AENC/resampling_chain Changeset - 82e14757bd1a · Centrum Wiskunde & Informatica (CWI)

Changeset - 82e14757bd1a

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Tom Bannink - 8 years ago 2017-09-07 15:22:53
tom.bannink@cwi.nl

Update proof of claim 10

3 files changed with 60 insertions and 58 deletions:

diagram_groups.pdf

bin+mod

diagram_groups.tex

main.tex

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diagram_groups.pdf

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diagram_groups.tex

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@@ @@ -15,7 +15,7 @@ @@
 \begin{document}
 \begin{tikzpicture}[scale=0.8]
-    \def\r{4};
+    \def\r{3};
     % Line and circles
     \foreach \a in {0,...,23} {
@@ @@ -25,10 +25,10 @@ @@
     % Red dots
     \foreach \a in {-3,0,2,3,4,5,6} {
-        \draw[fill,red] ({\a*15}:\r) circle (0.07);
+        \draw[fill,red] ({\a*15}:\r) circle (0.12);
+    }
     % Red label
-    \draw [fill,red] ({1*15}:\r/2) circle (0.07);
+    \draw [fill,red] ({1*15}:\r/2) circle (0.12);
     \draw ({1*15}:\r/2) +(0,-0.3) node {zeroes of $b_2$};
     % Blue squares
@@ @@ -43,8 +43,8 @@ @@
     \draw[->] ({7*15}:\r-1) -- ({7*15}:\r-0.2);
     \draw[->] ({20*15}:\r-1) -- ({20*15}:\r-0.2);
     % Labels j1 j2
     \draw ({7*15}:\r-1) +(0,-0.2) node {$j_1$};
     \draw ({20*15}:\r-1) +(0,+0.2) node {$j_2$};
     \draw ({7*15}:\r-1) +(0,-0.2) node {$v$};
     \draw ({20*15}:\r-1) +(0,+0.2) node {$w$};
 \end{tikzpicture}
 \end{document}

