Suppose I have $X,Y$ as bounded, nonnegative discrete random variables which have nonnegative covariance, $\text{Cov}(X,Y)\ge0$. Must it be the case that $X,Y^k$ for $k\ge 1$ have nonnegative covariance, $\text{Cov}(X,Y^k)\ge0$? As a throw in, we can say $X,Y\ge 1$.
This seems false to me, and I'm hoping someone has a ready example (or proof to the contrary.)
The assertion is in general wrong:
Suppose that $X$ takes each of the values $1,2,3$ with equal probability and define $Y=-(X-1)(X-3)=-X^2+4X-3$.
Then, by construction, $X$ and $Y$ are nonnegative bounded and discrete random variables with $cov(X,Y) = 0$ but $cov(X^2,Y)<0$ (should be $-2/9$?!). You can obviously add any constant to $Y$ to make its value at least $1$ without changing the covariances.
Note that "discrete" is not necessary here, as one should get a similar result when using the uniform distribution in the continuous case.