This shows you the differences between two versions of the page.
Both sides previous revision Previous revision Next revision | Previous revision | ||
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) | ||
Line 56: | Line 56: | ||
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. | ||