AENC/resampling_chain Changeset - 0d0911198b0d · Centrum Wiskunde & Informatica (CWI)

@@ -618,16 +618,15 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i

    Without loss of generality, we can assume that $0=j < i_* < s < n$. We will now consider intervals around vertex 0.

    For $l,r\geq 1$ and $l+r\leq n$, define the event ``zeroes patch'' $\mathrm{ZP}^{[-l,r]_0}$ as the event of getting zeroes inside the interval $[-l,r]_0$ but not on the boundary, i.e.

    $$\mathrm{ZP}^{[-l,r]_0} = \NZ{-l} \cap \Z{-l+1} \cap \cdots \cap \Z{0} \cap \cdots \cap \Z{r-1} \cap \NZ{r}$$

    Note that there are $r+l-1$ `zeroes' in this event, so $\P_{J}(\mathrm{ZP}^{[-l,r]_0}) = \mathcal{O}(p^{r+l-1-|J|})$ for $J\subseteq[-l,r]_0$.

    Note that there are $r+l-1$ `zeroes' in this event, so $\P_{J}(\mathrm{ZP}^{[-l,r]_0}) = \mathcal{O}(p^{r+l-1-|J|})$ for $J\subseteq[-l,r]_0$ is a lower bound on the order of $p$.\\

    Claim:

    \begin{align*}

        \P_{I}(\mathrm{ZP}^{[-l,r]_0}) &= \P_{I'}(\mathrm{ZP}^{[-l,r]_0})

        + \mathcal{O}(p^{\min\left( \dist{r}(i_*,-l) , d(i_*,r)\right)+l+r-|I|})

        + \mathcal{O}(p^{d(i_*,0)+1-|I|})

    \end{align*}

    \todo{These special cases.}

    If $r\geq i_*$ or $l\geq n-i_*$ then $\P_{I}(\mathrm{ZP}^{[-l,r]_0}) = \mathcal{O}(p^{d(i_*,0) + 1 - |I|})$.

    If $-l\in I$ or $r\in I$ then the left hand side is zero so the claim holds.

    If $[-l,r]_0$ has no overlap with $I$ then both sides of the above expression are zero so it also holds. We are left with the case where, $-l,r,\notin I$ and $[-l,r]_0 \cap I \neq \emptyset$ and $i_*\notin[-l,r]_0$.

    If $r\geq i_*$ or $l\geq n-i_*$ then $\P_{I}(\mathrm{ZP}^{[-l,r]_0}) = \mathcal{O}(p^{d(i_*,0) + 1 - |I|})$ and also $\P_{I'}(\mathrm{ZP}^{[-l,r]_0}) = \mathcal{O}(p^{d(i_*,0) + 1 - |I|})$ so then the claim holds.

    If $-l\in I$ or $r\in I$ (and $-l,r$ are both not $i_*$ because of the previous point) then the probability of $\mathrm{ZP}^{[-l,r]_0}$ is zero for both $I$ and $I'$ so the claim holds.

    If $[-l,r]_0$ has no overlap with $I$ then both sides are also zero so it also holds. We are left with the case where: $-l,r,\notin I$ and $[-l,r]_0 \cap I \neq \emptyset$ and $i_*\notin[-l,r]_0$.

    The following diagram illustrates the situation

    \begin{center}

        \includegraphics{diagram_circle_lemma.pdf}

@@ -664,9 +663,12 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i

        &= \P_{I'}(\mathrm{ZP}^{[-l,r]_0})

        + \mathcal{O}(p^{\min\left( \dist{r}(i_*,-l) , d(i_*,r)\right)+l+r-|I|})

    \end{align*}

    Case separation shows that $\min\left( \dist{r}(i_*,-l) , d(i_*,r)\right) \geq d(i_*,0) + 1$ \todo{prove.}

    Where we used Claim~\ref{claim:eventindependence} again.

    Case separation shows that

    $$\min\left( \dist{r}(i_*,-l) , d(i_*,r)\right) + l +r \geq d(i_*,0) + 1$$

    which proves the claim.

    Now we can prove the required equalities:

    The first equality that we have to prove now follows from the fact that the ``zeroes patch'' events are a partition of $\Z{0}$:

    \begin{align*}

        \P_{I}(\Z{0})

        &=\sum_{\substack{l,r\geq 1\\l+r\leq n}}

@@ -674,7 +676,8 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i

        \tag{the events are a partition of $\Z{0}$}\\

        &=\sum_{\substack{l,r\geq 1\\l+r\leq n}}

        \P_{I'}(\mathrm{ZP}^{[-l,r]_0})

        + \mathcal{O}(p^{\min\left( \dist{r}(i_*,-l) , d(i_*,r)\right)+l+r-|I|}) \\

        + \mathcal{O}(p^{d(i_*,0)+1-|I|})

        \tag{by claim} \\

        &= \P_{I'}(\Z{0}) + \mathcal{O}(p^{d(i_*,0)+1-|I|})

    \end{align*}

    Similarly, we have

@@ -683,13 +686,35 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i

        &=\sum_{l=1}^{n-s}\sum_{r=1}^{s}

        \P_{I}(\mathrm{ZP}^{[-l,r]_0},\NZ{s})

        \tag{partition of $\Z{0}$}\\

        &=\sum_{l=1}^{n-s}\sum_{r=1}^{i_*-1}

        \P_{I}(\mathrm{ZP}^{[-l,r]_0},\NZ{s})

        +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|}) \\

        &=\sum_{l=1}^{n-s}\sum_{r=1}^{i_*-1}

        \P_{I\cap[s,r]_0}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) \cdot

        \P_{I\setminus [s,r]_0}(\NZ{r},\NZ{s})

        +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|})

        \tag{Claim~\ref{claim:eventindependence}}\\

        &=\sum_{l=1}^{n-s}\sum_{r=1}^{s}

        \P_{I'\cap[s,r]_0}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) \cdot

        \P_{I\setminus [s,r]_0}(\NZ{r},\NZ{s})

        +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|})

        \tag{$i_*\in I \setminus[s,r]_0$}\\

        &=\sum_{l=1}^{n-s}\sum_{r=1}^{s}

        \P_{I'\cap[s,r]_0}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) \cdot

        \P_{I'\setminus [s,r]_0}(\NZ{r},\NZ{s}) \\

        &\qquad +\mathcal{O}(p^{\min\left( \dist{r}(i_*,s) , d(i_*,r)\right)+l+r-|I|})

        +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|})

        \tag{same argument as before}\\

        &=\sum_{l=1}^{n-s}\sum_{r=1}^{s}

        \P_{I'}(\mathrm{ZP}^{[-l,r]_0},\NZ{s})

        + \mathcal{O}(p^{something}) \\

        \P_{I'\cap[s,r]_0}(\mathrm{ZP}^{[-l,r]_0},\NZ{s}) \cdot

        \P_{I'\setminus [s,r]_0}(\NZ{r},\NZ{s}) \\

        &\qquad

        +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|})

        \tag{case separation \todo{does not seem to work?}}\\

        &= \P_{I'}(\Z{0} , \NZ{s})

        + \sum_{l=1}^{n-s}\sum_{r=1}^{s} + \mathcal{O}(p^{something})

        +\mathcal{O}(p^{\dist{s}(i_*,0)+1-|I|})

    \end{align*}

    \todo{finish details}

    This finishes the proof.

\end{proof}

\begin{definition}[Connected patches]