AENC/resampling_chain Changeset - f66b48a711a7 · Centrum Wiskunde & Informatica (CWI)

@@ -245,42 +245,26 @@

\newpage

\section{Proving that $a_k^{(k+1)}=a_k^{(n)}$ for all $n>k$}

It is useful to introduce some new notation. We will consider variations of the Markov Chains:

We consider $R^{(n)}(p)$ as a power series in $p$ and our main aim in this section is to show that $R^{(n)}(p)$ and $R^{(n+k)}(p)$ are the same up to order $n-1$.

The proof will consider variations of the Markov Chain:

\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 only resample its single neighbour so it draws only two new bits.

The process on the finite chain has the following modification at the boundary: if a boundary site is resampled, it can only resample itself and its single neighbour so it draws only two new bits.

We use the notation $\E^{(n)}$,$\E^{[a,b]}$ and $\E^{[n]}$ similarly for denoting expectations.

\begin{definition}[Paths]

	We define a \emph{path} of the Markov Chain as a sequence of states and resampling choices $\xi=((b_0,r_0),(b_1,r_1),...,(b_k,r_k)) \in (\{0,1\}^n\times[n])^k$ indicating that at time $t$ Markov Chain was in state $b_t\in\{0,1\}^n$ and then resampled site $r_t$. We denote by $\mathbb{P}[\xi]$ the probability that the process followed this path.

	We denote by $\paths{b}$ the set of all valid paths $\xi$ that start in state $b$ and end in state $\mathbf{1} := 1^n$.

\end{definition}

We can write the expected number of resamplings per site $R^{(n)}(p)$ as

\begin{align}

R^{(n)}(p) &= \frac{1}{n}\sum_{b\in\{0,1\}^{n}} \rho_b \; R_b(p) \label{eq:originalsum} ,

\end{align}

where $R_b(p)$ is the expected number of resamplings when starting from configuration $b$

\begin{align*}

R_b(p) &= \sum_{\xi \in \paths{b}} \mathbb{P}[\xi] \cdot |\xi| .

\end{align*}

We consider $R^{(n)}(p)$ as a power series in $p$ and our main aim in this section is to show that $R^{(n)}(p)$ and $R^{(n+k)}(p)$ are the same up to order $n-1$.

%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 $v\in[n]$ we define

    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 $A$ and any $v\in[n]$, define

    \begin{align*}

        \P^{[n]}_{b_v=1}(A) &= \P^{[n]}(A \;|\; v\text{ is initialized to }1),

        \P^{(n)}_b(A) &= \P^{(n)}(A \;|\; \start{b}) \\

        \P^{[n]}_{b_v=1}(A) &= \P^{[n]}(A \;|\; v\text{ is initialized to }1) \\

        \P^{[n]}_{b_v=b_w=1}(A) &= \P^{[n]}(A \;|\; v\text{ and }w\text{ are initialized to }1) ,

    \end{align*}

    and we define similarly $\P^{[n]}_{b_v=b_w=1}(A)$ for $v,w\in[n]$.

    The last two probabilities are not conditioned on any other bits of the starting state.

\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}

@@ -292,13 +276,13 @@ We consider $R^{(n)}(p)$ as a power series in $p$ and our main aim in this secti

%    \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}

\begin{wrapfigure}[7]{r}{0.25\textwidth} % The first [] argument is number of lines that are narrowed

    \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

    Let $b=b_1\land b_2\in\{0,1\}^n$ be a state with two separated groups of zeroes 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)

&=

@@ -393,19 +377,6 @@ The following lemma considers two vertices $v,w$ that are never ``crossed'' so t

        \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)},A_2).

    \end{align*}

    The second equality follows directly from $\mathbb{P}(A\mid B)=\mathbb{P}(A,B)/\mathbb{P}(B)$ and setting $A_1,A_2$ to the always-true event.

    %For the third equality, by the same reasoning we can decompose the paths

    %\begin{align*}

    %    \P^{(n)}_b(\mathrm{NZ}^{(v,w)},A_1,A_2) R_{b,\mathrm{NZ}^{(v,w)},A_1,A_2}

    %    &\equiv \sum_{\substack{\xi\in\start{b}\\\xi \in \mathrm{NZ}^{(v,w)}\cap A_1\cap A_2}} \P^{(n)}[\xi] |\xi| \\

    %    &= \sum_{\substack{\xi_1\in\start{b_1}\\\xi_1 \in \mathrm{NZ}^{(v,w)}\cap A_1}}

    %      \sum_{\substack{\xi_2\in\start{b_2}\\\xi_2 \in \mathrm{NZ}^{(v,w)}\cap A_2}}

    %    \P^{(n)}[\xi_1]\P^{(n)}[\xi_2] (|\xi_1| + |\xi_2|) \\

    %    &=

    %    \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)},A_2) \P^{(n)}_{b_1}(\mathrm{NZ}^{(v,w)},A_1) R_{b_1,\mathrm{NZ}^{(v,w)},A_1} \\

    %    &\quad +

    %    \P^{(n)}_{b_1}(\mathrm{NZ}^{(v,w)},A_1) \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)},A_2) R_{b_2,\mathrm{NZ}^{(v,w)},A_2} .

    %\end{align*}

    %Dividing by $\P^{(n)}_b(\mathrm{NZ}_{(v,w)},A_1,A_2)$ and using the first equality gives the desired result.

\end{proof}

\begin{lemma}[Conditional independence 2] \label{lemma:eventindependenceNew}

@@ -440,7 +411,7 @@ The following lemma considers two vertices $v,w$ that are never ``crossed'' so t

&=

        \P^{[v,w]}_{b|_{[v,w]}}(\mathrm{NZ}^{(v,w)} \cap A)

        \; \cdot \;

        \P^{[v,w]}_{b|_{[w,v]}}(\mathrm{NZ}^{(v,w)} \cap B)

        \P^{[w,v]}_{b|_{[w,v]}}(\mathrm{NZ}^{(v,w)} \cap B)

    \end{align*}

    Note that this also holds if $b$ has zeroes on the boundary (i.e. $b_v=0$ or $b_w=0$), because then both sides of the equations are zero.

    For the starting state we have the expression $\P^{(n)}(\start{b}) = (1-p)^{|b|} p^{n-|b|}$ so it splits into a product