%Paper: ewp-ge/9406002
%From: pklm@impa.br (Paulo Klinger Monteiro)
%Date: Fri, 10 Jun 94 13:39:41 GMT-3:00
%Date (revised): Thu, 20 Oct 1994 11:08:21 -0300 (EST)
%Date (revised): Thu, 20 Oct 1994 12:37:44 -0300 (EST)

\documentstyle{article}
\date{1994}
\author{Paulo Klinger Monteiro\\UFRJ/IMPA}
\title{ A concretization of Mas-Colell and Zame's counter-example}
\begin{document} 
\maketitle
\roman{enumi}
\newcommand{\ra}{\rightarrow}
\section{Introduction}
The incomplete markets equilibrium existence theorem for economies with
uncountably many states has been studied by several authors.However
the following hypothesis is made:
\[
\mbox{for every } \theta \geq v_{i} \,\, W^{i}_{s}+\theta A_{s}\geq 0
\mbox{ for almost every } s\in S
\]
where $S$ is the set of states,$W^{i}:S\ra \Re^{G}_{+} $ is the initial
endowments of consumer $i,1\leq i\leq I,\theta \in \Re^{J}$ is the vector
of portfolios whose uncovered sales lower bound is $v_{i}\in
-\Re^{J}_{++}$.Finally $A:S\ra (\Re^{G})^{J}$ is the matrix of real assets
returns .This
hypothesis is considered very strong.  Mas-Colell and Zame(1991) by means of 
a counter-example demonstrated its indispensability .In the construction of
theirs counter-example, they required the existence of 
utility functions  satisfying several 
properties.  However one may wonder which utilities functions satisfy their 
requirements.My purpose here is to specify such functions. The
utilities functions I've got doesn't fit exactly in Mas-Colell and Zame's 
hypotheses.But there is not an equilibrium as well.
\section{The utilities and endowments}
First choose number $(\bar{x},a,\beta,\phi)$: \begin{enumerate}
\item $0<\bar{x}<a<1$\\
\item   $\beta >(a+1)/a$ ,$\phi >\frac{1+a}{(a-\bar{x})(1+\bar{x})}$ \\
\item $\frac{\beta}{\phi}+\frac{1-\beta (1+\bar{x})}{1+\bar{x}}(a-\bar{x})>0$\\
\item $\phi e^{\phi  (\bar{x}-1)}>\beta e^{-\beta
\bar{x}}\frac{1+\bar{x}}{a-\bar{x}}$ \\
\item $-2+\beta (1+\bar{x})\geq 0$
\end{enumerate}
For example $\bar{x}=3/4$,$a=7/8$,$\phi=121/14$,$\beta=605/84$  \\
The utilities(state independent) are as follows: \\
$u^{1}(x,y)=-e^{-\beta x}+g(y)$ and $u^{2}(x,y)=w(x)-e^{-\phi y}$ where\\
\[ g(y)=\left\{ \begin{array}{ll}
		\beta \int_{0}^{y} e^{-\beta s}\frac{1+s}{a-s}ds & 0\leq y\leq \bar{x} \\
\beta \int_{0}^{\bar{x}} e^{-\beta s}\frac{1+s}{a-s}ds
+\beta e^{-\beta \bar{x}}\frac{1+\bar{x}}{a-\bar{x}}(y-\bar{x})  &  \bar{x}\leq
y \end{array} \right. \]
\mbox{}
\[ w(x)= \left\{ \begin{array}{ll}
\phi \int_{0}^{1-\bar{x}} e^{-\phi s}\frac{a-1+s}{2-s}ds +
\phi e^{-\phi (1-\bar{x})}\frac{a-\bar{x}}{1+\bar{x}}(x-1+\bar{x}) &
0\leq x\leq 1-\bar{x} \\
\phi \int_{0}^{x} e^{-\phi s}\frac{a-1+s}{2-s}ds & 1-\bar{x}\leq x\leq 1 \\
\phi \int_{0}^{1} e^{-\phi s}\frac{a-1+s}{2-s}ds+
\phi e^{-\phi}a(x-1) & 1\leq x \end{array} \right.\]
The Pareto  optimum set is as below.
\vspace{3cm}
We have that $u^{i}$  is concave and strictly monotone($u^{i}$ is concave if
(ii) is true).
\\ Define $U^{i}(x_{0},x,y)=x_{0}+\int_{0}^{1} u^{i}(x_{s},y_{s})ds,\,
x_{0},x_{s},y_{s} \in \Re_{+}$. $U^{i}$
is concave and strictly monotone.The endowments are $W^{1}_{s}=(s,a),
W^{2}_{s}=(1-s,1-a)$,$W^{1}+W^{2}=(1,1)$.There is only one asset $A=(1,0)$.

