Timeline for Divergence of general random series and a special case
Current License: CC BY-SA 3.0
15 events
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| Feb 9, 2016 at 10:12 | vote | accept | Sosha | ||
| Feb 5, 2016 at 2:30 | history | edited | Sosha | CC BY-SA 3.0 |
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| Feb 5, 2016 at 2:20 | vote | accept | Sosha | ||
| Feb 5, 2016 at 2:20 | |||||
| Feb 5, 2016 at 1:58 | history | edited | Sosha | CC BY-SA 3.0 |
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| Feb 4, 2016 at 22:31 | comment | added | Robert Israel | @AnthonyQuas You can exceed lower bounds, but not all conditions are lower bounds. | |
| Feb 4, 2016 at 22:25 | answer | added | Robert Israel | timeline score: 1 | |
| Feb 4, 2016 at 21:10 | answer | added | Alexander Shamov | timeline score: 3 | |
| Feb 4, 2016 at 20:53 | answer | added | Mark Fischler | timeline score: 1 | |
| Feb 4, 2016 at 16:13 | comment | added | Anthony Quas | Certainly not! If the $X_n$ take a value $A_n$ with probability $2^{-n}$ and 0 otherwise, then the series is convergent by Borel-Cantelli, no matter what the dependence structure is. At the same time, by building large $A_n$'s, you can exceed any desired moment bounds. | |
| Feb 4, 2016 at 15:30 | history | edited | Sosha | CC BY-SA 3.0 |
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| Feb 4, 2016 at 13:02 | comment | added | Ori Gurel-Gurevich | Do you mean that the sum diverges almost surely? If so, then having constant expectation is not enough. | |
| Feb 4, 2016 at 11:58 | comment | added | user83457 | Probably not. It's easy to make it converge by making it almost always 0, say, multiply anything bi independent bernouillis with $\sum p_i < \infty$, but that leaves you a lot of room to pick the moments. | |
| Feb 4, 2016 at 10:37 | comment | added | Carlo Beenakker | $E(X_n)$ = constant $\neq 0$ | |
| Feb 4, 2016 at 8:47 | history | edited | Sosha | CC BY-SA 3.0 |
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| Feb 4, 2016 at 8:36 | history | asked | Sosha | CC BY-SA 3.0 |