$$E[x_i\mid y_i] = 0 \implies \operatorname{cov}(x_i, y_i) = 0\,?$$
I am wondering if the above statement holds true. The LHS is saying that given any value of $y_i$, the expected value of $x_i$ is zero. This seems to suggest that the 2 have zero covariance. Is this true? If so, then it seems to be true as long as $E[x_i\mid y_i] = \text{constant}$, where the constant doesn't necessarily have to be zero?
In addition, I believe $E[x_i\mid y_i] = \text{constant} \implies E[x_i] = \text{constant}$, which implies that $x_i$ is independent of $y_i$.
Let me state it in the form of a proposition: Let $(\Omega,\mathcal{F},P)$ be a probability space. Let $X,Y$ be square-integrable random variables. If $E[X\mid Y]=0$ (i.e., conditional expectation $E[X\mid\sigma(Y)]=0$), then $Cov(X,Y)=0$.
Proof: Let $\mu_{x}=E[X]$ and $\mu_{y}=E[Y]$. Let $\mathcal{M}=\sigma(Y)$. Observe that $\mathcal{M=}\sigma(Y-\mu_{y})$. Now \begin{eqnarray*} Cov(X,Y) & = & E\left[(X-\mu_{x})(Y-\mu_{y})\right]\\ & = & E\left[E\left[(X-\mu_{x})(Y-\mu_{y})\mid\mathcal{M}\right]\right]\\ & = & E\left[(Y-\mu_{y})E\left[(X-\mu_{x})\mid\mathcal{M}\right]\right]\\ & = & E[(Y-\mu_{y})(-\mu_{x})]\\ & = & 0. \end{eqnarray*}