Timeline for When does iteration of an infinite Toeplitz matrix converge?
Current License: CC BY-SA 3.0
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| Nov 25, 2017 at 16:22 | comment | added | fedja | @Ian Post the actual setup then. Also, it looks like $\ell^1$ is the only problematic space (in the current simplified model $c_0$ and $\ell^p$ with $p>1$ present no trouble, as I said. | |
| Nov 25, 2017 at 13:42 | comment | added | Ian | @fedja Yes, I hang out on MSE a lot and hadn't received a notification on this question, so thanks for pinging me. | |
| Nov 25, 2017 at 13:37 | vote | accept | Ian | ||
| Nov 25, 2017 at 13:37 | comment | added | Ian | @oferzeitouni I see what you mean. It seems then that the approximations to my original problem were not quite good enough for me to use this to solve it. This is because actually $T,b$ in my real problem depend on $x$, and I was attempting to approximate them by simply $T(x^*),b(x^*)$. But the underlying random walk for $T(x^*)$ has mean zero, whereas other $T(x)$'s have positive mean and others have negative mean. | |
| Nov 25, 2017 at 3:22 | comment | added | fedja | @Ian This is just to attract your attention to Ofer's comments (MO doesn't automatically notify the poster of the question about comments to the answers as far as I know :-) | |
| Nov 24, 2017 at 9:16 | comment | added | ofer zeitouni | 3) If the mean vanishes and the walk is transient, it must oscillate. Indeed, if $S_n\to \infty$ with positive probability then necessarily by Kesten's lemma, $\liminf S_n/n>0$, which is impossible due to the ergodic theorem. (H. Kesten, Sums of stationary sequences cannot grow slower than linearly, Proc. AMS 49 (1975) pp. 205–211.) 4) When the walk is transient and the mean does not exist, P5 in page 181 of Spitzer says that $1-E(z^T)=e^{-\sum \frac{z^k}{k} P(S_k<0)}$ where $|z|<1$ and $T=\inf \{n: S_n<0\}$. Thus, everything is determined by the limit as $|z|\to 1$ of the sum. | |
| Nov 24, 2017 at 9:15 | comment | added | ofer zeitouni | Here is a classification for Fedja's last point: 1) If the mean $\sum k t_k$ exists and is positive, then the probability is positive. If it is negative, then it is zero. This follows from the strong law of large numbers. 2) In general there is a criterion for transience/recurrence: the walk is recurrent iff $Re[1/(1-\phi(\theta))]$ is not integrable in a neighborhood of 0 (here $\phi(\theta)$ is the characteristic function; this is Theorem in page 84 of Spitzer, principles of random walk, second edition). In case of recurrence, the probability you ask about is 0. | |
| Nov 18, 2017 at 5:35 | comment | added | fedja | @Ian No, you do not need to rewrite anything. Just think a bit of which paths in the random walk correspond to the admissible products in $T^k$. In the $\ell^p$ ($p\in(1,\infty)$) case just note that an arbitrary vector is a sum of a vector with finite support (that is killed by the above argument) and an arbitrarily small tail vector (whose norm cannot grow). The whole point of my post was just to brush off the trivialities and highlight the core of the question. | |
| Nov 18, 2017 at 5:05 | comment | added | Ian | I think in the $\ell^1$ case you are referring to the fact that $T$ can be written as $S D^{-1}$ where $S$ is row stochastic and $D$ is diagonal with $D_{ii}=\sum_k T_{ik}$, so powers of $T$ are more or less controlled by powers of $S$. Is that right? I think I see the idea you are referring to in the $\ell^p$ case (there is some sort of "spreading" that leads to decay in $\ell^p$ even if not in $\ell^1$). | |
| Nov 18, 2017 at 0:24 | history | answered | fedja | CC BY-SA 3.0 |