Is the relation of having positive covariance well behaved with respect to taking the inverse?

Question

Let $X$ and $Y$ be two random variables, $X$ strictly positive. Assume that Cov$(X,Y)>0$. Does this imply that Cov$(1/X, Y)<0$?

I know that being positively correlated is not a transitive relation; is something similar happening here? Thanks.

Answer 1

The fact that $\mathrm{Cov}(X,Y)\gt0$ does not imply that $\mathrm{Cov}(1/X,Y)\lt0$.

For a simple counterexample, assume that $X=1+U$ and $Z=U(1-U)$ with $U$ uniform on $(0,1)$. Since $(U,Z)$ and $(1-U,Z)$ are identically distributed, $\mathrm{Cov}(U,Z)=\mathrm{Cov}(1-U,Z)$. The sum of these two covariances is $\mathrm{Cov}(1,Z)=0$ hence each of them is zero, that is, $\mathrm{Cov}(X,Z)=0$. On the other hand, an explicit computation shows that $\mathrm{Cov}(1/X,Z)\lt0$.

Let $Y=xX-Z$ for some small positive $x$, then $\mathrm{Cov}(X,Y)\gt0$ and $\mathrm{Cov}(1/X,Y)\gt0$.