From: couchot <jf.couchot@gmail.com>
Date: Sat, 14 Mar 2015 08:57:20 +0000 (+0100)
Subject: stoping
X-Git-Url: https://bilbo.iut-bm.univ-fcomte.fr/and/gitweb/rairo15.git/commitdiff_plain/935ffe48008761ee21d78930849f51e32c62fcec?hp=07fc9a29ed26c33dfe987821fc94b3afa4cdd3a5

stoping
---

diff --git a/stopping.tex b/stopping.tex
index 0ad3e8a..faab606 100644
--- a/stopping.tex
+++ b/stopping.tex
@@ -20,23 +20,23 @@ the probability function $p$ defined on the set of edges as follows:
 $$
 p(e) \left\{
 \begin{array}{ll}
-= \frac{2}{3} \textrm{ if $e=(v,v)$ with $v \in \Bool^3$,}\\
-= \frac{1}{6} \textrm{ otherwise.}
+= \frac{1}{3} \textrm{ if $e=(v,v)$ with $v \in \Bool^3$,}\\
+= \frac{1}{3} \textrm{ otherwise.}
 \end{array}
 \right.  
 $$
 The matrix $P$ of the Markov chain associated to the function $f^*$ and to its probability function $p$ is 
 \[
-P=\dfrac{1}{6} \left(
+P=\dfrac{1}{3} \left(
 \begin{array}{llllllll}
-4&1&1&0&0&0&0&0 \\
-1&4&0&0&0&1&0&0 \\
-0&0&4&1&0&0&1&0 \\
-0&1&1&4&0&0&0&0 \\
-1&0&0&0&4&0&1&0 \\
-0&0&0&0&1&4&0&1 \\
-0&0&0&0&1&0&4&1 \\
-0&0&0&1&0&1&0&4 
+1&1&1&0&0&0&0&0 \\
+1&1&0&0&0&1&0&0 \\
+0&0&1&1&0&0&1&0 \\
+0&1&1&1&0&0&0&0 \\
+1&0&0&0&1&0&1&0 \\
+0&0&0&0&1&1&0&1 \\
+0&0&0&0&1&0&1&1 \\
+0&0&0&1&0&1&0&1 
 \end{array}
 \right)
 \]
@@ -75,7 +75,7 @@ P=\dfrac{1}{6} \left(
 
 
 
-This section considers functions $f: \Bool^n \rightarrow \Bool^n $ 
+This section considers functions $f: \Bool^{\mathsf{N}} \rightarrow \Bool^{\mathsf{N}} $ 
 issued from an hypercube where an Hamiltonian path has been removed.
 A specific random walk in this modified hypercube is first 
 introduced. We further detail
@@ -86,45 +86,45 @@ which is sufficient to follow to get a uniform distribution.
 
 
 
-First of all, let $\pi$, $\mu$ be two distributions on $\Bool^n$. The total
+First of all, let $\pi$, $\mu$ be two distributions on $\Bool^{\mathsf{N}}$. The total
 variation distance between $\pi$ and $\mu$ is denoted $\tv{\pi-\mu}$ and is
 defined by
-$$\tv{\pi-\mu}=\max_{A\subset \Bool^n} |\pi(A)-\mu(A)|.$$ It is known that
-$$\tv{\pi-\mu}=\frac{1}{2}\sum_{X\in\Bool^n}|\pi(X)-\mu(X)|.$$ Moreover, if
-$\nu$ is a distribution on $\Bool^n$, one has
+$$\tv{\pi-\mu}=\max_{A\subset \Bool^{\mathsf{N}}} |\pi(A)-\mu(A)|.$$ It is known that
+$$\tv{\pi-\mu}=\frac{1}{2}\sum_{X\in\Bool^{\mathsf{N}}}|\pi(X)-\mu(X)|.$$ Moreover, if
+$\nu$ 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 $\Bool^n$. $P(X,\cdot)$ is the
+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
-$$d(t)=\max_{X\in\Bool^n}\tv{P^t(X,\cdot)-\pi}.$$
+$$d(t)=\max_{X\in\Bool^{\mathsf{N}}}\tv{P^t(X,\cdot)-\pi}.$$
 
 and
 
 $$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
-One can prove that
+% One can prove that
 
