bla bla:

Merge branch 'master' of ssh://bilbo.iut-bm.univ-fcomte.fr/16dcc

diff --combined stopping.tex
index e86176a9481d32da89949a0c30b6eb330bf6ee5c,aa13c9ab8812ba567820cbfe01c38196ecbbe83a..1ac699934cc93a9f5bd6be19b60b634d87c42596
--- 1/stopping.tex
--- 2/stopping.tex
+++ b/stopping.tex
@@@ -33,7 -33,7 +33,7 @@@ P=\dfrac{1}{6} \left
  0&0&0&0&1&0&4&1 \\
  0&0&0&1&0&1&0&4 
  \end{array}
- \right)
+ \right).
  \]
  \end{xpl}
  
@@@ -69,9 -69,13 +69,13 @@@ an
  
  $$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
  
- Intuitively speaking, $t_{\rm mix}$ is a mixing time 
- \textit{i.e.}, is the time until the matrix $X$ 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 staionary distribution, wherever
+ the chain starts. 
  
  
  
@@@ -115,7 -119,7 +119,7 @@@ $$\P_X(X_\tau=Y)=\pi(Y).$
  
  
  A stopping time $\tau$ is a {\emph strong stationary time} if $X_{\tau}$ is
- independent of $\tau$. 
+ independent of $\tau$. The following result will be useful~\cite[Proposition~6.10]{LevinPeresWilmer2006},
  
  
  \begin{thrm}\label{thm-sst}
@@@ -231,7 -235,8 +235,8 @@@ This probability is independent of 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
+ $\ell$-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}
@@@ -345,7 -350,7 +350,7 @@@ direct application of lemma~\ref{prop:l
  \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
  $t_{\rm mix}\leq 32N^2+16N\ln (N+1)=O(N^2)$.
@@@ -354,11 -359,11 +359,11 @@@
  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 $\Theta(N\ln
  N)$.
@@@ -422,7 -427,7 +427,7 @@@ is realistic according the graph of $x 
  % \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}
  % $$