# Integrating over the empirical distribution function (EDF)

Suposse I have $$F$$ as a CDF of a random variable $$X$$ and a sample $$(X_i)_{i = 1}^n$$ , $$X_i \sim F$$, iid.

$$F(x) = P(X \leq x) = E[ I(X\leq x) ]$$

I define my EDF

$$\hat{F}_n (x) = \frac{1}{n} \sum_{i = 1}^{n} I(X_i \leq x)$$

I would like, for example, calculate $$E_{\hat{F}_n}[X_i]$$:

$$$$\label{eq1} \begin{split} E_{\hat{F}_n}[X_i] & = \int xd\hat{F}_n(x) \\ & = \int xd \frac{1}{n} \sum_{i = 1}^{n} I(X_i \leq x)\\ & = \frac{1}{n} \sum_{i = 1}^{n} \int x \, \,d I(X_i \leq x) \end{split}$$$$

Supposedly, it is known that $$\int x \, \,d I(X_i \leq x) = X_i$$. But I can not see the triviality of it. Some help?

If $$H(x)=I_{a \leq x}$$ then $$\int f(x)dH(x)=f(a)$$ for any function $$f$$. [$$H$$ corresponds to the degenerate measure at $$a$$ and it corresponds to the constant random variable $$a$$]. Taking $$a=X_i$$ and $$f(x)=x$$ gives you what you want.
• What is the definition of a "degenerate measure at $a$"?
• Ca I say that a degenerate measure at $a$ is a measure $\mu$ such that $\mu(A) = a$ for every borelian set $A$?
• No. The degenerate measure at $a$ is defined by $\mu(A)=1$ if $a \in A$ and $0$ if $a \notin A$. Jun 18, 2019 at 23:03