-$$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}(\frac{1}{4})$$
+% $$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}(\frac{1}{4})$$
 
 
 
 
 % It is known that $d(t+1)\leq d(t)$. \JFC{references ? Cela a-t-il 
-% un intérêt dans la preuve ensuite.}
+% un intérêt dans la preuve ensuite.}
 
 
 
 %and
 % $$t_{\rm mix}(\varepsilon)=\min\{t \mid d(t)\leq \varepsilon\}.$$
-% One can prove that \JFc{Ou cela a-t-il été fait?}
+% One can prove that \JFc{Ou cela a-t-il été fait?}
 % $$t_{\rm mix}(\varepsilon)\leq \lceil\log_2(\varepsilon^{-1})\rceil t_{\rm mix}(\frac{1}{4})$$
 
 
 
-Let $(X_t)_{t\in \mathbb{N}}$ be a sequence of $\Bool^n$ valued random
+Let $(X_t)_{t\in \mathbb{N}}$ be a sequence of $\Bool^{\mathsf{N}}$ valued random
 variables. A $\mathbb{N}$-valued random variable $\tau$ is a {\it stopping
   time} for the sequence $(X_i)$ if for each $t$ there exists $B_t\subseteq
-(\Bool^n)^{t+1}$ such that $\{\tau=t\}=\{(X_0,X_1,\ldots,X_t)\in B_t\}$. 
+(\Bool^{\mathsf{N}})^{t+1}$ such that $\{\tau=t\}=\{(X_0,X_1,\ldots,X_t)\in B_t\}$. 
 In other words, the event $\{\tau = t \}$ only depends on the values of 
 $(X_0,X_1,\ldots,X_t)$, not on $X_k$ with $k > t$. 
  
@@ -140,44 +140,44 @@ $$\P_X(X_\tau=Y)=\pi(Y).$$
 
 
 \begin{Theo}
-If $\tau$ is a strong stationary time, then $d(t)\leq \max_{X\in\Bool^n}
+If $\tau$ is a strong stationary time, then $d(t)\leq \max_{X\in\Bool^{\mathsf{N}}}
 \P_X(\tau > t)$.
 \end{Theo}
 
 
 %Let $\Bool^n$ be the set of words of length $n$. 
 Let $E=\{(X,Y)\mid
-X\in \Bool^n, Y\in \Bool^n,\ X=Y \text{ or } X\oplus Y \in 0^*10^*\}$.
+X\in \Bool^{\mathsf{N}}, Y\in \Bool^{\mathsf{N}},\ X=Y \text{ or } X\oplus Y \in 0^*10^*\}$.
 In other words, $E$ is the set of all the edges in the classical 
-$n$-cube. 
-Let $h$ be a function from $\Bool^n$ into $\llbracket 1, n \rrbracket$.
+${\mathsf{N}}$-cube. 
+Let $h$ be a function from $\Bool^{\mathsf{N}}$ into $\llbracket 1, {\mathsf{N}} \rrbracket$.
 Intuitively speaking $h$ aims at memorizing for each node 
-$X \in \Bool^n$ which edge is removed in the Hamiltonian cycle,
-\textit{i.e.} which bit in $\llbracket 1, n \rrbracket$ 
+$X \in \Bool^{\mathsf{N}}$ which edge is removed in the Hamiltonian cycle,
+\textit{i.e.} which bit in $\llbracket 1, {\mathsf{N}} \rrbracket$ 
 cannot be switched.
 
 
 
 We denote by $E_h$ the set $E\setminus\{(X,Y)\mid X\oplus Y =
-0^{n-h(X)}10^{h(X)-1}\}$. This is the set of the modified hypercube, 
-\textit{i.e.}, the $n$-cube where the Hamiltonian cycle $h$ 
+0^{{\mathsf{N}}-h(X)}10^{h(X)-1}\}$. This is the set of the modified hypercube, 
+\textit{i.e.}, the ${\mathsf{N}}$-cube where the Hamiltonian cycle $h$ 
 has been removed.
 
