Let $(\Omega,\mathcal F, P)$ be such a probability space that for every event $A\in\mathcal F$, the events that are independent to $A$ form an algebra $\mathcal B_A$ ($\Omega\in\mathcal B_A$). Independence to $A$ implying (I hope I get this right) for every finite $\mathcal C\subseteq\mathcal B_A$ $$P\left (A\cap\bigcap_{C\in\mathcal C}C\right ) = P(A)\prod_{C\in\mathcal C} P(C)\tag{1}\label{eq:1} $$

Let $A_1,A_2,A_3\in\mathcal F$ be pairwise independent events. Show that they are in fact independent events. So it remains to verify $$P(A_1\cap A_2\cap A_3) = P(A_1)P(A_2)P(A_3). $$

A quick (perhaps insignificant) observation is that $A\notin\mathcal B_A$ (because $P(A) = P(A\cap A)\neq P(A)P(A)$).

An idea was to use the fact that $\Omega\in \mathcal B_{A_1\cap A_2}$ : by additivity $$P(A_1\cap A_2\cap (A_3\cup A_3^c)) = P(A_1\cap A_2\cap A_3)+P(A_1\cap A_2\cap A_3^c) = P(A_1)P(A_2) $$ Similarly we obtain:
$$P(A_1\cap A_2\cap A_3)+P(A_1\cap A_2^c\cap A_3) = P(A_1)P(A_3)\\ P(A_1\cap A_2\cap A_3)+P(A_1^c\cap A_2\cap A_3) = P(A_2)P(A_3) $$ We now could use additivity again and compute the probability of $$(A_1^c\cap A_2\cap A_3)\cup (A_1\cap A_2^c\cap A_3)\cup (A_1\cap A_2\cap A_3^c) $$ my set arithmetics is complete garbage, so I don't know if we even obtain what we need here. I haven't actually invoked $\mathcal B_{A}$ being an algebra anywhere, so I'm kind of stuck.

Are there any ideas, any hints on how to either proceed or rethink my strategy?


I think the definition of ${\mathcal B}_A$, the events that are independent to $A$, should be $$ {\mathcal B}_A:=\{ B\in {\mathcal F}: P(A\cap B)=P(A)P(B)\}.\tag{1} $$ If this is what is intended (and I may be wrong about that), then the assertion is straightforward to prove. We are given that $A_1, A_2, A_3$ are pairwise independent. In particular $A_1$ is independent of $A_2$, and $A_1$ is independent of $A_3$. By $(1)$, this means that both $A_2$ and $A_3$ are members of ${\mathcal B}_{A_1}$. But by assumption ${\mathcal B}_{A_1}$ is an algebra, which means $A_2\cap A_3$ is also a member of ${\mathcal B}_{A_1}$ and therefore by $(1)$ $$ P(A_1\cap (A_2\cap A_3))=P(A_1)P(A_2\cap A_3). $$ Finally $P(A_2\cap A_3)=P(A_2)P(A_3)$ follows from pairwise independence of $A_2$ and $A_3$.

  • $\begingroup$ I was overthinking. Your $\mathcal B_A$ makes much more sense. $\endgroup$ – Alvin Lepik Oct 11 '17 at 6:58

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