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Suppose that $X$ and $Y$ are positive and square-integrable random variables such that $X$ and $Y$ are positively correlated, i.e., $\mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y] \geq 0$. Let $f: \mathbb{R} \to \mathbb{R}_+$ be a positive, measurable, and increasing function.

My question is: are $f(X)$ and $f(Y)$ still positively correlated?

Thanks a lot.

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    $\begingroup$ Look up FKG inequalities. $\endgroup$ – Abdelmalek Abdesselam Jun 2 '17 at 15:33
  • $\begingroup$ I think FKG is related to inequalities with the same random variable but with different functions. $\endgroup$ – Richie Jun 2 '17 at 15:35
  • $\begingroup$ It's not because the statement of a theorem involves two functions $f,g$ that it does not hold if the two functions happen to be equal. I will give a more detailed answer in a sec. $\endgroup$ – Abdelmalek Abdesselam Jun 2 '17 at 15:54
  • $\begingroup$ Relevant (on Math.SE): math.stackexchange.com/questions/2306873/… $\endgroup$ – Clement C. Jun 5 '17 at 14:37
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Here is an easy counterexample. Let $(X,Y)$ be uniformly distributed among $(1,1); (1,9); (9,1); (9,9); (a,b)$. The correlation has the same sign as $(a-5)(b-5)$. So choose, e.g. $a=7, b=8$, and then choose $f$ which fixes $1,8,9$ but has $f(7)=4$.

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The result is true if the pair of random variables $(X,Y)$ is an FKG system: see Lecture 8 from the course given by David Brydges at the 2009 PIMS Probability Summer School. For example if the joint law has density $e^{V(x,y)}$ with $$ \forall x,y,\ \ \frac{\partial^2 V}{\partial x\partial y}(x,y)\ge 0 $$ then for all increasing functions of two variables $F,G$ one has $$ \mathbb{E} F(X,Y)G(X,Y)\ge \mathbb{E} F(X,Y)\ \mathbb{E} G(X,Y)\ . $$ All you need to do is take $$ F(x,y)=f(x) \ {\rm and}\ G(x,y)=f(y) $$ with your notations.

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  • $\begingroup$ I see. Thanks very much for your explanation and reference. $\endgroup$ – Richie Jun 2 '17 at 16:18
  • $\begingroup$ a downvote for my answer, what did I do wrong? $\endgroup$ – Abdelmalek Abdesselam Jun 5 '17 at 14:52

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