Timeline for For a martingale $f_0,f_1,\ldots $ how can we bound $P(\frac{1}{n} \|f_n\| \le 1$ for all $ n \ge N)$?
Current License: CC BY-SA 4.0
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| Jun 11, 2019 at 15:44 | comment | added | Daron | Thanks for the feedback everyone. I'm surprised no one has seen something like this before. I'd imagine if the answer was no, then you'd be able to demonstrate that using the normal coin-flip martingale, but that would involve some very clever recursion tricks. | |
| Jun 1, 2019 at 13:27 | comment | added | Daron | Hey Yuval, sure thing! | |
| Jun 1, 2019 at 13:27 | history | edited | Daron | CC BY-SA 4.0 |
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| Jun 1, 2019 at 1:54 | comment | added | Yuval Peres | Daron, Can you make your question more precise? The title line has parenthesis in the wrong place and later on when you write "The crudest thing we can do is take a union bound to get $$P\bigg (\frac{1}{n}\|f_n\| \ge 1\text{ for all }n\ge N\bigg) \le \sum_{n=N}^\infty\exp\left (-\frac{n}{2L^2} \right) ",$$ you are using a union bound for an intersection... Thanks. | |
| May 31, 2019 at 18:41 | comment | added | Iosif Pinelis | @MateuszKwaśnicki : I too think one cannot do much better than that, but proving rigorously "cannot do" statements is usually too hard and not very gratifying. | |
| May 30, 2019 at 8:06 | comment | added | Mateusz Kwaśnicki | Iosif Pinelis is a user here, @IosifPinelis , he may have some insightful comments. (My intuition is that one cannot do much better than in your first approach, but that is just intuition). | |
| May 30, 2019 at 6:17 | history | edited | Michael Hardy | CC BY-SA 4.0 |
missing parenthesis supplied
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| May 29, 2019 at 19:36 | history | edited | Daron | CC BY-SA 4.0 |
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| May 29, 2019 at 10:22 | history | edited | Daron | CC BY-SA 4.0 |
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| May 29, 2019 at 10:17 | history | asked | Daron | CC BY-SA 4.0 |