Question:
Take a probability space ($\Omega$,$\mathcal{F}$,P) and suppose that $\mathcal{I}_{i}$, $i=1,2,3$ are $\pi$-systems. Assume that
\begin{equation}
P(I_{1}\cap I_{2}\cap I_{3}) = P(I_{1})P(I_{2})P(I_{3}), \ \ \ \forall I_i \in \mathcal{I_i}, \ i=1,2,3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\end{equation}
Find a sufficient condition on the $\pi$-systems $\mathcal{I_i}$ such that the sigma algebras $\sigma{(\mathcal{I_1})}$, $\sigma{(\mathcal{I_2})}$, $\sigma{(\mathcal{I_3})}$ are independent.
My Attempt:
To be honest I thought that eq. (1) was already enough to ensure independence of the $\pi$-systems and thus independence of the sigma algebras generated from them.
I know it is probably an easy question but I don't understand what I'm missing here…
Best Answer
You're (almost) right. Only thing that is required is that $(1)$ holds for the case $I_1 = \Omega$ ,$I_2 = \Omega$, or $I_3 = \Omega$. This may seem trivial, but I suspect that there is some case where $P(I_1\cap I_2) \neq P(I_1)P(I_2)$ for some $I_1,I_2$, even though $(1)$ holds true.(It is related to the concept of mutual independence.) Assuming this, apply Dynkin's $\pi$-$\lambda$ argument here. To be specific, let us define $$ D_1 := \{I \in\mathcal{F}\;|\; P(I\cap I_2\cap I_3) = P(I)P(I_2)P(I_3),\;\forall I_i\in \mathcal{I}_i, i=2,3\}. $$Then we can easily see that $\Omega \in D_1$, $I\in D_1$ implies $I^c \in D_1$, and if $A_n\in D_1$ ,$n\in\mathbb{N}$ is a disjoint family, then $\cup_n A_n \in D_1$ also. That is, $D_1$ is a $\lambda$-system containing $\mathcal{I}_1$. By Dynkin's theorem, we have $\sigma(\mathcal{I}_1)\subset D_1$. Next, set $$ D_2 := \{I \in\mathcal{F}\;|\; P(I_1\cap I_2\cap I_3) = P(I_1)P(I)P(I_3),\;\forall I_1 \in \sigma(\mathcal{I}_1), \forall I_3\in \mathcal{I}_3\}. $$ By almost the same argument, we can show $\sigma(\mathcal{I}_2)\subset D_2$. It can also be shown that $\sigma(\mathcal{I}_3) \subset D_3$ in a very similar way.