Expectation of truncated distribution Consider the random variables $X,Y$ and assume that
$$
E(X|Y)=0
$$
Does this imply that $$E(X|X\geq A,Y)\neq 0 ?$$
I think this holds for the truncated Normal, for example. But does it hold generically?

Using the law of total probability and $E_{p(x|x<a, y)} \leq a \leq E_{p(x|x\geq a, y)}$, we have
\begin{align*}
E_{p(x|y)}x
&= p(x\geq a) E_{p(x|x\geq a, y)}x + p(x<a) E_{p(x|x<a, y)}x
\\&\leq p(x\geq a) E_{p(x|x\geq a, y)}x + p(x<a) E_{p(x|x\geq a, y)}x
\\&= (p(x\geq a)+p(x<a)) E_{p(x|x\geq a, y)}x
\\&= E_{p(x|x\geq a, y)}x.
\end{align*}
The assumption $E_{p(x|y)}x = 0$ is not needed for this result, but it implies $E_{p(x|x\geq a, y)}x \geq 0$.
