Timeline for Azuma's Inequality when the conditions hold with high probability?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Feb 21, 2020 at 0:51 | comment | added | Hafees Adebayo Yusuff | Can someone please tell me the statement of azuma inequality | |
| Aug 15, 2012 at 8:22 | vote | accept | Patt Geffrey | ||
| Aug 14, 2012 at 23:02 | answer | added | Brendan McKay | timeline score: 1 | |
| Aug 14, 2012 at 21:02 | comment | added | Bill Johnson | The absolute values $|d_k|$ of the martingale differences can have any joint distribution subject to the boundedness constraint, so you can make the exceptional sets disjoint for a long time and $|d_k|$ large on the $k$-th exceptional set $A_k$. Piecing things together, you get a counterexample if $\Bbb{P}(A_k)\to 0$ but is not summable. | |
| Aug 14, 2012 at 18:17 | comment | added | Mark Meckes | @Patt: As fuzzytron says, "almost surely" means with probability equal to 1. You apparently mean what people sometimes call "asymptotically almost surely". I doubt there is a nice result under such a weak assumption as you stated in the comment above. If you don't really need the martingale structure, Hoeffding's inequality for independent random variables extends to a subgaussian tail condition. | |
| Aug 14, 2012 at 17:41 | comment | added | Patt Geffrey | The sequence is not necessarily non-increasing as for two distinct values of $N$ and fixed $k$, $P(|X_k - X_{k-1}|)$ is not necessarily equal. | |
| Aug 14, 2012 at 17:00 | comment | added | Mateusz Wasilewski | But this sequence of probabilities is clearly nonincreasing, so it has to be constantly equal to 1; thus, assumptions of Azuma's inequality are satisfied. | |
| Aug 14, 2012 at 16:54 | comment | added | Patt Geffrey | If the martingales considered is $(X_k)_{k=1}^{N}$, what I mean is that $\mathbb{P}(\forall k, |X_k - X_{k-1}| < c_k) \to 1$ as $N$ goes to $\infty$. | |
| Aug 14, 2012 at 16:38 | comment | added | Mateusz Wasilewski | What do you mean by high probability? Do you just want $|X_k - X_{k-1}|< c_{k}$ to hold with probability tending to $1$, as $k \to \infty$? I guess, that when this convergence is fast enough, then some sort of Azuma's inequality still holds, but only for tails distant enough and, of course, with worse constants. I haven't thought about it too long, so I may be wrong. | |
| Aug 14, 2012 at 16:31 | comment | added | Patt Geffrey | Yes, I mean almost surely instead of surely. | |
| Aug 14, 2012 at 16:26 | comment | added | TerronaBell | Do you mean almost surely vs. surely? Almost surely implies probability 1... | |
| Aug 14, 2012 at 16:18 | history | asked | Patt Geffrey | CC BY-SA 3.0 |