Timeline for Convergence of stochastic process $X_n$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 4, 2020 at 15:59 | comment | added | Jimmy R. | @DieterKadelka This is not $G_n$. I give a specific definition. No $U_n$ or any other pathology there. | |
| Nov 4, 2020 at 15:43 | comment | added | Dieter Kadelka | You need at least additional assumptions. You get simple counterexamples if $G_n = \lambda \cdot (1-F(X_n)) + U_n$ with $(U_n)_{n \in \mathbb{N}}$ a sequence of independent (not iid) random variables with $\mathbb{E}U_n = 0$, but $|U_n|$ "large". | |
| Nov 4, 2020 at 13:57 | comment | added | Jimmy R. | @DieterKadelka Yes, the model is Markovian. It only depends on the previous state. Plese use the notation that feels correct. Thank you. | |
| Nov 4, 2020 at 13:25 | comment | added | Dieter Kadelka | The difference is that with $\mathcal{F}_n$ you get a Markovian model, I think. This simplifies investigations of convergence. | |
| Nov 4, 2020 at 11:53 | comment | added | Jimmy R. | @DieterKadelka Yes, you also know $\mathcal F_n$. But since you only use $X_n$, does this make a difference? Sorry for asking, I am just not too familiar with these notions. | |
| Nov 4, 2020 at 11:18 | comment | added | Dieter Kadelka | Is it correct that you only have $\mathbb{E}[G_n | X_n] = \ldots$ or do you have $\mathbb{E}[G_n | \mathcal{F}_n] = \ldots$ (same expression on the right side) with $\mathcal{F}_n = \sigma(X_0,G_0,X_1,G_1,\ldots,G_{n-1},X_n)$? | |
| Nov 4, 2020 at 10:27 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo, added tag
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| Nov 4, 2020 at 10:24 | history | asked | Jimmy R. | CC BY-SA 4.0 |