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Let continuous random variable $X \ge 0$ and $E(X) \lt \infty$.

Let cumulative density function $F(X)$ is differentiable at all points of X,

show that

$E(X) = \int_0^{\infty}(1-F(x))dx$

Since $F'(x) = f(x)$,

$E(X) = \int_0^{\infty}xf(x)dx = \int_0^{\infty} xF'(x)dx$

Using integration by parts, we get

$\int_0^{\infty} xF'(x)dx = 1- \int_0^{\infty} F(x)dx = \int_0^{\infty}(1-F(x))dx$

Beverlie
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    Not correct. In the last equation check the first term of integration by parts. It is not $1$ but $xF(x)\bigg\vert_{0}^{\infty}$ – Dhruv Kohli Jun 25 '17 at 05:00

2 Answers2

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HINT As someone wrote in the comments IBP doesn't work here cause of the boundary term.

Write $$X = \int_0^\infty I(X > x)dx$$

where $I$ is the indicator function.

spaceisdarkgreen
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A lot of probability problems boil down to writing the number "1" in the right way. Here, we recognize that $1 = \lim_{x \to\infty} F(x)$ and write $$ \int_0^\infty (1 - F(x)) \,\mathrm{d}x = \int_0^\infty \int_x^\infty f(t) \,\mathrm{d}t \,\mathrm{d}x = \int_0^\infty \int_0^t f(t) \,\mathrm{d}x \,\mathrm{d}t = \int_0^\infty tf(t) \,\mathrm{d}t = \mathbb{E}[X], $$ where the interchange of integrals is permissible by the positivity of $f$.

Edit: this equality also holds if $\mathbb{E}[X] = \infty$; that is, one need not know that $\mathbb{E}[X]$ exists a priori before computing the integral. This is because Fubini's theorem does not demand an $L^1$ condition when the integrand is positive.

user217285
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  • $$\int_0^\infty \int_x^\infty f(t) ,\mathrm{d}t ,\mathrm{d}x = \int_0^\infty \int_0^t f(t) ,\mathrm{d}x ,\mathrm{d}t$$

    How could one derive above equality?

     – Beverlie Jun 25 '17 at 05:32
  • It follows from changing the order of integration. Check out Fubini's Theorem https://en.wikipedia.org/wiki/Fubini%27s_theorem#The_Fubini.E2.80.93Tonelli_theorem – tsm Jun 25 '17 at 05:35
  • @first thanks for your link, but only I can get from above link(maybe because of my shortage of knowledge) is the fact that we can interchange the order of integrals- but I still can't get it how $x$ to $\infty$ becomes $0$ to $t$, the integration interval. – Beverlie Jun 25 '17 at 05:42
  • Draw the region of integration on the $x$-$t$ plane. – user217285 Jun 25 '17 at 06:03