Identity for the expected value

Question

I want to show the following identity

$$\mathbb{E}X=\int_0^\infty \Pr(X>t) dt$$

where $X$ is a continuous random variable taking positive values.

We have \begin{align}\int_0^\infty \Pr(X>t) dt&=\int_0^\infty \int_t^\infty1P_X(ds)dt\\&=\int_{-\infty}^\infty1_{(0,\infty)}(t)\int_t^\infty P_X(ds)dt \\ &=\int1_{(0,\infty)}(t)\int1_{(t,\infty)}(s)P_X(ds)dt \end{align}

I am not sure how to continue from here

Switch order of integration. Btw, equality is only valid if $X\geq0$ a.s. — drhab, Nov 11 '18 at 11:57
What is $P_X(s)ds?$ The image measure is $P_X(ds)$ I dont know why you edited that. — , Nov 11 '18 at 12:04
This comment is to link this post as one of the (abstract) duplicates to the current choice of mother post. — Lee David Chung Lin, Nov 13 '18 at 12:30

Kenny Wong · Accepted Answer · 2018-11-11T12:44:32.837

2

Edit: First, I'll deal with the case where $P$ is absolutely continuous, with a density $p_X(s)$. We have shown that $$ \int_0^\infty \Pr(X > t) dt = \iint_R p_X(s) \ dsdt,$$ where $p_X(s)$ is the density for $X$ and $R$ is the region $$ R = \{ (s, t) \in \mathbb R^2 \ : \ 0 < t < \infty, \ t < s < \infty \}.$$ Now notice that the region $R$ can also be described like this: $$ R =\{ (s, t) \in \mathbb R^2 \ : \ 0 < s < \infty, \ 0 < t < s \} .$$

So an alternative, equivalent way of writing the double integral is: $$ \iint_R p_X(s) \ ds dt = \int_0^\infty ds \int_0^s dt p_X(s) = \int_0^\infty ds \left( p_X(s) \int_0^s 1 dt \right) = \int_0^\infty ds \ s p_X(s) = \mathbb E[X].$$

However, if we don't assume that $X$ is absolutely continuous (so we don't have a density $p_X(s)$), we can still proceed as follows: \begin{align} \mathbb E[X] & = \int_\Omega X(s) dP(s) \\ &= \int_\Omega \mu \left( [0, X(s))\right) \ dP(s) \\ &= (P \times \mu) \left( \left\{ (s,t) \in \Omega \times [0, \infty) : 0 \leq t < X(s) \right\}\right) \\ &= \int_{[0, \infty)} \ P (\{ s \in \Omega : X(s) > t \}) \ d\mu(t) \\ &= \int_0^\infty \Pr(X > t) dt \end{align}

Here, $\Omega$ is the probability space, and $\mu$ is the Lebesgue measure, and I'm assuming that $X(s) \geq 0$ on $\Omega$.

edited Nov 11 '18 at 12:44

answered Nov 11 '18 at 12:00

Kenny Wong

32,192

$0<t<0$ ? or $<s$? – Nov 11 '18 at 12:05
@itry sorry, typo – Kenny Wong Nov 11 '18 at 12:06
$P_X(s)$ is the density of $X$? – Daniel Camarena Perez Nov 11 '18 at 12:13
@DanielCamarenaPerez yes – Kenny Wong Nov 11 '18 at 12:14
But $X$ doesn't need to be absolutely continuous. – Nov 11 '18 at 12:15
Review "Alternative formula for expected value" in https://en.wikipedia.org/wiki/Expected_value – Daniel Camarena Perez Nov 11 '18 at 12:21
@DanielCamarenaPerez Yes, I'm editing my post about the case where $X$ isn't absolutely continuous. – Kenny Wong Nov 11 '18 at 12:23
your first method still works even if its not absolutely continuous. But thanks for the alternative formulation – Nov 11 '18 at 12:38
Oh, fair enough! :) sorry about the confusion! – Kenny Wong Nov 11 '18 at 12:39

score 1 · Answer 2 · answered Nov 12 '18 at 10:42

If the integrand is a nonnegative function then switching the order of integration is allowed, so:

$$\begin{aligned}\mathbb{E}X & =\int_{0}^{\infty}sP_{X}\left(ds\right)\\ & =\int_{0}^{\infty}\int_{0}^{\infty}\mathbf{1}_{\left(0,s\right)}\left(t\right)dtP_{X}\left(ds\right)\\ & =\int_{0}^{\infty}\int_{0}^{\infty}\mathbf{1}_{\left(0,s\right)}\left(t\right)P_{X}\left(ds\right)dt\\ & =\int_{0}^{\infty}\int_{0}^{\infty}\mathbf{1}_{\left(t,\infty\right)}\left(s\right)P_{X}\left(ds\right)dt\\ & =\int_{0}^{\infty}P\left(X>t\right)dt \end{aligned} $$

Identity for the expected value

2 Answers2