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I want to prove that if $X$ is a random variable, then $$X=\int_0^\infty\textbf{1}_{\{X>x\}}dx-\int_0^\infty\textbf{1}_{\{-X>x\}}dx$$ This answer and this answer use this result and I don't understand why can we write $X$ this way. Can someone tell me where to find proof of this? Or help me with it?

kubo
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    By the way, the fact that $X$ is a random variable is irrelevant here. The same fact holds for any real number $X$. – James Martin Apr 11 '21 at 13:55
  • @JamesMartin a real number is still a random variable, but a constant one – kubo Apr 11 '21 at 14:23
  • Sure. I am just pointing out that the property you are asking about doesn't rely on anything about random variables or probability in any way. It's true for any number. – James Martin Apr 11 '21 at 14:30

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Hint: if $X\geq 0$ then $\int_0^\infty\textbf{1}_{\{-X>x\}}dx=0$ and

$$ \int_0^\infty\textbf{1}_{\{X>x\}}dx = \int_0^{X}\,dx =X $$ if $X<0$ then $\int_0^\infty\textbf{1}_{\{X>x\}}dx=0$ and $$ \int_0^\infty\textbf{1}_{\{-X>x\}}dx = \int_0^{-X}dx =-X. $$

Suman Chakraborty
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  • Thank you for you answer. Could you help me write $|X|$ in this format, please? – kubo Apr 11 '21 at 12:35
  • $|X|$ is always non-negative, so you can just use the first equation. – Suman Chakraborty Apr 11 '21 at 12:37
  • I know, but I would like to express the indicator function in terms of $X$ instead of $|X|$. I guess I would have to split the integral in two, since here I don't know if $X$ is non-negative. But I'm not sure how to do it. I think it would be: $$|X|=X_+ + X_-=\int_0^\infty \textbf{1}{{X>x}}+\int_0^\infty \textbf{1}{{-X>x}}$$ And also: $$X-c= \int_{-c}^\infty \textbf{1}{{X>x}}-\int{-c}^\infty \textbf{1}_{{-X>x}}$$ Is this right? – kubo Apr 11 '21 at 12:44
  • The first one is certainly right but the second one does not seem correct (think about the ranges of these integral a little bit more). – Suman Chakraborty Apr 11 '21 at 12:51
  • It should be this right?: $$X-c=\int_0^\infty \textbf{1}{{X-c>x}}-\int_0^\infty \textbf{1}{{-X+c>x}}=\int_0^\infty \textbf{1}{{X>x+c}}-\int_0^\infty \textbf{1}{{-X>x-c}}=\int_c^\infty \textbf{1}{{X>x}}-\int_c^\infty \textbf{1}{{-X>x}}$$ – kubo Apr 11 '21 at 12:57
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    The last integral should have a range $-c$ to $\infty$. – Suman Chakraborty Apr 11 '21 at 13:02