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I am clustering probability distributions using the Affinity Propagation algorithm, and I plan to use Jensen-Shannon Divergence as my distance metric.

Is it correct to use JSD itself as the distance, or JSD squared? Why? What differences would result from choosing one or the other?

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1 Answer 1

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I think it depends on how it is to be used.

Just for reference for other readers, if $P$ and $Q$ are probability measures, then the Jensen-Shannon Divergence is $$ J(P,Q) = \frac{1}{2} \big( D(P \mid\mid R) + D(Q\mid\mid R) \big) $$ where $R = \frac{1}{2} (P + Q)$ is the mid-point measure and $D(\cdot\mid\mid\cdot)$ is the Kullback-Leibler divergence.

Now, I would be tempted to use the square root of the Jensen-Shannon Divergence since it is a metric, i.e. it satisfies all the "intuitive" properties of a distance measure.

For more details on this, see

Endres and Schindelin, A new metric for probability distributions, IEEE Trans. on Info. Thy., vol. 49, no. 3, Jul. 2003, pp. 1858-1860.

Of course, in some sense, it depends on what you need it for. If all you are using it for is to evaluate some pairwise measure, then any monotonic transformation of JSD would work. If you're looking for something that's closest to a "squared-distance", then the JSD itself is the analogous quantity.

Incidentally, you might also be interested in this previous question and the associated answers and discussions.

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  • $\begingroup$ Cool, I will read "a new metric for probability distribution" as soon as possible. Txh $\endgroup$
    – ocram
    Feb 27, 2011 at 7:04
  • $\begingroup$ Thanks! I didn't realize that JSD itself is already analogous to dist**2 $\endgroup$ Feb 27, 2011 at 10:53
  • $\begingroup$ Thanks for the great explanation! Just a quick question. I know J-Divergence is symmetric in that J(P,Q) = J(Q,P). I read that JS divergence is symmetric in P and Q. Does this mean JS(P,Q) = JS(Q,P)? I am asking this because I am using the KLdiv function from the flexmix package in R. For my two distributions, the matrix output from KLdiv is not symmetric. I was expecting JS to correct this but the output from JS (computed using KL) is not symmetric. $\endgroup$
    – Legend
    Jan 16, 2013 at 20:33
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    $\begingroup$ @Legend: Yes, the JS divergence is symmetric. Hopefully that is easy to see from the equation given in the answer. Make sure that you are taking the KL divergence between $P$ and the midpoint measure and $Q$ and the midpoint measure for each of the two terms. Separately, neither will be symmetric, necessarily. $\endgroup$
    – cardinal
    Jan 18, 2013 at 17:33
  • $\begingroup$ What about bounds? Article on wikipedia claims that "For log base e, or ln, which is commonly used in statistical thermodynamics, the upper bound is ln(2)". What happen to the bounds when we use sqrt(J(P,Q)) ? $\endgroup$ Dec 28, 2021 at 19:14

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