%Paper: ewp-mac/9902010
%From: ed.green@mpls.frb.org
%Date: Fri, 19 Feb 1999 17:15:32 -0600 (CST)

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\begin{document}

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\begin{center}
{{\LARGE Monetary Equilibrium from an Initial State: \par
\smallskip
the Case Without Discounting}\footnote{The views expressed herein are
those of the authors and not necessarily those of the Federal Reserve
Bank of Minneapolis or the Federal Reserve System.}\par}

\bigskip\bigskip

{\large Edward J.~Green \par 
Federal Reserve Bank of Minneapolis \par
90 Hennepin Avenue \par
Minneapolis, MN 55480-0291 \par
ejg@res.mpls.frb.fed.us \par

\bigskip
and
\bigskip

Ruilin Zhou \par
University of Pennsylvania and \par
Federal Reserve Bank of Minneapolis \par
90 Hennepin Avenue \par
Minneapolis, MN 55480-0291 \par
ruilin@ysidro.sas.upenn.edu \par

\bigskip
\bigskip
May, 1998 \par
(Preliminary)}

\end{center}

\vfill

\begin{abstract}
This paper studies the existence of single-price price equilibrium
from a given initial distribution of money holdings in a
search-theoretic model of money where agents have no time
preference. The model is similar to recent models (Green and Zhou
[2] and Zhou [5]) of search economies with no constraints on
money inventories, except that here money is modeled as indivisible
and traders are assumed to have overtaking-criterion preferences
rather than discounting. The equilibrium concept is dynamic
equilibrium from an initial distribution of money holdings rather than
steady-state equilibrium (which possibly might not be reachable from
an initial state) which was studied earlier. In this environment,
under some mild conditions on the initial distribution, single-price
equilibrium always exists. More precisely, there is an equilibrium
path, along which agents trade at the same price, and the
money-holdings distribution converges asymptotically to a unique
geometric distribution.

\bigskip
J.E.L. Classification: D51, E40
\end{abstract}

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\Section a{Introduction}

Models of environments where traders are randomly matched in pairs,
in which ``double coincidence of wants'' is absent, have recently
undergone rapid development. They have already become an important
class of models in the foundations of monetary economics.

Early versions of random-matching models specified that a trader can
hold only a single unit of money, which is taken to come in
indivisible units. This assumption makes the model highly artificial,
and in particular it is responsible for some very peculiar results
about welfare. In previous work (Green and Zhou [2], Zhou [5]), we
have developed a random-matching model with divisible money and without
inventory constraint. Surprisingly, we have found that the model
possesses a continuum of steady-state, single-price, equilibria (for
essentially any parameter values where equilibrium exists at all) that
support distinct real allocations yielding different levels of
welfare.

Our primary reason for undertaking the present research is to
understand more clearly the significance of this indeterminacy of
steady-state, single-price, equilibrium. On one view, the
indeterminacy might be unimportant because possibly only one (or
perhaps finitely many) of the equilibria could be reached from a given
initial state of the economy. To examine this view, we need to define
and analyze monetary equilibrium from an initial state. Moreover, to
examine the possibility that several distinct equilibria from the same
initial state might exist, we need to study nonstationary
equilibria. This study is begun in the present paper, which reports
research in progress.

In particular, in the current draft of this paper, we study
equilibrium with indivisible money, although without an inventory
constraint on money holdings. Also, we consider only the limiting case
of an economy with perfectly patient traders, whose preferences we
model in an overtaking-criterion framework. We restrict attention to
this framework for two reasons. It is analytically more tractable than
the analogous framework with discounting studied in Zhou [5], and it
has the feature that all stationary Markov-perfect equilibria are
single-price equilibria (which we establish in a companion paper,
Green and Zhou [3]).

We find that single-price equilibrium from an initial state exists
under a mild condition about the initial money-holdings distribution,
and that this equilibrium is asymptotically stationary. Furthermore,
for a given single-price steady state, we characterize a range of
initial distributions from which single-price equilibria converge to
that steady state. We conjecture that the results we obtain here can
be generalized to a divisible-money environment. If that is so, then
we will be able to show that single-price equilibrium from an initial
state, as well as steady-state equilibrium, is indeterminate in this
environment. The reason is that there would be a distinct equilibrium
path corresponding to each alternative level of aggregate real money
balances in the economy.

Even in its present interim state, our research has some intrinsic
interest. The exclusive focus on stationary equilibria in study of
random-matching monetary economies to date has made it difficult to
draw comparisons between results about these models and results about
models in a Walrasian spirit, in which equilibrium is defined with
respect to an initial situation that generically is not stationary.
Moreover, since adoption of a new monetary policy would typically be
expected to move the economy from one steady state to another, the
eventual usefulness of random-matching models to address policy
questions in monetary economics is likely to be enhanced greatly by
being able to characterize and analyze nonstationary equilibria
beginning from an exogenous initial situation. For these reasons, the
study of nonstationary equilibrium in this paper enhances the
usefulness of the random-matching framework for money.

\Section b{The Environment}

Economic activity occurs at dates $\,0,1,2,\ldots$. Agents are
infinitely lived, and they are nonatomic. For convenience, we assume
that the measure of the set of all agents is one. Each agent has a
type in $(0,1]$. The mapping from the agents to their types is a
uniformly distributed random variable, independent of all other random
variables in the model. Similarly, there is a continuum of
differentiated goods, each indexed by a number $j\in (0,1]$. These
goods are indivisible and nonstorable. Production is specialized such
that only the agents of type $i$ can produce one unit of ``brand'' $i$
good at a cost $c$ each period. Each agent consumes half of the brands
in the economy; agent $i$ consumes goods $j\in (i$, mod$(i+{1\over
2})]$ (for example, agent $0.3$ consumes goods $j\in (0.3, 0.8]$, and
agent $0.7$ consumes goods $j\in (0.7,1]\cup (0,0.2]$). From consuming
one unit of a desired good, an agent derives an instantaneous utility
$u$. In addition to the production/consumption goods, there is a fiat
money.\note{Logically, fiat money is an economy-wide accounting system
that satisfies restrictions such as we now describe. It is customary
in the money/search literature, but not logically necessary, to
interpret fiat money as some physical object.} Money comes in
indivisible units, and an agent can costless hold any number of units
of money. The total nominal stock of money remains constant at $M$
units per capita.  We assume that agents do not discount. Their
preferences are characterized by an overtaking criterion with respect
to expected utility, which will be formalized below.

Agents randomly meet pairwise each period. By the assumed pattern of
specialization in production and consumption, there is no double
coincidence of wants in any pairwise meeting. Each agent meets a
producer of his consumption goods with probability one-half, and a
consumer of his production goods with probability one-half. So, every
meeting is between a potential buyer and seller.  Consumption goods
cannot be used as a commodity money because they are nonstorable, so
money is the only medium of exchange available. An agent is
characterized by his type and the amount of money he holds. Within a
pairwise meeting, each agent observes the other's type, but not the
trading partner's money holding and trading history. They cannot
communicate about this information either. However, the economy-wide
money-holdings distribution is common knowledge. For simplicity, we
assume that each transaction occurs according to the following
simultaneous-move game. The potential buyer and seller submit a bid
and offer respectively. Trade occurs if and only if the bid is at
least as high as the offer, and in that case, the buyer pays the
seller's offer price.\footnote{In Green and Zhou [2] and Zhou [4], the
offer was assumed to be made before the bid. Equilibrium in that game
corresponds exactly to equilibrium in the simultaneous-move game. The
refinement of equilibrium in undominated strategies, introduced by
Zhou [4], also applies straightforwardly to the simultaneous-move game.}

\Section c{The Definition of Equilibrium}

The domain of agents' money holdings is $\Nat$. Let $\Delta$ be the
space of probability measures on infinite-dimensional probability
simplex, $\Delta= \{p\,|\,p=(p_0, p_1,\dots),\, \forall k\in\Nat \
p_k\ge 0, \, \sum_{k=0}^{\infty} p_k= 1\}$.  Suppose that the initial
money-holdings distribution is given by $p^0$.


At each date, the set of agents is randomly partitioned into
pairs. Within each pair, one of the agents desires to consume the good
that the other is able to produce. Thus, a bid and offer are
associated with each pair.

