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\title{Criticality of resampling on the cycle / in the evolution model}

%\author{?\thanks{QuSoft, CWI and University of Amsterdam, the Netherlands. \texttt{?@cwi.nl} }

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\begin{document}

	\maketitle

	\begin{abstract}

		The model we consider is the following~\cite{ResampleLimit}: We have a cycle of length $n\geq 3$. Initially we set each site to $0$ or $1$ independently at each site, such that we set it $0$ with probability $p$. After that in each step we select a random vertex with $0$ value and resample it together with its two neighbours assigning $0$ with probability $p$ to each vertex just as initially. The question we try to answer is what is the expected number of resamplings performed before reaching the all $1$ state. 

		We present strong evidence for a remarkable critical behaviour. We conjecture that there exists some $p_c\approx0.62$, such that for all $p\in[0,p_c)$ the expected number of resamplings is bounded by a $p$ dependent constant times $n$, whereas for all $p\in(p_c,1]$ the expected number of resamplings is exponentially growing in $n$.

	\end{abstract}

	%Let $R(n)$ denote this quantity for a length $n\geq 3$ cycle.

	We can think about the resampling procedure as a Markov chain. To describe the corresponding matrix we introduce some notation. For $b\in\{0,1\}^n$ let $r(b,i,(x_{-1},x_0,x_1))$ denote the bit string which differs form $b$ by replacing the bits at index $i-1$,$i$ and $i+1$ with the values in $x$, interpreting the indices $\!\!\!\!\mod n$. Also for $x\in\{0,1\}^k$ let $p(x)=p((x_1,\ldots,x_k))=\prod_{i=1}^{k}p^{(1-x_i)}(1-p)^{x_i}$. Now we can describe the matrix of the Markov chain. We use row vectors for the elements of the probability distribution indexed by bitstrings of length $n$. Let $M_{(n)}$ denote the matrix of the leaking Markov chain:

	$$

		M_{(n)}=\sum_{b\in\{0,1\}^n\setminus{\{1\}^n}}\sum_{i\in[n]:b_i=0}\sum_{x\in\{0,1\}^3}E_{(b,r(b,i,x))}\frac{p(x)}{n-|b|},

	$$

	where $E_{(i,j)}$ denotes the matrix that is all $0$ except $1$ at the $(i,j)$th entry.

	We want to calculate the average number of resamplings $R^{(n)}$, which we define as the expected number of resamplings divided by $n$. For this let $\rho,\mathbbm{1}\in[0,1]^{2^n}$ be indexed with elements of $\{0,1\}^n$ such that $\rho_b=p(b)$ and $\mathbbm{1}_b=1$. Then we use that the expected number of resamplings is just the hitting time of the Markov chain:

	\begin{align*}

		R^{(n)}:&=\mathbb{E}(\#\{\text{resampling before termination}\})/n\\

		&=\sum_{k=1}^{\infty}P(\text{at least } k \text{ resamplings are performed})/n\\

		&=\sum_{k=1}^{\infty}\rho M_{(n)}^k \mathbbm{1}/n\\

		&=\sum_{k=0}^{\infty}a^{(n)}_k p^k

	\end{align*}

	\begin{table}[]

	\centering

	\caption{Table of the coefficients $a^{(n)}_k$}

\begin{figure}

	\begin{center}

    	\includegraphics{diagram_gap.pdf}

    \end{center}

    \caption{\label{fig:diametergap} Illustration of Definition \ref{def:diameter}. A set $C=\{1,2,4,7,9\}\subseteq[n]$ consisting of 5 positions is shown by the red dots. The smallest interval containing $C$ is $[1,9]$, so the diameter is $\diam{C}=9$. The blue squares denote the set $\gaps{C} = \{3,5,6,8\}$. The dotted line at the top depicts the rest of the cycle which may be much larger. The largest gap of $C$ is $\maxgap{C}=2$ which is the largest connected component of $\gaps{C}$.}

\end{figure}

\begin{claim}[Strong cancellation claim] \label{claim:strongcancel}

	The lowest order term in

    \begin{align*}

        \sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R_{C(f)} ,

    \end{align*}

	is $p^{\diam{C}}$ when $n$ is large enough. All lower order terms cancel out.

\end{claim}

Example: for $C_0=\{1,2,4,7,9\}$ (the configuration shown in Figure \ref{fig:diametergap}) we computed the quantity up to order $p^{20}$ in an infinite system:

\begin{align*}

	\sum_{f\in\{0,1'\}^{|C_0|}} \rho_{C_0(f)} R_{C_0(f)} &= 0.0240278 p^{9} + 0.235129 p^{10} + 1.24067 p^{11} + 4.71825 p^{12} \\

    &\quad + 14.5555 p^{13} + 38.8307 p^{14} + 93.2179 p^{15} + 206.837 p^{16}\\

    &\quad + 432.302 p^{17} + 862.926 p^{18} + 1662.05 p^{19} + 3112.9 p^{20} + \mathcal{O}(p^{21})

\end{align*}

and indeed the lowest order is $\diam{C}=9$.

~

A weaker version of the claim is that if $C$ contains a gap of size $k$, then the sum is zero up to and including order $p^{|C|+k-1}$.

\begin{claim}[Weak cancellation claim] \label{claim:weakcancel}

	For $C\subseteq[n]$ a configuration of slot positions, the lowest order term in

    \begin{align*}

        \sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R_{C(f)} ,

    \end{align*}

    is at least $p^{|C|+\maxgap{C}}$ when $n$ is large enough. All lower order terms cancel out.

\end{claim}

This weaker version would imply \ref{it:const} but for $\mathcal{O}(k^2)$ as opposed to $k+1$.

\newpage

The reason that claim \ref{claim:strongcancel} would prove \ref{it:const} is the following: to know the value of $a_k^{(n)}$, for any $n\geq k+1$ it is enough to look at configurations $C$ with diameter at most $k$, since larger configurations do not contribute to $a_k^{(n)}$.

For a starting state $b\in\{0,1\}^n$ that \emph{does} give a nonzero contribution, you can take that same starting configuration and translate it to get $n$ other configurations that give the same contribution. (An exception is a starting state like $1010101010...$ which you can only translate twice, but we only have to consider configurations with small diameter, in which case you can make exactly $n$ translations.)

