Timeline for Obtaining Chebyshev bound based thresholds for a particular tail probability using higher order moments
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 11, 2020 at 19:43 | comment | added | Venkatraman Renganathan | Thank you so much. That explanation was crystal clear. | |
| Dec 11, 2020 at 19:39 | comment | added | James Martin | To be precise I should have said "does not increase" rather than "decreases". And then indeed you get that $\alpha_k$ is non-decreasing in $k$. To answer your specific question, let $\mathcal{M}_k$ be the set of distributions matching the first $k$ moments. Then $\mathcal{M}_k \subseteq \mathcal{M}_2$, and so $\sup_{P\in\mathcal{M}_k} P(q>\alpha_2) \leq \sup_{P\in\mathcal{M}_2} P(q>\alpha_2) \leq \mathcal{A}$. That gives you that the threshold $\alpha_k$ is no higher than $\alpha_2$. | |
| Dec 11, 2020 at 18:18 | comment | added | Venkatraman Renganathan | I understand your point with the argument using sup definition on a smaller subset of distributions. Also, by "obtained maximum decreases", do you refer to the threshold $\alpha_{k}$ decreasing for a given same tail probability? Does that mean that my intuition is correct ? Sorry for troubling you. I am a newbie to this forum. | |
| Dec 11, 2020 at 17:26 | comment | added | James Martin | I'm having a hard time understanding what's at issue here. Any distribution agreeing with the first $k$ moments for $k>2$ automatically agrees with the first $2$ moments. So the "worst-case" distribution agreeing with the first $k$ moments can't be worse than the "worst-case" agreeing with the first $2$ only. As $k$ increases, you're maximising over a smaller set, so the obtained maximum decreases. | |
| Dec 11, 2020 at 17:24 | answer | added | Iosif Pinelis | timeline score: 1 | |
| Dec 11, 2020 at 16:51 | review | First posts | |||
| Dec 11, 2020 at 17:16 | |||||
| Dec 11, 2020 at 16:48 | history | asked | Venkatraman Renganathan | CC BY-SA 4.0 |