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world:knn [2023/11/05 17:59] rdouc [Proof] |
world:knn [2023/11/05 18:04] (current) rdouc [Second proof] |
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Note that on $\{h(X_0)=1\}\cap \{X_0 \in A_n\} $, | Note that on $\{h(X_0)=1\}\cap \{X_0 \in A_n\} $, | ||
\begin{align*} | \begin{align*} | ||
- | 1&=h(X_0)=\PP\lr{ \| X_0- X_{1} \| \geq \epsilon | X_0 } \leq \PP\lr{ \| X_0-a_n\| + \| a_n-X_{1} \| \geq \epsilon | X_0 } \\ | + | 1=h(X_0)=\PP\lr{ \| X_0- X_{1} \| \geq \epsilon | X_0 } &\leq \PP\lr{ \| X_0-a_n\| + \| a_n-X_{1} \| \geq \epsilon | X_0 } \\ |
&\leq \PP\lr{ \epsilon/2 + \| a_n-X_{1} \| \geq \epsilon | X_0 }= \PP\lr{\| a_n-X_{1} \|\geq \epsilon/2 | X_0 }\\ | &\leq \PP\lr{ \epsilon/2 + \| a_n-X_{1} \| \geq \epsilon | X_0 }= \PP\lr{\| a_n-X_{1} \|\geq \epsilon/2 | X_0 }\\ | ||
&=\PP\lr{\| a_n-X_{1} \| \geq \epsilon/2 }=\PP(X_1 \notin A_n)=\PP(X_0 \notin A_n) | &=\PP\lr{\| a_n-X_{1} \| \geq \epsilon/2 }=\PP(X_1 \notin A_n)=\PP(X_0 \notin A_n) |