From fcbfed677dc9860ddfe0b44250b904ead894685d 2017-06-10 16:26:20 From: Tom Bannink Date: 2017-06-10 16:26:20 Subject: [PATCH] Shorten notation of conditional independence lemma --- diff --git a/main.tex b/main.tex index 0e2ad6e4fb928aaf6062a2d09a86d687f2bbae9d..99efbebcbbbfe427b05ceb54fb0492d99dc576c4 100644 --- a/main.tex +++ b/main.tex @@ -416,7 +416,7 @@ It is useful to introduce some new notation: \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. + 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} @@ -427,49 +427,47 @@ It is useful to introduce some new notation: \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} , \mathrm{NZ}_{j_2}, A_1, A_2) + \mathbb{P}_b(\mathrm{NZ}^{(j_1,j_2)}, A_1, A_2) &= - \mathbb{P}_{b_1}(\mathrm{NZ}_{j_1} , \mathrm{NZ}_{j_2}, A_1) + \mathbb{P}_{b_1}(\mathrm{NZ}^{(j_1,j_2)}, A_1) \; \cdot \; - \mathbb{P}_{b_2}(\mathrm{NZ}_{j_1} , \mathrm{NZ}_{j_2}, A_2) \\ - R_{b,\mathrm{NZ}_{j_1},\mathrm{NZ}_{j_2},A_1,A_2} + \mathbb{P}_{b_2}(\mathrm{NZ}^{(j_1,j_2)}, A_2) \\ + R_{b,\mathrm{NZ}^{(j_1,j_2)},A_1,A_2} &= - R_{b_1,\mathrm{NZ}_{j_1},\mathrm{NZ}_{j_2},A_1} + R_{b_1,\mathrm{NZ}^{(j_1,j_2)},A_1} \; + \; - R_{b_2,\mathrm{NZ}_{j_1},\mathrm{NZ}_{j_2},A_2} + 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 circle are independent. \begin{proof} - For clarity we do the proof for the infinite line, when there is only one index. Simply replace $\mathrm{NZ}_j$ by $\mathrm{NZ}_{j_1}\cap\mathrm{NZ}_{j_2}$ for the case of the circle. - - Note that any path $\xi\in\paths{b} \cap \mathrm{NZ}_j$ can be split into paths $\xi_1\in\paths{b_1}\cap \mathrm{NZ}_j$ and $\xi_2\in\paths{b_2}\cap\mathrm{NZ}_j$. This can be done by taking all resampling positions $r_i$ in $\xi$ and if $r_i$ is ``on the $b_1$ side of $j_1,j_2$'' then add it to $\xi_1$ and if its ``on the $b_2$ side of $j_1,j_2$'' then add it to $\xi_2$. Note that now $\xi_1$ is a 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 $j_1,j_2$ 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, all paths $\xi_1\in\paths{b_1}\cap \mathrm{NZ}_j$ and $\xi_2\in\paths{b_2}\cap\mathrm{NZ}_j$ also induce a path $\xi\in\paths{b} \cap \mathrm{NZ}_j$ by simply concatenating the resampling positions. By the same reasoning as in the proof of claim \ref{claim:expectationsum}, we obtain + Note that any path $\xi\in\paths{b} \cap \mathrm{NZ}^{(j_1,j_2)}$ can be split into paths $\xi_1\in\paths{b_1}\cap \mathrm{NZ}^{(j_1,j_2)}$ and $\xi_2\in\paths{b_2}\cap\mathrm{NZ}^{(j_1,j_2)}$. This can be done by taking all resampling positions $r_i$ in $\xi$ and if $r_i$ is ``on the $b_1$ side of $j_1,j_2$'' then add it to $\xi_1$ and if its ``on the $b_2$ side of $j_1,j_2$'' then add it to $\xi_2$. Note that now $\xi_1$ is a 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 $j_1,j_2$ 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, all paths $\xi_1\in\paths{b_1}\cap \mathrm{NZ}^{(j_1,j_2)}$ and $\xi_2\in\paths{b_2}\cap\mathrm{NZ}^{(j_1,j_2)}$ also induce a path $\xi\in\paths{b} \cap \mathrm{NZ}^{(j_1,j_2)}$ by simply concatenating 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 concatenating the resamplings in $\xi_1$ and $\xi_2$. However, all these paths have smaller weight, and by the same reasoning as in the proof of claim \ref{claim:expectationsum} these weights sum to exactly $1$, so we obtain \begin{align*} - \mathbb{P}_b(\mathrm{NZ}_j,A_1,A_2) - = \sum_{\substack{\xi\in\paths{b}\\\xi \in \mathrm{NZ}_j\cap A_1\cap A_2}} \mathbb{P}[\xi] - &= \sum_{\substack{\xi_1\in\paths{b_1}\\\xi_1 \in \mathrm{NZ}_j\cap A_1}} - \sum_{\substack{\xi_2\in\paths{b_1}\\\xi_2 \in \mathrm{NZ}_j\cap A_2}} + \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,A_1) + \mathbb{P}_{b_1}(\mathrm{NZ}^{(j_1,j_2)},A_1) \; \cdot \; - \mathbb{P}_{b_2}(\mathrm{NZ}_j,A_2). + \mathbb{P}_{b_2}(\mathrm{NZ}^{(j_1,j_2)},A_2). \end{align*} For the second equality, note that again by the same reasoning as in the proof of claim \ref{claim:expectationsum} we have \begin{align*} - \mathbb{P}_b(\mathrm{NZ}_j,A_1,A_2) R_{b,\mathrm{NZ}_j,A_1,A_2} - &:= \sum_{\substack{\xi\in\paths{b}\\\xi \in \mathrm{NZ}_j\cap A_1\cap A_2}} \mathbb{P}[\xi] |\xi| \\ - &= \sum_{\substack{\xi_1\in\paths{b_1}\\\xi_1 \in \mathrm{NZ}_j\cap A_1}} - \sum_{\substack{\xi_2\in\paths{b_2}\\\xi_2 \in \mathrm{NZ}_j\cap A_2}} + \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} + &:= \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,A_2) \mathbb{P}_{b_1}(\mathrm{NZ}_j,A_1) R_{b_1,\mathrm{NZ}_j,A_1} \\ + \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,A_1) \mathbb{P}_{b_2}(\mathrm{NZ}_j,A_2) R_{b_2,\mathrm{NZ}_j,A_2} . + \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 $\mathbb{P}_b(\mathrm{NZ}_j,A_1,A_2)$ and using the first equality gives the desired result. + Dividing by $\mathbb{P}_b(\mathrm{NZ}_{j_1}\cap\mathrm{NZ}_{j_2},A_1,A_2)$ and using the first equality gives the desired result. \end{proof} \begin{comment}