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I want to show that for a sequence of independent random variables $(X_i)_{i \in \mathbb{N}}$ we have that for any two disjoint sets $A,B \subset \mathbb{N}$ we have that $ \sigma(X_i : i \in A)$ and $ \sigma(X_i : i \in B)$ are independent.

I have heard that one can use the $\pi - \lambda$ theorem to prove it. (See here)

I am not quite sure on how to do so. As far as I know, $ \sigma(X_i : i \in A) = \sigma ( \cup_{i \in A} X_i^{-1} (\mathcal{B}))$. Where $ \mathcal{B}$ is the Borel-$\sigma$-algebra on $\mathbb{R}$.

I guess my $\pi$-system would be $ \sigma(X_i : i \in A) \cup \sigma(X_i : i \in B)$ and my Dynkin system would be $ \{ \sigma(X_i) : i \in \mathbb{N}$.

But I do not see how I could make the prove work. I would appreciate some help.

Snoop
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  • To profit from the $\pi$-$\lambda$ theorem you should take the $\pi$-systems to be as small as you can get away with. You should argue that $\sigma(X_i : i \in A)$ is generated by events of the form $X_i^{-1}(S)$ for $i \in A$ and a measurable set $S$, and similar for $B$. Then say these generate the $\pi$-systems $\mathcal A_A$ and $\mathcal A_B$, and try to show that these two $\pi$-systems are independent. Then in general you can prove that if $\mathcal A_1$ and $\mathcal A_2$ are independent $\pi$-systems, then their generated $\sigma$-algebras are also independent. – Izaak van Dongen May 03 '24 at 12:29
  • @IzaakvanDongen I think I can show that the sets $X_i ^{-1} (S)$ generate $\sigma(X_i : i \in A)$. However, I do not see why $ { X_i ^{-1} (S) : S \ is \ measurable, i \in A }$ is a $\pi$-system. Furthermore, what exactly do you mean by $\mathcal{A}_A$ and $\mathcal{A}_1$. –  May 03 '24 at 13:00
  • That set isn't a $\pi$-system, but it generates a $\pi$-system which I am calling $\mathcal A_A$. You should work out what this $\pi$-system is, and show that any event in this $\pi$-system is independent of any event in the corresponding $\pi$-system for $B$. This is enough to establish the independence of the $\sigma$-algebras you're interested in, which is where you need the $\pi$-$\lambda$ theorem. – Izaak van Dongen May 03 '24 at 13:06
  • In it generates a $\pi$-system do you mean the sigma-algebra generated is a $\pi$-system? –  May 03 '24 at 13:11
  • If so then that should be $\sigma(X_i : i \in A)$ –  May 03 '24 at 13:13
  • @IzaakvanDongen would be very glad if you can give me another hint or start formulating a solution –  May 03 '24 at 14:28

2 Answers2

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I will show the argument when the subsets of $\mathbb{N}$ are finite. Let $A=\{a_1,...,a_n\}\subset \mathbb{N}$. Let $(X_n)_{n \in \mathbb{N}}$ be random variables. Then, $(X_{a_1},...,X_{a_n}):\Omega\to \mathbb{R}^n$ is a $\mathscr{F}/\mathscr{B}(\mathbb{R}^n)$-random variable.

