Timeline for Recurrence of Poisson binomial distributed random walk
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 20, 2015 at 8:30 | comment | added | Dan Romik | @user45947 no problem, I was actually baffled that you accepted my answer and thought at the time this was premature. Glad to see I was right :-) | |
| Oct 20, 2015 at 8:28 | comment | added | user45947 | @DanRomik: I did a change on the accepted answer, as Serguei's answer (and your comment) resolved what I was after. Greatly appreciate the input from both. | |
| Oct 20, 2015 at 8:21 | vote | accept | user45947 | ||
| Oct 19, 2015 at 22:14 | comment | added | Dan Romik | To add a bit to @SergueiPopov's last comment, the condition for a.s. convergence is given by the Kolmogorov three-series theorem, which in this situation reduces to the condition $\sum_n \textrm{Var}(X_n) = \sum_n p_n(1-p_n)<\infty$. | |
| Oct 19, 2015 at 21:41 | comment | added | Serguei Popov | For example, if $\sum p_n<\infty$, then a.s. there will be only a finite number of 1's (so that $\sum X_n$ converges), and (since $E X_n=p_n$) $\sum E X_n$ converges as well. The other case is analogous (or consider $Y_n=1-X_n$, and observe that $X_n-EX_n=-(Y_n-EY_n)$). | |
| Oct 19, 2015 at 21:32 | comment | added | user45947 | Great! That seems to be exactly what I'm looking for. I'm just wondering: I get intuitively why $\sum p_n = \sum (1-p_n) = \infty$ would lead to recurrence. But how would you argue more strictly that this statement rules out convergence to a finite limit of $S_n$? Perhaps the answer is obvious, but I'd be grateful for some help along the way. | |
| Oct 19, 2015 at 21:08 | comment | added | Serguei Popov | Ah, seems that he changed chapters' order. In that version, it's Theorem 5.3.1. | |
| Oct 19, 2015 at 20:58 | comment | added | user45947 | Just to be sure, do you mean Theorem 4.1.2 in the 2013 version? (It can be found here: math.duke.edu/~rtd/PTE/PTE4_1.pdf) This theorem holds only when $Y_i := X_i - \textrm{E}X_i$ are iid. However, while $\textrm{E}Y_i= \textrm{E}Y_j = 0$, in general $Y_i$ and $Y_j$ do not have the same variance and hence are not iid, so the theorem does not apply. Do you know if the theorem generalizes so that your argument holds? | |
| Oct 19, 2015 at 19:39 | review | First posts | |||
| Oct 19, 2015 at 19:43 | |||||
| Oct 19, 2015 at 19:37 | history | answered | Serguei Popov | CC BY-SA 3.0 |