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R: Given a set of random numbers drawn from a continuous univariate distribution, find the distribution

Answer 1

+10 A:

My first approach would be to generate qq plots of the given data against the possible distributions.

x <- c(15.771062,14.741310,9.081269,11.276436,11.534672,17.980860,13.550017,13.853336,11.262280,11.049087,14.752701,4.481159,11.680758,11.451909,10.001488,11.106817,7.999088,10.591574,8.141551,12.401899,11.215275,13.358770,8.388508,11.875838,3.137448,8.675275,17.381322,12.362328,10.987731,7.600881,14.360674,5.443649,16.024247,11.247233,9.549301,9.709091,13.642511,10.892652,11.760685,11.717966,11.373979,10.543105,10.230631,9.918293,10.565087,8.891209,10.021141,9.152660,10.384917,8.739189,5.554605,8.575793,12.016232,10.862214,4.938752,14.046626,5.279255,11.907347,8.621476,7.933702,10.799049,8.567466,9.914821,7.483575,11.098477,8.033768,10.954300,8.031797,14.288100,9.813787,5.883826,7.829455,9.462013,9.176897,10.153627,4.922607,6.818439,9.480758,8.166601,12.017158,13.279630,14.464876,13.319124,12.331335,3.194438,9.866487,11.337083,8.958164,8.241395,4.289313,5.508243,4.737891,7.577698,9.626720,16.558392,10.309173,11.740863,8.761573,7.099866,10.032640)
> qqnorm(x)

For more info see link

Another possibility is based on the fitdistr function in the MASS package. Here is the different distributions ordered by their log-likelihood

> library(MASS)
> fitdistr(x, 't')$loglik
[1] -252.2659
Warning message:
In log(s) : NaNs produced
> fitdistr(x, 'normal')$loglik
[1] -252.2968
> fitdistr(x, 'logistic')$loglik
[1] -252.2996
> fitdistr(x, 'weibull')$loglik
[1] -252.3507
> fitdistr(x, 'gamma')$loglik
[1] -255.9099
> fitdistr(x, 'lognormal')$loglik
[1] -260.6328
> fitdistr(x, 'exponential')$loglik
[1] -331.8191
Warning messages:
1: In dgamma(x, shape, scale, log) : NaNs produced
2: In dgamma(x, shape, scale, log) : NaNs produced

midtiby 2010-04-18 07:36:55

You just beat me to it! That is exactly what I would do. A loop in order to find the min logLik. Beware! The OP has changed the values so your results aren't correct anymore. Using another software I actually found out that the best fit is given by an inverse normal distribution with parameters mu=9.976 and lambda = 42.411. Does fitdistr accept such a distribution?

gd047 2010-04-18 08:28:44

gd047: The number of observations was increased from 100 to 1000, but the underlying distribution is still the same.

knorv 2010-04-18 08:35:41

Answer 2

+7 A:

Another similar approach is using the fitdistrplus package

library(fitdistrplus)

Loop through the distributions of interest and generate 'fitdist' objects. Use either "mle" for maximum likelihood estimation or "mme" for matching moment estimation, as the fitting method.

f1<-fitdist(x,"norm",method="mle")

Use bootstrap re-sampling in order to simulate uncertainty in the parameters of the selected model

b_best<-bootdist(f_best)
print(f_best)
plot(f_best)
summary(f_best)

The fitdist method allows for using custom distributions or distributions from other packages, provided that the corresponding density function dname, the corresponding distribution function pname and the corresponding quantile function qname have been defined (or even just the density function).

So if you wanted to test the log-likelihood for the inverse normal distribution:

library(ig)
fitdist(x,"igt",method="mle",start=list(mu=mean(x),lambda=1))$loglik

You may also find Fitting distributions with R helpful.

gd047 2010-04-18 09:10:55

Answer 3

+3 A:

You could try using the Kolmogorov-Smirnov tests (ks.test in R).

If you have time-to-event data, here's software that does a Bayesian chi squared test against a list of common distributions to report the best fit.

John D. Cook 2010-04-18 12:10:11

+1 for Kolmogorov-Smirnov reference.

knorv 2010-04-19 23:06:34

Answer 4

+5 A:

I find it hard to imagine a realistic situation where this would be useful. Why not use a non-parametric tool like a kernel density estimate?

hadley 2010-04-18 13:56:52

Answer 5

A:

I don't think just comparing likelihoods is necessarily going to make sense, since some distributions have more parameters than others.

Glen_b 2010-05-02 05:33:59

Are you suggesting that extra parameters may allow one to "over-fit" the data, and thus achieve an undesireably low loglik?

unutbu 2010-05-02 14:17:06

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R: Given a set of random numbers drawn from a continuous univariate distribution, find the distribution

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