Therefore the coefficient in the expected number of resamplings is a multiple of $n$ which Andr\'as already divided out in the definition of $R^{(n)}(p)$. To show \ref{it:const} we argue that this is the \emph{only} dependency on $n$. This is because there are only finitely many (depending on $k$ but not on $n$) configurations where the $k$ slots are nearby regardless of the value of $n$. So there are only finitely many nonzero contributions after translation symmetry was taken out. For example, when considering all starting configurations with 5 slots one might think there are $\binom{n}{5}$ configurations to consider which would be a dependency on $n$ (more than only the translation symmetry). But since most of these configurations have a diameter larger than $k$, they do not contribute to $a_k$. Only finitely many do and that does not depend on $n$.

~

Section \ref{sec:computerb} shows how to compute $R_b$ (this is not relevant for showing the claim) and the section after that shows how to prove the weaker claim.

\newpage

\subsection{Computation of $R_b$} \label{sec:computerb}

By $R_{101}$ we denote $R_b(p)$ for a $b$ that consists of only $1$s except for a single zero. We compute $R_{101}$ up to second order in $p$. This requires the following transitions.

\begin{align*}

    \framebox{$1 0 1$} &\to \framebox{$1 1 1$} & (1-p)^3 = 1-3p+3p^2-p^3\\

    \hline

    \framebox{$1 0 1$} &\to

        \begin{cases}

            \framebox{$0 1 1$}\\

            \framebox{$1 0 1$}\\

            \framebox{$1 1 0$}

        \end{cases}

        & 3p(1-p)^2 = 3p-6p^2+3p^3\\

    \hline

    \framebox{$1 0 1$} &\to \framebox{$0 1 0$} & p^2(1-p) = p^2-p^3\\

    \framebox{$1 0 1$} &\to

        \begin{cases}

            \framebox{$1 0 0$}\\

            \framebox{$0 0 1$}

        \end{cases}

        & 2p^2(1-p) = 2p^2 - 2p^3\\

    \hline

    \framebox{$1 0 1$} &\to \framebox{$0 0 0$} & p^3

\end{align*}

With this we can write a recursive formula for the expected number of resamples from $101$:

\begin{align*}

    R_{101} &= (1-3p+3p^2 - p^3)(1) + (3p -6p^2 +3p^3) (1+R_{101}) \\

            &\quad + (p^2 - p^3) (1+R_{10101}) + (2p^2-2p^3) (1+R_{1001}) + p^3(1+R_{10001}) \\

			&= 1 + 3 p + 7 p^2 + 14.6667 p^3 + 29 p^4 + 55.2222 p^5 + 102.444 p^6 + 186.36 p^7 \\

            &\quad + 333.906 p^8 + 590.997 p^9 + 1035.58 p^{10} + 1799.39 p^{11} + 3104.2 p^{12} \\

            &\quad+ 5322.18 p^{13} + 9075.83 p^{14} + 15403.6 p^{15} + 26033.4 p^{16} + 43833.5 p^{17} \\

            &\quad+ 73555.2 p^{18} + 123053 p^{19} + 205290 p^{20} + 341620 p^{21} + 567161 p^{22} \\

            &\quad+ 939693 p^{23} + 1.5537\cdot10^{6} p^{24} + 2.56158\cdot10^{6} p^{25} + \mathcal{O}(p^{26})

\end{align*}

where the recursion steps were done with a computer for an infinite line (or a cirlce where $n$ is assumed to be much larger than the largest power of $p$ considered).

Note: in the first line at the second term it uses that with probability $(3p-6p^2 + 3p^3)$ the state goes to $\framebox{$101$}$ and then the expected number of resamplings is $1+R_{101}$. Note that the actual term in the recursive formula should be

$$(3p-6p^2+3p^3)\cdot\left( \sum_{\xi\in\paths{101}} \mathbb{P}[\xi] \cdot \left( 1 + |\xi|\right) \right) = (3p-6p^2+3p^3)\left( p_\mathrm{tot} + R_{101} \right)$$

where $p_\mathrm{tot} := \sum_{\xi\in\paths{b}} \mathbb{P}[\xi]$. However, since the state space is finite (for finite $n$) and there is always a non-vanishing probability to go to $\mathbf{1}$, we know that $p_\mathrm{tot}=1$, i.e. the process terminates almost surely.

\newpage

\subsection{Weak cancellation proof}

Here we prove claim \ref{claim:weakcancel}, the weaker version of the claim. We require the following definition

\begin{definition}[Path independence] \label{def:independence}

	We say two paths $\xi_i\in\paths{b_i}$ ($i=1,2$) of the Markov Chain are \emph{independent} if $\xi_1$ never resamples a site that was ever zero in $\xi_2$ and the other way around. It is allowed that $\xi_1$ resamples a $1$ to a $1$ that was also resampled from $1$ to $1$ by $\xi_2$ and vice versa. If the paths are not independent then we call the paths \emph{dependent}.

\end{definition}

\begin{definition}[Path independence - alternative] \label{def:independence2}

    Equivalently, on the infinite line $\xi_1$ and $\xi_2$ are independent if there is a site `inbetween' them that was never zero in $\xi_1$ and never zero in $\xi_2$. On the cycle $\xi_1$ and $\xi_2$ are independent if there are \emph{two} sites inbetween them that are never zero.

\end{definition}

\begin{claim}[Sum of expectation values] \label{claim:expectationsum}

\end{claim}

\textbf{Example}: For $b_1 = 0111111$ and $b_2 = 1111010$ we have $b=0111010$ and $k=3$. The claim says that the expected time to reach $\mathbf{1}$ from $b$ is the time to make the first group $1$ plus the time to make the second group $1$, as if they are independent. Simulation shows that

\begin{align*}

    R_{b_1} &= 1 + 3p + 7p^2 + 14.67p^3 + 29p^4 + \mathcal{O}(p^5)\\

    R_{b_2} &= 2 + 5p + 10.67p^2 + 21.11p^3+40.26p^4 + \mathcal{O}(p^5)\\

    R_{b} &= 3 + 8p + 17.67p^2 + 34.78p^3+65.27p^4 + \mathcal{O}(p^5)\\

    R_{b_1} + R_{b_2} &= 3 + 8p + 17.67p^2+35.78p^3 + 69.26p^4 +\mathcal{O}(p^5)

\end{align*}

and indeed the sums agree up to order $p^{k-1}=p^2$. When going up to order $p^{k}$ or higher, there will be terms where the groups interfere so they are no longer independent.

