I understand that $\int_0^\infty P(X>x)dx=E[x]$, and also the logic behind the discrete version here. What I don't understand is how the limits of integration change as is seen here, from $(x, \infty)$ to $(0, t)$. Thank you.
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$$ \int_0^\infty \int_x^\infty f_X(t) \,dt \, dx = \int_0^\infty \int_0^t f_X(t) \, dx \, dt $$
This transformation is true no matter what the context is (as long as the function is Lebesgue integrable). Both double integrals are taken over the region where $0 \leq x \leq t$ (which is an angle-like unbounded slice of the plane). Changing the order of integration usually requires some consideration of how the limits of the region are described in a different way.
The left side has $t \geq x$, where $x$ stays fixed for the integration with respect to $t$. But on the right side, $t$ stays fixed for the integration with respect to $x$, and we need to state the limits of $x$ in terms of $t$: $x \leq t$.
aschepler
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that is not true, $f$ must satisfy the assumptions of Fubini's theorem otherwise the other of integration may matter. In particular, if $f_X$ is a prob. density function there is no problem but $f_X$ can not be any arbitrary function. – Sergio Parreiras Dec 11 '13 at 04:54
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Good catch. Fixed, I think. – aschepler Dec 11 '13 at 05:10
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I'm still not getting it. So if we had the bounds as x = 5, what would t be? – Impossibility Dec 11 '13 at 05:37
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$x=5$ does not describe a region of the $x$-$t$ plane with positive area. – aschepler Dec 11 '13 at 05:40
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What do you mean? – Impossibility Dec 11 '13 at 05:49