Timeline for Asymptotic bound of quotient of absolute and squared deviation from mean
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 1, 2021 at 7:12 | vote | accept | AlbertRapp | ||
| Jun 30, 2021 at 16:05 | comment | added | Iosif Pinelis | @AlbertRapp : (iii) "When bounding the denominator, you started $k$ at $3N/4$. Heuristically, I understand that then $x_k-\bar x>0$ which we need to use the inequality we found due to Jensen. However, is that clear formally?" -- Here we are using the following: if $3N/4\le k\le N$, then $x_k-\bar x\ge\ln(S+k-1)-\ln(S+(N-1)/2)=\ln(1+\frac{k-1-(N-1)/2}{S+(N-1)/2})\ge0$. | |
| Jun 30, 2021 at 16:05 | comment | added | Iosif Pinelis | @AlbertRapp : (i) "How did you arrive at $N-k$ in the maximum?" -- We have $x_N-x_k=\ln(1+\frac{N-K}{S+k-1})\le\ln(1+\frac{N-K}S)$. (ii) "Are the $o(1)$ necessary?" -- They are needed already because I only assume that $N\sim(m-1)S$. Also, if $3N/4\le k\le N$, then $k-1-(N-1)/2\ge N/4-1/2=N/(4+o(1))$. | |
| Jun 30, 2021 at 15:42 | comment | added | AlbertRapp | When bounding the denominator, you started $k$ at $3N/4$. Heuristically, I understand that then $x_k - \bar{x} > 0$ which we need to use the inequality we found due to Jensen. However, is that clear formally? | |
| Jun 30, 2021 at 15:40 | comment | added | AlbertRapp | Are the $o(1)$ necessary? I do not understand the purpose. | |
| Jun 30, 2021 at 15:39 | comment | added | AlbertRapp | Great solution. Thanks. However, a few comments: How did you arrive at $N - k$ in the maximum? I could only bound $x_N - x_k$ by $\log(1 + (N - 1)/S$. In this case, it does not matter though since this bounds $\log(1 + (k - 1)/S$ from above, so we can bound the maximum accordingly. I lose a factor 1/2 though because then I only bound the sum by $N(N-1)$ instead of calculating the sum explicitly. | |
| Jun 30, 2021 at 14:08 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |