Suppose both random variables $X$ and $Y$ are non-negative. In particular, $Y$ satisfies $Y\leq c$, a.s., where $c$ is some positive constant. Now I want to formally show that \begin{equation} E((X-EX)Y)=0. \end{equation} .
I tried the following argument Writing $X-EX=Z$, \begin{align} E(ZY)=E(Z1_{\{|Z|\geq 0\}}Y)+E(Z1_{\{|Z|<0\}}Y). \end{align} Then, since $(Z1_{\{|Z|\geq 0\}}Y)\leq cZ$, it follows that $E(Z1_{\{|Z|\geq 0\}}Y)\leq cEZ=0$ by the domination property of the expectation. Further, because $0\leq (Z1_{\{|Z|\geq 0\}}Y)$, we have $E(Z_{1\{|Z|\geq 0\}}Y)\geq0$. Similarly for the second term.
Any comments/suggestions? Thanks!
It's not necessarily true that $\mathbb{E}[(X-\mathbb{E}[X])Y]=0$, even under the stated hypotheses.
For instance, suppose that $X$ takes the values $0$ and $1$, each with probability $\frac{1}{2}$, and that $Y=X$. Then $$\mathbb{E}[(X-\mathbb{E}[X])Y]=\mathbb{E}[XY]-\frac{1}{2}\mathbb{E}[Y]=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\neq 0$$