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Problem statement - Let $X$ be any random variable and $g(x)$ and $h(x)$ be any functions such that $E(g(X)), E(h(X))$ and $E(g(X)h(X))$ exist. If $g(x)$ is non-decreasing and $h(x)$ is non-increasing then prove that $E(g(X)h(X)) \le E(g(X))E(h(X))$.

I started from $E(g(X)h(X)) = \int_{-\infty}^\infty g(x)h(x)f(x) dx$

I have absolutely no idea how to proceed from here. I know that I need to use the fact that $g(x)$ is non-decreasing and $h(x)$ is non-increasing but don't know how to do it. Please tell me how to proceed.

Ankit Seth
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1 Answers1

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Let $Y$ be another random variable such rthat $X$ and $Y$ are independent and $Y$ has same distribution as $X$. If $X \geq Y$ then $(g(X)-g(Y))(h(Y)-h(X)) \geq 0$ (because both factors are non-negative). Check that the same inequality holds if $X \leq Y$. Hence $(g(X)-g(Y))(h(Y)-h(X)) \geq 0$ always and $E(g(X)-g(Y))(h(Y)-h(X)) \geq 0$. Expand this product and use the fact that $\{X,Y\}$ is i.i.d to complete the proof.

Kavi Rama Murthy
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