Now we provide an intuitive discussion of the distributions of bids
and offers, and we state some formal assumptions about those
distributions. Our assumptions are in the spirit if a ``continuum law of
large numbers.''\footnote{That is, we believe that they are logically
consistent with the results from probability theory that we will apply
in our analysis, although they cannot be derived from those
results. See Green [1] and Gilboa and Matsui [4] for further
discussion.}  For each random partition $\pi$ of the agents into pairs
at date $t$, there is a sample distribution $B_t^\pi$ of bids and a
sample distribution $O_t^\pi$ of offers. We assume that these sample
distributions do not depend on the partition. That is, there are bid
and offer distributions $B_t$ and $O_t$ such that for all partitions
$\pi$, $B_t^\pi=B_t$ and $O_t^\pi=O_t$. Moreover, because each agent
has a trading partner assigned at random, the probability distribution
of the trading partner's bid and offer should be identical to the
sample distribution. That is, $B_t$ and $O_t$ are the probability
distributions of bid and offer respectively that are received at date
$t$ by each individual agent, as well as being the sample distribution
in each random pairing of the population of agents.

Now let the probability space $(\O,\B,\P)$ represent the stochastic
process of encounters faced by a generic agent. This agent faces a
sequence $\o$ of random encounters, one at each date. His date-$t$
encounter, with some agent of type $j$, is characterized by her
trading type (buyer or seller) in the meeting and her bid/offer,
denote it by $\o_t= (\o_{t1}, \o_{t2})$,
\begin{eqnarray}
&\hbox{if $i$ meets a buyer,}& \o_{t1}=b, \;
\o_{t2} \;\hbox{is her bid} \nonumber \\
&\hbox{if $i$ meets a seller,}& \o_{t1}=s, \; \o_{t2}\; \hbox{is her
offer}. \nonumber 
\end{eqnarray}
The encounters $\{\o_t\}_{t=0}^\infty \equiv\o$ are independent across
time. $\O$ is the set of all possible sequences of encounters that an
arbitrary agent in the economy faces. 

At each date $t$, pairwise meetings are independent across the
population. That is, for each agent, $\o_{t1}$ follows a Bernoulli
distribution, a potential buyer's bid price $\o_{t2}$ is drawn from
the bid distribution $B_t$, and a potential seller's offer price
$\o_{t2}$ is drawn from the offer distribution $O_t$. For $t\ge 1$,
let $\B_t$ be the smallest $\s$-algebra on $\O$ that makes the vector of the first
$t$ coordinates, $\o^t=(\o_0, \o_1, \ldots, \o_{t-1})$,
measurable, and $\B_0=\{\phi, \O\}$. Let $\P_t$ be the probability
measure defined on $\B_t$. Then, for all $t\ge 0$, and $k\in \Nat$,
\begin{eqnarray}
\label{eqn:c1}
&&\P_t\{\o_{t1}=b \} = \P_t\{\o_{t1}=s \} ={1\over 2} \\
\label{eqn:c2}
&&\P_t\Bigl\{\o_{t2}= k\,|\, \o_{t1}=b \Bigr\} =B_{tk}\\
\label{eqn:c3}
&&\P_t\Bigl\{\o_{t2}= k\,|\, \o_{t1}=s \Bigr\} =O_{tk}.
\end{eqnarray}
Define $\B=\B_\infty$ and $\P=\P_\infty$.

Let $\s$ be the trading strategy of a generic agent of type $i$ with
initial money holding $\h_0$. His date-$t$ strategy $\s_t$ specifies
his bid $\s_{t1}$---his maximum willingness to pay if he
is paired with a seller of his consumption goods---and his offer
$\s_{t2}$---the price he is willing to sell if he meets a consumer of
his production good---as a function of his initial money holding and
his encounter history $\o$. The strategy $\s_t$ is measurable with
respect to $\B_t$. As a buyer, the agent has to be able to pay his
bid. Let $\h_t^\s$ denote the agent's money holding at the
beginning of date $t$ by adopting strategy $\s$. Then 
\begin{equation}
\label{eqn:c4}
\s_{t1}(\h_0,\o)\le \h_t^\s(\h_0,\o). 
\end{equation}
Given the agent's initial money holding $\h_0$, encounter history
$\o$, and strategy $\s=\{\s_t\}_{t=0}^\infty$, his money holding
evolves recursively as follows: $\h_0^\s(\h_0,\o)=\h_0$ and, for $t\ge 0$,
\begin{equation}
\label{eqn:c5}
\h_{t+1}^\s(\h_0,\o)=\left\{\begin{array}{lcl} 
\h_t^\s(\h_0,\o) + \s_{t2}(\h_0,\o) &\qquad& \hbox{ if } \;\o_{t1}=b
\; \hbox{ and }\; \s_{t2}(\h_0,\o) \le \o_{t2} \\
\h_t^\s(\h_0,\o) -\o_{t2} &\qquad& \hbox{ if } \; \o_{t1}=s
\;\hbox{ and }\; \s_{t1}(\h_0,\o) \ge \o_{t2} \\ 
\h_t^\s(\h_0,\o)&\qquad& \hbox{ otherwise}
\end{array} \right.
\end{equation}
Let $v_t^\s$ denote the agent's date-$t$ utility from his date-$t$ 
trading by adopting strategy $\s$. Then
\begin{equation}
\label{eqn:c6}
v_t^\s(\h_0,\o)=\left\{\begin{array}{lcl} 
-c &\qquad& \hbox{ if } \;\o_{t1}=b \; \hbox{ and }\;
\s_{t2}(\h_0,\o) \le \o_{t2} \\ 
u &\qquad& \hbox{ if } \; \o_{t1}=s \;\hbox{ and }\;
\s_{t1}(\h_0,\o) \ge \o_{t2} \\ 
0 &\qquad& \hbox{ otherwise}
\end{array} \right.
\end{equation}
Then, strategy $\s$ {\it overtakes} another strategy $\hat \s$ if for
all $\h_0\in \Nat$,
\begin{equation}  
\label{eqn:c7}
\liminf_{t\to \infty} \;\mean\Bigl[ \sum_{\t=0}^t v_\t^\s(\h_0,\o) -
\sum_{\t=0}^t v_\t^{\hat \s}(\h_0,\o) \Bigr] >0  
\end{equation}
where $\mean$ is the expectation operator with respect to the
probability measure $\P$.

We are going to focus on symmetric equilibrium at which an agent's
strategy is only a function of his own trading history and initial
money holding. In particular, the trading strategy does not depend on
an agent's type.

At the beginning of date $t$, given all agents' trading strategy
$\s_t$ and the initial money-holdings distribution $p^0$, rational
expectation requires that agents' belief regarding the
bid distribution $B_t$ and the offer distribution $O_t$
that prevail during date-$t$ trading confirm with the actual
distributions implied by the strategy. That is, for all $k\in \Nat$,
\begin{eqnarray}
\label{eqn:c8}
&&B_{tk}=\sum_{l=0}^\infty p_l^0 \P\Bigl\{\s_{t1}(l,\o) = k \Bigr\} \\
\label{eqn:c9} 
O_{tk}=\sum_{l=0}^\infty p_l^0 \P\Bigl\{\s_{t2}(l,\o) = k \Bigr\}
\end{eqnarray}
where $I$ is an indicator function; $I(\alpha,\beta)$ equals $1$ if
$\alpha=\beta$ and $0$ otherwise. 

The equilibrium concept that we adopt is Bayesian Nash equilibrium
with respect to the overtaking criterion.

