Timeline for Ratio of expectation involving random unit vectors
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 3, 2019 at 2:27 | vote | accept | neverevernever | ||
| Jul 1, 2019 at 1:50 | comment | added | fedja | @neverevernever "Comparable" means "bounded from both sides". I have estimated the probability from both above and below. | |
| Jul 1, 2019 at 1:43 | comment | added | neverevernever | I understand that we only need to look at $[0,t_0]$. I also know that on this interval, $\beta^{-q}$ is the upper bound of the integral. But why is $\beta^{-q}$ also the lower bound? | |
| Jul 1, 2019 at 1:19 | comment | added | fedja | @neverevernever By the integral identity I wrote. It does not matter what to integrate from $t_0$ to $\infty$: with either $t^q$ or some probability, the integral is exponentially small in $\beta$. And on $[0,t_0]$ the quantities are comparable. | |
| Jul 1, 2019 at 0:54 | comment | added | neverevernever | And how do we see $\mathbb{E}[\exp(-\beta Z)]$ is comparable to $\beta^{-q}$? | |
| Jul 1, 2019 at 0:07 | comment | added | fedja | @neverevernever The whole point is that it is the same. | |
| Jun 30, 2019 at 23:55 | comment | added | neverevernever | If $\mathbb{E}[\exp(-\beta Z)]$ is comparable to $\beta^{-q}$, the $q$ for the nominator and the denominator should be different right? | |
| Jun 28, 2019 at 18:58 | history | answered | fedja | CC BY-SA 4.0 |