Solved – How to get Bhattacharyya Distance in Excel (or Matlab, or R)

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I have several histograms which I would like to compare to one reference histogram to see which one is the most similar to the reference in terms of the shape of the distribution.

Kologorov-Smirnov gives unhelpful answers as often the largest difference in y-value is not indicative of worst fit. From inspection-by-eye, the shape of some this method says to be the worst fit are actually the most similar in shape, but with the distribution shifted up or down the x-axis.

Chi squared: not sure how I'd implement this as my data are continuous (continuous numerical data on x-axis, frequency of these, binned, on y-axis), so offput by the fact that the number of bins I arbitrarily choose to include has such a large effect, and don't know about degrees of freedom.

I don't need a good P-value, just to know qualitatively which of a variety of histograms are most similar to an initial histogram. None of the histograms follow a normal distribution or an approximation of a normal distribution.

From reading, the Bhattacharyya distance seems like a good way of getting what I am looking for, but I don't know how to get this from my excel histograms and data.

A description of how someone with excel and very limited matlab skills might compute bhattacharyya distance, or any other suggestions of how to qualitatively say which of several histograms is most similar in terms of shape to a reference histogram would be greatly appreciated.

Best Answer

Use the fpc package: https://cran.r-project.org/web/packages/fpc/fpc.pdf

library(fpc)
bhattacharyya.dist(mu1, mu2, covarianceMatrix, covarianceMatrix2)

Can use cov to help calculate covariance.

Related Solutions

Chi-Square Test – Comparing Two Histograms Using Chi-Square Distance

@Silverfish asked for an expansion of the answer by PolatAlemdar, which was not given, so I will try to expand on it here.

Why the name chisquare distance? The chisquare test for contingency tables is based on $$ \chi^2 = \sum_{\text{cells}} \frac{(O_i-E_i)^2}{E_i} $$ so the idea is to keep this form and use it as a distance measure. This gives the third formula of the OP, with $x_i$ interpreted as observation and $y_i$ as expectation, which explains PolatAlemdar's comment "It is used in discrete probability distributions", as for instance in goodness of fit testing. This third form is not a distance function, as it is asymmetric in the variables $x$ and $y$. For histogram comparison, we will want a distance function which is symmetric in $x$ and $y$, and the two first forms give this. The difference between them is only a constant factor $\frac12$, which is unimportant as long as you just chooses one form consistently (though the version with extra factor $\frac12$ is better if you want to compare with the asymmetric form). Note the similarity in these formulas with squared euclidean distance, that is not coincidence, chisquare distance is a kind of weighted euclidean distance. For that reason, the formulas in the OP is usually put under a root sign to get distances. In the following we follow this.

Chisquare distance is used also in correspondence analysis. To see the relationship to the form used there, let $x_{ij}$ be the cells of a contingency table with $R$ rows and $C$ columns. Denote the row totals be $x_{+j}=\sum_i x_{ij}$ and the column totals by $x_{i+}=\sum_j x_{ij}$. The the chisquare distance between rows $l,k$ is given by $$ \chi^2(l,k) = \sqrt{\sum_j \frac1{x_{+j}}\left(\frac{x_{lj}}{x_{l+}}-\frac{x_{kj}}{x_{k+}} \right)^2 } $$ For the case with only two rows (the two histograms) these recovers the OP's first formula (modulo the root sign).

EDIT

Answering to question in comments below: A book with long discussions of the chisquare distance is "CORRESPONDENCE ANALYSIS in PRACTICE (Second Edition)" by Michael Greenacre (Chapman & Hall). It is a well established name, coming from its similarity to chisquare as used with contingency tables. What distribution does it have? I have never studied that, but probably (under some conditions ...) it would have some chisquare distribution, approximately. Proofs should be similar to what is done with contingency tables, most literature about correspondence analysis does not go into distribution theory. A paper having some, maybe relevant such theory is ALTERNATIVE METHODS TO MULTIPLE CORRESPONDENCE ANALYSIS IN RECONSTRUCTING THE RELEVANT INFORMATION IN A BURT'S TABLE. Also see this other posts for some other relevant posts on this site.

Solved – the purpose of multiplying by the difference between the midpoints of two bins in this recipe

Since the histogram is a bar chart with area = height (yfit from dnorm) times base ”diff(h$mids[1:2])” the area converts the bar chart area to a probability area so final yfit (which is freq of occurrence) becomes probability (or area) times number of observations classical formula is

$prob = \frac{freq occurrence}{total possible occurrence}$

Here is the mapping to classical formula

yfit             =     yfit * diff(hmids[1:2]) *   length(x) 
freq occurrence  =     probability area         *   total occurrences

Best Answer

Related Solutions

Chi-Square Test – Comparing Two Histograms Using Chi-Square Distance

Solved – the purpose of multiplying by the difference between the midpoints of two bins in this recipe

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