\medskip
{\sc Definition.} A Bayesian Nash equilibrium is a four-tuple $\langle
\s, p^0, \{B_t\}_{t=0}^\infty, \{O_t\}_{t=0}^\infty \rangle$
that satisfies 
\newcounter{def}
\begin{list}{(\roman{def})}{\usecounter{def}
\setlength{\labelwidth}{1cm} \setlength{\labelsep}{0.3cm}
\setlength{\leftmargin}{1.5cm} \setlength{\topsep}{0.15cm} 
\setlength{\parsep}{0pt} \setlength{\itemsep}{0cm}} 
\item $p^0$ is the initial money-holdings distribution in the
environment.  
\item Given the bid distributions $\{B_t\}_{t=0}^\infty$
and the offer distributions $\{O_t\}_{t=0}^\infty$, and given that all
other agents adopt strategy $\s$, it is optimal for an arbitrary agent
to adopt strategy $\s$ as well, that is, there is no strategy that
overtakes strategy $\s$.
\item Given that all agents adopt trading strategy $\s$, for each
$t\ge 0$, $B_t$ and $O_t$ satisfy equations \eqn {c8} and \eqn {c9}.
\end{list}
\medskip

We are going to study one particular equilibrium at which at all
dates, all traders offer to buy their desired consumption goods at
price $1$ as long as they have money, and accept price $1$ in exchange
for their production goods, hence, all trades occur at price $1$. We
call it price-$1$ equilibrium. This equilibrium is markovian in the
sense that the dependence of agents' strategy on time and trading
history is only through their own current money holdings to satisfy
feasibility condition \eqn {c4}, despite the dynamic environment.
Formally, define the strategy $\ts$ as follows, for all $\h_0\in
\Nat$, encounter history $\o\in\O$, and $t\ge 0$,
\begin{equation}
\label{eqn:c10}
\ts_{t1}(\h_0,\o)=\min\{\h_t^{\ts}(\h_0,\o),1\}, \qquad
\ts_{t2}(\h_0,\o)=1. 
\end{equation}
Let $\tp^t\in \Delta$ denote the money-holdings distribution at the
beginning of date $t$ induced by strategy $\ts$. The bid distribution
implied by strategy $\ts$ puts measure $1-\tp_0^t$ on price $1$ to
reflect the bid of each agent with money, and it puts measure $\tp_0^t$
on price $0$ to reflect the agents who have no money and cannot buy.
The offer distribution implied by $\ts$ is stationary and degenerate
with mass at price $1$. That is
\begin{eqnarray}
\label{eqn:c11}
&&\tB_{t0}=\tp_0^t, \qquad \tB_{t1}=1-\tp_0^t \\
\label{eqn:c12}
&&\tO_{t0}=0, \qquad\; \tO_{t1}=1. 
\end{eqnarray}
The evolution of the money-holdings distribution $\tp^t$ is specified
in the next section.  In the next two sections, we are going to show
that $\langle \ts,p^0,\{\tB_t\}_{t=0}^\infty, \{\tO_t\}_{t=0}^\infty
\rangle$ is an equilibrium.

\Section d{The Convergence of Money-Holdings Distribution at Price-1
Equilibrium} 

In this section, we assume that $\langle \ts, p^0,
\{\tB_t\}_{t=0}^\infty, \{\tO_t\}_{t=0}^\infty \rangle$ defined above
is an equilibrium. We show that if all agents adopt the stationary
strategy $\ts$, and if the initial money-holdings distribution $p^0$
satisfies certain condition, then the economy asymptotically converges
to a unique stationary equilibrium at which the money-holdings
distribution is geometric. This is an important fact that will be used
to show that the conjectured equilibrium is indeed an equilibrium.

From the specification of the conjectured price-1 equilibrium, for each
agent, there are two decision-relevant objects at any date: the
agent's own money holding which determines his feasible bid price,
and the economywide money-holdings distribution which determines the
bid distribution. Hence, the decision-relevant state at date $t$ can
be represented by his money holding $\h_t$ and the money holdings
distribution $p^t$ instead of initial money holding $\h_0$, encounter
history $\o$, bid distribution $B_t$ and offer distribution
$O_t$. Given that all agents adopt strategy $\ts$, given the initial
distribution $p^0$, the money-holdings distribution evolves as
follows: for any $t\ge 0$,
\begin{eqnarray}
\label{eqn:b1}
&&p_0^{t+1}=(1-{m(p^t)\over 2}) p_0^t+{1\over 2} p_1^t \\
\label{eqn:b2}
\forall k\ge 1 &&p_k^{t+1}=(1-{m(p^t)\over 2}-{1\over 2}) p_k^t +
{1\over 2} p_{k+1}^t + {m(p^t)\over 2} p_{k-1}^t
\end{eqnarray}
where $m(p^t)$ is the measure of agents who have money, $m(p^t)=
\sum_{k=1}^\infty p_k^t$. The sequence $\{p^t\}_{t=0}^\infty$ of
money-holdings distributions can be obtained by applying \eqn {b1} and
\eqn {b2} recursively. It is easy to check that at any point of time
$t\ge 0$, the distribution satisfies the aggregation condition: the
nominal money stock remains at $M$, that is, $\sum_{k=1}^\infty k p_k^t=M$.

For technical convenience, we work with a transformation of the
probability measure $p$ instead of $p$ itself. 
Define a mapping $L\colon \Delta \to [0,1]^\infty$ as follows, 
\begin{equation}
\label{eqn:b3}
\forall p\in\Delta \quad \forall k\in\Nat \qquad L_k(p)=\sum_{j\ge
k} p_j.
\end{equation}
Obviously, $L_0(p)=1$, $L_k(p)\in [0,1]$ and $L_k(p)\ge L_{k+1}(p)$
for all $k\in \Nat$. Let $\Gamma=L(\Delta)$. Then, for any
$x\in\Gamma$, $x$ satisfies that $x_0=1$, $x_k\in [0,1]$ and $x_k\ge
x_{k+1}$ for all $k\in \Nat$. By definition, $L$ is a one-to-one
linear mapping from $\Delta$ to $\Gamma$. Therefore, to prove that the
sequence of probability measure $\{p^t\}_{t=0}^\infty$ converges in distribution, we
need only to show that the corresponding sequence $\{L(p^t)
\}_{t=0}^\infty$ converges in the $\ell^1$ metric.  The aggregation condition for the
distribution can be written as $\sum_{k=1}^\infty L_k(p^t)=M$, for any
$t\ge 0$. Define $S$ to be the space of all transformations of
probability measure defined by \eqn {b3} that satisfies the
aggregation condition,
\begin{equation}
\label{eqn:b4}
S=\Bigl\{x\,|\, x\in \Gamma,\; \sum_{k=1}^\infty x_k=M\Bigr\}. 
\end{equation}
The set $S$ is the space we are going to work with primarily in this
section.  It is easy to show the following.

\medskip
\lemma a{Both $S$ and $\Gamma$ are convex. That is, for $X = S$ or $X
= \Gamma$, $\forall x,\,y\,\in X$ and $\forall \a\,\in [0,1]$, $\,\a x
+ (1-\a)y\,\in X$.}
\medskip

By equations \eqn {b1} and \eqn {b2}, the law of motion of the
transformation of money-holdings distribution $L(p)$ is a mapping
$T\colon S\to S$ such that for all $x\in S$, for all $k\ge 1$,
\begin{equation}
\label{eqn:d2}
T_k(x)={1-x_1\over 2} x_k + {1\over 2} x_{k+1} + {x_1\over 2} x_{k-1}
\end{equation}
It is easy to check that $T_0(x)=1$ and $T(x)\in S$. Given that
$x^0=L(p^0)$, $x^t=T(x^{t-1})=L(p^t)$ for all $t\ge 1$. The following
lemma states that the mapping $T$ has a unique fixed point.

\medskip
\lemma b{The mapping $T$ has a unique fixed point $\bx\in S$ such that
$\bx=T(\bx)$: 
\begin{equation}
\label{eqn:d3}
\forall k\in \Nat\quad \bx_k=\bm^k, \qquad where \quad 
\bm={M\over M+1}.
\end{equation}} 
\medskip

\proof For all $x\in S$, by equation \eqn {d2}, $T(x)=x$ requires
that for all $k\ge 1$,
\begin{equation}
x_k=({1-x_1\over 2}) x_k + {1\over 2} x_{k+1} + {x_1\over 2} x_{k-1}
\nonumber 
\end{equation}
and $x_0=1$. This system of equations has a unique solution $\bx$,
$\forall k \in \Nat\; \bx_k=(\bx_1)^k$. Since $\bx\in S$,
$M=\sum_{k=1}^\infty \bx_k= \sum_{k=1}^\infty (\bx_1)^k$, which implies that
$\bx_1={M\over M+1}=\bm$.  \endproof

The unique fixed point $\bx$ of $T$ given in \lem b corresponds to the
geometric money-holdings distribution with parameter $\bm$:
$\bp=L^{-1}(\bx)$, $\bp_k=(1-\bm)\bm^k$ for all $k\in \Nat$.  We want
to show that starting from a given initial state $x^0$, the economy as
a dynamic system evolving according to mapping $T$, converge
asymptotically to the steady state characterized by $\bx$. Toward this
objective, we construct a Liapunov function which is a function
of the state of the dynamic system. We show that the Liapunov function
decreases over time and approaches to its minimum
asymptotically. Therefore, by a standard argument of dynamical-systems
theory, the economy asymptotically approaches a steady state, which is
represented by the unique fixed point of $T$, $\bx$, and does not
depend on the initial state.

