\section{Introduction.}

Soon after its early developments in probability theory [1] the study
of random walks and their applications has gradually carved its way into
the concerns of many topics in exact or applied science. In various
recent developments, the notion of ``walk'' in the proper sense of the
term has gradually blurred, giving way to a more general concept where
a ``step'' of the walk is not necessarily an algebraic distance between
two locations but may become a time interval between two consecutive
events of a chronological series, or even a sequence of real numbers.

Before mentioning a few significant fields of application, let us quote
more specifically the notion of ``correlated random walk,'' which underlies
the possibility for a step to depend on the values taken by some (or all) of
the previous steps. We will address the step correlations and their
consequences at length in the course of this paper, and will consider
therefore the ``direct'' problem, in contradistinction with the usual concern
prevailing in applied science which asks for answers to the ``inverse'' problem,
\textit{i.e.}, can a given set of data be suitably described by a (correlated or
uncorrelated) random walk formalism? Examples of fields resorting to the
inverse problem are neuromuscular [2] and cardiovascular medicine [3],
structural genetics [4-6], physiology [7], regulation of biological
rhythms [8]; to these fields pertaining essentially to the medical domain can be
added the soaring irruption of financial analysis [9-11] in view of market
predictions.

In this paper we will focus on the formalism of correlated random walks on
one-dimensional lattices and discuss the r\^ole of memory in some typical models
which can be solved exactly. With the help of a direct evaluation of the variance
of the walks, we will be able to prove that all walks with short-range memory
have a Hurst exponent equal to $\frac{1}{2}$  in the strict limit of an infinite
number of steps, although the apparent Hurst exponent may look different from
$\frac{1}{2}$. In particular, we will endeavor to produce a simple argument for
a necessary condition yielding a Hurst exponent larger than $\frac{1}{2}$.

\section{Correlated walks and memory}

The transition probabilities in a simple random walk on a
one-dimensional infinite lattice are constant and independent of the
course of the walk. If these transition probabilities depend upon the
$M$ previous steps, we have a correlated random walk whose memory range
is $M$. Correlated random walks were introduced by Taylor [12] in an analysis of
diffusion by continuous motions. The Taylor's model is a one-dimensional
random walk on a lattice, in which  steps can be made to nearest neighbors
only. The walker has a probability $\alpha$ to repeat his previous move,
and a probability $1-\alpha$ to move in the opposite direction. The resulting
correlated random walk has a memory range equal to 1. If
$\alpha > \frac{1}{2}$ (resp. $\alpha < \frac{1}{2}$) the
walk is said to be {\it persistent} (resp.  {\it antipersistent}). A complete
analysis of the Taylor's model has been given by Goldstein [13]. More examples
of persistent random walks have been studied by many other authors [14-16].
A detailed bibliography may be found in Weiss [17].

In this paper we study a class of correlated random walks with memory
range $M>1$. Our model may be described as follows. Let $(S_k)$ be
an infinite sequence of random variables such that, for $k>M$,
$$
S_k =
\begin{cases}
1 & \text{with probability $\frac{1}{2}p$}\\
-1 & \text{with probability $\frac{1}{2}p$}\\
f_M(S_{k-1}, S_{k-2},\ldots,S_{k-M}) & \text{with probability $q=1-p$,}\\
\end{cases}
$$
where $f_M$, called the {\it memory function}, is a Boolean function from
$\{-1,\!1\}^M$ onto $\{-1,\!1\}$.
For $k \leq M$, the $S_k$'s are independent, identically distributed (\textit{iid})
Bernoulli random variables, each
$S_k$ taking the values $\pm 1$ with equal probabilities. The values of these
$M$ random variables define the initial conditions for the walk. The position
of the random walker after $n$ steps is
$$
R_n = S_1 + S_2 + \cdots + S_n.
$$
The set $\mathcal{F}_M$ of all bounded functions defined on $\{-1,1\}^M$ is a
Euclidean space of dimension $2^M$ for the dot product defined by
\begin{eqnarray*}
\lefteqn{<f_M^{(1)}\mid f_M^{(2)}>\;\; = }\\
& & \left(\frac{1}{2}\right)^M
\tr f_M^{(1)}(S_{k-1}, S_{k-2},\ldots,S_{k-M})
f_M^{(2)}(S_{k-1}, S_{k-2},\ldots,S_{k-M}),\\
\end{eqnarray*}
where the trace operator $\tr$ is a sum over all the possible values of
the $M$ variables $S_{k-1}, S_{k-2},\ldots,S_{k-M}$.
The set of the following $2^M$ functions
$$\left\lbrace
\begin{array}{l}
1,\\
S_{k-1}, S_{k-2}, \ldots , S_{k-M},\\
S_{k-1}S_{k-2}, S_{k-1}S_{k-3}, \ldots , S_{k-M+1}S_{k-M},\\
S_{k-1}S_{k-2}S_{k-3}, \ldots , S_{k-M+2}S_{k-M+1}S_{k-M}, \\
\cdots,\\
S_{k-1}S_{k-2}\cdots S_{k-M},\\
\end{array}
\right.$$
is a complete orthonormal system, and any function in $\mathcal{F}_M$ can be
written as a linear combination of these $2^M$ functions. The most general
correlated random walk with a finite memory of range $M$ is, therefore,
defined by a sequence $(S_k)$ of random variables such that,
for $k>M$,
\begin{equation}
S_k =
\begin{cases}
1 & \text{with probability $\frac{1}{2}p$}\\
-1 & \text{with probability $\frac{1}{2}p$}\\
\varepsilon_{j_1j_2\cdots j_r}S_{k-j_1}S_{k-j_2}\cdots,S_{k-j_r}
& \text{with probability $qa_{j_1j_2\cdots j_r}$,}
\end{cases}
\label{general-model}
\end{equation}
where $p+q=1$, $1\leq j_1<j_2<\cdots<j_r\leq M$, and
$\varepsilon_{j_1j_2\cdots j_r}$, which are \textit{non-random}, are equal either to
$1$ or to $-1$. The $a$'s are conditional probabilities (some of them possibly
equal to zero) satisfying the completeness relation:
\begin{eqnarray*}
&  & \sum_{1\leq j_1\leq M}a_{j_1} + \sum_{1\leq j_1<j_2\leq M}a_{j_1j_2}
+\cdots\\
& & \qquad + \sum_{1\leq j_1<j_2\leq<j_r\leq M} a_{j_1j_2\cdots j_r}
+ \cdots + a_{12\cdots M} = 1.\\
\end{eqnarray*}

As above, for $k \leq M$, $S_k$ is a simple Bernoulli random variable
which takes the values $\pm 1$ with equal probabilities.

In order to simplify this model, we require that, as in the case of
a simple symmetric random walk, the average value $<R_n>$ of the position
of the random walker after $n$ steps should be equal to zero. While this
condition is automatically satisfied if $n\leq M$, a tedious inspection
shows that, for $n>M$, the above requirement is satisfied if, and only if,
the memory function $f_M$ has the following simple forms:

\begin{itemize}
\item $f_M(S_{k-1}, S_{k-2},\ldots,S_{k-M})$ is a linear combination of
multilinear terms of odd degree in the $S_k$'s;
\item $f_M(S_{k-1}, S_{k-2},\ldots,S_{k-M})$ reduces to a single
multilinear term containing an even number of $S_k$'s.
\end{itemize}

For example, if $M=3$, the most general form for the memory function $f_3$,
satisfying the condition $<R_n>=0$, is either
\begin{eqnarray*}
f_3(S_{k-1},S_{k-2},S_{k-3})
& = & a_1\varepsilon_1S_{k-1}+a_2\varepsilon_2S_{k-2} + a_3\varepsilon_3S_{k-3}\\
&   &\qquad + a_{123}\varepsilon_{123}S_{k-1}S_{k-2}S_{k-3},
\end{eqnarray*}
or
$$
f_3(S_{k-1},S_{k-2},S_{k-3}) = a_{ij}\varepsilon_{ij}S_{k-i}S_{k-j},
$$
with $1\leq i\leq 3$ and $1\leq j\leq 3$.

In the sequel we will only consider two cases: either $f_M$ is a linear
function of the $M$ previous steps, or the product of the $M$ previous steps.

Let us briefly examine these two cases.

In what follows $(X_k)$ denotes an infinite sequence of \textit{iid}
Bernoulli random variables such that, for all positive integers $k$, we have
$$
P(X_k=1) = P(X_k=-1) = \frac{1}{2},
$$

\textbf{1.} \textit{$f_M$ is linear.} If $(S_k)$ is an infinite sequence of
random variables defined by
\begin{equation}
S_k =
\begin{cases}
X_k & \text{if $k\le M$},\\[0.1cm]
X_k & \text{with probability $p$, if $k>M$},\\
\varepsilon_j S_{k-j} & \text{with probability $qa_j$, if $k>M$,}
\end{cases}
\label{general-linear}
\end{equation}
where, for $j=1,2,\ldots,M$, the $\varepsilon_j$ have fixed values chosen
\textit{a priori\/} equal either to $1$ or $-1$, and the $a_j$ are $M$
nonnegative real numbers such that
$$
a_1+a_2+\cdots+a_M = 1.
$$

It is clear that, for $k\leq M$, $<S_k>=0$. Hence
\begin{eqnarray*}
<S_{M+1}> & = & p <X_{M+1}>+qa_1\varepsilon_1<S_M>
+\cdots+qa_M\varepsilon_M<S_1>\\
& = & 0\\
<S_{M+2}> & = & p <X_{M+2}>+qa_1\varepsilon_1<S_{M+1}>
+\cdots+qa_M\varepsilon_M<S_2>\\
& = & 0\\
&   & \cdots\\
<S_{M+\ell}> & = & p <X_{M+\ell}>+qa_1\varepsilon_1<S_{M+\ell-1}>
+\cdots+qa_M\varepsilon_M <S_\ell>\\
& = & 0,
\end{eqnarray*}
and, therefore, $<R_n>=0$, for all positive integers $n$.

