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Let $X$ be a real continuous random variable with distribution $F$ with finite moments. I want to calculate

$$E[\vert X \vert] = \int_{-\infty}^{\infty} \vert x\vert dF(x)= -\int_{-\infty}^{0} x dF(x) + \int_{0}^{\infty} x dF(x).$$ But I want to obtain an alternative expression in order to get rid of the absolute value. I tried to split this variable using integration by parts $U=-x$, $dV = dF(x)$, then $dU= -dx$, $V=F(x)$ and

$$ -\int_{-\infty}^{0} x dF(x) = UV \Big\vert_{-\infty}^0 - \int_{-\infty}^0VdU$$

Then, $$ -\int_{-\infty}^{0} x dF(x) = \int_{-\infty}^0F(x)dx.$$

For the second integral this trick does not work as I get an infinite integral. How can I solve the second part? I think I should get something like $$ \int_{0}^{\infty} x dF(x) = \int_0^{\infty}1-F(x)dx,$$ but I am not sure how to prove it.

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  • $\begingroup$ This result is demonstrated in many posts here, usually incidental to solving some other problem. For instance, you can find the (one-line) proof in the middle of my post at stats.stackexchange.com/a/209193/919 after the statement "the expectation of any positive distribution F is the integral of its survival function." $\endgroup$ – whuber May 6 '19 at 14:02
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How about the following partial integration (formally written as an ordinary integral, not as a Stieltjes integral): $$\int_0^{\infty}x\cdot f(x)\;dx = -\int_0^{\infty}x\cdot(1-F(x))'\;dx = \underbrace{-x\cdot (1-F(x))\Big|_0^{\infty}}_{=0} + \int_0^{\infty}1-F(x) \; dx $$

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  • $\begingroup$ I like your answer, although I am not specifying that the density actually exists. But, in such case, your answer would apply. $\endgroup$ – expo May 6 '19 at 20:33
  • $\begingroup$ When the derivative is interpreted in terms of distributions (aka generalized functions), this even holds for the discrete case. Remember that $\theta'=\delta$, i.e., the derivative of Heaviside's step-function is Diracs delta-distribution, which is the density of probability distribution with mass one at zero. $\endgroup$ – cdalitz May 7 '19 at 11:46

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