~

\begin{proof}

    Consider a path $\xi_1\in\paths{b_1}$ and a path $\xi_2\in\paths{b_2}$ such that $\xi_1$ and $\xi_2$ are independent (Definition \ref{def:independence} or \ref{def:independence2}). The paths $\xi_1,\xi_2$ induce $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ different paths of total length $|\xi_1|+|\xi_2|$ in $\paths{b_1\land b_2}$. In the sums $R_{b_1}$ and $R_{b_2}$, the contribution of these paths are $\mathbb{P}[\xi_1]\cdot |\xi_1|$ and $\mathbb{P}[\xi_2]\cdot |\xi_2|$. The next diagram shows how these $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths contribute to $R_{b_1\land b_2}$. Point $(i,j)$ in the grid indicates that $i$ steps of $\xi_1$ have been done and $j$ steps of $\xi_2$ have been done. At every point (except the top and right edges of the grid) one has to choose between doing a step of $\xi_1$ or a step of $\xi_2$. The number of zeroes in the current state determine the probabilities with which this happens (beside the probabilities associated to the two original paths already). The grid below shows that at a certain point one can choose to do a step of $\xi_1$ with probability $p_i$ or a step of $\xi_2$ with probability $1-p_i$. These $p_i$ could in principle be different at every point in this grid. The weight of such a new path $\xi\in\paths{b_1\land b_2}$ is $p_\mathrm{grid}\cdot\mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2]$ where $p_\mathrm{grid}$ is the weight of the path in the diagram. By induction one can show that the sum over the $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ different terms $p_\mathrm{grid}$ is $1$.

\begin{center}

\includegraphics{diagram_paths.pdf}

\end{center}

 Hence the contribution of all $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths together to $R_{b_1\land b_2}$ is given by

\[

\mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2]\cdot(|\xi_1|+|\xi_2|) = \mathbb{P}[\xi_2]\cdot\mathbb{P}[\xi_1]\cdot|\xi_1| \;\; + \;\; \mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2]\cdot|\xi_2|.

\]

Ideally we would now like to sum this expression over all possible paths $\xi_1,\xi_2$ and use $p_\mathrm{tot}:=\sum_{\xi\in\paths{b_i}} \mathbb{P}[\xi] = 1$ (which also holds up to arbitrary order in $p$). The above expression would then become $R_{b_1} + R_{b_2}$. However, not all paths in the sum would satisfy the independence condition so it seems we can't do this. We now argue that it works up to order $p^{k-1}$.

For all $\xi\in\paths{b_1\land b_2}$ we have that \emph{either} $\xi$ splits into two independent paths $\xi_1,\xi_2$ as above, \emph{or} it does not. In the latter case, when $\xi$ can not be split like that, we know $\mathbb{P}[\xi]$ contains a power $p^k$ or higher because there is a gap of size $k$  and the paths must have moved at least $k$ times `towards each other' (for example one path moves $m$ times to the right and the other path moves $k-m$ times to the left). So the total weight of such a combined path is at least order $p^k$. Therefore we have

\[

	R_{b_1\land b_2} = \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1| + \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_1]\mathbb{P}[\xi_2]|\xi_2| + \sum_{\mathclap{\xi\;\mathrm{dependent}}} \mathbb{P}[\xi]|\xi|.

\]

where last sum only contains only terms of order $p^{k}$ or higher. Now for the first sum, note that

\[

	\sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|

    = \sum_{\xi_1\in\paths{b_1}} \sum_{\substack{\xi_2\in\paths{b_2}\\ \text{independent of }\xi_1}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|

\]

where the sum over independent paths could be empty for certain $\xi_1$. Now we replace this last sum by a sum over \emph{all} paths $\xi_2\in\paths{b_2}$. This will change the sum but only for terms where $\xi_1,\xi_2$ are dependent. For those terms we already know that $\mathbb{P}[\xi_1]\mathbb{P}[\xi_2]$ contains a factor $p^k$ and hence we have 

\begin{align*}

    \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|

    &= \sum_{\xi_1\in\paths{b_1}} \sum_{\xi_2\in\paths{b_2}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1| + \mathcal{O}(p^k) \\

    &= \sum_{\xi_1\in\paths{b_1}} \mathbb{P}[\xi_1]|\xi_1| + \mathcal{O}(p^k) \\

    &= R_{b_1} + \mathcal{O}(p^k)

\end{align*}

we can do the same with the second term and this proves the claim.

\end{proof}

~\\

\textbf{Proof of claim \ref{claim:weakcancel}}: We can assume $C$ consists of a group on the left with $l$ slots and a group on the right with $r$ slots (so $r+l=|C|$), with a gap of size $k=\mathrm{gap}(C)$ between these groups. Then on the left we have strings in $\{0,1'\}^l$ as possibilities and on the right we have strings in $\{0,1'\}^r$. The combined configuration can be described by strings $f=(a,b)\in\{0,1'\}^{l+r}$. The initial probability of such a state $C(a,b)$ is $\rho_{C(a,b)} = (-1)^{|a|+|b|} p^{r+l}$ and by claim \ref{claim:expectationsum} we know $R_{C(a,b)} = R_{C(a)} + R_{C(b)} + \mathcal{O}(p^k)$ where $C(a)$ indicates that only the left slots have been filled by $a$ and the other slots are filled with $1$s. The total contribution of these configurations is therefore

\begin{align*}

    \sum_{f\in\{0,1'\}^{|C|}} \rho_{C(f)} R_{C(f)}

    &= \sum_{a\in\{0,1'\}^l} \sum_{b\in\{0,1'\}^r} (-1)^{|a|+|b|}p^{r+l} \left( R_{C(a)} + R_{C(b)} + \mathcal{O}(p^k) \right) \\

    &=\;\;\; p^{r+l}\sum_{a\in\{0,1'\}^l} (-1)^{|a|} R_{C(a)} \sum_{b\in\{0,1'\}^r} (-1)^{|b|} \\

    &\quad + p^{r+l}\sum_{b\in\{0,1'\}^r} (-1)^{|b|} R_{C(b)} \sum_{a\in\{0,1'\}^l} (-1)^{|a|}

        + \mathcal{O}(p^{r+l+k})\\

    &= 0 + \mathcal{O}(p^{|C|+k})

\end{align*}

where we used the identity $\sum_{a\in\{0,1\}^l} (-1)^{|a|} = 0$.

