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I’ve encountered following problem: Let $X$ be a positive random variable with distribution $P^X$, cumulative distribution function $F_X$ and quantile function $q_X$. Show that $$E(X)=\int_0^\infty1-F_X(x)\,dx$$ I’ve gotten so far: $$\int_0^\infty1-F_X(x)\,dx=\int_0^\infty P^X((x,\infty[)\,dx=\int_0^\infty\int1_{(x,\infty)}(y)\,dP^X(y)\,dx\overset{(*)}{=}\int\int_0^\infty1_{(x,\infty)}(y)\,dx\,dP^X(y)$$ where $(*)$ follows from Tonelli’s theorem. Now, I have already proven that $$E(X)=\int_0^1q_X(u)\,du$$ and my best guess is that the idea is to somehow connect these two results by showing $$\int\int_0^\infty1_{(x,\infty)}(y)\,dx\,dP^X(y)=\int_0^1q_X(u)\,du$$ but I’ve got literally no idea about how to do this. I’d be very grateful for any tips or hints.

Papillus
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    $$E[X] = \int_0^\infty x P(dx) = \int_0^\infty \int_0^x dy P(dx) = \int_0^\infty \int_y^\infty P(dx)dy = \int_0^\infty P(X\geq y)dy$$ Of course, the same argument also shows $E[X] = \int_0^\infty P(X>y)dy$ – Andrew Dec 23 '23 at 19:01
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    Define $Q(x)=P[X>x]$. Then $$Q(x)=\int_0^1 1{Q(x)> u}du$$ and so $$\int_0^{\infty} Q(x)dx = \int_0^{\infty}\int_0^1 1{Q(x)> u}dudx$$ – Michael Dec 23 '23 at 20:44
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    Having said that, it seems much easier to prove the $E[X]$ formula for $X\geq 0$ without using quantile functions using $E[X] = \int X dP$: $$\left(X=\int_{0}^{\infty} 1{X>u}du\right) \implies E[X] = \int \left(\int_0^{\infty} 1{X>u} du\right)dP$$ – Michael Dec 23 '23 at 20:53

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