Timeline for Showing $o(1)$ convergence for ratio of successive binomial tail probabilities
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 21, 2019 at 19:59 | comment | added | Fedor Petrov | @stats134711 looks ok, yes | |
| Jan 21, 2019 at 19:10 | comment | added | stats134711 | Thank you. I've tried to write my own understanding of the solution below in a separate answer. Could you comment if it is rigorous enough? | |
| Jan 19, 2019 at 15:12 | vote | accept | stats134711 | ||
| Jan 18, 2019 at 23:37 | vote | accept | stats134711 | ||
| Jan 19, 2019 at 15:12 | |||||
| Jan 18, 2019 at 23:37 | history | bounty ended | stats134711 | ||
| Jan 18, 2019 at 19:02 | comment | added | Fedor Petrov | It is essentially the same argument. We fix $M$ and prove that $\liminf \frac{P(X>c)}{P(X>c-1)}\geqslant 1-\frac1M$. After proving this, we may remember that $M$ could be fixed arbitrary, thus $\liminf \frac{P(X>c)}{P(X>c-1)}=1$. You suggest to say it another way: denote $\liminf \frac{P(X>c)}{P(X>c-1)}=1-\varepsilon$, then choose $M>\varepsilon^{-1}$ and get a contradiction. | |
| Jan 18, 2019 at 18:58 | comment | added | stats134711 | Would it be more rigorous to write, let $\varepsilon>0$, and choose $M>0$, such that $M>1/\varepsilon$? Then $(M+o(1))^{-1}<1/M<\varepsilon$. This would then result in $P(X=c)/P(X\geq c)<\varepsilon$. I guess I am hung up on the fact that say if $M$ is fixed at the beginning, then how can we make it arbitrary at the end? | |
| Jan 18, 2019 at 17:21 | comment | added | Fedor Petrov | Fixed $M$ proves that the lower limit is not less than $1-1/M$. Since $M$ is arbitrary, the lower limit equals 1. | |
| Jan 18, 2019 at 15:19 | comment | added | stats134711 | If $M>0$ is fixed, then isn't the conclusion that the ratio is $O(1)$ instead of the desired $o(1)$? | |
| Jan 17, 2019 at 19:54 | history | answered | Fedor Petrov | CC BY-SA 4.0 |