\newpage

\subsection{Proving the strong cancellation claim}

It is useful to introduce some new notation. 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 $\textsc{start}(b)$ as the event that the starting state of the chain is the state $b$. For any event $A$, define

    \begin{align*}

        \mathbb{P}_b(A) &= \mathbb{P}(A \;|\; \textsc{start}(b)) \\

        R_{b,A} &= \mathbb{E}( \#resamples \;|\; A \; , \; \textsc{start}(b))

    \end{align*}

\end{definition}

\begin{definition}[Vertex visiting event] \label{def:visitingResamplings}

    Denote by $\mathrm{Z}^{(j)}$ the event that site $j$ becomes zero at any point in time before the Markov Chain terminates. Denote the complement by $\mathrm{NZ}^{(j)}$, i.e. the event that site $j$ does \emph{not} become zero before it terminates. Furthermore define $\mathrm{NZ}^{(j_1,j_2)} := \mathrm{NZ}^{(j_1)} \cap \mathrm{NZ}^{(j_2)}$, i.e. the event that \emph{both} $j_1$ and $j_2$ 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 $j_1,j_2$.}

\end{figure}

\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 ($b_1\lor b_2 = 1^n$) of zeroes that are separated by at least one site inbetween, as in Figure \ref{fig:separatedgroups}. Let $j_1$, $j_2$ 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 $j_1,j_2$'', and similar for $A_2$ (for example $\mathrm{Z}^{(i)}$ for an $i$ on the correct side). Then we have

    \begin{align*}

        \mathbb{P}_b(\mathrm{NZ}^{(j_1,j_2)}, A_1, A_2)

        &=

        \mathbb{P}_{b_1}(\mathrm{NZ}^{(j_1,j_2)}, A_1)

        \; \cdot \;

        \mathbb{P}_{b_2}(\mathrm{NZ}^{(j_1,j_2)}, A_2) \\

        \mathbb{P}_b(A_1, A_2 \mid \mathrm{NZ}^{(j_1,j_2)})

        &=

        \mathbb{P}_{b_1}(A_1 \mid \mathrm{NZ}^{(j_1,j_2)})

        \; \cdot \;

        \mathbb{P}_{b_2}(A_2 \mid \mathrm{NZ}^{(j_1,j_2)}) \\

        R_{b,\mathrm{NZ}^{(j_1,j_2)},A_1,A_2}

        &=

        R_{b_1,\mathrm{NZ}^{(j_1,j_2)},A_1}

        \; + \;

        R_{b_2,\mathrm{NZ}^{(j_1,j_2)},A_2}

    \end{align*}

    up to any order in $p$.

\end{lemma}

The lemma says that conditioned on $j_1$ and $j_2$ not being crossed, the two halves of the cycle are independent. 

\begin{proof}

        \frac{z_2 }{z_2+z_2'} \;

        \frac{z_2'}{z_3+z_2'}

        \cdots

        \P_{b_1}[\xi_1] \; \P_{b_2}[\xi_2] \tag{definition of $\P_{b_i}[\xi_i]$} \\

        &= (1-p_{1,1}) \; p_{1,2} \; p_{2,2} \; (1-p_{3,2}) \; \P_{b_1}[\xi_1] \; \P_{b_2}[\xi_2] \tag{definition of $p_{i,j}$} \\

        &= \P(\text{path in grid}) \; \P_{b_1}[\xi_1] \; \P_{b_2}[\xi_2]

    \end{align*}

    In the grid we see that at every point the probabilities sum to 1, and we always reach the end, so we know the sum of all paths in the grid is 1. This proves the required equality.

    We obtain

    \begin{align*}

        \mathbb{P}_b(\mathrm{NZ}^{(j_1,j_2)},A_1,A_2)

        &= \sum_{\substack{\xi\in\paths{b} \cap \\ \mathrm{NZ}^{(j_1,j_2)}\cap A_1\cap A_2}} \mathbb{P}[\xi] \\

        &= \sum_{\substack{\xi_1\in\paths{b_1} \cap \\ \mathrm{NZ}^{(j_1,j_2)}\cap A_1}} \;\;

          \sum_{\substack{\xi_2\in\paths{b_1} \cap \\ \mathrm{NZ}^{(j_1,j_2)}\cap A_2}}

        \mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2] \\

        &=

        \mathbb{P}_{b_1}(\mathrm{NZ}^{(j_1,j_2)},A_1)

        \; \cdot \;

        \mathbb{P}_{b_2}(\mathrm{NZ}^{(j_1,j_2)},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*}

        \mathbb{P}_b(\mathrm{NZ}^{(j_1,j_2)},A_1,A_2) R_{b,\mathrm{NZ}^{(j_1,j_2)},A_1,A_2}

        &\equiv \sum_{\substack{\xi\in\paths{b}\\\xi \in \mathrm{NZ}^{(j_1,j_2)}\cap A_1\cap A_2}} \mathbb{P}[\xi] |\xi| \\

        &= \sum_{\substack{\xi_1\in\paths{b_1}\\\xi_1 \in \mathrm{NZ}^{(j_1,j_2)}\cap A_1}}

          \sum_{\substack{\xi_2\in\paths{b_2}\\\xi_2 \in \mathrm{NZ}^{(j_1,j_2)}\cap A_2}}

        \mathbb{P}[\xi_1]\mathbb{P}[\xi_2] (|\xi_1| + |\xi_2|) \\

        &=

        \mathbb{P}_{b_2}(\mathrm{NZ}^{(j_1,j_2)},A_2) \mathbb{P}_{b_1}(\mathrm{NZ}^{(j_1,j_2)},A_1) R_{b_1,\mathrm{NZ}^{(j_1,j_2)},A_1} \\

        &\quad +

        \mathbb{P}_{b_1}(\mathrm{NZ}^{(j_1,j_2)},A_1) \mathbb{P}_{b_2}(\mathrm{NZ}^{(j_1,j_2)},A_2) R_{b_2,\mathrm{NZ}^{(j_1,j_2)},A_2} .

