# Given two probability density functions find a number that satisfies a given equation

I have a problem for which I either need a proof or a counterexample.

We are given two discrete random variables $$x_1$$ and $$x_2$$ in $$[0, n]$$ where $$F_1(x)$$ is the probability of $$x_1\leq x$$, and similarly $$F_2(x)$$ is the probability of $$x_2 \leq x$$. My goal is to show that there always exists a number $$c > 0$$ that satisfies $$\int_{0}^{\infty}x{f_2(x)}(1-F_1(x))\leq c\int_{c}^{\infty}f_1(x)F_2(x),$$ where $$f_1(x)= \Pr[x_1=x]$$ and $$f_2(x)=\Pr[x_2=x]$$.

Thanks for any help.

However, there is no essential difference here between the absolutely continuous and discrete versions, and counterexamples for the two versions can be quite similar. For instance, suppose that $$x_1$$ and $$x_2$$ are random variables taking only nonnegative integral values such that $$$$F_1(x)=1-\frac1{\ln x}\quad\text{and}\quad f_2(x)=\frac C{x\ln^2 x}$$$$ for integers $$x\ge3$$ and $$F_1(x)=f_2(x)=0$$ for integers $$x<3$$, where $$C$$ is a positive real constant such that $$f_2$$ is a pmf. Then $$\begin{multline} \int_{0}^{\infty}x{f_2(x)}(1-F_1(x))=\sum_{0}^{\infty}x{f_2(x)}(1-F_1(x)) =\sum_{3}^{\infty}\frac C{\ln^3 x}=\infty \\ >c\ge c\int_{c}^{\infty}f_1(x) \ge c\int_{c}^{\infty}f_1(x)F_2(x) \end{multline}$$ for any real $$c>0$$, so that we have the opposite of the desired inequality.
The functions $$F_1(x) = 1-\frac{1}{(x+1)^3}$$, $$F_2(x) = 1-e^{-x}$$ (so that $$f_1(x) = \frac3{(x+1)^4}$$, $$f_2(x) = e^{-x}$$) seem to provide a counterexample.