# If $X,Y$ and $Z$ are independent, then $\sigma(X,Y)$ and $\sigma(Z)$ are independent

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! :)

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

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).

• 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. May 8, 2021 at 1:44
• 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, $\sigma(X,Y)=\sigma\left(\sigma(X)\cup\sigma(Y)\right)\subseteq\sigma(\Pi_1)$. May 8, 2021 at 17:56
• @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. May 8, 2021 at 18:06