Expectation of muliplication E(xf(x)) vs E(x)E(f(x)

Question

Assume $X$ is a random variable with realization always positive. $f(x)$ is increasing in $x$ and also positive.

Is there any proposition regarding $E(X\cdot f(X))$ and $E(X)E(f(X))$?

My naive intuition is that $X$ and $f(X)$ are two positively correlated random variables so $cov(X,f(X))$ should be positive. Then $E(Xf(X))>E(X)E(f(X))$.

But it falls into a loop, since $$ cov(X,Y)=E(XY)-E(X)E(Y) $$ is the definition of covariance...

Any clue or counter-example? Really appreciate!

Answer 1

Saying that $f$ is nondecreasing is equivalent to saying that $$ (f(x)-f(x'))(x-x')\ge0 $$ for all $x,x'\in\mathbb R$. Consider then a random variable $X'$ independent of $X$ but with the same distribution. Then $$ \mathbb E[(f(X)-f(X'))(X-X')]\ge0, $$ that is $$ \mathbb E[f(X)X]\ge\mathbb E[f(X)]\mathbb E[X]. $$

Notes:

• The fact that $X>0$ was not used. It is valid regardless of its sign.
• We even have the strict inequality $\mathbb E[f(X)X]>\mathbb E[f(X)]\mathbb E[X]$ if $f$ is strictly increasing and $X$ is not almost surely constant. To see that, just notice that my first inequality is then strict as soon as $x\neq x'$.