0&0&0&0&1&0&4&1 \\
0&0&0&1&0&1&0&4
\end{array}
-\right)
+\right).
\]
\end{xpl}
$\nu$ 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 $B^N$ ?}
+%\ANNOT{incohérence de notation $X$ : entier ou dans $B^N$ ?}
and
$$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
-Intuitively speaking, $t_{\rm mix}$ is a mixing time
-\textit{i.e.}, is the time until the matrix $X$ \ANNOT{pas plutôt $P$ ?} of a Markov chain
-is $\epsilon$-close to a stationary distribution.
+%% Intuitively speaking, $t_{\rm mix}$ is a mixing time
+%% \textit{i.e.}, is the time until the matrix $X$ of a Markov chain
+%% 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 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 $\tau$.
+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},
\begin{thrm}\label{thm-sst}
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
+$\ell$-th bit of $X_t$ is $0$ or $1$ with the same probability, and
+independently of the value of the other bits, proving the
lemma.\end{proof}
\begin{thrm} \label{prop:stop}
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{proof}
Now using Markov Inequality, one has $\P_X(\tau > t)\leq \frac{E[\tau]}{t}$.
-With $t=32N^2+16N\ln (N+1)$, one obtains: $\P_X(\tau > t)\leq \frac{1}{4}$.
+With $t_n=32N^2+16N\ln (N+1)$, one obtains: $\P_X(\tau > t_n)\leq \frac{1}{4}$.
Therefore, using the defintion of $t_{\rm mix)}$ and
Theorem~\ref{thm-sst}, it follows that
$t_{\rm mix}\leq 32N^2+16N\ln (N+1)=O(N^2)$.
Notice that the calculus of the stationary time upper bound is obtained
under the following constraint: for each vertex in the $\mathsf{N}$-cube
there are one ongoing arc and one outgoing arc that are removed.
-The calculus does not consider (balanced) Hamiltonian cycles, which
+The calculus doesn't consider (balanced) Hamiltonian cycles, which
are more regular and more binding than this constraint.
Moreover, the bound
-is obtained using Markov Inequality which is frequently coarse. For the
-classical random walkin the $\mathsf{N}$-cube, without removing any
+is obtained using the coarse Markov Inequality. For the
+classical (lazzy) random walk the $\mathsf{N}$-cube, without removing any
Hamiltonian cylce, the mixing time is in $\Theta(N\ln N)$.
We conjecture that in our context, the mixing time is also in $\Theta(N\ln
N)$.
% \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}
