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Heading ##I am wondering whether these two properties are equivalent:

  • $X$ is conditionally independent of $Y$ given $Z$

  • $X$ is conditionally independent of $Y$ given $a^T Z$, $\forall a \in R^p$

Thank you.

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  • $\begingroup$ Just my guess: This is equivalently asking the sigma algebra generated by them is identical or not. Since both $a$ and $Z$ are $p$-dimensional, you may pick a group of $\{a_1, a_2, \ldots, a_p\}$ which are linearly independent, then the set $\{a_1^TZ = z_1, a_2^TZ = z_2, \ldots a_3^TZ = z_3\}$ will uniquely gives the solution of the form $Z = z$, so they should be equivalent. $\endgroup$ – BGM Nov 21 '18 at 9:55
  • $\begingroup$ I also have thought of this, but I cannot proof the result strictly.Thank you. $\endgroup$ – Aaron Nov 22 '18 at 15:36
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Actually they are not eqivalent.

Suppose $X = Z_1+Z_2+\epsilon_1,Y = Z_1+Z_2+\epsilon_2, Z=(Z_1,Z_2)^T,\epsilon_1 \bot Z,\epsilon_2 \bot Z$. It's obvious that $X \bot Y |Z$. But if we take $a=(1,0)^T$ , $X \not \bot Y|a^TZ$. So maybe the first property is a lot weaker than the second property.

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