The Liapunov function we choose to use can be interpreted as the
expected hazard rate for the corresponding distribution. Define
$Z\colon \Gamma\to \Re_+$, for all $x\in \Gamma$, 
\begin{equation}
\label{eqn:d4}
Z(x)=\sum_{k=0}^\infty {(x_k-x_{k+1})^2 \over x_k}.
\end{equation}
For technical reasons, we define the function $Z$ on the larger space
$\Gamma$ instead of on $S$.  For $Z$ to be a Liapunov function, it
should be continuous in some metric, it should be decreasing along
the trajectory of the system defined by $T$, and it should have a
unique minimum on $S$ where it is applied. We will show that $Z$ has
these properties one by one.

\medskip
\lemma c{The function $Z$ is strictly convex on $\Gamma$.}
\medskip

\proof Take arbitrary $x,\,y\,\in \Gamma$, $x\not=y$, and $\a\in
(0,1)$. Let $w(\a)=(1-\a)x + \a y=x + \a (y-x)$.
The set $\Gamma$ is convex, hence $w(\a)\in \Gamma$. For all
$k\in \Nat$, define $\d_k\equiv y_k-x_k$ and
$$z_k(w(\a))\equiv {(w_k(\a)-w_{k+1}(\a))^2\over w_k(\a)}= {(x_k -x_{k-1} + \a
(\d_k-\d_{k+1}))^2 
\over x_k + \a \d_k}.$$
Direct computation reveals that
$${d^2Z(w(\a))\over d\a^2}=\sum_{k=0}^\infty {d^2 z_k(w(\a)) \over
d\a^2} = \sum_{k=0}^\infty {2\over x_k+\a \d_k} \Bigl(\d_k-\d_{k+1} -
{\d_k(x_k-x_{k+1} +\a (\d_k-\d_{k+1}) \over x_k+\a \d_k}\Bigr)^2 \ge 0.$$
Moreover ${d^2Z(w(\a))\over d\a^2}=0$ if an only if $\forall k\in
\Nat\; {d^2 z_k(w(\a)) \over d\a^2}=0$, which is equivalent to
$\forall k\in \Nat\; y_{k+1} x_k=y_k x_{k+1}$, that is (since $x_0 =
y_0 = 1$), $\forall k\in \Nat\; x_k=y_k$. Given that $x\not=y$,
${d^2Z(w(\a))\over d\a^2}> 0$. Hence, $Z$ is strictly convex on
$\Gamma$.  \endproof

Using \lem c, the following proposition show that the function $Z$
satisfies a crucial criterion of a Liapunov function: it is strictly
decreasing along the trajectory of the system defined by $T$.

\medskip
\proposition d{For all $x\in S$, $Z(T(x))< Z(x)$, unless $x=T(x)$.}
\medskip

\proof 
Define mappings $\l\colon S\to \Gamma$ and $\r\colon S\to \Gamma$ as
follows: for all $x\in S$, $k\in \Nat$,
$$\l_k(x)={x_{k+1} \over x_1}, \qquad\qquad 
\r_0(x)=1, \quad \r_{k+1}(x)=x_1 x_k.$$ 
It is easy to check that $\l(x)\in \Gamma$ and $\r(x)\in
\Gamma$.\footnote{In general, for any $x\in S$, $\l(x)\not\in S$ and
$\r(x)\not\in S$. This is the reason we define $Z$ on $\Gamma$
instead of on $S$.} The measure $\l$ is a normalized left-shift of $x$,
and $\r$ is a normalized right-shift of $x$. Then by \eqn {d2}, we
can rewrite $T(x)$ as convex combinations of $x$, $\l(x)$ and
$\r(x)$. That is,  
$$T(x)={x_1\over 2} \l(x) + {1\over 2} \r(x) + {1-x_1\over 2} x.$$
Since $Z$ is strictly convex on $\Gamma$ by \lem c, unless
$\l(x)=\r(x)=x$, 
\begin{eqnarray}
Z(T(x)) &<& {x_1\over 2} Z(\l(x)) + {1\over 2} Z(\r(x)) +
{1-x_1\over 2} Z(x) \nonumber \\ 
&=& {x_1\over 2}{1\over x_1}\Bigl(Z(x)-(1-x_1)^2\Bigr) + {1\over 2}
\Bigl( (1-x_1)^2+ x_1Z(x)\Bigr) + {1-x_1\over 2} Z(x) \nonumber \\
&=& Z(x). \nonumber 
\end{eqnarray}
It is easy to verify that $\l(x)=x$ if and only if $x=T(x)$. Therefore,
unless $x=T(x)$, we have $Z(T(x))<Z(x)$.  
\endproof

Because of the aggregation condition ($\sum_{k=1}^\infty x_k=M$), $S$ is a
subset of the complete metric space $(\X, d)$, where
$\X= \{x\in [0,1]^\infty \,|\,\sum_{k=0}^\infty |x_k|<\infty\}$ and
$d$ is the usual metric associated with $\X$, for any $x,y\in \X$,
\begin{equation}
d(x,y)=\sum_{k=0}^\infty |x_k-y_k|.
\end{equation} 
By standard argument, $(S,d)$ is a complete metric subspace of
$(\X, d)$. Next, we show that both $Z$ and $T$ are continuous
mappings in metric $d$.
 
\medskip
\proposition i{The function $Z$ is continuous on $S$.}
\medskip

\proof We need to show that for any given $\e>0$, for any $x\in S$,
there exists a $\d$-neighborhood of $x$ such that for all $y$ satisfying
$d(x,y)<\d$, $|Z(x)-Z(y)| <\e$.

Fix an arbitrary $\e>0$, and an arbitrary $x\in S$. Since $x_k$ is
decreasing in $k$ and $\sum_{k=1}^\infty x_k= M$, there exists an
$I\ge 1$ such that 
\begin{equation}
\label{eqn:i1}
x_I < \e/8. 
\end{equation}
Let $J=\max\,\{j\,|\,j\le I,\, x_j>0\}$. So $x_J>0$. Without loss of
generality, assume $J\ge I-1$. Let $\d=(\e/8) x_J>0$. (Otherwise
$x_{J+1} = 0$, so $J+1$ satisfies $x_{J+1} < \e/8$.) Then for any $y$
such that $d(x,y)<\d$,
\begin{equation}
\label{eqn:i3}
y_I \le |y_I-x_I|+x_I\le d(x,y) +x_I < (\e/8) x_J+\e/8 \le \e/4.
\end{equation}
For all $k\in \Nat$, define $\xi_k(x)\equiv (x_k-x_{k+1})/ x_k\le 1$,
and $\xi_k(y)\equiv (y_k-y_{k+1})/y_k \le 1$. Then, for $k\le I-1$,
$x_k\ge x_J$, 
\begin{equation}
\label{eqn:i4}
|\xi_k(x)-\xi_k(y)|\le {1\over x_k}\Bigl(|x_{k+1}-y_{k+1}| + |x_k-y_k|
{y_{k+1}\over y_k}\Bigr) < {2\over x_J} {\e\over 8} x_J  ={\e \over 4}. 
\end{equation}
Now, applying \eqn {i1}--\eqn {i4}, we have
\begin{eqnarray}
&&|Z(x)-Z(y)|\nonumber \\ 
&=& \sum_{k=0}^{I-1} \Bigl|(x_k-x_{k+1}) \xi_k(x)-
(y_k-y_{k+1}) \xi_k(y)\Bigr| + \sum_{k\ge I} (x_k-x_{k+1}) \xi_k(x)+
\sum_{k\ge I} (y_k-y_{k+1}) \xi_k(y) \nonumber \\
&\le & \sum_{k=0}^{I-1} \Bigl((x_k-x_{k+1}) |\xi_k(x)- \xi_k(y)|
+|x_k- y_k| \xi_k(y) +|x_{k+1} -y_{k+1}| \xi_k(y)\Bigr) + x_I + y_I
\nonumber \\  
&<& \sum_{k=0}^{I-1} (x_k-x_{k+1}){\e \over 4} + \sum_{k=0}^{I-1}
|x_k- y_k| +  \sum_{k=0}^{I-1} |x_{k+1} -y_{k+1}| +
\e/8 +\e/4  \nonumber \\ 
&<& \e/4 + \e/8 +\e/8 + \e/8 +\e/4 <\e. \nonumber
\end{eqnarray}
We have shown that for any given $\e>0$, for any $x\in S$, there is
$\d>0$ such that for all $y$ satisfying $d(x,y)<\d$, $|Z(y)-Z(x)|<
\e$. Hence, $Z$ is continuous on $S$. \endproof

