On independence of collection of random variables

Question

This question comes from the proof of Blumenthal's 0-1 law: as part of the proof, one need to show that $A$ is independent of $\sigma(B_{t_{1}},\dots,B_{t_{p}})$. The author claimed that it suffices to show that for any bounded continuous function $f$, the following holds $$ \mathbb{E}\left[1_{A}f(B_{t_{1}},\dots,B_{t_{p}})\right] = \mathbb{P}(A)\mathbb{E}\left[f(B_{t_{1}},\dots,B_{t_{p}})\right] $$

So my question is why is such claim true?

Answer 1

There are several definitions of a measurable set $A$ being independent from a $\sigma$-algebra $\mathcal F$. Of course they are all equivalent. For instance:

$A$ is independent from $\mathcal F$ iff for all $B\in\mathcal F$, $\mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B)$.

Are you random variables $B_{t_i}$ real-valued? This does not matter. I assume that for all $i$, we have $B_{t_i}:\Omega\to E_i$, where $E_i$ is a separable metric space equipped with its Borel $\sigma$-algebra $\mathcal B(E_i)$.

By definition, $$ \sigma(B_{t_1},\cdots,B_{t_p})=\{\{(B_{t_1},\cdots,B_{t_p})\in H\}\mid H\in\bigotimes_{i=1}^p\mathcal B(E_i)\} $$

Let $B\in\sigma(B_{t_1},\cdots,B_{t_p})$. There exist $H\in\bigotimes_{i=1}^p\mathcal B(E_i)$ such that $B=\{(B_{t_1},\cdots,B_{t_p})\in H\}$. We have

\begin{align*} \mathbb P(A\cap B)&=\mathbb E[\mathbb 1_A1_H(B_{t_1},\cdots,B_{t_p})]\\ \mathbb P(A)\mathbb P(B)&=\mathbb P(A)\mathbb E[1_H(B_{t_1},\cdots,B_{t_p})]. \end{align*}

Let $$ \mathcal H=\{K\in\bigotimes_{i=1}^p\mathcal B(E_i)\mid\mathbb E[1_A1_K(B_{t_1},\cdots,B_{t_p})]=\mathbb P(A)\mathbb E[1_K(B_{t_1},\cdots,B_{t_p})]\} $$

So $\mathbb P(A\cap B)=\mathbb P(A)\mathbb P(B)$ iff $H\in\mathcal H$. So if we show that $\mathcal H=\bigotimes_{i=1}^p\mathcal B(E_i)$, then the proof is complete.

Let us then show that $\mathcal H=\bigotimes_{i=1}^p\mathcal B(E_i)$.

Let $K$ be an open subset of $\prod_{i=1}^pE_i$. Let $\varphi:\mathbb R\to\mathbb R$ be defined for all $x\in\mathbb R$ by $\varphi(x)=1$ if $x\le 0$, $\varphi(x)=1-x$ if $0\le x\le1$ and $\varphi(x)=0$ if $x\ge1$. For all $n\in\mathbb N$, let $f_n(x)=1-\varphi(n\times d(x,K^\complement))$, where $d$ denotes a metric which induces the product topology on $\prod_{i=1}^pE_i$. $(f_n)_{n\in\mathbb N}$ is a sequence of continuous and bounded functions increasing to $1_K$, so we deduce by the monotone convergence theorem that $K\in\mathcal H$.

It is easy to see that $\mathcal H$ is a Dynkin system. Moreover, the spaces $E_i$ are separable so $\bigotimes_{i=1}^p\mathcal B(E_i)=\mathcal B(\prod_{i=1}^pE_i)$ and is therefore generated by the open subsets of $\prod_{i=1}^pE_i$, which is a $\pi$-system. By Dynkin $\pi$-$\lambda$'s theorem, we deduce that $\mathcal H=\bigotimes_{i=1}^p\mathcal B(E_i)$.

Hence $A$ is independent from $\sigma(B_{t_1},\cdots,B_{t_p})$.

Answer 2

I think the claim can be understood in the following way:

For simplicity we only show that if $f,g$ being bounded continuous implies $\mathbb{E}\left[f(X)g(Y)\right] = \mathbb{E}\left[f(X)\right]\mathbb{E}\left[g(Y)\right]$, then $Y$ is independent of $Y$.

First observe that to verify $X$ is independent of $Y$ one only needs to check $\mathbb{E}\left[e^{isX}e^{itY}\right] = \mathbb{E}\left[e^{isX}\right]\mathbb{E}\left[e^{itY}\right]$. But $e^{isX}=\cos{sX}+i\sin{sX}$, then by the fact that $\sin,\cos$ are continuous and bounded we have \begin{align} &\mathbb{E}\left[(\cos{sX}+i\sin{sX})(\cos{tY}+i\sin{tY})\right] \\ =& \mathbb{E}\left[\cos{sX}\cos{tY}-\sin{sX}\sin{tY}\right] + i\mathbb{E}\left[\sin{sX}\cos{tY}+\cos{sX}\sin{tY}\right] \\ (\text{by assumption})=& \mathbb{E}\cos(sX)\mathbb{E}\cos{tY}-\mathbb{E}\sin{sX}\mathbb{E}\sin{tY} + i\left(\mathbb{E}\sin{sX}\mathbb{E}\sin{tY}+\mathbb{E}\cos(sX)\mathbb{E}\cos{tY}\right) \\ =& \mathbb{E}e^{isX}\mathbb{E}e^{itY} \end{align} An alternative idea is to approximate $1_{B}$ by a sequence of continuous bounded functions where $B$ is closed.