Timeline for A Markov process which is not a strong markov process?
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 7, 2014 at 23:54 | answer | added | Fantastic | timeline score: 6 | |
| