AENC/resampling_chain Changeset - b09c769c096e · Centrum Wiskunde & Informatica (CWI)

Changeset - b09c769c096e

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Tom Bannink - 8 years ago 2017-09-06 14:16:20
tom.bannink@cwi.nl

Add better explanation of independence claim

3 files changed with 145 insertions and 45 deletions:

diagram_paths2.pdf

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diagram_paths2.tex

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main.tex

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 \documentclass{standalone}
 \usepackage[T1]{fontenc}
 \usepackage{amsmath}
 \usepackage{amsfonts}
 \usepackage{parskip}
 \usepackage{marvosym} %Lightning symbol
 \usepackage[usenames,dvipsnames]{color}
 \usepackage[hidelinks]{hyperref}
 \renewcommand*{\familydefault}{\sfdefault}
 \usepackage{bbm} %For \mathbbm{1}
 %\usepackage{bbold}
 \usepackage{tikz}
 \begin{document}
 \begin{tikzpicture}
     \def\height{4};
     \draw[step=1cm,gray,dotted] (-0.9,-0.9) grid (8.9,\height+0.9);
+    %
     % Red line through grid
+    %
     \draw [line width=3.0,red] (0,0) -- (0,1) -- (1,1) -- (2,1) -- (2,2);
+    %
     % Arrows of the grid
+    %
     \foreach \x in {0,...,7} {
         \foreach \y in {1,...,\height} {
             \draw[->] (\x,\y-1) -- (\x+0.9,\y-1);
             \draw[->] (\x,\y-1) -- (\x,\y-1+0.9);
+        }
         \draw [->] (\x,\height) -- (\x+0.9,\height);
+    }
     \foreach \y in {1,...,\height} %somehow the loop cant go to '\height-1'
         \draw [->] (8,\y-1) -- (8,\y-1+0.9); % so we fix it like this with '\y-1'
+    %
     % Move labels
+    %
     \foreach \y in {1,...,\height} {
         \draw (-1, \y - 0.5) node {$(z_\y',s_\y',r_\y')$};
+    }
     \foreach \x in {1,...,8} {
         \draw (\x-0.6, -1.4) node[rotate=70] {$(z_\x,s_\x,r_\x)$};
+    }
+    %
     % bitstring labels
+    %
     \draw(-0.1,-0.4) node {$b_1\land b_2$};
     \draw(8.2,-0.4) node {$\mathbf{1} \land b_2$};
     \draw (-0.2,\height+0.3) node {$b_1\land\mathbf{1}$};
     \draw (8.2,\height+0.3) node {$\mathbf{1}$};
+    %
     % -> steps of xi
+    %
     \draw (4,-2.5) node {$\to$ steps of $\xi_1$};
     \node[rotate=90,anchor=south,xshift=2cm,yshift=1.9cm] {$\to$ steps of $\xi_2$};
+    %
     % (Red) circles
+    %
     \draw[fill,red] (0,0) circle (0.08);
     \draw[fill    ] (8,0) circle (0.05);
     \draw[fill    ] (0,\height) circle (0.05);
     \draw[fill,red] (8,\height) circle (0.08);
+    %
     % Probability labels
+    %
     \def\x{6};
     \def\y{3};
     \draw[fill,black] (\x,\y) circle (0.07);
     \draw[fill=white,draw=black] (\x+0.23,\y-0.26) rectangle +(0.5,0.5);
     \draw[fill=white,draw=black] (\x-0.55,\y+0.26) rectangle +(1.1,0.5);
     \draw (\x+0.5,\y) node {$p_{ij}$};
     \draw (\x,\y+0.5) node {$1-p_{ij}$};
     \def\x{2};
     \def\y{1};
     \draw[fill,black] (\x,\y) circle (0.07);
     \draw[fill=white,draw=black] (\x+0.12,\y-0.26) rectangle +(0.7,0.5);
     \draw[fill=white,draw=black] (\x-0.75,\y+0.26) rectangle +(1.5,0.5);
     \draw (\x+0.5,\y) node {$p_{3,2}$};
     \draw (\x,\y+0.5) node {$1-p_{3,2}$};
     \def\x{8};
     \def\y{1};
     \draw[fill,black] (\x,\y) circle (0.07);
     \draw[fill=white,draw=black] (\x-0.25,\y+0.26) rectangle +(0.5,0.5);
     \draw (\x,\y+0.5) node {$1$};
     \def\x{3};
     \def\y{\height};
     \draw[fill,black] (\x,\y) circle (0.07);
     \draw[fill=white,draw=black] (\x+0.25,\y-0.25) rectangle +(0.5,0.5);
     \draw (\x+0.5,\y) node {$1$};
 \end{tikzpicture}
 \end{document}

