diff --git a/main.tex b/main.tex index 6119dd84d3ca3a2ae4fca124879e10d5963820f6..d3fb97508419c25186a0804b3ecbae40578a4e79 100644 --- a/main.tex +++ b/main.tex @@ -413,20 +413,22 @@ The proof of claim \ref{claim:expectationsum} also proves the following claim \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}_b(\mathrm{NZ}_j) + \mathbb{P}_b(\mathrm{NZ}_{j_1} , \mathrm{NZ}_{j_2}) &= - \mathbb{P}_{b_1}(\mathrm{NZ}_j) + \mathbb{P}_{b_1}(\mathrm{NZ}_{j_1} , \mathrm{NZ}_{j_2}) \; \cdot \; - \mathbb{P}_{b_2}(\mathrm{NZ}_j) \\ - R_{b,\mathrm{NZ}_j} + \mathbb{P}_{b_2}(\mathrm{NZ}_{j_1} , \mathrm{NZ}_{j_2}) \\ + R_{b,\mathrm{NZ}_{j_1},\mathrm{NZ}_{j_2}} &= - R_{b_1,\mathrm{NZ}_j} + R_{b_1,\mathrm{NZ}_{j_1},\mathrm{NZ}_{j_2}} \; + \; - R_{b_2,\mathrm{NZ}_j} + R_{b_2,\mathrm{NZ}_{j_1},\mathrm{NZ}_{j_2}} \end{align*} - up to any order in $p$. Furthermore the equalities also hold when $\mathrm{NZ}_j$ is replaced by any subset $A\subseteq\mathrm{NZ}_j$. + up to any order in $p$. \end{claim} \begin{proof} + For clarity we do the proof for the infinite line, when there is only one index. Simply replace $\mathrm{NZ}_j$ by $\mathrm{NZ}_{j_1}\cap\mathrm{NZ}_{j_2}$ for the case of the circle. + 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) @@ -456,32 +458,52 @@ The proof of claim \ref{claim:expectationsum} also proves the following claim ~ -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 $\sum_{j=1}^k \mathbb{P}_b(\mathrm{PZ}_j) = 1 - \mathbb{P}_b(\mathrm{AZ}) = 1 - \mathcal{O}(p^k)$. +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-zeros'' 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-zeros'' 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. +~ + +Furthermore, if site $j$ becomes zero when starting 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}) + \mathbb{P}_{b_2}(\mathrm{NZ}_j) &= 1 - \mathbb{P}_{b_2}(\mathrm{Z}_j) = 1 - \mathcal{O}(p^{k-j+1}) +\end{align*} +Following the proof of claim \ref{claim:eventindependence} we also have +\begin{align*} + \mathbb{P}_b(\mathrm{PZ}_{j}) + &= + \mathbb{P}_{b_1}(\mathrm{PZ}_{j}) + \; \cdot \; + \mathbb{P}_{b_2}(\mathrm{NZ}_{j}) \\ + R_{b,\mathrm{PZ}_{j}} + &= + R_{b_1,\mathrm{PZ}_{j}} + \; + \; + R_{b_2,\mathrm{NZ}_{j}} \end{align*} + + Now observe that \begin{align*} 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} + &= \sum_{j=1}^k \mathbb{P}_{b_2}(\mathrm{NZ}_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{NZ}_j) R_{b_2,\mathrm{NZ}_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} + + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{NZ}_j) R_{b_2,\mathrm{NZ}_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} + + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{NZ}_j) R_{b_2,\mathrm{NZ}_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} + + \sum_{j=1}^k \mathbb{P}_{b_1}(\mathrm{PZ}_j)\mathbb{P}_{b_{2}}(\mathrm{NZ}_j) R_{b_2,\mathrm{NZ}_j} + \mathcal{O}(p^k) \\ - &= R_{b_1} + R_{b_2} + \mathcal{O}(p^k) + &\overset{???}{=} R_{b_1} + R_{b_2} + \mathcal{O}(p^k) \end{align*} \newpage