Timeline for Distribution of running maximum of a local martingale
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
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| Sep 29, 2010 at 16:29 | history | edited | Steve Huntsman |
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| Jul 31, 2010 at 10:30 | vote | accept | kenneth | ||
| Jul 30, 2010 at 20:08 | comment | added | Jeff Schenker | Doh! Of course this has nothing to do with the behavior near zero. I read it all too quickly and imagined the question was about the typical time to reach zero. Thanks for explaining the notation. | |
| Jul 30, 2010 at 19:55 | answer | added | George Lowther | timeline score: 4 | |
| Jul 30, 2010 at 19:21 | comment | added | George Lowther | The bound $\mathbb{P}(X^*_T>K)\le x/K$ follows from the local martingale property. Stopped at the first time it hits K, its expectation is x, but is equal to K with probability at least $\mathbb{P}(X^*_\infty>K)$, giving the inequality (actually, it is Doob's maximal inequality). It is not possible to achieve this bound, so the question can be understood as asking if we can get very close to it in some sense. | |
| Jul 30, 2010 at 19:12 | comment | added | George Lowther | Actually, it is not going to depend on what $\sigma$ is like near zero much at all. It is a local martingale and, once it gets very close to zero, it is unlikely to escape. It depends more on how fast $\sigma$ grows as x goes to infinity. $C^{0,1/2}$ is the class of Holder continuous functions of exponent 1/2. This does guarantee a unique strong solution, but I don't think that has much bearing on the question. | |
| Jul 30, 2010 at 18:50 | comment | added | Jeff Schenker | It is pretty clear that the estimate you are after will depend strongly on the behavior of $\sigma$ near 0. Can you be more specific about what $C^{0,1/2}_{loc}$ is? | |
| Jul 30, 2010 at 17:08 | history | asked | kenneth | CC BY-SA 2.5 |