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I have a real-valued random variable $X$ that takes on positive and negative values, and

$$E(X)=0 \tag{A}$$

There is also another real-valued random variable $Y$, not independent from $X$, neither $X$-measurable. The underlying probability space is $(\Omega,\mathscr{F},P)$. I considered having

$$E(X\mid Y=y) \neq 0 ,\;\;\; \forall y \in S_Y \subseteq \mathbb R \tag{B}$$

alongside $E\big[E(X\mid Y)\big] = E(X)=0$, where $S_Y$ is the range of $Y$.

Initially I thought that the two conditions together could not hold, but that was a knee-jerk reaction, I could not prove anything to that effect, and eventually, it appears natural that they can hold together indeed.

Nevertheless, I still sense that there may be not-infrequent, and not "artificial" situations in which, in order for conditions $(A), (B)$ to hold together, some non-trivial additional conditions must also hold. I also conjecture that whether the variables are each discrete or continuous may make a difference.

EXAMPLES
A) Since we want $E(X\mid Y=y)$ to not equal zero for all values that $Y$ can take, it means that $0$ is not in its range. Moreover, since $E\big[E(X\mid Y)\big] =E(X) = 0$, it must take on positive and negative values. The two together imply that if the random variable $E(X\mid Y)$ has a probability density function, the latter will have a point of discontinuity at $0$.

B) If the bivariate joint distribution of $\{X,Y\}$ belongs to the elliptical class (or to the Pearson family I think), the conditional expectation $E(X\mid Y)$ will be a linear function of $Y-E(Y)$ (not affine, since $E(X) =0$), and then we would have to have that $E(Y)$ does not belong in the range of $Y$ in order to have $E(X\mid Y=y) \neq 0, \forall y \in S_Y$.

Are there any other situations of some generality, where additional conditions must hold, in order to have both properties $(A), (B)$?

Note: my motive is not a search for theoretical oddities. I have encountered the above situation in the context of my research, as a requirement for the fundamental validity/usefulness of an estimator.

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  • $\begingroup$ $E[X\mid Y]$ is a random variable $Z$ that is a function of the random variable $Y$. It just so happens that the expected value of $Z$ is equal to $E[X]$, the expected value of $X$. Now, we are told that $E[X]=0$ and so $Z$ is a zero-mean random variable. Why can't $Y$ be a random variable whose function $Z$ has expected value $0$? Even if $Y$ is nonnegative, its function $Z$ might still have mean $0$. $\endgroup$ – Dilip Sarwate Jun 3 '15 at 13:36
  • $\begingroup$ @DilipSarwate I don't disagree - and I already wrote that the two conditions together appear to be able to hold. I just added an example which is not too narrow, since it holds for the elliptical class of bivariate distributions. $\endgroup$ – Alecos Papadopoulos Jun 3 '15 at 14:04
  • $\begingroup$ Intuitively, I do not see any paradox in $\{ \ E(X)=0, \ \ E(X|Y) \neq 0, \ \ \forall Y=y \ \}$. Do you have an example where this combination is paradoxical or somehow counterintuitive? Edit: OK, I see the example at the end of the post. $\endgroup$ – Richard Hardy Jun 3 '15 at 14:11
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    $\begingroup$ Alecos: please, don't get pissed off. Try to make your formulation a little more precise. Let the underlying probability space be $(\Omega,\mathscr{F},P)$. The support of an r.v. $Y:\Omega\to\mathbb{R}$ is the smallest closed Borel set $S_Y$ such that $P(Y\in S_Y)=1$. The conditional expectation $\mathrm{E}(X\mid Y):\Omega\to\mathbb{R}$ is also an r.v. You may talk about $\mathrm{E}(X\mid Y)(\omega)$ for all $\omega$ (or for every $\omega$ outside a set of zero probability). But your "$\forall y\in S_Y$" makes no sense, because $S_Y$ is a subset of $\mathbb{R}$; it's not a subset of $\Omega$. $\endgroup$ – Zen Jun 3 '15 at 18:10
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    $\begingroup$ @Zen Or possibly the question is whether (assuming $E[X]=0$) it is possible that $E[X|Y=y]\ne 0$ for every $y\in\mathbb{R}$? Yes, it is. Alecos, this stark difference in solutions that is made possible by the ambiguity of your notation is why I am asking for clarification. It's not to give you a hard time, but intended only to create a question that has a reasonable chance of being understood in the same way by you, the people who answer, and all other readers. $\endgroup$ – whuber Jun 3 '15 at 19:22

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