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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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	%?%

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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 $a^{(n)}_k$}

	\label{tab:coeffs}

	\resizebox{\columnwidth}{!}{%

		\begin{tabular}{c|ccccccccccccccccccccc}

			\backslashbox[10mm]{$n$}{$k$} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\		\hline

			3 &	0 & 1 & \cellcolor{blue!25}2 & 3+1/3 & 5.00 & 7.00 & 9.33 & 12.00 & 15.00 & 18.33 & 22.00 & 26.00 & 30.33 & 35.00 & 40.00 & 45.333 & 51.000 & 57.000 & 63.333 & 70.000 & 77.000 \\

			4 &	0 & 1 & 2 & \cellcolor{blue!25}3+2/3 & 6.16 & 9.66 & 14.3 & 20.33 & 27.83 & 37.00 & 48.00 & 61.00 & 76.16 & 93.66 & 113.6 & 136.33 & 161.83 & 190.33 & 222.00 & 257.00 & 295.50 \\

			5 &	0 & 1 & 2 & 3+2/3 & \cellcolor{blue!25}6.44 & 10.8 & 17.3 & 26.65 & 39.43 & 56.48 & 78.65 & 106.9 & 142.2 & 185.8 & 238.7 & 302.41 & 378.05 & 467.13 & 571.14 & 691.69 & 830.44 \\

			6 &	0 & 1 & 2 & 3+2/3 & 6.44 & \cellcolor{blue!25}11.0 & 18.5 & 30.02 & 47.10 & 71.68 & 106.0 & 152.9 & 215.4 & 297.4 & 403.1 & 537.21 & 705.25 & 913.31 & 1168.2 & 1477.4 & 1849.1 \\

 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!\P^{[k]}([\ell],[r,k]\in\mathcal{P})

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

 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!\P^{[k]}([\ell],[r,k]\in\mathcal{P})

 		+\bigO{p^{k}} \tag*{$\left(\P^{[k]}([k]\in\mathcal{P})=\bigO{p^{k}}\right)$}\\	

 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!

 		\P^{[\ell+1]}_{b_{\ell+1}=1}([\ell]\in\mathcal{P})

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

 		\P^{[r-1,k]}_{b_{r-1}=1}([r,k]\in\mathcal{P})

 		+\bigO{p^{k}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\

 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!

 		\P^{[\ell+1]}_{b_{\ell+1}=1}([\ell]\in\mathcal{P})

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

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

 		\P^{[r-1,k]}_{b_{r-1}=1}([r,k]\in\mathcal{P})

 		+\bigO{p^{k}} \tag{by induction}\\

 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!

 		\P^{[\ell+1]}_{b_{\ell+1}=1}([\ell]\in\mathcal{P})

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

 		\P^{[1,r-1]}_{b_{r-1}=1}(\NZ{r-1})\right)

 		\P^{[r-1,k]}([r,k]\in\mathcal{P})

 		+\bigO{p^{k}} \tag{by Corrolary~\ref{cor:probIndepNew}}\\

 		&=\!\!\!\sum_{\ell, r\in [k]: \ell<r-1}\!\!\!

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

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

 		+\bigO{p^{k}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\

 		&=\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)

 		+\bigO{p^{k}} \tag*{$\left(\P^{[k]}([\ell]\in\mathcal{P})=\bigO{p^{\ell}}\right)$}\\	

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

 		+\bigO{p^{k}}.	

 		\end{align*}

 	\end{proof}

	Again the intuition of the final theorem is simmilar to the previous lemmas. A site can only realise the length of the cycle after an interaction chain was formed around the cycle, implying that every vertex was resampled to $0$ at least once.

	\begin{theorem} $R^{(n)}=\E^{[-m,m]}(\Res{0})+\bigO{p^{n}}$ for all $m\geq n \geq 3$, thus

		$R^{(n)}-R^{(m)}=\bigO{p^{n}}$.

	\end{theorem}

	\begin{proof} In the proof we identify the sites of the $n$-cycle with the$\mod n$ remainder classes.

