# Is this statement true: $\mathsf{Cov}(X,Y)\geq 0$, $\mathsf{P}(Y>0)=1$, show that $\mathsf{Cov}\Big(X,\dfrac{1}{Y}\Big)\leq 0$

I was trying to prove a problem and I got stuck at a point. The problem leads to a point where I have to show:

Let, $X$ and $Y$ are two r.v. If $\mathsf{Cov}(X,Y)\geq 0$ and $\mathsf{P}(Y>0)=1$, then show that $\mathsf{Cov}\Big(X,\dfrac{1}{Y}\Big)\leq 0$.

But I could not find a way to go from $Y$ to $\dfrac{1}{Y}$, also I have a feeling that there may be a counterexample of the problem. Any help would be appreciated.

Thanks to Einar Rødland, but in the problem I did not have $(X,Y)$ and $(X, 1/Y)$ same in distribution.

The problem was to show if $\mathsf{E}\Big(\dfrac{Z_1-Z_2}{Z_1+Z_2}\Big)\geq 0$, then $\mathsf{E}(Z_1-Z_2)\geq 0$, where $Z_1$ and $Z_2$ are different positive r.v. and $\mathsf{Cov}(Z_1+Z_2,Z_1-Z_2)\geq 0$

• There is some confusion as to what your problem really is. The two versions don't seem to match. Could you clarify which version you actually intended? – Vim Aug 31 '17 at 12:27
• Also, by Einar's example, the statement in your question's title is simply wrong. You would need to impose extra conditions to make it correct. Einar's example is totally valid, because it doesn't violate any of the conditions you give. – Vim Aug 31 '17 at 12:33
• @Vim why do not match? If $\mathsf{Cov}(1/(Z_1+Z_2),Z_1-Z_2)\leq 0$, then I can have $E(Z_1-Z_2)E(1/(Z_1+Z_2))\geq \mathsf{E}\Big(\dfrac{Z_1-Z_2}{Z_1+Z_2}\Big)\geq 0$. Now $E(1/(Z_1+Z_2))>0$, then $\mathsf{E}(Z_1-Z_2)\geq 0$. Where am I wrong? – Stat_prob_001 Aug 31 '17 at 12:38
• @Vim I know his counterexample is right. I did not add at first the second part. That is why I mentioned him. – Stat_prob_001 Aug 31 '17 at 12:41
• @Stat_prob_001: The new question in the text and the one in the comment above are different: eg, the covariances are not the same. Which is the correct one? And it's easier if you formulate the question first stating the assumptions clearly, and then the (assumed) consequence, rather than the result in the middle with assumptions before and after, as it can become unclear which are assumtions and which are consequences. – Einar Rødland Aug 31 '17 at 15:21

Let $\Pr(X=-1, Y=1)=1/2$ and $\Pr(X=1, Y=2)=\Pr(X=1, Y=1/2)=1/4$.
Note that $(X,Y)$ and $(X,1/Y)$ have the same distribution, both with $\text{E}[X]=0$, which made the example a little easier to make and explain.
That makes $\text{Cov}[X,Y]=\text{Cov}[X,1/Y]=1/8$.
• I think $cov[X,Y]= 13/8$ since $cov[X,Y]=E[XY]$ in this case. – user64066 Aug 31 '17 at 11:50
• can you give an example where $(X,Y)\neq (X,1/Y)$ in distribution? – Stat_prob_001 Aug 31 '17 at 11:52
• @Stat_prob_001: Just modify the probabilities slightly, and the inequalities will still hold but the two distributions no longer be identical: eg, use probabilities $1/4+a$ and $1/4-a$ for $(X,Y)=(1,2), (1,1/2)$. – Einar Rødland Aug 31 '17 at 12:44
• @user64066: Not sure how you got $13/8$. I got $\text{E}[XY]=-1/2+1/2+1/8=1/8$ adding the three cases. – Einar Rødland Aug 31 '17 at 12:46