Mathematica Stack Exchange is a question and answer site for users of Mathematica. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let's say I have some data from a real world system:

data = {0.026666156, 1.27421*^-6, 0.027878597, 0.017035598, 
   0.011036215, 0.038684388, 0.025161479, 0.021353902, 0.027487123, 
   0.021190747, 0.029328752, 0.014554109, 0.040037348, 0.020063044, 
   0.018363514, 0.0116034, 0.02609979, 0.024298555, 0.020568345, 
   0.017299039, 0.005427133, 0.023453297, 0.037443984, 0.031152865, 
   0.020158735, 0.014362383, 0.034687449, 0.01472421, 1.29928*^-6, 
   0.023887209, 0.014531724, 0.03055792, 0.027206875, 0.02227913, 
   0.01618167, 0.032540959, 1.26651*^-6, 0.021810796, 0.038698191, 
   0.014228248, 0.020261827, 0.031292827, 0.01142724, 0.039466342, 

If I generate histograms for this data it looks like this:

    Histogram[data, {.005}, "Probability", ImageSize -> Large], 
    Histogram[data, {0.005}, "CDF", ImageSize -> Large]


My question is twofold.

  1. Is there a way to "fit" this data with a distribution? Or make a pseudo-distribution with this data? I've found d = EmpiricalDistribution[data]; DiscretePlot[CDF[d, x], {x, 0, .05, .0001}], but is this the best way?

  2. How do you programmatically calculate the P50 value from this data? (50% confidence that the value will be realized. y = 0.5 on CDF plot)

share|improve this question
I'd also suggest looking at howto/DoStatisticalAnalysis (and keep a handy link to the reference). It's very useful. – R. M. Aug 30 '12 at 1:40
@R.M, I'll keep the link handy. And thanks for the quality editing. – kale Aug 30 '12 at 2:10
See empirical distribution, introduced by Kolmogorov: – alancalvitti Aug 30 '12 at 3:07
Related:… – Joe Aug 30 '12 at 7:34
up vote 19 down vote accepted

There are basically four ways you can do this:

  1. Discretely, using EmpiricalDistribution, as you suggest in your question
  2. Also discretely, but in a bucketed way, using HistogramDistribution
  3. Smoothly, using SmoothKernelDistribution or KernelMixtureDistribution.
  4. By fitting an assumed distribution, e.g. by using EstimatedDistribution, FindDistributionParameters and related commands.

Which you choose depends on the nature of your data and what you know about it. If you have reason to believe the data are distributed according to some known distribution, choose #4. Since your data are real-valued, it is likely that it comes from a continuous distribution and so choices #3 or #4 seem most appropriate.

Once you have your estimated distribution, you can calculate the P50 value using Quantile.

It is worth noting, though, that particular settings will give different estimates of the quantile. For example there is more than one way to define the kernel used in SmoothKernelDistribution. Using the data in your question:

Quantile[SmoothKernelDistribution[data ,  Automatic, #], 0.5] & /@
 {"Biweight", "Cosine", "Epanechnikov", "Gaussian", 
  "Rectangular", "SemiCircle", "Triangular", "Triweight"}

{0.0222671, 0.0223012, 0.0223522, 0.0223533, 0.0223458, 0.0223374,  0.0222653, 0.0222342}

Of course, in some fields, four decimal places is close enough.

share|improve this answer
"...using Quantile[]." - or InverseCDF[]. – J. M. Aug 30 '12 at 1:44
@Verbeia, Thanks for the answer. And yes, I usually work on stuff a mile or more underground, 4 decimals is plenty. – kale Aug 30 '12 at 2:08
@kale - you are most welcome. I count my own field (economics) as one where four significant figures is plenty. Indeed, more than three or four usually invites mockery and (justified) jibes of providing false precision. – Verbeia Aug 30 '12 at 2:16
@Verbeia, in addition to SmoothKernelDistribution, Quantile itself has a 4-dimensional parameter space (so much for "non-parameteric") that is not well-documented. Certain points in this parameter space are associated with named methods, but it's difficult to relate them or interpolated values. – alancalvitti Aug 30 '12 at 15:14

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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