main.tex

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 \begin{definition}[Events conditioned on starting state] \label{def:conditionedevents}
     For any state $b\in\{0,1\}^n$, define $\textsc{start}(b)$ as the event that the starting state of the chain is the state $b$. For any event $A$, define
     \begin{align*}
         \mathbb{P}_b(A) &= \mathbb{P}(A \;|\; \textsc{start}(b)) \\
         R_{b,A} &= \mathbb{E}( \#resamples \;|\; A \; , \; \textsc{start}(b))
     \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. 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}
     	\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 $j_1,j_2$.}
 \end{figure}
 \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,j_2)}, A_1, A_2)
         &=
         \mathbb{P}_{b_1}(\mathrm{NZ}^{(j_1,j_2)}, A_1)
         \; \cdot \;
         \mathbb{P}_{b_2}(\mathrm{NZ}^{(j_1,j_2)}, A_2) \\
         \mathbb{P}_b(A_1, A_2 \mid \mathrm{NZ}^{(j_1,j_2)})
         &=
         \mathbb{P}_{b_1}(A_1 \mid \mathrm{NZ}^{(j_1,j_2)})
         \; \cdot \;
         \mathbb{P}_{b_2}(A_2 \mid \mathrm{NZ}^{(j_1,j_2)}) \\
         R_{b,\mathrm{NZ}^{(j_1,j_2)},A_1,A_2}
         &=
         R_{b_1,\mathrm{NZ}^{(j_1,j_2)},A_1}
         \; + \;
         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 cycle are independent.
 \begin{proof}
     From any path $\xi\in\paths{b} \cap \mathrm{NZ}^{(j_1,j_2)}$ we can construct paths $\xi_1\in\paths{b_1}\cap \mathrm{NZ}^{(j_1,j_2)}$ and $\xi_2\in\paths{b_2}\cap\mathrm{NZ}^{(j_1,j_2)}$ as follows. Let us write the path $\xi$ as
     $$\xi=\left( (z_1, s_1, r_1), (z_2, s_2, r_2), ..., (z_{|\xi|}, s_{|\xi|}, r_{|\xi|}) \right)$$
     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_b[\xi] &= \P(\text{choose }s_1) \P(r_1) \P(\text{choose }s_2) \P(r_2) \cdots \P(\text{choose }s_{|\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 empty sequences $\xi_1,\xi_2$ and for each step $(z_i,s_i,r_i)$ in $\xi$ do the following: if $s_i$ is ``on the $b_1$ side of $j_1,j_2$'' then add $(z^{(1)}_i,s_i,r_i)$ to $\xi_1$ and if its ``on the $b_2$ side of $j_1,j_2$'' then add $(z^{(2)}_i,s_i,r_i)$ 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*}
     %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 $j_1,j_2$ 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''.
     The probability of $\xi_1$, started from $b_1$, is given by
     \begin{align*}
         \P_{b_1}[\xi_1] &= \P(\text{choose }s_{a_1}) \P(r_{a_1}) \P(\text{choose }s_{a_2}) \P(r_{a_2}) \cdots \P(\text{choose }s_{a_{|\xi|}}) \P(r_{a_{|\xi|}}) \\
                 &= \frac{1}{z^{(1)}_{a_1}} \P(r_{a_1}) \frac{1}{z^{(1)}_{a_2}} \P(r_{a_2}) \cdots \frac{1}{z^{(1)}_{a_{|\xi|}}} \P(r_{a_{|\xi|}})
     \end{align*}
     and similar for $\xi_2$.
     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 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 $\xi_2$. The following diagram illustrates all possible interleavings:
     \begin{center}
         \includegraphics{diagram_paths.pdf}
     \end{center}
