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I saw this equation in my book, but I don't know where it comes from. $F$ is the distribution function of a RV.

$$\int_a^b (x - a)dF(x) = \int_a^b (1 - F(x)) dx$$

$a > 0$ is arbitrary but $F(b) = 1$, not sure if that plays a part here.

Martin Sleziak
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saei
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    The left hand side is $E[(X-a)I_{a<X<b}]$ where $I$ is an indicator variable. Now compare this with https://math.stackexchange.com/questions/172841/explain-why-ex-int-0-infty-1-f-x-t-dt-for-every-nonnegative-rando?noredirect=1&lq=1. – StubbornAtom Dec 25 '19 at 19:47

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This is just a corollary of the main result here.

For any random variable $X$ and $a>0$,

\begin{align} (X-a)\mathbf1_{a<X<b}&=\int_0^{X-a}\,dt\cdot\mathbf1_{a<X<b} \\&=\int_0^\infty \mathbf1_{X-a>t\,,\,a<X<b}\,dt \\&=\int_0^\infty \mathbf1_{a+t<X<b}\,dt \\&=\int_0^{b-a}\mathbf1_{X>a+t}\,dt \end{align}

Therefore,

\begin{align} \mathbb E[(X-a)\mathbf1_{a<X<b}]&=\color{blue}{\int (x-a)\mathbf1_{a<x<b}\,dF(x)}\qquad\qquad\,\,\small(\text{by definition}) \\&=\int \left(\int_0^{b-a}\mathbf1_{x>a+t}\,dt\right)dF(x) \\&=\int_0^{b-a} \underbrace{\int \mathbf1_{x>a+t}\,dF(x)}_{\large\mathbb E[\mathbf1_{X>a+t}]}\,dt\qquad\quad\small(\text{by Fubini's theorem}) \\&=\int_0^{b-a}\mathbb P(X>a+t)\,dt \\&=\int_0^{b-a} (1-F(a+t))\,dt \\&=\color{red}{\int_a^b(1-F(x))\,dx} \end{align}

StubbornAtom
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