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Let $X$ be a random variable, and let $X_1:=g_1(X)$ and $X_2:=g_2(X)$. Does it hold that $X\perp \!\!\! \perp X_1 | (X_1, X_2)$? (This statement is made in the proof of Proposition 1 in the appendix of Tschannen et al. https://openreview.net/pdf?id=rkxoh24FPH)

Ričards Marcinkevičs
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  • Which statement in the appendix are you interpreting as $X\perp !!! \perp X_1 | (X_1, X_2)$? I don't see that there. – joriki Jan 30 '20 at 13:39
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    At the spot where "The second step is to observe that $X\rightarrow (X_1, X_2)\rightarrow X_1$". This is a Markov chain for which $X\perp !!! \perp X_1 | (X_1, X_2)$ holds. – Ričards Marcinkevičs Jan 30 '20 at 13:48
  • I see. Isn't $X_1$ independent of everything given $X_1$ (and thus also given $(X_1,X_2)$), because it's constant? (see e.g. https://math.stackexchange.com/questions/1585379) -- or am I misunderstanding something? – joriki Jan 30 '20 at 14:01
  • That is exactly what I was wondering about, thanks – Ričards Marcinkevičs Jan 30 '20 at 14:03

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As already discussed in the comment, a constant random variable is independent of all random variables (see e.g. Independence between a constant random variable and another random variable.), and $X_1\mid(X_1,X_2)$ is constant.

joriki
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