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E.g. let us imagine we have $X_3 := X_1 \text{XOR} X_2$ with both $X_1, X_2$ being sampled from $\{0, 1\}$ with $p=0.5$. Then $X_1 \perp X_2, X_3$ but $X_1 \not \perp X_2 | X_3$.

Are there conditions that guarantee that pairwise independence implies conditional independence?

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One needs to be carefull when writing $X_1 \bot X_2, X_3$. In your example this means that $X_1 \bot X_2$ and $X_1 \bot X_3$, but it could also mean that $X_1 \bot (X_2, X_3)$ which is not the case here. You have that $X_1$ is independent of $X_2$ and is independent from $X_3$, but $X_1$ is not independant from the couple $(X_2, X_3)$.

If you had that $X_1 \bot (X_2, X_3)$, then you would have that conditionally on any $X_3$ value, $X_1$ and $X_2$ would be independant (this is quite easy to show).

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$$ P(X_1 \mid X_2) = P(X_1 \mid X_3) = P(X_1) $$ When does this imply $P(X_1) = P(X_1 \mid X_2, X_3$)?

One case is when $(X_1 \perp X_2) \mid X_3$ and $X_2 \perp X_3$ since then $$ P(X_1 \mid X_2, X_3) = \frac{P(X_1, X_2 \mid X_3)P(X_3)}{P(X_2, X_3)} = \frac{P(X_1)P(X_2)P(X_3)}{P(X_2) P(X_3)} = P(X_1) $$ Another case is the same as above but switch 2 and 3.

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  • $\begingroup$ Are these the only scenarios where this is possible? $\endgroup$ – gcsk Jun 10 '20 at 16:13
  • $\begingroup$ No, but it is perhaps the least constrained case. Another case, not really relevant for your situation, is when all three variables are independent. $\endgroup$ – Hunaphu Jun 11 '20 at 11:15

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