I am searching for the name of the distribution associated with this density on $\mathbb{R}_+$:

$$p(r|\lambda) = \frac{2\lambda r\exp\left(\lambda\exp\left(-r^{2}\right)-r^{2}\right)}{\exp\left(\lambda\right)-1}.$$

It arises from the mixture distribution of

($n\in \mathbb N_+$) with mixture weights from a "positive" Poisson distribution on $n>0$ $$p(n|\lambda)=\frac{\lambda^n \exp(-\lambda)}{n! (1-\exp(-\lambda))}$$

I first thought it looked like a Gumbel distribution, but I couldn't get it into the right shape.

  • 3
    $\begingroup$ There are plenty of ways this can be transformed into a "named" distribution. E.g., writing $r^2=-\log(\log(z))$ entails $1\lt z \lt e,$ $|dr|=dz/(z\log(z))$, and $$2 r\exp\left(\lambda\exp\left(-r^{2}\right)-r^{2}\right)|dr|=z^{\lambda-1}\, \mathrm{d}z.$$ This exhibits the distribution as a transformation of a truncated power law. $\endgroup$
    – whuber
    Commented Nov 25, 2014 at 17:22
  • $\begingroup$ I completely agree. I was actually searching for the name of the distribution in that parametrization. Otherwise we could call all distributions uniform on $[0,1]$. $\endgroup$
    – fabee
    Commented Nov 26, 2014 at 14:31
  • 2
    $\begingroup$ The situation is not that trivial: the transformation to uniformity you refer to captures all the information about the distribution and typically is specific to that distribution. When you can find a fixed, simple transformation that--when applied to all distributions within a parameterized family--produces nice formulas, then you have accomplished something. That is why @Xi'an proposes a square and why I have pointed out the log-log transformation. These are perfectly analogous to the relationship between Normal and Lognormal distributions, for instance. $\endgroup$
    – whuber
    Commented Nov 26, 2014 at 15:53
  • $\begingroup$ Hmm, I guess I misunderstood your previous point. But I see your last point. $\endgroup$
    – fabee
    Commented Nov 26, 2014 at 16:37

1 Answer 1


I think it relates to a Gumbel distribution:

if I define $S=R^2$, when $R\sim p(r|\lambda)$, the density of $S$ is given by the Jacobian formula: \begin{align*} q(s|\lambda) &= p(\sqrt{s}|\lambda)\times \left|\frac{\text{d}r}{\text{d}s}\right|\\ &= p(\sqrt{s}|\lambda)\times \frac{1}{2\sqrt{s}}\\ &= \frac{2\lambda \sqrt{s}\exp\left(\lambda\exp\left(-s\right)-s\right)}{\exp\left(\lambda\right)-1}\,\frac{1}{2\sqrt{s}}\\ &= \frac{\lambda \exp\left(\lambda\exp\left(-s\right)-s\right)}{\exp\left(\lambda\right)-1}\\ &= \frac{\exp\left(\exp\left(\log\{\lambda\}-s\right)+\log\{\lambda\}-s\right)}{\exp\left(\lambda\right)-1}\\ &= \frac{\exp\left(\exp\left(-z\right)-z\right)}{\exp\left(\lambda\right)-1}\\ \end{align*} where $z=s-\log\{\lambda\}$

So $S$ is almost distributed as a Gumbel distribution with parameter $(\log\{\lambda\},1)$ except that (a) its support is truncated to $(0,+\infty)$ and (b) there is a missing - in front of the exponential inside the exponential...

This means that $S-\log\{\lambda\}$ has a fixed distribution with the above density and cdf $F$. From this representation, there exists a transform $G^{-1}\circ F$ (with $G$ being the cdf of the Gumbel distribution) that turns $ S-\log\{\lambda\}$ into a standard Gumbel, but this is not very useful!

Note that, in the mixture representation, $S=R^2$ is then distributed as an Exponential $\mathcal{E}(n)$ variate.

  • 1
    $\begingroup$ Thanks for the answer Xi'an, but if it were a Gumbel distribution, shouldn't there be a "-" in front of the second exp, i.e. $\exp(-\exp(-x)-x)$. $\endgroup$
    – fabee
    Commented Nov 25, 2014 at 16:00
  • $\begingroup$ Right! So this is an imaginary Gumble! (grumble, grumble) $\endgroup$
    – Xi'an
    Commented Nov 25, 2014 at 16:11
  • 2
    $\begingroup$ I think Grumble distribution would be an awesome name. :) $\endgroup$
    – fabee
    Commented Nov 25, 2014 at 16:25

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