\medskip
\proposition h{The mapping $T$ is continuous on $S$.}
\medskip

\proof We need to show that for any given $\e>0$ and $x\in S$, there
is a $\d>0$ such that for all $y$ satisfying $d(x,y)<\d$,
$d(T(x),T(y)) <\e$.

Fix an arbitrary $\e>0$ and an arbitrary $x\in S$. By \eqn {d2}, for
all $y\in S$, for all $k\ge 1$, 
$$T_k(y)={1-y_1\over 2} y_k + {1\over 2} y_{k+1} + {y_1\over 2} y_{k-1}.$$ 
Take $\d=\e/3>0$, and let $y$ be such that $d(y,x)<\d$. Then,
\begin{eqnarray}
\label{eqn:h1}
&&d(T(x),T(y)) \;=\;\sum_{k=1}^\infty |T_k(y)-T_k(x)| \nonumber \\ 
&=&\sum_{k=1}^\infty {1\over 2}\Bigl(|x_k-y_k| + |x_{k+1}-y_{k+1}| +
x_1 |x_k-y_k| + x_1 |x_{k+1}-y_{k+1}| + (y_k+y_{k+1})
|x_1-y_1|\Bigr) \nonumber \\ 
&<&{1\over 2}\Bigl(\e/3+\e/3 +\e/3 +\e/3 +\e/3 +\e/3\Bigr)=\e \nonumber
\end{eqnarray}
Therefore, $T$ is continuous on $S$.  \endproof

The set $S$ we have been working with is unfortunately not compact. To
insure the convergence of the system from some initial state, we
introduce a subset of $S$ that is compact. Define an ordering
relation between two vectors $x$ and $y$: $y$ dominates $x$, denote 
$x\preceq y$, if and only if for all $k\in\Nat\,$ $x_k\le y_k$. 
For a given strictly positive vector $\pi\in \X$, let $S_{\pi}$ be
the set of vectors in $S$ that are dominated by $\pi$,
\begin{equation}
\label{eqn:h2}
S_\pi=\{x\in S\,|\,x\preceq \pi\}.
\end{equation}

\medskip
\proposition f{For a given strictly positive vector $\pi\in\X$, the
set $S_\pi$ is compact.}
\medskip

\proof To prove $S_\pi$ is compact, we need to show that $S_\pi$
is complete and totally bounded subset of $\X$. The completeness of
$S_\pi$ is trivial given that $S$ is complete, and the proof is omitted
here. To show that $S_\pi$ is totally bounded, we need to show that
there exist a finite $\e$-net for $S_\pi$ in $\X$ for any $\e>0$.

Fix an arbitrary $\e>0$. Since $\pi$ is strictly positive and $\pi\in
\X$, $\sum_{k=0}^\infty \pi_k <\infty$. Hence, there exists an $I>0$
such that $\sum_{k>I} \pi_k <\e/2$. For any $x\in S_\pi$, let $\hx$ be
the vector of $x$ truncated at $I$, $\hx=(x_0, x_1, \ldots, x_I, 0,
0,\ldots)$. Then $d(x,\hx)=\sum_{k>I} x_k\le \sum_{k>I} \pi_k
<\e/2$. Let $\hat S_\pi$ be the set of $\hx$ associated with $x\in
S_\pi$. The set $\hat S_\pi$ is a totally bounded in $I$-dimensional
Euclidean space (with the usual metric). Let $A$ be a finite
$\e/2$-net for $\hat S_\pi$. Then $A$ is a finite $\e$-net for
$S_\pi$.  \endproof

The vector $\pi$ can be any strictly positive element of $\X$. In
particular, let $\pi^\te$ denote the geometric vector defined by some
$\te\in (0,1)$: for all $k\in \Nat$,
\begin{equation}
\pi_k^\te=\te^k.
\end{equation}
The vector $\pi^\te$ as defined above is an element of $\X$ as well as
$\Gamma$. Also, it is a fixed point of $T$.  The following lemma
states that for $\pi^\te$, $S_{\pi^\te}$ is closed under $T$.

\medskip
\lemma g{For any $x\in S$ and any $\te\in(0,1)$, if 
$x\preceq \pi^\te$, then $T(x)\preceq \pi^\te$.}
\medskip

\proof Suppose that there exists a $\te\in(0,1)$ such that
$x\preceq \pi^\te$. By definition, $x\preceq \pi^\te$ implies that
$x_k\le \te^k$ for all $k\in \Nat$. By equation \eqn {d2}, for all
$k\ge 1$, 
$$T_k(x)={1\over 2}\Bigl((1-x_1)x_k + x_{k+1} + x_1 x_{k-1}\Bigr) \le
{1\over 2}\Bigl((1-x_1)\te^k + \te^{k+1} + x_1 \te^{k-1}\Bigr).$$ 
Since the expression in the righthand side of the above inequality is
an increasing function of $x_1$ and by assumption, $x_1\le \te$, 
$$T_k(x) \le {1\over 2}\Bigl((1-\te) \te^k + \te^{k+1} + \te
\te^{k-1}\Bigr)=\te^k.$$ 
By definition, $T_0(x)=1=\pi^\te_0$. Therefore, $T(x)\preceq
\pi^\te$. \endproof 

By \lem g, if a given initial state $x^0$ satisfies the following
condition, 

\medskip
$(\star)$ \enspace {\it there exists a $\te\in (0,1)$
and a $t>0$ such that $T^t(x^0)\preceq \pi^\te$}
\medskip
 
\noindent then all the subsequent states of the dynamic system
$T^n(x^0)$, $n\ge t$, are dominated by $\pi^\te$ as well, hence,
they are elements of $S_{\pi^\te}$.

\medskip
\proposition j{If the initial state $x^0$ satisfies condition
$(\star)$, then the economy as a dynamic system evolving from $x^0$
according to mapping $T$ converges asymptotically to the steady state
characterized by distribution $\bx$ which uniquely satisfies
$T(\bx)=\bx$ and $d(\bx,0)=d(x^0,0)$.}
\medskip

\proof Suppose that condition $(\star)$ holds, that is, there exists
$\te\in (0,1)$ and a $t>0$ such that $T^t(x^0)\preceq \pi^\te$, then
$T^n(x^0)\in S_{\pi^\te}$ for all $n\ge t$. By \prop f,
$S_{\pi^\te}$ is a compact set, and by \prop i, the function $Z$ is
continuous on $S$, hence on $S_{\pi^\te}$, $Z$ achieves its minimum
on $S_{\pi^\te}$. Furthermore, by \lem c, $Z$ is strictly convex on
$S$, hence on $S_{\pi^\te}$, $Z$ has a unique minimum on
$S_{\pi^\te}$. Last, $Z$ is strictly decreasing along 
the trajectory of the system defined by $T$ by \prop d. Therefore,
$Z$ is a Liapunov function. With this Liapunov function, we show next
the convergence of the system from the initial state $x^0$.

