Changeset - f99db951a095
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Tom Bannink - 8 years ago 2017-09-07 17:27:27
tom.bannink@cwi.nl
Change figure position to wrapfigure
1 file changed with 13 insertions and 8 deletions:
main.tex
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main.tex
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\documentclass[a4paper,11pt,english,final]{article}
 
\pdfoutput=1
 

	
 
\usepackage[utf8]{inputenc}
 
\usepackage[english]{babel}
 
\usepackage{fullpage}
 

	
 
\usepackage{graphics}
 
\usepackage{diagbox}
 
\usepackage[table]{xcolor}% http://ctan.org/pkg/xcolor
 
\usepackage{graphicx}
 
\usepackage{wrapfig}
 
\usepackage{caption}
 
\captionsetup{compatibility=false}
 
\graphicspath{{./}}
 

	
 

	
 
\usepackage{tikz}
 
\usepackage{amssymb}
 
\usepackage{mathtools}
 
\usepackage{bm}
 
\usepackage{bbm}
 
%\usepackage{bbold}
 
\usepackage{verbatim}
 

	
 
%for correcting large brackets spacing
 
\usepackage{mleftright}\mleftright
 

	
 
\usepackage{algorithm}
 
\usepackage{algorithmic}
 
\usepackage{enumitem}
 
\usepackage{float}
 

	
 
%\usepackage{titling}
 

	
 
%\setlength{\droptitle}{-5mm}  
 
@@ -432,76 +433,80 @@ where we used the identity $\sum_{a\in\{0,1\}^l} (-1)^{|a|} = 0$.
 
It is useful to introduce some new notation. We will consider variations of the Markov Chains:
 
\begin{itemize}
 
    \item $\P^{(n)}$ refers to the original process on the length-$n$ cycle.
 
    \item $\P^{[a,b]}$ or $\P^{[n]}$ refers to a similar Markov Chain but on a finite chain ($[a,b]$ or $[1,n]$).
 
\end{itemize}
 
The process on the finite chain has the following modification at the boundary: if a boundary site is resampled, it can not resample one of its neighbors so it ignores it and only draws two new bits.
 

	
 
%Note that an \emph{event} is a subset of all possible paths of the Markov Chain.
 
\begin{definition}[Events conditioned on starting state] \label{def:conditionedevents}
 
    For any state $b\in\{0,1\}^n$, define $\start{b}$ as the event that the starting state of the chain is the state $b$. For any event $A$, define
 
    \begin{align*}
 
        \P^{(n)}_b(A) &= \P^{(n)}(A \;|\; \start{b}) %\\
 
        %R_{b,A} &= \mathbb{E}( \#resamples \;|\; A \; , \; \start{b})
 
    \end{align*}
 
    Furthermore, for the Markov Chain on the finite chain, define
 
    \begin{align*}
 
        \P^{[n]}_{\partial=1}(A) &= \P^{[n]}(A \;|\; \text{boundary is initialized to }1)
 
    \end{align*}
 
    where the boundary of $[n]$ is site $1$ and site $n$, and the boundary of $[a,b]$ are $a$ and $b$.
 
\end{definition}
 
%Note that we have $\P^{(n)}(\start{b}) = (1-p)^{|b|}p^{n-|b|}$ by definition of our Markov Chain.
 
\begin{definition}[Vertex visiting event] \label{def:visitingResamplings}
 
    Denote by $\mathrm{Z}^{(v)}$ the event that site $v$ becomes zero at any point in time before the Markov Chain terminates. Denote the complement by $\mathrm{NZ}^{(v)}$, i.e. the event that site $v$ does \emph{not} become zero before it terminates. Furthermore define $\mathrm{NZ}^{(v,w)} := \mathrm{NZ}^{(v)} \cap \mathrm{NZ}^{(w)}$, i.e. the event that \emph{both} $v$ and $w$ do not become zero before termination.
 
