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world:knn [2023/11/05 17:58]
rdouc [Proof]
world:knn [2023/11/05 18:04] (current)
rdouc [Second proof]
Line 41: Line 41:
 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)   
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 which proves \eqref{eq:​h} and the proof of the Proposition is completed. ​ which proves \eqref{eq:​h} and the proof of the Proposition is completed. ​
  
-Amelioration 
- 
-Note that since $n \mapsto \|X_0-X^k_{(n)}\|$ is non-increasing,​ $\lim_n X^k_{(n)}=X_0$ means that for any $\epsilon \in \bbQ_+^*$, the open ball $\ball{X_0,​\epsilon}$ contains at least $k$ points among $(X_n)_{n \geq 1}$. Conversely, if $\lim_n X^k_{(n)}=X_0$ does not hold, it means that for some $\epsilon \in \bbQ_+^*$, the ball $\ball{X_0,​\epsilon}$ contains at most $k-1$ points among $(X_n)_{n \geq 1}$. This, in turn, implies that $\| X_0- X_{m} \| \geq \epsilon$ for any sufficiently large $m$. Hence, 
-\[ 
-\{\lim_{n \to \infty} X^k_{(n)}=X_0\}^c \subset ​ \bigcup_{(\epsilon,​n) \in \bbQ^*_+ \times \nset^*} A_{\epsilon,​n} \quad \text{where} \quad A_{\epsilon,​n} = \{\forall m \geq n, \ \| X_0- X_{m} \| \geq \epsilon\} 
-\] 
- 
-From this inclusion property, in order to show that $\PP$-a.s., $\lim_{n \to \infty} X^k_{(n)}=X_0$ and hence $\PP\lr{\{\lim_{n \to \infty} X^k_{(n)}=X_0\}^c}=0$,​ we only need to show that $\PP(A_{\epsilon,​n})=0$ for any $(\epsilon,​n) \in \bbQ^*_+ \times \nset^*$. Using that the $(X_m)_{m \geq 0}$ are iid, 
-\begin{align*} 
-    \PP(A_{\epsilon,​n})&​=\PP(A_{\epsilon,​1})=\PP(\cap_{m=1}^\infty \{\| X_0- X_{m} \| \geq \epsilon\})=\PE\lrb{\lim_{\ell \to \infty} \PE\lrb{\left. \prod_{m=1}^\ell\indiacc{\| X_0- X_{m} \| \geq \epsilon} \right| X_0 }}= \PE\lrb{\lim_{\ell \to \infty} h(X_0)^\ell} 
-\end{align*} 
-where $h(X_0)=\PP\lr{ \| X_0- X_{1} \| \geq \epsilon | X_0 }$. To complete the proof, we only need to show that $h(X_0) \in $[0,1)$, $\PP$-a.s. Equivalently,​ since we already have $h(X_0) \in [0,1]$, we only need to show that  
-\begin{equation} \label{eq:​h} 
-   ​\PP\lr{ h(X_0)=1}=0 
-\end{equation} 
- 
-Choose a $\epsilon/​2$-net countable covering of $\rset^p$, that is $\rset^p \subset \cup_{n \in \nset} A_n$ where $A_n = \ball{a_n,​\epsilon/​2}$. ​ 
- 
-Note that on $\{h(X_0)=1\}\cap \{X_0 \in A_n\}$, ​ 
-\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 }  \\  
-    &​\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)    
-\end{align*} 
-Hence, $\{h(X_0)=1\}\cap \{X_0 \in A_n\} \subset \{1=\PP(X_0 \notin A_n)\}\cap \{X_0 \in A_n\}$ 
-\[ 
-\PP\lr{h(X_0)=1,​X_0 \in A_n} \leq \PP\lr{1 = \PP(X_0 \notin A_n),X_0 \in A_n}=\indiacc{1 = \PP(X_0 \notin A_n)} \PP(X_0 \in A_n) =0 
-\] 
- 
-This implies that  
-\[ 
-\PP\lr{ h(X_0)=1} = \PP\lr{ h(X_0)=1, X_0 \in \cup_{n ​ \in \nset}A_n } \leq \sum_{n \in \nset} \PP\lr{h(X_0)=1,​X_0 \in A_n}=0 
-\] 
- 
-which proves \eqref{eq:​h} and completes the proof of the Proposition. 
  
world/knn.1699203510.txt.gz ยท Last modified: 2023/11/05 17:58 by rdouc