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Suppose that we have 4 random variables. Assume them to be discrete without loss of generality. I want to check whether ($X_1$ conditionally independent of $X_2$ given $X_4$) and ($X_1$ conditionally independent of $X_3$ given $X_4$) imply that $X_1$ conditionally independent of the pair $X_2,X_3$ given $X_4$. Let's denote the pmf of an r.v. $X$ by $p(X)$ instead of $p_X(x)$ to simplify things here. By the two assumptions we have that $$p(X_1,X_2|X_4)=p(X_1|X_4)p(X_2|X_4)$$ $$p(X_1,X_3|X_4)=p(X_1|X_4)p(X_3|X_4)$$ I am trying to check if $p(X_1,X_2,X_3|X_4)=p(X_1|X_4)p(X_2,X_3|X_4)$ or equivalently $p(X_1|X_2,X_3,X_4)=p(X_1|X_4)$. Therefore \begin{eqnarray} p(X_1|X_4)p(X_2,X_3|X_4)&=&\frac{p(X_1,X_3|X_4)}{p(X_3|X_4)}\sum\limits_{X_1}p(X_1,X_2,X_3|X_4)\\ &=&\frac{p(X_1,X_3|X_4)}{p(X_3|X_4)}\sum\limits_{X_1}p(X_3|X_1,X_2,X_4)p(X_1,X_2|X_4)\\ &=&\frac{p(X_1,X_3|X_4)}{p(X_3|X_4)}p(X_2|X_4)\sum\limits_{X_1}p(X_3|X_1,X_2,X_4)p(X_1|X_4) \end{eqnarray} Any ideas?

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    $\begingroup$ If this was true conditionally for every $(X_1,X_2, X_3,X_4)$, it would also be true without any conditioning on $X_4$ (set $X_4$ to be $1$ with probability $1$). Now use any standard example of $3$ random variables that are pairwise independent but not mutually independent to get a contradiction (for instance example $1$ here ) $\endgroup$ – stochasticboy321 Apr 21 '17 at 20:07
  • $\begingroup$ Yeah, I've edited the comment for a simpler counterexample. Essentially, since everything is conditioned on $X_4$, and your hypothesis doesn't care about $X_4$ beyond the fact that conditional on it $X_1 \perp X_2$ and $X_1 \perp X_3$, if the statement is true, then it's also true when $X_4$ is degenerate, but it's well known that it's not true in that case. In case the prvious example is still visible to you, it's simply a condition-ified version of the example linked in the updated comment. $\endgroup$ – stochasticboy321 Apr 21 '17 at 20:37

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