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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.

Lee David Chung Lin
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Jon Gan
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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 R. 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
  • A proof explicitly using Fubini and integrating $dP$: https://math.stackexchange.com/questions/536442/intuitive-explanation-for-mathbbex-int-0-infty-1-fx-dx – D.R. Nov 20 '19 at 08:30

3 Answers3

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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, by applying Tonelli's Theorem,

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

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

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

JEdwards
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Did
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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 02 '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 02 '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 02 '14 at 22:25
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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 03 '14 at 10:09
  • Thank you. I see. And then the second step follows from changing Lebesgue measure to Probability measure, right? – Cupitor Jun 03 '14 at 10:56
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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 03 '14 at 15:14
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    @Did Can you tell me how to do this formally? I've been trying to understand your argument for hours now. First: What arguments does the function $1_{X>t}$ take? It's a function of $t$ isn't it. I will write this explicitly in the following calculation. By integration I get: $E(X)=\int_{-\infty}^{\infty}X(\omega)P(d\omega)=\int_{-\infty}^\infty\int_0^{\infty}1_{X(\omega)\geq t}(t)dtP(d\omega)=\int_{0}^\infty\int_{-\infty}^{\infty}1_{X(\omega)\geq t}(t)P(d\omega)dt$, where the last step I suppose is Fubini. I know that $\int_{-\infty}^{\infty}1_{X(\omega)\geq t}(\omega)dP(\omega)=P(X>t)$. – azureai Feb 12 '17 at 21:34
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    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? – azureai Feb 12 '17 at 21:34
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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
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    I do not quite see how the "hence" follows from the definition. Could you elaborate more? It would be enough in the first part. – user2316602 Oct 21 '19 at 11:22
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    Maybe Fubini-Tonelli deserves a mention – Maximilian Janisch Jan 19 '20 at 23:23
  • To get an answer requires going through comments is not good one. – Kuo Feb 23 '24 at 06:19
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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.

Community
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Henry
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    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. – pre-kidney Jul 19 '19 at 05:13
  • @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. – Henry Jul 21 '19 at 17:21
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    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...) – pre-kidney Jul 21 '19 at 21:17
  • I think it should be $dS(t)$ but not $dF(t)$ in the left equation as the function of the curve is $S(t)$. – mxdxzxyjzx Aug 11 '20 at 00:49
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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, $$ \mathbb EX=\int_0^{\infty}\mathbb P(X>t)\ dt. $$

pre-kidney
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    How do we get $$\int_0^{x}1[t>0]\ dt=\int_0^{\infty}1[x>t]\ dt,\qquad \forall x\in\mathbb R. $$ – ExcitedSnail Apr 23 '21 at 19:41
  • @T34driver They are both $\int\limits_0^{x}1\ dt =x$ when $x>0$ and are both $0$ when $x \le 0$ – Henry Nov 12 '21 at 03:34
  • Would you write the formula of ${\mathbb E} X$ by using the identity $X$ with indicator function, explicitly? Interchange integrals is not the essential part, but how to introduce the integrands and interpret the structure. – Kuo Feb 23 '24 at 07:50