		\vskip-3mm

		\begin{align*}

			R^{(n)}

			&= \E^{(n)}(\Res{0}) \tag{by translation invariance}\\

			&= \sum_{k=1}^{\infty}\P^{(n)}(\Res{0}\!\geq\! k) \\		

			&= \sum_{k=1}^{\infty}\sum_{\underset{v+w\leq n+1}{v,w\in [n]}}\P^{(n)}(\Res{0}\!\geq\! k\,\&\, \underset{P_{v,w}:=}{\underbrace{[-v\!+\!1,w\!-\!1]}}\in\mathcal{P}) \tag{partition}\\[-1mm]

			&= \sum_{k=1}^{\infty}\sum_{\underset{v+w\leq n}{v,w\in [n]}}\P^{(n)}(\Res{0}\!\geq\! k\,\&\, P_{v,w}\!\in\!\mathcal{P}) +\bigO{p^{n}}\\[-1mm]

			&= \sum_{k=1}^{\infty}\smash{\sum_{\underset{v+w\leq n}{v,w\in [n]}}}\P^{[-v,w]}_{b_{-v}=b_{w}=1}(\Res{0}\!\geq\! k\,\&\, P_{v,w}\!\in\!\mathcal{P}) \P^{[w,n-v]}(\NZ{w,n-v}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\

			&= \sum_{k=1}^{\infty}\smash{\sum_{\underset{v+w\leq n}{v,w\in [n]}}}\P^{[-v,w]}_{b_{-v}=b_{w}=1}(\Res{0}\!\geq\! k\,\&\, P_{v,w}\!\in\!\mathcal{P})  \left(\left(\P^{[w,n-v]}(\NZ{w})\right)^{\!\!2}\!+\!\bigO{p^{n-v-w+1}}\right) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:independenetSidesNew}}\\

			&= \sum_{k=1}^{\infty}\smash{\sum_{\underset{v+w\leq n}{v,w\in [n]}}}\P^{[-v,w]}_{b_{-v}=b_{w}=1}(\Res{0}\!\geq\! k\,\&\, P_{v,w}\!\in\!\mathcal{P})  \left(\P^{[-m,-v]}(\NZ{-v})\P^{[w,m]}(\NZ{w})\!+\!\bigO{p^{n-v-w+1}}\right) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:independenetSidesNew}}\\	

			&= \sum_{k=1}^{\infty}\smash{\sum_{\underset{v+w\leq n}{v,w\in [n]}}}\P^{[-v,w]}_{b_{-v}=b_{w}=1}(\Res{0}\!\geq\! k\,\&\, P_{v,w}\!\in\!\mathcal{P}) \P^{[-m,-v]}(\NZ{-v})\P^{[w,m]}(\NZ{w}) +\bigO{p^{n}} \tag{$|P_{v,w}|=v+w-1$}\\

			&= \sum_{k=1}^{\infty}\sum_{\underset{v+w\leq n}{v,w\in [n]}}\P^{[-m,m]}(\Res{0}\!\geq\! k\,\&\, P_{v,w}\!\in\!\mathcal{P}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\[-1mm]

			&= \sum_{k=1}^{\infty}\sum_{\underset{|P|<n}{P\text{ patch}:0\in P}}\P^{[-m,m]}(\Res{0}\!\geq\! k\,\&\, P\in\mathcal{P}) +\bigO{p^{n}} \\[-1mm]

			&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:0\in P}\P^{[-m,m]}(\Res{0}\!\geq\! k\,\&\, P\in\mathcal{P}) +\bigO{p^{n}} \\

			&= \E^{[-m,m]}(\Res{0})+\bigO{p^{n}}.\\[-3mm]										

		\end{align*}  

		\noindent Repeating the same argument with $m$ and comparing the results completes the proof.

	\end{proof} 	

\begin{comment}

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

		\begin{align*}

		R^{(n)}

		&= \E^{(n)}(\Res{1}) \tag{by translation invariance}\\

		&= \sum_{k=1}^{\infty}\P^{(n)}(\Res{1}\geq k) \\

		%&= \sum_{k=1}^{\infty}\sum_{\underset{\ell\geq r-1}{\ell,r\in[n]}}\P^{(n)}(\Res{1}\geq k\,\&\, [\ell+1,r-1]\in\mathcal{P}) \tag{partition}\\

		%&= \sum_{k=1}^{\infty}\sum_{\underset{\ell\geq r}{\ell,r\in[n]}}\P^{(n)}(\Res{1}\geq k\,\&\, [\ell+1,r-1]\in\mathcal{P})  +\bigO{p^{n}} \\	

		%&= \sum_{k=1}^{\infty}\sum_{\underset{\ell\geq r}{\ell,r\in[n]}}\P^{[l,r]}_{b_{\ell}=b_{r}=1}(\Res{1}\geq k\,\&\, [\ell+1,r-1]\in\mathcal{P}) \P^{[r,\ell]}(\NZ{\ell,r}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\				

		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}\P^{(n)}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) \tag{partition}\\

		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{(n)}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) +\bigO{p^{n}}\\

