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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

2 Answers 2

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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.2012-06-19
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    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 $J$.2012-06-19
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    You see it now?2012-06-19
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    @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.2012-07-23
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    @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 $\sigma(\mathcal C) \subseteq \sigma(Y_j, j \in J)$.2012-07-23
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    @martini Sorry for bothering you, but why does it follows from the fact, that $\sigma (Y_j;j\in J)$ makes the $Y_j$ measurable, we have $C\in\sigma (Y_j;j\in J)$ ? Since a typical set $C\in \mathcal{C}$ involves generally different $Y_j$'s. Sry, I guess this is a stupid question, but I do not see right now this implication2012-07-23
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    @math Let $C \in \mathcal C$, then there are a finite $I \subseteq J$ and $B_j \subseteq \mathbb R$ Borel with $\bigcap_{j\in I} Y_j^{-1}[B_j] = C$. For $j \in I$: As $Y_j$ is $\sigma(Y_j, j \in J)$-$\mathrm{Bor}(\mathbb R)$-measurable, we have $Y_j^{-1}[B_j] \in \sigma(Y_j, j\in J)$ for each $j \in I$. As a $\sigma$-algebra is closed under finite intersections, $C\in \sigma(Y_j, j \in J)$.2012-07-23
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    @martini Thx a lot!2012-07-23
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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.