main.tex

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@@ @@ -57,6 +57,7 @@ @@
 \newcommand{\diam}[1]{\mathcal{D}\left(#1\right)}
 \newcommand{\paths}[1]{\mathcal{P}\left(#1\to\mathbf{1}\right)}
 \newcommand{\start}[1]{\textsc{start}\left(#1\right)}
 \newcommand{\maxgap}[1]{\mathrm{maxgap}\left(#1\right)}
 \newcommand{\gaps}[1]{#1_{\mathrm{gaps}}}
 \renewcommand{\P}{\mathbb{P}}
 \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))
         \P^{(n)}_b(A) &= \P^{(n)}(A \;|\; \textsc{start}(b)) %\\
         %R_{b,A} &= \mathbb{E}( \#resamples \;|\; A \; , \; \textsc{start}(b))
     \end{align*}
 \end{definition}
 %Note that we have $\P^{(n)}(\textsc{start}(b)) = (1-p)^{|b|}p^{n-|b|}$ by definition of our Markov Chain.
 \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.
+    Denote by $\mathrm{Z}^{(v)}$ the event that site $v$ becomes zero at any point in time before the Markov Chain terminates. Denote the complement by $\mathrm{NZ}^{(v)}$, i.e. the event that site $v$ does \emph{not} become zero before it terminates. Furthermore define $\mathrm{NZ}^{(v,w)} := \mathrm{NZ}^{(v)} \cap \mathrm{NZ}^{(w)}$, i.e. the event that \emph{both} $v$ and $w$ 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 $j_1,j_2$.}
+    \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 $v,w$.}
 \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
+    Let $b=b_1\land b_2\in\{0,1\}^n$ be a state with two groups of zeroes that are separated by at least one site inbetween, as in Figure \ref{fig:separatedgroups}. Let $v$, $w$ 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 $v,w$'', 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)
+        \P^{(n)}_b(\mathrm{NZ}^{(v,w)}, A_1, A_2)
         &=
-        \mathbb{P}_{b_1}(\mathrm{NZ}^{(j_1,j_2)}, A_1)
+        \P^{(n)}_{b_1}(\mathrm{NZ}^{(v,w)}, 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)})
         \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)}, A_2) \\
         \P^{(n)}_b(A_1, A_2 \mid \mathrm{NZ}^{(v,w)})
         &=
-        \mathbb{P}_{b_1}(A_1 \mid \mathrm{NZ}^{(j_1,j_2)})
+        \P^{(n)}_{b_1}(A_1 \mid \mathrm{NZ}^{(v,w)})
         \; \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 $p$.
         \P^{(n)}_{b_2}(A_2 \mid \mathrm{NZ}^{(v,w)}) .%\\
         %R_{b,\mathrm{NZ}^{(v,w)},A_1,A_2}
         %&=
         %R_{b_1,\mathrm{NZ}^{(v,w)},A_1}
         %\; + \;
         %R_{b_2,\mathrm{NZ}^{(v,w)},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.
+The lemma says that conditioned on $v$ and $w$ 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
     $$\xi=\left( (z_1, s_1, r_1), (z_2, s_2, r_2), ..., (z_{|\xi|}, s_{|\xi|}, r_{|\xi|}) \right)$$
     From any path $\xi\in\start{b} \cap \mathrm{NZ}^{(v,w)}$ we can construct paths $\xi_1\in\start{b_1}\cap \mathrm{NZ}^{(v,w)}$ and $\xi_2\in\start{b_2}\cap\mathrm{NZ}^{(v,w)}$ as follows. Let us write the path $\xi$ as
     $$\xi=\left( (\text{initialize }b), (z_1, s_1, r_1), (z_2, s_2, r_2), ..., (z_{|\xi|}, s_{|\xi|}, r_{|\xi|}) \right)$$
     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{pick }s_1) \P(r_1) \P(\text{pick }s_2) \P(r_2) \cdots \P(\text{pick }s_{|\xi|}) \P(r_{|\xi|}) \\
+        \P^{(n)}_b[\xi] &= \P(\text{pick }s_1 | z_1) \P(r_1) \P(\text{pick }s_2 | z_2) \P(r_2) \cdots \P(\text{pick }s_{|\xi|} | z_{|\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 $z_i = z^{(1)}_i + z^{(2)}_i$.
+    To construct $\xi_1$ and $\xi_2$, start with $\xi_1 = \left( (\text{initialize }b_1) \right)$ and $\xi_2 = \left( (\text{initialize }b_2) \right)$. For each step $(z_i,s_i,r_i)$ in $\xi$ do the following: if $s_i$ is ``on the $b_1$ side of $v,w$'' then append $(z^{(1)}_i,s_i,r_i)$ to $\xi_1$ and if its ``on the $b_2$ side of $v,w$'' then append $(z^{(2)}_i,s_i,r_i)$ 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*}
     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''.
