# independence of random variables for finite subfamilies

Let $X$ be a random variable and $\{Y_j\}, j\in J$ a family of random variables. $J$ should be an index set, perhaps uncountable. My question is, if $X$ is independent to every finite subfamily of $\{Y_j\}$, i.e. for every $I \subset J$ and $|I|\in \mathbb{N}$ the family $\{Y_j;j\in I\}$ and $X$ are independent. Could we conclude that $X$ is independent to the whole family $\{Y_j; j\in J\}$?

cheers

math

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Yes, as $\mathcal C := \left\{\bigcap_{j \in I} Y_j^{-1}[B_j] \biggm| I \subseteq J\text{ finite}, B_j \subseteq \mathbb R\text{ Borel} \right\}$ is a $\cap$-stable generator of $\sigma(Y_j, j \in J)$ and for $A \in \sigma(X)$ and $C \in \mathcal C$ we have independence of $C$ and $A$.

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The $\cap$-stable (and non empty) classes are often called $\pi$-systems. – Did Jun 19 '12 at 15:42
Ah....Thanks a lot. I knew this result, but I didn't see that $\mathcal{C}$ is a generator of $\sigma (Y_j;j\in J)$ for probably uncountable $J$. – math Jun 19 '12 at 15:47
You see it now? – martini Jun 19 '12 at 15:50
@martini : sorry for this late answer. One inclusion is obvious, i.e. $\sigma (\mathcal{C})\supset \sigma (Y_j;j\in J)$. Can just give a short argument why the other direction is true? That would be very helpful. – math Jul 23 '12 at 13:11
@math As $\sigma(Y_j, j \in J)$ makes the $Y_j$ measurable, we have $C \in \sigma(Y_j, j \in J)$ for $C \in \mathcal C$. As $\sigma(Y_j, j \in J)$ is a $\sigma$-algebra, we have $\sigma(\mathcal C) \subseteq \sigma(Y_j, j \in J)$. – martini Jul 23 '12 at 13:54

Yes, we can. It all relies on the following result:

If $\mathcal{G}_1,\ldots,\mathcal{G}_n$ are systems of events that are closed under intersection such that $\mathcal{G}_1,\ldots,\mathcal{G}_n$ are independent, then $\sigma(\mathcal{G}_1),\ldots,\sigma(\mathcal{G}_n)$ are also independent.

Since $\sigma((Y_j)_{j\in J})=\sigma((Y_i)_{i\in I}\mid I\subseteq J, |I|\in \mathbb{N})$ the above result gives what you want.

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