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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    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
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    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
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    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
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    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
up vote 27 down vote accepted

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

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

  • 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
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    @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
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    $U=\mathbf 1_{X\geqslant t}$ 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
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    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
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    @see Yes, your reading of these formulas and the proof in your first comment are both correct. – Did Feb 12 '17 at 21:35

Copied from Cross Validated / stats.stackexchange:

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

Another way is that we know: $X=F^{-1}(U)$ where $F$ is the CDF of $X$. So the expected value will be $$\int_{0}^{1} F^{-1}(U) 1 du.$$ If we look at this region, we notice that it is equivalent to the area above the CDF bounded by 1. So we get $$\int_{0}^{1} F^{-1}(U) 1 du = \int_{-\infty}^{\infty} (1-F(U)) du = \int_{-\infty}^{\infty} P(X \geq x) dx$$

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