Timeline for Upper bound of the waiting time of a sum process
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 22, 2015 at 12:16 | vote | accept | Fisher | ||
| May 19, 2015 at 9:54 | comment | added | Fisher | @James Martin: The exact value is $\frac{143}{108} + \sum_{i=3}^\infty \frac{i!}{(i+1)^{i+1}}$, but it seems to coincide with your integral. The idea to choose one large $x_1$ and exponential decreasing small $x_i$ is the same. | |
| May 19, 2015 at 1:18 | comment | added | Michael | The Lorden's inequality (a bound on the overshoot of Wald-type problems) looks relevant here: en.wikipedia.org/wiki/Lorden%27s_inequality | |
| May 19, 2015 at 1:11 | answer | added | Michael | timeline score: 1 | |
| May 18, 2015 at 21:58 | comment | added | James Martin | Is your $\approx 1.36$ in fact $\int_0^\infty \frac{e^{-x}}{1-xe^{-x}} dx$? This is what I get in the limit with $n$ large, $\epsilon$ small, and $x_i=(1-\epsilon)\epsilon^{i-1}$. I also haven't seen anything bigger.... | |
| May 18, 2015 at 12:07 | history | asked | Fisher | CC BY-SA 3.0 |