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Is there a way to interpret the Jensen-Shannon divergence which is normalized to be between 0 and 1 between two probability distributions as a "percent difference", i.e., there is a x% difference between the two probability distributions.

I am asking this possibly strange question because I have to use the J-S distance in a business application, and the audience which is non-technical understands percentages better than "distances". :)

Thanks.

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    $\begingroup$ Interesting question, +1. Do you have a bunch of J-S distances in your presentation? $\endgroup$ – Dave Nov 26 '20 at 6:06
  • $\begingroup$ Yes, for example: 0.10389, 0.0897, etc... $\endgroup$ – Thomas Moore Nov 26 '20 at 6:06
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    $\begingroup$ One idea to consider is comparing all of the distances to some reference, maybe the biggest difference or the distance between to common distributions in your field. $\endgroup$ – Dave Nov 26 '20 at 6:08
  • $\begingroup$ okay, great, I can try that. Thanks. $\endgroup$ – Thomas Moore Nov 26 '20 at 6:12

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