\documentstyle{article}
\begin{document}
\title{Some properties of $k$-step exclusion processes }
\author{H. Guiol\thanks{Supported by FAPESP grant
number 96/04859-9}\\
Instituto de Matem\'atica e Estat\'\i stica,\\
Universidade de S\~ao Paulo,\\
PB 66281, 05315-970 S\~ao Paulo, SP, Brasil,\\
e-mail: herve@ime.usp.br}
\maketitle
\begin{abstract}
We introduce $k$-step exclusion processes as generalizations of the
simple exclusion process. We state their main equilibrium properties
when the
underlying stochastic matrix corresponds to a random walk or is positive
recurrent and reversible. Finally we prove laws of large numbers for
tagged and second class particles.\\
\\
Keywords: Interacting particle systems; Invariant measures; Tagged
particle;
second class particle
\end{abstract}
\newcommand{\carn}{\hfill\rule{0.25cm}{0.25cm}}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\section{Introduction}
Let ${\bf S}$ be a finite or countable set, and $p(.,.)$ a transition
matrix of a discrete Markov chain on ${\bf S}$.
For $k\in{\bf N}$, a $k$-step exclusion process is a natural
generalization
of simple exclusion process. It is a
continuous time Markov process on state space ${\bf X}=\{0,1\}^{\bf S}$,
the
set of configurations of particles distributed on ${\bf S}$ with at
most one particle per site.
Informally this process can be describe in the following way: on each
site
of ${\bf S}$ we have one clock that rings (independently of others
clocks)
at random exponential times with parameter 1. If a particle is present
at
site $x$ when the clock associated to this site rings,
then the particle occupies the first vacant site encountered in the
sequence $(X_n)_{1\leq n\leq k}$, where $\{X_n\}_n$ is the Markov chain
with
probability transition $p(.,.)$ starting at $x$ (site $x$ itself is
consider
vacant for the attempt, {\it i.e.}, if the chain returns to $x$ before
encountering an empty site then the particle stays at $x$). If no empty
site
is encountered during these $k$
attempts then the movement is cancelled and the particle stays at $x$
(waiting
for the next of its clock ring).
When $k=1$ the process is exactly the simple exclusion process
introduced by
Spitzer
(1970) \cite{Spitzer}. This process has been intensively studied since
then;
we refer the reader to Liggett's book (1985) \cite{LiggettBook} Chapter
VIII
to an overview of its principal properties.
When $k$ goes to infinity it is plausible that these processes will
approach the Long Range Exclusion Process (see Guiol (1995)\cite{Guiol0}
or
Andjel \& al (1998)\cite{anfegu}). In the Long Range Exclusion Process a
particle that moves, follows the chain $X_n$ until it finds an empty
site,
such that particles can travel long distance in short time. This process
was also introduced by Spitzer (1970) in \cite{Spitzer} and studied
systematically by Liggett (1980) in \cite{liggett1} and in a lesser
proportion by Zheng (1988)\cite{Xiaogu} and Guiol (1997)\cite{Guiol}. A
version
of the $k$-step exclusion processes was used by Liggett in
\cite{liggett1} to
approximate the long range exclusion process, but in that case particles
disappear if they do not encounter a vacant site in their attempt to
move.
Recent developments in the studies of long range processes such as the
Long
Range
Exclusion Process \cite{Guiol,anfegu}, the Hammersley Process (see
Aldous \&
Diaconis (1995)\cite{alddia}), Self-organizing Particle Systems (see
Carlson
\& al (1993)\cite{gaswto}), the Toom model (see Lebowitz \& al
(1996)\cite{lenera})
shows the existence of a real interest to have (Feller) approximations
for
these processes.
The aim of this paper is to give an overview of the principal properties
of
$k$-step exclusion processes.
After a short review in section 2 of definitions and properties, we
describe in
section 3 some invariant measures for the $k$-step exclusion process.
We denote by ${\cal I}_k$ the set of invariant measure of the $k$-step
Exclusion Process.
Let $\rho:{\bf S}\rightarrow [0,1]$, we denote by $\nu_{\rho}$ the
product
measure such that
\[
\nu_{\rho}\{\eta(x_i)=1, 1\leq i\leq n\}=\prod_{i=1}^n\rho(x_i)
\]
for all $n\in{\bf N}$, $x_i\in{\bf S}$, $1\leq i\leq n$.
Suppose that $p(.,.)$ is reversible. Then we have
\begin{theorem} \label{produto}
Let $p(.,.)$ be reversible with $\pi$ i.e.
\[
\pi(x)p(x,y)=\pi(y)p(y,x)\mbox{ for all }x,y\in{\bf S},
\]
and
\begin{equation}\label{defrho}
\rho(x)=\frac{\pi(x)}{ 1+\pi (x)}.
\end{equation}
Then for all $k\geq 1$
\[
\nu _{\rho} \in {\cal I}_k.
\]
\end{theorem}
This result is absolutely similar to the one in the simple exclusion
case (see
Liggett
(1985) Chapter VIII, p. 380). We recall that for this last process the
double
stochasticity of the transition matrix $p(.,.)$ was also a condition to
show
that Bernoulli product measures with
constant densities were invariant measures too. At the beginning of
section 3 we
give a counter example for which this result is no more true for
$2$-step
exclusion process.
Nevertheless, for a random walk on ${\bf Z}^d$ we have the (familiar)
results.
\begin{theorem} \label{produto2}
Suppose $p(x,y)=p(0,y-x)$ for all $x,y\in{\bf Z}^d$ then
\[
\nu_{\alpha}\in{\cal I}_k,
\]
for every constant $\alpha\in[0,1]$.
\end{theorem}
Furthermore in this case Bernoulli measures with constant densities are
the only (ergodic) invariant measures that are translation invariant. We
denote by ${\cal S}$ the set of translation invariant measure, $({\cal
S})_e$ the set of ergodic elements of ${\cal S}$ and $({\cal
I}_k)_e$ the set of extremal elements of ${\cal I}_k$.
\begin{theorem}\label{ergodic}
Suppose $p(x,y)=p(0,y-x)$ for all $x,y\in{\bf Z}^d$ and $p(.,.)$
irreducible
then
\[
({\cal I}_k\cap {\cal S})_e=\left\{ \nu _\alpha :\alpha \in \left[
0,1\right] \right\}.
\]
\end{theorem}
When the transition matrix $p(.,.)$ is reversible and positive recurrent
(ergodic), as for the simple exclusion process, we are able to
characterize
all the invariant measures. In this case there is a unique reversible
probability stationary measure (for the chain $p(.,.)$), $\pi$, so that,
according to theorem \ref{produto}, $\nu _{\rho} \in {\cal I}_k$, with
$\rho$
defined as in (\ref{defrho}). Denote by $\nu^{(n)}$ the measure
\[
\nu^{(n)}(.)=\nu_{\rho}\left(.\left\vert\sum_x\eta(x)=n\right.\right).
