AENC/resampling_chain Changeset - 17575ab3f75c · Centrum Wiskunde & Informatica (CWI)

@@ -611,13 +611,13 @@ Consider the chain (instead of the cycle) for simplicity with vertices identifie

    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}

	Let $i\neq j\in [n]$, and let $A_1$ be any event that depends only on the sites $[i,j]$ and similarly $A_2$ an event that depends only on the sites $[j,i]$. (For example $\mathrm{Z}^{(s)}$ or ``$s$ has been resampled $k$ times before termination'' for an $s$ on the correct side). Then we have

	Let $i\neq j\in [n]$, and let $A_1$ be any event that depends only on the sites $[i,j]$ (meaning the initialization and resamples) and similarly $A_2$ an event that depends only on the sites $[j,i]$. (For example $\mathrm{Z}^{(s)}$ or ``$s$ has been resampled at least $k$ times'' for an $s$ on the correct interval). Then we have

	\begin{align*}

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

&=

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

	\; \cdot \;

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

@@ -629,37 +629,32 @@ New:

	\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}

	$\P^{[n]}(\Z{1})-\P^{[n-1]}(\Z{1}) = O(p^{n})$.

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

\end{lemma}

\begin{proof}

	\begin{figure}

		\begin{center}

			\includegraphics{diagram_proborders.pdf}

		\end{center}

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

	\end{figure}

	The proof uses induction on $n$. For $n=1$ the statement is easy, since $\P^{\emptyset}(\Z{1})=0$ and $\P^{[1]}(\Z{1})=p$.

	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 $n-1$

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

	\begin{align*}

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

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

	\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]}(\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-1]}(\NZ{1})+O(p^{n-k})\right)/(1-p)+ O(p^{n}) \tag{by induction} \\

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

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

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

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

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

\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:

@@ -668,13 +663,13 @@ The intuition of the following lemma is that the far right can only affect the z

 		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 for all $\ell\leq k$, 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)

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

 		\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})\\

@@ -716,14 +711,15 @@ The intuition of the following lemma is that the far right can only affect the z

		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}}\\

		&= \sum_{s=1}^{n-1}\E^{[-N,N]}(\# \text{ resamples in }[s]|[s]\in \mathcal{P})+\mathcal{O}(p^{n}) \tag{by Lemma~\ref{lemma:eventindependenceNew},~\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+\mathcal{O}(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+\mathcal{O}(p^{n}) \tag{by Corollary~\ref{cor:probIndepNew}}\\

		&= \sum_{s=1}^{n-1}\E^{[-N,N]}(\# \text{ resamples in }[s]|[s]\in \mathcal{P})+\mathcal{O}(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})+\mathcal{O}(p^{n}).

		\end{align*}

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

	\end{proof}

Old: