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 $E_{\hat{F}_n}[X_i]$:

\begin{equation} \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} \end{equation}

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?


1 Answer 1


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.

  • $\begingroup$ What is the definition of a "degenerate measure at $a$"? $\endgroup$
    – Fam
    Jun 18, 2019 at 17:05
  • $\begingroup$ Ca I say that a degenerate measure at $a$ is a measure $\mu$ such that $\mu(A) = a$ for every borelian set $A$? $\endgroup$
    – Fam
    Jun 18, 2019 at 17:06
  • 1
    $\begingroup$ No. The degenerate measure at $a$ is defined by $\mu(A)=1$ if $a \in A$ and $0$ if $a \notin A$. $\endgroup$ Jun 18, 2019 at 23:03

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