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Jun 15, 2020 at 7:27 history edited CommunityBot
Commonmark migration
Apr 20, 2020 at 21:36 history edited dohmatob CC BY-SA 4.0
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Apr 20, 2020 at 7:11 history edited dohmatob CC BY-SA 4.0
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Apr 20, 2020 at 7:05 history edited dohmatob CC BY-SA 4.0
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Apr 20, 2020 at 6:56 history edited dohmatob CC BY-SA 4.0
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Apr 19, 2020 at 20:51 history edited dohmatob CC BY-SA 4.0
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Apr 19, 2020 at 20:46 history edited dohmatob CC BY-SA 4.0
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Apr 19, 2020 at 17:23 history edited dohmatob CC BY-SA 4.0
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Apr 19, 2020 at 16:54 comment added dohmatob Oops, seems I went through all the trouble of establishing the bound (*) for nothing. Indeed, by symmetry of the sphere, for every unit vector $z \in \mathbb R^d$, the random variable $U^Tz$ has the same distribution as $U_1$ (the first coordinate of the random $n$-dimensional vector $U$). Thus, we can use the argument in mathoverflow.net/a/315232/78539 to get the (slightly more precise) bound $P(|U^Tz| > \delta) = P(|U_1| > \delta) = P(|\mathcal N(0, 1)| > \sqrt{n}\delta) + \mathcal O(1/n)$, and there are well-known tail bounds for the standard normal distribution $\mathcal N(0,1)$.
Apr 19, 2020 at 15:46 history edited dohmatob CC BY-SA 4.0
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Apr 19, 2020 at 15:40 history edited dohmatob CC BY-SA 4.0
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Apr 19, 2020 at 15:32 history edited dohmatob CC BY-SA 4.0
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Apr 19, 2020 at 15:08 history edited dohmatob CC BY-SA 4.0
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Apr 19, 2020 at 15:02 history edited dohmatob CC BY-SA 4.0
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Apr 19, 2020 at 14:38 history edited dohmatob CC BY-SA 4.0
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Apr 19, 2020 at 14:15 history edited dohmatob CC BY-SA 4.0
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Apr 19, 2020 at 13:54 history edited dohmatob CC BY-SA 4.0
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Apr 19, 2020 at 13:48 history edited dohmatob CC BY-SA 4.0
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Apr 19, 2020 at 13:40 history edited dohmatob CC BY-SA 4.0
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Apr 19, 2020 at 13:24 history edited dohmatob CC BY-SA 4.0
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Apr 19, 2020 at 12:57 history edited dohmatob CC BY-SA 4.0
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Apr 19, 2020 at 12:42 history asked dohmatob CC BY-SA 4.0