\]
It is not difficult to prove that if $p(.,.)$ is irreducible, then these
measures are invariants and extremal.
\begin{theorem}\label{posrecrev}
Suppose that $p(x,y)$ is a positive recurrent, reversible, and
irreducible matrix on ${\bf S}$. Then
(a) $({\cal I}_k)_e=\{\nu_n,0\leq n\leq \infty \}$;
(b) If $\mu \{\eta :\sum_x\eta (x)=n\}=1,$ then $\lim
_{t\rightarrow \infty }\mu S_k(t)=\nu_n$ for $0\leq n\leq \infty$,
\noindent
where $\nu_{\infty}$ is the measure concentrated on $\eta\equiv 1$.
\end{theorem}
This result is also a generalization to the $k$-step exclusion of a
simple
exclusion result (see Liggett (1985) \cite{LiggettBook}, p. 384).
In section 4 we state some properties for a tagged particle a the second
class particle involve in a $k$-step exclusion processes. We
suppose ${\bf S}={\bf Z}^d$ and that $p(x,y)=p(0,y-x)$ ({\it i.e.}
corresponds
to a random walk on ${\bf Z}^d$).
Informally a tagged particle behaves exactly like a regular particle of
the
system. A second class particle moves like other particles (with the
exclusion
rule) except that it has to exchange position when a regular (first
class)
particle tries to jump over it.
We first concentrate on the $k$-step exclusion process as seen from a
tagged
particle. Tagged particle for
the simple exclusion process was introduced by Spitzer
(1970)\cite{Spitzer},
and was studied by Arratia (1983) \cite{arratia},
Ferrari (1986)\cite{Ferrari} and Saada (1987)\cite{Saada}.
We generalize to $k$-step exclusion some results of invariance and
ergodicity
of product Bernoulli measure as seen from a tagged particle obtain from
\cite{Ferrari}. For the notation of the following results we refer the
reader
to section 4.1.
\begin{theorem}\label{marque}
(a) The Palm measure $\widehat{\nu_{\rho}}$ of $\nu_{\rho}$ is invariant
for
the tagged particle process,
\[
\mbox{\it i.e.
}\widehat{\nu_{\rho}}\widehat{S}_k(t)=\widehat{\nu_{\rho}}.
\]
(b) Furthermore if $p(.,.)$ is irreducible
\[
(\widehat{\cal I}_k \cap\widehat{\cal S})_e=\{\widehat{\nu}_\rho : \
0\leq \rho \leq 1\}.
\]
\end{theorem}
The second class particle was introduced to analyze the shock structure
and its
fluctuation for
the simple exclusion process (see Ferrari \& Fontes (1993)\cite{ferfon}
for a
review on the topic). We show that all product Bernoulli measures with
constant densities $\rho\in]0,1]$ as seen from a second class particle
are
invariant if and only if $p(.,.)$ is symmetric.
\begin{theorem}\label{SecondClass}
For all $\rho\in]0,1]$, $\widehat{\nu}_\rho\in\overline{\cal I}_k$ if
and only
if $p(.,.)$ is symmetric.
\end{theorem}
In section 5 we prove laws of large numbers for the tagged particle and
for the second class particle showing that $X_t/t$ (where $X_t$ is the
position of the Tagged particle at time $t$) and $Y_t/t$ (where $Y_t$ is
the
position of the second class particle at time $t$) converge {\it a.s.}
and in
$L_{1}$ (theorems \ref{TagAsymp} and \ref{wlsc}).
Although there are strong links and similarities between simple and
$k$-step exclusion they do not always behave in the same way. For the
symmetric case duality relations fail for the $k$-step exclusion
processes.
This technique was a key ingredient for the study of the symmetric
simple exclusion process (see \cite{LiggettBook} Chapter VIII
Section 1). Furthermore, results obtained in the symmetric case have
also some
relevance for asymmetric case. For instance the method to prove that
Bernoulli product measures with constant densities are extremal for the
asymmetric simple
exclusion process (see the end of the proof of Proposition 2 p. 377 in
Saada (1987)\cite{Saada}) relies
highly on the result for the symmetric
case. In particular this last fact was one ingredient to prove the
strong laws
of large numbers for the tagged particle for the simple exclusion case
in
Saada(1987) \cite{Saada}.
\section{Definitions and Remarks}
Let $k\in{\bf N}^*$,
under the mild hypothesis :
\[
\sup _{y\in {\bf S}}\sum_{x\in {\bf S}}p(x,y)<+\infty
\]
one can defined a continuous semi-group $S_k(t)$ on $C({\bf X})$
(see \cite{LiggettBook} page 30) with infinitesimal generator $\Omega_k$
given by :
for all cylindric function $f$,
\begin{equation} \label{generateur}
\Omega _kf(\eta )=\sum_{\eta (x)=1,\eta (y)=0}q_k(x,y,\eta )\left[
f(\eta
_{xy})-f(\eta )\right],
\end{equation}
where $q_k(x,y,\eta )=
{\bf E}^x\left[ \prod_{i=1}^{\sigma _y-1}\eta (X_i),\sigma _y\leq
\sigma
_x,\sigma _y\leq k\right] $ is the intensity for moving from $x$ to
$y$ on configuration $\eta $ and $\sigma _y=\min \left\{ n\geq
1:X_n=y\right\} $ is the first (non zero) arrival time at site $y$ of
the
chain starting at site $x$.
By Hille-Yosida's theorem the closure of $\Omega_{k}$ corresponds to
a continuous Markov semi-group, which corresponds to the $k$-step
exclusion process.
\begin{remark}
It is also possible to construct this process via the graphical
construction
due to Harris (1978)\cite{harris}.
\end{remark}
Let ${\cal P}$ be the set of probability measures on ${\bf X}$. When
${\bf
S}={\bf Z}^d$ we will denote by
${\cal S}$ the set of elements of ${\cal P}$ which are translation
invariants.
For every $\mu $ $\in {\cal P}$ we denote $\mu S_k(t)$ the
probability measure given by
\[
\int f\ d\mu S_k(t)=\int S_k(t)f\ d\mu \mbox{,}
\]
for all $f\in C({\bf X})$.
Let ${\cal D}$ be the set of cylindric functions on ${\bf X}$, and let
${\cal I}_{k}$ denotes the set of elements of ${\cal
P}$ which are invariant by the $k$-step exclusion process, {\it i.e.}
\begin{equation}\label{invafeller}
\begin{array}{ll}
{\cal I}_{k}&=\{\mu\in{\cal P}:\mu S_{k}(t)=\mu\mbox{ for all }t\}\\
&=\{\mu\in{\cal P}:\int \Omega_{k}f\ d\mu=0\mbox{ for all
}f\in{\cal D}\}.\\
\end{array}
\end{equation}
\begin{remark}
For all $k\in{\bf N}^*$, $k$-step exclusion process has the Feller
property:
\[
\mbox{for all }f\in C({\bf X})\mbox{ and all }t>0,\mbox{ we have
}S_k(t)f\in
C({\bf X}).