 We define the Markov matrix $P_h$ for each line $X$ and 
 each column $Y$  as follows:
 $$\left\{
 \begin{array}{ll}
-P_h(X,X)=\frac{1}{2}+\frac{1}{2n} & \\
+P_h(X,X)=\frac{1}{{\mathsf{N}}} & \\
 P_h(X,Y)=0 & \textrm{if  $(X,Y)\notin E_h$}\\
-P_h(X,Y)=\frac{1}{2n} & \textrm{if $X\neq Y$ and $(X,Y) \in E_h$}
+P_h(X,Y)=\frac{1}{{\mathsf{N}}} & \textrm{if $X\neq Y$ and $(X,Y) \in E_h$}
 \end{array}
 \right.
 $$ 
 
-We denote by $\ov{h} : \Bool^n \rightarrow \Bool^n$ the function 
-such that for any $X \in \Bool^n $, 
-$(X,\ov{h}(X)) \in E$ and $X\oplus\ov{h}(X)=0^{n-h(X)}10^{h(X)-1}$. 
-The function $\ov{h}$ is said {\it square-free} if for every $X\in \Bool^n$,
+We denote by $\ov{h} : \Bool^{\mathsf{N}} \rightarrow \Bool^{\mathsf{N}}$ the function 
+such that for any $X \in \Bool^{\mathsf{N}} $, 
+$(X,\ov{h}(X)) \in E$ and $X\oplus\ov{h}(X)=0^{{\mathsf{N}}-h(X)}10^{h(X)-1}$. 
+The function $\ov{h}$ is said {\it square-free} if for every $X\in \Bool^{\mathsf{N}}$,
 $\ov{h}(\ov{h}(X))\neq X$. 
 
 \begin{Lemma}\label{lm:h}
@@ -186,24 +186,25 @@ If $\ov{h}$ is bijective and square-free, then $h(\ov{h}^{-1}(X))\neq h(X)$.
 
 \begin{proof}
 Let $\ov{h}$ be bijective.
-Let $k\in \llbracket 1, n \rrbracket$ s.t. $h(\ov{h}^{-1}(X))=k$.
+Let $k\in \llbracket 1, {\mathsf{N}} \rrbracket$ s.t. $h(\ov{h}^{-1}(X))=k$.
 Then $(\ov{h}^{-1}(X),X)$ belongs to $E$ and 
-$\ov{h}^{-1}(X)\oplus X = 0^{n-k}10^{k-1}$.
+$\ov{h}^{-1}(X)\oplus X = 0^{{\mathsf{N}}-k}10^{k-1}$.
 Let us suppose $h(X) = h(\ov{h}^{-1}(X))$. In such a case, $h(X) =k$.
 By definition of $\ov{h}$, $(X, \ov{h}(X)) \in E $ and 
-$X\oplus\ov{h}(X)=0^{n-h(X)}10^{h(X)-1} = 0^{n-k}10^{k-1}$.
+$X\oplus\ov{h}(X)=0^{{\mathsf{N}}-h(X)}10^{h(X)-1} = 0^{{\mathsf{N}}-k}10^{k-1}$.
 Thus $\ov{h}(X)= \ov{h}^{-1}(X)$, which leads to $\ov{h}(\ov{h}(X))= X$.
 This contradicts the square-freeness of $\ov{h}$.
 \end{proof}
 
 Let $Z$ be a random variable that is uniformly distributed over
-$\llbracket 1, n \rrbracket \times \Bool$.
-For $X\in \Bool^n$, we
-define, with $Z=(i,b)$,  
+$\llbracket 1, {\mathsf{N}}$.
+For $X\in \Bool^{\mathsf{N}}$, we
+define, with $Z=i$,  
 $$
 \left\{
 \begin{array}{ll}
-f(X,Z)=X\oplus (0^{n-i}10^{i-1}) & \text{if } b=1 \text{ and } i\neq h(X),\\
+%f(X,Z)=X\oplus (0^{{\mathsf{N}}-i}10^{i-1}) & \text{if } b=1 \text{ and } i\neq h(X),\\
+f(X,Z)=X\oplus (0^{{\mathsf{N}}-i}10^{i-1}) & \text{if $i\neq h(X)$},\\
 f(X,Z)=X& \text{otherwise.} 
 \end{array}\right.
 $$
@@ -215,18 +216,18 @@ $$
 