\textbf{2.} \textit{$f_M$ is a single product of $r$ terms.} If $(S_k)$ is
an infinite sequence of random variables defined by
\begin{equation}
S_k =
\begin{cases}
X_k & \text{if $k\le M$},\\[0.1cm]
X_k & \text{with probability $p$ if $k>M$},\\
\varepsilon_{j_1j_2\cdots j_r}S_{k-j_1}S_{k-j_2}\cdots S_{k-jr}
& \text{with probability $q$ if $k>M$},
\end{cases}
\label{general-product}
\end{equation}
where $q=1-p$, $1\leq j_1<j_2<\cdots<j_r\leq M$, and
$\varepsilon_{j_1j_2\cdots j_r}$ is given and equal to either $1$ or $-1$.
Here again, for $k\leq M$, $<S_k>=0$. Hence, assuming first that $r=M$,
we have
\begin{eqnarray*}
<S_{M+1}>& = & p<X_{M+1}>+q\varepsilon_{12\cdots M}<S_1S_2\cdots S_M>\\
         & = & p<X_{M+1}>+q\varepsilon_{12\cdots M}<S_1><S_2>\cdots <S_M>\\
         & = & 0\\
<S_{M+2}>& = & p<X_{M+2}>+q\varepsilon_{12\cdots M}
                        <S_2S_3\cdots S_M S_{M+1}>\\
         & = & pq\varepsilon_{12\cdots M}<S_2S_3\cdots S_M X_{M+1}>\\
         &   & \qquad + q^2\varepsilon_{12\cdots M}<S_1S_2^2S_3^2\cdots S_M^2>\\
         & = & q^2\varepsilon_{12\cdots M}<S_1>\;=\;0
\end{eqnarray*}
where in the expression of $S_{M+2}$ we have replaced $S_{M+1}$ by its
expression. More generally,
\begin{eqnarray*}
<S_{M+\ell}> & = & p<X_{M+\ell}>+q<S_\ell S_{\ell+1}\cdots S_{M+\ell-1}>\\
             & = & pq<S_\ell S_{\ell+1}\cdots X_{M+\ell-1}>+
            q^2<S_{\ell-1}S_\ell^2 S_{\ell+1}^2\cdots S_{M+\ell-2}^2>\\
             & = & q^2<S_{\ell-1}>,
\end{eqnarray*}
which shows that for all positive integers $k$, $<S_k>=0$,
and, consequently, $<R_n>=0$.

Along similar lines, it can be shown that the result  $<R_n>=0$ remains
valid for $r<M$, although its proof requires a larger number of substitutions.

In the following sections we will show that, for any finite value of $M$,
we can determine exactly the probability distribution of $<R_n>$,
using a transfer matrix method. It will turn out that the rank of the matrix
is equal to $2^M$. Due to the increasing complexity of the eigensystem problem
with increasing $M$, this method becomes rapidly cumbersome, although it
rather easily provides a numerical solution if the number of steps $n$ is
not too large. In order to illustrate the method, we will first briefly review
the Goldstein's solution of the Taylor's model.

Since the asymptotic behavior of the variance $<R_n^2>$ is of particular
interest to natural scientists,
we will directly derive its exact expression through a recursive method
that applies at any finite space dimensionality. From this expression we
will be able to show that the asymptotic behavior $<R_n^2>$,
when $n$ goes to infinity, exhibits a crossover phenomenon.

\section{Taylor's model.}

The Taylor's model corresponds to Model~\ref{general-linear} for $M=1$,
that is,
$$
S_1 = X_1,
$$
and, for all $k>1$,
$$
S_k=
\begin{cases}
X_k, & \text{with probability $p$}\\
\varepsilon S_{k-1}, & \text{with probability $q=1-p$,}
\end{cases}
$$
where $(X_k)$ is a sequence of \textit{iid} Bernoulli variables such that,
for all positive integers $k$,
$$
P(X_k=-1) = P(X_k=1) = \frac{1}{2},
$$
and $\varepsilon$ is a \textit{non-random number} equal to $\pm 1$.

If $p_n(k)$ denotes the probability that the walker will be at site $k$ after
the $n$th step has been completed, we have
$$
p_n(k) = p_n^+(k)+p_n^-(k),
$$
where the superscript $\pm$ refers to the sign of the $n$-th step, that is
$$
p^{\pm}(k,n) = P((R_n=k)\cap (S_n = \pm 1)).
$$
Taking into account the conditional probabilities:
\begin{eqnarray*}
P(S_n=1\mid S_{n-1}=1)  =  P(S_n=-1\mid S_{n-1}=-1) & = &
\frac{1}{2}\left(p+(1+\varepsilon)q\right)\\
P(S_n=1\mid S_{n-1}=-1) = P(S_n=-1\mid S_{n-1}=1) & = &
\frac{1}{2}\left(p+(1-\varepsilon)q\right),
\end{eqnarray*}
we obtain the following recursion relations
\begin{eqnarray}
p_n^+(k) & = & \frac{1}{2}\left(p+(1+\varepsilon)q\right)p_{n-1}^+(k-1)+
            \frac{1}{2}\left(p+(1-\varepsilon)q\right)p_{n-1}^-(k-1)\nonumber\\
         & = &  \frac{1}{2}\left(1+\varepsilon q\right)p_{n-1}^+(k-1)+
            \frac{1}{2}\left(1-\varepsilon q\right)p_{n-1}^-(k-1)\label{pplus}\\
p_n^-(k) & = & \frac{1}{2}\left(p+(1-\varepsilon)q\right)p_{n-1}^+(k+1)+
             \frac{1}{2}\left(p+(1+\varepsilon)q\right)p_{n-1}^-(k+1)\nonumber\\
         & = &  \frac{1}{2}\left(1-\varepsilon q\right)p_{n-1}^+(k+1)+
            \frac{1}{2}\left(1+\varepsilon q\right)p_{n-1}^-(k+1)\label{pminus},
\end{eqnarray}
or, in a more condensed form which will be useful when we consider
the case $M>1$,
\begin{equation}
p_n^\sigma(k)=  \frac{1}{2}\left(1-\varepsilon q\right)p_{n-1}^+(k-\sigma)+
              \frac{1}{2}\left(1+\varepsilon q\right)p_{n-1}^-(k-\sigma),
\label{psigma}
\end{equation}
with $\sigma=\pm1$.
The probability $\alpha$ in Goldstein's notations coincides here with
$ \frac{1}{2}\left(1+\varepsilon q\right)$.

The generating function of the probability distribution of the
random walk is defined by
$$
f_n(x) = f_n^+(x) + f_n^-(x),
$$
with
\begin{eqnarray*}
f_n^+(x) & = & \sum_k p_n^+(k)\; x^{k}\\
f_n^-(x) & = & \sum_k p_n^-(k)\; x^{k},
\end{eqnarray*}
where the summation index $k$ runs from $-n$ to $n$ with step 2.
Let
$$
\mathbf{f}_n(x) =
\left(\begin{array}{c}
f_n^+(x)\\[0.2cm]
f_n^-(x)
\end{array}\right)
$$
and
$$
{\mathbf M}(x) =
\left(\begin{array}{cc}
  \frac{1}{2}\left(1+\varepsilon q\right)\; x
&  \frac{1}{2}\left(1-\varepsilon q\right) \; x\\
\\
  \frac{1}{2}\left(1-\varepsilon q\right) \; \frac{1}{x}
&   \frac{1}{2}\left(1+\varepsilon q\right)\; \frac{1}{x}\\
\end{array}\right).
$$
Then
\begin{equation}
\mathbf{f}_n(x) = \mathbf{M}(x) \; \mathbf{f}_{n-1}(x).
\label{1fiter}
\end{equation}
Iterating this recursion relation, we obtain
\begin{equation}
\mathbf{f}_n(x) = \mathbf{M}^{n-1}(x) \; \mathbf{f}_1(x),
\label{recur-1fn}
\end{equation}
where
$$
\mathbf{f}_1(x) =
\left(\begin{array}{c}
f_{1}^+(x)\\[0.2cm]
f_{1}^-(x)
\end{array}\right)=
\left(\begin{array}{c}
\frac{x}{2}\\[0.2cm]
\frac{1}{2x}
\end{array}\right).
$$

Let $\mathbf{R}(x)$ be the transformation matrix such that
$$
\mathbf{D}(x) = \mathbf{R}^{-1}(x) \; \mathbf{M}(x) \; \mathbf{R}(x)
=\left(\begin{array}{cc}\lambda_1(x) & 0\\[0.2cm]
                         0 & \lambda_2(x)
\end{array}\right),
$$
where $\lambda_1(x)$ and $\lambda_2(x)$ are the eigenvalues of $\mathbf{M}(x)$.
Then
$$
\mathbf{f}_n(x) =
\mathbf{R}(x)\; \mathbf{D}^{n-1}(x)\; \mathbf{R}^{-1}(x)\; \mathbf{f}_1(x)
$$

