I'd like to fit a distribution (any you like) based on these requirements:

  1. Produces integer values (preferable but not required)
  2. Mean: $\mu=100$
  3. Std $=114$
  4. Quantiles $( 25\%, 50\%, 75\%)=(6,39,200)$
  5. $\min=0$; $\max\approx300$ (but $\infty$ is acceptable);

Poisson fits criteria 1,2 and nearly 3, but not 4 by far. Lognormal... maybe.

EXTRA info: value $0$ is produced $8\%$ of the time and $300$ is produced $18\%$ of the time.

Is it possible to do something like this?


1 Answer 1


It's certainly possible, but you have more unknowns (301) than constraints (9), which means that you'll have a lot of freedom. One way of restricting that freedom is to use the maximum entropy principle. It will give you the most likely distribution under the constraints.

What you do is maximize $-\sum_i p_i \ln p_i$ under the constraints you have given and where the $p_i$ are the probabilities of all discrete events $i=0,\ldots,300$. You'd best do this numerically.

  • $\begingroup$ Do I know all the $p_i$ values though a-priori? We worked this out in a thread here: stats.stackexchange.com/questions/21638/distribution-fitting/… $\endgroup$
    – HCAI
    Feb 10, 2012 at 12:45
  • $\begingroup$ Eventually we used a graph digitizer software to manually extract the data points from a printed graph and create a distribution. The results were similar (but significantly different) but I'm interested in the maximum entropy principle non-the less $\endgroup$
    – HCAI
    Feb 10, 2012 at 12:47
  • $\begingroup$ I see, my suggestion is the same as Xi'an's. The maximum entropy method is a method to generate a priori distributions. But seeing the comments you got on CrossValidated, it seems that you actually have the full data set, so ME is not really the approach you need. The comments of shujaa seem to me more to the point. $\endgroup$ Feb 10, 2012 at 12:51
  • $\begingroup$ Since the answers on Cross Validated were more complete, I'd suggest closing the question here. $\endgroup$ Feb 10, 2012 at 12:52

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