 
 
-%%%%%%%%%%%%%%%%%%%%%%%%%%%ù
+%%%%%%%%%%%%%%%%%%%%%%%%%%%ù
 %\section{Stopping time}
 
-An integer $\ell\in \llbracket 1,n \rrbracket$ is said {\it fair} 
+An integer $\ell\in \llbracket 1,{\mathsf{N}} \rrbracket$ is said {\it fair} 
 at time $t$ if there
-exists $0\leq j <t$ such that $Z_{j+1}=(\ell,\cdot)$ and $h(X_j)\neq \ell$.
-In other words, there exist a date $j$ before $t$ where 
-the first element of the random variable $Z$ is exactly $l$ 
+exists $0\leq j <t$ such that $Z_{j+1}=\ell$ and $h(X_j)\neq \ell$.
+In other words, there exist a date $j$ before $t$ where
+the random variable $Z$ is  $l$ 
 (\textit{i.e.}, $l$ is the strategy at date $j$) 
 and where the configuration $X_j$ allows to traverse the edge $l$.  
  
-Let $\ts$ be the first time all the elements of $\llbracket 1, n \rrbracket$
+Let $\ts$ be the first time all the elements of $\llbracket 1, {\mathsf{N}} \rrbracket$
 are fair. The integer $\ts$ is a randomized stopping time for
 the Markov chain $(X_t)$.
 
@@ -237,10 +238,11 @@ The integer $\ts$ is a strong stationary time.
 
 \begin{proof}
 Let $\tau_\ell$ be the first time that $\ell$ is fair. The random variable
-$Z_{\tau_\ell}$ is of the form $(\ell,b)$ %with $\delta\in\{0,1\}$ and
-such that 
-$b=1$ with probability $\frac{1}{2}$ and $b=0$ with probability
-$\frac{1}{2}$. Since $h(X_{\tau_\ell-1})\neq\ell$ the value of the $\ell$-th
+$Z_{\tau_\ell}$ is of the form $\ell$ %with $\delta\in\{0,1\}$ and
+% such that 
+% $b=1$ with probability $\frac{1}{2}$ and $b=0$ with probability
+% $\frac{1}{2}$.
+Since $h(X_{\tau_\ell-1})\neq\ell$ the value of the $\ell$-th
 bit of $X_{\tau_\ell}$ 
 is $0$ or $1$ with the same probability ($\frac{1}{2}$).
 
@@ -252,15 +254,15 @@ lemma.\end{proof}
 
 \begin{Theo} \label{prop:stop}
 If $\ov{h}$ is bijective and square-free, then
-$E[\ts]\leq 8n^2+ n\ln (n+1)$. 
+$E[\ts]\leq {\mathsf{N}}^2+ (\mathsf{N}+2)(\ln(\mathsf{N})+2)$. 
 \end{Theo}
 
-For each $X\in \Bool^n$ and $\ell\in\llbracket 1,n\rrbracket$, 
+For each $X\in \Bool^{\mathsf{N}}$ and $\ell\in\llbracket 1,{\mathsf{N}}\rrbracket$, 
 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
-$$S_{X,\ell}=\min \{t \geq 1\mid h(X_{t-1})\neq \ell\text{ and }Z_t=(\ell,.)\text{ and } X_0=X\}.$$
+$$S_{X,\ell}=\min \{t \geq 1\mid h(X_{t-1})\neq \ell\text{ and }Z_t=\ell \text{ and } X_0=X\}.$$
 
  We denote by
 $$\lambda_h=\max_{X,\ell} S_{X,\ell}.$$
@@ -268,39 +270,39 @@ $$\lambda_h=\max_{X,\ell} S_{X,\ell}.$$
 
 \begin{Lemma}\label{prop:lambda}
 If $\ov{h}$ is a square-free bijective function, then the inequality 
-$E[\lambda_h]\leq 8n^2$ is established.
+$E[\lambda_h]\leq 2{\mathsf{N}}^2$ is established.
 
