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Suppose I have two real valued random variables $X$ and $Y$. Let $cov(\cdot)$ denote the covariance operator. What is $cov(XY,X)$? Is it zero if $cov(Y,X)=0$?

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No, a counterexample can be constructed if we choose $Y$ to be degenerate, that is $Y\equiv \alpha$ for some constant $0\neq \alpha\in \Bbb R$. In this case $cov(Y,X)=cov(\alpha,X)=0$, but $$cov(XY,X)=\alpha\cdot cov(X,X)=\alpha\cdot Var(X)\neq 0$$ if $X$ is a random variable such that $Var(X)\neq 0$.

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In general, not much can be said. Let $X,Y$ be two independent, identically distributed random variables. Then $\def\cov{\operatorname{cov}}\cov(X,Y) = 0$, and \begin{align*} \cov(XY,X) &= \def\E{\mathbf E}\E[X^2Y] - \E[X]^2\E[Y]\\ &= \E[X^2]\E[Y] - \E[X]^2\E[Y]\\ &= \mathrm{Var}(X)\E Y \end{align*} which is non-zero in general.

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  • $\begingroup$ No, in general $\def\E{\mathbf E}\E[X^2] - \E[X]^2 = {\operatorname{Var}}(X) \ne 0$. $\endgroup$ – martini Mar 22 '16 at 10:19
  • $\begingroup$ No, the second term is $$\def\E{\mathbf E}\E[XY]\E[X] = \E[X]\E[Y]\E[X] = \E[X]^2 \E[Y]$$ $\endgroup$ – martini Mar 22 '16 at 10:21

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