Timeline for Bounding random process
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 11 at 19:27 | history | edited | tony | CC BY-SA 4.0 |
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| Jul 14, 2023 at 21:04 | vote | accept | tony | ||
| Jul 13, 2023 at 16:38 | comment | added | tony | ah ok. if I understood correctly: the goal is to derive bound on $\sup |X_s|$, am I correct? but then how does this relate to $\sup X_s$? | |
| Jul 13, 2023 at 15:45 | comment | added | Thomas Kojar | I didn't write $sup_{t}X_{t}$. I fixed $t_{0}$ because usually in a stochastic process the pointwise tail estimate is known (otherwise we can't even use chaining to estimate the supremum). I simply bounded $|X_{s}|\leq |X_{s}-X_{t_{0}}|+|X_{t_{0}}|$ and then took supremum over only the first one (which might be unnecessary). For the first you have a bound by $zero$ (and you can use modulus estimates). | |
| Jul 13, 2023 at 12:17 | answer | added | Iosif Pinelis | timeline score: 1 | |
| Jul 13, 2023 at 7:05 | comment | added | tony | @ThomasKojar Thanks! by using $\sup |X_s|\leq \sup_{s,t}|X_s-X_t|+\sup_tX_t$ we can only get information on $\sup |X_s|$ right? Then how can we know about $\sup X_s$? | |
| Jul 13, 2023 at 7:03 | history | edited | tony | CC BY-SA 4.0 |
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| Jul 13, 2023 at 0:12 | comment | added | Alf | The Lemma in question is stated for centered random variables, so the expectation of the supremum is always nonnegative. | |
| Jul 12, 2023 at 22:18 | comment | added | Thomas Kojar | in the case of a Gaussian process, you can just use symmetry $w=-w$ to get absolute value ie $P(\sup |w_{t}|>r)\leq 2P(\sup w_{t}>r)$. | |
| Jul 12, 2023 at 22:17 | comment | added | Thomas Kojar | $\sup |X_{s}|\leq \sup_{s,t} |X_{s}-X_{t}|+|X_{t_{0}}|$ | |
| Jul 12, 2023 at 22:12 | history | edited | tony | CC BY-SA 4.0 |
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| Jul 12, 2023 at 21:56 | history | asked | tony | CC BY-SA 4.0 |