Suppose $X$ and $Z$ are random variables. Can the covariance of $X^2$ with $Z$ be expressed in terms of the means, population variances, and covariance of $X$ and $Z$ alone?


My attempts at solving this problem end in something of a recursive loop: the term I'm hoping to simplify, $cov(X^2,Z)$, comes back:


$$
cov(X^2,Z)=E[(X^2-E(X^2))(Z-E(Z))]
\\
=E(X^2Z)-E(X^2)E(Z)
\\
=\overline{X^2}\overline{Z}+cov(X^2,Z)-(\overline{X}^2+variance(X))\overline{Z}
$$


Is there a way out of this? Thanks!



Express $cov(X^2,Z)$ in terms of means, variances, and covariance of $X$ and $Z$?
You can simplify anything involving covariances, variances, and means to an expression with only means. But using $cov(XZ),var(X),var(Z),E(X),E(Z)$ you can only get quadratic or smaller terms inside expected values, the general thing you can calculate is functions of $E(XZ), E(X^2), E(Z^2), E(X), E(Z), 1$. In other words, you can't get $E(X^2Z)$ which is necessary for $cov(X^2Z)$.


As another illustration, I bet you can come up with discrete random variables $X$ and $Z$ which have given mean, variance, and covariance, but $cov(X^2Z)$ is arbitrary.


