Timeline for Do i.i.d. sums concentrate any faster than martingales?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 29, 2019 at 10:58 | comment | added | Daron | Thanks. The new question can be found here: mathoverflow.net/questions/332756/… | |
| May 29, 2019 at 0:02 | comment | added | Iosif Pinelis | @Daron : Yes, I think it is always best, in more ways than one, to ask additional questions separately. | |
| May 28, 2019 at 19:01 | comment | added | Daron | Doing the union bound gives something like $C e^{-cN}$ for some constants $C,c>0$. I tried to be clever and instead force some subsequence of $\|f_n\|/n$ to be less than $1/2$ say because this forces the elements in between to be less than $1$. That leads to integrating something like $\exp(ab^x)$ and I can only get a bound like $C e^{-cN}/N$. I can write this out as a full question if you like? | |
| May 28, 2019 at 18:57 | comment | added | Daron | I wonder could you provide any reference for the following type of Martingale problem? Suppose $f_0,f_1,f_2,\ldots$ is an (infinite) martingale starting at $f_0=0$ and we want to bound the chance that after some big $N \in \mathbb N$ all the normalised values $\|\frac{f_n}{n}\|$ are less than $1$ say. We can of course take the exponential bound from one of your theorems, do a union bound, and estimate that by integrating the exponential. But that seems a pretty crude way to go about things. Are there any more sophisticated techniques available? | |
| May 26, 2019 at 17:41 | comment | added | Iosif Pinelis | @Daron : Concerning your latest comment: that is right. | |
| May 26, 2019 at 16:18 | comment | added | Daron | To write out the restriction in full, should it be $\sup\{\|X(x)\|_2: x \in \Omega\} \le 1/2$ where $\Omega$ is the probability space and $\|\cdot\|_2$ is the $2$-norm on $\mathbb R^d$? | |
| May 26, 2019 at 16:16 | vote | accept | Daron | ||
| May 26, 2019 at 16:14 | comment | added | Daron | Thank you for pointing this out! Clearly I have been staring at these things for too long to realize the generalsiation of $-1/2 \le X_n \le 1/2$ is not in fact $\|X_n\| \le 1$ but $\|X_n\| \le 1/2$. | |
| May 26, 2019 at 14:31 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |