Is "Covariance is 0" equivalent to "Conditional Expectation equals Unconditional Expectation"?

Question

In formulas: \begin{align} Cov(X,Y)=0 \quad \Leftrightarrow\quad \mathbb{E}[Y|X]=\mathbb{E}[Y] \end{align} is this true?

Answer 1

Only "$\Leftarrow$" is true:

\begin{align} Cov(X,Y)&=\mathbb{E}[(X-\mathbb{E}[X])(Y-\mathbb{E}[Y])] \\ &=\mathbb{E}[XY]-\mathbb{E}[X]\mathbb{E}[Y]\\ &=\mathbb{E}[\mathbb{E}[XY|X]]-\mathbb{E}[X]\mathbb{E}[Y]\\ &=\mathbb{E}[X\mathbb{E}[Y|X]]-\mathbb{E}[X]\mathbb{E}[Y]\\ &=\mathbb{E}[X\mathbb{E}[Y]]-\mathbb{E}[X]\mathbb{E}[Y]\\ &=0 \end{align}

While "$\Rightarrow$" is false in general:

Be $X\sim N(0,1)$ and $Y:=|X|$ then $\mathbb{E}[X|Y]=0$ which implies that $Cov(X,Y)=0$ as shown above. But $\mathbb{E}[Y|X]=Y$