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I have a probability mass function of some experimental data who's log looks like the following: (please ignore the fact that it is not normalized) ![alt text][1]

[image shack image removed]

(meaning if p(x) is the pmf, this is log(p(x)) ) Does anyone know what parametric family it might belong to? (note that this is a discrete distribution)

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    $\begingroup$ The graph kind of looks like a rotated Fermi-Dirac distribution... Can you give some information about where this data is coming from? Do you have any theoretical model to compare it to? $\endgroup$
    – Alicia Garcia-Raboso
    Feb 9, 2010 at 22:14
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    $\begingroup$ if anything it is more of an inverse sigmoid. It came from the distribution of certain image features after they have been quantized (it is the distribution of a sort of 'visual vocabulary'). The distribution is from a large database of images with no special characteristic (natural + synthetic images). $\endgroup$
    – liza
    Feb 9, 2010 at 22:30
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    $\begingroup$ So the y axis is log-number of features falling in a particular bin, and X is just feature index sorted by popularity? $\endgroup$
    – user3035
    Feb 9, 2010 at 22:44
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    $\begingroup$ Alas, without the image, it's pretty much impossible to tell what this old question is asking. The only answer below suffers from the same problem too. $\endgroup$
    – Ilmari Karonen
    Aug 29, 2015 at 21:27
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    $\begingroup$ Since without the image the question is meaningless, and since the OP has visited the site for the last time 4 years ago, I am voting to close as "unclear what you're asking". $\endgroup$
    – Stefan Kohl ♦
    Aug 29, 2015 at 22:35

1 Answer 1

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I tried taking the logarithm to the pdf of a $\beta(0.5, 1.5)$ (see Beta distribution) and it gave me this

[image shack image removed]

Maybe this can be fitted for your data.

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    $\begingroup$ The Beta-family seems a really good guess, though at this point it is just what it is: an educated guess. If liza could explain to us what and how has been measured, we might say more :). $\endgroup$
    – fedja
    Feb 22, 2010 at 15:59
  • $\begingroup$ Very interesting! I'll try it $\endgroup$
    – liza
    Feb 23, 2010 at 20:37

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