Timeline for Does strong stochastic ordering exist?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 4, 2023 at 20:18 | comment | added | Jinxiang Yao | Yes, that's great! I misunderstood symbols. | |
| Nov 4, 2023 at 19:58 | comment | added | Dieter Kadelka | My result is even stronger! Note that $\tilde \mu \not\leq \nu$ implies that $\tilde \mu \leq \tilde \nu$ cannot be true for all $\tilde \mu \in U$ and $\tilde \nu \in V$. | |
| Nov 4, 2023 at 19:51 | comment | added | Jinxiang Yao | It is right. But we have to prove that there are no such two subsets $U$ and $V$ of Probability measure space, such that $\mu\leq\nu$ for any $\mu\in U$ and $\nu\in V$. It seems not equivalent to the fact that for any neighborhood of any $\mu$, there is $\nu\in U$ such that $\mu$ is not order related to $\nu$. | |
| Nov 4, 2023 at 19:31 | comment | added | Dieter Kadelka | We have a metric $d$ and $d(\mu,U^c) = \inf \{d(\mu,\rho) \colon \rho \not\in U\}$. Note that my conctruction can be adapted (with some caution) to measures with bounded support. | |
| Nov 4, 2023 at 19:23 | comment | added | Jinxiang Yao | Thank you for your answer! What does "$\epsilon$ be the distance of $\mu$ and $U^c$" mean? | |
| Nov 4, 2023 at 18:27 | history | edited | Dieter Kadelka | CC BY-SA 4.0 |
added 48 characters in body
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| Nov 4, 2023 at 18:11 | history | edited | Dieter Kadelka | CC BY-SA 4.0 |
added 41 characters in body
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| Nov 4, 2023 at 18:05 | history | answered | Dieter Kadelka | CC BY-SA 4.0 |