From the given $x^0$, construct a sequence $\{x^n\}_{n=1}^\infty$ by
applying $T$ recursively, $x^n=T^n(x^0)$. Consider the sequence
excluding the first $t$ elements, $\{x^n\}_{n=t}^\infty$, which is in
$S_{\pi^\te}$ by assumption. By \prop d, the corresponding
sequence $\{Z(x^n)\}_{n=t}^\infty$ is monotonically decreasing. Since
$S_{\pi^\te}$ is compact, there exists a subsequence $\{x^{n_k}\}$
that converges to some $\hx\in S_{\pi^\te}$. Suppose that $\hx$ is not
a fixed point of $T$. Then by \prop d,
$Z(T(\hx)) < Z(\hx)$. Since $Z$ is continuous and $T$ is
continuous, there exists a $\d>0$ such that for all $y$ satisfying
$d(\hx,y)<\d$, $Z(T(y)) < Z(\hx)$. Since $\{x^{n_k}\}$ converges
to $\hx$, there exists an $K$ such that for all $k\ge K$, $d(\hx,
x^{n_k})<\d$, hence, $Z(T(x^{n_k})) < Z(\hx)$, or,
\begin{equation}
\label{eqn:j1}
Z(x^{n_k+1}) < Z(\hx).
\end{equation} 
But since $\{Z(x^n)\}_{n=t}^\infty$ is monotonically decreasing,
and since $\hx$ is the limit of $x^{n_k}$, regardless of the
arrangement of the subsequence, 
\begin{equation}
\label{eqn:j2}
Z(x^{n_k+1}) \ge Z(\hx)
\end{equation} 
which contradicts \eqn {j1}. Therefore, the limit $\hx$ has to be a
fixed point of $T$. Since $T$ has a unique fixed point
$\bx$ in $S$ by \lem b, $\bx=\hx\in S_{\pi^\te}$. Hence, for the given
initial state $x^0$, $T^n(x^0)\to \bx$ as $n\to \infty$. This
strengthened statement, that the entire sequence (rather than only the
subsequence selected above) converges to $\bx$, follows from a
standard argument involving the Liapunov function $Z$.
\endproof  

The assumption made in \prop j, that the initial state $x^0$, a
linear transformation of the initial distribution $p^0$, satisfies
condition $(\star)$,  is not a strong assumption as it seems. The
following proposition states a class of initial state that
satisfies the condition.

\medskip
\proposition e{If the initial money-holdings distribution $p^0$ has a
thin tail that is dominated by the tail of a geometric distribution,
that is, there is a $J>0$ and an $\a\in (0,1)$ such that
$p_j^0\le(1-\a)\a^j$ for all $j>J$, then $x^0=L(p^0)$ satisfies
condition $(\star)$.}
\medskip

\proof The condition states that there is a $J>0$ and an $\a\in (0,1)$
such that $p_j^0\le (1-\a) \a^j$ for all $i>J$, or equivalently,
$x_j^0\le \a^j$ for all $j>J$. Let $K$ be the minimum of the support
of $p^0$, $K=\min\{j\,|\, p_j^0\not=0\}$.  Then by definition,
$x_j^0=1$ for all $j\le K$.

First, suppose that $x_1^0<1$ ($p_0^0>0$ and $K=0$). Then, there
exists a $\te\in [\a,1)$ such that $x_1^0\le \te^J$. Since $x_j^0$ is
decreasing in $j$, for $j\le J$, $x_j^0 \le x_1^0 \le \te^J\le
\te^j$. For $j>J$, $x_j^0\le \a^j\le \te^j$ since $\a\le
\te$. Therefore, $x^0\preceq \pi^\te$. That is, condition $(\star)$
holds for $x^0$.

Next, consider the case where $x_1^0=1$ ($p_0^0=0$ and $K\ge
1$). Then, $x_j^0=1$ for all $j\le K$ and $x_{K+1}^0<1$. By equation
\eqn {d2}, $T_K^1(x^0)=(x_{K+1}^0+1)/2<1$, and $T_{j+1}^1(x^0) \le
\a^j$ for all $j>J$. Similarly, after $K$ repeated operations of $T$
on $x^0$, we have $T_1^K(x^0)<1$, $\;T_{j+K}^K(x^0)\le \a^j$ for all
$j>J$. Now we can treat $T^K(x^0)$ as the $x^0$ in the case
above. Specifically, there exists a $\te\in [\a^{J\over J+K},1)$ such
that $T_1^K(x^0)<\te^{J+K}$. For $j\le J+K$, $T_j^K(x^0) \le
T_1^K(x^0) \le \te^{J+K}\le \te^j$. For $j>J+K$, $x_j^0\le \a^{j-K}\le
\te^j$ given $\te\ge \a^{J\over J+K}$. Therefore, $T^K(x^0)\preceq
\pi^\te$, which implies that condition $(\star)$ holds.  \endproof

As a practical matter, economists are not likely to find that the
condition in \prop e is a restrictive one. An initial money-holdings
distribution $p^0$ with finite support (that is, there is a $J>0$ such
that $p_j^0=0$ for all $j>J$) satisfies the condition. Distributions
with finite support are dense in the space of probability simplex
$\Delta$. The condition is also satisfied if one is to increase the
nominal money stock in an economy from a steady-state geometric
distribution by distributing a finite amount of money to people whose
money holdings are less than certain finite amount (e.g. ``poor''
people), in other words, if the money injection has finite support. We
can conclude now that if the initial money-holdings distribution
satisfies the condition given in \prop e, and if all agents adopt
strategy $\ts$, then by \prop j, the economy asymptotically converges
to a unique stationary equilibrium at with the money-holdings
distribution is geometric.

\Section e{The Existence of Price-1 Equilibrium}

In this section, we show that from an initial distribution $p^0$ such
that the assumption in \prop e is satisfied, the price-1 equilibrium
defined in Section 3 is a Bayesian Nash equilibrium. In particular, we
show that for an arbitrary agent, given that all other agents in the
economy adopt the strategy $\ts$ defined in \eqn {c10} (hence the bid
and offer distributions are given by $\{\tB_t\}_{t=0}^\infty$ and
$\{\tO_t\}_{t=0}^\infty$ defined in \eqn {c11}--\eqn {c12}), it is
optimal for the agent in question to adopt strategy $\ts$ as well,
that is, no strategy overtakes $\ts$.

Consider an arbitrary agent of type $i$. Suppose that the agent's
initial money holdings is $\h_0$. Since $\h_0$ is fixed and is taken
as given when we compare different strategies, for notational
convenience, we will suppress $\h_0$ as an argument of all functions
such as $\s$ and $\h_t^s$ in the rest of the section, and write them
as functions of $\o$ alone. Also note that given all the other agents
adopt strategy $\ts$ and the agent in question has measure $0$,
although his trading history will be determined by his strategy $\s$,
his encounter history $\o$ is independent of the strategy he adopts.  

Let $\h_t^\s(\o)$ denote the agent's money holdings at the beginning
of date $t$ in encounter history $\o$ if he adopts strategy $\s$,
$\h_0^\s(\o)=\h_0$.  Define the agent's {\it achievement function} at
the beginning of date $t$ if he adopts strategy $\s$, $A_t^\s\colon
\O\to \Re_+$, to be the sum of his total utility up to date $t$ and
the future utility that will be brought by the money accumulated up to
date $t$, $\h_t^\s$, given that the agent buys his future consumption
goods at price 1. That is, for any encounter history $\o\in \O$,
\begin{equation}
\label{eqn:e1}
A_t^\s(\o)=\sum_{\t=0}^{t-1} v_\t^\s (\o)  + \h_t^\s(\o) u
\end{equation}
where $v_\t^\s(\o)$ is defined in \eqn {c6}, and $\h_t^\s(\o)$ is
defined recursively by \eqn {c5}. For notational
convenience, define for all $t\ge 0$,
\begin{equation}
\label{eqn:e2}
\tA_t=A_t^{\ts},\qquad \tv_t=v_t^{\ts}, \qquad \th_t=\h_t^{\ts}.
\end{equation}

Note that by the definition \eqn {c7} of the overtaking criterion, given
that all other agents adopt strategy $\ts$, any strategy that
specifies at any time to offer to sell at price $0$ is obviously
overtaken by some strategy since the seller in transaction gains
nothing but suffers a loss in production cost $c$. In the rest of the
paper when we compare strategies with $\ts$, we exclude those
strategies with $0$ offer price at any time.  The following lemma
shows that strategy $\ts$ is associated with the highest achievement
function of any strategy.

