Independence of collections of events

Question

Suppose we have three random variables $X_1,X_2,X_3$ on the same probability space such that the collections $\{X_1,X_2 \},\{X_2,X_3\}$ and $\{ X_3,X_1 \}$ are mutually independent.Note that this is different from saying that each collection is independent,i.e. they are not pairwise independent. So, can we say $\{X_1,X_2,X_3\}$ is independent?

I feel that they are not independent but I can't find a valid counter example. Any help is appreciated. Thanks!

Answer 1

The mutual independence of $\{X_1, X_2\}$, $\{X_2, X_3\}$, and $\{X_3, X_1\}$ refers to, by definition, the fact that $\sigma$-algebras $\sigma(X_1, X_2)$, $\sigma(X_2, X_3)$, and $\sigma(X_3,X_1)$ are mutually independent. Now it is easy to verify:

Lemma. If the $\sigma$-algebras $\mathcal{F}_1, \dots, \mathcal{F}_n$ are mutually independent and $\mathcal{G}_i\subseteq\mathcal{F}_i$ are $\sigma$-sugalgebras, then $\mathcal{G}_1, \dots, \mathcal{G}_n$ are also mutually independent.

From this, we know that $\sigma(X_1) \subseteq \sigma(X_1, X_2)$, $\sigma(X_2) \subseteq \sigma(X_2, X_3)$, and $\sigma(X_3) \subseteq \sigma(X_3,X_1)$ are mutually independent, which in turn implies the mutual independence of $X_1, X_2, X_3$.

Answer 2

It seems that maybe you are looking for the answer to a more elementary question: Pairwise independence of random variables in general does not imply joint independence. Consider the following counter example: Suppose two fair six sided dice are independently rolled, and let $X_1=1$, if the first die is 1, and zero otherwise, $X_2=1$ if the second die is 1, and zero otherwise, and $X_3=1$ if the sum of the rolls is 7, and zero otherwise. $X_1$ and $X_2$ are clearly independent. It is a little work, but a good exercise, to show $X_1$ and $X_3$ are independent, and similarly $X_2$ and $X_3$ are independent. However $P(X_1=1,X_2=1,X_3=1)=0 \ne P(X_1=1)P(X_2=1)P(X_3=1)>0$, so $X_1,X_2$ and $X_3$ are not mutually independent.