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Here are some definitions of Independence. Let a random variable be defined as $X:\Omega\rightarrow\mathbb{R}$.

$X_1,X_2,\dots,$ random variables are independent iff $\forall$ Borel sets $B_1,B_2,\ldots,\in\mathbb{R}$ the events {$X_1\in B_1$}, {$X_2\in B_2$}, $\ldots$, are independent.

$X$ a random variable, $\mathcal{F}$ $\sigma$-algebra are independent if $\forall\mathcal{B} \subseteq \mathbb{R}$ Borel and $\forall F \in \mathcal{F}, \{X \in B \}$ and $F$ are independent.

Questions:

  1. {$X_1\in B_1$} means the sets in $\Omega$, that are pullbacks of $X_1$ from $B_1$ that form a Borel $\sigma$-algebra, yeah?
  2. Each {$X_i\in B_i$} forms a family of sets. Say we have {$X_1\in B_1$} and {$X_2\in B_2$}, call it $\mathcal{F}_1$ and $\mathcal{F}_2$ for convenience. When do we say these 2 are independent? Is it when we take any event from the $\mathcal{F}_1$, it is independent ($\mathbb{P}(A_1 \cap A_2) = \mathbb{P}(A_1)\cdot\mathbb{P}(A_2)$), from all events in $\mathcal{F}_2$?
  3. Just to be sure, I think the second definition is just a re-wording of the first definition no?
AyamGorengPedes
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1 Answers1

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  1. Yes, the collection of subsets of $\Omega$ defined by $$\big\{\{X_1\in B\} : B\in \mathcal{B}(\mathbb R)\big\}$$ forms a $\sigma$-algebra, it is called the $\sigma$-algebra generated by $X_1$.

  2. Yes, more precisely, using your notation the set $\big\{\{X_1\in B_1\} : B_1\in \mathcal{B}(\mathbb R)\big\}$ is a $\sigma$-algebra, not just $\{X_1\in B_1\}$.

  3. One definition is for the independence of random variables, and the other is for independence between a random variable and a $\sigma$-algebra. Not all $\sigma$-algebra can be generated by a (Borel measurable) random variable, see here.

Xiao
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