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Consider the random variables $Y,X$. Can we always write $$ Y=E(Y|X)+\epsilon\tag 1 $$ with $\epsilon$ independent of $X$?

Note: from this answer here, we know that we can always write (1) with $E(\epsilon|X)=0$. However, here I am asking if we strengthen the relationship between $\epsilon $ and $X$ to stochastic independence.

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Suppose $$ Y=X^2+u $$ where $u|X\sim(0,X^2)$ has conditional heteroskedasticity. Then, $$ \epsilon=Y-E(Y|X)=X^2+u-E(Y|X)=u, $$ which has conditional mean zero but is not independent of $X$, as its second moment depends on $X$.

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    $\begingroup$ I'm guessing there is an N missing in $u|X\sim(0,X^2)$ $\endgroup$
    – Stef
    Commented Dec 21, 2022 at 11:14
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    $\begingroup$ Not necessarily, the notation is to indicate that $u$ is, conditionally on $X$, distributed with mean 0 and variance $X^2$. It does not have to be a normal distribution. $\endgroup$
    – Christoph Hanck
    Commented Dec 22, 2022 at 7:45

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