\P^{[i,j]}(\mathrm{NZ}^{(i,j)}\cap A_1)

	\; \cdot \;

	\P^{[j,i]}(\mathrm{NZ}^{(i,j)}\cap A_2)/(1-p)^2 \\

	\P^{(n)}(A_1\cap A_2|\mathrm{NZ}^{(i,j)})

	&=

	\P^{[i,j]}(A_1|\mathrm{NZ}^{(i,j)})

	\; \cdot \;

	\P^{[j,i]}(A_2|\mathrm{NZ}^{(i,j)})

	\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}

	$\forall n\in \mathbb{N}_+:\P^{[n]}(\Z{1})-\P^{[n+1]}(\Z{1}) = O(p^{n})$. (Should be true with $O(p^{n+1})$ as well.)

\end{lemma}

\begin{proof}

	The proof uses induction on $n$. For $n=1$ the statement is easy, since $\P^{[1]}(\Z{1})=p$ and $\P^{[2]}(\Z{1})=p+p^2+O(p^{3})$.

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

	\begin{align*}

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

	&=\sum_{k=1}^{n+1}\P^{[n]}([k]\in\mathcal{P}) \tag{the events are a partition}\\

	&=\sum_{k=1}^{n-1}\P^{[n]}([k]\in\mathcal{P}) + O(p^{n})\\

	&=\sum_{k=1}^{n-1}\P^{[k+1]}([k]\in\mathcal{P})\cdot \P^{[n-k+1]}(\NZ{1})/(1-p)+ O(p^{n}) \tag{by Claim~\ref{lemma:eventindependenceNew}}\\

	&=\sum_{k=1}^{n-1}\P^{[k+1]}([k]\in\mathcal{P})\cdot \left(\P^{[n-k]}(\NZ{1})+O(p^{n-k})\right)/(1-p)+ O(p^{n}) \tag{by induction} \\	

	&=\sum_{k=1}^{n-1}\P^{[k+1]}([k]\in\mathcal{P})\cdot \P^{[n-k]}(\NZ{1})/(1-p)+ O(p^{n}) \\	

	&=\sum_{k=1}^{n-1}\P^{[n]}([k]\in\mathcal{P})+ O(p^{n}) \tag{by Claim~\ref{lemma:eventindependenceNew}}\\

	&=\sum_{k=1}^{n}\P^{[n]}([k]\in\mathcal{P})+ O(p^{n}) \\

	&=\P^{[n]}(\Z{1})	+ O(p^{n}) 

	\end{align*}

\end{proof}

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

	$\P^{[n]}(\Z{1})-\P^{[m]}(\Z{1}) = O(p^{\min(n,m)})$. (Should be true with $O(p^{\min(n,m)+1})$ too.)

\end{corollary}

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

 		$$\P^{[k]}(\Z{1}\cap \Z{k})=\P^{[k]}(\Z{1})\P^{[k]}(\Z{k})+\mathcal{O}(p^{k})=\left(\P^{[k]}(\Z{1})\right)^2+\mathcal{O}(p^{k}).$$

 	\end{lemma}   

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

 	$$\P^{[k]}(\NZ{1}\cap \NZ{k})=\P^{[k]}(\NZ{1})\P^{[k]}(\NZ{k})+\mathcal{O}(p^{k}).$$

 	\begin{proof}

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

 		Now observe that:

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

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

 		\begin{align*}

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

 		&=\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})

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

 		&\overset{Lemma~\ref{lemma:eventindependenceNew}}{=}\sum_{\ell, r\in [k]: \ell<r-1}

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

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

 		\P^{[r-1,k]}([r,k]\in\mathcal{P})/(1-p)^2

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

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

 		\P^{[\ell+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]}([r,k]\in\mathcal{P})/(1-p)^2

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

 		&\overset{Corrolary~\ref{cor:probIndepNew}}{=}\sum_{\ell, r\in [k]: \ell<r-1}

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

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

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

 		\P^{[r-1,k]}([r,k]\in\mathcal{P})/(1-p)^2

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

 		&\overset{Lemma~\ref{lemma:eventindependenceNew}}{=}\sum_{\ell, r\in [k]: \ell<r-1}

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

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

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

 		&=\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)

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

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

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

 		\end{align*}

 	\end{proof}

	\begin{theorem}

		$R^{(n)}-R^{(m)}=\mathcal{O}(p^{\min(n,m)})$.

	\end{theorem}

	\begin{proof}

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

		\begin{align*}

		R^{(n)} &=  \\

		&= \frac{1}{n}\sum_{v\in[n]}\sum_{t=1}^{\infty}t\cdot \P^{(n)}(v \text{ is resampled in }t\text{ times})\\

		&= \frac{1}{n}\sum_{v\in[n]}\sum_{t=1}^{\infty}\sum_{P\text{ patch}}t\cdot\P^{(n)}(v \text{ is resampled in }t\text{ times and }P\text{ is a patch})\\	

		&= \frac{1}{n}\sum_{P\text{ patch}}\E^{(n)}(\# \text{ resamples in }P|P\in \mathcal{P})\\			

		&= \sum_{s=1}^{n-1}\E^{(n)}(\# \text{ resamples in }[s]|[s]\in \mathcal{P})+\mathcal{O}(p^{n})\tag{by translation symmetry}\\   		

		&= \sum_{s=1}^{n-1}\E^{[0,s+1]}(\# \text{ resamples in }[s]|[s]\in \mathcal{P})\P^{[s+1,n]}(\NZ{s+1}\cap\NZ{n})/(1+p)^2+\mathcal{O}(p^{n}) \tag{by Lemma~\ref{lemma:eventindependenceNew}}\\   

		&= \sum_{s=1}^{n-1}\E^{[0,s+1]}(\# \text{ resamples in }[s]|[s]\in \mathcal{P})\left(\P^{[s+1,n]}(\NZ{s+1})\right)^2/(1+p)^2+\mathcal{O}(p^{n}) \tag{by Lemma~\ref{lemma:independenetSidesNew}}\\