Timeline for Generalized density functions on the natural numbers
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 4, 2015 at 0:01 | comment | added | Anthony Quas | I think roughly the same argument works for other values of $t$. Define $Y_n$ to be 1 if $a_1+\ldots+a_n>t\sqrt n$ and 0 otherwise. When you do the Brownian motion approximation argument with $m<n$, you end up with Cov$(Y_m,Y_n)$ $\approx \mathbb P(N_1>t;N_2>t\sqrt{n/(n−m)}−N_1\sqrt{m/(n−m)}−\mathbb P(N_1>t)^2$ where $N_1$ and $N_2$ are independent standard normals. The weighting means you don't have to worry about terms with $m$ and $n$ within a bounded factor of each other. Outside this range, this is close to 0 as before so that you get the desired convergence. | |
| Oct 3, 2015 at 19:17 | comment | added | James Propp | Yes, I believe this would work. So that settles the case $t=0$. But it's less clear to me how to handle other values of $t$ (from the original problem). | |
| Oct 3, 2015 at 19:15 | vote | accept | James Propp | ||
| Oct 3, 2015 at 14:50 | comment | added | Anthony Quas | So I was thinking about this. I think the point is that the sequence $S_N$ is very slowly varying indeed. Nothing much can happen between times $e^{(1+\epsilon)^k}$ and $e^{(1+\epsilon)^{k+1}}$. So if the sequence goes to 0 with probability 1 along that sequence of times for each $\epsilon$, you get convergence for the full sequence. Now work with a countable sequence of $\epsilon$ going to 0. An argument of this type (for standard weights) appears in notes of Wierdl and Rosenblatt in the Cambridge Volume "Convergence in Ergodic Theory" | |
| Oct 3, 2015 at 13:49 | comment | added | James Propp | Modulo Brownian motion calculations that I haven't checked, this looks good, as far as it goes. But how do we get to "converges to 1/2 with probability 1"? Given how slowly 1/log $N$ goes to 0, I don't see why the outliers in the sequence $S_1,S_2,...$ are still constrained to go to 0. There's probably a simple argument for this, but off the top of my head I don't see it. | |
| Oct 3, 2015 at 11:58 | history | edited | Anthony Quas | CC BY-SA 3.0 |
added 2 characters in body
|
| Oct 3, 2015 at 5:08 | history | answered | Anthony Quas | CC BY-SA 3.0 |