Let $(\mathcal{X},d)$ a metric space with its Borel $\sigma$-algebra $\mathcal{F}_{\mathcal{X}}$.
Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space.
Let $m\in\mathbb{N}$ with $m\ge2$ and $$X,X_1,…,X_m:(\Omega,\mathcal{F})\to(\mathcal{X},\mathcal{F}_{\mathcal{X}})$$ be $\mathbb{P}$-i.i.d. random variables.
Let $\sigma^1:[0,+\infty)^m\to\{1,…,m\}$ be a measurable function such that
$$\forall r_1,…,r_m\ge0, \sigma^1(r_1,…,r_m)\in \operatorname{argmin}_{k\in\{1,…,m\}} r_k$$
and $\sigma^2:[0,+\infty)^m\to\{1,…,m\}$ be a measurable function such that
$$\forall r_1,…,r_m\ge0, \sigma^2(r_1,…,r_m)\in \operatorname{argmin}_{k\in\{1,…,m\}\backslash \{\sigma^1(r_1,…,r_m)\}} r_k.$$
Let $x\in\mathcal{X}$.
Define $$\pi^1:\mathcal{X}^m\to\{1,…,m\}, (x_1,…,x_m)\mapsto\sigma^1(d(x,x_1),…,d(x,x_m))$$
and
$$\pi^2:\mathcal{X}^m\to\{1,…,m\}, (x_1,…,x_m)\mapsto\sigma^2(d(x,x_1),…,d(x,x_m)).$$
Define
$$X^1:\Omega\to\mathcal{X}, \omega \mapsto X_{\pi^1(X_1(\omega),…,X_m(\omega))},\\
X^2:\Omega\to\mathcal{X}, \omega \mapsto X_{\pi^2(X_1(\omega),…,X_m(\omega))},\\
W:\Omega\to[0,+\infty), \omega \mapsto d(x,X^2(\omega))$$
Intuitively $X^1$ and $X^2$ are respectively the first and the second random variables chosen from $X_1,…,X_m$ that are closer to $x$, and $W$ is the distance of $X^2$ from $x$.
Is it true that the distribution of $X^1$ given $W$ is equal to the distribution of $X$ given that the distance of $X$ from $x$ is less or equal then $W$? I.e.
is it true that
$$\forall A\in\mathcal{F}_{\mathcal{X}}, \mathbb{P}(X^1\in A | W) = \mathbb{P}(X\in A | d(x,X)\le W)?$$
Intuitively, it seems obvious: the closer random variable has to belong to the closed ball centered in $x$ of radius $W$ and, since the other random variables are banned from this ball, we have only one chance distributed as $\mathbb{P}_X$ bounded to this ball to hit $A$.
However I'm a bit in trouble trying to formalize this argument…
Any help?
Best Answer
It seems that the claim is false in general since problems arise when there's a non-null probability of hitting a shell. What follows is a counter-example (at least if I haven't made any mistake).
Let $m=2$ and define \begin{equation*} \sigma^1:[0,+\infty)^2\to\{1,2\}, (r_1,r_2)\mapsto \min\left({\operatorname{argmin}_{k\in\{1,2\}}}r_k\right) \end{equation*} so that
\begin{equation*} \forall r_1,r_2\ge 0, \sigma ^1(r_1,r_2)= \begin{cases} 1, &\text{if $r_1 \le r_2$;}\\ 2, &\text{otherwise;}\\ \end{cases} \end{equation*}
and
\begin{equation*} \forall r_1,r_2\ge 0, \sigma ^2(r_1,r_2)= \begin{cases} 2, &\text{if $r_1 \le r_2$;}\\ 1, &\text{otherwise.}\\ \end{cases} \end{equation*}
Define $\mathcal{X}=\{0,1\}$ and let $d$ be the discrete metric on $\mathcal{X}$. Let $p_0\in (0,1)$ and let $\mathbb{P}_X$ be the unique probability measure on $2^{\{0,1\}}$ such that $\mathbb{P}_{X}(\{0\})=p_0$. Let $x=0$ and $A=\{x\}$. Let's show that
\begin{equation*} \mathbb{P}\left(X^1 \in A | W=1\right) \neq \mathbb{P}\left(X \in A | d(x,X)\le 1\right) \end{equation*} First, notice that
\begin{equation*} \mathbb{P}\left(X \in A | d(x,X)\le 1\right) = \mathbb{P}\left(X \in A\right) = \mathbb{P}(X=0) = p_0. \end{equation*} On the other hand \begin{equation*} \{W=1\} = \{X_2 = 1\} \cup \left(\{X_1 = 1\}\cap \{X_2 = 0\}\right) \end{equation*} and \begin{align*} \{X^1\in A\} \cap \{W=1\} &= \left(\{X^1=0\}\cap\{X_2 = 1\}\right) \cup \left(\{X^1=0\}\cap\{X_1 = 1\}\cap \{X_2 = 0\}\right)\\ &= \left(\{X_1=0\}\cap\{X_2 = 1\}\right) \cup \left(\{X_1 = 1\}\cap \{X_2 = 0\}\right),\\ \end{align*} so \begin{align*} \mathbb{P}\left(X^1\in A | W=1\right) &= \frac{\mathbb{P}\left(\{X^1\in A\}\cap \{W=1\}\right)}{\mathbb{P}\left(\{W=1\}\right)}\\ &= \frac{\mathbb{P}\left(\left(\{X_1=0\}\cap\{X_2 = 1\}\right) \cup \left(\{X_1 = 1\}\cap \{X_2 = 0\}\right)\right)}{\mathbb{P}\left(\{X_2 = 1\} \cup \left(\{X_1 = 1\}\cap \{X_2 = 0\}\right)\right)}\\ &= \frac{2 p_0(1-p_0)}{1-p_0+p_0(1-p_0)}= \frac{2 p_0(1-p_0)}{1-p_0^2}\\ &= \frac{2 p_0}{1+p_0}\neq p_0=\mathbb{P}\left(X \in A | d(x,X)\le 1\right). \end{align*}