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Consider 3 random variables $X$, $Y$, $Z$. Suppose:

  1. $E(X)=0$
  2. $E(X|Y) =0$
  3. $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? $\endgroup$
    – Henry
    Commented Sep 20, 2023 at 8:49
  • $\begingroup$ 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 :) $\endgroup$
    – Nidjsi
    Commented Sep 20, 2023 at 8:57
  • $\begingroup$ Though failure to prove may suggest looking for a counter-example could be worthwhile $\endgroup$
    – Henry
    Commented Sep 20, 2023 at 9:09

1 Answer 1

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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 $E[X \mid Z=0]=\frac12$

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  • $\begingroup$ Beautifully simple. Thanks $\endgroup$
    – Nidjsi
    Commented Sep 20, 2023 at 12:57

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