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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.

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to motivate the further the question; X and 1/X are always negatively correlated, so the intuitive answer would be that if X and Y are positvely correlated, Y and 1/X are negatively correlated. But this "transitive" line of reasoning spectacularily fails for example when checking the positive correlation property across three variables. – Mr_3_7 Apr 15 '13 at 13:08
up vote 1 down vote accepted

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 $\mathrm{Cov}(1/X,Y)\gt0$.

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