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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).

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1 Answer 1

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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.

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  • $\begingroup$ What is the ingredient we need to add to conclude (4.)? $\endgroup$
    – 0xbadf00d
    Nov 2, 2020 at 6:47
  • $\begingroup$ 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 $\mathcal F_2$ . $\endgroup$ Nov 2, 2020 at 6:51
  • $\begingroup$ 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 $i=2,\ldots,k$? $\endgroup$
    – 0xbadf00d
    Nov 2, 2020 at 7:07
  • $\begingroup$ 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? $\endgroup$
    – 0xbadf00d
    Nov 2, 2020 at 7:16
  • $\begingroup$ Please take a look at the specific problem I'm trying to solve: math.stackexchange.com/q/3890356/47771. $\endgroup$
    – 0xbadf00d
    Nov 2, 2020 at 13:54

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