    \end{align*}

    Dividing by $\mathbb{P}_b(\mathrm{NZ}_{(j_1,j_2)},A_1,A_2)$ and using the first equality gives the desired result.

\end{proof}

\begin{comment}

TEST: Although a proof of claim \ref{claim:expectationsum} was already given, I'm trying to prove it in an alternate way using claim \ref{claim:eventindependence}.

~

Assume that $b_1$ ranges up to site $0$, the gap ranges from sites $1,...,k$ and $b_2$ ranges from site $k+1$ and onwards. For $j=1,...,k$ define the ``partial-zeros'' event $\mathrm{PZ}_j = \mathrm{Z}_1 \cap \mathrm{Z}_2 \cap ... \cap \mathrm{Z}_{j-1} \cap \mathrm{NZ}_j$ i.e. the first $j-1$ sites of the gap become zero and site $j$ does not become zero. Also define the ``all-zeros'' event $\mathrm{AZ} = \mathrm{Z}_1 \cap ... \cap \mathrm{Z}_k$, where all sites of the gap become zero. Note that these events partition the space, so we have for all $b$ that $\sum_{j=1}^k \mathbb{P}_b(\mathrm{PZ}_j) = 1 - \mathbb{P}_b(\mathrm{AZ}) = 1 - \mathcal{O}(p^k)$.

~

Furthermore, if site $j$ becomes zero when starting from $b_1$ it means all sites to the left of $j$ become zero as well. Similarly, from $b_2$ it implies all the sites to the right of $j$ become zero.

Because of that, we have

\begin{align*}

    \mathbb{P}_{b_1}(\mathrm{PZ}_j) &= \mathbb{P}_{b_1}(\mathrm{Z}_{j-1} \cap \mathrm{NZ}_j) = \mathcal{O}(p^{j-1}) \\

    \mathbb{P}_{b_2}(\mathrm{NZ}_j) &= 1 - \mathbb{P}_{b_2}(\mathrm{Z}_j) = 1 - \mathcal{O}(p^{k-j+1})

\end{align*}

Following the proof of claim \ref{claim:eventindependence} we also have

\begin{align*}

    \mathbb{P}_b(\mathrm{PZ}_{j})

    &=

    \mathbb{P}_{b_1}(\mathrm{PZ}_{j})

    \; \cdot \;

    \mathbb{P}_{b_2}(\mathrm{NZ}_{j}) \\

    R_{b,\mathrm{PZ}_{j}}

    &=

    R_{b_1,\mathrm{PZ}_{j}}

    \; + \;

    R_{b_2,\mathrm{NZ}_{j}}

\end{align*}

Now observe that

\begin{align*}

    R_b &= \sum_{j=1}^k \mathbb{P}_b(\mathrm{PZ}_j) R_{b,\mathrm{PZ}_j} + \mathbb{P}_b(\mathrm{AZ}) R_{b,\mathrm{AZ}} \\

        &= \sum_{j=1}^k \mathbb{P}_{b_2}(\mathrm{NZ}_j)\mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}

        + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{NZ}_j) R_{b_2,\mathrm{NZ}_j}

        + \mathcal{O}(p^k) \\

        &= \sum_{j=1}^k \mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}

        - \sum_{j=1}^k \mathbb{P}_{b_2}(\mathrm{Z}_j)\mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}

        + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{NZ}_j) R_{b_2,\mathrm{NZ}_j}

        + \mathcal{O}(p^k) \\

        &= \sum_{j=1}^k \mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}

        + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{NZ}_j) R_{b_2,\mathrm{NZ}_j}

        + \mathcal{O}(p^k) \\

        &= R_{b_1}

        + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{NZ}_j) R_{b_2,\mathrm{NZ}_j}

        + \mathcal{O}(p^k) \\

        &\overset{???}{=} R_{b_1} + R_{b_2} + \mathcal{O}(p^k)

\end{align*}

\end{comment}

Consider the chain (instead of the cycle) for simplicity with vertices identified by $\mathbb{Z}$.

\begin{definition}[Starting state dependent probability distribution.]

	Let $I\subset\mathbb{Z}$ be a finite set of vertices.

    Let $b_I$ be the initial state where everything is $1$, apart from the vertices corresponding to $I$, which are set $0$. Define $P_I(A)=P_{b_I}(A)$ where the latter is defined in Definition \ref{def:conditionedevents}, i.e. the probability of seeing a resample sequence from $A$ when the whole procedure started in state $b_I$. 

\end{definition}

New:

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

	\begin{align*}

	\end{align*}

	up to any order in $p$.

\end{lemma}

The intuition of the following lemma is that the far right can only affect the zero vertex if there is an interaction chain forming, which means that every vertex should get resampled to $0$ at least once.

\begin{lemma}\label{lemma:probIndepNew}

\end{lemma}

\begin{proof}

	Induction step: suppose we proved the claim for $n-1$, then

	\begin{align*}

	\P^{[n+1]}(\Z{1})

	\end{align*}

\end{proof}

\begin{corollary}\label{cor:probIndepNew}

\end{corollary}

 	\begin{lemma}\label{lemma:independenetSidesNew}	

 	\end{lemma}   

 	Note that using De Morgan's law and the inclusion-exclusion formula we can see that this is equivalent to saying:

 	\begin{proof}

 		We proceed by induction on $k$. For $k=1,2$ the statement is trivial.

