Timeline for Exponential inequality for the sum of martingale differences $X_1, \dots, X_n$ when $\sum_{i=1}^{n} \operatorname{Var}(X_i) \leq B^2$
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Sep 26, 2020 at 20:23 | answer | added | Yuval Peres | timeline score: 2 | |
| Sep 25, 2020 at 11:23 | vote | accept | Siam | ||
| Sep 25, 2020 at 1:51 | answer | added | Iosif Pinelis | timeline score: 2 | |
| Sep 24, 2020 at 17:00 | history | edited | YCor | CC BY-SA 4.0 |
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| Sep 24, 2020 at 14:49 | comment | added | Siam | @BillJohnson, Azuma's inequality doesn't take into consideration the variance and therefore could be sub-optimal when the variance is small. In particular, if $B^2 \leq x$, then the upper bound in the question grows as $\exp(-x)$ instead of $\exp(-x^2)$ as suggested by Azuma's inequality | |
| Sep 24, 2020 at 14:30 | comment | added | Bill Johnson | Google "Azuma's inequality". | |
| Sep 24, 2020 at 13:32 | comment | added | Siam | @IosifPinelis, thank you. I have made the changes to my question so as to involve y in the bound. Could you please clarify if Corollary 2 can be extended to martingale difference sequences with $B^2$ being the upper bound on $\sum_{i=1}^{n} Var(X_i)$? | |
| Sep 24, 2020 at 13:27 | history | edited | Siam | CC BY-SA 4.0 |
As pointed out in the comments, I had missed adding "y" in the bound earlier. So corrected that now.
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| Sep 24, 2020 at 13:12 | comment | added | Iosif Pinelis | @Siam : For positive $y$, the right result would be Theorem 3 (or Corollary 2) in the paper you read. | |
| Sep 24, 2020 at 9:35 | comment | added | Siam | @BillJohnson, thanks. Could you please elaborate on that? My main concern is having the bound in terms of $Var(X_i)$ and not $Var(X_i|\mathcal{F_{i-1}})$. | |
| Sep 24, 2020 at 9:26 | comment | added | Siam | @IosifPinelis, thanks for your comment. I had misread Theorem 7 in epubs.siam.org/doi/abs/10.1137/1134032 earlier. I think it doesn't involve y (i.e. the bound on the value of individual $X_i$'s), because y is assumed to be 0. I see that you are the author of that paper. Could you please elaborate what Theorem 7 would look like if y is some positive number? I suppose the denominator in the RHS exponential will be something akin to 2(B^2 + yx). | |
| Sep 23, 2020 at 18:34 | comment | added | Bill Johnson | Under a stronger hypothesis you can get the conclusion you want from Azuma's inequality. | |
| Sep 23, 2020 at 17:12 | comment | added | Iosif Pinelis | Your bound will not hold even when the $X_i$'s are independent, because your bound does not involve $y$. | |
| Sep 23, 2020 at 17:05 | review | First posts | |||
| Sep 23, 2020 at 17:41 | |||||
| Sep 23, 2020 at 17:01 | history | asked | Siam | CC BY-SA 4.0 |