# If generators of $\sigma$-algebra independent, then $\sigma$-algebras are independent

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

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.

• "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 $J$? – EpsilonDelta Nov 24 '19 at 11:19
• 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. – p4sch Nov 24 '19 at 15:12