 		Now observe that:

 		$$\P^{[k]}(\Z{1})=\sum_{P\text{ patch}\,:\,1\in P}\P^{[k]}(P\in\mathcal{P})$$

 		$$\P^{[k]}(\Z{k})=\sum_{P\text{ patch}\,:\,k\in P}\P^{[k]}(P\in\mathcal{P})$$

 		Suppose we proved the statement up to $k-1$, then we proceed using induction similarly to the above

 		\begin{align*}

 		&\P^{[k]}(\Z{1}\cap \Z{k})=\\

 		+\P^{[k]}([k]\in\mathcal{P})\\

 		\P^{[\ell+1,r-1]}(\NZ{\ell+1}\cap \NZ{r-1})

 		\left(\P^{[\ell+1,r-1]}(\NZ{\ell+1})

		\P^{[\ell+1,r-1]}(\NZ{r-1})\right)

 		\left(\P^{[\ell+1,k]}(\NZ{\ell+1})

 		\P^{[k]}([\ell]\in\mathcal{P})

 		\P^{[k]}([r,k]\in\mathcal{P})

 		&=\left(\sum_{\ell\in [k]}\P^{[k]}([\ell]\in\mathcal{P})\right)

 		\left(\sum_{r\in [k]}\P^{[k]}([r,k]\in\mathcal{P})\right)

 		&=\P^{[k]}(\Z{1})\P^{[k]}(\Z{k})

 		\end{align*}

 	\end{proof}

	\begin{theorem}

	\end{theorem}

	\begin{proof}

        Some notation: let $P$ be an interval $[a,b]$. We say $P$ is a \emph{patch} when the $\Z{i}$ event holds for all $i \in [a,b]$ and $\NZ{a-1}$ and $\NZ{b+1}$ holds. We denote this event by $P\in\mathcal{P}$, so

        \begin{align*}

            P\in\mathcal{P} \equiv \NZ{a-1} \cap \Z{a} \cap \Z{a+1} \cap \cdots \cap \Z{b-1} \cap \Z{b} \cap \NZ{b+1} .

        \end{align*}

        Note that we have the following partition of the event $\Z{v}$ for any vertex $v\in[n]$:

        \begin{align*}

            \Z{v} = \dot\bigcup_{P : v\in P} (P\in\mathcal{P})

        \end{align*}

		Let $N\geq \max(2n,2m)$, then

		\begin{align*}

		Repeating the same calculation with $m$, and comparing the two expressions completes the proof.

	\end{proof} 	

Old:

The intuition of the following lemma is that the far right can only affect the zero vertex if there is an interaction chain forming, which means that every vertex should get resampled to $0$ at least once.

\begin{lemma}\label{lemma:probIndep}

	Suppose we have a finite set $I\subset\mathbb{N}_+$ of vertices.

    Let $I_{\max}:=\max(I)$ and $I':=I\setminus\{I_{\max}\}$, and similarly let $I_{\min}:=\min(I)$. These definitions are illustraded in Figure \ref{fig:lemmaillustration}.

	Then $\P^\infty_{I}(\Z{0})-\P^\infty_{I'}(\Z{0}) = O(p^{I_{\max}-|I'|})$.

\end{lemma}

\begin{proof}

\begin{figure}

	\begin{center}

    	\includegraphics{diagram_proborders.pdf}

    \end{center}

    \caption{\label{fig:lemmaillustration} Illustration of setup of Lemma \ref{lemma:probIndep}.}

\end{figure}

	The proof uses induction on $|I|$. For $|I|=1$ the statement is easy, since every resample sequence that resamples vertex $0$ to zero must produce at least $I_{\max}$ zeroes in-between.

    Induction step: For an event $A$ and $k>0$ let us denote $A_k = A\cap\left(\cap_{j=0}^{k-1} \mathrm{Z}^{(j)}\right)\cap \NZ{(k)}$, i.e. $A_k$ is the event $A$ \emph{and} ``Each vertex in $0,1,2,\ldots, k-1$ becomes $0$ at some point before termination (either by resampling or initialisation), but vertex $k$ does not''. Observe that these events form a partition, so $\Z{(0)}=\dot{\bigcup}_{k=1}^{\infty}\Z{(0)}_k$.

    Let $I_{<k}:=I\cap[1,k-1]$ and similarly $I_{>k}:=I\setminus[1,k]$, finally let $I_{><}:=\{I_{\min}+1,I_{\max}-1]\}\setminus I$ (note that $I_{><} = \gaps{I}$ as shown in Figure \ref{fig:diametergap}). Suppose we have proven the claim up to $|I|-1$, then the induction step can be shown by

	\begin{align*}

		\P^\infty_{I}(\Z{(0)})

		&=\sum_{k=1}^{\infty}\P^\infty(\Z{(0)}_k) \tag{the events are a partition}\\

        &=\sum_{k\in \mathbb{N}\setminus I}\P^\infty(\Z{(0)}_k) \tag{$\mathbb{\P^\infty}(A_k)=0$ for $k\in I$}\\

        &=\sum_{k\in\mathbb{N}\setminus I}\P^\infty_{I_{<k}}(\Z{(0)}_k)\cdot \P^\infty_{I_{>k}}(\NZ{(k)}) \tag{by Claim~\ref{claim:eventindependence}}\\

        &=\sum_{k\in I_{><}}\P^\infty_{I_{<k}}(\Z{(0)}_k)\cdot \P^\infty_{I_{>k}}(\NZ{(k)})+\mathcal{O}(p^{I_{\max}+1-|I|})

		\tag{$k<I_{\min}\Rightarrow \P^\infty_{I_{<k}}(\Z{(0)}_k)=0$}\\

        &=\sum_{k\in I_{><}}\P^\infty_{I'_{<k}}(\Z{(0)}_k)\cdot \P^\infty_{I_{>k}}(\NZ{(k)})+\mathcal{O}(p^{I_{\max}+1-|I|})	

		\tag{$k< I_{\max}\Rightarrow I_{<k}=I'_{<k}$}\\

		&=\sum_{k\in I_{><}}\P^\infty_{I'_{<k}}(\Z{(0)}_k)\cdot

        \left(\P^\infty_{I'_{>k}}(\NZ{(k)})+\mathcal{O}(p^{I_{\max}-k+1-|I_{>k}|})\right) +\mathcal{O}(p^{I_{\max}+1-|I|})	\tag{by induction, since for $k>I_{\min}$ we have $|I_{<k}|<|I|$}\\

		&=\sum_{k\in I_{><}}\P^\infty_{I'_{<k}}(\Z{(0)}_k)\cdot

        \P^\infty_{I'_{>k}}(\NZ{(k)}) +\mathcal{O}(p^{I_{\max}+1-|I|})	

