Consider 3 random variables $X$, $Y$, $Z$. Suppose:
- $E(X)=0$
- $E(X|Y) =0$
- $Z\perp Y $
Does this imply $E(X|Y,Z) = E(X|Z)$?
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$\begingroup$ No it does not. What have you tried? $\endgroup$– HenrySep 20 at 8:49
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$\begingroup$ I've tried to show that $p(X|Y,Z) = p(X|Z)$ given the assumptions, and playing with the conditional probability/Bayes rule. Could not make it work, but my inability to prove it does not imply it does not hold :) $\endgroup$– NidjsiSep 20 at 8:57
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$\begingroup$ Though failure to prove may suggest looking for a counter-example could be worthwhile $\endgroup$– HenrySep 20 at 9:09
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1 Answer
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If you have
X Y Z prob
0 0 0 1/4
0 0 1 1/4
+1 1 0 1/4
-1 1 1 1/4
then the conditions are satisfied
while $E[X \mid Y=0, Z=0] = 0$ while $E[X \mid Z=0]=\frac12$
