the Grumble distribution and an ODE

Posted in Books, Kids, R, Statistics, University life with tags , , , , , , on December 3, 2014 by xi'an

As ‘Og’s readers may have noticed, I paid some recent visits to Cross Validated (although I find this too addictive to be sustainable on a long term basis!, and as already reported a few years ago frustrating at several levels from questions asked without any preliminary personal effort, to a lack of background material to understand hints towards the answer, to not even considering answers [once the homework due date was past?], &tc.). Anyway, some questions are nonetheless great puzzles, to with this one about the possible transformation of a random variable R with density

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

into a Gumble distribution. While the better answer is that it translates into a power law,

$V=e^{e^{-R^2}}\sim q(v|\lambda)\propto v^{\lambda-1}\mathbb{I}_{(1,e)}(v)$,

I thought using the S=R² transform could work but obtained a wrong sign in the pseudo-Gumble density

$W=S-\log(\lambda)\sim \eth(w)\propto\exp\left(\exp(-w)-w\right)$

and then went into seeking another transform into a Gumbel rv T, which amounted to solve the differential equation

$\exp\left(-e^{-t}-t\right)\text{d}t=\exp\left(e^{-w}-w\right)\text{d}w$

As I could not solve analytically the ODE, I programmed a simple Runge-Kutta numerical resolution as follows:

solvR=function(prec=10^3,maxz=1){
z=seq(1,maxz,le=prec)
t=rep(1,prec) #t(1)=1
for (i in 2:prec)
t[i]=t[i-1]+(z[i]-z[i-1])*exp(-z[i-1]+
exp(-z[i-1])+t[i-1]+exp(-t[i-1]))
zold=z
z=seq(.1/maxz,1,le=prec)
t=c(t[-prec],t)
for (i in (prec-1):1)
t[i]=t[i+1]+(z[i]-z[i+1])*exp(-z[i+1]+
exp(-z[i+1])+t[i+1]+exp(-t[i+1]))
return(cbind(c(z[-prec],zold),t))
}


Which shows that [the increasing] t(w) quickly gets too large for the function to be depicted. But this is a fairly useless result in that a transform of the original variable and of its parameter into an arbitrary distribution is always possible, given that  W above has a fixed distribution… Hence the pun on Gumble in the title.