\]
Second equality in (\ref{invafeller}) is due to that fact.
\end{remark}
\begin{remark}\label{intensite}$\quad$
One can write the rates in the following way.
For all $k\in{\bf N}$, $x,y\in{\bf S}$ and $\eta\in{\bf X}$ we
have
\[
q_k(x,y,\eta)
=p(x,y)+\sum_{z\neq x,y} p(x,z)p(z,y)\eta(z)+...
\]
\[
...+\sum_{z_1\neq x,y,...,z_{k-1}\neq x,y}
p(x,z_1)...p(z_{k-1},y)\eta(z_1)...\eta(z_{k-1}).
\]
In particular we have for any $k,x,y,\eta$
\[
q_k(x,y,\eta)\geq p(x,y).
\]
\end{remark}
\begin{remark}
Last remark is the lack of monotonicity in $k$ of these processes.
{\it i.e} $f\in C({\bf X})$ monotone increasing function does not
imply $S_{k+1}(t)f\geq S_{k}(t)f$.
\end{remark}
This can be seen in the following simple example.
Let ${\bf S}=\{0,1,2\}$, $p(0,1)=p(1,2)=p(2,0)=1$.
Denote by $\eta$ the initial configuration such that
$\eta(0)=\eta(1)=1$, $\eta(2)=0$. Then construct a coupling
$(\xi^{\eta}_t,\zeta^{\eta}_t)$ with $\xi^{\eta}_t$ and $\zeta^{\eta}_t$
be respectively the $2$-step exclusion process and the simple exclusion
process starting from configuration $\eta$. Then
$\xi^{\eta}_0=\zeta^{\eta}_0=\eta$, but with a positive probability
particles
on site $0$ will try to jump (before particle at $1$) saying at time
$\delta^-$
then $\xi^{\eta}_{\delta}\not\leq\zeta^{\eta}_{\delta}$ and
$\xi^{\eta}_{\delta}\not\geq\zeta^{\eta}_{\delta}$. \carn
\section{Equilibrium results}
\begin{remark}
The correspondence between simple exclusion process
and $k$-step exclusion processes is not as easy as expected. For
instance the
condition $p(.,.)$ doubly stochastic does not assure anymore the
invariance of
the product measure with constant densities when $k\geq 2$ as one can
see in
the following example.
Let ${\bf S}=\{-1,0,1\}$ suppose that $p(-1,0)=p(-1,1)=p(0,1)=1/2$,
$p(0,-1)=p(1,0)=1/3$, $p(-1,-1)=p(1,1)=0$, $p(0,0)=1/6$, $p(1,-1)=2/3$.
Then $p(.,.)$ is doubly stochastic so
that the product measure $\nu_{\alpha}$ is invariant for the simple
exclusion
process for every constant $\alpha\in[0,1]$ (cf. theorem 2.1 (a) Chap.
VIII in
\cite{LiggettBook}). But a simple calculation using the generator of the
$2$-step exclusion process gives
\[
\int\Omega_2 {\bf 1}_\{\eta\in {\bf X}:\eta(-1)=\eta(0)=1\}\
d\nu_{\alpha}=\frac {\alpha^2(1-\alpha)} {36}\not=0
\]
for every constant $\alpha\in ]0,1[$. This means that these measures are
not
invariant for the $2$-step exclusion process for that particular
$p(.,.)$.
\end{remark}
Nevertheless characterize all the invariant and translation invariant
measures
most of the time it is an easy generalization of the simple exclusion
case.
The following properties are some generalization of well known results
for the simple exclusion process.
\subsection{Proofs of Theorem \ref{produto} and \ref{produto2}}
We will prove the following proposition which contains both results.
\begin{proposition} \label{rever} For all $k\geq 1$,
(a) Suppose it exists $\pi (.):{\bf S}\rightarrow {\bf R}^+$ such that
\begin{equation} \label{revers}
\pi (x)p(x,y)=\pi (y)p(y,x)\qquad \mbox{ for all }x,y\in {\bf S}.
\end{equation}
Then,
\[
\nu _\rho \in {\cal I}_k,
\]
where $\rho (x)=\pi(x)/\left[ 1+\pi (x)\right] $.
(b) If $\{p(.,.)\}$ is translation invariant then for all constant
$\alpha
\in\left[ 0,1\right] $, we have
\[
\nu _\alpha \in {\cal I}_k.
\]
\end{proposition}
{\bf Proof :}
For all $A\subset {\bf Z}^d,\left| A\right| =n$ finite and $f_A(\eta
)=\prod_{u\in A}\eta (u)$ we need to show that
\[
\int \Omega _kf_A\ d\nu _\alpha =0.
\]
Applying (\ref{generateur}) to $f_A$ :
\[
\int \Omega _kf_A(\eta )\ d\nu _\alpha =\sum_{y\in A}\sum_{x\in
A^c}\int
q_k(x,y,\eta )\eta (x)\left[ 1-\eta (y)\right] \prod_{u\in A,u\neq
y}\eta
(u)\ d\nu _\alpha
\]
\begin{equation} \label{caracinvar}
-\sum_{y\in A}\sum_{x\in A^c}\int q_k(y,x,\eta )f_A(\eta )\left[
1-\eta
(x)\right] \ d\nu _\alpha \ .
\end{equation}
Note that for all $1\leq l\leq k-1$
\[
\alpha(x)\left[1-\alpha(y)\right]p(x,z_1)p(z_1,z_2)...p(z_l,y)
\]
\[
=\left[1-\alpha(y)\right]\frac{\pi(z_1)}{1+\pi(x)}p(z_1,x)p(z_1,z_2)...p(z_l,y)
\]
\[
...
\]
\[
=\left[1-\alpha(y)\right]\left[1-\alpha(x)\right]p(z_l,z_{l-1})...p(z_1,x)
\pi(z_l)p(z_l,y)
\]
\begin{equation}\label{reversible}
=\alpha(y)\left[1-\alpha(x)\right]p(y,z_l)p(z_l,z_{l-1})...p(z_1,x)
\end{equation}
Therefore using remark \ref{intensite}, integrating in
(\ref{caracinvar}) and summing terms with the same power in $\alpha(u)$
for
$u\neq x,y$ we obtain $2^k-1$ terms of the form :
for any $1\leq l\leq k-1$ and $(i_1,i_2,...,i_l)$ permutation of
$1,...,l$
\[
\begin{array}{c}
\sum_{
\begin{array}{c}
\small{y\in A,z_{i_1},...,z_{i_j}\in A\setminus\{y\}}\\
\small{x\in A^c,z_{i_{j+1}},...,z_{i_l}\in A^c\setminus\{x\}}\\
\end{array}
}
\left( \prod_{y\neq u\in A\cup\{z_{i_{j+1}},...,z_{i_l}\}}\alpha(u)
\right)\times \\
\left\{ \alpha (x)\left( 1-\alpha (y)\right)
p(x,z_1)...p(z_l,y) -\alpha (y)\left[ 1-\alpha (x)\right]
p(y,z_l)...p(z_1,x)\right\}
\end{array}
\]
The term in brackets is zero by (\ref{reversible}) this gives (a).
