For digital images (assume 2D gray scale image), the normalised intensity histogram is often treated as the probability distribution function of the intensity ie., intensity is treated as a random variable (RV).

  1. Where is the randomness in intensity coming from ?
  2. Should I treat this RV as the intensity value at any pixel in the overall image ie., can I infer this statement from the PDF - “choose any pixel. Probability that the intensity at that pixel has a value of 100 is 0.35” ? If so, it would seem that all pixels have the same PDF. Shouldn’t the overall structure of the image structure have a bearing on the PDF? eg., if the image is black at a pixel, shouldn't we would expect the PDF to be concentrated only around 0 at that pixel and zero everywhere else.

1 Answer 1


Well, If you model your image as a realization of a random variable generator then the Histogram is the best estimation (Assuming no other information like prior, etc..) you have for the PDF of the random variable.

For instance, you can see this model is used when doing Histogram Equalization (Transforming the realization into a realization of Uniformity Generator).

Pay attention that this is a very simple model.
For instance it doesn't take care of the correlation between adjacent pixels in image.

Indeed your interpretation is correct given the model.

  • $\begingroup$ Thanks for the answer. Actually, the reason for asking this question was Histogram Equalization - every explanation of it seems to involve PDF. Is there a reason why this simplistic probability model is chosen when explaining Histogram equalization ? Can we explain Histogram Equalization without involving probabilities ? $\endgroup$
    – shunya
    May 22, 2018 at 13:32
  • $\begingroup$ Because in those Element Wise operations we look at image as an array of numbers without even caring about their spatial relationship. For instance if you shuffle the pixel the Histogram will be the same and even the value of each pixel after Equalization will be the same. The current most effective way of dealing with large set of numbers is through Statistics. Hence we utilize those tools. $\endgroup$
    – Royi
    May 22, 2018 at 13:36
  • $\begingroup$ Apologies for not being clear in my earlier comment . My doubt is this - why do we explain Histogram equalization using probabilities ? Is there another intuitive way of understanding Histogram equalization without using probabilities ? $\endgroup$
    – shunya
    May 22, 2018 at 15:25

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