AENC/resampling_chain Changeset - 8b8013ea2793 · Centrum Wiskunde & Informatica (CWI)

@@ -609,7 +609,7 @@ Let $G=(V,E)$ be an undirected graph with vertex set $V$ and edge set $E$. We de

            The condition does not apply to subsequent resamplings of vertices in $S$, it only specifies the initial assignment.

        \item Define $G\setminus S$ as the graph obtained by removing from $G$ all vertices in $S$ and edges adjacent to $S$.

        \item Define the $d$-neighbourhood $B^G(S;d)$ of $S$ as the set of vertices reachable from $S$ within $d$ steps.

        \item Define the boundary $\partial S$ of $S$ as the set of vertices adjacent to $S$, excluding $S$ itself. In other words $\partial S = B(S;1) \setminus S$.

        \item Define the boundary $\overline{\partial} S$ of $S$ as the set of vertices adjacent to $S$, excluding $S$ itself. In other words $\overline{\partial} S = B(S;1) \setminus S$.

    \end{itemize}

\end{definition}

@@ -640,8 +640,8 @@ The following Lemma says that if a set $S$ splits the graph in two, then those t

    \begin{align*}

        \P^{G}_S(\xi^G) &=

        \P(\text{initialize }b \mid \text{initialize $S$ to }1)

        \P(\text{pick }s_1 \mid z_1) \P(r_1)

        \P(\text{pick }s_2 \mid z_2) \P(r_2) \cdots \\

        \P(\text{pick }v_1 \mid z_1) \P(r_1)

        \P(\text{pick }v_2 \mid z_2) \P(r_2) \cdots \\

        &= \frac{(1-p)^{|b|} p^{|V|-|b|}}{(1-p)^{|S|}} \cdot

        \frac{1}{z_1} \P(r_1) \cdot

        \frac{1}{z_2} \P(r_2) \cdots

@@ -661,58 +661,84 @@ The following Lemma says that if a set $S$ splits the graph in two, then those t

        &= \P^{G\setminus Y}_S(\xi^{G\setminus Y}) \cdot \P^{G\setminus X}_S(\xi^{G\setminus X})

    \end{align*}

    where both sums are over $\binom{|\xi^{G\setminus Y}|+|\xi^{G\setminus X}|}{|\xi^{G\setminus Y}|}$ terms.

    This is best explained by an example. Lets consider the following fixed $\xi^{G\setminus Y},\xi^{G\setminus X}$ and an example interleaving where we choose steps from $\xi^{G\setminus X},\xi^{G\setminus Y},\xi^{G\setminus Y},\xi^{G\setminus X},\cdots$:

    \todo{from here}

    This is best explained by an example. Lets consider the following fixed $\xi^{G\setminus Y},\xi^{G\setminus X}$ and an example interleaving where we choose vertices from $Y,X,X,Y,\cdots$:

    \begin{align*}

        \xi^{G\setminus Y} &= \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^{G\setminus X} &= \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)

        \xi^{G\setminus Y} &= \left( (\text{initialize to }b^X\;1^S),

        (z^X_1, v^X_1, r^X_1),

        (z^X_2, v^X_2, r^X_2),

        (z^X_3, v^X_3, r^X_3),

        (z^X_4, v^X_4, r^X_4),

        \cdots  \right) \\

        \xi^{G\setminus X} &= \left( (\text{initialize to }1^S\;b^Y),

        (z^Y_1, v^Y_1, r^Y_1),

        (z^Y_2, v^Y_2, r^Y_2),

        (z^Y_3, v^Y_3, r^Y_3),

        (z^Y_4, v^Y_4, r^Y_4),

        \cdots  \right) \\

        \xi^G             &= \big( (\text{initialize to }b^X \; 1^S \; b^Y),

        (z^X_1+z^Y_1, v^Y_1, r^Y_1),

        (z^X_1+z^Y_2, v^X_1, r^X_1), \\

        &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad

        (z^X_2+z^Y_2, v^X_2, r^X_2),

        (z^X_3+z^Y_2, v^Y_2, r^Y_2),

        \cdots \big)

    \end{align*}

    The probability of $\xi^{G\setminus Y}$, started from $b_1$, is given by

    Here $b^X\in \{0,1\}^{X}$ and $b^Y\in\{0,1\}^Y$. Since we condition on the event that $S$ is initialized to ones, we know the initial state is of the form $b^X\;1^S$ in $\xi^{G\setminus Y}$. Similarly, since these paths satisfy the $\NZ{S}$ event, we know all the vertices $v_i$ resampled in $\xi^{G\setminus Y}$ are vertices in $X$, and the resampled bits $r_i$ are bits corresponding to vertices in $X$.

    In the newly constructed path $\xi^G$ the number of zeroes is the number of zeroes in $X$ and $Y$ together, so this starts as $z^X_1 + z^Y_1$. Then in this example, after the first step the number of zeroes is $z^X_1+z^Y_2$ since a step of $\xi^{G\setminus X}$ was done (so a vertex in $Y$ was resampled).

