Timeline for On the 1/2 assumption on concentration of measure for continuous cube
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 7, 2018 at 21:15 | vote | accept | random_shape | ||
| Feb 7, 2018 at 20:50 | answer | added | Iosif Pinelis | timeline score: 6 | |
| Feb 7, 2018 at 16:48 | comment | added | random_shape | I meant to say, it is uniform probability measure. I tend to call uniform probability measure on a geometric objects with normalized volume. Sorry for the confusion. In here, the $\infty$ notation is just an annotation to contrast the cube with the $L^p$ case. On your second point, I think you got my inequality reversed. The inequality also appears at: www-users.math.umn.edu/~bobko001/papers/… Let me know if my question is still confusing | |
| Feb 7, 2018 at 16:46 | history | edited | random_shape | CC BY-SA 3.0 |
added 2 characters in body
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| Feb 7, 2018 at 8:30 | comment | added | Adrien Hardy | I'm puzzled by the statement: What is $\mu_\infty$? More precisely, since the (Lebesgue) volume of $[0,1]^n$ is one, what do you mean by normalized uniform measure? If it's the usual Lebesgue measure, then by taking $A:=\{(x_1,\ldots,x_n)\in[0,1]^n:\;0\leq x_1\leq \alpha\}$ with $\alpha\geq 1/2$, I obtain $\mu_\infty(A_\epsilon)=\alpha+\epsilon\leq 1-e^{-\pi\epsilon^2}$... | |
| Feb 7, 2018 at 1:16 | history | asked | random_shape | CC BY-SA 3.0 |