We first prove that $\sigma(X_{a_k},a_k \in A)=\sigma((X_{a_1},...,X_{a_n}))$. We have $(\subseteq )$ because - for each $k\in\{1,...,n\}$ - $X_{a_k}=\pi_k((X_{a_1},...,X_{a_n}))$ (the $k$-th coordinate projection, which is a measurable function); indeed, by monotonicity of smallest $\sigma$-algebras $$\cup_{a_k\in A}\sigma(X_{a_k})\subseteq \sigma((X_{a_1},...,X_{a_n}))\implies \sigma(X_{a_k},a_k \in A)\subseteq \sigma((X_{a_1},...,X_{a_n}))$$ Now to prove $(\supseteq)$ we argue as following. We have that $\mathscr{B}(\mathbb{R}^n)=\sigma(\times_{n}\mathscr{B}(\mathbb{R}))$ where the sets in $\times_{n}\mathscr{B}(\mathbb{R})$ are the 'rectangles' $B_1\times ...\times B_n$, $B_\ell \in \mathscr{B}(\mathbb{R})$. Also note that for any such rectangle, $$(X_{a_1},...,X_{a_n})^{-1}(B_1\times ...\times B_n)=X_{a_1}^{-1}(B_1)\cap...\cap X_{a_n}^{-1}(B_n)\in \sigma(X_{a_k},a_k \in A)$$ so $(X_{a_1},...,X_{a_n})^{-1}(\times_{n}\mathscr{B}(\mathbb{R}))\subseteq \sigma(X_{a_k},a_k \in A)$. Therefore, again by monotonicity and commutativity of smallest $\sigma$-algebras: $$\begin{aligned}\sigma((X_{a_1},...,X_{a_n}))&:=(X_{a_1},...,X_{a_n})^{-1}(\mathscr{B}(\mathbb{R}^n))\\ &=(X_{a_1},...,X_{a_n})^{-1}(\sigma(\times_{n}\mathscr{B}(\mathbb{R})))\\ &=\sigma((X_{a_1},...,X_{a_n})^{-1}(\times_{n}\mathscr{B}(\mathbb{R})))\\ &\subseteq \sigma(X_{a_k},a_k \in A) \end{aligned}$$

At this point, suppose $(X_n)_{n \in \mathbb{N}}$ are independent. Let $C=\{c_1,...,c_k\}$ s.t. $A\cap C=\emptyset$. We have $$\begin{aligned} &P((X_{a_1},...,X_{a_n})\in B_1\times ...\times B_n,(X_{c_1},...,X_{c_k}) \in D_1\times ...\times D_k)\\ &=P(X_{a_1}\in B_{1},...,X_{a_n}\in B_n,X_{c_1}\in D_1,...,X_{c_k}\in D_k)\\ &=P(X_{a_1}\in B_{1},...,X_{a_n}\in B_n)P(X_{c_1}\in D_1,...,X_{c_k}\in D_k)\\ &=P((X_{a_1},...,X_{a_n})\in B_1\times ...\times B_n)P((X_{c_1},...,X_{c_n})\in D_1\times ...\times D_k) \end{aligned}$$ Now fix $D_1\times ...\times D_k$. We have: $$\begin{aligned}\times_n\mathscr{B}(\mathbb{R})\subseteq &\{E \in \mathscr{B}(\mathbb{R}^n):P((X_{a_1},...,X_{a_n})\in E)P((X_{c_1},...,X_{c_n})\in D_1\times ...\times D_k)\\ &=P((X_{a_1},...,X_{a_n})\in E,(X_{c_1},...,X_{c_n})\in D_1\times ...\times D_k)\} \end{aligned}$$ The rhs is a Dynkin $\lambda$-system. The lhs is a $\cap$-stable family (a $\pi$-system). The $\pi$-$\lambda$ theorem yields $$\mathscr{B}(\mathbb{R}^n)=\textrm{ rhs}$$ Now note $D_1\times ...\times D_k$ was an arbitrary rectangle from $\times_k\mathscr{B}(\mathbb{R})$. We repeat the previous step. We have $$\begin{aligned}\times_k\mathscr{B}(\mathbb{R})\subseteq &\{F \in \mathscr{B}(\mathbb{R}^k):P((X_{a_1},...,X_{a_n})\in E)P((X_{c_1},...,X_{c_n})\in F)\\ &=P((X_{a_1},...,X_{a_n})\in E,(X_{c_1},...,X_{c_n})\in F),\forall E \in \mathscr{B}(\mathbb{R}^n)\} \end{aligned}$$ Again, the $\pi$-$\lambda$ theorem yields $$\mathscr{B}(\mathbb{R}^k)=\textrm{ rhs}$$ But then this means: $$\begin{aligned} &P((X_{a_1},...,X_{a_n})\in E)P((X_{c_1},...,X_{c_n})\in F)\\ &=P((X_{a_1},...,X_{a_n})\in E,(X_{c_1},...,X_{c_n})\in F),\\ &\forall E \in \mathscr{B}(\mathbb{R}^n),\forall F \in \mathscr{B}(\mathbb{R}^k) \end{aligned}$$ Which means that $\sigma((X_{a_1},...,X_{a_n}))$ is independent of $\sigma((X_{c_1},...,X_{c_n}))$. We conclude.