The above relation involves the eigenvalues and eigenvectors of
$\mathbf{M}(x)$ which can be readily obtained.
However, the exact expression
for the generating function of $p_n(k)$ is cumbersome to manipulate,
and does not easily yield the variance. Therefore, we resort to an
alternative and more direct method, which allows for a recursive
evaluation of $<R_n^2>$:
\begin{eqnarray*}
<R_n^2> & = & <(S_1+S_2+\cdots+S_n)^2>\\
        & = & <(S_1+S_2+\cdots+S_{n-1})^2> \\
        &  & \qquad + 2<S_n\;(S_1+S_2+\cdots+S_{n-1})> + <S_n^2>\\
        & = & <R_{n-1}^2> + 2<S_n\;(S_1+S_2+\cdots+S_{n-1})> + <S_n^2>.\\
\end{eqnarray*}
As will appear below, the correlation coefficient $<S_n S_{n-k}>$ does not
depend explicitly upon $n$, thus we use the notation
$$
c_k = <S_n S_{n-k}>,
$$
and obtain
$$
<R_n^2> = <R_{n-1}^2> + 2 (c_1+c_2 + \cdots + c_{n-1}) + 1,
$$
where we have taken into account that $<S_n^2>=1$. Evaluating the $c_k$'s,
we get
\begin{eqnarray*}
c_1 & = & <S_nS_{n-1}> = p\;<X_nS_{n-1}> + \;\varepsilon q\;<S_{n-1}^2>
=\varepsilon q\\
c_2 & = & <S_nS_{n-2}> = p\;<X_nS_{n-2}> + \; \varepsilon q\;<S_{n-1}S_{n-2}>
=\varepsilon qc_1=q^2\\
    & \cdots & \\
c_k & = & <S_nS_{n-k}> = p\;<X_nS_{n-k}> + \varepsilon q\;<S_{n-1}S_{n-k}>
=\varepsilon qc_{k-1}=(\varepsilon q)^k.
\end{eqnarray*}
Substituting the $c_k$'s in the expression of $<R_n^2>$ we obtain
\begin{eqnarray*}
<R_n^2> & = & <R_{n-1}^2> +\; 2\;(\varepsilon q+q^2+\cdots
+(\varepsilon q)^{n-1}) + 1\\
        & = & <R_{n-1}^2> +
\; 2\;\frac{\varepsilon q-(\varepsilon q)^n}{1-\varepsilon q}+1,\\
\end{eqnarray*}
and iteration of this recursion relation yields
\begin{eqnarray}
<R_n^2> & = & n + \frac{2}{(1-\varepsilon q)^2}
\;\left((n-1)\varepsilon q - nq^2+(\varepsilon q)^{n+1}\right)\nonumber\\
        & = & n + \frac{2}{(1-\varepsilon (1-p))^2}\times\nonumber\\
        &   & \quad \left((n-1)\varepsilon (1-p) - n(1-p)^2
+(\varepsilon)^{n+1}(1-p)^{n+1}\right).\label{varofp}
\end{eqnarray}

It is quite remarkable that for a similar random walk on a $d$-dimensional
simple cubic lattice, the above result is preserved. Of course, in this case,
the sequence of one-dimensional $X_k$'s is replaced by a sequence of
$d$-dimensional random vectors ${\mathbf X}_k$'s, and each elementary step
can be performed along each of the $d$ axes, either in the positive or the
negative direction.
For any fixed nonzero value of $p$, when $n$ goes to infinity, the term
$(1-p)^{n+1}$ goes to zero, and $<R^2_n>$ behaves as
$$
\frac{1+\varepsilon q}{1-\varepsilon q}\;n,
$$
and the random walk is eventually Gaussian at very large $n$.
However, when $q$ goes to 1, according to whether $\varepsilon$ is equal to
$+1$ or $-1$, the prefactor of $n$ goes, respectively, either to infinity
or to zero. Consequently, if $p$ is very small, the asymptotic behavior
of $<R_n^2>$ should be carefully analyzed, distinguishing the two cases
$\varepsilon=1$ and $\varepsilon=-1$.

\subsubsection{Persistent random walk ($\varepsilon=1$).} Replacing $\varepsilon$ by $+1$ in
\ref{varofp}, and expanding the resulting expression in powers of $p$,
we obtain
$$
<R_n^2> = n + \frac{2}{p^2}\left(\frac{n^2p^2}{2}-\frac{np^2}{2}
-\frac{n(n^2-1)p^3}{6}+O(p^4)\right),
$$
which shows that, as expected, when $p$ tends to zero, $<R_n^2>$ tends to
$n^2$. Therefore, for a small fixed value of $p$, when $n$ increases,
we expect, for the variance, a crossover from a $n^2$ behavior to an $n$
behavior. This crossover is controlled by the term $(1-p)^{n+1}$ in
\ref{varofp}, and the above expansion of $<R_n^2>$ shows that the relevant
parameter is the variable $a=n p$. To characterize this behavior, let
define the exponent
$$
E(n,p) = \frac{\partial\log<R_n^2>}{\partial\log n},
$$
which is twice the Hurst exponent. A simple calculation yields
$$
\lim_{p\to 0} E\left(\frac{a}{p}, p\right)
= \frac{a(1-\mathrm{e}^{-a})}{a+\mathrm{e}^{-a}-1}.
$$
This expression shows that, if $n$ is large but small compared to $1/p$ (\textit{i.e.},
$a$ is small), the exponent $E$ tends to 2, whereas if $n$ is large compared to
$1/p$ ($a$ is large), $E$ tends to 1. This result is illustrated in
Figure~\ref{fig:M1H5m5+} for $p=0.00005$. Note that the crossover takes
place around $n=\mathrm{e}^{11}\approx 6\,10^4$, $np$ being of the order of unity.

\begin{figure}[h]
\centerline{\includegraphics*[scale=0.8]{M1H5m5+.eps}}
\caption{\label{fig:M1H5m5+}\textit{Exponent $E(n,p)$ of the persistent walk
as a function of $\log n$, for $p=5\,10^{-5}$ and $n_{\max}=1.6\,10^7$.}}
\end{figure}

The crossover from $E(n,p)=2$ to $E(n,p)=1$, clearly observed in the above
figure, where $p$ has a small fixed value and $n$ varies, can also be evidenced
from the figures representing
the probability distribution for a fixed value of $p$:

\begin{itemize}
\item Figure~~\ref{fig:M1n100p2m3+}: at small $p$ ($np\ll 1$) most of the
weight of the distribution is concentrated at the edges of the distribution range.
\item Figure~\ref{fig:M1n100p2m1+}: at large $p$ ($np\gg 1$) the significant
weight of the distribution spreads out in a Gaussian-like profile, with a
width roughly proportional to $\sqrt{n}$.
\end{itemize}

\vspace{0.3cm}
\begin{figure}[h]
\noindent
\begin{minipage}{0.46\linewidth}
\centering\epsfig{figure=M1n100p2m3+.eps, width=\linewidth}
\caption{\label{fig:M1n100p2m3+}\textit{$p_n(k)$ for $M=1$,
$\varepsilon=1$, $p=0.002$ and $n=100$.}}
\end{minipage}\hfill
\begin{minipage}{0.46\linewidth}
\centering\epsfig{figure=M1n100p2m1+.eps, width=\linewidth}
\caption{\label{fig:M1n100p2m1+}\textit{$p_n(k)$ for $M=1$,
$\varepsilon=1$, $p=0.2$ and $n=100$.}}
\end{minipage}
\end{figure}

The fulfillment of both conditions (with perhaps the possibility that, in
the case corresponding to $np\ll 1$, the localization takes place at various
points of the distribution range, and not necessarily at the edges,
as, for example, for the antipersistent walk discussed below) will be from
now on considered as a sufficient condition for the occurrence of a crossover.

\subsubsection{Antipersistent random walk ($\varepsilon=-1$).} Clearly, if
$p=0$ the walk will be a zigzag
walk, and $<R_n^2>=0$ (resp. $<R_n^2>=1$) if $n$ is even (resp. odd). If $p$
is small, a calculation similar to the previous one shows that:

$\bullet$ If $n p\gg 1$, $<R_n^2>$ behaves as $\frac{1}{2}n p$, and the
oscillatory terms related to the parity of $n$ are negligible.

$\bullet$ If $n p\ll 1$, we find that $<R_n^2>$ behaves as $n p$ when $n$ is
even, and as $1$ when $n$ is odd. Both expressions are the order of unity or
less, so that $<R_n^2>$ behaves as $n^0$.

Figure\ref{fig:M1H5m5-} represents the exponent characterizing the behavior
of the variance. In order to avoid, for values of $n$ such that $np\ll 1$,
the spurious behavior of $<R_n^2>$ in $n$ for even values of $n$, we have
only considered the odd values of $n$.

\begin{figure}[h]
\centerline{\includegraphics*[scale=0.8]{M1H5m5-.eps}}
\caption{\label{fig:M1H5m5-}\textit{Exponent $E(n,p)$ of the antipersistent walk
as a function of $\log n$ ($n$ odd),
for $p=5\,10^{-5}$ and $n_{\max}=1.6\,10^7$.}}
\end{figure}

\section{Linear memory function.}

We first address explicitly the case $M=2$, as a significant illustration
of the complexity which manifests as soon as $M>1$. Even in this case, a closed
form for the probability distribution is  already difficult to write
down. However, we provide a formalism that can be fruitfully applied, at least
for numerical purposes, to larger values of $M$.

