modifications de la partie stopping time et calcul de la borne dans les exp

[rairo15.git] / stopping.tex

diff --git a/stopping.tex b/stopping.tex
index 9cc720cb5344b94c9062a31a4edfd4d0767f86b3..3a07e06a67fc073968d7355d2bff8196d485c29f 100644 (file)
--- a/stopping.tex
+++ b/stopping.tex
@@ -81,7 +81,9 @@ A specific random walk in this modified hypercube is first
 introduced. We further detail
 a theoretical study on the length of the path 
 which is sufficient to follow to get a uniform distribution.
 introduced. We further detail
 a theoretical study on the length of the path 
 which is sufficient to follow to get a uniform distribution.

- 
+Notice that for a general references on Markov chains
+see~\cite{LevinPeresWilmer2006}
+, and particularly Chapter~5 on stopping times.  

 
 
 
 
 
 


@@ -138,6 +140,9 @@ randomized stopping time (possibly depending on the starting position $X$),
 such that  the distribution of $X_\tau$ is $\pi$:
 $$\P_X(X_\tau=Y)=\pi(Y).$$
 
 such that  the distribution of $X_\tau$ is $\pi$:
 $$\P_X(X_\tau=Y)=\pi(Y).$$
 

+A stopping time $\tau$ is a {\emph strong stationary time} if $X_{\tau}$ is
+independent of $\tau$. 
+

 
 \begin{thrm}
 If $\tau$ is a strong stationary time, then $d(t)\leq \max_{X\in\Bool^{\mathsf{N}}}
 
 \begin{thrm}
 If $\tau$ is a strong stationary time, then $d(t)\leq \max_{X\in\Bool^{\mathsf{N}}}


@@ -254,7 +259,7 @@ lemma.\end{proof}
 
 \begin{thrm} \label{prop:stop}
 If $\ov{h}$ is bijective and square-free, then
 
 \begin{thrm} \label{prop:stop}
 If $\ov{h}$ is bijective and square-free, then

-$E[\ts]\leq 8{\mathsf{N}}^2+ {\mathsf{N}}\ln ({\mathsf{N}}+1)$. 
+$E[\ts]\leq 8{\mathsf{N}}^2+ 4{\mathsf{N}}\ln ({\mathsf{N}}+1)$. 

 \end{thrm}
 
 For each $X\in \Bool^{\mathsf{N}}$ and $\ell\in\llbracket 1,{\mathsf{N}}\rrbracket$, 
 \end{thrm}
 
 For each $X\in \Bool^{\mathsf{N}}$ and $\ell\in\llbracket 1,{\mathsf{N}}\rrbracket$, 


@@ -264,14 +269,16 @@ from $X$ until we reach a configuration where
 $\ell$ is fair. More formally
 $$S_{X,\ell}=\min \{t \geq 1\mid h(X_{t-1})\neq \ell\text{ and }Z_t=(\ell,.)\text{ and } X_0=X\}.$$
 
 $\ell$ is fair. More formally
 $$S_{X,\ell}=\min \{t \geq 1\mid h(X_{t-1})\neq \ell\text{ and }Z_t=(\ell,.)\text{ and } X_0=X\}.$$
 

- We denote by
-$$\lambda_h=\max_{X,\ell} S_{X,\ell}.$$
+%  We denote by
+% $$\lambda_h=\max_{X,\ell} S_{X,\ell}.$$

 
 
 \begin{lmm}\label{prop:lambda}
 
 
 \begin{lmm}\label{prop:lambda}

-If $\ov{h}$ is a square-free bijective function, then the inequality 
-$E[\lambda_h]\leq 8{\mathsf{N}}^2$ is established.
-
+Let $\ov{h}$ is a square-free bijective function. Then
+for all $X$ and 
+all $\ell$, 
+the inequality 
+$E[S_{X,\ell}]\leq 8{\mathsf{N}}^2$ is established.

 \end{lmm}
 
 \begin{proof}
 \end{lmm}
 
 \begin{proof}


@@ -284,7 +291,7 @@ Indeed,
 $\P(S_{X,\ell}=1)=\frac{1}{2{\mathsf{N}}}\geq \frac{1}{4{\mathsf{N}}^2}$. 
 \item otherwise, $h(X)=\ell$, then
 $\P(S_{X,\ell}=1)=0$.
 $\P(S_{X,\ell}=1)=\frac{1}{2{\mathsf{N}}}\geq \frac{1}{4{\mathsf{N}}^2}$. 
 \item otherwise, $h(X)=\ell$, then
 $\P(S_{X,\ell}=1)=0$.

-But in this case, intutively, it is possible to move
+But in this case, intuitively, 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,
 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,


@@ -315,11 +322,10 @@ $$E[S_{X,\ell}]\leq 1+1+2
 which concludes the proof.
 \end{proof}
 
 which concludes the proof.
 \end{proof}
 

-Let $\ts^\prime$ be the first time that there are exactly ${\mathsf{N}}-1$ fair
-elements. 
+Let $\ts^\prime$ be the time used to get all the bits but one fair.

 
 \begin{lmm}\label{lm:stopprime}
 
 \begin{lmm}\label{lm:stopprime}

-One has $E[\ts^\prime]\leq {\mathsf{N}} \ln ({\mathsf{N}}+1).$
+One has $E[\ts^\prime]\leq 4{\mathsf{N}} \ln ({\mathsf{N}}+1).$