We prove (b) for $k=2$. The general case consists in gathering terms
that
annihilates each other. As it is quite long and tedious no more detail
will be
given.
Let $\alpha (x)\equiv \alpha $ we the have the sum of three
following terms (multiplied by $\alpha ^n(1-\alpha )$)
\begin{equation} \label{egalun}
\left\{ \sum_{y\in A}\sum_{x\in
A^c}p(x,y)-\sum_{y\in
A^c}\sum_{x\in A}p(x,y)\right\}
\end{equation}
\begin{equation} \label{egaldeux}
\left\{ \sum_{y\in A}\sum_{z\in
A\setminus\{y\}}\sum_{x\in
A^c}p(x,z)p(z,y)-\sum_{x\in A}\sum_{z\in A\setminus\{x\}}\sum_{y\in
A^c}p(x,z)p(z,y)\right\}
\end{equation}
\begin{equation} \label{egaltrois}
\alpha \left\{ \sum_{x\in A^c}\sum_{z\in
A^c\setminus\{x\}}\sum_{y\in
A}p(x,z)p(z,y)-\sum_{y\in A^c}\sum_{z\in A^c\setminus\{y\}}\sum_{x\in
A}p(x,z)p(z,y)\right\} .
\end{equation}
Those three terms are zero :
for (\ref{egalun}) we only have to add and subtract the same
quantity in the brackets
$\sum_{y\in A}\sum_{x\in A}p(x,y)$
\[
\sum_{y\in A}\sum_{x\in {\bf S}}p(x,y)=\left| A\right| =\sum_{y\in
{\bf S}
}\sum_{x\in A}p(x,y).
\]
For (\ref{egaldeux}) we add and subtract $\sum_{y\in
A}\sum_{z\in
A}\sum_{x\in A}p(x,z)p(z,y)$, and since
\[
\begin{array}{c}
\sum_{y\in A}\sum_{z\in A}\sum_{x\in {\bf S}}p(x,z)p(z,y)=\sum_{y\in
A}\sum_{z\in A}p(z,y) \\
=\sum_{z\in A}\sum_{x\in A}p(x,z)=\sum_{y\in {\bf S}}\sum_{z\in
A}\sum_{x\in
A}p(x,z)p(z,y);
\end{array}
\]
denoting $p=p(x,x)$ we have
\[
p\left\{ \sum_{y\in A}\sum_{x\in A^c}p(x,y)-\sum_{y\in
A^c}\sum_{x\in A}p(x,y)\right\}
\]
which is null by the preceding argument.
Finally for (\ref{egaltrois}) we add and subtract
\noindent
$\sum_{y\in
A}\sum_{z\in A^c}\sum_{x\in A}p(x,z)p(z,y)$ and since
\[
\begin{array}{c}
\sum_{y\in A}\sum_{z\in A^c}\sum_{x\in {\bf S}}p(x,z)p(z,y)=\sum_{y\in
A}\sum_{z\in A^c}p(z,y) \\
=\sum_{z\in A^c}\sum_{x\in A}p(x,z)=\sum_{y\in {\bf S}}\sum_{z\in
A^c}\sum_{x\in A}p(x,z)p(z,y).
\end{array}
\]
it remains
\[
p\left\{ \sum_{y\in A^c}\sum_{x\in A}p(x,y)-\sum_{y\in
A}\sum_{x\in A^c}p(x,y)\right\}
\]
which is null as in the first step. This concludes (b).
$\carn$
\subsection{Ergodicity}
We will only sketch the proof of theorem \ref{ergodic} as it is very
standard
(see Liggett's book Chapter VIII, for instance).
As these processes are Feller it is possible to show that if $\mu_1$ and
$\mu_2$ are in $({\cal I}_k\cap{\cal S})_e$ then one can find a coupled
measure $\tilde{\nu}\in(\widetilde{\cal I}_k\cap\widetilde{\cal S})_e$
with
marginals $\mu_1$ and $\mu_2$.
Using basic coupling then one as to show that if
$\tilde{\nu}\in(\widetilde{\cal I}_k\cap\widetilde{\cal S})$ then
$\tilde{\nu}\{\eta\geq\xi\mbox{ or }\eta\leq\xi\}=1$.
Those two points are sufficient to conclude (see Andjel
(1981)\cite{Andjel} for
instance).
(A complete detailed proof can be found in Guiol (1995)\cite{Guiol0}).
\subsection{Reversible cases}
In this paragraph we will suppose that $p(.,.)$ is reversible ({\it
i.e.} there exists a positive $\pi(.)$ on ${\bf S}$ such that
$\pi(x)p(x,y)=\pi(y)p(y,x)$ for all $x,y\in{\bf S}$).
We have shown in Proposition \ref{rever} (a) that (changing notations
a little)
$\nu^{(\alpha)}\in{\cal I}_{k}$ for $0\leq\alpha\leq\infty$ and all
$k\geq
1$, where $\nu^{(\alpha)}$ is the product measure on $U$ with
$\nu^{(\alpha)}\{\eta(x)=1\}=\alpha\pi(x)/(1+\alpha\pi(x))$ for all
$x\in{\bf S}$.
{\bf Proof of Theorem \ref{posrecrev}}
Let
\[
\nu^{(n)}(.)=\nu^{(\alpha)}\left(.\vert\sum_x \eta(x)=n\right).
\]
{\bf Proof :}
Suppose $p(.,.)$ positive recurrent, reversible with stationary measure
$\pi $ :
\[
\pi (x)p(x,y)=\pi (y)p(y,x),\qquad \sum_x\pi (x)=1.
\]
Hence by proposition \ref{rever} (a) we have
$\nu^{(\alpha)} \in {\cal I}_{k\mbox{ }}$ where
$\nu^{(\alpha)}\{\eta(x)=1\}=\alpha\pi
(x)/\left[
1+\alpha\pi (x)\right] $ for $\alpha>0$.
As $\sum_x\pi (x)=1$, these measures are concentrated on
$\left\{ \eta :\sum_x\eta (x)<\infty \right\}$
because
\[
\int\sum_x \eta(x)\ d\nu^{(\alpha)}=\sum_x \nu^{(\alpha)}\{\eta (x)=1\}
=\sum_x \frac {\alpha\pi (x)} {1+\alpha\pi (x)}<\infty
\]
and $\nu^{(\alpha)}\{\eta :\ \sum_x \eta(x)=n\}>0$ for all $0\leq n
<\infty$.
