# Covariance sign

Hi,

Is it true that $Cov[f(X),g(X)]>0$ where $X$ is a random variable of unbounded support and $f,g$ are two strictly increasing real functions? I think by Chebyschev integral inequality I must assume that the random variable support must be bounded, otherwise I can not assure the sign of this covariance. Am I right?

Thanks

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There is a short proof based on coupling. Let $Y$ be a r.v. with the same distribution as $X$ and independent of $X$. Then $f(X)-f(Y)$ and $g(X)-g(Y)$ are centered r.v.'s.

On the one hand, $Cov(f(X)-f(Y), g(X)-g(Y))= \mathsf{E} (f(X)-f(Y)) (g(X)-g(Y))\ge 0$. (actually $>0$ unless $g(X)$ or $f(X)$ is constant a.s.)

On the other hand, $Cov(f(X)-f(Y), g(X)-g(Y))=$

$= Cov(f(X),g(X))-Cov(f(X),g(Y))-Cov(f(Y),g(X))+Cov(f(Y),g(Y))=2Cov(f(X),g(X))$

Combining these two lines we get $2Cov(f(X),g(X))>0$.

A couple of relevant remarks: 1.This statement is also due to Chebyshev. 2. The property can be formulated as "One r.v. forms an associated family". In general, r.v.'s $X_1,\ldots, X_n$ are called associated if for any two bounded coordinatewise nondecreasing functions $f$ and $g$, $Cov(f(X_1,\ldots,X_n),g(X_1,\ldots,X_n))\ge 0$. There is a recent book by Bulinskiy and Shashkin on association.

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This, and much more, is explained in Lectures on the Coupling Method by Torgny Lindvall store.doverpublications.com/0486421457.html. – Did Jun 3 '11 at 17:42

I think you are looking for this inequality: http://en.wikipedia.org/wiki/FKG_inequality

However, the strict hypothesis requires that the random variable $X$ lives in a finite set, which doesn't seem to be what you are looking for. That said, I don't see why this can't be generalized to the case when the ambient space is infinite as long as appropriate moment convergence conditions are applied.

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Here is a simple proof I think. Let $\mu_f=E[f(X)]$ and $\mu_g = E[g(X)]$. Then, by expanding the interior of the expectation, it is easy to see for any $c \in \mathbb{R}$ that $$Cov(f(X),g(X)) = E[(f(X)-\mu_f)(g(X)-\mu_g)]=E[(f(X)-c)(g(X)-\mu_g)].$$ Now, let $x_0$ be the infimum of all $x$ such that $g(x) \geq \mu_g$ and choose $c=f(x_0)$. Therefore, $(f(x)-c)(g(x)-\mu_g) \geq 0$ for all $x$ and so the covariance is non-negative.

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