\end{definition}
 
\begin{figure}
 
	\begin{center}
 
    	\includegraphics{diagram_groups.pdf}
 
    \end{center}
 
    \caption{\label{fig:separatedgroups} Illustration of setup of Lemma \ref{lemma:eventindependence}. Here $b_1,b_2\in\{0,1\}^n$ are bitstrings such that all zeroes of $b_1$ and all zeroes of $b_2$ are separated by two indices $v,w$.}
 
\end{figure}
 
\begin{lemma}[Conditional independence] \label{lemma:eventindependence} \label{claim:eventindependence}
 
%\begin{figure}
 
%	\begin{center}
 
%    	\includegraphics{diagram_groups.pdf}
 
%    \end{center}
 
%    \caption{\label{fig:separatedgroups} Illustration of setup of Lemma \ref{lemma:eventindependence}. Here $b_1,b_2\in\{0,1\}^n$ are bitstrings such that all zeroes of $b_1$ and all zeroes of $b_2$ are separated by two indices $v,w$.}
 
%\end{figure}
 
\begin{wrapfigure}{r}{0.25\textwidth}
 
    \centering
 
    \includegraphics{diagram_groups.pdf}
 
    \caption{\label{fig:separatedgroups} Lemma \ref{lemma:eventindependence}.}
 
\end{wrapfigure}
 
The following lemma considers two vertices $v,w$ that are never ``crossed'' so that two halves of the cycle become independent.\begin{lemma}[Conditional independence] \label{lemma:eventindependence} \label{claim:eventindependence}
 
    Let $b=b_1\land b_2\in\{0,1\}^n$ be a state with two groups of zeroes that are separated by at least one site inbetween, as in Figure \ref{fig:separatedgroups}. Let $v$, $w$ be any indices inbetween the groups, such that $b_1$ lies on one side of them and $b_2$ on the other, as shown in the figure. Furthermore, let $A_1$ be any event that depends only on the sites ``on the $b_1$ side of $v,w$'', and similar for $A_2$ (for example $\mathrm{Z}^{(i)}$ for an $i$ on the correct side). Then we have
 
    \begin{align*}
 
        \P^{(n)}_b(\mathrm{NZ}^{(v,w)}, A_1, A_2)
 
        &=
 
        \P^{(n)}_{b_1}(\mathrm{NZ}^{(v,w)}, A_1)
 
        \; \cdot \;
 
        \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)}, A_2) \\
 
        \P^{(n)}_b(A_1, A_2 \mid \mathrm{NZ}^{(v,w)})
 
        &=
 
        \P^{(n)}_{b_1}(A_1 \mid \mathrm{NZ}^{(v,w)})
 
        \; \cdot \;
 
        \P^{(n)}_{b_2}(A_2 \mid \mathrm{NZ}^{(v,w)}) .%\\
 
        %R_{b,\mathrm{NZ}^{(v,w)},A_1,A_2}
 
        %&=
 
        %R_{b_1,\mathrm{NZ}^{(v,w)},A_1}
 
        %\; + \;
 
        %R_{b_2,\mathrm{NZ}^{(v,w)},A_2}
 
    \end{align*}
 
    %up to any order in $p$.
 
\end{lemma}
 
The lemma says that conditioned on $v$ and $w$ not being crossed, the two halves of the cycle are independent. 
 

	
 
\begin{proof}
 
    From any path $\xi\in\start{b} \cap \mathrm{NZ}^{(v,w)}$ we can construct paths $\xi_1\in\start{b_1}\cap \mathrm{NZ}^{(v,w)}$ and $\xi_2\in\start{b_2}\cap\mathrm{NZ}^{(v,w)}$ as follows. Let us write the path $\xi$ as
 
    $$\xi=\left( (\text{initialize }b), (z_1, s_1, r_1), (z_2, s_2, r_2), ..., (z_{|\xi|}, s_{|\xi|}, r_{|\xi|}) \right)$$
 
    where $z_i\in[n]$ denotes the number of zeroes in the state before the $i$th step, $s_i\in [n]$ denotes the site that was resampled and $r_i\in \{0,1\}^3$ is the result of the three resampled bits. We have
 
    \begin{align*}
 
        \P^{(n)}_b[\xi] &= \P(\text{pick }s_1 | z_1) \P(r_1) \P(\text{pick }s_2 | z_2) \P(r_2) \cdots \P(\text{pick }s_{|\xi|} | z_{|\xi|}) \P(r_{|\xi|}) \\
 
