$\newcommand{\erf}{\operatorname{erf}}$ I have tried out to define a new distribution lie in [0,1] using my special function which is montioned in this paper, The new special function is defined as: $$I(a)=\int_{0}^{a}{(e^{-x²})}^{\erf(x)}dx, \tag{01}$$ take $t=\erf(x)$ then $x=\erf^{-1}(t)$ such that $ dx=\dfrac{\sqrt{\pi}}{2}e^{(\erf^{-1}(t))^2 }dt$ by substitution in $\int_{0}^{a}{(e^{-x²})}^{\erf(x)}dx$ yield to get : $$\dfrac{\sqrt{\pi}}{2}\int_{0}^{b}{(e^{(\erf^{-1}(t))^2-t(\erf^{-1}(t))^2})}dt, \tag{02}$$ using this approximation for small $t$ we have :$\erf^{-1}(t) \sim t\dfrac{\sqrt{\pi}}{2}$ , Replace this approixmation gives the following :$$\dfrac{\sqrt{\pi}}{2}\int_{0}^{b}{e^{\dfrac{\pi}{4}(t^2-t^3)}}dt .\tag{03}$$ Now for the latter equality I want to define a new distribution since we are now in [0,1], because the precedent variable change $t$ vary from $0$ to $1$ because $ \erf^{-1}(\infty)=1 $, Then let us define a Unit distribution with $3$ parametre : $\beta$ and $\mu$ and $\theta $ such that: $$ f(t,\beta ,\mu , \theta )=e^{\theta(t-\mu)^{\beta}} \text{and} f(t,\beta ,\mu , \theta )=0, \text{for}: t >1 \tag{04}$$ , Now my question here is this a known distribution and if it is what is it ?

Edit: I have edited parameters for generalisation to be clear

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    $\begingroup$ 1. Your title equation has $t$ on the left hand side and $x$ on the right (as per your equation 04 as well); this will be a problem. Best to stick to one variable if you want it to look like an equality. 2. It's not clear what you mean by "unite" here. Do you intend "univariate" instead? $\endgroup$
    – Glen_b
    Commented Aug 18, 2019 at 23:47
  • $\begingroup$ Thanks it is a wrong typo now it fixed $\endgroup$ Commented Aug 18, 2019 at 23:48
  • $\begingroup$ @Glen I mean by unite the distribution defined on (0,1) like Beta distribution $\endgroup$ Commented Aug 18, 2019 at 23:49
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    $\begingroup$ Okay, that's not how you say that in English. "distribution on the unit interval" would work but may be too long for the title. You should explain the bounds on the variables explicitly in the question where you define the density such that it's clear you intend the density to be 0 outside that interval $\endgroup$
    – Glen_b
    Commented Aug 18, 2019 at 23:50
  • $\begingroup$ I've cleaned up the title a little. You will need similar changes in the body text. $\endgroup$
    – Glen_b
    Commented Aug 18, 2019 at 23:54

2 Answers 2


The distribution seems to be close to a truncated form of the generalized Gaussian distribution which has the same formula (except it has support on entire $\mathbb{R}$ and some parameters are in a different place, for instance $\theta = -1/\alpha^\beta$ )

$$f(x) \propto e^{-(\vert x-\mu \vert /\alpha)^\beta} $$

One difference is that the function uses the absolute value of $x-\mu $. But , this might be necessary for your distribution as well. In the cases $\mu>0$ and $\beta \notin \mathbb{Z}$ then you the term $(x-\mu)^\beta$ is a non-integer power of a negative number which is problematic. So I guess that your distribution should use the restriction $\mu<0$ in which case the difference with the absolute sign is not relevant.

  • $\begingroup$ Probably you meant $x \notin (0,1) $ not $\mu \notin (0,1) $ $\endgroup$ Commented Aug 19, 2019 at 8:09
  • $\begingroup$ The generalized Gaussian distribution has a symmetry around $\mu $. If that symmetry boundary $\mu $ is within zero and one than it will not coincide with your distribution (which does not have this absolute value). $\endgroup$ Commented Aug 19, 2019 at 8:34
  • $\begingroup$ look your expoenent and your setting of theta , alpha is always positive skew for generalized Gaussian distribution , then we can't get a positive theta as the case i want $\endgroup$ Commented Aug 19, 2019 at 12:54
  • $\begingroup$ @zeraouliarafik could you clarify this in the question that you are looking for solutions in the case of positive theta. In the comments you suggested that you are looking for cases with theta=-1 $\endgroup$ Commented Aug 19, 2019 at 13:17
  • $\begingroup$ I'm ready to add a new question for more clarification $\endgroup$ Commented Aug 19, 2019 at 13:18

$f$, in your formulation, is not a probability density function as it does not integrate to 1 on the interval $t\in(0,1)$ for all values of the parameters. You would need to normalize it, but then the expression would not be that nice as it will depend on the parameters through some special functions (actually the cdf of the Generalized Normal distribution).

One alternative is to use the logit transformation applied to the Generalized Normal distribution, which leads to

$$f(t;\alpha,\beta,\mu) = {\frac {\beta }{2\alpha \Gamma (1/\beta )}} \frac{1}{t(1-t)}\;e^\left\{-\left(\left\vert\mbox{logit}(t)-\mu \right\vert/\alpha \right)^{\beta }\right\}, $$ $t\in(0,1)$, $\mu\in{\mathbb R}$, $\alpha,\beta>0$.

You can call it the Generalized Logit-Normal distribution, which generalizes the logit normal distribution:



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