\subsection{Probability distribution for $M=2$.}

Let $p^{\sigma_1\sigma_2}(k,n)$ denote the probability that, after
$n$ steps, the random walker is at site $k$ and the steps $S_{n-1}$
and $S_n$ are , respectively, equal to $\sigma_1$ and $\sigma_2$, that is
$$
p_n^{\sigma_1\sigma_2}(k)=P((R_n=k)\cap ((S_{n-1}=\sigma_1)\cap(S_n=\sigma_2))).
$$
If $S_{n-1}=\sigma_1$ and $S_n=\sigma_2$, then $R_{n-1}=k-\sigma_2$, and
$p_n^{\sigma_1\sigma_2}(k)$ is a linear combination of two probabilities
of the form $p_{n-1}^{\sigma\sigma_1}(k-\sigma_2)$ where $\sigma=\pm 1$:
\begin{eqnarray}
p_n^{\sigma_1\sigma_2}(k) & = & \frac{1}{2}
\left(1+q(a_1\sigma_1\sigma_2\varepsilon_1+a_2\sigma_2\varepsilon_2)\right)
p_{n-1}^{+\sigma_1}(k-\sigma_2)\nonumber\\
&  & \quad + \frac{1}{2}
\left(1+q(a_1\sigma_1\sigma_2\varepsilon_1-a_2\sigma_2\varepsilon_2)\right)
p_{n-1}^{-\sigma_1}(k-\sigma_2).
\label{p12}
\end{eqnarray}
To this equation we associate the equation expressing
$p_n^{\sigma_1{\overline{\sigma_2}}}(k)$, where ${\overline\sigma_2}=-\sigma_2$,
since this probability is a linear combination of the two probabilities
$p_{n-1}^{\sigma\sigma_1}(k-\overline{\sigma_2}) =
p_{n-1}^{\sigma\sigma_1}(k+\sigma_2)$ where $\sigma=1$ or $-1$.
We have
\begin{eqnarray}
p_n^{\sigma_1{\overline\sigma_2}}(k) & = & \frac{1}{2}
\left(1+q(a_1\sigma_1{\overline\sigma_2}\varepsilon_1
+a_2{\overline\sigma_2}\varepsilon_2)\right)
p_{n-1}^{+\sigma_1}(k-{\overline\sigma_2})\nonumber\\
&  & \quad + \frac{1}{2}
\left(1+q(a_1\sigma_1{\overline\sigma_2}\varepsilon_1
-a_2{\overline\sigma_2}\varepsilon_2)\right)
p_{n-1}^{-\sigma_1}(k-{\overline\sigma_2}).
\label{p12bar}
\end{eqnarray}
The generating function of the probability distribution of $<R_n>$ is defined by
$$
f_n(x) = f_n^{++}(x) + f_n^{+-}(x) + f_n^{-+}(x) + f_n^{--}(x),
$$
with
$$
f_n^{\sigma_1\sigma_2}(x) =  \sum_k p_n^{\sigma_1\sigma_2}(k)\; x^{k},
$$
where the summation index $k$ runs from $-n$ to $n$ with step 2.
From Equations~\ref{p12} and \ref{p12bar}, replacing $\sigma_1$ and
$\sigma_2$ by all their possible respective values, we obtain
\begin{equation}
\left(\begin{array}{c}
f_n^{++}(x)\\[0.2cm]
f_n^{+-}(x)
\end{array}\right)
=
\mathbf{M^+}
\left(\begin{array}{c}
f_{n-1}^{++}(x)\\[0.2cm]
f_{n-1}^{-+}(x)
\end{array}\right),
\label{fMplus}
\end{equation}
where
\begin{equation}
\mathbf{M^+}(x) = \left(\begin{array}{cc}
\frac{1}{2}
\left(1+q(a_1\varepsilon_1+a_2\varepsilon_2)\right)\; x
& \frac{1}{2}
\left(1+q(a_1\varepsilon_1-a_2\varepsilon_2)\right)\;x\\[0.2cm]
\frac{1}{2}
\left(1-q(a_1\varepsilon_1
+a_2\varepsilon_2)\right)\; \frac{1}{x}
&\frac{1}{2}
\left(1-q(a_1\varepsilon_1
-a_2\varepsilon_2)\right)\;\frac{1}{x}
\end{array}\right),
\label{Mplus}
\end{equation}
and
\begin{equation}
\left(\begin{array}{c}
f_n^{-+}(x)\\[0.2cm]
f_n^{--}(x)
\end{array}\right)
=
\mathbf{M^-}
\left(\begin{array}{c}
f_{n-1}^{+-}(x)\\[0.2cm]
f_{n-1}^{-+}(x)
\end{array}\right),
\label{fMminus}
\end{equation}
where
\begin{equation}
\mathbf{M^-}(x) = \left(\begin{array}{cc}
\frac{1}{2}\left(1-q(a_1\varepsilon_1-a_2\varepsilon_2)\right)\; x
&\frac{1}{2}\left(1-q(a_1\varepsilon_1+a_2\varepsilon_2)\right)\; x\\[0.2cm]
\frac{1}{2}\left(1+q(a_1\varepsilon_1-a_2\varepsilon_2)\right)\;\frac{1}{x}
&\frac{1}{2}\left(1+q(a_1\varepsilon_1+a_2\varepsilon_2)\right)
\;\frac{1}{x}
\end{array}\right).
\label{Mminus}
\end{equation}

Equations~\ref{fMplus} and \ref{fMminus} can be grouped together in a unique
equation that reads
$$
\left(\begin{array}{c}
f_n^{++}(x)\\[0.18cm]
f_n^{+-}(x)\\[0.18cm]
f_n^{-+}(x)\\[0.18cm]
f_n^{--}(x)
\end{array}\right)
=
\left(\begin{array}{cc}
\mbox{\huge{$\mathbf{M^+}(x)$}} &  \mbox{\huge{$\mathbf{0}$}}\\[0.8cm]
\mbox{\huge{$\mathbf{0}$}} & \mbox{\huge{$\mathbf{M^-}(x)$}}
\end{array}\right)
\left(\begin{array}{c}
f_{n-1}^{++}(x)\\[0.18cm]
f_{n-1}^{-+}(x)\\[0.18cm]
f_{n-1}^{+-}(x)\\[0.18cm]
f_{n-1}^{--}(x)
\end{array}\right).
$$
Since the column vector on the right hand side of the above equation is not
correctly ordered, we have to introduce a $4\times 4$ permutation matrix, and
write this equation under the final form
\begin{equation}
\mathbf{f}_n(x)  =  \mathbf{M}(x)\;\mathcal{P}_4\; \mathbf{f}_{n-1}(x)
\label{2fiter}
\end{equation}
where
$$
\mathbf{f}_n(x) =
\left(\begin{array}{c}
f_n^{++}(x)\\[0.2cm]
f_n^{+-}(x)\\[0.2cm]
f_n^{-+}(x)\\[0.2cm]
f_n^{--}(x)\\
\end{array}\right),
$$

$$
\mathbf{M}(x) =
\left(\begin{array}{cc}
\mathbf{M^+}(x) &  \mathbf{0}\\[0.2cm]
\mathbf{0} & \mathbf {M^-}(x)
\end{array}\right)
$$
and
$$
\mathcal{P}_4=
\left(\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 1
\end{array}\right).
$$

Iterating Relation~\ref{2fiter}, we obtain
\begin{equation}
\mathbf{f}_n(x)
= \left(\mathbf{M}(x)\;\mathcal{P}_4\right)^{n-2} \; \mathbf{f}_2(x),
\label{recur-2fn}
\end{equation}
where
$$
\mathbf{f}_2(x) =
\left(\begin{array}{c}
f_{2}^{++}(x)\\[0.2cm]
f_{2}^{+-}(x)\\[0.2cm]
f_{2}^{-+}(x)\\[0.2cm]
f_{2}^{--}(x)\\
\end{array}\right)=
\left(\begin{array}{c}
\frac{x^2}{4}\\[0.2cm]
\frac{1}{4}\\[0.2cm]
\frac{1}{4}\\[0.2cm]
\frac{1}{4x^2}
\end{array}\right).
$$

\vspace{0.3cm}
\begin{figure}[t]
\noindent
\begin{minipage}{0.46\linewidth}
\centering\epsfig{figure=LM2n50p5m3++.eps, width=\linewidth}
\caption{\label{fig:LM2n50p5m3++}\textit{$p_n(k)$ for
$M=2$, $a_1=0.5$, $a_2=0.5$, $\varepsilon_1=1$, $\varepsilon_2=1$, $p=0.005$,
and $n=50$.}}
\end{minipage}\hfill
\begin{minipage}{0.46\linewidth}
\centering\epsfig{figure=LM2n50p5m1++.eps, width=\linewidth}
\caption{\label{fig:LM2n50p5m1++}\textit{$p_n(k)$ for
$M=2$, $a_1=0.5$, $a_2=0.5$, $\varepsilon_1=1$, $\varepsilon_2=1$, $p=0.5$,
and $n=50$.}}
\end{minipage}
\end{figure}

Further progress toward the expression of the generating function $f_n(x)$
requires the explicit diagonalization of the $4\times 4$ matrix
${\mathbf M}(x)\mathcal{P}_4$, followed by the procedure outlined in the $M=1$
case. The formal derivation of $f_n(x)$ is extremely complicated, and would
reveal untractable for $M>2$ as will be seen below.
However, the formalism is well suited to numerically evaluate the probability
distribution of $R_n$ recursively using~\ref{recur-2fn}.
Figures~\ref{fig:LM2n50p5m3++} and \ref{fig:LM2n50p5m1++}
show the probability distribution $p_n(k)$ when $a_1=a_2=0.5$ and
$\varepsilon_1=\varepsilon_2=1$
for two values of $p$ to illustrate cases $np\ll 1$ and $np\gg 1$.