\medskip
\lemma k{If all other agents adopt strategy $\ts$, then for an
arbitrary agent facing any encounter history $\o\in \O$, adopting a
strategy $\s$, for all $t\ge 0$, $A_t^\s(\o) \le \tA_t(\o).$}
\medskip

\proof Consider an agent of type $i$ with a history $\o\in \O$. 
Obviously, $A_0^\s(\o) = \tA_0(\o)=\h_0 u$.  We compare an 
arbitrary strategy $\s$ with $\ts$ at the begining of date $t+1$,
$t\ge 0$.

Case (1). $\o_{t1}=s$ and $\o_{t2}=1$. In this case, regardless the
agent's strategy (including $\ts$), $A_{t+1}^\s(\o) =A_t^\s(\o)$.

Case (2). $\o_{t1}=b$ and $\o_{t2}=0$. This is a case that the buyer
encountered has no money, hence, has bid price $0$. By remark
above, $\s_{t2}(\o)> 0$. So regardless of the strategy (including
$\ts$), no trade can take place, 
$A_{t+1}^\s(\o)=A_t^\s(\o)$.

Case (3). $\o_{t1}=b$ and $\o_{t2}=1$. If $\s_{t2}(\o)=1=\ts(\o)$,  
$A_{t+1}^\s(\o)-A_t^\s(\o)=-c+u=\tA_{t+1}(\o)-\tA_t(\o).$ 
If $\s_{t2}(\o)>1$, the encountered buyer is not able to buy, hence
trade does not take place with $\s$, but it does take place with
$\ts$, 
$A_{t+1}^\s(\o)-A_t^\s(\o)=0<-c+u=\tA_{t+1}(\o)-\tA_t(\o).$

Combine the above three cases, we conclude that for any strategy $\s$,
for all history $\o\in \O$, $A_0^\s(\o) - \tA_0(\o)=0$, and for all
$t\ge 0$,  
$$A_{t+1}^\s(\o)-\tA_{t+1}(\o)\le A_t^\s(\o)-\tA_t(\o).$$
Hence, by induction for all $t\ge 0$, $A_t^\s(\o)\le \tA_t(\o)$. 
\endproof
 
In Section 4, we have shown that if all agents adopt strategy $\ts$,
for a given initial distribution $p^0$, if the assumption in \prop e
is satisfied, the economy conveges to a unique stationary equilibrium
at which the money-holdings distribution is the geometric $\bp$. For
each agent using strategy $\ts$, when all other agents adopt
$\ts$ also, the expected money holdings at this limit is
$M$.\footnote{This is a continuum-law-of-large-number type of
assertion, so \eqn {e3} must be assumed as an axiom.}  In other
words,
\begin{equation}
\label{eqn:e3}
\lim_{t\to\infty} \mean [\th_t(\o)]=M.
\end{equation}
The next lemma is about the expected money holdings if an agent adopt
some other strategy $\s$.

\medskip
\lemma l{Under the assumption in \prop e, given that an arbitrary
agent adopts strategy $\s$ while all other agents adopt strategy
$\ts$, if $\,\mean [A_t^\s(\o) - \tA_t(\o)] \not\to -\infty$ as
$t\to\infty$, then $\,\liminf_{t\to\infty} \mean[\h_t^\s(\o)]\ge M$.}
\medskip

\proof For strategy $\s$, for all $\o\in\O$, define $\d^\s(\o)$ to be
the set of dates at which the agent who adopts strategy $\s$ meets a
buyer, but his offer price is above $1$, $\d^\s(\o) =\{t\,|\,
\o_{t1}=b,\, \s_{t2}(\o)>1\}$. For any set $D$, let $\# D$ denote the
cardinality of $D$.

\medskip 
{\it Claim 1. If $\,\mean [A_t^\s(\o) - \tA_t(\o)] \not\to -\infty$ as
$t\to\infty$,$\;\#\d^\s <\infty$ a.s.}

\noindent To prove this, consider an arbitrary encounter sequence
$\o\in \O$. For $t\in \d^\s(\o)$, given that the agent adopts
strategy $\s$, there is no trade takes place ($\s_{t2}(\o)>1
=\o_{t2}$), hence, $A_{t+1}^\s(\o) - A_t^\s(\o)=0$, while
if the agent adopts strategy $\ts$, then trade takes place at price $1$, 
$\tA_{t+1}(\o)-\tA_t(\o)=u-c>0$. Therefore, 
\begin{equation}
\label{eqn:e4}
\forall t\in \d^\s(\o)\quad A_{t+1}^\s(\o)-\tA_{t+1}(\o)=
A_t^\s(\o)-\tA_t(\o)-(u-c). 
\end{equation}  
For $t\not\in \d^\s(\o)$, it is easy to check, for cases (1)--(3) as in the proof of
\lem k, that $A_{t+1}^\s(\o) - \tA_{t+1}(\o)= A_t^\s(\o) -
\tA_t(\o)$. Hence, by \eqn {e4}, if $\# \d^\s(\o) =\infty$,
$\lim_{t\to\infty} [A_t^\s(\o) -\tA_t(\o)] =
- -\infty$, which implies that if $\P\{\o\,|\, \#\d^\s(\o)= \infty\}>0$,
$\lim_{t\to\infty} \mean [A_t^\s(\o) - \tA_t(\o)]=-\infty$, 
which contradicts to the assumption. That is, the claim holds.

Given claim 1, for any $\e>0$, there exists a $t_\e>0$ such that
$\P\{\o\,|\,\max\, \d^\s(\o)\le t_\e\}>1-\e/2$. Recall that for all
$t\in \d^\s(\o)$, $\ts_{t2}(\o)=1$. Define
$\mu_\e(\o)\equiv\min\, \{t\,|\, t\ge t_\e,\, \th_t(\o)=0\}$.

\medskip 
{\it Claim 2. For all $\o\in \O$ such that $\max\, \d^\s(\o)\le
t_\e$, for all $t\ge \mu_\e(\o)$, $\th_t(\o)\le \h_t^\s(\o)$.}

\noindent This claim can be proved by induction. For $t=\mu_\e(\o)$,
the claim holds automatically since $\th_t(\o)=0\le \h_t^\s(\o)$.
Suppose that it holds for some $t\ge \mu_\e(\o)$, consider
date-$(t+1)$ transaction. 
\begin{eqnarray}
&\hbox{if }\, \o_{t1}=b,\; \o_{t2}=1,\qquad\qquad\qquad &
\quad\hbox{then }\; \th_{t+1}(\o) = \th_t(\o)+1 \le \h_t^\s(\o)+1 =
\h_{t+1}^\s(\o) \nonumber \\
&\hbox{if }\,\o_{t1}=b, \; \o_{t2}=0, \qquad\qquad\qquad &
\quad\hbox{then }\; \th_{t+1}(\o) = \th_t(\o) \le \h_t^\s(\o) =
\h_{t+1}^\s(\o) \nonumber \\
&\hbox{if }\, \o_{t1}=s,\; \th_t(\o)=0, \qquad\qquad\quad &
\quad\hbox{then }\; \th_{t+1}(\o) = 0 \le \h_{t+1}^\s(\o) \nonumber \\
&\hbox{if }\, \o_{t1}=s,\; \th_t(\o)\ge 1,\;\s_{t1}(\o)\ge 1, &
\quad\hbox{then }\; \th_{t+1}(\o) =\th_t(\o)-1 \le \h_t^\s(\o)-1 =
\h_{t+1}^\s(\o)  \nonumber \\
&\hbox{if }\, \o_{t1}=s,\; \th_t(\o)\ge 1,\;\s_{t1}(\o)=0, &
\quad\hbox{then }\; \th_{t+1}(\o) =\th_t(\o)-1 < \h_t^\s(\o) =
\h_{t+1}^\s(\o)  \nonumber  
\end{eqnarray}
That is, $\th_{t+1}(\o)\le \h_{t+1}^\s(\o)$. Hence, the
claim holds for all $t\ge \mu_\e(\o)$.  

