What is the integral of a cumulative distribution function?

Question

I cannot find what is the integral of a cumulative distribution function

$$\int G(\xi)d\xi$$

I think it should be simple, but I have no idea where else to look for it.

Answer 1

Integrating by parts $$\int G(x)dx= x G(x) - \int x \,G'(x) dx =x G(x) - \int x \,g(x) dx $$

where $G'(x)=g(x)$ is the corresponding density function.

In particular, if $g(x)$ is well behaved ($X$ is continuous) then you could write: $$ \int_a^b x \,g(x) dx=(G(b)-G(a)) E[X_{a,b}]=P(X\in [a, b])\, E[X_{a,b}] $$ where $E[]$ denotes expected value and $ X_{a,b}$ represents the original r.v. $X$ truncated (conditioned) to the interval $[a,b]$.

Answer 2

Let $(\Omega, \mathcal{F},P)$ be a probability space and $X \in \mathcal{F}$ an integrable random variable.

Notice that

$\int_{-\infty}^{x}F_{X}(z)dz=\int_{-\infty}^{x}P(X \le z)dz=\int_{-\infty}^{x}\int_{\Omega}1_{(X(w) \le z)}(w,z)dPdz\overset{\text{Fubini}}{=}\int_{\Omega}\int_{-\infty}^{x}1_{(X(w) \le z)}(w,z)dzdP=E[\int_{-\infty}^{x}1_{(X(w) \le z)}(w,z)dz]$.

Now, $\int_{-\infty}^{x}1_{(X \le z)}dz=\begin{cases} \int_{X}^{x}dz=x-X &\mbox{if}\quad x \ge X,\\ 0 &\mbox{if}\quad x<X. \end{cases}$

Hence $\int_{-\infty}^{x}1_{(X \le z)}dz=(x-X)^{+}$ and $\int_{-\infty}^{x}F_{X}(z)dz=E[(x-X)^{+}]$