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I have the following histogram of count data. And I would like to fit a discrete distribution to it. I am not sure how I should go about this. enter image description here

Should I first superimpose a discrete distribution, say Negative Binomial distribution, on the histogram so that I would obtain the parameters of the discrete distribution and then run a Kolmogorov–Smirnov test to check the p-values?

I am not sure if this method is correct or not.

Is there a general method to tackle a problem like this?

This is a frequency table of the count data. In my problem, I am only focusing on non-zero counts. 

  Counts:     1    2    3    4    5    6    7    9   10 
 Frequency: 3875 2454  921  192   37   11    1    1    2

UPDATE: I would like to ask: I used the fitdistr function in R to obtain the parameters for fitting the data.

fitdistr(abc[abc != 0], "Poisson")
     lambda  
  1.68147852 
 (0.01497921)

I then plot the probability mass function of Poisson distribution on top of the histogram. enter image description here

However, it seems like the Poisson distribution fails to model the count data. Is there anything I can do?

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    $\begingroup$ A general method is to use maximum likelihood to fit a candidate distribution. What you mean by superimposing a distribution to obtain the parameters isn't clear, but if you mean guessing parameter values until you get a good fit that's a lousy method. Kolmogorov-Smirnov isn't useful here. You need decent software that gives you inferential results, so you need to indicate your software of choice so that people using that can try to help you. Your histogram isn't clear, but if there gaps then no distribution is likely to fit well. $\endgroup$ – Nick Cox Jul 31 '13 at 22:05
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    $\begingroup$ While using a KS test in that manner is a lousy method (and in any case the KS test is not for discrete distributions), it would be possible to estimate parameters by minimizing the KS statistic over all possible parameter values; but if you're going that way (optimizing some goodness of fit), minimum chi-square would be the more typical approach. As Nick Cox suggests ML would be the obvious thing to do, almost certainly more efficient, easier to get standard errors out of, and more readily accepted by others. (There are other possibilities, like method of moments, but ML is the main thing.) $\endgroup$ – Glen_b Jul 31 '13 at 22:15
  • $\begingroup$ I am using R. When you say estimating MLE, is there any algorithms that you will recommend for the job? And after finding the ML, what should I do next ? $\endgroup$ – user1769197 Jul 31 '13 at 22:32
  • $\begingroup$ I'd start here ?MASS::fitdistr, since it's already in your R distribution (see the final example at the bottom; see rnegbin for more information about that parameterization of the Negative Binomial). .... "And after finding the ML, what should I do next?" -- well at that point you have parameter estimates and standard errors. Beyond that, what do you want to achieve? -- I can't guess. $\endgroup$ – Glen_b Jul 31 '13 at 23:25
  • $\begingroup$ It occurs to me that you may have been trying to ask 'how do I assess the fit of my model?'. If that's the case, could you update your question to reflect that? $\endgroup$ – Glen_b Aug 1 '13 at 0:31
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Methods of fitting discrete distributions

There are three main methods* used to fit (estimate the parameters of) discrete distributions.

1) Maximum Likelihood

This finds the parameter values that give the best chance of supplying your sample (given the other assumptions, like independence, constant parameters, etc)

2) Method of moments

This finds the parameter values that make the first few population moments match your sample moments. It’s often fairly easy to do, and in many cases yields fairly reasonable estimators. It’s also sometimes used to supply starting values to ML routines.

3) Minimum chi-square

This minimizes the chi-square goodness of fit statistic over the discrete distribution, though sometimes with larger data sets, the end-categories might be combined for convenience. It often works fairly well, and it even arguably has some advantages over ML in particular situations, but generally it must be iterated to convergence, in which case most people tend to prefer ML.

The first two methods are also used for continuous distributions; the third is usually not used in that case.

These by no means comprise an exhaustive list, and it would be quite possible to estimate parameters by minimizing the KS-statistic for example – and even (if you adjust for the discreteness), to get a joint consonance region from it, if you were so inclined. Since you’re working in R, ML estimation is quite easy to achieve for the negative binomial. If your sample were in x, it’s as simple as library(MASS);fitdistr (x,"negative binomial"):

> library(MASS) 
> x <- rnegbin(100,7,3)
> fitdistr (x,"negative binomial")
     size         mu    
  3.6200839   6.3701156 
 (0.8033929) (0.4192836)

Those are the parameter estimates and their (asymptotic) standard errors.

In the case of the Poisson distribution, MLE and MoM both estimate the Poisson parameter at the sample mean.

If you'd like to see examples, you should post some actual counts. Note that your histogram has been done with bins chosen so that the 0 and 1 categories are combined and we don't have the raw counts.