Snoop
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  • Thank you. Would you mind pointing out the idea for infinite subsets? – user007 May 03 '24 at 15:30
  • @user007 you're welcome. As first impression, starting from the result for finite sets, we could maybe extend it to infinite sets by using e.g. a sequence of finite sets $(A_\ell)_\ell$ that approach an infinite set $A\subseteq \mathbb{N}$. I will think about it. – Snoop May 03 '24 at 15:54
  • I don't think there are any problems with infinite sets (even uncountable ones). The $\sigma$-algebra $\sigma(X_i: i \in I)$ will always still be generated by the $\pi$-system of events which are finite intersections of events of the form $X_i^{-1}(S)$. I think it's probably easier to keep track of if you just prove as an entirely separate lemma: "if $\mathcal A_1$ and $\mathcal A_2$ are independent $\pi$-systems, then $\sigma(\mathcal A_1)$ and $\sigma(\mathcal A_2)$ are independent". – Izaak van Dongen May 03 '24 at 16:39
  • @Snoop I will have a look at it later. Thanks for your input – user007 May 03 '24 at 17:14
  • @IzaakvanDongen it is not entirely clear to me as of why the set of finite intersections generates the sigma-algebras – user007 May 03 '24 at 17:16
  • @user007, what is your definition of $\sigma(X_i : i \in I)$? Whatever it is, it shouldn't be hard to show that this $\sigma$-algebra is generated by events of the form $X_i^{-1}(S)$ (which is sometimes even the definition). – Izaak van Dongen May 03 '24 at 17:22
  • @IzaakvanDongen I have defined it as $\sigma (X_i : i \in I) = \sigma (\cup_{i\in I} \sigma(X_i))$ –  May 03 '24 at 17:42
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I think I understand how to prove it now. Please let me know, if my reasoning is correct. Big thanks to @IzzakvanDongen.

Let $\mathcal{A}$ and $\mathcal{B}$ denote the sets of all possible finite intersections of sets of the form $X_i^{-1} (S)$ ($S$ is measurable) for $i \in A$ or $i \in B$ respectively. First we will prove that the set $\mathcal{A}$ and $ \mathcal{B}$ generate the sigma-algebras $\sigma(X_i : i \in A)$ and $\sigma(X_i : i \in B)$ respectively. ($A\cap B = \emptyset$).

The first inclusion $ \mathcal{A} \subset \sigma(X_i : i \in A)$ is pretty straightforward, since $ \sigma(X_i : i \in A) := \sigma(\cup_{i \in A} \sigma(X_i))$, thus $\sigma(X_i : i \in A)$ contains certainly every finite intersection of elements in $\mathcal{A}$.

The other inclusion$ \sigma(X_i : i \in A) \subset \mathcal{A}$ can be shown as follows: $\sigma(\mathcal{A})$ contains especially $\sigma(X_i)$ for some $i \in A$ since every set $X_i^{-1} (S)$ which form $\sigma(X_i)$ is contained in $\mathcal{A}$. Thus $ \sigma(X_i) \subset \sigma(\mathcal{A})$ and furhtermore $ \sigma(X_i : i \in A) = \sigma(\cup_{i \in A} \sigma(X_i)) \subset \sigma(\mathcal{A})$. Therefore, we can conclude $ \sigma(\mathcal{A}) = \sigma(X_i : i \in A)$. The same arguments show $ \sigma(\mathcal{B}) = \sigma(X_i : i \in B)$

Now obviously, $\mathcal{A}$ and $\mathcal{B}$ are closed under intersection. And since, $(X_i)_{i \in \mathbb{N}}$ is independent it follows that all finite subsets $X_{i_1}^{-1} (S_1),..., X_{i_N}^{-1} (S_N)$ are independent. Thus, since $A\cap B = \emptyset$ especially two sets $S_1 \in \mathcal{A}$ and $S_2 \in \mathcal{B}$ are independent. Therefore, $\mathcal{A}$ and $\mathcal{B}$ independent and closed under intersections so it follows that $\sigma(\mathcal{A}) = \sigma(X_i : i \in A)$ and $\sigma(\mathcal{B}) = \sigma(X_i : i \in B)$ are independent. Which concludes the proof.