Independent $\sigma$-algebras using $\pi$-$\lambda$-theorem

Solution 1:

Fix $A_i,\forall i=2,\cdots,n$, define $G=\{B\in \sigma(\mathcal{E}_1)|P(B\cap\cdots\cap\ A_n)=P(B)\cdot \cdots \cdot P(A_n)\}$,

By assumption $\mathcal{E}_1\subset G$, it suffices to show $G$ is a $\lambda$-system by definition.

Then by $\pi-\lambda$ theorem, we have $\sigma(\mathcal{E}_1)\subset G$, this shows given $\mathcal{E}_1, \cdots,\mathcal{E}_n$ independent, we have $\sigma(\mathcal{E}_1), \cdots,\mathcal{E}_n$ independent, hence $\mathcal{E}_2, \cdots,\mathcal{E}_n,\sigma(\mathcal{E}_1)$ independent. THen repeart

Solution 2:

Since $\mathcal E_i$ are $π$-systems and independent, then you know (if not, there is a sketch of the proof in the end of this answer) that also the induced Dynkin systems $δ(\mathcal E_i)$ are independent. Now, since the $\mathcal E_i$'s are $π$-systems, Dynkin's theorem states that $$δ(\mathcal E_i)=σ(\mathcal E_i)$$ which completes the proof.

Statement: If $\mathcal E_i,\ i \in I$ are independent in $(Ω,\mathcal F, P)$, then the induced Dynkin systems $δ(\mathcal E_i), i\in I$ are also independent.

Sketch of the proof: Define $$\mathsf E_1=\{A\in \mathcal F:P(AA_2\dots A_n)\}=P(A)P(A_2)\dots P(A_n), \forall A_i\in\mathcal E_i \cup Ω, i=2,\dots, n\}$$ i.e. $\mathsf E_1$ is the set of all the "good sets" that are independent of $\mathcal E_2, \dots, \mathcal E_n$ (so $\mathsf E_1$ is the maximal class such that $\mathsf E_1$ and $\mathcal E_2, \dots, \mathcal E_n$ are independent. Now, show that $\mathsf E_1$ is a Dynkin system (about two pages of calculations in my textbook...) which by definition contains $\mathcal E_1$. Hence the minimality of the induced Dynkin system $δ(\mathcal E_1)$ yields $δ(\mathcal E_1)\subseteq \mathsf E_1$ which completes the proof.