Let $(\Omega, \mathcal{A}, P)$ be a probability space and $\mathcal{E}_i\subset \mathcal{A},\ \forall i\in I$. If $(\mathcal{E}_i \cup \{\emptyset\})$ is $\cap$-stable, then

$(\mathcal{E}_i)_{i\in I}\text{ independent} \Leftrightarrow\left (\sigma(\mathcal{E}_i)\right )_{i\in I}\text{ independent}$

Every proof I have seen is quite long, so I am not sure if mine is correct.

For my proof I use the principle of good sets and the $\pi$-$\lambda$ theorem.

Let $\mathcal{G}:=\{A\in\sigma(\mathcal{E}_1)\colon P\left (\bigcap_{i\in I\setminus{\{1\}}}(E_i)\cap A \right )=\prod_{i\in I\setminus{\{1\}}}P(E_i)\cdot P(A),\ E_i\in \mathcal{E}_i\ \forall i\in I\setminus{\{1\}}\}$

Since $\mathcal{E}_1\subset \mathcal{G}$, if I can show that $\mathcal{G}$ is a $\lambda$-system, I have by the $\pi$-$\lambda$ theorem that $\sigma(E_1),(\mathcal{E}_i)_{i\in I\setminus{\{1\}}}$ are independent. This I want to repeat for every $ \sigma(E_i)$ from which the conclusion follows.

Thus, the only thing I need to show now is that $\mathcal{G}$ is a $\lambda$-system.

First axiom ($\Omega\in\mathcal{G}):$ I will not show this, as it immediately follows.

Second axiom ($A,B\in\mathcal{G}, A\subset B$ then $B\setminus A\in\mathcal{G}$):

$P\left (\bigcap_{i\in I\setminus\{1\}}E_i\cap(B\setminus A)\right)=P\left (\bigcap_{i\in I\setminus\{1\}}E_i\cap(B\cap A^c)\right)=\prod_{i\in I\setminus\{1\}}P(E_i)P(A^c)P(B)=\prod_{i\in I\setminus\{1\}}P(E_i)P(B\setminus A)$

This follows from the fact (would even follow immediately) that if sets are independent, their complement and every combination of taking complement are independent, too.

Third axiom ($\bigcup_{n\in\mathbb{N}}A_n\in\mathcal{G}$ for any disjoint sets $A_1,...\in\mathcal{G}$):

$P\left ((\bigcap_{i\in I\setminus\{1\}}E_i)\cap(\dot\bigcup_{n\in\mathbb{N}}A_n)\right) = P\left (\dot\bigcup_{n\in \mathbb{N}}\left (\bigcap_{i\in I\setminus\{1\}}E_i\right) \cap A_n \right) \\ =\sum_{n\in \mathbb{N}}\prod_{i\in I\setminus\{1\}}P(E_i)P(A_n)=\prod_{i\in I\setminus\{1\}}P(E_i)\cdot \left( \sum_{n\in \mathbb{N}}P(A_n)\right)\\ =\prod_{i\in I\setminus\{1\}}P(E_i)P(\dot\bigcup_{n\in\mathbb{N}}A_n)$

where I have only used distributivity between intersection and union and that $\sigma$-additivity.


1 Answer 1


Your proof is almost correct. At first, note that $I$ can be an infinite set. Therefore you need to start with fixing a finite subset $J \subset I$. For a fixed $J = \{i_1,\ldots,i_m\}$ you start with your first step.

In the repeation of your argument you have to be more careful: You have to consider in the $j$-th step the set systems $$\sigma(\mathcal{E}_{i_1}), \ldots , \sigma(\mathcal{E}_{i_{j-1}}), \mathcal{E}_{i_j},\ldots,\mathcal{E}_{i_m}$$ in the definition of $\mathcal{G}$ to conclude after $m$ steps that $\sigma(\mathcal{E}_{i_1}), \ldots , \sigma(\mathcal{E}_{i_m})$ are independent.

However, I am not sure what kind of proofs you have seen, since your argument is - in my optionen - the 'standard way' to prove this statement.

  • $\begingroup$ "In the repeation of your argument you have to be more careful" Do you mean because I excluded the index $\{1\}$ as the first step and since, $J$ might be arbitrary, the index $\{1\}$ is i.g. not in $J$? $\endgroup$ Nov 24, 2019 at 11:19
  • 1
    $\begingroup$ I wanted only underline that the set systems in the definition of $\mathcal{G}$ have to be replaced. You have to choose $\mathcal{G}_i:=\{A\in\sigma(\mathcal{E}_{i_j})\colon P\left (\bigcap_{i\in I\setminus{\{1\}}}(E_i)\cap A \right )=\prod_{i\in I\setminus{\{1\}}}P(E_i)\cdot P(A),\ E_1 \in \sigma(\mathcal{E}_{i_1}), \ldots E_{j-1} \in \sigma (\mathcal{E}_{i_{j-1}}), E_{j+1} \in \mathcal{E}_{i_{j+1}}, \ldots, E_m \in \mathcal{E}_{i_m} \}$, i.e. the first $j-1$ systems $\mathcal{E}_{i_1}, \ldots, \mathcal{E}_{i_{j-1}}$ have to be replaced by the $\sigma$-algebras generated by them. $\endgroup$
    – p4sch
    Nov 24, 2019 at 15:12

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