Since $p(.,.)$ is irreducible by hypothesis the restriction of the
$k$-step exclusion process to $\left\{ \eta
:\sum_x\eta (x)=n\right\} $ is a positive recurrent, irreducible Markov
chain,
with stationary measure
\[
\nu^{(n)}(.)=\nu^{(\alpha)} ( .|\sum_x\eta (x)=n)
\]
which is independent of $\alpha$.
Then we have
\[
\begin{array}{ll}
\displaystyle{\frac {\alpha\pi (x)} {1+\alpha\pi
(x)}}&=\nu^{(\alpha)}\{\eta :\ \eta(x)=1\}\\
&=\sum_{n=0}^{\infty}\nu^{(n)}\{\eta :\ \eta(x)=1\}\nu^{(\alpha)}\{\eta
:\
\sum_x \eta(x)=n\}.\\
\end{array}
\]
Since $\eta_n\leq\eta_{n+1}$ (from monotonicity of $\eta_t$), letting
$\alpha\uparrow\infty$ in the last equality we obtain
\begin{equation}\label{limun}
\lim_{n\rightarrow\infty}\nu_n=\nu^{\infty},
\end{equation}
where $\nu^{\infty}=\delta_1$ is the Dirac mass on $\varpi_1\equiv
1$.
Now let $\mu$ be such that
\[
\mu\{\eta :\ \sum_x \eta(x)=\infty\}=1
\]
and for $1\leq n<\infty$ choose a probability measure $\mu^{(n)}$ such
that
$\mu^{(n)}\leq\mu$ and $\mu^{(n)}\{\eta :\ \sum_x \eta(x)=n\}=1$.
We have
\begin{equation}\label{croiss}
\mu^{(n)}S_k(t)\leq\mu S_k(t)
\end{equation}
for all $t>0$.
Therefore by the convergence theorem of positive recurrent Markov chains
:
\begin{equation}\label{limdeux}
\lim_{n\rightarrow\infty}\mu^{(n)} S_k(t)=\nu^{(n)}
\end{equation}
for all $n$. From this we deduce (b) using (\ref{limun}), (\ref{croiss})
and (\ref{limdeux}).
(a) is a consequence of (b). $\carn$
\section{Tagged and Second Class particles in the $k$-step exclusion
process}
We are interested in the $k$-step exclusion process as seen
from a tagged particle (resp. a second class particle).
For all these processes the origin will always be occupied.
For the tagged particle, after a random exponential time with mean 1, a
transition of the system will occur moving the origin to $X_\tau$,
where $\{X_n\}$ is a Markov chain with probability transition $p(.,.)$,
$X_0=0$ and $\tau=\inf\{1\leq n\leq k: \eta(X_n)=0\}$. When $\{1\leq
n\leq k:
\eta(X_n)=0\}=\emptyset$ we set $X_{\infty}=0$ and the system does not
move.
The other particles move as in the $k$-step exclusion process.
For the second class particle we add to the preceding rules that after
a random exponential time mean 1, the system is translated putting the
origin on $y$ if $\eta(y)=1$ with probability $q_k(y,0,\eta)$.
So the second particle moves like a standard particle but when an
other particle arrive on its site (here the origin) then the two
particles exchange their positions.
\subsection{Notations and definitions}
We will suppose ${\bf S}={\bf Z}^d$ and $p(x,y)=p(0,y-x)$.
The state space will be a little modified :
\[
\widehat{\bf X}={\bf X}\cap\{\eta : \ \eta(0)=1\}.
\]
The $k$-step exclusion process as seen from a tagged particle
has for state space $\widehat{\bf X}$ and for generator :
\[
\widehat{\Omega}_k f(\eta)=\sum_{ \eta (x)=1,\eta(y)=0 \\ x,y\neq
0 }
q_k(x,y,\eta)\left[f(\eta_{xy})-f(\eta)\right]
\]
\[
+\sum_{ \eta(y)=0 \\ y\neq 0 }
q_k(0,y,\eta)\left[f(\eta_{0y}-y)-f(\eta)\right].
\]
Where configuration $\eta-y$ corresponds to $(\eta-y)(z)=\eta(z+y)$.
We will also consider the $k$-step exclusion process as seen from a
second class particle on $\widehat{\bf X}$ with generator :
\[
\overline{\Omega}_k f(\eta)=\widehat{\Omega}_k f(\eta)+\breve\Omega_k
f(\eta)
\]
where
\[
\breve\Omega_k f(\eta)=\sum_{ \eta(x)=1 \\ x\neq 0 }
q_k(x,0,\eta)\left[f(\eta_{0x}-x)-f(\eta)\right].
\]
By Liggett's criterion :
\[
\sum_{ \eta (x)=1,\eta(y)=0 \\ x,y\neq 0 }
q_k(x,y,\eta)\left[f(\eta_{xy})-f(\eta)\right]
\]
is a generator of a contractive semi-group and
\[
\sum_{ \eta(y)=0 \\ y\neq 0 }
q_k(0,y,\eta)\left[f(\eta_{0y}-y)-f(\eta)\right]
\]
(resp.
\[
\sum_{ \eta(x)=1 \\ x\neq 0 }
q_k(x,0,\eta)\left[f(\eta_{0x}-x)-f(\eta)\right])
\]
is a bounded operator. Hence by theorem 2 of Gustafson
\cite{Gustafson}
$\widehat{\Omega}_k$ and $\overline{\Omega}_k$ are infinitesimal
generator of contractive semi-group that we denote respectively
$\widehat{S}(t)$ et $\overline{S}(t)$. To each of these semi-group
corresponds a unique Markov process on $\widehat{\bf X}$ with generators
$\widehat{\Omega}_k$ and $\overline{\Omega}_k$.
Let ${\cal P}(\widehat{\bf X})$ be the set of probability measure on
$\widehat{\bf X}$, and
$\widehat{\cal I}_k=\{\mu\in{\cal P}(\widehat{\bf X}):\ \mu \widehat
S_k(t)=\mu\}$
(resp.
$\overline{\cal I}_k=\{\mu\in{\cal P}(\widehat{\bf X}):\ \mu \overline
S_k(t)=\mu\}$) the set of invariant measures for the $k$-step exclusion
process as seen
from the tagged particle (resp. the second class particle).
\subsection{Invariant measures}
Let $\mu\in{\cal S}$ such that $\beta(\mu)=\mu\{\eta(0)=1\}>0$.
We define the Palm measure of $\mu$, by $\widehat{\mu}$ a measure
on $\widehat{\bf X}$ such that
\[
\widehat{\mu}=\mu(.\vert \eta(0)=1).
\]
(Let us recall that
if $\mu$ puts all its mass on empty configuration ({\it i.e.}
$\mu=\delta_0$) one define
$\widehat\mu=\delta_{\left\{\eta=\{0\}\right\}}$
as the measure that puts all its mass on configuration with only one
particle at $0$).
