Timeline for Can we find such $k$ so that the following inequality holds?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| Dec 8, 2022 at 20:38 | comment | added | Iosif Pinelis | @Hermi : This follows by (i) what is oftentimes called the Bernstein--Chernoff inequality (en.wikipedia.org/wiki/Chernoff_bound#The_generic_bound) $P(Y\ge y)\le e^{-hy}\,Ee^{hY}$ for any random variable $Y$ and any real $h\ge0$ and (ii) the iid property of the $X_i$'s. Here, we take $Y:=\sum_{i=2}^k(1-X_i^2)$. | |
| Dec 8, 2022 at 18:54 | comment | added | Hermi | I mean the third line in $(5)$. | |
| Dec 8, 2022 at 15:44 | comment | added | Iosif Pinelis | @Hermi : (5) is a multi-line display. Which of the lines is/are unclear to you? | |
| Dec 8, 2022 at 4:19 | comment | added | Hermi | Thanks! Can I ask how to get inequality (5)? | |
| Dec 7, 2022 at 22:42 | comment | added | Iosif Pinelis | @Hermi : Of course, not the same, but similar results are possible. | |
| Dec 7, 2022 at 18:11 | comment | added | Hermi | Thank you! Can we also prove this same result if $X_i$ is replaced by an asymptotic normal distribution? For example, consider a sequence of n-dimensional random vectors $u, v_1, v_2,\dots, v_k$ (independent) uniformly distributed on the sphere. Let $X_i:=\sqrt{n}u\cdot v_i$. So $X_i\to N(0,1)$ for $i=1,\dots, k$ as $n\to \infty$ are i.i.d. asymptotic normal. | |
| Dec 7, 2022 at 15:48 | vote | accept | Hermi | ||
| Dec 7, 2022 at 4:23 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 4 characters in body
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| Dec 7, 2022 at 1:48 | comment | added | Iosif Pinelis | @Hermi : Any $\delta\ge1$ will trivially do, since any probability is $\ge0$. So, without loss of generality $\delta<1$ (and hence $\delta<2$). | |
| Dec 7, 2022 at 0:55 | comment | added | Hermi | Thank you very much! So in my case, it seems that $k\ge (\log(\delta/2))(1+\epsilon^2)+1$. So we need $\delta<2$? | |
| Dec 6, 2022 at 23:38 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |