Timeline for Prove an anti-concentration inequality for a martingale
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Nov 4, 2017 at 12:02 | comment | added | T.T | My question wasn't really precise. I knew how to compute the probability, but I didn't know how to combine it with Doob's inequality. But that's fine now, too. One just has to define $\tau_M := \inf \{t \ge \tau^-:\dots \}$ where $\tau^-$ is the first hitting time of $-\epsilon \sqrt{vl}$. Then one can can combine these results to get a lower bound on the probability of $X_l$ smaller than $-\epsilon/2 \sqrt{vl}$ conditioned on $\tau \le l$ and $X_{\tau} \le -\epsilon \sqrt{vl}$ and get the wanted result. | |
| Nov 4, 2017 at 11:15 | vote | accept | T.T | ||
| Nov 3, 2017 at 16:37 | comment | added | Serguei Popov | No, on step (4) you only use the Optional Stopping Theorem and the boundedness of jumps. Just think how would you solve the Gambler's Ruin Problem for equally strong players using the fact that the one-dimensional SRW is a martingale; that argument easily generalizes to any martingale with bounded jumps (you'll obtain 2 inequalities instead of 1 equality, but that's still OK). | |
| Nov 3, 2017 at 14:46 | comment | added | T.T | Thanks for your helpful answer, Serguei. I get the idea of the proof and I'm able to comprehend each of the steps formally, but I do not understand how you make sure that the probability of staying left of $-1/2\epsilon\sqrt(vl)$ conditioned on having hit $- \epsilon\sqrt(vl)$ before time $l$ is strictly positive (so how you combine steps (3)-(5)). It seems to me that you want to use some Markov-like property in step 4 (without assuming a Markov process?). | |
| Nov 1, 2017 at 22:54 | history | answered | Serguei Popov | CC BY-SA 3.0 |