Skip to main content
11 events
when toggle format what by license comment
Jul 9, 2023 at 11:45 vote accept Math_Y
Jul 9, 2023 at 11:44 comment added Math_Y Yes. $\alpha$-stable distributions.
Jul 7, 2023 at 14:42 comment added usul Does there exist a distribution that satisfies this?
Jul 7, 2023 at 4:18 answer added Fedor Petrov timeline score: 4
Jul 7, 2023 at 3:19 history edited Math_Y CC BY-SA 4.0
added 12 characters in body
Jul 6, 2023 at 20:32 comment added Michael Hardy I would have written $F(x)= \Pr(X>x) = x^{-0.5} \text{ for } x\ge 1$ (being explicit about that lower bound). And also "a sample of $N$ independent observations from this distribution" rather than "$N$ samples from this distribution." $\qquad$
Jul 6, 2023 at 18:04 comment added Math_Y Thank you. You are right, I fixed it. I will be grateful if you write the answer in detail. I would like to know the details of such an analysis.
Jul 6, 2023 at 18:02 history edited Math_Y CC BY-SA 4.0
edited body
Jul 6, 2023 at 10:18 comment added Fedor Petrov If it should read as $X\geq x$, then we have a large number of order $n^2$ with high probability, thus the terms which are greater than $\sqrt{n} $ dominate the sum, and we have $o(n) $ such numbers
Jul 6, 2023 at 10:11 comment added Fedor Petrov $F(x)=\mathbb{P}[X\leq x]=x^{-0.5}$ is not here a typo?
Jul 6, 2023 at 6:41 history asked Math_Y CC BY-SA 4.0