AENC/resampling_chain Changeset - 29be8ea7ee3c · Centrum Wiskunde & Informatica (CWI)

Changeset - 29be8ea7ee3c

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András Gilyén - 8 years ago 2017-09-07 02:44:49
gilyen@cwi.nl

simplified proof

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main.tex

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  		We proceed by induction on $k$. For $k=1,2$ the statement is trivial.
  		Now observe that:
  		$$\P^{[k]}(\Z{1})=\sum_{P\text{ patch}\,:\,1\in P}\P^{[k]}(P\in\mathcal{P})$$
  		$$\P^{[k]}(\Z{k})=\sum_{P\text{ patch}\,:\,k\in P}\P^{[k]}(P\in\mathcal{P})$$
  		Suppose we proved the statement up to $k-$, then we proceed using induction similarly to the above
+ 		Suppose we proved the statement up to $k-1$, then we proceed using induction similarly to the above
  		\begin{align*}
  		&\P^{[k]}(\Z{1}\cap \Z{k})=\\
  		&=\sum_{\ell, r\in [k]: \ell<r-1}\P^{[k]}([\ell],[r,k]\in\mathcal{P})
  		+\P^{[k]}([k]\in\mathcal{P})\\
  		&=\sum_{\ell, r\in [k]: \ell<r-1}\P^{[k]}([\ell],[r,k]\in\mathcal{P})
  		+\mathcal{O}(p^{k})\\

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