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This is my first post in CrossValidated and hope you can solve my problem. I've tried to calculate the bimodality coefficient (b) of two frequency distributions by using this formula:

enter image description here

where g is de skewness, K is the Kurtosis and n is the sample size. The b of a given emprirical distribution is then compared to a bencmark value of 0.555: higher numbers point toward bimodality whereas lower numbers point toward unimodality. I've been working with two frequency distributions in order to detect the degree of bimodality for each one. Here I show the two plots:

1)

enter image description here

2)

enter image description here

So, to have a quick look, it seems that the second distribution is more bimodal (two peaks) than the first one (hump-shaped). However, the bimodality coefficients do not say the same when I run in the R programme:

Kurtosis Skewness b 1 4.046225 1.927172 0.6602856 2 4.064557 1.843587 0.6145607

Apparently, both distributions are bimodal due to their b values are higher than 0.55. Seems logic for plot2 but plot1 has a clear unimodal distribution. Furthermore, I'm afraid that the values of Kurtosis and Skewness were incorrect. In theory, skewness values higher than 0 means that the distribution is right skewed but I could not see in the original plots. For Kurtosis >3 the distribution is leptokurkic which means to be sharper than a normal distribution.

Maybe I'm doing something wrong in the methodology to calculate Skewness and Kurtosis. I've used the e1071 package in R programme to calculate these variables by using the following script:

library("e1071")
  kurt<-kurtosis(frequency,type=2)
  skew<-skewness(frequency,type=2)

Formula type 2 for kurtosis: G_2 = ((n+1) g_2 + 6) * (n-1) / ((n-2)(n-3))

Formula type 2 for skewness: G_1 = g_1 * sqrt(n(n-1)) / (n-2)

It would be very useful to me if someone could explain me these results. Maybe I miss something in the methodology.

Thank you very much for your help!

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    $\begingroup$ I suggest that you give a source for this procedure, which strikes me as singularly bizarre. I see no reason why a complicated composite of kurtosis and skewness should be an especially good indicator of bimodality and the idea that something special happens around 0.555 is triply mysterious. An old-school view is that bimodality is evident on almost any distribution graph when it genuinely occurs. But those prejudices aside, documenting something as unusual as this is surely good practice. $\endgroup$
    – Nick Cox
    Commented Mar 12, 2015 at 10:16
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    $\begingroup$ Moment-based skewness and kurtosis results are frequently not as simple to interpret as many introductions imply. It's easily possible to get (e.g.) positive skewness even if a major mode is not in the left half in the distribution. Persistent debate over whether kurtosis measures peakedness or heavy tails can be resolved by showing examples where either or both can pull up kurtosis. Many threads here on such matters, as the tags imply: did you search previous posts? $\endgroup$
    – Nick Cox
    Commented Mar 12, 2015 at 10:21
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    $\begingroup$ Finding puzzling results is not a reason to suspect bugs in software like this. The problem lies in texts and courses that simplify a very complicated area; they have to do that, as it's their job to do so! The field is full of strange puzzles: e.g. for the Poisson, a textbook example of a right-skewed distribution except in the limit, it's possible and common for the median to exceed the mean! $\endgroup$
    – Nick Cox
    Commented Mar 12, 2015 at 10:23
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    $\begingroup$ A comment on Nick's comment, "Persistent debate over whether kurtosis measures peakedness or heavy tails can be resolved by showing examples where either or both can pull up kurtosis." Kurtosis does measure heavy tails. The supposed "counterexamples" only involve a measure of "heavy tails" that involves distributions crossing at some point. It does not correspond to the practical application of "heavy tails" which concerns potential outliers, which is in fact measured by the kurtosis statistic. $\endgroup$
    – BigBendRegion
    Commented Nov 20, 2017 at 1:23

1 Answer 1

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I agree with @NickCox : I think the mistake is in the first line of your post, where you define "bimodality coefficient". I Googled and found Pfister et al (which references SAS/STAT from 1990). That paper indicates problems with BC that are quite similar to the ones you found and recommends Hartigan's dip test, instead of BC (or in addition to it). The dip test is available in R through the diptest package. In addition, the kurtosis in the formula is supposed to be excess kurtosis and you appear to not have adjusted for that (although I am not certain of this)

The SAS documentation also mentions problems with BC, in particular

Very heavy-tailed distributions have small values of regardless of the number of modes.

The long tail of your second distribution is probably lowering the value of BC.

In short, the problem is in the formula, not in your code. There is, as far as I know, no perfect measure of the number of modes.

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  • $\begingroup$ Yes, I read Pfister et al 2013 despite facing problems with BC. I will use dip test with diptest package but as Pfister et al 2013 said: "both measures have merit for assessing bimodality but neither statistic is perfectly sensitive and specific at the same time". Keep working $\endgroup$
    – Rinot
    Commented Mar 12, 2015 at 14:39

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