The consumer problem is \[
\max\,\, x_{0}+\int_{0}^{1} u^{i}(x_{s},y_{s})ds
\] subject to $w^{i}\geq x_{0}+\pi\theta$ , $\theta \geq v_{i}$ 
and to $(1,q_{s})(W^{i}_{s}+\theta A)\geq x_{s}+q_{s}y_{s}$
This problem is equivalent to \[
\max -\pi \theta +\int_{0}^{1} v^{i}_{s}((1,q_{s})(W^{i}_{s}+\theta A))ds
\] subject to $w^{i}\geq \pi\theta$ and $\theta \geq v_{i}$ 
and to $(1,q_{s})(W^{i}_{s}+\theta A)\geq 0$
\\where $v^{i}(R)$ is the indirect utility at prices $(1,q_{s})$ and income
$R$.
An equilibrium is a function $q:S\ra \Re_{+},\pi \in \Re$ and numbers $(\bar{
\theta}^{i})_{i=1,2}$ such that $\bar{\theta}^{1}+\bar{\theta}^{2}=0$ and
$W^{1}_{s}+W^{2}_{s}=\sum_{i=1}^{2}x_{i}((1,q_{s}),(1,q_{s})(W^{i}_{s}
+\bar{\theta}^{i}A))$ where $x_{i}(\cdot,\cdot),$ is the Marshallian
demand,solves $\max \, x^{0}+\int_{0}^{1}u^{i}(x_{s})ds$ subject
to $x^{0}+\pi \theta \leq w^{i},\theta \geq v_{i},(1,q_{s})(W^{i}_{s}+\theta
A)\geq 0$. 
\section{The Pareto optimum points of $(u^{1},u^{2})$. }
Suppose $\lambda \in (0,1)$.  I want to find the solutions of \[
\max_{0\leq x\leq 1,0\leq y \leq 1}
 \lambda (-e^{-\beta x}+g(y))+(1-\lambda)(w(1-x)-e^{-\phi (1-y)}) \]
Define $\phi (x)=-\lambda e^{-\beta x}+(1-\lambda)w(1-x)$ and $\psi (y)=
\lambda g(y)-(1-\lambda)e^{-\phi (1-y)}$. It is easy to check that both
functions are strictly concave.Suppose $\alpha$ is the solution of
$\max_{0\leq x\leq 1}\phi (x)$. We have:\\
$\alpha \leq \bar{x} \Leftrightarrow \phi^{'}(\bar{x})\leq 0 \Leftrightarrow
\frac{\lambda \beta}{(1-\lambda)\phi} \leq e^{(\beta+\phi)\bar{x}-\phi}\frac{
a-\bar{x}}{1+\bar{x}}$ \\
The first order conditions are the same for $\phi$ and $\psi$ if
$\frac{\lambda \beta}{(1-\lambda)\phi} \leq e^{(\beta+\phi)\bar{x}-\phi}\frac{
a-\bar{x}}{1+\bar{x}}$.Therefore in this case $\phi$ and $\psi$ have
the same maximizer $\alpha \leq \bar{x}$.If 
$\frac{\lambda \beta}{(1-\lambda)\phi} \geq e^{(\beta+\phi)\bar{x}-\phi}\frac{
a-\bar{x}}{1+\bar{x}}$ then call $u$ the maximizer of $\phi$ and $v$ the 
maximizer of $\psi$. We have that both are not smaller than $\bar{x}$.The first
order
condition imply: \\
$\lambda \beta e^{-\beta u}=(1-\lambda)\phi e^{-\phi
(1-\bar{x})}\frac{a-\bar{x}}{1+\bar{x}},\, \beta \lambda e^{-\beta
\bar{x}}\frac{1+\bar{x}}{a-\bar{x}}=
\phi (1-\lambda)e^{-\phi (1-v)}$.\\
Then dividing one by the other and cancelling terms in common and finally
taking logarithms we have \[
\frac{\beta}{\phi}(u-\bar{x})+\bar{x}=v \]
The supporting price of the allocation $(x,x)$ is 
$p(x)=\frac{1+x}{a-x}$(I take
the first price coordinate equal to one).For $x\geq \bar{x}$ we have
$p(x)=e^{\beta (x-\bar{x})}\frac{1+\bar{x}}{a-\bar{x}}$. For $x=-1$ let 
$Y$ be the meeting point of the line passing through $(u,v)$ with 
inclination $-p(u)$: $(1,p(u))(-1,Y)
=(1,p(u))(u,\frac{\beta}{\phi}(u-\bar{x})+\bar{x})$. 