 \end{lmm}
 
 \begin{proof}
 \end{lmm}
 
 \begin{proof}


@@ -328,21 +334,36 @@ random variable counting the number of moves done in the Markov chain while
 we had exactly $i-1$ fair bits. One has $\ts^\prime=\sum_{i=1}^{{\mathsf{N}}-1}W_i$.
  But when we are at position $X$ with $i-1$ fair bits, the probability of
  obtaining a new fair bit is either $1-\frac{i-1}{{\mathsf{N}}}$ if $h(X)$ is fair,
 we had exactly $i-1$ fair bits. One has $\ts^\prime=\sum_{i=1}^{{\mathsf{N}}-1}W_i$.
  But when we are at position $X$ with $i-1$ fair bits, the probability of
  obtaining a new fair bit is either $1-\frac{i-1}{{\mathsf{N}}}$ if $h(X)$ is fair,

- or  $1-\frac{i-2}{{\mathsf{N}}}$ if $h(X)$ is not fair. It follows that 
-$E[W_i]\leq \frac{{\mathsf{N}}}{{\mathsf{N}}-i+2}$. Therefore
-$$E[\ts^\prime]=\sum_{i=1}^{{\mathsf{N}}-1}E[W_i]\leq {\mathsf{N}}\sum_{i=1}^{{\mathsf{N}}-1}
- \frac{1}{{\mathsf{N}}-i+2}={\mathsf{N}}\sum_{i=3}^{{\mathsf{N}}+1}\frac{1}{i}.$$
+ or  $1-\frac{i-2}{{\mathsf{N}}}$ if $h(X)$ is not fair. 
+
+Therefore,
+$\P (W_i=k)\leq \left(\frac{i-1}{{\mathsf{N}}}\right)^{k-1} \frac{{\mathsf{N}}-i+2}{{\mathsf{N}}}.$
+Consequently, we have $\P(W_i\geq k)\leq \left(\frac{i-1}{{\mathsf{N}}}\right)^{k-1} \frac{{\mathsf{N}}-i+2}{{\mathsf{N}}-i+1}.$
+It follows that $E[W_i]=\sum_{k=1}^{+\infty} \P (W_i\geq k)\leq {\mathsf{N}} \frac{{\mathsf{N}}-i+2}{({\mathsf{N}}-i+1)^2}\leq \frac{4{\mathsf{N}}}{{\mathsf{N}}-i+2}$.
+
+
+
+It follows that 
+$E[W_i]\leq \frac{4{\mathsf{N}}}{{\mathsf{N}}-i+2}$. Therefore
+$$E[\ts^\prime]=\sum_{i=1}^{{\mathsf{N}}-1}E[W_i]\leq 
+4{\mathsf{N}}\sum_{i=1}^{{\mathsf{N}}-1} \frac{1}{{\mathsf{N}}-i+2}=
+4{\mathsf{N}}\sum_{i=3}^{{\mathsf{N}}+1}\frac{1}{i}.$$

 
 But $\sum_{i=1}^{{\mathsf{N}}+1}\frac{1}{i}\leq 1+\ln({\mathsf{N}}+1)$. It follows that
 $1+\frac{1}{2}+\sum_{i=3}^{{\mathsf{N}}+1}\frac{1}{i}\leq 1+\ln({\mathsf{N}}+1).$
 Consequently,
 
 But $\sum_{i=1}^{{\mathsf{N}}+1}\frac{1}{i}\leq 1+\ln({\mathsf{N}}+1)$. It follows that
 $1+\frac{1}{2}+\sum_{i=3}^{{\mathsf{N}}+1}\frac{1}{i}\leq 1+\ln({\mathsf{N}}+1).$
 Consequently,

-$E[\ts^\prime]\leq {\mathsf{N}} (-\frac{1}{2}+\ln({\mathsf{N}}+1))\leq {\mathsf{N}}\ln({\mathsf{N}}+1)$.
+$E[\ts^\prime]\leq 
+4{\mathsf{N}} (-\frac{1}{2}+\ln({\mathsf{N}}+1))\leq 
+4{\mathsf{N}}\ln({\mathsf{N}}+1)$.

 \end{proof}
 
 One can now prove Theorem~\ref{prop:stop}.
 
 \begin{proof}
 \end{proof}
 
 One can now prove Theorem~\ref{prop:stop}.
 
 \begin{proof}

-One has $\ts\leq \ts^\prime+\lambda_h$. Therefore,
+Since $\ts^\prime$ is the time used to obtain $\mathsf{N}-1$ fair bits.
+Assume that the last unfair bit is $\ell$. One has
+$\ts=\ts^\prime+S_{X_\tau,\ell}$, and therefore
+$E[\ts] = E[\ts^\prime]+E[S_{X_\tau,\ell}]$. Therefore,

 Theorem~\ref{prop:stop} is a direct application of
 lemma~\ref{prop:lambda} and~\ref{lm:stopprime}.
 \end{proof}
 Theorem~\ref{prop:stop} is a direct application of
 lemma~\ref{prop:lambda} and~\ref{lm:stopprime}.
 \end{proof}


@@ -350,7 +371,7 @@ lemma~\ref{prop:lambda} and~\ref{lm:stopprime}.
 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. 
 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 does not consider (balanced) Hamiltonian cycles, which 

 are more regular and more binding than this constraint.
 In this later context, we claim that the upper bound for the stopping time 
 should be reduced.
 are more regular and more binding than this constraint.
 In this later context, we claim that the upper bound for the stopping time 
 should be reduced.