AENC/resampling_chain Changeset - 82e90f72c8ad · Centrum Wiskunde & Informatica (CWI)

Changeset - 82e90f72c8ad

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Tom Bannink - 8 years ago 2017-05-23 12:18:58
tom.bannink@cwi.nl

Fix pathsplit proof

1 file changed with 9 insertions and 3 deletions:

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 \[
 	R_{b_1\land b_2} = \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1| + \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_1]\mathbb{P}[\xi_2]|\xi_2| + \sum_{\mathclap{\xi\;\mathrm{dependent}}} \mathbb{P}[\xi]|\xi|.
 \]
-The last sum only contains only terms of order $p^{k}$ or higher. Now for the first sum, note that
+where last sum only contains only terms of order $p^{k}$ or higher. Now for the first sum, note that
 \[
 	\sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|
     = \sum_{\xi_1\in\paths{b_1}} \sum_{\substack{\xi_2\in\paths{b_2}\\ \text{independent of }\xi_1}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|
 \]
 where the sum over independent paths could be empty.
 where the sum over independent paths could be empty for certain $\xi_1$. Now we replace this last sum by a sum over \emph{all} paths $\xi_2\in\paths{b_2}$. This will change the sum but only for terms where $\xi_1,\xi_2$ are dependent. For those terms we already know that $\mathbb{P}[\xi_1]\mathbb{P}[\xi_2]$ contains a factor $p^k$ and hence we have
 \begin{align*}
     \sum_{\mathclap{\substack{\xi_{1,2}\in\paths{b_{1,2}}\\ \mathrm{independent}}}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1|
     &= \sum_{\xi_1\in\paths{b_1}} \sum_{\xi_2\in\paths{b_2}} \mathbb{P}[\xi_2]\mathbb{P}[\xi_1]|\xi_1| + \mathcal{O}(p^k) \\
     &= \sum_{\xi_1\in\paths{b_1}} \mathbb{P}[\xi_1]|\xi_1| + \mathcal{O}(p^k) \\
     &= R_{b_1} + \mathcal{O}(p^k)
 \end{align*}
 we can do the same with the second term and this proves the claim.
 \end{proof}
 \begin{center}
 \includegraphics{diagram_paths.pdf}
 \end{center}

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