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Let $X$ be a non-negative random variable and $F_{X}$ the corresponding CDF. Show, $$E(X) = \int_0^\infty (1-F_X (t)) \, dt$$ when $X$ has : a) a discrete distribution, b) a continuous distribution.

I assumed that for the case of a continuous distribution, since $F_X (t) = \mathbb{P}(X\leq t)$, then $1-F_X (t) = 1- \mathbb{P}(X\leq t) = \mathbb{P}(X> t)$. Although how useful integrating that is, I really have no idea.

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    $\begingroup$ In the two cases, it's a rewritting of the sum. Start from the RHS, that you can express in the first case as an integral of a sum and in the second as a double integral, then switch them. This is allowed because all the quantities are non-negative. $\endgroup$ – Davide Giraudo Jul 19 '12 at 13:42
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    $\begingroup$ This question was asked here previously. Check and you will find a more detailed answer. Either here or on CV. $\endgroup$ – Michael R. Chernick Jul 19 '12 at 14:21
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    $\begingroup$ See for example, the answers to this question which include both formal proofs (by Didier, who has answered your question here) as well as more intuitive approaches to the problem. $\endgroup$ – Dilip Sarwate Jul 19 '12 at 15:38
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    $\begingroup$ As far as usefulness, this can be more numerically stable than differentiating $F$, mulitplying by $t$, and integrating. Actually, most random variables don't have pdfs, so differentiating $F$ may not even be possible. $\endgroup$ – cantorhead Oct 26 '15 at 21:12
  • $\begingroup$ A proof explicitly using Fubini and integrating $dP$: math.stackexchange.com/questions/536442/… $\endgroup$ – D.R. Nov 20 '19 at 8:30
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For every nonnegative random variable $X$, whether discrete or continuous or a mix of these, $$ X=\int_0^X\mathrm dt=\int_0^{+\infty}\mathbf 1_{X\gt t}\,\mathrm dt=\int_0^{+\infty}\mathbf 1_{X\geqslant t}\,\mathrm dt, $$ hence

$$ \mathrm E(X)=\int_0^{+\infty}\mathrm P(X\gt t)\,\mathrm dt=\int_0^{+\infty}\mathrm P(X\geqslant t)\,\mathrm dt. $$


Likewise, for every $p>0$, $$ X^p=\int_0^Xp\,t^{p-1}\,\mathrm dt=\int_0^{+\infty}\mathbf 1_{X\gt t}\,p\,t^{p-1}\,\mathrm dt=\int_0^{+\infty}\mathbf 1_{X\geqslant t}\,p\,t^{p-1}\,\mathrm dt, $$ hence

$$ \mathrm E(X^p)=\int_0^{+\infty}p\,t^{p-1}\,\mathrm P(X\gt t)\,\mathrm dt=\int_0^{+\infty}p\,t^{p-1}\,\mathrm P(X\geqslant t)\,\mathrm dt. $$

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    $\begingroup$ @Cupitor The left-hand-side, the middle side and the right-hand-side are all random variables, for example the value at $\omega$ of the right-hand-side is $$\int_0^{+\infty}\mathbf 1_{X(\omega)\geqslant t}\,\mathrm dt.$$ $\endgroup$ – Did Jun 2 '14 at 19:35
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    $\begingroup$ $U=\mathbf 1_{X\geqslant t}$ is the function defined on $\Omega$ by $U(\omega)=1$ if $X(\omega)\geqslant t$ and $U(\omega)=0$ otherwise. $\endgroup$ – Did Jun 3 '14 at 10:09
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    $\begingroup$ The second step is to consider the expectation of each side (that is, its integral with respect to $P$). $\endgroup$ – Did Jun 3 '14 at 15:14
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    $\begingroup$ If I had to take a guess, I would say that in truth the indicator-function takes both $\omega$ and $t$ as arguments instead of only one of those each, and by this the result follows. Is this correct? $\endgroup$ – azureai Feb 12 '17 at 21:34
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    $\begingroup$ @see Yes, your reading of these formulas and the proof in your first comment are both correct. $\endgroup$ – Did Feb 12 '17 at 21:35
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Copied from Cross Validated / stats.stackexchange:

enter image description here

where $S(t)$ is the survival function equal to $1- F(t)$. The two areas are clearly identical.

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    $\begingroup$ The two areas may be clearly identical, but what is unclear is why the integrals equal match the diagrams. Moreover, it appears that this proof does not apply in general: it only works when the random variable $X$ in question possesses a density. $\endgroup$ – pre-kidney Jul 19 '19 at 5:13
  • $\begingroup$ @pre-kidney In the left hand diagram I would have thought it was sensible to regard the white bordered slice as essentially having width $t$ and height $\delta F(t)$, so the area is $\int t\, dF(t)$ while the right hand diagram white bordered slice as essentially having width $\delta t$ and height $S(t)=1-F(t)$ so an area which is $\int S(t) \, dt$, with this duality working for all non-negative distributions discrete or continuous. You are correct that $\int t\, f(t) \, dt$ is only meaningful when there is a density function, since $f(t)$ is that density. $\endgroup$ – Henry Jul 21 '19 at 17:21
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    $\begingroup$ Certainly there are more issues than just the surface level you acknowledge - beyond just the fact that $\int_{t=0}^{\infty}t\ f(t)\ dt$ is meaningless in general, the way you bring in the integral $\int_{t=0}^{\infty} t\ dF(t)$ (which, by the way, I still don't know exactly what quantity you are referring to when $X$ lacks a density) seems to be via Riemann sum approximations to a Riemann integral, which cannot work for the general case (it would require a Lebesgue integral, at least in the standard formulations of probability theory...) $\endgroup$ – pre-kidney Jul 21 '19 at 21:17
  • $\begingroup$ I think it should be $dS(t)$ but not $dF(t)$ in the left equation as the function of the curve is $S(t)$. $\endgroup$ – mxdxzxyjzx Aug 11 at 0:49
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The function $x1[x>0]$ has derivative $1[x>0]$ everywhere except for $x=0$, so by a measurable version of the Fundamental Theorem of Calculus $$ x1[x>0]=\int_0^{x}1[t>0]\ dt=\int_0^{\infty}1[x>t]\ dt,\qquad \forall x\in\mathbb R. $$ Applying this identity to a non-negative random variable $X$ yields $$ X=\int_0^{\infty}1[X>t]\ dt,\quad a.s. $$ Taking expectations of both sides and using Fubini to interchange integrals, $$ \mathbb EX=\int_0^{\infty}\mathbb P(X>t)\ dt. $$

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