To compare distributions, it is common to use box blots.

I'm looking for a similarity measure that calculates whether distributions are the similar or not.

Ideally, given that e.g. four distributions are exactly the same, it would return 1. The more they are different, it would approach to 0.

To compute that, is it mandatory to have the same sample size?

Thanks for your help.

  • $\begingroup$ What do you mean by 'exactly the same characteristics'? $\endgroup$
    – Glen_b
    Commented Jan 13, 2014 at 11:31
  • $\begingroup$ I edited the question. $\endgroup$
    – semantium
    Commented Jan 13, 2014 at 11:39
  • $\begingroup$ So if I take a set of numbers and add the range (max-min) to every value to produce a second set which is identical in shape but doesn't overlap the first, under your criterion, the two would be very different (and so score near 0)? $\endgroup$
    – Glen_b
    Commented Jan 13, 2014 at 11:41

2 Answers 2


It's more common to measure discrepancy than similarity, but some of them can be converted easily to your way around.

Possible measures of discrepancy in distribution include (but are not limited to):

Kolmogorov-Smirnov distance. This distance between cdfs (or emprical cdfs), $D$, is small when the distributions are the same and close to 1 when they're very different, so $1-D$ should have the property you seek and doesn't require the same number of observations (indeed many of these measures don't).

Bhattacharyya distance. The Bhattacharyya coefficient, to which it is related (see the article) is a measure of similarity of distributions of the form you suggest.

Information-divergence. This is not symmetric (so D(x,y) is not D(y,x), and is not a metric), but it can be made symmetric (e.g. by looking at D(x,y)+D(y,x) for example) and there are some related metric distances to this divergence measure.

Chi-square distance: A variety of related measures get this name, used for discrete data (or discretized continuous data) -
$\quad$ I'll mention one: $d(x,y) = \frac{1}{2}\sum \frac{(x_i-y_i)^2 }{ x_i+y_i }$. This, as with the other chi-square distances, requires discretization into the same set of categories for both variables, and the x's and y's are proportions of their total category counts. This distance lies between 0 and 1, and is converted to a similarity by subtracting from 1.


You say "have exactly the same characteristics". So, if you can enumerate these characteristics that are important for you and the application at hand, then you can deploy a test for each one of those. Probably you would need to specify some tolerance level for each metric. You can decide that two distributions are the same if they pass all the tests. From a pair of distributions you can easily generalise to a set.

If you are after a single metric for comparing distributions you can go for the standard Kullback–Leibler divergence. Pay attention that the simple measure is not symmetric - although there is a symmetric form discussed in the wikipedia article.

  • 1
    $\begingroup$ Isn't KL Divergence valid only for probability distributions? $\endgroup$
    – otaku
    Commented Apr 30, 2020 at 21:18
  • $\begingroup$ "From a pair of distributions you can easily generalise to a set" - how can one do this? $\endgroup$
    – ForceBru
    Commented Mar 29 at 9:02

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