Timeline for Is entropy conserved under invertible mappings?
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 2 at 20:20 | comment | added | jbowman | @knrumsey - the accepted answer relies on the invertibility of a function on those values, but entropy isn't calculated using those values, it's calculated using the probabilities of those values. | |
| Feb 2 at 20:02 | comment | added | knrumsey | @jbowman, I agree with you that the values of the random variable are irrelevant. I do not readily see how that relates to a flaw in the accepted answer, however. | |
| Feb 2 at 19:51 | comment | added | jbowman | @knrumsey - the flaw is that entropy is calculated based on the probability distribution, not the values of the random variable, so the fact that the transform of the random variable is invertible is irrelevant. | |
| Feb 2 at 19:39 | comment | added | knrumsey | It appears that this answer is flawed, as pointed out in a new answer. For a linear transformation $f(x) = \mu + \sigma x$, it is easy to show that $H(f(X)) = \log\sigma + H(X)$, which contradicts this answer. I do not immediately see a flaw in your logic though -- do you have any thoughts on reconciling these contradictory claims? | |
| Feb 10, 2018 at 22:39 | vote | accept | CommunityBot | ||
| Feb 10, 2018 at 6:19 | history | answered | Ami Tavory | CC BY-SA 3.0 |