7
$\begingroup$

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
$$p(r|n)=2nr\cdot\exp\left(-nr^{2}\right)$$

($n\in \mathbb N_+$) 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.

$\endgroup$
4
  • 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

5
$\begingroup$

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.

$\endgroup$
3
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.