# 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. – 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 R. 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
• A proof explicitly using Fubini and integrating $dP$: math.stackexchange.com/questions/536442/… – D.R. Nov 20 '19 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

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

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 '14 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 '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
• 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
• @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: 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. – pre-kidney Jul 19 '19 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. – Henry Jul 21 '19 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...) – 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 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 $$\int_0^{x}1[t>0]\ dt=\int_0^{\infty}1[x>t]\ dt,\qquad \forall x\in\mathbb R.$$ – T34driver Apr 23 at 19:41