		\tag{as $\P^\infty_{I'_{<k}}(\Z{(0)}_k)=\mathcal{O}(p^{k-|I'_{<k}|})$}\\

		&=\sum_{k\in\mathbb{N}\setminus I}\P^\infty_{I'_{<k}}(\Z{(0)}_k)\cdot

        \P^\infty_{I'_{>k}}(\NZ{(k)}) +\mathcal{O}(p^{I_{\max}+1-|I|})\\

		&=\sum_{k\in\mathbb{N}\setminus I'}\P^\infty_{I'_{<k}}(\Z{(0)}_k)\cdot

        \P^\infty_{I'_{>k}}(\NZ{(k)}) +\mathcal{O}(p^{I_{\max}+1-|I|})	\tag{$k=I_{\max}\Rightarrow \P^\infty_{I'_{<k}}(\Z{(0)}_k)=\mathcal{O}(p^{I_{\max}-|I'|})=\mathcal{O}(p^{I_{\max}+1-|I|})$}\\

		&=\P^\infty_{I'}(\Z{(0)}) +\mathcal{O}(p^{I_{\max}-|I'|})	\tag{analogously to the beginning}			

	\end{align*}

\end{proof}

\begin{corollary}\label{cor:probIndep}

	Suppose $I,J\subset\mathbb{N}_+$ are finite sets of vertices, and let $m=\min(\Delta(I,J))$.

	Then $\P^\infty_{I}(\Z{0})-\P^\infty_{J}(\Z{0}) = O(p^{|[m]\cap I\cap J|})$.

\end{corollary}

	%The main insight that Lemma~\ref{lemma:probIndep} gives is that if we separate the slots to two halves, in order to see the cancellation of the contribution of the expected resamples on the right, we can simply pair up the left configurations by the particle filling the leftmost slot. And similarly for cancelling the left expectations we pair up right configurations based on the rightmost filling. 

	%Also this claim finally ``sees'' how many empty places are between slots. These properties make it possible to use this lemma to prove the sought linear bound. We show it for the infinite chain, but with a little care it should also translate to the cycle.

\begin{definition}[Connected patches]

	Let $\mathcal{P}\subset 2^{\mathbb{Z}}$ be a finite system of finite subsets of $\mathbb{Z}$. We say that the patch set of a resample sequence is $\mathcal{P}$,

	if the connected components of the vertices that have ever become $0$ are exactly the elements of $\mathcal{P}$. We denote by $A^{(\mathcal{P})}$ the event that the set of patches is $\mathcal{P}$. For a patch $P$ let $A^{(P)}=\bigcup_{\mathcal{P}:P\in \mathcal{P}}A^{(\mathcal{P})}$ denote the event that one of the patches is equal to $P$ but there can be other patches as well.

	As a shorthand we are going to use the notation $P\in \mathcal{P}$ for the event $A^{(P)}$.

\end{definition} 

\begin{definition}[Conditional expectations]

	Let $S\subset\mathbb{Z}$ be a finite slot configuration, and for $f\in\{0,1'\}^{|S|}$ let $I:=S(f)$ be the set of vertices filled with a particle (i.e. $1'$). 

	Then we define

	$$R_I:=\mathbb{E}[\#\{\text{resamplings when started from inital state }I\}],$$ 

	and similarly for a patch $P$ we define

	$$R^{(P)}_I:=\mathbb{E}[\#\{\text{resamplings inside }P\text{ when started from inital state }I\}|A^{(P)}].$$	

\end{definition} 

 	\begin{lemma}Suppose $I\subseteq [k]$, then on the infinite chain

		$$\P^\infty_I(\Z{1}\cap \Z{k})=\P^\infty_I(\Z{1})\P^\infty_I(\Z{k})+\mathcal{O}(p^{k-|I|}).$$

	\end{lemma}   

	Note that using De Morgan's law and the inclusion-exclusion formula we can see that this is equivalent to saying:

	$$\P^\infty_I(\NZ{1}\cap \NZ{k})=\P^\infty_I(\NZ{1})\P^\infty_I(\NZ{k})+\mathcal{O}(p^{k-|I|}).$$

	\begin{proof}

		We proceed by induction on $|I|$. For $|I|=0$ the statement is trivial.

		Now observe that:

		$$\P^\infty_I(\Z{1})=\sum_{P\text{ patch}\,:\,1\in P}\P^\infty_I(P\in\mathcal{P})$$

		$$\P^\infty_I(\Z{k})=\sum_{P\text{ patch}\,:\,k\in P}\P^\infty_I(P\in\mathcal{P})$$

		Suppose we proved the statement for all $\ell< |I|$, then we proceed using induction similarly to the above (let us use the notation $>I<:=I\cap \overline{P_l}\cap \overline{P_r}$ for simplicity)

		\begin{align*}

		&\P^\infty_I(\Z{1}\cap \Z{k})=\\

		&=\sum_{P_l\neq P_r\text{ patches}\,:\,1\in P_l,k\in P_r}

		\P^\infty_I(P_l,P_r\in\mathcal{P})

		+\sum_{P\text{ patch}\,:\,1,k\in P}\P^\infty_I(P\in\mathcal{P})\\

		&=\sum_{P_l\neq P_r\text{ patches}\,:\,1\in P_l,k\in P_r}

		\P^\infty_I(P_l,P_r\in\mathcal{P})

		+\mathcal{O}(p^{k-|I|})\\

		&\overset{Lemma~\ref{claim:eventindependence}}{=}\sum_{P_l\neq P_r\text{ patches}\,:\,1\in P_l,k\in P_r}

		\P^\infty_{I\cap P_l}(P_l\in\mathcal{P})

		\P^\infty_{>I<}(\NZ{P_l^{\max}+1}\cap \NZ{P_r^{\min}-1})

		\P^\infty_{I\cap P_r}(P_r\in\mathcal{P})

		+\mathcal{O}(p^{k-|I|})\\

		&\overset{\text{induction}}{=}\sum_{P_l\neq P_r\text{ patches}\,:\,1\in P_l,k\in P_r}

		\P^\infty_{I\cap P_l}(P_l\in\mathcal{P})

		\P^\infty_{> I <}(\NZ{P_l^{\max}+1})\P^\infty_{> I <}(\NZ{P_r^{\min}-1})

		\P^\infty_{I\cap P_r}(P_r\in\mathcal{P})

