If $E(Y)=0$, $Cov(Y, E(Y|X))\geq0$ for any random variable $X$

Question

Let $Y$ be a random variable with mean zero and finite variance, and let $X$ be any other random variable. Using the fact that $\mathbb{E}[Y]=0$ and the law of iterated expectation, we have $$\text{Cov}(Y, \mathbb{E}[Y|X])=\mathbb{E}[(Y-\mathbb{E}[Y])(\mathbb{E}[Y|X]-\mathbb{E}[\mathbb{E}[Y|X]])]=\mathbb{E}[Y\mathbb{E}[Y|X]].$$ I am tempted to say this is the expectation of $Y^2$, which would settle the question, but I don't know how to justify it (or even if it's true in the first place).

Answer 1

The lemma below comes handy:

Lemma. Assume $X$ and $Y$ are $L^1$-random variables on the probability space $(\Omega, \mathcal{F}, \mathbf{P})$. Let $\mathcal{G}$ be a sub-$\sigma$-algebra of $\mathcal{F}$. Then

$$ \mathbf{Cov}(\mathbf{E}[X \mid \mathcal{G}], Y) = \mathbf{Cov}(\mathbf{E}[X \mid \mathcal{G}], \mathbf{E}[Y \mid \mathcal{G}]). $$

Proof. We will make use of the law of total expectation and "pulling-out-what's-known" property. Also, for better readibility, we write $\color{navy}{\bar{X}} = \mathbf{E}[X \mid \mathcal{G}]$ and $\color{navy}{\bar{Y}} = \mathbf{E}[Y \mid \mathcal{G}]$. Then

\begin{align*} &\mathbf{Cov}(\color{navy}{\bar{X}}, Y) \\ &= \mathbf{E}[\color{navy}{\bar{X}}Y] - \mathbf{E}[\color{navy}{\bar{X}}]\mathbf{E}[Y] \\ &= \mathbf{E}[\mathbf{E}[\color{navy}{\bar{X}}Y \mid \mathcal{G}]] - \mathbf{E}[\color{navy}{\bar{X}}]\mathbf{E}[\mathbf{E}[Y \mid \mathcal{G}]] \tag{by LTE} \\ &= \mathbf{E}[\color{navy}{\bar{X}}\mathbf{E}[Y \mid \mathcal{G}]] - \mathbf{E}[\color{navy}{\bar{X}}]\mathbf{E}[\mathbf{E}[Y \mid \mathcal{G}]] \tag{pulling out} \\ &= \mathbf{E}[\color{navy}{\bar{X}}\color{navy}{\bar{Y}}] - \mathbf{E}[\color{navy}{\bar{X}}]\mathbf{E}[\color{navy}{\bar{Y}}] \\ &= \mathbf{Cov}(\color{navy}{\bar{X}}, \color{navy}{\bar{Y}}), \end{align*}

completing the proof. $\square$

Returning to OP's problem, we get

$$ \mathbf{Cov}(Y, \mathbf{E}[Y \mid X]) = \mathbf{Var}(\mathbf{E}[Y \mid X]) \geq 0. $$

Note that the assumption $\mathbf{E}[Y] = 0$ is redundant here. We only need to assume that $Y$ is $L^1$.

Answer 2

If $f(X)=E(YE(Y|X) $ then for any $A$ we have $$\int_Af(x)\mu(dx)=E(f(X)1_{X\in A})=E(Y^2)1_{X\in A})\geq 0$$ hence $f\geq 0 $ $\mu$ almost everywhere.