As near as I can guess, your data are roughly as follows:

    Count:  0&1   2   3   4   5   6  >6    
Frequency:  311 197  74  15   3   1   0

But the big numbers will be uncertain (it depends heavily on how accurately the low-counts are represented by the pixel-counts of their bar-heights) and it could be some multiple of those numbers, like twice those numbers (the raw counts affect the standard errors, so it matters whether they're about those values or twice as big)

The combining of the first two groups makes it a little bit awkward (it's possible to do, but less straightforward if you combine some categories. A lot of information is in those first two groups so it's best not to just let the default histogram lump them).

* Other methods of fitting discrete distributions are possible of course (one might match quantiles or minimise other goodness of fit statistics for example). The ones I mention appear to be the most common.

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  • $\begingroup$ +1, nice info. Just out of curiosity, why do you often seem to use 1), instead of the markup numbering that CV supports (ie, 1.--which leads to indenting)? $\endgroup$ – gung - Reinstate Monica Aug 1 '13 at 2:22
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    $\begingroup$ @gung Mostly, I don't think about it - I prefer my lists to look the way I type them; but when I do think about it I find markdown's editing of the numbers I type to what it thinks they should be greatly annoying (if I typed "36.", I really didn't mean "1.") ... so when it does occur to me to use "1." rather than "1)", I often end up typing "1\." explicitly to stop it from recognizing it and doing things I didn't intend. I got so attuned to this behavior from 5 years of reddit that I don't even think about it - I'd end up automatically interfering with it by force of habit. $\endgroup$ – Glen_b Aug 1 '13 at 2:29
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    $\begingroup$ @gung If the indenting was much smaller, I'd probably be inclined to put up with the renumbering-hazard and use it. As it stands, I usually find it distracting when I see it. But in this particular case I was looking to make them look more like subheadings, so I'd probably have avoided it even if I wasn't put off by the behavior. (On the other hand, on the occasions other people have decided to edit my lists to make it do the indenting, I generally leave them to please themselves. If they want to make it look a certain way, they can as long as it doesn't alter the meaning.) $\endgroup$ – Glen_b Aug 1 '13 at 2:32
  • $\begingroup$ I'll try to restrain myself ;-). $\endgroup$ – gung - Reinstate Monica Aug 1 '13 at 2:47
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In an edit, you gave some data, and added a new question:

"This is a frequency table of the count data. In my problem, I am only focusing on non-zero counts.

   Counts:     1    2    3    4    5    6    7    9   10 
Frequency:  3875 2454  921  192   37   11    1    1    2

Can someone give me an example of how you would carry out the chi-squared goodness of fit test here?"

This leads to further comments:

  1. Having zeros but wanting to ignore them can make sense, but generally statistical and subject-matter people would want to see a good reason for why.

  2. If you choose to ignore zeros, you are placing yourself in difficult territory, as you can't just fire up routines for e.g. Poisson or negative binomial if you leave out the zeros. Well, you can, but the answers would be wrong. You need special purpose functions or commands for distributions such as the zero-truncated Poisson or zero-truncated negative binomial. That's challenging stuff and needs dedicated reading to be clear on what you are doing.

  3. Asking how to do a chi-square test suggests to me that you have not really understood what I said very briefly and @Glen_b said in much more detail (and, to my mind, very clearly). Splitting that in two:

    • There can be no chi-square test without expected frequencies and there can be no expected frequencies without parameter estimates. It may be that you are most familiar with chi-square test routines in which independence of rows and columns in a two-way table is tested. Although that is the chi-square test most met in introductory courses, it is actually very unusual among chi-square tests in general in that the usual software in effect does the parameter estimation for you and thereby gets the expected frequencies. Beyond that, in most more complicated problems, such as yours, you have to get the parameter estimates first.

    • A chi-square test isn't wrong, but if you estimate parameters by maximum likelihood it's irrelevant as the fitting routine gives you estimates and standard errors and allows tests in their wake. @Glen_b gave an example already in his answer.

A side-issue is that it would be clearer to tweak your histograms to respect the discreteness of the variable and show probabilities, not densities. The apparent gaps are just artefacts of default bin choice not respecting the discreteness of the variable.

UPDATE: The supplementary question about a chi-square test has now been deleted. For the moment I am letting #3 above stand, in case someone else follows the same path of wanting a chi-square test.

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  • $\begingroup$ I have to ignore the zeros, because I am trying to model active counts. Counts = 0 is referred to as inactive counts. $\endgroup$ – user1769197 Aug 1 '13 at 9:08
  • $\begingroup$ That's a substantive choice. Note that in many fields there are so-called two part models, in which you model (in your terms) active versus inactive and then how active. $\endgroup$ – Nick Cox Aug 1 '13 at 9:11
  • $\begingroup$ I tried to tweak the histograms by doing "plot(table(abc), type = "h")". But, I am not sure how can I get it to show probabilities $\endgroup$ – user1769197 Aug 1 '13 at 9:47
  • $\begingroup$ I don't use R, but you can get advice on that. You might need to ask separately. $\endgroup$ – Nick Cox Aug 1 '13 at 9:48

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