-% \vspace{-0.5em}
-% It is easy to associate a Markov Matrix $M$ to such a graph $G(f)$
-% as follows:
-
-% $M_{ij} = \frac{1}{n}$ if there is an edge from $i$ to $j$ in $\Gamma(f)$ and $i \neq j$; $M_{ii} = 1 - \sum\limits_{j=1, j\neq i}^n M_{ij}$; and $M_{ij} = 0$ otherwise.
-
-% \begin{xpl}
-% The Markov matrix associated to the function $f^*$ is
-
-% \[
-% M=\dfrac{1}{3} \left(
-% \begin{array}{llllllll}
-% 1&1&1&0&0&0&0&0 \\
-% 1&1&0&0&0&1&0&0 \\
-% 0&0&1&1&0&0&1&0 \\
-% 0&1&1&1&0&0&0&0 \\
-% 1&0&0&0&1&0&1&0 \\
-% 0&0&0&0&1&1&0&1 \\
-% 0&0&0&0&1&0&1&1 \\
-% 0&0&0&1&0&1&0&1
-% \end{array}
-% \right)
-% \]
-%\end{xpl}
-
-
-It is usual to check whether rows of such kind of matrices
-converge to a specific
-distribution.
-Let us first recall the \emph{Total Variation} distance $\tv{\pi-\mu}$,
-which is defined for two distributions $\pi$ and $\mu$ on the same set
-$\Omega$ by:
-$$\tv{\pi-\mu}=\max_{A\subset \Omega} |\pi(A)-\mu(A)|.$$
-% It is known that
-% $$\tv{\pi-\mu}=\frac{1}{2}\sum_{x\in\Omega}|\pi(x)-\mu(x)|.$$
-
-Let then $M(x,\cdot)$ be the
-distribution induced by the $x$-th row of $M$. If the Markov chain
-induced by
-$M$ has a stationary distribution $\pi$, then we define
-$$d(t)=\max_{x\in\Omega}\tv{M^t(x,\cdot)-\pi}.$$
-Intuitively $d(t)$ is the largest deviation between
-the distribution $\pi$ and $M^t(x,\cdot)$, which
-is the result of iterating $t$ times the function.
-Finally, let $\varepsilon$ be a positive number, the \emph{mixing time}
-with respect to $\varepsilon$ is given by
-$$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
-It defines the smallest iteration number
-that is sufficient to obtain a deviation lesser than $\varepsilon$.
+
+Let thus be given such kind of map.
+This article focusses on studying its iterations according to
+the equation~(\ref{eq:asyn}) with a given strategy.
+First of all, this can be interpreted as walking into its iteration graph
+where the choice of the edge to follow is decided by the strategy.
+Notice that the iteration graph is always a subgraph of
+$n$-cube augemented with all the self-loop, \textit{i.e.}, all the
+edges $(v,v)$ for any $v \in \Bool^n$.
+Next, if we add probabilities on the transition graph, iterations can be
+interpreted as Markov chains.
+
+\begin{xpl}
+Let us consider for instance
+the graph $\Gamma(f)$ defined
+in \textsc{Figure~\ref{fig:iteration:f*}.} and
+the probability function $p$ defined on the set of edges as follows:
+$$
+p(e) \left\{
+\begin{array}{ll}
+= \frac{2}{3} \textrm{ if $e=(v,v)$ with $v \in \Bool^3$,}\\
+= \frac{1}{6} \textrm{ otherwise.}
+\end{array}
+\right.
