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In general, statistical dependence is not transitive. If $Y$ and $X_{1}$ are dependent, and $Y$ and $X_{2}$ are dependent, then $X_{1}$ and $X_{2}$ are NOT necessarily dependent.

However, in some cases they CAN be dependent. A question which occurred to me earlier this week on which I have not been able to make some headway, is: are there any random variables $Y$ such that any variables dependent on $Y$ must be dependent on each other?

Related is the idea of decomposability (whether the random variable can be expressed as the sum of independent random variables), but I was unable to find anything on this more general problem.

The simplest case which occurs to me is a Bernoulli random variable, but I have neither been able to prove it nor offer a counterexample. If someone could shed light on the issue, it would be much appreciated.

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Here's a counterexample. Consider a probability space with $4$ outcomes, and the following probabilities:

$$ \matrix{ Y & X_1 & X_2 & \text{probability} \cr 0 & 0 & 0 & (1-s)^2 \cr 1 & 0 & 1 & s - s^2 \cr 1 & 1 & 0 & s - s^2 \cr 1 & 1 & 1 & s^2 \cr} $$ where $0 < s < 1$. $Y$ is Bernoulli with parameter $p = 1 - (1-s)^2$, which can be anything in $(0,1)$. $X_1$ and $X_2$ are independent Bernoulli random variables with parameter $s$. However, $X_i$ and $Y$ are dependent, since ${\mathbb P}(X_i = 0 | Y = 0) = 1 \ne {\mathbb P}(X_i = 0)$.

EDIT: This can be generalized. Let $Y$ be any random variable that is not a.s. constant. Take $A$ so that $0 < p = {\mathbb P}(Y \in A) < 1$, and take $s = 1 - \sqrt{1-p}$ so that $p = 1 - (1-s)^2$. Let $U$ be independent of $Y$ with ${\mathbb P}(U=1) = {\mathbb P}(U=2) = (s -s^2)/p$, ${\mathbb P}(U=3) = s^2/p$. Let $$(X_1,X_2) = \cases{(0,0) & if $Y \notin A$\cr (0,1) & if $Y \in A$ and $U=1$\cr (1,0) & if $Y \in A$ and $U= 2$\cr (1,1) & if $Y \in A$ and $U = 3$\cr }$$ Then $X_1$ and $X_2$ are independent Bernoulli with parameter $s$, but $X_i$ and $Y$ are dependent.

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  • $\begingroup$ Thank you very much for your prompt response and excellent counterexample! $\endgroup$ Feb 18, 2016 at 0:25

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