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I am looking for a probability distribution with the shape of the CDF similar to the sinh or logit function. I need it to be centered at 0 and symmetric, meet the usual CDF restrictions. I understand that most likely it will need to be defined on a specific interval, that's okay. The important property is that F'(x) < 1, F''(x) < 0 close to the left of the mean, and F'(x)<1 , F''(x)>0 close to the right. For more farther away values, however defined, I need F'(x) > 1, F''(x) > 0. Needless to say, the easier the formula, the better! I tried to play around with the sinh function but I couldn't get it to work. Any ideas? Here is an example of the cdf I am looking for:

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  • $\begingroup$ So, after $k$ do you want it to constantly be $1$? Same with before $-k$ and being $0$. $\endgroup$
    – Mark
    Dec 10, 2016 at 6:30
  • $\begingroup$ $Erf(x)$, the CDF of the Gaussian looks like the logit function. EDIT: Nvm, thought you meant logistic. $\endgroup$
    – Demetri Pananos
    Dec 10, 2016 at 6:34
  • $\begingroup$ Mark, yes. Although I wouldn't mind if it defined on the entire real line, but I feel like this may be difficult given the shape. $\endgroup$
    – Paul
    Dec 10, 2016 at 6:41
  • $\begingroup$ There is a distribution corresponding to any function $F(x)$ which goes from $0$ to $1$, is weakly increasing and is right-continuous, and any cumulative distribution function must have these properties. Your specific requirements put further restrictions, and in particular your $F'(x) \gt 1$ can only be true for a limited number of values of $x$ $\endgroup$
    – Henry
    Dec 10, 2016 at 13:09

2 Answers 2

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Given the desire for simplicity, here are two suggestions:

U-quadratic pdf $\quad f(x) = \frac32 x^2 \quad \text {defined on } (-1,1)$

enter image description here

with cdf:

enter image description here

The cdf is: $F(x) = \frac12 (1 + x^3)$ on (-1,1)

Inverse Triangular pdf $\quad f(x) = |x| \quad \text {defined on } (-1,1)$

enter image description here

with cdf:

enter image description here

The cdf is:

$$F(x) = \left\{ \begin{array}{cc} 1 & \text{if } x\geq 1 \\ \frac{1}{2} \left(1+x^2\right) & \text{if } 0<x<1 \\ \frac{1}{2} \left(1-x^2\right) & \text{if } -1<x\leq 0 \\ 0 & \text{otherwise} \\ \end{array} \right. $$

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  • $\begingroup$ Thank you, this is great! I am wondering - is it possible to somehow incorporate an additional parameter k that would determine the range [-k,k] on which the function is defined (in case I need it to differ from [-1,1])? $\endgroup$
    – Paul
    Dec 10, 2016 at 20:43
  • $\begingroup$ Nevermind, I think I figured it out: $ F(x) = \frac{1}{2}((kx)^2 sgn(x) + 1) $ should be defined on $ (-\frac{1}{k}, \frac{1}{k}) $. Thanks! $\endgroup$
    – Paul
    Dec 10, 2016 at 20:55
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There is a natural (and important) distribution on $[0,1]$, the arcsine distribution (http://www.randomservices.org/random/special/Arcsine.html). Its cdf is

$$F(x)=\begin{cases}0 & \text{for} & x \leq 0 \\ \frac{2}{\pi}\arcsin(\sqrt{x})& \text{for} & 0<x<1 \\ 1 & \text{for} & x \geq 1 \end{cases}$$

Remarks:

  • "$\arcsin$" is now a less common notation - though more correct - than "$\sin^{-1}$".

  • The "arcsine" distribution is a particular case of a "beta" distribution.

  • Edit: cdf $F$ should be replaced by $F_a$ defined on $-\frac{1}{2a}<x<\frac{1}{2a}$ by

$$F_a(x):=\frac{2}{\pi}\arcsin\left(\sqrt{ax+\frac12}\right)$$

with $a$ arbitrary small in order that its curve is symmetrical with respect to point $(0,1/2)$, and that its derivative is arbitrarily small in the vicinity of $0$.

The graphical representation of $F$ is as follows:

enter image description here

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    $\begingroup$ You are going to have to adjust this slightly to make it symmetric about $0$ and to get $F'(x) \lt 1$ close to the mean $\endgroup$
    – Henry
    Dec 10, 2016 at 13:03
  • $\begingroup$ @Henry I have answered concerning the symmetry. Concerning the constraint $F'(x)<1$ it suffices to "spread" the distribution by taking $F_1(ax)$ with $a>0$ arbitrarily small. $\endgroup$
    – Jean Marie
    Dec 10, 2016 at 14:02
  • $\begingroup$ Thank you, this is very useful! $\endgroup$
    – Paul
    Dec 11, 2016 at 0:01

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