    The probability of $\xi^{G\setminus Y}$ is given by

    \begin{align*}

        \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|}) .

        \P^{G\setminus Y}_S(\xi^{G\setminus Y}) &=

        \P(\text{initialize }b^X\;1^S \mid \text{initialize $S$ to }1)

        \P(\text{pick }v^X_1 \mid z^X_1) \P(r^X_1)

        \P(\text{pick }v^X_2 \mid z^X_2) \P(r^X_2) \cdots \\

        &= (1-p)^{|b^X|} p^{|X|-|b^X|} \cdot

        \frac{1}{z^X_1} \P(r^X_1) \cdot

        \frac{1}{z^X_2} \P(r^X_2) \cdots

        \frac{1}{z^X_{|\xi^{G\setminus Y}|}} \P(r^X_{|\xi^{G\setminus Y}|}) .

    \end{align*}

    and similar for $\xi^{G\setminus X}$ but with primes.

    The following diagram illustrates all possible interleavings, and the red line corresponds to the particular interleaving $\xi$ in the example above.

    and similar for $\xi^{G\setminus X}$.

    Instead of choosing a step in $Y,X,X,Y,\cdots$ we could have chosen other orderings. The following diagram illustrates all possible interleavings, and the red line corresponds to the particular interleaving $Y,X,X,Y$ in the example above.

    \begin{center}

        \includegraphics{diagram_paths2.pdf}

        \includegraphics{diagram_paths3.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

    For the labels shown within the grid, define $p_{ij} = \frac{z^X_i}{z^X_i + z^Y_j}$.

    The probability of this particular interleaving $\xi^G$ is given by

    \begin{align*}

        \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{1}{z_2 } \P(r_2 ) \;

        \frac{z_2'}{z_3+z_2'} \frac{1}{z_2'} \P(r_2')

        \P^{G}_S(\xi^{G})

        &= (1-p)^{|b^X\; b^Y|} p^{|X\cup Y|-|b^X\;b^Y|} \quad

        \frac{1}{z^X_1+z^Y_1} \P(r^Y_1) \cdot

        \frac{1}{z^X_1+z^Y_2} \P(r^X_1) \cdots \\

        &= (1-p)^{|b^X|} p^{|X|-|b^X|} \cdot (1-p)^{|b^Y|} p^{|Y|-|b^Y|} \\

        &\qquad \cdot

        \frac{z^Y_1}{z^X_1+z^Y_1} \frac{1}{z^Y_1} \P(r^Y_1) \;

        \frac{z^X_1}{z^X_1+z^Y_2} \frac{1}{z^X_1} \P(r^X_1) \;

        \frac{z^X_2}{z^X_2+z^Y_2} \frac{1}{z^X_2} \P(r^X_2)

        \cdots \tag{rewrite fractions}\\

&=

        \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'}

        \frac{z^Y_1}{z^X_1+z^Y_1}

        \frac{z^X_1}{z^X_1+z^Y_2}

        \frac{z^X_2}{z^X_2+z^Y_2}

        \cdots

        \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]

        \P^{G\setminus Y}_S(\xi^{G\setminus Y}) \; \P^{G\setminus X}_S(\xi^{G\setminus X})

        \tag{definition} \\

        &= (1-p_{1,1}) \; p_{1,2} \; p_{2,2} \; (1-p_{3,2}) \; \P^{G\setminus Y}_S(\xi^{G\setminus Y}) \; \P^{G\setminus X}_S(\xi^{G\setminus X})

        \tag{definition of $p_{i,j}$} \\

        &= \P(\text{path in grid}) \; \P^{G\setminus Y}_S(\xi^{G\setminus Y}) \; \P^{G\setminus X}_S(\xi^{G\setminus X})

    \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*}

        \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) \\

&=

        \P^{(n)}_{b_1}(\mathrm{NZ}^{(v,w)},A_1)

        \; \cdot \;

        \P^{(n)}_{b_2}(\mathrm{NZ}^{(v,w)},A_2).

        \P^{G}_S(\NZ{S} \cap A^X \cap A^Y)

        &= \sum_{\xi^G \in \NZ{S}\cap A^X \cap A^Y} \P^{G}_S(\xi^G) \\

        &= \sum_{\xi^{G\setminus Y} \in \NZ{S}\cap A^X}

           \sum_{\xi^{G\setminus X} \in \NZ{S}\cap A^Y}

            \P^{G\setminus Y}_S(\xi^{G\setminus Y}) \cdot

            \P^{G\setminus X}_S(\xi^{G\setminus X}) \\

        &= \P^{G\setminus Y}_S(\NZ{S} \cap A^X) \; \cdot \; \P^{G\setminus X}_S(\NZ{S} \cap A^Y)

    \end{align*}

\end{proof}

\todo{rewrite from here}

\begin{lemma}[Conditional independence 2] \label{lemma:eventindependenceNewGen}

	Let $v,w \in [n]$, and let $A$ be any event that depends only on the sites $[v,w]$ (meaning the initialization and resamples) and similarly $B$ an event that depends only on the sites $[w,v]$. (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*}