     A particular interleaving is a path through the above grid. So for a fixed $\xi_1,\xi_2$ there are $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ possible interleavings $\xi$, and vice versa there are $\binom{|\xi_1|+|\xi_2|}{|\xi_1|}$ paths $\xi$ that decompose into the same $\xi_1$ and $\xi_2$. For a fixed $\xi_1,\xi_2$ we now show the following:
     Vice versa, any two 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 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\paths{b} \cap \mathrm{NZ}^{(j_1,j_2)} \text{ s.t.}\\ \xi \text{ decomposes into } \xi_1,\xi_2 }} \P_b[\xi] &=
         \sum_{\text{interleavings of }\xi_1,\xi_2} \P(\text{interleaving}) \cdot \P_{b_1}[\xi_1] \cdot \P_{b_2}[\xi_2] \\
         &= \P_{b_1}[\xi_1] \cdot \P_{b_2}[\xi_2]
     \end{align*}
     This is best explained by an example. We have, for an example interleaving:
     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*}
         \xi   &= \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_1 &= \left( (z^{(1)}_1, s_1, r_1), \phantom{(z^{(2)}_2, s_2, r_2), (z^{(1)}_3, s_3, r_3),} (z^{(1)}_4, s_4, r_4),\cdots  \right) \\
         \xi_2 &= \left( \phantom{(z^{(1)}_1, s_1, r_1),} (z^{(2)}_2, s_2, r_2), (z^{(2)}_3, s_3, r_3),\phantom{(z^{(2)}_4, s_4, r_4), } \cdots \right)
         \P_{b_1}[\xi_1] &= \P(\text{choose }s_1) \P(r_{a_1}) \P(\text{choose }s_2) \P(r_{a_2}) \cdots \P(\text{choose }s_{|\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*}
     Remember $z^{(1)}_i + z^{(2)}_i = z_i$.
     The probability associated to this interleaving is
     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_b[\xi] &= \frac{1}{z_1} \P(r_1) \frac{1}{z_2} \P(r_2) \frac{1}{z_3} \P(r_3) \frac{1}{z_4} \P(r_4) \cdots \\
+        \P_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}\\
         &=
         \frac{z^{(1)}_1}{z_1} \frac{1}{z^{(1)}_1} \P(r_1) \;
         \frac{z^{(2)}_2}{z_2} \frac{1}{z^{(2)}_2} \P(r_2) \;
         \frac{z^{(2)}_3}{z_3} \frac{1}{z^{(2)}_3} \P(r_3) \;
         \frac{z^{(1)}_4}{z_4} \frac{1}{z^{(1)}_4} \P(r_4)
         \cdots \\
         \frac{z_1'}{z_1+z_1'} \frac{1}{z_1'} \P(r_1') \;
         \frac{z_1 }{z_1+z_2'} \frac{1}{z_1 } \P(r_1 ) \;
         \frac{z_2 }{z_2+z_2'} \frac{1}{z_2 } \P(r_2 ) \;
         \frac{z_2'}{z_3+z_2'} \frac{1}{z_2'} \P(r_2')
         \cdots \tag{rewrite fractions}\\
         &=
         \frac{z^{(1)}_1}{z_1}
         \frac{z^{(2)}_2}{z_2}
         \frac{z^{(2)}_3}{z_3}
         \frac{z^{(1)}_4}{z_4}
         \frac{z_1'}{z_1+z_1'} \;
         \frac{z_1 }{z_1+z_2'} \;
         \frac{z_2 }{z_2+z_2'} \;
         \frac{z_2'}{z_3+z_2'}
         \cdots
         \P_{b_1}[\xi_1] \P_{b_2}[\xi_2]
     \end{align*}
     \todo{write more}
     \begin{align*}
         \P(\text{interleaving}) = \P(\text{choose step of }\xi_1) \P(\text{choose step of }\xi_2) \P(\text{choose step of }\xi_2) \P(\text{choose step of }\xi_1) \cdots
     \end{align*}
     These choices depend on the number of zeroes present in the state:
     \begin{align*}
         \P(\text{choose step of }\xi_1) &=
         \frac{z^{(1)}_1}{z^{(1)}_1 + z^{(2)}_1} \\
         \P(\text{choose step of }\xi_2) &=
         \frac{z^{(2)}_1}{z^{(1)}_1 + z^{(2)}_1}
         \P_{b_1}[\xi_1] \; \P_{b_2}[\xi_2] \tag{definition of $\P_{b_i}[\xi_i]$} \\
