$\nu$ is a distribution on $\Bool^{\mathsf{N}}$, one has
$$\tv{\pi-\mu}\leq \tv{\pi-\nu}+\tv{\nu-\mu}$$
-Let $P$ be the matrix of a Markov chain on $\Bool^{\mathsf{N}}$. $P(X,\cdot)$ 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
+Let $P$ be the matrix of a Markov chain on $\Bool^{\mathsf{N}}$. For any
+$X\in \Bool^{\mathsf{N}}$, let $P(X,\cdot)$ be the distribution induced by the
+${\rm bin}(X)$-th row of $P$, where ${\rm bin}(X)$ is the integer whose
+binary encoding is $X$. If the Markov chain induced by $P$ has a stationary
+distribution $\pi$, then we define
$$d(t)=\max_{X\in\Bool^{\mathsf{N}}}\tv{P^t(X,\cdot)-\pi}.$$
+%\ANNOT{incohérence de notation $X$ : entier ou dans $B^N$ ?}
and
$$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
%% is $\epsilon$-close to a stationary distribution.
Intutively speaking, $t_{\rm mix}(\varepsilon)$ is the time/steps required
-to be sure to be $\varepsilon$-close to the staionary distribution, wherever
+to be sure to be $\varepsilon$-close to the stationary distribution, wherever
the chain starts.
\subsection{Upper bound of Stopping Time}\label{sub:stop:bound}
-
A stopping time $\tau$ is a {\emph strong stationary time} if $X_{\tau}$ is
independent of $\tau$. The following result will be useful~\cite[Proposition~6.10]{LevinPeresWilmer2006},
random variable that counts the number of steps
from $X$ until we reach a configuration where
$\ell$ is fair. More formally
-$$S_{X,\ell}=\min \{t \geq 1\mid h(X_{t-1})\neq \ell\text{ and }Z_t=(\ell,.)\text{ and } X_0=X\}.$$
+\[
+\begin{array}{rcl}
+S_{X,\ell}&=&\min \{t \geq 1\mid h(X_{t-1})\neq \ell\text{ and }Z_t=(\ell,.) \\
+&& \qquad \text{ and } X_0=X\}.
+\end{array}
+\]
% We denote by
% $$\lambda_h=\max_{X,\ell} S_{X,\ell}.$$
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
-$$E[S_{X,\ell}]=\sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq i)\leq
-\P(S_{X,\ell}\geq 1)+\P(S_{X,\ell}\geq 2)+2 \sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq 2i).$$
+\[
+\begin{array}{rcl}
+ E[S_{X,\ell}]&=&\sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq i)\\
+&\leq&
+\P(S_{X,\ell}\geq 1) +\P(S_{X,\ell}\geq 2)\\
+&& \qquad +2 \sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq 2i).
+\end{array}
+\]
Consequently,
$$E[S_{X,\ell}]\leq 1+1+2
\sum_{i=1}^{+\infty}\left(1-\frac{1}{4{\mathsf{N}}^2}\right)^i=2+2(4{\mathsf{N}}^2-1)=8{\mathsf{N}}^2,$$
\end{algorithm}
Practically speaking, for each number $\mathsf{N}$, $ 3 \le \mathsf{N} \le 16$,
-10 functions have been generaed according to method presented in section~\ref{sec:hamilton}. For each of them, the calculus of the approximation of $E[\ts]$
+10 functions have been generated according to method presented in section~\ref{sec:hamilton}. For each of them, the calculus of the approximation of $E[\ts]$
is executed 10000 times with a random seed. The Figure~\ref{fig:stopping:moy}
summarizes these results. In this one, a circle represents the
approximation of $E[\ts]$ for a given $\mathsf{N}$.
% \hline
% \mathsf{N} & 4 & 5 & 6 & 7& 8 & 9 & 10& 11 & 12 & 13 & 14 & 15 & 16 \\
% \hline
-% \mathsf{N} & 21.8 & 28.4 & 35.4 & 42.5 & 50 & 57.7 & 65.6& 73.5 & 81.6 & 90 & 98.3 & 107.1 & 16 \\
+% \mathsf{N} & 21.8 & 28.4 & 35.4 & 42.5 & 50 & 57.7 & 65.6& 73.5 & 81.6 & 90 & 98.3 & 107.1 & 115.7 \\
% \hline
% \end{array}
% $$
\begin{figure}
\centering
-\includegraphics[scale=0.5]{complexity}
+\includegraphics[width=0.49\textwidth]{complexity}
\caption{Average Stopping Time Approximation}\label{fig:stopping:moy}
\end{figure}
