Is $Y - E[Y|X]$ independent from $X$? For any two random variables $X$ and $Y$, is $Y - E[Y|X]$ independent from $X$? Intuitively I think it should be the case, since conditioning should include all the aspects of $X$ which are tangled with $Y$.
 A: Those are always uncorrelated, indeed
$$ Cov(X,Y - \mathbb E[Y|X]) = Cov(X,Y) - Cov(X,\mathbb E[Y|X]) $$
Rewritting as expectations and using tower property and $X$ measurability of $X$, we get $$Cov(X,Y - \mathbb E[Y|X])= \mathbb E[XY] - \mathbb E[X]\mathbb E[Y] - \mathbb E[X\mathbb E[Y|X]] + \mathbb E[X]\mathbb E[\mathbb E[Y|X]]=0$$
But taking $X \sim \mathcal N(0,1)$, $\varepsilon = \pm 1$ with probability $\frac{1}{2}$ independent of $X$ and $Y=\varepsilon X$, you'll get $\mathbb E[Y|X] = 0$ almost surely, but $X$ and $Y$ are not independent (consider $X \in (0,1) , Y \in (1,2)$)
A: Suppose $Y \sim U(-1,1)$ and $X=Y^2$ then $E[Y \mid X] \equiv 0$, but $Y$ and $X$ are not themselves independent.
What is true is that $X$ and $Y-E[Y \mid X]$ are uncorrelated.
By the way, one can cook up a large family of examples just like this by considering any strictly increasing function $f : [0,\infty) \to \mathbb{R}$ (which are automatically Borel measurable), essentially any $Y$ with symmetric distribution about $0$, and $X=f(|Y|)$. (I use the word "essentially" because one must of course assume that $E[Y]$ exists.)
