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What are some examples of random variables X, A, B such that X is independent to A, and to B, but not to A and B jointly, i.e., X is not independent to (A,B). In other words, $X \perp A$ and $X \perp B$ but not $X \perp A, B$

I got curious while reading http://en.wikipedia.org/wiki/Conditional_independence

It is enough to find examples such that $X \perp A$ and $X \perp B$ but not $X \perp A \mid B$.

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closed as too localized by Yemon Choi, Bill Johnson, Noah Stein, Mark Meckes, Michael Greinecker Sep 27 '12 at 14:49

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2 Answers 2

Let $A$ and $B$ be independent random variables with $P(A=1)=P(A=-1)=P(B=1)=P(B=-1)=1/2$. Let $X = AB$. Then any two of $A$, $B$ and $X$ are independent, but the three are not. It's easy to generalize this to generate a sequence of $n$ random variables, any $n-1$ of which are independent, but where they aren't all independent.

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Let $A$ and $B$ be indipendent and uniformly distributed in $S^1$ (for example you can take $\Omega = S^1 \times S^1$ with the uniform probability and $A$ and $B$ as the two projection); then it is clear that $X=A+B$ is indipendent of $A$, it is indipendent of $B$, but it isn't indipendent of $(A,B)$.

In this reasoning it's crucial that the uniform law on $S^1$ is traslation invariant.

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...where $S^1 = \mathbf{R}/\mathbf{Z}$ and not $\lbrace z\in\mathbf{C}: |z|=1\rbrace$. (In the latter case $X=A+B$ is not what you mean.) –  Sean Eberhard Sep 27 '12 at 15:22
    
Sure... By the way, also this method could be generalized to $n$ variables. It seems to be true also for countable many variables... Think of this example: take $\{ X_i \}_{i \in \mathbb{N}}$ i.i.d. variables which take values in $\mathbb{Z} / 2 \mathbb{Z}$ uniformly. Then consider $Y_n = \sum_{ i \leq n} X_i$ (where the sum is also in $\mathbb{Z} / 2 \mathbb{Z}$), and then take $Z$ as the limit of the $Y_i$s along some non-principal ultrafilter. Then $Z$ is indipendent of any $A$ proper subset of $ \{ X_i \}_{i \in \mathbb{N}}$ but clearly not indipendent from all of them. –  Simo_the_Wolf Sep 27 '12 at 18:17
    
I've always taken a set of rvs to be independent iff every finite subset is independent. Also, I think that by the Kolmogorov 0-1 law, $Z$ isn't going to be measurable (since if $Z$ is measurable, then by symmetry $P(Z=1) = P(Z=0) = 1/2$). –  Alexander Pruss Mar 19 '13 at 15:36

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