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I am trying to compare two PDFs using Kullback-Leibler divergence but I am getting a value which means they are almost identical. Am I missing something? Here is my code.

import numpy
import scipy.stats as ss
import matplotlib.pyplot as plt

a = numpy.random.rand(1000)
b = numpy.random.rand(650)
ag = ss.gaussian_kde(a)
bg = ss.gaussian_kde(b)
bk = numpy.linspace(numpy.min(b), numpy.max(b), 1000)
ak = numpy.linspace(numpy.min(a), numpy.max(a), 1000)
agv = ag(ak)
bgv = bg(bk)
e = ss.entropy(agv, bgv)
0.008704001913773865

plt.plot(ak,agv)
plt.show()

a - PDF

plt.plot(bk,bgv)
plt.show()

b - PDF

These two seems not equal. But why am I getting such a low entropy?

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    $\begingroup$ What does ss.entropy do? I'm not familiar with Python. In R I would integrate numerically density(a)*log(density(a)/density(b)), in 0,1. $\endgroup$ – utobi Nov 12 '16 at 6:45
  • $\begingroup$ @utobi docs.scipy.org/doc/scipy/reference/generated/… $\endgroup$ – Math1000 Nov 12 '16 at 8:51
  • $\begingroup$ Seems to do the right thing. However, although the KL value here may seem pretty small, it is not easy to say "how much small is small" ... $\endgroup$ – utobi Nov 12 '16 at 11:29
  • $\begingroup$ Is it too small? I mean both $a$ and $b$ are being sampled from uniform(0,1) $\endgroup$ – sntx Apr 11 '17 at 17:44
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That's the problem with divergences in general. As there isn't a scale for them it's not easy to say it's value is small.

What I use to do is computing KL for two PDFs I know should be close. Results lower than this indicate the two PDFs are close. This can be understood as a "calibration" for the measure.

Be aware that bigger the sample size, lower the divergence value. For instance, see how the color bar changes in the figures below (code based on yours: https://github.com/luizfrias/ica/blob/master/KL%20divergence%20test.ipynb)

enter image description here

enter image description here

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    $\begingroup$ In information theory, KL divergence does have a scale of bits. So if a and b were two probability distributions of byte values, to convert from a to b you'd need additionally to use 0.0087 bits /byte of data. And it is also bounded so that 0 < KL < 8 bits. $\endgroup$ – Paul Uszak Dec 19 '16 at 21:37

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