-
-Let $\pi$, $\mu$ be two distribution on a same set $\Omega$. The total
-variation distance between $\pi$ and $\mu$ is denoted $\tv{\pi-\mu}$ and is
-defined by
-$$\tv{\pi-\mu}=\max_{A\subset \Omega} |\pi(A)-\mu(A)|.$$ It is known that
-$$\tv{\pi-\mu}=\frac{1}{2}\sum_{x\in\Omega}|\pi(x)-\mu(x)|.$$ Moreover, if
-$\nu$ is a distribution on $\Omega$, one has
-$$\tv{\pi-\mu}\leq \tv{\pi-\nu}+\tv{\nu-\mu}$$
-
-Let $P$ be the matrix of a markov chain on $\Omega$. $P(x,\cdot)$ 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
-$$d(t)=\max_{x\in\Omega}\tv{P^t(x,\cdot)-\pi},$$
-and
-
-$$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
-One can prove that
-
-$$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}(\frac{1}{4})$$
-
-It is known that $d(t+1)\leq d(t)$.
-
-
-
-Let $(X_t)_{t\in \mathbb{N}}$ be a sequence of $\Omega$ 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 $\{tau=t\}=\{(X_0,X_1,\ldots,X_t)\in B_t\}$.
-
-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
-$(Z_t)_{t\in\mathbb{N}}$. It he markov chain is irreductible and has $\pi$
-as stationary distribution, then a {\it stationay 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 $d(t)\leq \max_{x\in\Omega}
-\P_x(\tau > t)$.
-\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
-% $\Omega$ 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 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 $\varepsilon$.
