Timeline for Entropy conjecture for distributions over $\mathbb{Z}_n$

6 events

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Mar 6, 2012 at 10:08	history	edited	Aaron Meyerowitz	CC BY-SA 3.0	added 169 characters in body
Mar 6, 2012 at 10:02	comment	added	Aaron Meyerowitz		You are correct. OK I am convinced. For what it is worth, I don't think that you can do better than that 1.332599 if you try to put values (a,b,b,c,c,c,c,c,c) on (0,3,6,1,2,4,5,7,8) (i.e. best is to put (c=0,b=0.1135,a=0.7433).
Mar 6, 2012 at 9:32	comment	added	VSJ		I just checked your example.. say $a = 0.8607538$ and $b = 0.017405778$. I took $p = [a,b,b,b,b,b,b,b,b]$ and $q = p \circledast p$. This turned out to be $q = [0.7433, 0.0321, 0.0321, 0.0321, 0.0321, 0.0321, 0.0321, 0.0321, 0.0321 ]$ whose entropy is pretty big, $1.5917$ to be exact. I hope I'm wrong, perhaps you could recheck your answer? Thanks for your comments.
Mar 6, 2012 at 8:48	history	edited	Aaron Meyerowitz	CC BY-SA 3.0	added 651 characters in body; deleted 286 characters in body
Mar 6, 2012 at 5:37	comment	added	VSJ		I have a specific example to counter the intuition that one would want to take $X$ and $Y$ to be supported on $\{0,g\}$. If you take $n=9$ (odd and composite) and $H(X)=H(Y)=1$. If we support $X$ and $Y$ on $\{0,1\}$, we get $H(Z) = H(1/4 1/2 1/2) = 1.5$. However if we support $X$ and $Y$ on $\{0,3,6\}$ we can take $p_X = p_Y = (0.1135, 0.1135, 0.7730)$ to get $H(Z) = 1.3326 < 1.5$.
Mar 6, 2012 at 4:38	history	answered	Aaron Meyerowitz	CC BY-SA 3.0