I am looking for a nice counterexample of the statement: if $\mathbb{E}(X | Y, Z) = f(Y)$ for a measurable function $f$ implies that $X$ is independent of $Z$ given $Y$, i.e. $X \perp Z | Y$.
Any help is appreciated.
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asked Jun 25 at 15:46
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1 Answer
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Let $Z$ and $T$ be independent uniform on $\{-1, 1\}$, $Y = 0$ and $X = T(Z + 1)$.
Then $\mathbb E(X | Y, Z = 1) = \mathbb E(2T | Y, Z = 1) = 0$ and $\mathbb E(X | Y, Z = -1) = \mathbb E(0 | Y, Z = 1) = 0$, but $P(X = 2 | Z = 1, Y = 0) = \frac{1}{2} \neq 0 = P(X = 2 | Z = -1, Y = 0)$.
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2$\begingroup$ I don't follow $\mathbb{E}(X|Y, Z) = 0$ -- I calculate that it is uniform on $\{-1.5, 1.5\}$. $\endgroup$– hunterCommented Jun 25 at 16:17
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1$\begingroup$ I agree with @hunter, this conditional expectation is $\mathbb{E}[X \mid Y,Z] = \frac{3}{2} Z$ $\endgroup$ Commented Jun 25 at 16:23
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$\begingroup$ Yes, sorry, we need zero expectation from component independent of $Z$, not from $Z$ itself. Updated. $\endgroup$– mihaildCommented Jun 25 at 16:31
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$\begingroup$ @mihaild Thank you very much, this looks correct to me. To hunter? $\endgroup$ Commented Jun 25 at 16:42
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