Clearly, at the light of the comment at the end of subsection 3.0.1, the set
of these two figures is a sure sign of the existence of a crossover.

\subsection{Variance for $M=2$.}

Correlations are more tedious to evaluate than in the case $M=1$. Let
$$
c_{1,\ell} = <S_\ell S_{\ell+1}>;
$$
then
\begin{eqnarray*}
c_{1,1} & = & <S_1 S_2> = 0\\
c_{1,2} & = & <S_2 S_3> = <S_2 (p X_3 +
q(a_1\varepsilon_1S_2 + a_2\varepsilon_2 S_1))> = qa_1\varepsilon_1
\end{eqnarray*}
and, for $\ell\ge 3$,
\begin{eqnarray*}
c_{1,\ell} & = & <S_\ell S_{\ell+1}>
= <S_\ell (pX_{\ell+1} + qa_1\varepsilon_1 S_\ell
+ qa_2\varepsilon_2 S_{\ell-1})> \\
        & = & qa_1\varepsilon_1 + qa_2\varepsilon_2 c_{1,\ell-1}.
\end{eqnarray*}
Hence
\begin{equation}
c_{1,\ell} = qa_1\varepsilon_1
\frac{1-(qa_2\varepsilon_2)^{\ell-1}}{1-qa_2\varepsilon_2}\quad(\ell\ge 1)
\label{correl-1ell}
\end{equation}
Similarly, if
$$
c_{2,\ell} = <S_\ell S_{\ell+2}>;
$$
then, for all positive integers $\ell$,
\begin{eqnarray*}
c_{2,\ell} & = & <S_\ell S_{\ell+2}>
= <S_\ell (pX_{\ell+2} + q(a_1\varepsilon_1 S_{\ell+1}
+ a_2\varepsilon_2 S_\ell))>\\
        & = & qa_1\varepsilon_1 c_{1,\ell} + qa_2\varepsilon_2
\end{eqnarray*}
so that
\begin{eqnarray}
c_{2,\ell} = \frac{(qa_1\varepsilon_1)^2}
{1-qa_2\varepsilon_2}((1-qa_2\varepsilon_2)^{\ell-1}) + qa_2\varepsilon_2
\quad(\ell\ge 1).
\label{correl-2ell}
\end{eqnarray}
More generally, for $r>2$,
\begin{eqnarray}
c_{r,\ell} & = & <S_\ell S_{\ell+r}> = <S_\ell (p X_{\ell+r} +
    q(a_1\varepsilon_1 S_{k+r-1} + a_2\varepsilon_2 S_{k+r-2}))>\nonumber\\
        & = & q(a_1\varepsilon_1 c_{r-1,\ell} + a_2\varepsilon_2 c_{r-2,\ell}).
\label{correl-rell}
\end{eqnarray}
From this recursion relation, we obtain the following expression of $c_{r,\ell}$
\begin{equation}
c_{r,\ell} = \frac{c_{1,\ell}\lambda_2-c_{2,\ell}}{\lambda_2-\lambda_1}
\lambda_1^{r-1}
+ \frac{c_{2,\ell}-c_{1,\ell}\lambda_1}{\lambda_2-\lambda_1}
\lambda_2^{r-1},\label{correlofr}
\end{equation}
where $\lambda_1$ and $\lambda_2$ are the roots of
$$
\lambda^2-qa_1\varepsilon_1\lambda-qa_2\varepsilon_2=0.
$$
It is easy to verify that $|\lambda_1|\le 1$ and $|\lambda_2|\le 1$, the equal
sign being possible if $p=0$. For $p>0$, these conditions imply that the
correlations $c_{r,\ell}$ decrease exponentially in $r$, and this, in turn,
will ensure a linear dependence in $n$ for the variance.

Since the variance is given by
\begin{eqnarray*}
<R_n^2> & = & <(S_1+S_2+\cdots+S_n)^2>\nonumber\\
& = & n + 2\sum_{\ell=1}^{n-1}\sum_{r=1}^{n-\ell}c_{r,\ell},
\label{LMV}
\end{eqnarray*}
to find its its explicit expression, we need to evaluate the double summation.
Summing first over $r$, we obtain
\begin{small}
\begin{eqnarray*}
\sum_{r=1}^{n-\ell}c_{r,\ell} & = &
\frac{1-\lambda_1^{n-\ell}}{(\lambda_2-\lambda_1)(1-\lambda_1)}
\left(\frac{qa_1\varepsilon_1}{1-qa_2\varepsilon_2}
(1-(qa_2\varepsilon_2)^{\ell-1})(\lambda_2-qa_1\varepsilon_1)-qa_2\varepsilon_2\right)\\
& & + \frac{1-\lambda_2^{n-\ell}}{(\lambda_2-\lambda_1)(1-\lambda_2)}
\left(\frac{qa_1\varepsilon_1}{1-qa_2\varepsilon_2}
(1-(qa_2\varepsilon_2)^{\ell-1})(qa_1\varepsilon_1-\lambda_1)+qa_2\varepsilon_2\right),
\end{eqnarray*}
\end{small}
where we have replaced $c_{2,\ell}$ by its expression given by \ref{correl-2ell}.
Summing now over $\ell$ finally yields
\begin{small}
\begin{eqnarray*}
<R_n^2> & = & n + \frac{2(n-1)}{(\lambda_2-\lambda_1)(1-\lambda_1)}\left(
\frac{qa_1\varepsilon_1(\lambda_2-qa_1\varepsilon_1)}{1-qa_2\varepsilon_2}-qa_2\varepsilon_2
\right)\nonumber\\
& & - \frac{2}{(\lambda_2-\lambda_1)(1-\lambda_1)}\times
\frac{qa_1\varepsilon_1(\lambda_2-qa_1\varepsilon_1)}{1-qa_2\varepsilon_2}\times
\frac{1-(qa_2\varepsilon_2)^{n-1}}{1-qa_2\varepsilon_2}\nonumber\\
& & -\frac{2}{(\lambda_2-\lambda_1)(1-\lambda_1)}\left(
\frac{qa_1\varepsilon_1(\lambda_2-qa_1\varepsilon_1)}{1-qa_2\varepsilon_2}-qa_2\varepsilon_2
\right)\frac{\lambda_1(1-\lambda_1^{n-1})}{1-\lambda_1}\nonumber\\
& & + \frac{2}{(\lambda_2-\lambda_1)(1-\lambda_1)}\times
\frac{qa_1\varepsilon_1(\lambda_2-qa_1\varepsilon_1)}{1-qa_2\varepsilon_2}\times
\frac{\lambda_1(\lambda_1^{n-1}-(qa_2\varepsilon_2)^{n-1})}{\lambda_1-qa_2\varepsilon_2}
\nonumber\\
& & + \frac{2(n-1)}{(\lambda_2-\lambda_1)(1-\lambda_2)}\left(
\frac{qa_1\varepsilon_1(qa_1\varepsilon_1-\lambda_1)}{1-qa_2\varepsilon_2}+qa_2\varepsilon_2
\right)\nonumber\\
& & - \frac{2}{(\lambda_2-\lambda_1)(1-\lambda_2)}\times
\frac{qa_1\varepsilon_1(qa_1\varepsilon_1-\lambda_1)}{1-qa_2\varepsilon_2}\times
\frac{1-(qa_2\varepsilon_2)^{n-1}}{1-qa_2\varepsilon_2}\nonumber\\
& & -\frac{2}{(\lambda_2-\lambda_1)(1-\lambda_2)}\left(
\frac{qa_1\varepsilon_1(qa_1\varepsilon_1-\lambda_1)}{1-qa_2\varepsilon_2}+qa_2\varepsilon_2
\right)\frac{\lambda_2(1-\lambda_2^{n-1})}{1-\lambda_2}\nonumber\\
& & + \frac{2}{(\lambda_2-\lambda_1)(1-\lambda_2)}\times
\frac{qa_1\varepsilon_1(qa_1\varepsilon_1-\lambda_1)}{1-qa_2\varepsilon_2}\times
\frac{\lambda_2(\lambda_2^{n-1}-(qa_2\varepsilon_2)^{n-1})}{\lambda_2-qa_2\varepsilon_2}
\end{eqnarray*}
\end{small}

For our purpose, a complete analytic discussion of this expression
would take us too far. Instead, we will illustrate the behavior of
the variance with four typical choices for the set of the parameters
$\varepsilon_1,\varepsilon_2,a_1,a_2$ in Figures~\ref{fig:PM2a5m1p5m5++},
\ref{fig:PM2a5m1p5m5+-}, \ref{fig:PM2a5m1p5m5-+}, and
\ref{fig:PM2a5m1p5m5--}. Note that we obtain crossovers, respectively,
similar to those obtained for the persistent and antipersistent walks
of the Taylor's model when $\varepsilon_2=1$ and $\varepsilon_1$ is either equal
to $1$ or $-1$. On the contrary when $\varepsilon_2=-1$, There is no crossover,
the variance behaves as $n$ for both signs of $\varepsilon_1$.

