If $F_1$ and $F_2$ are both independent of $F_3$ and independent of each other, is $\sigma(F_1\cup F_2)$ independent of $F_3$?

Let $$(\Omega,\mathcal A,\operatorname P)$$ be a probability space and $$\mathcal F_i\subseteq\mathcal A$$.

Remember that $$\mathcal F_1$$ and $$\mathcal F_2$$ are called ($$\operatorname P$$)-independent if $$\operatorname P[A_1\cap A_2]=\operatorname P[A_1]\operatorname P[A_2]\;\;\;\text{for all }A_i\in\mathcal F_i\tag1.$$ If $$\mathcal F_2$$ is a $$\sigma$$-algebra, then $$(1)$$ is equivalent to $$\operatorname P\left[A_1\mid\mathcal F_2\right]=\operatorname P[A_1]\;\;\;\text{for all }A_1\in\mathcal F_1\tag2.$$

It's trivial to see that, if

1. $$\mathcal F_1$$ and $$\mathcal F_3$$ are independent; and
2. $$\mathcal F_2$$ and $$\mathcal F_3$$ are independent,

then 3. $$\mathcal F_1\cup\mathcal F_2$$ and $$\mathcal F_3$$ are independent.

On the other hand, (1.) and (2.) do not imply that

1. $$\sigma(\mathcal F_1\cup\mathcal F_2)$$ and $$\mathcal F_3$$ are independent.

Question: If we assume (1.), (2.) and additionally

1. $$\mathcal F_1$$ and $$\mathcal F_2$$ are independent,

can we then conclude (4.)? (Maybe, if necessary, assuming that $$\mathcal F_i$$ is a $$\sigma$$-algebra).

1 Answer

The answer is NO. There exist events $$A,B,C$$ such that any two of them are independent but they are not jointly independent. In this case $$C$$ is not indepdent of $$A \cap B$$ so we can take $$\mathcal F_1=\sigma (A),\mathcal F_2=\sigma (B), \mathcal F_3=\sigma (C)$$ for a counter-example.

In two independent tosses of a fair coin let $$A$$ be the event that the first toss results in Heads, $$B$$ the event that the second one results in Heads and $$C$$ the even that the outcomes are both heads or both tails. Then $$A,B,C$$ are pairwise independent but not jointly independent.

• What is the ingredient we need to add to conclude (4.)? Nov 2, 2020 at 6:47
• 4) is expressed by saying that $\mathcal F_3$ is jointly independent of $\mathcal F_1$ and $\mathcal F_2$. You cannot write a sufficient condiotion by separating $\mathcal F_1$ and $\mathcal F_2$ . Nov 2, 2020 at 6:51
• Couldn't we show that $\mathcal F_1,\ldots,\mathcal F_k$ are independent if and only if $\mathcal F_i$ is independent of $\mathcal F_1,\ldots,\mathcal F_{i-1}$ for all $i=2,\ldots,k$? Nov 2, 2020 at 7:07
• And similarly, if $(\mathcal F_1,\mathcal F_2,\mathcal F_3)$ is independent, then $(\mathcal F_1\cup\mathcal F_2,\mathcal F_3)$ should be independent as well. And the latter independence is equivalent to the independence of $(\sigma(\mathcal F_1\cup\mathcal F_2),\sigma(\mathcal F_3))$. Or am I missing something? Nov 2, 2020 at 7:16
• Please take a look at the specific problem I'm trying to solve: math.stackexchange.com/q/3890356/47771. Nov 2, 2020 at 13:54