     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 $\xi_2$.
     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 $v,w$ 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, any two paths $\xi_1\in\start{b_1}\cap \mathrm{NZ}^{(v,w)}$ and $\xi_2\in\start{b_2}\cap\mathrm{NZ}^{(v,w)}$ also induce a path $\xi\in\start{b} \cap \mathrm{NZ}^{(v,w)}$ 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]
         \sum_{\substack{\xi\in\start{b} \cap \mathrm{NZ}^{(v,w)} \text{ s.t.}\\ \xi \text{ decomposes into } \xi_1,\xi_2 }} \P^{(n)}_b[\xi] &=
         \sum_{\text{interleavings of }\xi_1,\xi_2} \P(\text{interleaving}) \cdot \P^{(n)}_{b_1}[\xi_1] \cdot \P^{(n)}_{b_2}[\xi_2] \\
         &= \P^{(n)}_{b_1}[\xi_1] \cdot \P^{(n)}_{b_2}[\xi_2]
     \end{align*}
     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 $\xi_2,\xi_1,\xi_1,\xi_2,\cdots$:
     \end{align*}
     The probability of $\xi_1$, started from $b_1$, is given by
     \begin{align*}
-        \P_{b_1}[\xi_1] &= \P(\text{pick }s_1) \P(r_1) \P(\text{pick }s_2) \P(r_2) \cdots \P(\text{pick }s_{|\xi_1|}) \P(r_{|\xi_1|}) \\
+        \P^{(n)}_{b_1}[\xi_1] &= \P(\text{pick }s_1|z_1) \P(r_1) \P(\text{pick }s_2|z_2) \P(r_2) \cdots \P(\text{pick }s_{|\xi_1|}|z_{|\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*}
     and similar for $\xi_2$ but with primes.
     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+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}\\
+        \P^{(n)}_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'}{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{z_2'}{z_3+z_2'}
         \cdots
         \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]
         \P^{(n)}_{b_1}[\xi_1] \; \P^{(n)}_{b_2}[\xi_2] \tag{definition of $\P^{(n)}_{b_i}[\xi_i]$} \\
         &= (1-p_{1,1}) \; p_{1,2} \; p_{2,2} \; (1-p_{3,2}) \; \P^{(n)}_{b_1}[\xi_1] \; \P^{(n)}_{b_2}[\xi_2] \tag{definition of $p_{i,j}$} \\
         &= \P(\text{path in grid}) \; \P^{(n)}_{b_1}[\xi_1] \; \P^{(n)}_{b_2}[\xi_2]
     \end{align*}
     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] \\
         \P^{(n)}_b(\mathrm{NZ}^{(v,w)},A_1,A_2)
         &= \sum_{\substack{\xi\in\start{b} \cap \\ \mathrm{NZ}^{(v,w)}\cap A_1\cap A_2}} \P^{(n)}_b(\xi) \\
         &= \sum_{\substack{\xi_1\in\start{b_1} \cap \\ \mathrm{NZ}^{(v,w)}\cap A_1}} \;\;
           \sum_{\substack{\xi_2\in\start{b_1} \cap \\ \mathrm{NZ}^{(v,w)}\cap A_2}}
         \P^{(n)}_{b_1}(\xi_1)\cdot\P^{(n)}_{b_2}(\xi_2) \\
         &=
-        \mathbb{P}_{b_1}(\mathrm{NZ}^{(j_1,j_2)},A_1)
+        \P^{(n)}_{b_1}(\mathrm{NZ}^{(v,w)},A_1)
         \; \cdot \;
-        \mathbb{P}_{b_2}(\mathrm{NZ}^{(j_1,j_2)},A_2).
+        \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)},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.
     %For the third equality, by the same reasoning we can decompose the paths
     %\begin{align*}
     %    \P^{(n)}_b(\mathrm{NZ}^{(v,w)},A_1,A_2) R_{b,\mathrm{NZ}^{(v,w)},A_1,A_2}
     %    &\equiv \sum_{\substack{\xi\in\start{b}\\\xi \in \mathrm{NZ}^{(v,w)}\cap A_1\cap A_2}} \P^{(n)}[\xi] |\xi| \\
     %    &= \sum_{\substack{\xi_1\in\start{b_1}\\\xi_1 \in \mathrm{NZ}^{(v,w)}\cap A_1}}
     %      \sum_{\substack{\xi_2\in\start{b_2}\\\xi_2 \in \mathrm{NZ}^{(v,w)}\cap A_2}}
     %    \P^{(n)}[\xi_1]\P^{(n)}[\xi_2] (|\xi_1| + |\xi_2|) \\
     %    &=
     %    \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)},A_2) \P^{(n)}_{b_1}(\mathrm{NZ}^{(v,w)},A_1) R_{b_1,\mathrm{NZ}^{(v,w)},A_1} \\
     %    &\quad +
     %    \P^{(n)}_{b_1}(\mathrm{NZ}^{(v,w)},A_1) \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)},A_2) R_{b_2,\mathrm{NZ}^{(v,w)},A_2} .
     %\end{align*}
     %Dividing by $\P^{(n)}_b(\mathrm{NZ}_{(v,w)},A_1,A_2)$ and using the first equality gives the desired result.
 \end{proof}
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