\vspace{0.3cm}
\begin{figure}[h]
\noindent
\begin{minipage}{0.46\linewidth}
\centering\epsfig{figure=LM2a5m1p5m5++.eps, width=\linewidth}
\caption{\label{fig:PM2a5m1p5m5++}\textit{Log-log plot of $<R_n^2>$ vs.n
for $M=2$, $\varepsilon_1=1$, $\varepsilon_2=1$, $a_1=a_2=0.5$, $p=5\,10^{-5}$,
$n_{\max}=1.6\,10^8$.}}
\end{minipage}\hfill
\begin{minipage}{0.46\linewidth}
\centering\epsfig{figure=LM2a5m1p5m5+-.eps, width=\linewidth}
\caption{\label{fig:PM2a5m1p5m5+-}\textit{Log-log plot of $<R_n^2>$ vs.n
for $M=2$, $\varepsilon_1=1$, $\varepsilon_2=-1$, $a_1=a_2=0.5$, $p=5\,10^{-5}$,
$n_{\max}=1.6\,10^8$.}}
\end{minipage}
\end{figure}

\vspace{0.3cm}
\begin{figure}[h]
\noindent
\begin{minipage}{0.46\linewidth}
\centering\epsfig{figure=LM2a5m1p5m5-+.eps, width=\linewidth}
\caption{\label{fig:PM2a5m1p5m5-+}\textit{Log-log plot of $<R_n^2>$ vs.n
for $M=2$, $\varepsilon_1=-1$, $\varepsilon_2=1$, $a_1=a_2=0.5$, $p=5\,10^{-5}$,
$n_{\max}=1.6\,10^8$.}}
\end{minipage}\hfill
\begin{minipage}{0.46\linewidth}
\centering\epsfig{figure=LM2a5m1p5m5--.eps, width=\linewidth}
\caption{\label{fig:PM2a5m1p5m5--}\textit{Log-log plot of $<R_n^2>$ vs.n
for $M=2$, $\varepsilon_1=-1$, $\varepsilon_2=-1$, $a_1=a_2=0.5$, $p=5\,10^{-5}$,
$n_{\max}=1.6\,10^8$.}}
\end{minipage}
\end{figure}

\subsection{Linear memory for $M>2$.}

The matrix method used for the case $M=2$ is formally easy to generalize. It
yields, for the generating function, a recursion relation similar to
\ref{2fiter}, where $\mathbf{f}_n$ is a $2^M$-component vector,
$\mathbf{M}$ a $2^M\times 2^M$ matrix and $\mathcal{P}_4$ is replaced by a
$2^M\times 2^M$ permutation matrix $\mathcal{P}_{2^M}$. With increasing values
of $M$ it becomes rapidly untractable. Unfortunately, due to the large number
of parameters when $M>2$, even the variance $<R_n^2>$ is unmanageable.
This is not the case for the single product memory, as will be seen
in the next section. However, there is, at least, one case for which we can
predict that the variance exhibits the crossover observed in the persistent
walk of the Taylor's model. This occurs when all the $\varepsilon_j$
($j=1,2,\ldots,M$) are equal to 1, since in this case, it is easy to verify
that for $p=0$, the variance behaves as $n^2$.

\section{Single product memory function.}

Here again we will first consider the case $M=2$ as an illustration
of the complexity which manifests as soon as $M>1$. Since we will
only consider the case of a single product, the notation
$\varepsilon_{j_1j_2\cdots j_r}$ will be replaced by $\varepsilon$.

\subsection{Probability distribution for $M=2$.}

As in the linear memory function case,
$p_n^{\sigma_1\sigma_2}(k)$ denotes the probability
$P((R_n=k)\cap((S_{n-1}=\sigma_1)\cap(S_n=\sigma_2)))$,
and as above, if $S_{n-1}=\sigma_1$ and $S_n=\sigma_2$, then
$R_{n-1}=k-\sigma_2$, which implies that $p_n^{\sigma_1\sigma_2}(k)$
is a linear combination of the two probabilities
$p_{n-1}^{\sigma\sigma_1}(k-\sigma_2)$, where $\sigma = \pm 1$.
Since, with a probability $q$, $S_n=\varepsilon S_{n-1}S_{n-2}$, we have
$\sigma_2=\varepsilon\sigma\sigma_1$ or $\sigma=\varepsilon\sigma_1\sigma_2$.
Therefore, using notations similar to those of the linear case, we have
$$
\left(\begin{array}{c}
f_n^{++}(x)\\[0.2cm]
f_n^{+-}(x)
\end{array}\right)
=
\mathbf{M^+}
\left(\begin{array}{c}
f_{n-1}^{++}(x)\\[0.2cm]
f_{n-1}^{-+}(x)
\end{array}\right),
$$
where
\begin{equation}
\mathbf{M^+}(x) = \left(\begin{array}{cc}
\frac{1}{2}
(1+q\varepsilon)\; x
& \frac{1}{2}
(1-q\varepsilon)\;x\\[0.2cm]
\frac{1}{2}
(1-q\varepsilon)\; \frac{1}{x}
&\frac{1}{2}
(1+q\varepsilon)\;\frac{1}{x}
\end{array}\right),
\label{PMplus}
\end{equation}
and
$$
\left(\begin{array}{c}
f_n^{-+}(x)\\[0.2cm]
f_n^{--}(x)
\end{array}\right)
=
\mathbf{M^-}
\left(\begin{array}{c}
f_{n-1}^{+-}(x)\\[0.2cm]
f_{n-1}^{-+}(x)
\end{array}\right),
$$
where
\begin{equation}
\mathbf{M^-}(x) = \left(\begin{array}{cc}
\frac{1}{2}
(1-q\varepsilon)\; x
& \frac{1}{2}
(1+q\varepsilon)\;x\\[0.2cm]
\frac{1}{2}
(1+q\varepsilon)\; \frac{1}{x}
&\frac{1}{2}
(1-q\varepsilon)\;\frac{1}{x}
\end{array}\right).
\label{PMminus}
\end{equation}

The above equations can be grouped together in a unique
equation that reads
$$
\left(\begin{array}{c}
f_n^{++}(x)\\[0.18cm]
f_n^{+-}(x)\\[0.18cm]
f_n^{-+}(x)\\[0.18cm]
f_n^{--}(x)
\end{array}\right)
=
\left(\begin{array}{cc}
\mbox{\huge{$\mathbf{M^+}(x)$}} &  \mbox{\huge{$\mathbf{0}$}}\\[0.8cm]
\mbox{\huge{$\mathbf{0}$}} & \mbox{\huge{$\mathbf{M^-}(x)$}}
\end{array}\right)
\left(\begin{array}{c}
f_{n-1}^{++}(x)\\[0.18cm]
f_{n-1}^{-+}(x)\\[0.18cm]
f_{n-1}^{+-}(x)\\[0.18cm]
f_{n-1}^{--}(x)
\end{array}\right).
$$
Since the column vector on the right hand side of the above equation is not
correctly ordered, we have to introduce a $4\times 4$ permutation matrix, and
write this equation under the final form
\begin{equation}
\mathbf{f}_n(x)  =  \mathbf{M}(x)\;\mathcal{P}_4\; \mathbf{f}_{n-1}(x)
\label{P2fiter}
\end{equation}
where
$$
\mathbf{f}_n(x) =
\left(\begin{array}{c}
f_n^{++}(x)\\[0.2cm]
f_n^{+-}(x)\\[0.2cm]
f_n^{-+}(x)\\[0.2cm]
f_n^{--}(x)\\
\end{array}\right),
$$

$$
\mathbf{M}(x) =
\left(\begin{array}{cc}
\mathbf{M^+}(x) &  \mathbf{0}\\[0.2cm]
\mathbf{0} & \mathbf {M^-}(x)
\end{array}\right)
$$
and
$$
\mathcal{P}_4=
\left(\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 1
\end{array}\right).
$$

Iterating Relation~\ref{P2fiter}, we obtain
\begin{equation}
\mathbf{f}_n(x)
= \left(\mathbf{M}(x)\;\mathcal{P}_4\right)^{n-2} \; \mathbf{f}_2(x),
\label{recur-P2fn}
\end{equation}
where
$$
\mathbf{f}_2(x) =
\left(\begin{array}{c}
f_{2}^{++}(x)\\[0.2cm]
f_{2}^{+-}(x)\\[0.2cm]
f_{2}^{-+}(x)\\[0.2cm]
f_{2}^{--}(x)\\
\end{array}\right)=
\left(\begin{array}{c}
\frac{x^2}{4}\\[0.2cm]
\frac{1}{4}\\[0.2cm]
\frac{1}{4}\\[0.2cm]
\frac{1}{4x^2}
\end{array}\right).
$$

