# Explain why $E(X) = \int_0^\infty (1-F_X (t)) \, dt$ for every nonnegative random variable $X$

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.

• 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. Jul 19, 2012 at 13:42
• This question was asked here previously. Check and you will find a more detailed answer. Either here or on CV. Jul 19, 2012 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. Jul 19, 2012 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. Oct 26, 2015 at 21:12
• A proof explicitly using Fubini and integrating $dP$: math.stackexchange.com/questions/536442/…
– D.R.
Nov 20, 2019 at 8:30

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

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

• @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, 2014 at 19:35
• $U=\mathbf 1_{X\geqslant t}$ is the function defined on $\Omega$ by $U(\omega)=1$ if $X(\omega)\geqslant t$ and $U(\omega)=0$ otherwise.
– Did
Jun 3, 2014 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, 2014 at 15:14
• @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 $\int_{-\infty}^{\infty}1_{X(\omega)\geq t}(\omega)dP(\omega)=P(X>t)$. Feb 12, 2017 at 21:34
• Maybe Fubini-Tonelli deserves a mention Jan 19, 2020 at 23:23

Copied from Cross Validated / stats.stackexchange: where $S(t)$ is the survival function equal to $1- F(t)$. The two areas are clearly identical.

• 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. Jul 19, 2019 at 5:13
• @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. Jul 21, 2019 at 17:21
• 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...) Jul 21, 2019 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 $S(t)$. Aug 11, 2020 at 0:49

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, $$\mathbb EX=\int_0^{\infty}\mathbb P(X>t)\ dt.$$

• How do we get $$\int_0^{x}1[t>0]\ dt=\int_0^{\infty}1[x>t]\ dt,\qquad \forall x\in\mathbb R.$$ Apr 23, 2021 at 19:41
• @T34driver They are both $\int\limits_0^{x}1\ dt =x$ when $x>0$ and are both $0$ when $x \le 0$ Nov 12, 2021 at 3:34