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Aug 25, 2021 at 17:32 history closed user44191
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Aug 24, 2021 at 13:35 comment added TheGrandDuke @user44191 Classical definition of probability is acceptable. However, does the u.d. mod 1 property and the probability in that matter hold for not fixed numbers?
Aug 24, 2021 at 12:47 comment added user44191 Presumably, $P(\{a_n\} > \frac{1}{n})$ is meant to be something like $\frac{|\{1 \leq i \leq n | a_i > \frac{1}{n}\}|}{n}$. Or do you specifically want to match the term $a_n$ with the denominator of the fraction $\frac{1}{n}$?
Aug 24, 2021 at 12:43 history edited YCor
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Aug 24, 2021 at 12:24 comment added TheGrandDuke @RonniePavlov thank you very much for the comment. I edited the question and tried to clarify the confusion.
Aug 24, 2021 at 12:22 history edited TheGrandDuke CC BY-SA 4.0
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Aug 23, 2021 at 16:39 comment added Ronnie Pavlov In case this is the point of confusion: what do you mean by "probability that $a_n > \epsilon$" here? A sequence doesn't inherently come with a probability! Generally, one would associate the Lebesgue measure, since it's the limit of the empirical measures $\frac{1}{n} \sum_{i=0}^{n-1} \delta_{a_n}$ when $(a_n)$ is uniformly distributed. Then, indeed, the answer is trivial, since $P(x > \epsilon) = 1-\epsilon$.
Aug 23, 2021 at 13:51 review Close votes
Aug 25, 2021 at 17:32
Aug 23, 2021 at 13:49 comment added TheGrandDuke Can you give me a reference to this fact?
Aug 23, 2021 at 13:47 comment added Random I don't think this really appropriate for mathoverflow, but the answer is yes: for each $\frac{1}{2} > \varepsilon > 0$ we have $\mathbb{P} \left( \{ a_n \} > \varepsilon \right) \to 1 - 2 \varepsilon$. Taking $\varepsilon$ arbitrarily small we get the desired result.
Aug 23, 2021 at 13:38 comment added TheGrandDuke My intuition says yes, but I don't want to reinvent the wheel, if someone already did it.
Aug 23, 2021 at 13:34 history edited TheGrandDuke CC BY-SA 4.0
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Aug 23, 2021 at 13:30 comment added user44191 This seems like it should be trivial from the definition of "u.d. mod 1". No? (I assume you mean something equivalent to "equidistributed sequence").
Aug 23, 2021 at 13:04 review First posts
Aug 23, 2021 at 13:30
Aug 23, 2021 at 13:03 history edited TheGrandDuke CC BY-SA 4.0
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Aug 23, 2021 at 12:57 history asked TheGrandDuke CC BY-SA 4.0