 \end{Lemma}
 
 \begin{proof}
-For every $X$, every $\ell$, one has $\P(S_{X,\ell})\leq 2)\geq
-\frac{1}{4n^2}$. 
+For every $X$, every $\ell$, one has $\P(S_{X,\ell}\leq 2)\geq
+\frac{1}{{\mathsf{N}}^2}$. 
 Let $X_0= X$.
 Indeed, 
 \begin{itemize}
 \item if $h(X)\neq \ell$, then
-$\P(S_{X,\ell}=1)=\frac{1}{2n}\geq \frac{1}{4n^2}$. 
+$\P(S_{X,\ell}=1)=\frac{1}{{\mathsf{N}}}\geq \frac{1}{{\mathsf{N}}^2}$. 
 \item otherwise, $h(X)=\ell$, then
 $\P(S_{X,\ell}=1)=0$.
 But in this case, intutively, it is possible to move
-from $X$ to $\ov{h}^{-1}(X)$ (with probability $\frac{1}{2N}$). And in
+from $X$ to $\ov{h}^{-1}(X)$ (with probability $\frac{1}{N}$). And in
 $\ov{h}^{-1}(X)$ the $l$-th bit can be switched. 
 More formally,
 since $\ov{h}$ is square-free,
 $\ov{h}(X)=\ov{h}(\ov{h}(\ov{h}^{-1}(X)))\neq \ov{h}^{-1}(X)$. It follows
 that $(X,\ov{h}^{-1}(X))\in E_h$. We thus have
-$P(X_1=\ov{h}^{-1}(X))=\frac{1}{2N}$. Now, by Lemma~\ref{lm:h},
+$P(X_1=\ov{h}^{-1}(X))=\frac{1}{{\mathsf{N}}}$. Now, by Lemma~\ref{lm:h},
 $h(\ov{h}^{-1}(X))\neq h(X)$. Therefore $\P(S_{x,\ell}=2\mid
-X_1=\ov{h}^{-1}(X))=\frac{1}{2N}$, proving that $\P(S_{x,\ell}\leq 2)\geq
-\frac{1}{4N^2}$.
+X_1=\ov{h}^{-1}(X))=\frac{1}{{\mathsf{N}}}$, proving that $\P(S_{x,\ell}\leq 2)\geq
+\frac{1}{{\mathsf{N}}^2}$.
 \end{itemize}
 
 
 
 
-Therefore, $\P(S_{X,\ell}\geq 3)\leq 1-\frac{1}{4n^2}$. By induction, one
+Therefore, $\P(S_{X,\ell}\geq 3)\leq 1-\frac{1}{{\mathsf{N}}^2}$. By induction, one
 has, for every $i$, $\P(S_{X,\ell}\geq 2i)\leq
-\left(1-\frac{1}{4n^2}\right)^i$.
+\left(1-\frac{1}{{\mathsf{N}}^2}\right)^i$.
  Moreover,
 since $S_{X,\ell}$ is positive, it is known~\cite[lemma 2.9]{proba}, that
 $$E[S_{X,\ell}]=\sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq i).$$
@@ -309,32 +311,57 @@ $$E[S_{X,\ell}]=\sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq i)\leq
 \P(S_{X,\ell}\geq 1)+\P(S_{X,\ell}\geq 2)+2 \sum_{i=1}^{+\infty}\P(S_{X,\ell}\geq 2i).$$
 Consequently,
 $$E[S_{X,\ell}]\leq 1+1+2
-\sum_{i=1}^{+\infty}\left(1-\frac{1}{4n^2}\right)^i=2+2(4n^2-1)=8n^2,$$
+\sum_{i=1}^{+\infty}\left(1-\frac{1}{{\mathsf{N}}^2}\right)^i=2+2({\mathsf{N}}^2-1)=2{\mathsf{N}}^2,$$
 which concludes the proof.
 \end{proof}
 
-Let $\ts^\prime$ be the first time that there are exactly $n-1$ fair
+Let $\ts^\prime$ be the first time that there are exactly ${\mathsf{N}}-1$ fair
 elements. 
 
