## Cumulative distribution function

### Set

 definiendum $F\in\mathrm{CDF}$ inclusion $F:\mathbb R\to\mathbb R$ … right-continuous function, monotonically increasing postulate $\lim_{x\to\ -\infty} F(x)=0$ postulate $\lim_{x\to\ +\infty} F(x)=1$

$P=F'$
Let $S:({\mathbb D}\to {\mathbb R}_{\ge 0})\to{\mathbb R}_{\ge 0}$ be linear.
If $f$ with $\infty>Sf>0$, then $\bar{f}:=\frac{1}{Sf}\cdot f$ has $S\bar{f}=\frac{Sf}{Sf}=1$.
So we can use such $S$ to normalize functions.

For ${\mathbb D}={\mathbb N}$ the general case is $Sf:=\sum_{n=0}^\infty (L_nf)(n)$, where $(L_n)$ is a suitable sequence of linear operations (e.g. differential operators). For $L_n={\mathrm{id}}$ we get the standard sum (see below).
For ${\mathbb D}\subseteq{\mathbb R}^m$ we have integrals.

Let $a:{\mathbb N}\to{\mathbb R}_{\ge 0}$ be a sequence, then

$\bar{a}:{\mathbb N}\to[0,1]$

$\bar{a}(n):=\frac{1}{\sum_{k=0}^\infty a(k)}\cdot a(n)$

has
$\sum_{n=0}^\infty \bar{a}(n)=1$
The “monomial bump” on $[-d,d]$, which goes against the constant probability $\frac{1}{2d}$ for large $n$:
$P_{n,d}(x):=\frac{1}{2d}\left(1+\frac{1}{2n}\right)\cdot\left(1-\left(\frac{x}{d}\right)^{2n}\right)$
$\int_{-d}^dP_{n,d}(x)\,{\mathrm d}x=1$
Die Funktion hängt mit dem sog. Epanechnikov-Kern zusammen.