We denote
\[
\widehat{\cal S}=\left\{\widehat\mu :\ \mu\in{\cal S}\right\}.
\]
The next property will be useful :
\begin{lemma}\label{criterePalm}
Let $\mu, \nu\in{\cal S}$. If $\widehat\mu=\widehat\nu$ and
$\beta=\beta(\mu)=\beta(\nu)$ then
\[
\mu=\nu.
\]
\end{lemma}
{\bf Proof :}
Suppose $0<\beta<1$.
Note that for all measurable $A$
\[
\mu(A)=\beta\mu\left\{A\vert \eta(0)=1\right\}
+(1-\beta)\mu\left\{A\vert\eta(0)=0\right\}
\]
and since $\widehat\mu=\widehat\nu$
\[
=\beta\nu\left\{A\vert \eta(0)=1\right\}
+(1-\beta)\mu\left\{A\vert\eta(0)=0\right\}.
\]
Furthermore
\[
\mu\left\{A\vert\eta(0)=0\right\}=\frac
{\mu\left\{A,\eta(0)=0\right\}} {1-\beta},
\]
we only have to show that
\begin{equation}
\mu\left\{A,\eta(0)=0\right\}=\nu\left\{A,\eta(0)=0\right\}
\end{equation}
to conclude $\mu=\nu$.
Now take a one to one mapping :
\[
\begin{array}{c}
{\bf N}\setminus\{0\}\longrightarrow{\bf Z}^d\setminus\{0\}\\
i \longmapsto x_i
\end{array}
\]
we have
\begin{equation}\label{decomposition}
\begin{array}{c}
\mu\left\{A,\eta(0)=0\right\}\\
=\sum_{i=1}^\infty\mu\left\{A,\eta(0)=0,\eta(x_1)=...=\eta(x_{i-1})=0,\eta(x_i)=1
\right\}\\
+\mu(A,\eta\equiv 0).
\end{array}
\end{equation}
But by translation invariance of $\mu$ one has for all $i$
\[
\mu\left\{A,\eta(0)=0,\eta(x_1)=...=\eta(x_{i-1})=0,\eta(x_i)=1\right\}
\]
\[
=\mu\left\{\tau_{x_i}A,\eta(-x_i)=0,\eta(x_1-x_i)=...=\eta(x_{i-1}-x_i)=0,\eta(0)
=1\right\}
\]
\[
=\mu\left\{\tau_{x_i}A,\eta(-x_i)=0,\eta(x_1-x_i)=...=\eta(x_{i-1}-x_i)=0\vert
\eta(0)=1\right\}\beta
\]
and since $\widehat\mu=\widehat\nu$
\[
=\nu\left\{\tau_{x_i}A,\eta(-x_i)=0,\eta(x_1-x_i)=...=\eta(x_{i-1}-x_i)=0\vert
\eta(0)=1\right\}\beta
\]
\[
=\nu\left\{\tau_{x_i}A,\eta(-x_i)=0,\eta(x_1-x_i)=...=\eta(x_{i-1}-x_i)=0,
\eta(0)=1\right\}
\]
and by $\nu$'s translation invariance
\[
=\nu\left\{A,\eta(0)=0,\eta(x_1)=...=\eta(x_{i-1})=0,\eta(x_i)=1\right\}.
\]
On the other hand as
\[
\mu(\eta\equiv 0)=1-\sum_{i=0}^\infty\mu\left\{\eta(x_0)=
\eta(x_1)=...=\eta(x_{i-1})=0,\eta(x_i)=1\right\},
\]
we deduce by the same type of argument that
\[
\mu(\eta\equiv 0)=\nu(\eta\equiv 0).
\]
Therefore (\ref{decomposition})
\[
\mu\left\{A,\eta(0)=0\right\}=\nu\left\{A,\eta(0)=0\right\}
\]
Which allows us to conclude.
$\carn$
\noindent
{\bf Proof of theorem \ref{marque}}
The proof is decomposed into the two following theorems.
\begin{theorem}\label{TaggedPart}
(a) If $\nu\in{\cal S}$ then for all $t\geq 0$ we have
\[
\widehat{\nu}\widehat{S}_k(t)=\widehat{\nu S_k(t)}.
\]
(b) The Palm measure $\widehat{\nu_{\rho}}$ of $\nu_{\rho}$ is invariant
for
the tagged particle process.
\[
\mbox{\it i.e.
}\widehat{\nu_{\rho}}\widehat{S}_k(t)=\widehat{\nu_{\rho}}.
\]
\end{theorem}
{\bf Proof:}
(b) is an immediate consequence of (a) and theorem \ref{produto2}.
Define $T:\ C(\widehat{\bf X})\rightarrow C(U)$ an operator by
\[
Tf(\eta)=\left\{\begin{array}{ll}
\eta(0)f(\eta) & \\
0 & \mbox{if }\eta(0)=0.
\end{array}\right.
\]
Note that if $\mu\in{\cal S}$ and $f\in C(\widehat{\bf X})$
\[
\int Tf\ d\mu=\beta(\mu)\int f\ d\widehat\mu.
\]
Then for all measure $\mu\in{\cal S}$ and all $f\in C(\widehat{\bf X})$
we have
\begin{equation}\label{invmarq}
\int T\widehat{\Omega}_k f\ d\mu=\int \Omega_k Tf\ d\mu
\end{equation}
for all cylindric $f$.
Indeed :
\[
\Omega_k Tf(\eta)=\sum_{ \eta(x)=1 \\ \eta(y)=0}
q_k(x,y,\eta)\left[\eta_{xy}(0)f(\eta_{xy})-\eta(0)f(\eta)\right]
\]
\[
=\sum_{ \eta (x)=1,\eta(y)=0 \\ x,y\neq 0 }
q_k(x,y,\eta)\eta(0)\left[f(\eta_{xy})-f(\eta)\right]
\]
\[
+\sum_{ \eta(y)=0 \\ \eta(0)=1 }
q_k(0,y,\eta)\left[\eta(y)f(\eta_{0y})-\eta(0)f(\eta)\right].
\]
\[
+\sum_{ \eta(x)=1 \\ \eta(0)=0 }
q_k(x,0,\eta)\left[\eta(x)f(\eta_{0x})-\eta(0)f(\eta)\right].
\]
\[
=\sum_{ \eta (x)=1,\eta(y)=0 \\ x,y\neq 0 }
q_k(x,y,\eta)\eta(0)\left[f(\eta_{xy})-f(\eta)\right]
\]
\[
+\sum_{ \eta(0)=0 \\ x\neq 0 } q_k(x,0,\eta)\eta(x)f(\eta_{0x})
-\sum_{ \eta(y)=0 \\ y\neq 0 } q_k(0,y,\eta)\eta(0)f(\eta).
