AENC/resampling_chain Changeset - 83372538681f · Centrum Wiskunde & Informatica (CWI)

Changeset - 83372538681f

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András Gilyén - 8 years ago 2017-09-07 02:37:14
gilyen@cwi.nl

simplified proof

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 	&=\sum_{k=1}^{n-1}\P^{[n]}([k]\in\mathcal{P})+ O(p^{n}) \tag{by Claim~\ref{lemma:eventindependenceNew}}\\
 	&=\sum_{k=1}^{n}\P^{[n]}([k]\in\mathcal{P})+ O(p^{n}) \\
 	&=\P^{[n]}(\Z{1})	+ O(p^{n})
 	\end{align*}
 \end{proof}
 \begin{corollary}\label{cor:probIndepNew}
-	$\P^{[n]}(\Z{1})-\P^{[m]}(\Z{1}) = O(p^{\min(n,m)})$. (Should be true with $O(p^{\min(n,m)+1})$ as well.)
+	$\P^{[n]}(\Z{1})-\P^{[m]}(\Z{1}) = O(p^{\min(n,m)})$. (Should be true with $O(p^{\min(n,m)+1})$ too.)
 \end{corollary}
  	\begin{lemma}\label{lemma:independenetSidesNew}
  		$$\P^{[k]}(\Z{1}\cap \Z{k})=\P^{[k]}(\Z{1})\P^{[k]}(\Z{k})+\mathcal{O}(p^{k})=\left(\P^{[k]}(\Z{1})\right)^2+\mathcal{O}(p^{k}).$$
  	\end{lemma}
  	Note that using De Morgan's law and the inclusion-exclusion formula we can see that this is equivalent to saying:

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