\vspace{0.3cm}
\begin{figure}[b]
\noindent
\begin{minipage}{0.46\linewidth}
\centering\epsfig{figure=PM2n50p5m3.eps, width=\linewidth}
\caption{\label{fig:PM2n50p5m3}\textit{$p_n(k)$ for
$M=2$, $\varepsilon=1$, $p=0.005$, and $n=50$.}}
\end{minipage}\hfill
\begin{minipage}{0.46\linewidth}
\centering\epsfig{figure=PM2n50p5m1.eps, width=\linewidth}
\caption{\label{fig:PM2n50p5m1}\textit{$p_n(k)$ for
$M=2$, $\varepsilon=1$, $p=0.5$, and $n=50$.}}
\end{minipage}
\end{figure}

\vspace{0.3cm}
\begin{figure}[t]
\noindent
\begin{minipage}{0.46\linewidth}
\centering\epsfig{figure=PM2n50p5m3-.eps, width=\linewidth}
\caption{\label{fig:PM2n50p5m3-}\textit{$p_n(k)$ for
$M=2$, $\varepsilon=-1$, $p=0.005$, and $n=50$.}}
\end{minipage}\hfill
\begin{minipage}{0.46\linewidth}
\centering\epsfig{figure=PM2n50p5m1-.eps, width=\linewidth}
\caption{\label{fig:PM2n50p5m1-}\textit{$p_n(k)$ for
$M=2$, $\varepsilon=-1$, $p=0.5$, and $n=50$.}}
\end{minipage}
\end{figure}

As in the linear case, further progress toward the expression of the generating
function $f_n(x)$ is extremely complicated, and would
reveal untractable for $M>2$.
However the formalism is well suited to numerically evaluate
the probability distribution of $R_n$ recursively using~\ref{recur-P2fn}.
Figures~\ref{fig:PM2n50p5m3}, \ref{fig:PM2n50p5m1},  \ref{fig:PM2n50p5m3-},
\ref{fig:PM2n50p5m1-} show the probability distribution $p_n(k)$
for two values of $p$ to illustrate cases $np\ll 1$ and $np\gg 1$ when
$\varepsilon=\pm1$.

At the light of the comment at the end of subsection 3.0.1,
Figures~\ref{fig:PM2n50p5m3}, \ref{fig:PM2n50p5m1}, \ref{fig:PM2n50p5m3-},
\ref{fig:PM2n50p5m1-} reveal the existence of crossovers. This property
will be confirmed directly in the next subsection.

\subsection{Variance for $M=2$.}

In order to evaluate recursively the variance, we need to determine
the correlations. We have
\begin{eqnarray*}
<S_\ell S_{\ell+1}> & = & \varepsilon q<S_\ell (S_{\ell}S_{\ell-1})>\\
             & = & \varepsilon q <S_{\ell-1}>=0;
\end{eqnarray*}
\begin{eqnarray*}
<S_\ell S_{\ell+2}> & = & \varepsilon q<S_\ell (S_{\ell+1}S_\ell)>\\
             & = & \varepsilon q <S_{\ell+1}>=0;
\end{eqnarray*}
\begin{eqnarray*}
<S_\ell S_{\ell+3}> & = & \varepsilon q<S_\ell (S_{\ell+2}S_{\ell+1})>\\
             & = & \varepsilon^2 q^2 <S_\ell((S_{\ell+1}S_\ell)S_{\ell+1}>\\
             & = & q^2;
\end{eqnarray*}
and, for $m>\ell+3$,
\begin{eqnarray*}
<S_\ell S_m> & = & \varepsilon q<S_\ell (S_{m-1}S_{m-2})>\\
             & = & \varepsilon^2 q^2 <S_\ell((S_{m-2}S_{m-3})S_{m-2}>\\
             & = & q^2 <S_\ell S_{m-3}>.
\end{eqnarray*}

These results provide all the correlations:
\begin{equation}
<S_\ell S_m> =
\begin{cases}
0 & \text{if $m-\ell$ is not a multiple of 3},\\[0.1cm]
q^{2(m-\ell)/3} & \text{if $m-\ell$ is a multiple of 3}.
\end{cases}
\label{correlPM2}
\end{equation}
Note that the correlations do not depend upon $\varepsilon$.
On the other hand, the variance satisfies the recursion relation
$$
<R_n^2> = <R_{n-1}^2> + 1 + 2<S_1S_n + S_2S_n + \cdots + S_{n-1}S_n>.
$$
Iterating this relation and taking into account \ref{correlPM2}, we
finally obtain
\begin{eqnarray}
<R_n^2> & = & n + (n-3)q^2+(n-6)q^4+\cdots+
(n-3\lfloor n/3\rfloor) q^{2\lfloor n/3\rfloor}\nonumber\\
& = & n + n q^2\frac{1-q^{2\lfloor n/3\rfloor}}{1-q^2}\nonumber\\
&  &\quad - 3q^2\frac{\lfloor n/3\rfloor q^{2\lfloor n/3\rfloor+2}-
\left(\lfloor n/3\rfloor+1\right)q^{2\lfloor n/3\rfloor} +1}
{(1-q^2)^2},\label{varPM2}
\end{eqnarray}
where $\lfloor x\rfloor$ denotes the largest integer less or equal to $x$.
On this expression of $<R_n^2>$ we readily verify that for $q=0$ the
variance is equal to $n$, while for $q=1$ it takes the value
$$
n + (2n-3)\lfloor n/3\rfloor - 3 (\lfloor n/3\rfloor)^2 = \lceil n^2/3\rceil,
$$
where $\lceil x\rceil$ denotes the smallest integer greater or equal to $x$.
Since the functions floor and ceiling are not differentiable, we define the
crossover exponent $E(n,p)$ by
$$
E(n,p)= \frac{n}{2}\frac{<R_{n+1}^2>(p)-<R_{n-1}^2>(p)}{<R_n^2>(p)}.
$$
Figure~\ref{fig:PM2H5m5} represents the variation of $E(n,p)$ as a function of
of $\log n$ for $p=5\,10^{-5}$.

\vspace{0.3cm}
\begin{figure}[h]
\centerline{\includegraphics*[scale=0.8]{PM2H5m5.eps}}
\caption{\label{fig:PM2H5m5}\textit{Exponent $E(n,p)$ as a function of
$\log n$, for $M=2$, $p=5\,10^{-5}$ and $n_{\max}=1.6\,10^7$.}}
\end{figure}

\subsection{Probability distribution for $M>2$.}

The matrix method used for the case $M=2$ is formally easy to generalize. It
yields, for the generating function, a recursion relation similar to
\ref{P2fiter}, where $\mathbf{f}_n$ is a $2^M$-component vector,
$\mathbf{M}$ a $2^M\times 2^M$ matrix and $\mathcal{P}_4$ is replaced by a
$2^M\times 2^M$ permutation matrix $\mathcal{P}_{2^M}$. With increasing values
of $M$ it becomes rapidly untractable. However, the variance $<R_n^2>$ can be
exactly evaluated as we will show in next subsection.

\subsection{Variance for $M>2$.}

In order to evaluate recursively the variance, we need to determine
the correlations. By definition, $<S_\ell S_m>=0$ if $\ell<m\le M$. This is
also true if $m=M+1$:
\begin{eqnarray*}
<S_\ell S_{M+1}> & = & q\varepsilon<S_\ell(S_1S_2\cdots S_\ell\cdots S_M)>\\
                 & = &q\varepsilon <S_1><S_2>\cdots<S_M>=0,
\end{eqnarray*}
due the fact that the random variables $S_1, S_2, \ldots, S_M$
are independent, and $<S_\ell^2>=1$.

Considering now the case: $<S_\ell S_m>$ (any $\ell$ and $m>\ell$), we first
note that
\begin{eqnarray*}
<S_\ell S_{\ell+M+1}> & = & q\varepsilon<S_\ell(S_{\ell+1}S_{\ell+2}\cdots S_{\ell+M})>\\
                 & = & (q\varepsilon)^2<(S_\ell S_{\ell+1}\cdots S_{\ell+M-1})^2>\\
                 & = & q^2,
\end{eqnarray*}
whereas, in the general case,
\begin{eqnarray}
<S_\ell S_{\ell+r}> & = & q\varepsilon
<S_\ell(S_{\ell+r-M}S_{\ell+r-M+1}\cdots S_{\ell+r-1})>\nonumber\\
                 & = & q^2<S_\ell S_{\ell+r-(M+1)}>,
\label{correliterPM}
\end{eqnarray}
where it is assumed that $\ell+r-(M+1)$ is positive but may be smaller
than $\ell$. Iterated application of~\ref{correliterPM} in the presence
of the initial conditions shows that, for
$m-\ell = \big\lfloor{\frac{m-\ell}{M+1}}\big\rfloor + b$ ($0\leq b<M+1$),
\begin{equation}
<S_\ell S_m> =
\begin{cases}
0 & \text{if $b>0$}\\
q^{2\big\lfloor\frac{m-\ell}{M+1}\big\rfloor} & \text{if $b=0$}\\
\end{cases}
\label{correlPMgeneral}
\end{equation}
Substitution of~\ref{correlPMgeneral} in the recursion relation satisfied by
the variance:
$$
<R_n^2> = <R_{n-1}^2> + 1 + 2<S_1S_n + S_2S_n + \cdots + S_{n-1}S_n>.
$$
yields
\begin{eqnarray}
<R_n^2> & = & n + (n-(M+1))q^2 + (n -2(M+1))q^4+\cdots\nonumber\\
& & +\left(n-\Big\lfloor\frac{n}{M+1}\Big\rfloor(M+1)\right)
q^{2\lfloor\frac{n}{M+1}\rfloor}\nonumber\\
& = & n + n q^2\frac{1-q^{2\lfloor\frac{n}{M+1}\rfloor}}{1-q^2}
-(M+1)q^2\nonumber\\
& & \times\frac{1 - \left(\big\lfloor\frac{n}{M+1}\big\rfloor+1\right)
q^{2\lfloor\frac{n}{M+1}\rfloor}+\big\lfloor\frac{n}{M+1}\big\rfloor
q^{2\lfloor\frac{n}{M+1}\rfloor+2}}{(1-q^2)^2},\label{varPMofq}
\end{eqnarray}
which is the generalization for any $M$ of the result~\ref{varPM2}
obtained for $M=2$. Note that $M=1$ is not a particular case of~\ref{varPMofq}.