\]
In the same way we write
\[
T\widehat{\Omega}_kf(\eta)=\sum_{ \eta (x)=1,\eta(y)=0 \\ x,y\neq
0 }
q_k(x,y,\eta)\eta(0)\left[f(\eta_{xy})-f(\eta)\right]
\]
\[
+\sum_{ \eta(y)=0 \\ y\neq 0 }
q_k(0,y,\eta)\eta(0)\left[f(\eta_{0y}-y)-f(\eta)\right].
\]
The first terms and the end of the second terms of $\Omega_k
Tf(\eta)$
and $T\widehat{\Omega}f(\eta)$ are corresponding, then we only have
to show that
\[
\sum_{x\neq 0}\mu\{ q_k(x,0,\eta)\left[ 1-\eta(0)
\right]\eta(x)f(\eta_{0x})\}
\]
\[
=\sum_{y\neq 0}\mu\{
q_k(0,y,\eta)\eta(0)\left[1-\eta(y)\right]f(\eta_{0y}-y)\}
\]
Which is true because of the translation invariance of $\mu$ :
\[
\mu\{ q_k(x,0,\eta)\left[ 1-\eta(0) \right]\eta(x)f(\eta_{0x})\}
\]
\[
=\mu\{ q_k(0,-x,\eta)\left[ 1-\eta(-x) \right]\eta(0)f(\eta_{-x0}+x)\}
\]
\[
=\mu\{ q_k(0,y,\eta)\eta(0)\left[1-\eta(y)\right]f(\eta_{0y}-y)\}
\]
letting $y=-x$.
Now compute for $f\in C(\widehat{\bf X})$
\[
T\widehat{S}_k(t)f-S_k(t)Tf=\int_0 ^t\frac d {ds}
S_k(t-s)T\widehat{S}_k(s)f\ ds
\]
\[
=\int_0 ^t S_k(t-s)[T\widehat{\Omega}_k-\Omega_k T]\widehat{S}_k(s)f\
ds
\]
It follows
\[
\int [T\widehat{S}_k(t)f-S_k(t)Tf]\ d\mu
=\int_0 ^t \int [T\widehat{\Omega}_k-\Omega_k T]\widehat{S}_k(s)f\
d\mu S_k(t-s)\ ds
\]
which is zero by (\ref{invmarq}) because $\mu S_k(t)\in{\cal S}$
for all $t>0$.
Then
\begin{equation}
\int T\widehat{S}_k(t)f\ d\mu=\int S_k(t)Tf\ d\mu
\end{equation}
for all measure $\mu\in{\cal S}$. Therefore
\[
\beta(\mu)\int f\ d\widehat{\mu}\widehat{S}_k(t)
=\beta(\mu S_k(t))\int f \ d\widehat{\mu S_k(t)}.
\]
Since $\mu\in{\cal S}$ we have $\beta(\mu)=\beta(\mu S_k(t))$ (see
Lemma 2.3 section 1) and the result follows. $\carn$
\begin{remark}
This proof is a very general one, we never use explicitly the
expression of the coefficients $q_k(x,y,\eta)$.
As quoted in \cite{Ferrari} part (a) is and extension of theorems of
Harris
\cite{harris1} and Port and Stone \cite{porsto}.
\end{remark}
The next result is similar to \cite{Ferrari}
\begin{theorem}
Suppose $p(.,.)$ is irreducible, then
\[
(\widehat{\cal I}_k \cap\widehat{\cal S})_e=\{\widehat{\nu}_\rho : \
0\leq \rho \leq 1\}.
\]
\end{theorem}
{\bf Proof :} The proof is the same as in \cite{Ferrari}
page 1282. $\carn$
This ends the proof of theorem \ref{marque}.
\noindent
{\bf Proof of theorem \ref{SecondClass}}
For this we prove the following proposition that contains the theorem.
\begin{proposition}
For all $k\geq 1$,
(a) If $p(.,.)$ is symmetric and $\mu\in{\cal S}$ then
$\widehat{\mu}\in\overline{\cal I}_k$.
(b) Let $\rho\in ]0,1]$, $\widehat{\nu}_\rho\in\overline{\cal I}_1$ if
and only if
$p(.,.)$ is symmetric.
(c) Let $\rho\in ]0,1]$, for ${\bf S}={\bf Z}$ if
$p(x,x+1)=1-p(x,x-1)=p$ then
$\widehat{\nu}_\rho\in\overline{\cal I}_k$ if and only if
$p(.,.)$ is symmetric ({\it i.e.} $p=1/2$).
(d) If f $\widehat{\nu}_\rho\in\overline{\cal I}_k$ then $p(.,.)$ is
symmetric except eventually for a finite number $(\leq k)$ of $\rho$.
\end{proposition}
Note that if $\mu\in{\cal S}$ then
\begin{equation}
\int T\overline{\Omega}_k f(\eta)\ d\mu
=\int \Omega_kTf(\eta)\ d\mu+\int T\breve\Omega_kf(\eta)\ d\mu
\end{equation}
where
\[
\breve\Omega_kf(\eta)=\sum_{ \eta(x)=1 \\ x\neq 0 }
q_k(x,0,\eta)\left[f(\eta_{0x}-x)-f(\eta)\right].
\]
We try to find conditions for which for all $f\in{\cal D}$
\begin{equation}
\int T\breve\Omega_kf(\eta)\ d\mu=0.
\end{equation}
\[
\int T\breve\Omega_kf(\eta)\ d\mu=\sum_{x\neq 0}
\int q_k(x,0,\eta)\eta(0)\eta(x)\left[f(\eta_{0x}-x)-f(\eta)\right]\
d\mu
\]
and since $\eta_{0x}=\eta$ when $\eta(x)=\eta(0)$, by translation
invariance of $\mu$ we obtain
\[
=\sum_{x\neq 0}\left[\int q_k(0,-x,\eta)\eta(-x)\eta(0)f(\eta)\ d\mu
-\int q_k(x,0,\eta)\eta(0)\eta(x)f(\eta)\ d\mu\right]
\]
\begin{equation}\label{transinvar}
=\sum_{x\neq 0}\int \left[q_k(0,x,\eta)-q_k(x,0,\eta)\right]
\eta(0)\eta(x)f(\eta)\ d\mu
\end{equation}
which is zero if $p(.,.)$ is symmetric and shows part (a)
applying the same ideas as for the end of the proof of theorem
\ref{TaggedPart}.
Now let $\mu=\nu_\rho$, for $\rho\in]0,1[$.
For all $y\in{\bf S}\setminus\{0\}$ let $f_y=\eta(y)$.
Then writing (\ref{transinvar}) for $f_y$ and developing
coefficients $q_k(.,.,.)$ we obtain the three following cases :
- {\bf 1st case} if $k=1$
\[
\int T\breve\Omega_kf_y(\eta)\
d\mu=\rho^2\left(\left[p(0,y)-p(y,0)\right]
+\rho\sum _{ x\neq 0 \\
x\neq y}
\left[p(0,x)-p(x,0)\right]\right)
\]
and as $p(.,.)$ is doubly stochastic
\[
=\rho^2(1-\rho)\left[p(0,y)-p(y,0)\right].
