Timeline for Sum of RVs satisfying Bernstein condition on moments
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Oct 18 at 15:05 | comment | added | Yauhen Yakimenka | I agree with (ii). It can also be used earlier already (after "Observe that if $J$...") Thanks! | |
| Oct 18 at 15:01 | comment | added | Iosif Pinelis | I remove my objection (i). As for (ii), the result will be the same, but obtained more simply. | |
| Oct 18 at 14:40 | comment | added | Yauhen Yakimenka | (i) I guess I am not entirely sure I understood that point. Do you mean it's a mistake or just "it can be done earlier because it's obvious"? Can you please elaborate? (ii) It will give the same result, right? (iii) I will add more justification a bit later (I don't have my notes with me). (iv) Good point, I will change that, thanks | |
| Oct 18 at 13:35 | comment | added | Iosif Pinelis | This is very nice. However: (i) my previous point (i) has not been taken into account. (ii) After (3), you can just use $\sigma_i\sigma_j\le(\sigma_i^2+\sigma_j^2)/2$. (iii) How do you prove that $\frac{1}{w!} \binom{2l-w}{w-1} \le 2^{l-w}$? (iv) Instead of "we continue"'s, you can denote the expression by a symbol and then use that symbol. | |
| Oct 18 at 0:33 | comment | added | Yauhen Yakimenka | I changed it to be more explicit in the summations | |
| Oct 18 at 0:25 | history | edited | Yauhen Yakimenka | CC BY-SA 4.0 |
using $w_\mathrm{H}(i_1,\dotsc,i_n)$ in the summations
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| Oct 17 at 23:56 | comment | added | Yauhen Yakimenka | Not sure if it was clear, but $w=w_H(i_1,\dotsc,i_n)$ is just a notational shorthand in the expressions when we sum over the choices of $i_1,\dotsc,i_n$. Later, however, $w$ becomes just an iterator in the sums... | |
| Oct 17 at 23:46 | history | edited | Yauhen Yakimenka | CC BY-SA 4.0 |
typo: $\mu_i$ should be $\mu_j$
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| Oct 17 at 23:42 | comment | added | Yauhen Yakimenka | And I exclude $i_j=1$ in my second display after I explicitly mention the reason for that ($\mathbb E X_j =0$). Some places in my proof are a bit too detailed perhaps but well... Or do you mean it is a mistake to not exclude them in the very first display? (Then I don't see what the problem is...) | |
| Oct 17 at 23:38 | history | edited | Yauhen Yakimenka | CC BY-SA 4.0 |
1. Special case $k=3$ mentioned. 2. Notation $w_H$ explained.
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| Oct 17 at 23:35 | comment | added | Yauhen Yakimenka | $w_H$ is a Hamming weight (or support), i.e., the number of non-zeros among $i_1,\dotsc, i_n$. I will add an explanation about this, thanks! | |
| Oct 17 at 16:03 | comment | added | Iosif Pinelis | Looks intriguing. Two things so far: (i) You should exclude $i_j=1$ already in your first display. (ii) What is $w_H$? | |
| Oct 17 at 15:31 | history | answered | Yauhen Yakimenka | CC BY-SA 4.0 |