         &= (1-p_{1,1}) \; p_{1,2} \; p_{2,2} \; (1-p_{3,2}) \; \P_{b_1}[\xi_1] \; \P_{b_2}[\xi_2] \tag{definition of $p_{i,j}$} \\
         &= \P(\text{path in grid}) \; \P_{b_1}[\xi_1] \; \P_{b_2}[\xi_2]
     \end{align*}
-    These are the $p_i$ and $1-p_i$ shown in the grid diagram. In $\P_{b_1}[\xi_1]$ we have a factor $\frac{1}{z^{(1)}_1}$, so together with the probability above this gives $\frac{z^{(1)}_1}{z^{(1)}_1 + z^{(2)}_1} \frac{1}{z^{(1)}_1} = \frac{1}{z_1}$, as in the original expression for $\P_b[\xi]$. In the grid we see that at every point the probabilities sum to 1, and we always reach the end, so we know the sum of all paths in the grid is 1. This means the sum of all $\P(\text{interleaving})$ is equal to 1.
+    In the grid we see that at every point the probabilities sum to 1, and we always reach the end, so we know the sum of all paths in the grid is 1. This proves the required equality.
     We obtain
     \begin{align*}
         \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_1,j_2)},A_1)
         \; \cdot \;
         \mathbb{P}_{b_2}(\mathrm{NZ}^{(j_1,j_2)},A_2).
     \end{align*}
     The second equality follows directly from $\mathbb{P}(A\mid B)=\mathbb{P}(A,B)/\mathbb{P}(B)$ and setting $A_1,A_2$ to the always-true event.
     For the third equality, by the same reasoning we can decompose the paths
     \begin{align*}
         \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}
         &\equiv \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_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_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 $\mathbb{P}_b(\mathrm{NZ}_{(j_1,j_2)},A_1,A_2)$ and using the first equality gives the desired result.
 \end{proof}
 \begin{comment}
 TEST: Although a proof of claim \ref{claim:expectationsum} was already given, I'm trying to prove it in an alternate way using claim \ref{claim:eventindependence}.
+~
 Assume that $b_1$ ranges up to site $0$, the gap ranges from sites $1,...,k$ and $b_2$ ranges from site $k+1$ and onwards. For $j=1,...,k$ define the ``partial-zeros'' event $\mathrm{PZ}_j = \mathrm{Z}_1 \cap \mathrm{Z}_2 \cap ... \cap \mathrm{Z}_{j-1} \cap \mathrm{NZ}_j$ i.e. the first $j-1$ sites of the gap become zero and site $j$ does not become zero. Also define the ``all-zeros'' event $\mathrm{AZ} = \mathrm{Z}_1 \cap ... \cap \mathrm{Z}_k$, where all sites of the gap become zero. Note that these events partition the space, so we have for all $b$ that $\sum_{j=1}^k \mathbb{P}_b(\mathrm{PZ}_j) = 1 - \mathbb{P}_b(\mathrm{AZ}) = 1 - \mathcal{O}(p^k)$.
+~
 Furthermore, if site $j$ becomes zero when starting from $b_1$ it means all sites to the left of $j$ become zero as well. Similarly, from $b_2$ it implies all the sites to the right of $j$ become zero.
 Because of that, we have
 \begin{align*}
     \mathbb{P}_{b_1}(\mathrm{PZ}_j) &= \mathbb{P}_{b_1}(\mathrm{Z}_{j-1} \cap \mathrm{NZ}_j) = \mathcal{O}(p^{j-1}) \\
     \mathbb{P}_{b_2}(\mathrm{NZ}_j) &= 1 - \mathbb{P}_{b_2}(\mathrm{Z}_j) = 1 - \mathcal{O}(p^{k-j+1})
 \end{align*}
 Following the proof of claim \ref{claim:eventindependence} we also have
 \begin{align*}
     \mathbb{P}_b(\mathrm{PZ}_{j})
     &=

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