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If we have 3 independent random variables $X,Y$ and $Z$ on a probability space $(\Omega,\mathscr{F},\mathbb{P})$, then how do we show that $\sigma(X,Y)$ and $\sigma(Z)$ are independent?

It feels intuitively obvious, but I'm having real difficulty making it precise. I feel like using the fact that if two measures $\mu _1$ and $\mu _2$ are equal on a $\pi$ system $\mathscr{A}$ then they are equal on $\sigma (\mathscr{A})$ could be useful, but am unsure exactly what measures/$\pi$ systems to use.

Any help would be really appreciated - thanks! :)

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  • $\begingroup$ Where did you find this exercise? $\endgroup$
    – Snoop
    May 7, 2021 at 21:39
  • $\begingroup$ This came up as a past paper question for an upcoming exam I'm revising for :) $\endgroup$ May 7, 2021 at 21:44

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A similar proof for what follows can be found on page 39 of the book Probability with Martingales (1991) by David Williams. Define the following collections of sets:

\begin{align*}\Pi_1&=\Big\{\lbrace \omega\in\Omega: X(\omega)\le x\rbrace\cap \lbrace \omega\in\Omega: Y(\omega)\le y\rbrace:x,y\in\mathbb{R}\Big\},\\ \Pi_2&=\Big\{\lbrace \omega\in\Omega: Z(w)\le z\rbrace:z\in\mathbb{R}\Big\}. \end{align*}

You can readily show that $\Pi_1$ and $\Pi_2$ are $\pi$-systems such that $$\sigma(\Pi_1)=\sigma(X,Y)\quad\text{and}\quad \sigma(\Pi_2)=\sigma(Z).$$

To prove that $\sigma(X,Y)$ and $\sigma(Z)$ are independent, by definition we must show that $P(A\cap B)=P(A)P(B)$ for all $A\in \sigma(X,Y)$ and $B\in \sigma(Z)$.

To that end, fix $A\in \Pi_1$ and define the following two measures (you can prove that they are measures) $\mu_{1},\mu_{2}:\mathcal{F}\rightarrow [0,\infty)$ as follows: $$\mu_1(B)=P(A\cap B)\quad\text{and} \quad \mu_2(B) = P(A)P(B).$$ Since $A\in \Pi_1$, we can write $A=\lbrace X\le x\rbrace\cap \lbrace Y\le y\rbrace$ for some $x,y\in\mathbb{R}$. By independence of the random variables $X$, $Y$, and $Z$, if $B=\lbrace{Z\le z\rbrace}\in \Pi_2$ then we find that \begin{align*}\mu_1(B) &= P(A\cap B) \\&= P\left(\lbrace X\le x\rbrace\cap \lbrace Y\le y\rbrace\cap \lbrace{Z\le z\rbrace}\right) \\&= P\left(\lbrace X\le x\rbrace\right)\cdot P\left(\lbrace Y\le y\rbrace\right)\cdot P\left(\lbrace{Z\le z\rbrace}\right)\\&=P\left(\lbrace X\le x\rbrace\cap \lbrace Y\le y\rbrace\right)\cdot P\left(\lbrace{Z\le z\rbrace}\right)\\&=P(A) P(B)\\&=\mu_2(B). \end{align*} This proves that the measures $\mu_1$ and $\mu_2$ agree on the $\pi$-system $\Pi_2$. Furthermore, note that $\mu_1(\Omega)=\mu_2(\Omega) = P(A)<\infty$. Now we can use the following fact: finite measures that agree on a $\pi$-system also agree on the $\sigma$-algebra generated by the $\pi$-system. Therefore, $$P(A\cap B) = P(A)P(B)\quad\text{for all}\quad A\in \Pi_1\quad\text{and}\quad B\in \sigma(Z).$$ Now fix $B\in \sigma(Z)$ and define two more measures $\mu_{3},\mu_{4}:\mathcal{F}\rightarrow [0,\infty)$ as follows: $$\mu_3(A)=P(A\cap B)\quad\text{and} \quad \mu_2(A) = P(A)P(B).$$ Our previous argument showed that $\mu_3$ and $\mu_4$ agree on the $\pi$-system $\Pi_1$. Furthermore, $\mu_3(\Omega)=\mu_4(\Omega)=P(B)<\infty$. Using the same fact as before, we conclude that $\mu_3$ and $\mu_4$ agree on $\sigma(X,Y)$. This completes the proof.

This argument can be generalized as follows: Let $X_1,X_2,\dots$ be a sequence of independent random variables. Then $\sigma(X_1,\dots, X_n)$ and $\sigma(X_{n+1}, X_{n+2}, \dots)$ are independent for each $n\in\mathbb{N}$; see page 47 in Williams (1991).

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  • $\begingroup$ Could you show me how $\sigma(\Pi_1)=\sigma(X,Y)$? I don't really get how to extend Williams' argument for the individual random variable to two or more. $\endgroup$
    – Snoop
    May 8, 2021 at 1:44
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    $\begingroup$ By definition, $\sigma(X,Y)$ is the smallest $\sigma$-algebra that makes both $X$ and $Y$ measurable. Since it (obviously) must contain the sets in $\Pi_1$, we have $\sigma(\Pi_1)\subseteq \sigma(X,Y)$. Conversely, note that $\lbrace X\le a\rbrace = \bigcup_{n=1}^{\infty}(\lbrace X\le a\rbrace\cap\lbrace Y\le n\rbrace\in\sigma(\Pi_1)$ for all $a\in\mathbb{R}$. Similarly, $\lbrace Y\le b\rbrace\in\sigma(\Pi_1)$ for all $b\in\mathbb{R}$. So $\sigma(X)\subseteq \sigma(\Pi_1)$ & $\sigma(Y)\subseteq\sigma(\Pi_1)$. Hence, $\sigma(X,Y)=\sigma\left(\sigma(X)\cup\sigma(Y)\right)\subseteq\sigma(\Pi_1)$. $\endgroup$
    – Satana
    May 8, 2021 at 17:56
  • $\begingroup$ @Snoop If any of these measure-theoretic concepts are unfamiliar, my best advice is invest in an analysis book. Two good options are Real & Complex Analysis by Walter Rudin or Real Analysis: Modern Techniques and Their Applications by Gerald B. Folland. They are written for a graduate student in math and assume basic knowledge from undergraduate analysis (elementary set theory, limits, etc.), but are otherwise self-contained. Even measure-theoretic probability texts usually devote a chapter (or an appendix) to analysis. This type of argument is pretty standard. $\endgroup$
    – Satana
    May 8, 2021 at 18:06

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