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Jun 19, 2013 at 14:35 vote accept user1234383
Jun 12, 2013 at 10:58 comment added user1234383 Ok, now I understand and thanks for the suggestion. Indeed I was trying to varying the size of the grid and I will try the rotation, now I get your point. I think that continuous limit of the entropy does not exist, at least does not exist trivially in fact for continuous distribution the generalization of the definition sum->integral result to be ill-defined.
Jun 12, 2013 at 9:24 comment added Has QUIT--Anony-Mousse Grids cause artifacts. If you change the grid, your get different artifacts.
Jun 12, 2013 at 9:02 comment added user1234383 why continuous notion of entropy? I don't get the point with this observation.
Jun 12, 2013 at 9:01 comment added user1234383 Although it's not perfectly clear to me which rotation you are proposing, in my case given the fact that I don't have many observations for building the empirical distribution on which I calculate the entropy, I fear that the rotation will produce an almost equal situation, however I will try. Thank you.
Jun 12, 2013 at 8:56 history edited Has QUIT--Anony-Mousse CC BY-SA 3.0
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Jun 12, 2013 at 8:50 comment added Has QUIT--Anony-Mousse Your grid will change due to the rotation, and I'm not aware of a continuous notion of entropy.
Jun 12, 2013 at 8:33 comment added user1234383 Thank you for your answer but entropy should be invariant under the rotation of the system that you propose and in general under permutation/rearrangement of the cells. $H(p1,p2,p3)= H(p2,p3,p1) = H(p2,p1,p3)$ and etc, even if I rotate the system the $p_i$ which enter in the sum defining the entropy are the same, therefore I would find the same result.
Jun 11, 2013 at 20:07 history answered Has QUIT--Anony-Mousse CC BY-SA 3.0