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I have a data set with approximately $10\ 000$ samples of a random variable which takes values in the interval $[0,1]$. I am interested in finding out if this variables follows any well-known distribution, even approximately.

I started by using SciPy's gaussian_kde() function to get a feel for what the pdf may look like (I have no problem assuming the pdf exists in this case). I also made some histograms--they looked similar in form to what the kde function produced.

I'm trying to figure out what function may fit this. The "head" of the distribution (near 0) can be well-approximated by a Gaussian, while the tail can be well approximated by a function of the form $$ \exp(-x^\gamma) $$ for some $0<\gamma<1$ (this curve is labeled ``log fit'' in the plots below—$\gamma\approx0.48$ in these plots). Does this look familiar to anyone?

I tried some subexponential distributions, including: Burr Type XII, Weibull, Levy, and log-normal, as well as a Beta distribution, but none give a satisfying fit. I know that these functions have support on $[0,\infty)$ (basically) not $[0,1]$, but I was thinking that there may be some way to re-normalize.

If it helps, when you look at the data, it seems that the gaussian_kde() function may be over-estimating the function at 0. It would be reasonable for the function $\rho$ that I'm looking for to satisfy $\rho(0)=0$. (I admit this is imprecise but I am looking for anything useful here.)

Fit of head of distribution Fit of tail of distribution

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  • $\begingroup$ It looks Gamma like except the support is wrong. Try transforming the data with $tanh$ or similar, and see what it looks like with semi-infinite support. $\endgroup$ – user121049 Dec 22 '17 at 9:19
  • $\begingroup$ Thanks for novel suggestion. I tried fitting gamma and loggamma by hand but neither one is seeming to do the trick. For gamma, the parameter has to be very large to agree with the peak of the distribution, but then it doesn’t fit the characteristics of the right tail at all. $\endgroup$ – David M. Dec 22 '17 at 22:51
  • $\begingroup$ I need to think more about the transform to semi-infinite support. $\endgroup$ – David M. Dec 22 '17 at 22:52

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