By the definition of $\mu_\e$, $\mu_\e<\infty$ a.s. (since $\th_t(\o)$
is a random walk with nonpositive drift, and with reflecting barrier
at $0$, $\th_t(\o)$ hits $0$ in finite time a.s.). Then for the $\e$
chosen above, there exists a $\xi_\e>0$ such that
$\P\{\o\,|\,\mu_\e(\o)\le \xi_\e\} >1-\e/2$. Define 
$$\O_1(\e)=\{\o\,|\,\max\, \d^\s(\o)\le t_\e\;\hbox{ and }\;
\mu_\e(\o)\le \xi_\e\}$$
$$\O_2(\e)=\{\o\,|\,\max\, \d^\s(\o)> t_\e\;\,\hbox{ or }\;\,
\mu_\e(\o)> \xi_\e\}.$$
Then $\O=\O_1(\e) \cup \O_2 (\e)$, $\P(\O_1(\e))>1-\e$ and
$\P(\O_2(\e))\le \e$. Take $\e=1/n^2$. For $\o\in \O_1(1/n^2)$,
$t_{1/n^2}\le \mu_{1/n^2}(\o) \le \xi_{1/n^2}$. For a fixed $n$,
consider the sequence $\{\h_t^\s(\o) - \th_t(\o)\}$ for
$t\ge \xi_{1/n^2}$. Let $\O_{1n}=\O_1(1/n^2)$ and $\O_{2n}=\O_2(1/n^2)$.
\begin{eqnarray}
\label{eqn:e5}
\liminf_{t\to\infty} \mean [\h_t^\s(\o) - \th_t(\o)]&\ge& \liminf_{t
\to\infty} \int_{\O_{1n}} (\h_t^\s(\o) - \th_t(\o)) d\P(\o) \nonumber \\
&& + \liminf_{t\to\infty} \,\Bigl(\int_{\O_{2n}}
\h_t^\s(\o)d\P(\o) -  \int_{\O_{2n}}\th_t(\o) d\P(\o)\Bigr).
\end{eqnarray} 
The first term of the right hand side of \eqn {e5} is nonnegative by
claim 2 since $\o\in \O_{1n}$. Since $\th_t$ is ergodic, $\O_{2n}$
is a fixed set for a given $n$, and the limit of $\mean[\th_t(\o)]$
exists and equals $M$ by \eqn {e3}, the second $\liminf$ on the right
hand side of \eqn {e5} can be broken down to two terms,
\begin{equation}
\label{eqn:e6}
\liminf_{t\to\infty} \,\Bigl(\int_{\O_{2n}} \h_t^\s(\o)d\P(\o) -
\int_{\O_{2n}}\th_t(\o) d\P(\o)\Bigr) = \liminf_{t\to\infty}
\int_{\O_{2n}} \h_t^\s(\o)d\P(\o) - M \P(\O_{2n}). 
\end{equation}
The first term of the right hand side of \eqn {e6} is
nonnegative. Combine \eqn {e5} and \eqn {e6}, we have
\begin{equation}
\label{eqn:e7}
\liminf_{t\to\infty} \mean [\h_t^\s(\o) - \th_t(\o)]\ge - M
P(\O_{2n}). 
\end{equation}
Take limit of $n\to \infty$ for inequality \eqn {e7}, the left hand
side is unrelated to $n$, hence not affected, and the right hand side
goes to $0$ since $\P(\O_{2n})\le 1/n^2 \to 0$. Therefore,
$\liminf_{t\to\infty} \mean [\h_t^\s(\o)] \ge \lim_{t\to\infty} \mean
[\th_t(\o)] = M$. \endproof

Now, we are ready to prove the main proposition of the paper.

\medskip
\proposition m{Under the assumption in \prop e, given that all other
agents adopt strategy $\ts$, it is optimal for an arbitrary agent to
take strategy $\ts$ as well. That is, there is no strategy $\s$ that
overtakes $\ts$.}
\medskip

\proof For an arbitrary strategy $\s$, consider the following two
cases.

\medskip

Case 1. $\,\limsup_{t\to\infty} \mean[\h_t^\s(\o)]\ge M$. \enspace
Then, for any $\e>0$, there exists an inifinite set $G_\e^\s=\{t\, |\,
\mean[\h_t^\s(\o)]\ge M-\e/2\}$. Since $\lim_{t\to\infty} \mean
[\th_t(\o)] = M$ by \eqn {e3}, the set $J_\e^\s=\{t\, |\,
\mean[\th_t (\o)]< M+\e/2\}$ is also infinite. For all $t\in G_\e^\s
\cap J_\e^\s$, by \lem k,
$$0\ge\mean [A_t^\s-\tA_t]= \mean[\sum_{\t=0}^{t-1} v_\t^\s -
\sum_{\t=0}^{t-1} \tv_\t] + \mean[\h_t^\s-\th_t] u \ge
\mean[\sum_{\t=0}^{t-1} v_\t^\s -\sum_{\t=0}^{t-1} \tv_\t] - u\e$$
(the $\o$ is suppressed for convenience). Since $\e$ can be
arbitrarily small, the above inequality implies that 
$$\liminf_{t\to\infty}\, \mean[\sum_{\t=0}^{t-1} v_\t^\s -
\sum_{\t=0}^{t-1} \tv_\t] \le 0.$$
By definition of overtaking criterion \eqn {c7}, strategy $\s$ does
not overtake $\ts$.  

\medskip

Case 2. $\,\limsup_{t\to\infty} \mean[\h_t^\s(\o)]< M$. \enspace By
the proof of \lem k, for all $\o\in \O$, $\{A_t^\s(\o) -
\tA_t(\o)\}_{t=0}^\infty$ is a weakly decreasing sequence.  
If $\mean[A_t^\s(\o) - \tA_t(\o)]\not\to -\infty$ as $t\to\infty$,
by \lem l, $\,\liminf_{t\to\infty} \mean[\h_t^\s(\o)]\ge M$, which
contradicts to the assumption. If $\mean[A_t^\s - \tA_t]\to -\infty$
as $t\to\infty$, and since
$$\mean [A_t^\s-\tA_t]= \mean[\sum_{\t=0}^{t-1} v_\t^\s -
 \sum_{\t=0}^{t-1} \tv_\t] + \mean[\h_t^\s-\th_t] u $$ 
we have 
$$\liminf_{t\to\infty}\, \mean[\sum_{\t=0}^{t-1} v_\t^\s -
\sum_{\t=0}^{t-1} \tv_\t] \le \liminf_{t\to\infty}\, \mean
[A_t^\s-\tA_t] + u M - u \liminf_{t\to\infty}\, \mean[ \h_t^\s] =-\infty.$$
Again by the definition \eqn {c7}, strategy $\s$ does not overtake
$\ts$.  \endproof 

By \prop m, strategy $\ts$ is a Bayesian Nash strategy according to
the overtaking criterion. This proves (ii) of the equilibrium
definition, and (iii) is evidently satisfied.  Hence, the price-1
equilibrium always exists.

\Section f{Conclusion}

We have shown that a price-1 equilibrium from an initial state exists
(under a mild assumption) and is asymptotically stationary. In fact,
there can be other equilibria as well. For example, suppose that the
initial distribution is distributed on the lattice in multiples of
six. Then there would be equilibria with asymptotic distributions on
the the lattices in multiples of one, two, three and six. Because the
asymptotic distributions differ, the equilibria clearly are
distinct. It will be apparent from our earlier paper (Green and Zhou
[2]) that the equilibrium asymptotically distributed on the finest
lattice is the one that achieves the highest level of welfare for the
economy, since it facilitates the greatest amount of trade.

\newpage

\def\Jet{{\it Journal of Economic Theory\/ }}
\def\Ier{{\it International Economic Review\/ }}

\noindent{\bf References}

\bigskip

\newcounter{ref}
\begin{list}{\arabic{ref}.}{\usecounter{ref}
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E.~J.~Green (1994). ``Individual Level Randomness in a Nonatomic
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\item 
E.~J.~Green and R.~Zhou (1998). ``A rudimentary random-matching model
with divisible money and prices,'' \Jet {\bf 81}, 252-271.
\item 
E.~J.~Green and R.~Zhou (1998). ``Money and the Law of One Price: the
Case Without Discounting,'' manuscript, University of Pennsylvania.   
\item 
I.~Gilboa and A.~Matsui (1992). ``A model of random matching,'' {\it
Journal of Mathematical Economics} {\bf 21}, 185-97. 
\item
R.~Zhou. ``Individual and Aggregate Real Balances in a Random
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\end{list}

\end{document}
