X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/16dcc.git/blobdiff_plain/70f3455e44e96f0ad00501af3d6f396bc09ef436..2600bcf75b0329f96238df06abcbfbfeb7feedab:/stopping.tex?ds=inline

diff --git a/stopping.tex b/stopping.tex
index 1ac6999..284fc49 100644
--- a/stopping.tex
+++ b/stopping.tex
@@ -65,6 +65,7 @@ distribution induced by the $X$-th row of $P$. If the Markov chain induced by
 $P$ has a stationary distribution $\pi$, then we define
 $$d(t)=\max_{X\in\Bool^{\mathsf{N}}}\tv{P^t(X,\cdot)-\pi}.$$
 
+\ANNOT{incohérence de notation $X$ : entier ou dans $B^N$ ?}
 and
 
 $$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
@@ -74,7 +75,7 @@ $$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
 %% 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. 
 
 
@@ -117,7 +118,6 @@ $$\P_X(X_\tau=Y)=\pi(Y).$$
 
 \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},
 
@@ -249,7 +249,12 @@ let $S_{X,\ell}$ be the
 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
 % $$\lambda_h=\max_{X,\ell} S_{X,\ell}.$$
@@ -296,8 +301,14 @@ has, for every $i$, $\P(S_{X,\ell}\geq 2i)\leq
 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,
 $$E[S_{X,\ell}]\leq 1+1+2
 \sum_{i=1}^{+\infty}\left(1-\frac{1}{4{\mathsf{N}}^2}\right)^i=2+2(4{\mathsf{N}}^2-1)=8{\mathsf{N}}^2,$$
@@ -406,7 +417,7 @@ $\textit{fair}\leftarrow\emptyset$\;
 \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 $\mathsf{N}$.
@@ -436,7 +447,7 @@ is realistic according the graph of $x \mapsto 2x\ln(2x+8)$.
 
 \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}