                &= \frac{1}{z_1} \P(r_1) \frac{1}{z_2} \P(r_2) \cdots \frac{1}{z_{|\xi|}} \P(r_{|\xi|}) .
 
    \end{align*}
 
    To construct $\xi_1$ and $\xi_2$, start with $\xi_1 = \left( (\text{initialize }b_1) \right)$ and $\xi_2 = \left( (\text{initialize }b_2) \right)$. For each step $(z_i,s_i,r_i)$ in $\xi$ do the following: if $s_i$ is ``on the $b_1$ side of $v,w$'' then append $(z^{(1)}_i,s_i,r_i)$ to $\xi_1$ and if its ``on the $b_2$ side of $v,w$'' then append $(z^{(2)}_i,s_i,r_i)$ to $\xi_2$. Here $z^{(1)}_i$ is the number of zeroes that were on the $b_1$ side and $z^{(2)}_i$ is the number of zeroes on the $b_2$ side so we have $z_i = z^{(1)}_i + z^{(2)}_i$.
 
    %Let the resulting paths be
 
    %\begin{align*}
 
    %    \xi_1 &= \left( (z^{(1)}_{a_1}, s_{a_1}, r_{a_1}), (z^{(1)}_{a_2}, s_{a_2}, r_{a_2}), ..., (z^{(1)}_{a_{|\xi_1|}}, s_{a_{|\xi_1|}}, r_{a_{|\xi_1|}}) \right) \\
 
    %    \xi_2 &= \left( (z^{(2)}_{b_1}, s_{b_1}, r_{b_1}), (z^{(2)}_{b_2}, s_{b_2}, r_{b_2}), ..., (z^{(2)}_{b_{|\xi_1|}}, s_{b_{|\xi_1|}}, r_{b_{|\xi_1|}}) \right)
 
    %\end{align*}
 
    Now $\xi_1$ is a valid (terminating) path from $b_1$ to $\mathbf{1}$, because in the original path $\xi$, all zeroes ``on the $b_1$ side'' have been resampled by resamplings ``on the $b_1$ side''. Since the sites $v,w$ inbetween never become zero, there can not be any zero ``on the $b_1$ side'' that was resampled by a resampling ``on the $b_2$ side''.
 
    Vice versa, any two paths $\xi_1\in\start{b_1}\cap \mathrm{NZ}^{(v,w)}$ and $\xi_2\in\start{b_2}\cap\mathrm{NZ}^{(v,w)}$ also induce a path $\xi\in\start{b} \cap \mathrm{NZ}^{(v,w)}$ by simply interleaving the resampling positions. Note that $\xi_1,\xi_2$ actually induce $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths $\xi$ because of the possible orderings of interleaving the resamplings in $\xi_1$ and $\xi_2$.
 
    For a fixed $\xi_1,\xi_2$ we will now show the following:
 
    \begin{align*}
 
        \sum_{\substack{\xi\in\start{b} \cap \mathrm{NZ}^{(v,w)} \text{ s.t.}\\ \xi \text{ decomposes into } \xi_1,\xi_2 }} \P^{(n)}_b[\xi] &=
 
        \sum_{\text{interleavings of }\xi_1,\xi_2} \P(\text{interleaving}) \cdot \P^{(n)}_{b_1}[\xi_1] \cdot \P^{(n)}_{b_2}[\xi_2] \\
 
        &= \P^{(n)}_{b_1}[\xi_1] \cdot \P^{(n)}_{b_2}[\xi_2]
 
    \end{align*}
 
    where both sums are over $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ terms.
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