super-differential $p_{n} \in \partial f\left( {\mu \left( T \right) - \frac{1}{n}\omega_{0} } \right)$ we have:

$$ x_{a} + x_{0} = f\left( {\mu \left( T \right)} \right) \le f\left( {\omega_{a} + \left( {1 - \frac{1}{n}} \right)\omega_{0} } \right) + p_{n} \left( {\frac{1}{n}\omega_{0} } \right) $$

(2)

Therefore, by (1) and (2) we have:

$$ x_{a} + x_{0} \le x_{a} + \left( {1 - \frac{1}{n}} \right)x_{0} + p_{n} \left( {\frac{1}{n}\omega_{0} } \right) $$

yielding $\frac{1}{n}x_{0} \le \frac{1}{n}p_{n} \omega_{0}$. Thus, $x_{0} \le p_{n} \omega_{0}$ for all $n = 2,3, \ldots$. Letting $n \to \infty$, then by the concavity of $f\left( \cdot \right)$ and its continuity, we obtain by the u.h.c. of $\partial f\left( \cdot \right)$ on $R_{ + + }^{M}$ that there is $p \in \partial f(\mu \left( T \right))$ with $x_{0} \le p\omega_{0}$. $\square$

Appendix 2

Assume there is $\alpha \in \left( {0,1} \right)$ with $f\left( {\omega_{a} + \alpha \omega_{0} } \right) > x_{a} + \alpha x_{0}$, then for all $\beta \in [1,\infty )$ we have $\left( {\omega_{a} + \beta \omega_{0} } \right) \le x_{a} + \beta x_{0}$.

Proof

Assume by negation $f\left( {\omega_{a} + \beta \omega_{0} } \right) > x_{a} + \beta x_{0}$ for some $\beta > 1.$

Then by convexity of the interval $[\alpha , \beta ]$ containing 1 in its interior, there are $(a,b) \in R_{ + + }^{2}$ with $a + b = 1$ such that $a\alpha + b\beta = 1.$ Therefore, by the identities,

$$ a\left( {\omega_{a} + \alpha \omega_{0} } \right) + b(\omega_{a} + \beta \omega_{0} ) = \left( {a + b} \right)\omega_{a} + \left( {a\alpha + b\beta } \right)\omega_{0} = \omega_{a} + \omega_{0} , $$

$$ a\left( {x_{a} + \alpha x_{0} } \right) + b(x_{a} + \beta x_{0} ) = \left( {a + b} \right)x_{a} + \left( {a\alpha + b\beta } \right)x_{0} = x_{a} + x_{0} , $$

We obtain from the concavity of the production function:

$$ \begin{aligned} f\left( {\omega_{a} + \omega_{0} } \right) & = f(a\left( {\omega_{a} + \alpha \omega_{0} } \right) + b(\omega_{a} + \beta \omega_{0} )) \\ & \ge af\left( {\omega_{a} + \alpha \omega_{0} } \right) + bf\left( {\omega_{a} + \beta \omega_{0} } \right) \\ & > a\left( {x_{a} + \alpha x_{0} } \right) + b\left( {x_{a} + \beta x_{0} } \right) = x_{a} + x_{0} . \\ \end{aligned} $$

This strict inequality contradicts the efficiency of the imputation. $\square$