		+\mathcal{O}(p^{k-|I|})\\

		&\overset{Corrolary~\ref{cor:probIndep}}{=}\sum_{P_l\neq P_r\text{ patches}\,:\,1\in P_l,k\in P_r}

		\P^\infty_{I\cap P_l}(P_l\in\mathcal{P})

		\P^\infty_{I\setminus P_l}(\NZ{P_l^{\max}+1})\P^\infty_{I\setminus P_r}(\NZ{P_r^{\min}-1})

		\P^\infty_{I\cap P_r}(P_r\in\mathcal{P})

		+\mathcal{O}(p^{k-|I|})\\

		&\overset{Lemma~\ref{claim:eventindependence}}{=}\sum_{P_l\neq P_r\text{ patches}\,:\,1\in P_l,k\in P_r}

		\P^\infty_{I}(P_l\in\mathcal{P})

		\P^\infty_{I}(P_r\in\mathcal{P})

		+\mathcal{O}(p^{k-|I|})\\

		&=\left(\sum_{P_l\text{ patch}\,:\,1\in P_l}\P^\infty_{I}(P_l\in\mathcal{P})\right)

		\left(\sum_{P_r\text{ patch}\,:\,k\in P_r}\P^\infty_{I}(P_r\in\mathcal{P})\right)

		+\mathcal{O}(p^{k-|I|})\\

		&=\P^\infty_I(\Z{1})\P^\infty_I(\Z{k})

		+\mathcal{O}(p^{k-|I|}).	

		\end{align*}

	\end{proof}

	\begin{corollary}\label{cor:independenetSides}

		Let $k>0$ and $d:=\lfloor\frac{k}{2}\rfloor$, then 

		\begin{align*}

		\sum_{S\subseteq [k]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)} \P^\infty_{S(f)}(\NZ{1}\cap \NZ{k})

		=\left(\sum_{\underset{|S|<\infty}{S\subseteq \mathbb{N}_+}}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)} \P^\infty_{S(f)}(\NZ{1})\right)^{\!\!\!2}

		+\mathcal{O}(p^{d}).

		\end{align*}

	\end{corollary}

	(Question: is the above is true with $d=k$?)

	\begin{proof}

		Let $[-d]:=\{k-d+1,\ldots, k\}$ and $[\pm d]:=[d]\cup [-d]$. By the above statement we know that

		\begin{align}\label{eq:intervalIndep}

		\sum_{S\subseteq [k]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)} \P^\infty_{S(f)}(\NZ{1}\cap \NZ{k})

		\end{align}

		Now suppose $s\in S\subseteq[k]$ such that $s\notin [\pm d]$, then 

		\begin{align*}

		&=\sum_{f_{\overline{s}}\in\{0,1'\}^{|S|-1}}\rho_{S\setminus\{s\}(f_{\overline{s}})} \left(p\P^\infty_{S(f_{\overline{s}},f_s=0)}(\NZ{1})\P^\infty_{S(f_{\overline{s}},f_s=0)}(\NZ{k})-p\P^\infty_{S(f_{\overline{s}},f_s=1)}(\NZ{1})\P^\infty_{S(f_{\overline{s}},f_s=1)}(\NZ{k})\right)\\

		&=\mathcal{O}(p^{d}). \tag{by Corollary~\ref{cor:probIndep}}

		\end{align*}

		Therefore we can see that 

		\begin{align*}

		\eqref{eq:intervalIndep}&=\sum_{S\subseteq [\pm d]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)} \P^\infty_{S(f)}(\NZ{1})\P^\infty_{S(f)}(\NZ{k})+\mathcal{O}(p^{d})\\

		&=\sum_{S\subseteq [d]}\sum_{f\in\{0,1'\}^{|S|}}\sum_{S'\subseteq [d]}\sum_{f'\in\{0,1'\}^{|S|}}\rho_{S(f)} \P^\infty_{S(f)}(\NZ{1})\rho_{S'(f')}\P^\infty_{S'(f')}(\NZ{k})+\mathcal{O}(p^{d}) \tag{Corollary \ref{cor:probIndep}}\\

		&=\left(\sum_{S\subseteq [d]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)} \P^\infty_{S(f)}(\NZ{1})\right)

		\left(\sum_{S'\subseteq [-d]}\sum_{f'\in\{0,1'\}^{|S'|}}\rho_{S'(f')} \P^\infty_{S'(f')}(\NZ{k})\right)

		+\mathcal{O}(p^{d})\\

		&=\left(\sum_{S\subseteq [d]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)} \P^\infty_{S(f)}(\NZ{1})\right)^{\!\!2}

		+\mathcal{O}(p^{d})\tag{by symmetry}\\		

		&=\left(\sum_{\underset{|S|<\infty}{S\subseteq \mathbb{N}_+}}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)} \P^\infty_{S(f)}(\NZ{1})\right)^{\!\!2}

		+\mathcal{O}(p^{d}).\tag{Corollary \ref{cor:probIndep}}

		\end{align*}

	\end{proof}

	\begin{remark}\label{rem:cycleContained}

		For all $n\geq k$ and $I\subseteq [k]$ we have that 

		$$\P^n_I(\NZ{1}\cap \NZ{k})=\P^\infty_I(\NZ{1}\cap \NZ{k}).$$

	\end{remark}

	\begin{theorem}

	\end{theorem}

	\begin{proof}

		\begin{align*}

		R^{(n)} &= \frac{1}{n}\sum_{b\in\{0,1,1'\}^{n}} \rho_b \; R_{\bar{b}}\\

		&= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)} R_{S(f)}\\

		&= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)} \sum_{v\in[n]}\sum_{t=1}^{\infty}\P_{S(f)}(v \text{ is resampled in the }t\text{-th step})\\

		&= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{f\in\{0,1'\}^{|S|}}\rho_{S(f)} \sum_{v\in[n]}\sum_{t=1}^{\infty}\sum_{P\text{ patch}}\P_{S(f)}(v \text{ is resampled in the }t\text{-th step and }P\text{ is a patch})\\

		&= \frac{1}{n}\sum_{S\subseteq [n]}\sum_{f\in\{0,1'\}^{|S|}}

		\rho_{S(f)} \sum_{P\text{ patch}} R^{(P)}_{S(f)}\mathbb{P}_{S(f)}(A^{(P)}) \tag{by definition}\\