Independence under regular conditional probability

Question

Let $\{X_j\}_{j=1}^{n}$ be a independent identically distributed random variables taking values in $\mathbb{R}^d$. We write $\mu$ for the distributions. We assume moreover that $\mu$ is absolutely continuous to the $d$-dim Lebesgue measure.

For $x \in \mathbb{R}^d$ and $r>0$, we denote by $B_x(r) \subset \mathbb{R}^d$ the closed ball centered at $x$ with radius $r>0$. Then For $x \in \mathbb{R}^d$ and $A \in \mathcal{F}$, we can consider $ P\left[A \mid X_1=x \right] $, the regular conditional probability given $X_1=x$:

My question. In a published paper, the authors conclude that \begin{align*} &P\left[\bigcap_{j=2}^{n-1}\{X_{j} \notin B(X_n,|x-X_n|)\} \mid X_1=x \right]=P\left[\bigcap_{j=2}^{n-1}X_{j} \notin B(X_n,|x-X_n|) \right]. \end{align*}

I do not know the reason why this identity is valid... Why is this true? The left-hand side should be interpreted as $P[A_x \mid X_1=x]$. Here, $A_x=\bigcap_{j=2}^{n-1}\{X_{j} \notin B(X_n,|x-X_n|)\} \in \mathcal{F}$ is a measurable set. From the definition of the regular conditional probability, can we obtain the following identity?: for any measurable $C$, \begin{align*} &\int_{C}P\left[\bigcap_{j=2}^{n-1}X_{j} \notin B(X_n,|x-X_n|) \mid X_1=x \right]\,d\mu(x) =E\left[\bigcap_{j=2}^{n-1}X_{j} \notin B(X_n,|X_1-X_n|) ,\,X_1 \in C \right]. \end{align*} I feel like it's impossible because the event $A_x$ is dependent on $x$.

Answer 1

$\newcommand\ov\overline\newcommand\R{\mathbb R}$This is just an application of Tonelli's theorem. Indeed, let $X:=X_1$, $Y:=(Y_2,\dots,Y_n)$, $Y_i:=X_i$ for $i\in\ov{2,n}$, where $\ov{k,l}:=[k,l]\cap\mathbb Z$. Let \begin{align*} A&:=\{(x,y):=(x,y_2,\dots,y_n)\in(\R^d)^n \colon\\ &\qquad\qquad\qquad\forall j\in\ov{2,n-1}\ y_j\notin B_{y_n}(|x-y_n|)\} \\ &:=\{(x,y)\in(\R^d)^n \colon\forall j\in\ov{2,n-1}\ |y_j-y_n|>|x-y_n|\}, \end{align*} so that $A$ is a Borel set. Let
\begin{equation*} f:=1_A. \end{equation*} Then for any Borel $C\subseteq\R^d$ \begin{align*} P(\{(X,Y)\in A\}\cap\{X\in C\})&=Ef(X,Y)1(X\in C\} \\ &=\int_{\R^d} P(X\in dx)\int_{(\R^d)^{n-1}} P(Y\in dy)f(x,y)1(x\in C) \\ &=\int_{\R^d} P(X\in dx)1(x\in C)\,Ef(x,Y) \\ &=\int_C P(X\in dx)\,P(Y\in A_x), \end{align*} where $A_x:=\{y\in(\R^d)^{n-1}\colon(x,y)\in A\}$; the second equality in the above display follows by Tonelli's theorem and the independence of $X$ and $Y$.

So, the regular conditional probability of the event $\{(X,Y)\in A\}$ given $X=x$ is $P(Y\in A_x)$, as desired.

Answer 2

More general: Let $(\Omega,\mathcal{A},\mathbb{P})$ a probability space and $X \colon \Omega \to E$ and $Y \colon \Omega \to F$ be independent random variables with values in Polish spaces $E$ and $F$ and $\mathcal{B}(E)$ be the Borel space of $E$, similarly for $F$. Then $\mathcal{B}(E \times F) = \mathcal{B}(E) \otimes \mathcal{B}(F)$. Let $A \in \mathcal{B}(E) \otimes \mathcal{B}(F)$ be arbitrary. Then for all $B \in \mathcal{B}(E)$ first by the theorem of Fubini and second by the definition of conditional expectation

$$\mathbb{P}((X,Y) \in A \cap B \times F) = \mathbb{P}^X \otimes \mathbb{P}^Y(A \cap B \times F) = \int_B \mathbb{P}^X(dx) \mathbb{P}^Y(A_x) = \int \mathbb{P}_B^X(dx) \mathbb{P}(A_x|X=x).$$ It follows that $\mathbb{P}^Y(A_x) = \mathbb{P}(A_x|X=x)$ for $\mathbb{P}^X$-a.e. $x \in E$.