AENC/resampling_chain Changeset - ffbfb3763633 · Centrum Wiskunde & Informatica (CWI)

@@ -601,21 +601,25 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i

    Then

    \begin{align*}

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

        \P_{I}(\Z{j},\NZ{s}) &= \P_{I'}(\Z{j},\NZ{s}) + \mathcal{O}(p^{\dist{s}(i_*,j) + 1 - |I|}) .

        \P_{I}(\Z{j},\NZ{s}) &= \P_{I'}(\Z{j},\NZ{s}) + \mathcal{O}(p^{\min\left( \dist{s}(i_*,j), \dist{j}(i_*,s) \right) + 1 - |I|}) .

    \end{align*}

\end{lemma}

\begin{proof}

    We will prove both statements inductively on $|I|$. For $|I|=1$ we have $I=\{i_*\}$ and $I'=\emptyset$, so $\P_{I'}(\Z{j})=0$ and

    Without loss of generality, we can assume that $j=0$ and  $0 < i_* < s < n$ (because we can shift $j$ to $0$ and switch the direction to get the correct ordering). Therefore, we have to prove:

    \begin{align*}

        \P_{I}(\Z{j})       &= \mathcal{O}(p^{d(i_*,j)}) \\

        \P_{I}(\Z{j},\NZ{s}) &= \mathcal{O}(p^{\dist{s}(i_*,j)})

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

        \P_{I}(\Z{0},\NZ{s}) &= \P_{I'}(\Z{0},\NZ{s}) + \mathcal{O}(p^{\min\left( i_*, s-i_* \right) + 1 - |I|}) .

    \end{align*}

    simply because a chain of zeroes has to be formed between $i_*$ and $j$, and in the second case this chain can not go through $s$. Now assume both statements hold up to $|I|-1$, then we prove them both for sets of size $|I|$.

    We will prove both statements inductively on $|I|$. For $|I|=1$ we have $I=\{i_*\}$ and $I'=\emptyset$, so $\P_{I'}(\Z{0})=0$ and

    \begin{align*}

        \P_{I}(\Z{0})        &= \mathcal{O}(p^{d(i_*,0)}) \\

        \P_{I}(\Z{0},\NZ{s}) &= \mathcal{O}(p^{i_*}) = \mathcal{O}(p^{\min\left( i_*, s-i_* \right)})

    \end{align*}

    simply because a chain of zeroes has to be formed between $i_*$ and $0$, and in the second case this chain can not go through $s$ so the shortest path has length $i_*$. Now assume both statements hold up to $|I|-1$, then we prove them both for sets of size $|I|$.

    When we refer to an interval $[a,b]$ on the circle we could be referring to two possible intervals because of the periodicity of the circle. Define $[a,b]_j$ as the interval with vertex $j$ on the \emph{inside}. Furthermore by $-a$ we mean the vertex $n-a$, as one would expect modulo $n$.

    %If we refer to only $[a,b]$ then we mean $\{a,a+1,...,b\}$ where numbers are considered modulo $n$. So $[a,b]$ and $[b,a]$ are different intervals that cover the circle together.

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

 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 $J\subseteq[-l,r]_0$ is a lower bound on the order of $p$.\\

@@ -633,7 +637,7 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i

    \end{center}

    Note that by Claim~\ref{claim:eventindependence} we have

    \begin{align*}

        \P_{I}(\mathrm{ZP}^{[-l,r]_0}) = \P_{I \cap [-l,r]_0}(\mathrm{ZP}^{[-l,r]_0}) \;\cdot\; \P_{I\setminus [-l,r]_0}(\NZ{a},\NZ{b})

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

    \end{align*}

    We have $i_*\in I \setminus[-l,r]_0$, and $I\cap[-l,r]_0 = I' \cap [-l,r]_0$. Define $J=I\setminus[-l,r]_0$ and $J'=I'\setminus[-l,r]_0$. We have $|J|<|I|$ so we can apply the induction hypothesis to $J$:

    \begin{align*}

@@ -647,7 +651,7 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i

        - \P_{J'}(\Z{-l},\NZ{r})

        - \P_{J'}(\Z{r}) \\

        &\quad + \mathcal{O}(p^{\dist{r}(i_*,-l)+1-|J|})

        &\quad + \mathcal{O}(p^{\min\left( \dist{r}(i_*,-l), \dist{-l}(i_*,r) \right) +1-|J|})

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

&=

        \P_{J'}(\NZ{-l},\NZ{b})

@@ -666,7 +670,7 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i

    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.

    for $l,r\geq 1$ which proves the claim.

    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*}

@@ -688,31 +692,31 @@ Here, I (Tom) tried to set do the same Lemma but for the circle instead of the i

        \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|}) \\

        +\mathcal{O}(p^{i_*+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|})

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

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

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

        &=\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|})

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

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

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

        &=\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}) \\

        &\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|})

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

        \tag{same argument as before}\\

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

        &=\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}) \\

        &\qquad

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

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

        +\mathcal{O}(p^{\min\left( i_* , s-i_* \right) +1-|I|})

        \tag{case separation}\\

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

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

        +\mathcal{O}(p^{\min\left( i_* , s-i_* \right) +1-|I|})

    \end{align*}

    This finishes the proof.

\end{proof}