Done. But I find it surprising that tex has \cot but not \cov which is far more useful. Sometimes I drop \operatorname just to convey my annoyance! — DirkGently Oct 4, 2015 at 5:01

Yes for the word "symmetric", i.e. $A$ is a symmetric martix you can say $\cov(X_i, X_j) = \cov(X_j, X_i)$. — Supriyo Oct 13, 2013 at 11:49

$\newcommand{\cov}{\operatorname{cov}}$ $\newcommand{\E}{\mathbb E}$ Possibly the identity $\E(XY) = \E(\cov(X,Y\mid Z)) + \cov(\E(X\mid Z),\E(Y\mid Z))$ would be useful here. This holds without any assumptions about whether $\E(X)$ or $\E(Y)$ is $0$. ${}\qquad{}$ — Michael Hardy Nov 21, 2013 at 3:50

$\newcommand{\E}{\operatorname{E}} \newcommand{\cov}{\operatorname{cov}} \newcommand{\var}{\operatorname{var}}$ Let's look at their conditional distributions given $X$: \begin{align} & \E(X\mid X)\E(XY^2\mid X) - \E(XY\mid X)\E(XY\mid X) \\ = {} & X^2 \Big( \E(Y^2\mid X) - \left( \E(Y\mid X) \right)^2 \Big) \\ = {} & X^2 \Big( \var(Y\mid X) \Big) \ge 0. \end{align} Hence with probability $1$, we have $$ \E(X\mid X)\E(XY^2\mid X) \ge \E(XY\mid X)\E(XY\mid X). $$ More later, maybe$\,\ldots\qquad$ — Michael Hardy Jun 20, 2016 at 22:05

If they were $\DeclareMathOperator{\cov}{cov}\cov(Y,W)=\cov(X_n,X_n)=\DeclareMathOperator{\var}{var}\var(X_n)=\sigma^2$, because $\cov(X_i,X_j)=0$. — Patricio Aug 20, 2021 at 9:08