\]
This last expression is null for all $y$ only if $p(.,.)$
is symmetric. This ends part (b);
- {\bf 2nd case} if ${\bf S}={\bf Z}$ et $p(x,x+1)=1-p(x,x-1)=p$ and
$q=1-p$; without loss of generality we take $y$ such that $\vert
y\vert>k$,
then (\ref{transinvar}) is a polynomial in $\rho$ degree $k+2$ whose
order $2l+1$ coefficients, $3\leq 2l+1\leq k+2$ are
\[
(p^{2l+1}-q^{2l+1})+C_{2l+1}^{2l}pq(p^{2l-1}-q^{2l-1})+...+C_{2l+1}^{l+1}p^{l}q^{l}(p-q),
\]
et and order $2l$, $3\leq 2l\leq k+2$ are
\[
(p^{2l}-q^{2l})+C_{2l}^{2l-1}pq(p^{2l-2}-q^{2l-2})+...+C_{2l}^{l+2}p^{l}q^{l}(p^2-q^2).
\]
Since these coefficients have the same sign then (\ref{transinvar})
is null only if $p=q=1/2$. This ends part (c);
- {\bf 3rd case} For part (d) without loss of generality we
may suppose that $p(x,x)=0$, we then obtain a
polynomial in $\rho$ of degree $k+2$ for which 0 is a root of order two
and so it has at most $k$ non null roots.
Its lowest degree term is
\[
\rho^{2}\left[ p(0,y)-p(y,0)\right].
\]
This term is constantly null only if $p(.,.)$ is symmetric.
$\carn$
\section{Weak law of large numbers}
\subsection{Introduction}
We still suppose that $p(.,.)$ is translation invariant and ${\bf
S}={\bf
Z}^d$.
We are interested in the asymptotic behavior of a tagged particle
(resp. a second class particle) originally at $x\in {\bf
S}$, when the other particles of the system are distributed according to
$\nu_\rho$ the Bernoulli product measure, $(0\leq\rho\leq 1)$.
For $k=1$, supposing that $p(.,.)$ has a finite first moment, F.
Spitzer \cite{Spitzer} has compute ${\bf E}X_t$, where $X_t$ is the
position of the tagged particle at time $t$, and has shown the
existence of an almost sure limit for $X_t/t$. C. Kipnis \cite{CKipnis}
(in the nearest neighbors case in dimension one) and E. Saada
\cite{Saada}
in the other cases have shown that this limit is constant and equals to
\[
\lim_{t\rightarrow\infty}\frac{X_t} t=\sum_{y\in {\bf Z}^d}yp(0,y)
\mbox{ {\it a.s.}}
\]
\subsection{Tagged Particle}
For the tagged particle in the $k$-step exclusion process we have
\begin{theorem}\label{TagAsymp}
Suppose $p(.,.)$ with bounded first moment.
If the $k$-step exclusion process as seen from the tagged particle
has for initial law $\widehat\nu_\rho$ then,
\[
{\bf E}X_t=x+(1-\rho)t\sum_y y p(0,y)+...
\]
\[
...+(1-\rho)t
\sum_y \sum_{z_1,...,z_{k-1}\neq 0,y}\rho^\gamma
yp(0,z_1)...p(z_{k-1},y)
\]
\[
\mbox{where }\gamma(z_1,...,z_{k-1})=Card\{z_1,...,z_{k-1}\}
\]
\[
\mbox{ and}
\]
\[
\lim_{t\rightarrow\infty}\frac {X_t} t \ \mbox{ exists {\it a.s} and
in $L^1$.}
\]
\end{theorem}
{\bf Proof :}
It's a corollary of theorem \ref{TaggedPart}
(see Liggett \cite{LiggettBook} p. 396). By this theorem
$X_t$ is stationary increasing. Then ${\bf E}X_t$ is a linear
function of $t$.
On the other hand
\[
\lim_{t\downarrow 0}\frac{{\bf E}X_t-x} t =(1-\rho)\sum_y yp(0,y)
+...
\]
\[
...+(1-\rho)
\sum_y \sum_{z_1,...,z_{k-1}}\rho^{\gamma(z_1,...,z_{k-1})}
yp(0,z_1)...p(z_{k-1},y)
\]
we have the first result.
The {\it a.s. } convergence of $X_n/n$ is a consequence of the
Ergodic theorem. Noting that
\[
{\bf E}\sup_{0\leq t\leq 1}\|X_t\|<\infty,
\]
and using a Borel-Cantelli argument
we deduce the convergence for all $t$.
$\carn$
We then state a natural conjecture :
\begin{conjecture}
For $k\geq 2$,
Under the hypothesis of theorem \ref{TagAsymp}
\[
\lim_{t\rightarrow\infty}\frac {X_t} t =(1-\rho)\sum_y y p(0,y)+...
\]
\[
...+(1-\rho)
\sum_y \sum_{z_1,...,z_{k-1}}\rho^{\gamma(z_1,...,z_{k-1})}
yp(0,z_1)...p(z_{k-1},y)\ \mbox{\it a.s.}
\]
\end{conjecture}
\begin{remark}
The major difficulty to prove this is to show that
\[
\nu_\rho\in({\cal I}_k)_e
\]
when $k\geq 2$.
Using coupling arguments of Liggett (1976) in \cite{Liggett} it is
possible,
at least for the nearest neighbor asymmetric case ({\it i.e. $d=1$,
$p(x,x+1)=p$, $p(x,x-1)=1-p$, $p\not=1/2$}), to show that
$\nu_\rho\in({\cal
I}_k)_e$.
\end{remark}
\subsection{Second class particle}
Let $Y_t$ the position at time $t$ of the second class particle
originally at $x\in{\bf S}$.
\begin{theorem}\label{wlsc}
With the hypothesis of theorem \ref{TagAsymp} and if $p(.,.)$ is
symmetric then
\[
{\bf E}Y_t=x\quad \mbox{ and}
\]
\[
\lim_{t\rightarrow \infty}\frac {Y_t} t \mbox{ exists {\it a.s.} and
in }L^1.
\]
\end{theorem}
{\bf Proof :}
The proof parallels the one for the tagged particle.
$\carn$
\noindent
{\bf Acknowledgments}
This research is part of FAPESP's {\sl Projeto Tem\' atico} 95/0790-1
and
Pronex No 41.96.0923.00.
Part of this paper is an adaptation of my Ph.D. thesis at the CMI de
l' Universit\'e de Provence, Marseille. Thanks are given to my adviser,
Enrique Andjel. I would also like to thank Ellen Saada, Dominique Bakry,
Pablo Ferrari and Mihail Menshikov for valuable discussions. Special
thanks to
Krisnamurthi Ravishankar and Ellen Saada for encouraging me to publish
this
paper.
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\end{document}