Let $X$ be a positive random variable with distribution function $F$. Show that $E(X)=\int_0^\infty(1-F(x))dx$

Question

Let $X$ be a positive random variable with distribution function $F$. Show that $$E(X)=\int_0^\infty(1-F(x))dx$$

Attempt

$\int_0^\infty(1-F(x))dx= \int_0^\infty(1-F(x)).1dx = x (1-F(x))|_0^\infty + \int_0^\infty(dF(x)).x $ (integration by parts)

$=0 + E(X)$ where boundary term at $\infty$ is zero since $F(x)\rightarrow 1$ as $x\rightarrow \infty$

Is my proof correct?

This is a classical result that you can also prove using a double integral + Fubini (see for example (http://www.sosmath.com/CBB/viewtopic.php?t=36986)) — Jean Marie, Oct 02 '16 at 13:59

score 1 · Answer 1 · answered Oct 02 '16 at 13:20

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The fact that $x(1-F(x))\to0$ as $x\to\infty$ needs justification. What if $F(x)\sim 1-\dfrac{1}{\sqrt{x}}$ for large $x$? You need to show such things can't occur.

But the justification is simple: $xP(X>x)=x\int_x^\infty dF(t)\leq \int_x^\infty tdF(t)\to0$ as $x\to\infty$ since $E(X)=\int_0^\infty xdF(x)<\infty$.

answered Oct 02 '16 at 13:20

Landon Carter

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Why this $x\int_x^\infty dF(t)\leq \int_x^\infty tdF(t)$ is true? – User Oct 02 '16 at 13:27
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Because for all $t\in[x,\infty), t\geq x$. – Landon Carter Oct 02 '16 at 13:28
Where have you used $E(X)\lt \infty$ and why this is so? – User Oct 02 '16 at 13:42
You're right; I am not given finite expectation. If $E(X)<\infty$ what I wrote above holds. I will think over what happens if $EX=\infty$. – Landon Carter Oct 03 '16 at 05:00
It seems that when $EX = \infty$ it is really tricky. Consider PDF f(x) = 2/3x (for 0<x<1) and f(x) = 2/(3 x^2) (for x>=1). It will end up $EX = \infty$, and the part now needs more justification. – Weikeng Chen Jan 27 '19 at 06:53

Let $X$ be a positive random variable with distribution function $F$. Show that $E(X)=\int_0^\infty(1-F(x))dx$

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