Changeset - 62d60a9017a9
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Tom Bannink - 8 years ago 2017-09-07 17:14:59
tom.bannink@cwi.nl
1 file changed with 16 insertions and 12 deletions:
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@@ -663,6 +663,8 @@ The intuition of the following lemma is that the far right can only affect the z
 
	$\P^{[n]}(\Z{1})-\P^{[m]}(\Z{1}) = \bigO{p^{\min(n,m)}}$. (Should be true with $\bigO{p^{\min(n,m)+1}}$ too.)
 
\end{corollary}
 

	
 
	The intuition of the following lemma is simmilar to the previous. The events on the two sides should be independent unless an interaction chain is forming, implying that every vertex gets resampled to $0$ at least once.
 

	
 
 	\begin{lemma}\label{lemma:independenetSidesNew}	
 
 		$$\P^{[k]}(\Z{1}\cap \Z{k})=\P^{[k]}(\Z{1})\P^{[k]}(\Z{k})+\bigO{p^{k}}=\left(\P^{[k]}(\Z{1})\right)^2+\bigO{p^{k}}.$$
 
 	\end{lemma}   
 
@@ -710,6 +712,8 @@ The intuition of the following lemma is that the far right can only affect the z
 
 		+\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}
 
		$R^{(n)}-R^{(m)}=\bigO{p^{\min(n,m)}}$.
 
@@ -727,18 +731,18 @@ The intuition of the following lemma is that the far right can only affect the z
 
		\begin{align*}
 
			R^{(n)}
 
			&= \E^{(n)}(\Res{1}) \tag{by translation invariance}\\
 
			&= \sum_{k=1}^{\infty}\P^{(n)}(\Res{1}\geq 1) \\
 
			&= \sum_{k=1}^{\infty}\sum_{P\text{ patch}:1\in P}\P^{(n)}(\Res{1}\geq 1\& P\in\mathcal{P}) \tag{partition}\\
 
			&= \frac{1}{n}\sum_{v\in[n]}\sum_{t=1}^{\infty}\sum_{P\text{ patch}}t\cdot\P^{(n)}(v \text{ is resampled }t\text{ times and }v\in P | P\in\mathcal{P}) \; \P^{(n)}(P\in\mathcal{P})\\
 
			&= \frac{1}{n}\sum_{P\text{ patch}}\E^{(n)}(\# \text{ resamples in }P|P\in \mathcal{P}) \; \P^{(n)}(P\in\mathcal{P})\\
 
			&= \sum_{s=1}^{n-1}\E^{(n)}(\# \text{ resamples in }[s] \;|\; [s]\in \mathcal{P}) \; \P([s]\in\mathcal{P}) +\bigO{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+\bigO{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+\bigO{p^{n}} \tag{by Lemma~\ref{lemma:independenetSidesNew}}\\   
 
			&= \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+\bigO{p^{n}} \tag{by Corollary~\ref{cor:probIndepNew}}\\   			
 
			&= \sum_{s=1}^{n-1}\E^{[-N,N]}(\# \text{ resamples in }[s]|[s]\in \mathcal{P})+\bigO{p^{n}} \tag{by Lemma~\ref{lemma:eventindependenceNew}, Corollary~\ref{cor:probIndepNew}}\\   	
 
			&= \sum_{s=1}^{N}\E^{[-N,N]}(\# \text{ resamples in }[s]|[s]\in \mathcal{P})+\bigO{p^{n}}.
 
			&= \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*}		
 
		
 
		Repeating the same calculation with $m$, and comparing the two expressions completes the proof.
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