# Integral of CDF equals expected value

The question as below...

Let $X$ be a non-negative random variable and $F_{X}$ the corresponding CDF. Show,

$$E(X) = \int_0^\infty (1-F_X (t)) \, dt$$

in the case that, $X$ has a

a) discrete distribution b) 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.

Thanks for the help!

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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. – Davide Giraudo Jul 19 '12 at 13:42
This question was asked here previously. Check and you will find a more detailed answer. Either here or on CV. – Michael Chernick Jul 19 '12 at 14:21
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. – Dilip Sarwate Jul 19 '12 at 15:38
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. – cantorhead Oct 26 '15 at 21:12

For every nonnegative random variable $X$, whether discrete or continuous or a mix of these, $$X=\int_0^{+\infty}\mathbf 1_{X\gt t}\,\mathrm dt=\int_0^{+\infty}\mathbf 1_{X\geqslant t}\,\mathrm dt,$$ hence $$\mathrm E(X)=\int_0^{+\infty}\mathrm P(X\gt t)\,\mathrm dt=\int_0^{+\infty}\mathrm P(X\geqslant t)\,\mathrm dt.$$

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may I ask you how you derive the first equation? The left side is a function from sample space (possibly to $\mathbb{R}$) and the right side is an integral and therefore a number. Am I right? – Cupitor Jun 2 '14 at 15:26
@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 $$\int_0^{+\infty}\mathbf 1_{X(\omega)\geqslant t}\,\mathrm dt.$$ – Did Jun 2 '14 at 19:35
Thanks for the reply. But $X(\omega)$ is a real number right? Then if $\textbf{1}$ is characteristic function it should be defined on a set, but $X(\omega)>t$ is a condition. – Cupitor Jun 2 '14 at 22:25
$U=\mathbf 1_{X\geqslant t}$ is the function defined on $\Omega$ by $U(\omega)=1$ if $X(\omega)\geqslant t$ and $U(\omega)=0$ otherwise. – Did Jun 3 '14 at 10:09
The second step is to consider the expectation of each side (that is, its integral with respect to $P$). – Did Jun 3 '14 at 15:14

Copied from Cross Validated / stats.stackexchange:

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

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