I have to compare two bimodal distributions. So far, I have found the Bimodality Coefficient:

$$ BC = \frac{\mu_{3}^{2} + 1}{\mu_{4} + 3\frac{(n-1)^2}{(n-2)(n-3)}}, $$

with $\mu_{3}$ referring to the skewness of the distribution and $\mu_{4}$ referring to its excess kurtosis, with both moments being corrected for sample bias using the sample size $n$.

The BC of a given empirical distribution is then compared to a benchmark value of $BC_{crit} = 5/9 \approx 0.555$ that would be expected for a uniform distribution; higher values point towards bimodality, whereas lower values point toward unimodality.

Anything else?

  • $\begingroup$ Kurtosis has little to do with bimodality, see Figure 2 in Westfall (2014, The American Statistician). That said, your $BC$ seems to only compare one distribution against the uniform - how would you use it to compare two distribution against each other? $\endgroup$ – Stephan Kolassa Oct 7 '15 at 8:26
  • $\begingroup$ And what exactly are you interested in? Do you know a priori that your distributions are bimodal? Are you most interested in the positions of the modes? Would one unimodal distribution $F_1$ with a mode between the two modes of a bimodal distribution $G$ be "more similar" to $G$ than a bimodal distribution $F_2$ with modes that are far away from the two modes of $G$? $\endgroup$ – Stephan Kolassa Oct 7 '15 at 8:30
  • $\begingroup$ It's not "my" BC: journal.frontiersin.org/article/10.3389/fpsyg.2013.00700/full However, I obtain a BC for each distribution and then I see if they are similar. $\endgroup$ – stochazesthai Oct 7 '15 at 8:31
  • $\begingroup$ In Figure 1, the unimodal distribution $C$ is closer to the bimodal distributions $B$ and $D$ than those two are to each other by the $BC$ (see my previous comment). Is that what you want? $\endgroup$ – Stephan Kolassa Oct 7 '15 at 8:36
  • $\begingroup$ The Wasserstein metric, also known as the earth mover distance may be helpful. It all depends on the relative importance you place on "similarity in bimodality" versus "similarity on mode position(s)"... $\endgroup$ – Stephan Kolassa Oct 7 '15 at 8:38

Given there are continuous bimodal distributions with exactly the same skewness and kurtosis as the normal, and others which have the same skewness but with either lower or higher kurtosis than the normal, I doubt that this statistic can be of much value in general.

In very limited circumstances - within particular families perhaps - it may provide some sort of value.

Consider the collection of distributions described here:

enter image description here

They all have the same "bimodality coefficient" as the normal distribution, including that bimodal one!

It's trivial to construct bimodal distributions that have lower values of the bimodality coefficient than the normal (which distribution has BC = $\frac{_1}{^3}$). For example, here's a very similar looking pair of distributions to the normal and the above bimodal one, but these have a lower bimodality coefficient.

enter image description here

Which means -- according to BC as a similarity measure -- that the unimodal distribution just above is more similar to the bimodal distribution beside it than the two very similar bimodal distributions in my answer are to each other, and the bimodal one just above is "less bimodal" according to BC than the normal distribution.

So I don't think that's a very useful way of judging "similarity" ... and I really don't think this bimodality coefficient is much good at telling unimodality from bimodality.


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