When $q=0$, $<R_n^2>=n$, whereas, when $p$ goes to zero ($q\to 1$), $<R_n^2>$
tends to
$$
n + (2n-(M+1))\bigg\lfloor\frac{n}{M+1}\bigg\rfloor,
$$
which is of the order of $\frac{n^2}{M+1}$ for large $n$.
These two limiting behaviors are the signal of a crossover, which is evidenced
in Figure~\ref{fig:multiPMVp5m5}. This figure shows that the variance is a
decreasing function of $M$ at a given $n$.

\vspace{0.3cm}
\begin{figure}[h]
\centerline{\includegraphics*[scale=0.8]{multiPMVp5m5.eps}}
\caption{\label{fig:multiPMVp5m5}\textit{Log-log plot of the variance
as a function of $n$,
for $M=2,8,32,128$ (to be read downwards), $p=5\,10^{-5}$ and
$n_{\max}=1.6\,10^8$.}}
\end{figure}

The general expression of the variance is an even function of $q$, and,
therefore, does not depend upon $\varepsilon$. Such is not the case for the
probability distribution $p_n(k)$ of $R_n$ as in Figures~\ref{fig:PM5n13p0+} and
\ref{fig:PM5n13p0-}.

\vspace{0.3cm}
\begin{figure}[h]
\noindent
\begin{minipage}{0.46\linewidth}
\centering\epsfig{figure=PM5n13p0+.eps, width=\linewidth}
\caption{\label{fig:PM5n13p0+}\textit{$p_n(k)$ for
$M=5$, $\varepsilon=1$, $p=0$, and $n=13$. $<R_{13}^2>=29$.}}
\end{minipage}\hfill
\begin{minipage}{0.46\linewidth}
\centering\epsfig{figure=PM5n13p0-.eps, width=\linewidth}
\caption{\label{fig:PM5n13p0-}\textit{$p_n(k)$ for
$M=5$, $\varepsilon=-1$, $p=0$, and $n=13$. $<R_{13}^2>=29$.}}
\end{minipage}
\end{figure}

Here the probability distribution are symmetric. This is not always the case as
can be observed in Figures~\ref{fig:PM2n50p5m3}, \ref{fig:PM2n50p5m1},
\ref{fig:PM2n50p5m3-}, and \ref{fig:PM2n50p5m1-}.

\section{Conclusion and perspectives}

We have studied a family of correlated random walks with a finite memory
range. These walks are a generalization of the Taylor's walk.
We have established a matrix formalism for the generating functions of
the probability distributions of these walks. We have also derived in many
cases an analytic expression for the variance. As a function of the number
of steps $n$, for each of these walks, the variance becomes ultimately
linear in $n$ in the limit $n\to\infty$. However, at comparatively small $n$,
a different behavior of the variance can be observed leading to the existence
of a crossover phenomenon. This feature shows that, when analyzing experimental
results, one finds a Hurst exponent different from $\frac{1}{2}$, this does not
necessarily imply an infinite-range memory. If, experimentally feasible, one
should increase significantly the number of steps to cross-check the stability
of the exponent before concluding that the system under consideration exhibits
a non-Gaussian behavior.

The ultimate Gaussian behavior of the variance
at very large $n$ is a direct consequence of the exponential decay of the
step-step correlations as a function of the absolute value of the difference
between step indices. Therefore, in order to find a different behavior, one
has to consider step-step correlations decreasing more slowly than
exponentially. This is what may happen in the case of a power law decay
as can be shown as follows.
For a symmetric random walk, the variance is given by
\begin{eqnarray*}
<R_n^2> & = & n + 2 \sum_{\ell=1}^{n-1}\sum_{m=\ell+1}^n <S_\ell S_m>\\
& = & n + 2 \sum_{\ell=1}^{n-1}\sum_{r=1}^n <S_\ell S_{\ell+r}>\\
& = & n + 2 \sum_{\ell=1}^{n-1}\sum_{r=1}^n c_{r,\ell},
\end{eqnarray*}
where as above $c_{r,\ell}=<S_\ell S_{r+\ell}>$. If we now assume that,
for some reason,
$$
c_{r,\ell} = \frac{1}{r^\alpha}
$$
independently of $\ell$, then
$$
<R_n^2> = n + 2\sum_{r=1}^{n-1}\frac{n-r}{r^\alpha}.
$$

The possible non-Gaussian behavior of $<R_n^2>$ can only result from the
asymptotic behavior at large $n$ of the summation in the right hand side
of the above relation. According to the value of $\alpha$, the
asymptotic behaviors of
$$
S(n,\alpha) = \sum_{r=1}^{n-1}\frac{n-r}{r^\alpha}
$$
depends on $\alpha$ as follows:
\begin{itemize}
\item{if $\alpha>1$, $S(n,\alpha)\sim n\zeta(\alpha)$, where $\zeta$ is the Riemann
function.}
\item{if $\alpha=1$, $S(n,1)\sim n\log n$.}
\item{if $\alpha<1$, $S(n,\alpha)\sim\frac{1}{(1-\alpha)(2-\alpha)}n^{2-\alpha}$.}
\end{itemize}
Note that the power law decay for the step-step correlations is only a
sufficient condition to obtain a non-trivial Hurst exponent equal to $1-a/2$.

\section*{References}

{\parindent = 0pt

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An Introduction to Probability Theory and Its Applications,
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2388--2391, 1998.

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[5] C.-K. Peng, S. Buldyrev, A. Goldberger, S. Havlin, F. Sciortino,
M. Simons and E. Stanley, Long-range correlations in nucleotide sequences,
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{\bf 58,} 914--918, 1998.

[7] L. Liebovitch and W. Yang, Transition from persistent to antipersistent
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North-Holland, Amsterdam, 1994.

}

\end{document}

\begin{eqnarray*}
p^+(t,x) & = & \left(1-\frac{p}{2}\right) \; p^+(t-\tau, x-\xi) +
               \frac{p}{2} \; p_-(t-\tau, x-\xi)\\
p^-(t,x) & = & \frac{p}{2} \; p^+(t-\tau, x+\xi) +
               \left(1-\frac{p}{2}\right) \; p^-(t-\tau, x+\xi)\\
\end{eqnarray*}

\begin{eqnarray*}
p^+(t,x) & = & \left(1-\frac{p}{2}\right) \;\left(p^+(t,x) -
               \frac{\partial p^+}{\partial t}\;\tau -
               \frac{\partial p^+}{\partial x}\;\xi +
               \frac{1}{2}\;\frac{\partial^2 p^+}{\partial x^2}\;\xi^2\right)\\
          &  & \qquad +
               \frac{p}{2} \; \left(p^-(t,x) -
               \frac{\partial p^-}{\partial t}\;\tau -
               \frac{\partial p^-}{\partial x}\;\xi +
                \frac{1}{2}\;\frac{\partial^2 p^+}{\partial x^2}\;\xi^2\right)\\
p^-(t,x) & = & \frac{p}{2} \; \left(p^+(t,x) -
               \frac{\partial p^+}{\partial t}\;\tau +
               \frac{\partial p^+}{\partial x}\;\xi +
               \frac{1}{2}\;\frac{\partial^2 p^+}{\partial x^2}\;\xi^2\right)\\
          &  & \qquad +
               \left(1-\frac{p}{2}\right) \; \left(p_-(t,x) -
               \frac{\partial p^-}{\partial t}\;\tau +
               \frac{\partial p^-}{\partial x}\;\xi +
                \frac{1}{2}\;\frac{\partial^2 p^+}{\partial x^2}\;\xi^2\right)\\
\end{eqnarray*}

$$
\begin{array}{ccccccc}
<R_n^2> & = & <R_{n-1}^2> & + & 1 & + & 2<S_1S_n + S_2S_n
+ \cdots + S_{n-1}S_n>\\
<R_{n-1}^2> & = & <R_{n-2}^2> & + & 1 & + & 2<S_1S_{n-1}
+ \cdots + S_{n-2}S_{n-1}>\\
        & \cdots & \\
<R_2^2> & = & <R_1^2> & + & 1 & + & 2 <S_1S_2>\\
<R_1^2> & = & & & 1 &
\end{array}
$$