Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

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=~300 (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%.

Is it possible to do something like this?

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer
Do I know all the $p_i$ values though a-priori? We worked this out in a thread here:… – HCAI Feb 10 '12 at 12:45
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 – HCAI Feb 10 '12 at 12:47
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. – Raskolnikov Feb 10 '12 at 12:51
Since the answers on Cross Validated were more complete, I'd suggest closing the question here. – Raskolnikov Feb 10 '12 at 12:52

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.