Timeline for A Markov process which is not a strong markov process?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Jul 30, 2023 at 13:10 | comment | added | Sextus Empiricus | @Titti are you saying that this process is weak Markov? | |
| Jul 30, 2023 at 13:08 | comment | added | Sextus Empiricus | @Titti "to check the weak Markov property you only need to work with two fixed times s<t at the time" Shouldn't the Markov property satisfy the check for all s<t? Maybe one can define the check as something occuring at two points at the time, but the check should still be repeated for any time points and not just two. | |
| Jul 29, 2023 at 23:53 | comment | added | No-one | @SextusEmpiricus I think that the point here is that to check the weak Markov property you only need to work with two fixed times $s<t$ at the time, so you can do that trick. If you were to change the kernel for all times, then you would get a BM with absorption at $0$ as you say. In facts, this is exactly why the process doesn't satisfy the strong Markov property, since we can consider a stopping time $T$ with uncountable range such that $X_T=0$ (for example $T=$"first time the process crosses $0$"). | |
| Jul 29, 2023 at 14:22 | comment | added | Sextus Empiricus | @Titti I don't follow how that transition kernel follows from W being a standard Brownian motion. If $W_{t+1}|W_t \sim \text{constant}(0)$ when $W_t = 0$, then don't we have instead a Brownian motion with an absorption state at zero? | |
| Jul 29, 2023 at 13:10 | comment | added | No-one | @SextusEmpiricus I don't think so because if $W$ is a BM then for all $t$ $\mathbb{P}(W_t=0)=0$, so we can say $W_{t+1}|W_t$ is distributed as $N(W_t,1)$ if $W_t\neq 0$ and as the $0$ constant if $W_t=0$. Note that the same transition kernel also works when $Y$ is the $0$ process, and hence it works for any process $X$ constructed by "mixing" $W$ and $Y$. | |
| Nov 17, 2020 at 8:03 | comment | added | Sextus Empiricus | Isn't the issue that makes that this process does not satisfy the strong Markov property an issue that makes that it also does not satisfy the 'regular' Markov property? | |
| Dec 9, 2018 at 13:39 | comment | added | George Lowther | Ok, but by a slight modification, you can let the starting position $x$ be random. Say, $\mathbb{P}(X_0=0)=\mathbb{P}(X_0=1)=1/2$. Then, if $\tau$ is the first time at which $X$ hits zero, we do not have $\mathbb{P}(\cdots\vert F_{\le\tau})=\mathbb{P}(\cdots\vert X_\tau)$ as $X_\tau=0$ generates the trivial sigma-algebra but $F_{\le\tau}$ includes the event $\{X_0=0\}$ which tells us whether $X$ is stuck at 0 or not. | |
| Nov 18, 2018 at 12:23 | comment | added | Taro Tokyo | This is only a good example if we consider the semigroup version of Markov property. For the version $\mathbb{P}( \cdots | F_{ \le \tau}) = \mathbb{P}( \cdots | X_{\tau}) $, the strong Markov property still holds. | |
| Oct 28, 2010 at 11:46 | vote | accept | Simon Lyons | ||
| Oct 27, 2010 at 17:24 | history | answered | George Lowther | CC BY-SA 2.5 |