Integral w.r.t. a cumulative density function

Question

I was reading a paper and couldn't understand the following transition. Could someone tell me where the term of $p^k (\frac{1}{2} − c′)$ comes from in the following transition?

Def: Cumulative function of interest

$$ F(x) = \begin{cases} 0 & \text{for } x < -c' \\ p^k & \text{for } x \in [-c', 0) \\ (p + (1 - p) x)^k & \text{for } x \in [0, 1] \\ 1, & \text{for } x > 1 \end{cases} $$

Integral of interest

$$ \int^{\infty}_{-\infty} x d F(x) = p^k (\frac{1}{2} - c') + \int^{1}_{0} x d F(x) = \cdots $$

My attempts

First, I expanded the integral based on the sections as follows; $$ \int^{\infty}_{-\infty} x d F(x) = \int^{-c'}_{- \infty} x dF(x) + \int^{0}_{-c'} x dF(x) + \int^{1}_{0} x dF(x) + \int^{\infty}_{1} x dF(x) $$ Then, i was not sure how to compute each term... could someone tell me how to compute the integral over a CDF?

Reference

• Appendix B of this paper
• https://math.stackexchange.com/questions/4605847/integral-over-a-cumulative-density-function

Answer 1

Let us express the CDF in terms of the Heaviside function $H(\cdot )$, this function is defined by, $$ H(a) = \left\{ \begin{array}{ccc} 1 & \text{ if } & a\geq 0 \\ 0 & \text{ if } & a < 0 \end{array}\right. $$ Note, $H$ itself is a CDF, in some sense it is the "simplest" CDF. Now, we will do something extremely controversial, we claim that the derivative of the Heaviside function is the delta function, i.e., $H' = \delta$. Where $\delta(\cdot)$ is defined as follows, $$ \delta(a) = \left\{ \begin{array}{ccc} \infty & \text{ if } & a = 0 \\ 0 & \text{ if } & a\not = 0 \end{array}\right.$$ This is the delta "function", it has the additional property that, $\smallint_I (g\cdot \delta) = g(0)$ provided that $I$ is an interval that contains the singularity $0$ and $g$ is a continuous function at $0$. For instance, $\smallint_{(1,2)} (g\cdot \delta) = 0$ but $\smallint_{(-1,2)} (g\cdot \delta) = g(0)$. You can intuitively think of $\delta$ as a PDF, which integrates to $H$. Of course, no such function exists in the classical sense with such properties, but let us not be worried about that yet.

Now the function that you wish to integrate can be described by, $$ F(x) = p^k \bigg( H(x+c') - H(x) \bigg) + (p+(1-p)x)^k\bigg( H(x) - H(x-1) \bigg) + H(x-1)$$ Therefore, the "derivative" is equal to, $$ F'(x) = \delta(x-1) + p^k \bigg( \delta(x+c') - \delta(x) \bigg) + k(1-p)(p+(1-p)x)^{k-1}\bigg( H(x) - H(x-1) \bigg) + (p+(1-p)x)^k\bigg( \delta(x) - \delta(x-1) \bigg)$$ Therefore, the calculation you are looking for is equal to, $$ \int_{\mathbb{R}} x ~ d F(x) = \int_{\mathbb{R}} x F'(x) $$ Now substitute the above "derivative" and use the delta properties.