		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{[P\cup \partial P]}_{b_{\partial P}=1}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) \P^{[\overline{P}]}(\NZ{\partial P}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\

		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{[P\cup \partial P]}_{b_{\partial P}=1}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) \left(\left(\P^{[|\overline{P}|]}(\NZ{1})\right)^2+\bigO{p^{|\overline{P}|}}\right) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:independenetSidesNew}}\\

		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{[P\cup \partial P]}_{b_{\partial P}=1}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) \left(\left(\P^{[N]}(\NZ{1})\right)^2+\bigO{p^{|\overline{P}|}}\right) +\bigO{p^{n}} \tag{by Corollary~\ref{cor:probIndepNew}}\\

		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}^{|P|<n}\P^{[-N,N]}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\

		&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}\P^{[-N,N]}(\Res{1}\geq k\,\&\, P\in\mathcal{P}) +\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\

		&= \E^{[-N,N]}(\Res{1})+\bigO{p^{n}}.

		\end{align*}	

\end{comment}			

~

Questions:

\begin{itemize}

	\item Can we generalise the proof to other translationally invariant spaces, like the torus?

	\item In view of this proof, can we better characterise $a_k^{(k+1)}$?

	\item Why did Mario's and Tom's simulation show that for fixed $C$ the contribution coefficients have constant sign? Is it relevant for proving \ref{it:pos}-\ref{it:geq}?

\end{itemize} 

	%I think the same arguments would translate to the torus and other translationally invariant spaces, so we could go higher dimensional as Mario suggested. Then I think one would need to replace $|S_{><}|$ by the minimal number $k$ such that there is a $C$ set for which $S\cup C$ is connected. I am not entirely sure how to generalise Lemma~\ref{lemma:probIndep} though, which has key importance in the present proof.

\newpage

\section{General graphs proof}

We consider the following generalization of the Markov Chain.

Let $G=(V,E)$ be a graph with vertex set $V$ and edge set $E$. We define a Markov Chain $\mathcal{M}_G$ as the following process: initialize every vertex of $G$ independently to 0 with probability $p$ and 1 with probability $1-p$. Then at each step, select a uniformly random vertex that has value $0$ and resample it and its neighbourhood, all of them independently with the same probability $p$. The Markov Chain terminates when all vertices have value $1$. We use $\P^{G}$ to denote probabilities associated to this Markov Chain.

\begin{definition}[Events] \label{def:events}

    \begin{align*}

    \end{align*}

\end{definition}

$\NZ{S}$

$\Z{S}$

patch

$B(S;d)$

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 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{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 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)

        &=

        \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}

\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.

    This is best explained by an example. Lets consider the following fixed $\xi_1,\xi_2$ and an example interleaving where we choose steps from $\xi_2,\xi_1,\xi_1,\xi_2,\cdots$:

    \begin{align*}

        \xi_1 &= \left( (z_1, s_1, r_1), (z_2, s_2, r_2), (z_3, s_3, r_3), (z_4, s_4, r_4),\cdots  \right) \\

        \xi_2 &= \left( (z_1', s_1', r_1'), (z_2', s_2', r_2'), (z_3', s_3', r_3'), (z_4', s_4', r_4'),\cdots  \right) \\

        \xi   &= \left( (z_1 + z_1', s_1', r_1'), (z_1+z_2', s_1, r_1), (z_2+z_2', s_2, r_2), (z_3+z_2', s_2', r_2'), \cdots \right)

    \end{align*}

    The probability of $\xi_1$, started from $b_1$, is given by

    \begin{align*}

        \P^{(n)}_{b_1}[\xi_1] &= \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_1|}|z_{|\xi_1|}) \P(r_{|\xi_1|}) \\

                &= \frac{1}{z_1} \P(r_1) \frac{1}{z_2} \P(r_2) \cdots \frac{1}{z_{|\xi_1|}} \P(r_{|\xi_1|}) .

    \end{align*}

    and similar for $\xi_2$ but with primes.

    The following diagram illustrates all possible interleavings, and the red line corresponds to the particular interleaving $\xi$ in the example above.

    \begin{center}

        \includegraphics{diagram_paths2.pdf}

    \end{center}

    For the labels shown within the grid, define $p_{ij} = \frac{z_i}{z_i + z_j'}$.

    The probability of $\xi$ is given by

    \begin{align*}

        \P^{(n)}_b[\xi] &= \frac{1}{z_1+z_1'} \P(r_1') \frac{1}{z_1+z_2'} \P(r_1) \frac{1}{z_2+z_2'} \P(r_2) \frac{1}{z_3+z_2'} \P(r_2') \cdots \tag{by definition}\\

        &=