$\nu$ 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
+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{references ? 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
+
+$$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}(\frac{1}{4})$$
+
+
+
+
+% It is known that $d(t+1)\leq d(t)$. \JFC{references ? Cela a-t-il
+% un intérêt dans la preuve ensuite.}
$(X_0,X_1,\ldots,X_t)$, 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
$$\P_X(X_\tau=Y)=\pi(Y).$$
-\JFC{Ou ceci a-t-il ete prouvé. On ne définit pas ce qu'est un strong stationary time.}
\begin{Theo}
If $\tau$ is a strong stationary time, then $d(t)\leq \max_{X\in\Bool^n}
\P_X(\tau > t)$.
X_t= f(X_{t-1},Z_t)
$$
-The pair $(f,Z)$ is a random mapping representation of $P_h$.
-\JFC{interet de la phrase precedente}
-
%%%%%%%%%%%%%%%%%%%%%%%%%%%ù
$\frac{1}{2}$. Since $h(X_{\tau_\ell-1})\neq\ell$ the value of the $\ell$-th
bit of $X_{\tau_\ell}$
is $0$ or $1$ with the same probability ($\frac{1}{2}$).
-By symmetry, \JFC{Je ne comprends pas ce by symetry} for $t\geq \tau_\ell$, the
+
+ Moving next in the chain, at each step,
+the $l$-th bit is switched from $0$ to $1$ or from $1$ to $0$ each time with
+the same probability. Therefore, for $t\geq \tau_\ell$, the
$\ell$-th bit of $X_t$ is $0$ or $1$ with the same probability, proving the
-lemma.
-\end{proof}
+lemma.\end{proof}
\begin{Theo} \label{prop:stop}
If $\ov{h}$ is bijective and square-free, then
random variable that counts the number of steps
from $X$ until we reach a configuration where
$\ell$ is fair. More formally
-$$S_{X,\ell}=\min \{t \geq 1\mid h(X_{t-1})\neq \ell\text{ and }Z_t=(\ell,\.)\text{ and } X_0=X\}.$$
+$$S_{X,\ell}=\min \{t \geq 1\mid h(X_{t-1})\neq \ell\text{ and }Z_t=(\ell,.)\text{ and } X_0=X\}.$$
We denote by
$$\lambda_h=\max_{X,\ell} S_{X,\ell}.$$
\begin{Lemma}\label{prop:lambda}
-If $\ov{h}$ is a square-free bijective function, then one has $E[\lambda_h]\leq 8n^2.$
+If $\ov{h}$ is a square-free bijective function, then the inequality
+$E[\lambda_h]\leq 8n^2$ is established.
+
\end{Lemma}
\begin{proof}
For every $X$, every $\ell$, one has $\P(S_{X,\ell})\leq 2)\geq
-\frac{1}{4n^2}$. Indeed, if $h(X)\neq \ell$, then
-$\P(S_{X,\ell}=1)=\frac{1}{2n}\geq \frac{1}{4n^2}$. If $h(X)=\ell$, then
-$\P(S_{X,\ell}=1)=0$. Let $X_0=X$. Since $\ov{h}$ is square-free,
-$\ov{h}(\ov{h}^{-1}(X))\neq X$. It follows that $(X,\ov{h}^{-1}(X))\in E_h$.
-Therefore $P(X_1=\ov{h}^{-1}(X))=\frac{1}{2n}$. now,
-by Lemma~\ref{lm:h}, $h(\ov{h}^{-1}(X))\neq h(X)$. Therefore
-$\P(S_{X,\ell}=2\mid X_1=\ov{h}^{-1}(X))=\frac{1}{2n}$, proving that
-$\P(S_{X,\ell}\leq 2)\geq \frac{1}{4n^2}$.
-
-Therefore, $\P(S_{X,\ell}\geq 2)\leq 1-\frac{1}{4n^2}$. By induction, one
+\frac{1}{4n^2}$.
+Let $X_0= X$.
+Indeed,
+\begin{itemize}
+\item if $h(X)\neq \ell$, then
+$\P(S_{X,\ell}=1)=\frac{1}{2n}\geq \frac{1}{4n^2}$.
+\item otherwise, $h(X)=\ell$, then
+$\P(S_{X,\ell}=1)=0$.
+But in this case, intutively, it is possible to move
+from $X$ to $\ov{h}^{-1}(X)$ (with probability $\frac{1}{2N}$). And in
+$\ov{h}^{-1}(X)$ the $l$-th bit can be switched.
+More formally,
+since $\ov{h}$ is square-free,
+$\ov{h}(X)=\ov{h}(\ov{h}(\ov{h}^{-1}(X)))\neq \ov{h}^{-1}(X)$. It follows
+that $(X,\ov{h}^{-1}(X))\in E_h$. We thus have
+$P(X_1=\ov{h}^{-1}(X))=\frac{1}{2N}$. Now, by Lemma~\ref{lm:h},
+$h(\ov{h}^{-1}(X))\neq h(X)$. Therefore $\P(S_{x,\ell}=2\mid
+X_1=\ov{h}^{-1}(X))=\frac{1}{2N}$, proving that $\P(S_{x,\ell}\leq 2)\geq
+\frac{1}{4N^2}$.
+\end{itemize}
+
+
+
+
+Therefore, $\P(S_{X,\ell}\geq 3)\leq 1-\frac{1}{4n^2}$. By induction, one
has, for every $i$, $\P(S_{X,\ell}\geq 2i)\leq
\left(1-\frac{1}{4n^2}\right)^i$.
Moreover,
-since $S_{X,\ell}$ is positive, it is known~\cite[lemma 2.9]{}, that
+since $S_{X,\ell}$ is positive, it is known~\cite[lemma 2.9]{proba}, that
$$E[S_{X,\ell}]=\sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq i).$$
Since $\P(S_{X,\ell}\geq i)\geq \P(S_{X,\ell}\geq i+1)$, one has
$$E[S_{X,\ell}]=\sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq i)\leq
-\P(S_{X,\ell}\geq 1)+2 \sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq 2i).$$
+\P(S_{X,\ell}\geq 1)+\P(S_{X,\ell}\geq 2)+2 \sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq 2i).$$
Consequently,
-$$E[S_{X,\ell}]\leq 1+2
-\sum_{i=1}^{+\infty}\left(1-\frac{1}{4n^2}\right)^i=1+2(4n^2-1)=8n^2-2,$$
+$$E[S_{X,\ell}]\leq 1+1+2
+\sum_{i=1}^{+\infty}\left(1-\frac{1}{4n^2}\right)^i=2+2(4n^2-1)=8n^2,$$
which concludes the proof.
\end{proof}
$E[\ts^\prime]\leq n (-\frac{1}{2}+\ln(n+1))\leq n\ln(n+1)$.
\end{proof}
-One can now prove Theo~\ref{prop:stop}.
+One can now prove Theorem~\ref{prop:stop}.
\begin{proof}
One has $\ts\leq \ts^\prime+\lambda_h$. Therefore,
lemma~\ref{prop:lambda} and~\ref{lm:stopprime}.
\end{proof}
-\end{document}
+
