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I am trying to see whether the conditional expectaction of E(X|Y) is the same as conditional expectaction of E(X|f(Y)), where f(Y) is any function of Y which is smooth, e.g. polynomial, sin etc.

Martin
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    What do you mean by "is the same"? They would be the same functions only for $f(Y) = Y$ if you mean literally the same. – Tim Jan 10 '23 at 14:52
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    If $f$ is a 1-1 function then it could be reasonable to think $E[X \mid Y = y]=E[X \mid f(Y)=f(y)]$ but this is not sane same thing as $E[X \mid Y ]=E[X \mid f(Y)]$ when $f(Y)\not = Y$ – Henry Jan 10 '23 at 15:01
  • To specify, I am looking for independence. If E(X|Y) = 0 (X is independent of Y), will it also hold that E(X|f(Y)) = 0 ? That is, X is mean independent of f(Y)? – Martin Jan 10 '23 at 15:03
  • @Martin your question in the comment is a completely different one, as in the comment you seem to be asking if $X$ is independent of $Y$, would it also be independent of $f(Y)$. If that's the case, you should probably ask this as a separate question. – Tim Jan 10 '23 at 15:09
  • @Tim, I am sorry for this. I have asked the comment question as a separate one. Could you please answer that one too? – Martin Jan 10 '23 at 15:17
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    This question is equivalent to asking whether the sigma algebra generated by $f(Y)$ is the same as that of $Y.$ As soon as you start contemplating the simplest cases, such as a constant function (which is polynomial and infinitely smooth), the answer clearly is negative. – whuber Jan 10 '23 at 15:32

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It's not. Take a trivial example, where $Y$ is a random variable that takes the $\{-1,0,1\}$ values, and $X = Y + \varepsilon$. You would see different results if you look at $E[X|Y]$ vs $E[X|f(Y)]$ where $f(y) = y^2$, because in the second case, the right-hand side is equal to $1$ maps to "two times more" $X$'s: the cases where for the raw data $Y=1$ and $Y=-1$.

Tim
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