Welcome to Randal Douc's wiki

You've loaded an old revision of the document! If you save it, you will create a new version with this data.

====== Statement ======

Let $\seq{X_n}{n \in \nset}$ be a sequence of iid random vectors
taking values on $\rset^p$ on the same probability space
$(\Omega,\mcf, \PP)$.

For any $k \in  [1:n]$, define $X^k_{(n)}$ the $k$-th nearest neighboor of
$X_0$ among the set $\set{X_k}{k \in [1:n]}$, that is we have the properties 
  * $\|X_0-X^1_{(n)}\| \leq \|X_0-X^2_{(n)}\| \leq \ldots \leq \|X_0-X^n_{(n)}\|$. 
  * $\{X^1_{(n)}, \ldots, X^n_{(n)}\}=\{X_1,\ldots,X_n\}$.

===== Proof  =====

Note that since $n \mapsto \|X_0-X^k_{(n)}\|$ is non-increasing, $\lim_n X^k_{(n)}=X_0$ means that for any $\epsilon \in \bbQ_+^\star$, 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 $\epsilon \in \bbQ_+^\star$, 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, 
$$
\{\lim_{n \to \infty} X^k_{(n)}=X_0\}^c \subset  \cup_{(\epsilon,n) \in \bbQ^\star_+ \times \nset^*}A_{\epsilon,n} \quad \mbox{where} \quad A_{\epsilon,n}=\{\forall m \geq n, \ \| X_0- X_{m} \| \geq \epsilon\}
$$
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^\star_+ \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 the proof of the Proposition is completed.

Please fill all the letters into the box to prove you're human. P F E P W Please keep this field empty:

Edit summary

Welcome to Randal Douc's wiki

User Tools

Site Tools

Page Tools