 \begin{Lemma}\label{lm:stopprime}
-One has $E[\ts^\prime]\leq n \ln (n+1).$
+One has $E[\ts^\prime]\leq (\mathsf{N}+2)(\ln(\mathsf{N})+2)$.
 \end{Lemma}
 
 \begin{proof}
-This is a classical  Coupon Collector's like problem. Let $W_i$ be the
-random variable counting the number of moves done in the Markov chain while
-we had exactly $i-1$ fair bits. One has $\ts^\prime=\sum_{i=1}^{n-1}W_i$.
- But when we are at position $X$ with $i-1$ fair bits, the probability of
- obtaining a new fair bit is either $1-\frac{i-1}{n}$ if $h(X)$ is fair,
- or  $1-\frac{i-2}{n}$ if $h(X)$ is not fair. It follows that 
-$E[W_i]\leq \frac{n}{n-i+2}$. Therefore
-$$E[\ts^\prime]=\sum_{i=1}^{n-1}E[W_i]\leq n\sum_{i=1}^{n-1}
- \frac{1}{n-i+2}=n\sum_{i=3}^{n+1}\frac{1}{i}.$$
-
-But $\sum_{i=1}^{n+1}\frac{1}{i}\leq 1+\ln(n+1)$. It follows that
-$1+\frac{1}{2}+\sum_{i=3}^{n+1}\frac{1}{i}\leq 1+\ln(n+1).$
-Consequently,
-$E[\ts^\prime]\leq n (-\frac{1}{2}+\ln(n+1))\leq n\ln(n+1)$.
+This is a classical  Coupon Collector's like problem. Let $W_i$ 
+be the time to obtain the $i$-th fair bit
+after $i-1$ fair bits have been obtained.
+One has $\ts^\prime=\sum_{i=1}^{{\mathsf{N}}}W_i$.
+
+At position $X$ with $i-1$ fair bits,
+we  do not obtain a new fair if $Z$ is one of the $i-1$ already fair bits
+or if $Z$ is a new fair bit but $h(X)$ is $Z$.  
+This occures with probability 
+$p 
+= \frac{i-1}{{\mathsf{N}}} + \frac{n-i+1}{\mathsf{N}}.\frac{1}{\mathsf{N}}
+=\frac{i(\mathsf{N}-1) +1}{\mathsf{N^2}}
+$. 
+The random variable $W_i$ has a geometric distribution 
+\textit{i.e.}, $P(W_i = k) = p^{k-1}.(1-p)$ and 
+$E(W_i) = \frac{\mathsf{N^2}}{i(\mathsf{N}-1) +1}$.
+Therefore
+$$E[\ts^\prime]=\sum_{i=1}^{{\mathsf{N}}}E[W_i]
+=\frac{\mathsf{N^2}}{\mathsf{N}(\mathsf{N}-1) +1}  + \sum_{i=1}^{{\mathsf{N}}-1}E[W_i].$$
+
+A simple study of the function $\mathsf{N} \mapsto \frac{\mathsf{N^2}}{\mathsf{N}(\mathsf{N}-1) +1}$ shows that it is bounded by $\frac{4}{3} \leq 2$.
+For the second term, we successively have 
+$$
+\sum_{i=1}^{{\mathsf{N}}-1}E[W_i] 
+= \mathsf{N}^2\sum_{i=1}^{{\mathsf{N}}-1} \frac{1}{i(\mathsf{N}-1) +1} 
+\leq \mathsf{N}^2\sum_{i=1}^{{\mathsf{N}}-1} \frac{1}{i(\mathsf{N}-1)} 
+\leq \frac{\mathsf{N}^2}{\mathsf{N}-1}\sum_{i=1}^{{\mathsf{N}}-1} \frac{1}{i} 
+\leq (\mathsf{N}+2)\sum_{i=1}^{{\mathsf{N}}-1} \frac{1}{i} 
+$$
+
+
+It is well known that 
+$\sum_{i=1}^{{\mathsf{N}}-1}\frac{1}{i}\leq 1+\ln({\mathsf{N}}-1)$.
+It follows that
+$2+(\mathsf{N}+2)\sum_{i=1}^{{\mathsf{N}}-1}\frac{1}{i}
+\leq 
+2+(\mathsf{N}+2)(\ln(\mathsf{N}-1)+1)
+\leq 
+(\mathsf{N}+2)(\ln(\mathsf{N})+2)$.
 \end{proof}
 
 One can now prove Theorem~\ref{prop:stop}.