Timeline for Asymptotic behavior of a random geometric sum
Current License: CC BY-SA 4.0
10 events
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| Jun 21, 2020 at 2:17 | comment | added | Iosif Pinelis | @BenC. : The convergence of the expectations by itself may mean little about the convergence of the corresponding random variables, as it is the case here. Other than this, I have nothing to add to my answer and comments. | |
| Jun 20, 2020 at 20:13 | comment | added | MMM | I thought that this might be possible, because at least in expectation $E (q^{S_n}) / ((1+ q)/2)^n \to 1$. So I thought that maybe the same holds without the expectation... | |
| Jun 19, 2020 at 11:27 | comment | added | Iosif Pinelis | @BenC. : The obvious random $f(n)$ such that $q^{S_n}/f(n)\to1$ is $q^{S_n}$, and I don't think you can get anything more transparent than that. | |
| Jun 18, 2020 at 23:19 | vote | accept | MMM | ||
| Jun 18, 2020 at 23:19 | comment | added | MMM | Thanks. The part was with $q^{S_n} / q^{n/2} \to 1 $ was stupid. It seems that it is difficult to match the growth behavior of $q^{S_n}$. Would you have any idea for what (random) function $f(n)$, we have $q^{S_n} / f(n) \to 1$ a.s.? | |
| Jun 18, 2020 at 17:24 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jun 18, 2020 at 16:30 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jun 18, 2020 at 14:00 | comment | added | MMM | Thank you very much. My main question was still something more precise (I did not make it clear enough before). See my Edit. | |
| Jun 18, 2020 at 3:11 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Jun 18, 2020 at 2:29 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |