Changeset - 71d7c8f3b5ef
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Tom Bannink - 8 years ago 2017-05-30 16:45:55
tom.bannink@cwi.nl
Add claim on conditional independence
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@@ -402,24 +402,87 @@ where we used the identity $\sum_{a\in\{0,1\}^l} (-1)^{|a|} = 0$.
 

	
 
~
 

	
 
It is useful to introduce some new notation: for any event $A$ (where an event is a set of paths), define
 
\begin{align*}
 
    \mathbb{P}_b(A) &= \mathbb{P}(A \;|\; \text{start in }b) \\
 
    R_{b,A} &= \mathbb{E}( \#resamples \;|\; A\;,\; \text{start in }b)
 
\end{align*}
 
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.
 

	
 
The proof of claim \ref{claim:expectationsum} also proves the following claim
 
\begin{claim}[Probability independence] \label{claim:pathindependence}
 
    As in \ref{claim:expectationsum}, 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. Let $j_1$, $j_2$ be indices `inbetween' the groups (or only one index in case of the infinite line). Denote by $\mathrm{NZ}_j$ the event that site $j$ does not become zero before the Markov Chain terminates. Then we have
 
\begin{claim}[Conditional independence] \label{claim:eventindependence}
 
    As in \ref{claim:expectationsum}, 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. Let $j_1$, $j_2$ be indices `inbetween' the groups (or only one index in case of the infinite line). Then we have
 
    \begin{align*}
 
        \mathbb{P}[\mathrm{NZ}_{j_1} , \mathrm{NZ}_{j_2} |\;\text{start in }b]
 
        =
 
        \mathbb{P}[\mathrm{NZ}_{j_1} , \mathrm{NZ}_{j_2} \;|\;\text{start in }b_1]
 
        \mathbb{P}_b(\mathrm{NZ}_j)
 
        &=
 
        \mathbb{P}_{b_1}(\mathrm{NZ}_j)
 
        \; \cdot \;
 
        \mathbb{P}[\mathrm{NZ}_{j_1} , \mathrm{NZ}_{j_2} \;|\;\text{start in }b_2]
 
        \mathbb{P}_{b_2}(\mathrm{NZ}_j) \\
 
        R_{b,\mathrm{NZ}_j}
 
        &=
 
        R_{b_1,\mathrm{NZ}_j}
 
        \; + \;
 
        R_{b_2,\mathrm{NZ}_j}
 
    \end{align*}
 
up to any order in $p$.
 
    up to any order in $p$. Furthermore the equalities also hold when $\mathrm{NZ}_j$ is replaced by any subset $A\subseteq\mathrm{NZ}_j$.
 
\end{claim}
 
Since the left hand side is defined as
 
\begin{proof}
 
    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 $\xi_2\in\paths{b_2}\cap\mathrm{NZ}_j$ and by the same reasoning as in the proof of claim \ref{claim:expectationsum}, we obtain
 
    \begin{align*}
 
        \mathbb{P}_b(\mathrm{NZ}_j)
 
        = \sum_{\substack{\xi\in\paths{b}\\\xi \in \mathrm{NZ}_j}} \mathbb{P}[\xi]
 
        &= \sum_{\substack{\xi_1\in\paths{b_1}\\\xi_1 \in \mathrm{NZ}_j}}
 
          \sum_{\substack{\xi_2\in\paths{b_1}\\\xi_2 \in \mathrm{NZ}_j}}
 
        \mathbb{P}[\xi_1]\cdot\mathbb{P}[\xi_2] \\
 
        &=
 
        \mathbb{P}_{b_1}(\mathrm{NZ}_j)
 
        \; \cdot \;
 
        \mathbb{P}_{b_2}(\mathrm{NZ}_j).
 
    \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) R_{b,\mathrm{NZ}_j}
 
        := \sum_{\substack{\xi\in\paths{b}\\\xi \in \mathrm{NZ}_j}} \mathbb{P}[\xi] |\xi| 
 
        &= \sum_{\substack{\xi_1\in\paths{b_1}\\\xi_1 \in \mathrm{NZ}_j}}
 
          \sum_{\substack{\xi_2\in\paths{b_2}\\\xi_2 \in \mathrm{NZ}_j}}
 
        \mathbb{P}[\xi_1]\mathbb{P}[\xi_2] (|\xi_1| + |\xi_2|) \\
 
        &=
 
        \mathbb{P}_{b_2}(\mathrm{NZ}_j) \mathbb{P}_{b_1}(\mathrm{NZ}_j) R_{b_1,\mathrm{NZ}_j}
 
        \; + \;
 
        \mathbb{P}_{b_1}(\mathrm{NZ}_j) \mathbb{P}_{b_2}(\mathrm{NZ}_j) R_{b_2,\mathrm{NZ}_j} .
 
    \end{align*}
 
    Dividing by $\mathbb{P}_b(\mathrm{NZ}_j)$ and using the first equality gives the desired result.
 
\end{proof}
 

	
 
~
 

	
 
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-zero 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-zero 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 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{PZ}_j) &= \mathbb{P}_{b_2}(\mathrm{NZ}_j) = 1 - \mathbb{P}_{b_2}(\mathrm{Z}_j) = 1 - \mathcal{O}(p^{k-j+1})
 
\end{align*}
 
Now observe that
 
\begin{align*}
 
    \mathbb{P}[\mathrm{NZ}_{j_1} , \mathrm{NZ}_{j_2} |\;\text{start in }b]
 
    = \sum_{\substack{\xi\in\paths{b}\\j_1,j_2 \text{ not 0 in } \xi}} \mathbb{P}[\xi]
 
    R_b &= \sum_{j=1}^k \mathbb{P}_b(\mathrm{PZ}_j) R_{b,\mathrm{PZ}_j} + \mathbb{P}_b(\mathrm{AZ}) R_{b,\mathrm{AZ}} \\
 
        &= \sum_{j=1}^k \mathbb{P}_{b_2}(\mathrm{PZ}_j)\mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}
 
        + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{PZ}_j) R_{b_2,\mathrm{PZ}_j}
 
        + \mathcal{O}(p^k) \\
 
        &= \sum_{j=1}^k \mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}
 
        - \sum_{j=1}^k \mathbb{P}_{b_2}(\mathrm{Z}_j)\mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}
 
        + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{PZ}_j) R_{b_2,\mathrm{PZ}_j}
 
        + \mathcal{O}(p^k) \\
 
        &= \sum_{j=1}^k \mathbb{P}_{b_{1}}(\mathrm{PZ}_j) R_{b_1,\mathrm{PZ}_j}
 
        + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{PZ}_j) R_{b_2,\mathrm{PZ}_j}
 
        + \mathcal{O}(p^k) \\
 
        &= R_{b_1}
 
        + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{PZ}_j) R_{b_2,\mathrm{PZ}_j}
 
        + \mathcal{O}(p^k) \\
 
        &= R_{b_1} + R_{b_2} + \mathcal{O}(p^k)
 
\end{align*}
 
we see that all such paths $\xi$ can be split into paths $\xi_1\in\paths{b_1}$ and $\xi_2\in\paths{b_2}$ and by the same reasoning as in the proof of claim \ref{claim:expectationsum}, we obtain the right hand side.
 

	
 
\newpage
 
    \subsection{Sketch of the (false) proof of the linear bound \ref{it:const}}
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