+$$
+The matrix $P$ of the Markov chain associated to the function $f^*$ and to its probability function $p$ is
+\[
+P=\dfrac{1}{6} \left(
+\begin{array}{llllllll}
+4&1&1&0&0&0&0&0 \\
+1&4&0&0&0&1&0&0 \\
+0&0&4&1&0&0&1&0 \\
+0&1&1&4&0&0&0&0 \\
+1&0&0&0&4&0&1&0 \\
+0&0&0&0&1&4&0&1 \\
+0&0&0&0&1&0&4&1 \\
+0&0&0&1&0&1&0&4
+\end{array}
+\right)
+\]
+\end{xpl}
+
+
+
+More generally, let $\pi$, $\mu$ be two distribution on $\Bool^n$. The total
+variation distance between $\pi$ and $\mu$ is denoted $\tv{\pi-\mu}$ and is
+defined by
+$$\tv{\pi-\mu}=\max_{A\subset \Bool^n} |\pi(A)-\mu(A)|.$$ It is known that
+$$\tv{\pi-\mu}=\frac{1}{2}\sum_{x\in\Bool^n}|\pi(x)-\mu(x)|.$$ Moreover, if
+$\nu$ is a distribution on $\Bool^n$, one has
+$$\tv{\pi-\mu}\leq \tv{\pi-\nu}+\tv{\nu-\mu}$$
+
+Let $P$ be the matrix of a markov chain on $\Bool^n$. $P(x,\cdot)$ is the
+distribution induced by the $x$-th row of $P$. If the markov chain induced by
+$P$ has a stationary distribution $\pi$, then we define
+$$d(t)=\max_{x\in\Bool^n}\tv{P^t(x,\cdot)-\pi}.$$
+It is known that $d(t+1)\leq d(t)$. \JFC{référence ? Cela a-t-il
+un intérêt dans la preuve ensuite.}
+
+
+
+%and
+% $$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
+% One can prove that \JFc{Ou cela a-t-il été fait?}
+% $$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}(\frac{1}{4})$$
+
+
+
+Let $(X_t)_{t\in \mathbb{N}}$ be a sequence of $\Bool^n$ valued random
+variables. A $\mathbb{N}$-valued random variable $\tau$ is a {\it stopping
+ time} for the sequence $(X_i)$ if for each $t$ there exists $B_t\subseteq
+\Omega^{t+1}$ such that $\{\tau=t\}=\{(X_0,X_1,\ldots,X_t)\in B_t\}$.
+In other words, the event $\{\tau = t \}$ only depends on the values of
+$(X_0,X_1,\ldots,X_t)$, not on $X_k$ with $k > t$.
+
+
+\JFC{Je ne comprends pas la definition de randomized stopping time, Peut-on enrichir ?}
+
+Let $(X_t)_{t\in \mathbb{N}}$ be a markov chain and $f(X_{t-1},Z_t)$ a
+random mapping representation of the markov chain. A {\it randomized
+ stopping time} for the markov chain is a stopping time for
+$(Z_t)_{t\in\mathbb{N}}$. If the markov chain is irreductible and has $\pi$
+as stationary distribution, then a {\it stationary time} $\tau$ is a
+randomized stopping time (possibily depending on the starting position $x$),
+such that the distribution of $X_\tau$ is $\pi$:
+$$\P_x(X_\tau=y)=\pi(y).$$
+
+
+\JFC{Ou ceci a-t-il ete prouvé}
+\begin{Theo}
+If $\tau$ is a strong stationary time, then $d(t)\leq \max_{x\in\Bool^n}
+\P_x(\tau > t)$.
+\end{Theo}
+
+% % Let us first recall the \emph{Total Variation} distance $\tv{\pi-\mu}$,
+% % which is defined for two distributions $\pi$ and $\mu$ on the same set
+% % $\Bool^n$ by:
+% % $$\tv{\pi-\mu}=\max_{A\subset \Bool^n} |\pi(A)-\mu(A)|.$$
+% % It is known that
+% % $$\tv{\pi-\mu}=\frac{1}{2}\sum_{x\in\Bool^n}|\pi(x)-\mu(x)|.$$
+
+% % Let then $M(x,\cdot)$ be the
+% % distribution induced by the $x$-th row of $M$. If the Markov chain
+% % induced by
+% % $M$ has a stationary distribution $\pi$, then we define
+% % $$d(t)=\max_{x\in\Bool^n}\tv{M^t(x,\cdot)-\pi}.$$
+% Intuitively $d(t)$ is the largest deviation between
+% the distribution $\pi$ and $M^t(x,\cdot)$, which
+% is the result of iterating $t$ times the function.
+% Finally, let $\varepsilon$ be a positive number, the \emph{mixing time}
+% with respect to $\varepsilon$ is given by
+% $$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
+% It defines the smallest iteration number
+% that is sufficient to obtain a deviation lesser than $\varepsilon$.
