Timeline for Intuitive Proof of Cramer's Decomposition Theorem
Current License: CC BY-SA 2.5
10 events
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| Sep 22, 2010 at 0:47 | comment | added | Per Vognsen | No, you're of course right, that doesn't make sense. Do you see a simple way of repairing the argument? The whole idea is based on the picture I have in my head of what happens when $X$ has less than maximum entropy. Then it can be smoothed further towards a Gaussian which in turn further smoothes out $X + Y$. | |
| Sep 21, 2010 at 21:52 | comment | added | fedja | "If one function is kept fixed while the entropy of the other is increased, the entropy of the convolution also increases". This would imply that convolving one function with two functions of the same entropy results in functions of the same entropy and, in one more step, that the entropy of the convolution depends only on the entropies of the functions convolved. Do you really believe that? | |
| Sep 21, 2010 at 9:18 | history | edited | Per Vognsen | CC BY-SA 2.5 |
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| Sep 21, 2010 at 9:16 | comment | added | Per Vognsen | Ah, I see what you mean now. I need to strengthen the inequality I outlined in the first paragraph. | |
| Sep 21, 2010 at 9:11 | history | edited | Per Vognsen | CC BY-SA 2.5 |
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| Sep 21, 2010 at 9:05 | history | edited | Per Vognsen | CC BY-SA 2.5 |
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| Sep 21, 2010 at 8:57 | comment | added | Benoît Kloeckner | I guess you should make the above argument more qualitative to work (you need to compare entropies of $X+Y$ and $X'+Y$, not only of $X$ and $X+Y$), but this seems interesting. | |
| Sep 21, 2010 at 8:40 | history | edited | Per Vognsen | CC BY-SA 2.5 |
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| Sep 21, 2010 at 8:35 | history | edited | Per Vognsen | CC BY-SA 2.5 |
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| Sep 21, 2010 at 8:28 | history | answered | Per Vognsen | CC BY-SA 2.5 |