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please could you help me with this?

If a random variable X is independent of Y, that is E(X|Y) = 0 , does it also hold that X will be independent of any smooth function of Y? i.e. E(X|f(Y)) = 0 ? f(Y) being a polynomial, or exponential etc.

Thank you for your answer.

Martin
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    Is the function independent of $X$? E.g we do not have things like $f(Y) = Y+X$. – Sextus Empiricus Jan 10 '23 at 15:21
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    The part "that is $E(X|Y) = 0$" is confusing. It is not a definition of independence. Counterexample: Let $Y \sim N(0,1)$, $X \sim N(0,Y)$. Then $E(X|Y) = 0$, but $X$ is not independent from $Y$. – Sextus Empiricus Jan 10 '23 at 15:25
  • Indeed $E(X \mid Y)=0$ is not necessary either: $X,Y$ independently $\sim N(17,4)$ would have $E(X \mid Y)=17$ – Henry Jan 10 '23 at 17:12

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