Then \begin{eqnarray}
Y(u) & = & \frac{(u+1)(a-\bar{x})}{(1+\bar{x})}
e^{\beta (u-\bar{x})}+\frac{ \beta}{\phi}(u-\bar{x})+\bar{x}
 \nonumber \\
Y'(u) & = & \frac{a-\bar{x}}{1+\bar{x}}\{ 
e^{-\beta (u-\bar{x})}-(u+1)\beta e^{-
\beta (u-\bar{x})}\}+\beta/\phi \nonumber \\
Y''(u) & = & \frac{a-\bar{x}}{1+\bar{x}} \beta e^{-\beta 
(u-\bar{x})}\{-2+ \beta (u+1)\}>0 \nonumber
 \end{eqnarray}
Therefore since $Y'(\bar{x})>0$ we have that $Y(u)>Y(\bar{x})=a$ if
$u>\bar{x}$.So every budget line supporting a Pareto point after $\bar{x}$ do
not touch
 the line $x=-1$ below $a$\footnote{for $x=1$,$\frac{\beta}{\phi}(1-\bar{x})+
\bar{x}\leq y\leq 1,\mbox{ we have that } Y\geq Y(1)>a$.}

Now suppose $(\bar{x}^{i}_{0},\bar{x}^{i},\bar{y}^{i},\bar{\theta}^{i},\pi,
q_{s})$ is an equilibrium.We have that $\bar{\theta}^{1}+\bar{\theta}^{2}=0$,
$\bar{x}_{0}=w^{i}-\pi
\bar{\theta}^{i}$,$(1,q_{s})(\bar{x}^{i}_{s},\bar{y}^{i}_{s})
=(1,q_{s})(W^{i}_{s}+\bar{\theta}^{i}A) \mbox{ a.e. $s$}$. We have also
$u^{i}(\bar{x}^{i}_{s},\bar{y}^{i}_{s})=\nu^{i}_{s}((1,q_{s})(W^{i}_{s}+
\bar{\theta}^{i}A))$,and $q_{s}=\frac{\partial u^{i}/\partial y}{\partial
u^{i}/\partial x}$. Necessarily $W^{1}_{1s}+\bar{\theta}^{1}>-1 \mbox{ a.e.
$s$}$ since there are not supporting prices passing through $(x,a),x<-1$.
Therefore
$\bar{x}^{i}_{s}>\bar{x}$,$\bar{y}^{i}_{s}>\bar{x}_{i}\mbox{ a.e. $s$}$.
Define \[
h^{i}(\theta)=-\pi\theta+\int \nu^{i}_{s}((1,q_{s})(W^{i}_{s}+\theta A))ds \]
\mbox{}
\[ Dh^{i}(\theta)=-\pi +\int \frac{\partial \nu^{i}_{s}}{\partial R}ds
\]
\mbox{ }\[
Dh^{i}(\bar{\theta}^{i})=-\pi +\int \frac{\partial u^{i}}{\partial x}(
\bar{x}^{i}_{s})ds \]
At the optimum we have:\[
 Dh^{1}(\bar{\theta}^{1})=-\pi +\int \beta e^{-\beta \bar{x}^{1}_{s}}ds
\leq -\pi +\int \beta e^{-\beta \bar{x}}ds=-\pi +\beta e^{-\beta \bar{x}} 
\] and  \[
 Dh^{2}(\bar{\theta}^{2})=-\pi +\int w'(\bar{x}^{2}_{s})ds=-\pi+\phi e^{-\phi
(1-\bar{x})}\frac{a-\bar{x}}{1+\bar{x}}>Dh^{1}(\bar{\theta}^{1}) \]
The proof now is divided in two cases:\\
a) $\bar{\theta}^{1}\geq 0$\\
Then $Dh^{1}(\bar{\theta}^{1})=0 \mbox{ or }
Dh^{1}(\bar{\theta}^{1})\geq 0 \mbox{ and } w^{1}-\pi\bar{\theta}^{1}=0$.
In any case this implies 
$Dh^{2}(\bar{\theta}^{2})>0$.So consumer 2 wants to increase
$\bar{\theta}^{2}$.
This is possible if $\bar{\theta}^{2}\leq 0$. A contradiction.\\
b) $\bar{\theta}^{1}<0$.\\
The income inequality is not binding,since at $(\bar{x},\bar{x})$ the income
has a strictly positive lower bound.The inequality $w^{1}-\pi\bar{\theta}^{1}
\geq 0$ is not binding also.Therefore we have $Dh^{1}(\bar{\theta}^{1})=0$,
hence $\pi=\beta e^{-\beta\bar{x}}$. Also $Dh^{2}(\bar{\theta}^{2})>0$.So
$w^{2}-\pi\bar{\theta}^{2}=0$,so $\bar{\theta}^{1}=-w^{2}/(\beta e^{-\beta
\bar{x}})$. Now we choose $v_{1}<\frac{-w^{2}}{\beta e^{-\beta\bar{x}}}$ and
$w^{2}$ such that \[
 -\frac{w^{2}}{\beta e^{-\beta \bar{x}}}<-1\]
This contradicts the fact that $s+\bar{\theta}^{1}>-1 $ for almost
every $s$.
\begin{center}
\bf References
\end{center}
Mas-Colell,A,W.Zame(1991),The existence of security markets equilibria with a
non-atomic state space,to appear in Journal of Mathematical Economics.

\end{document}
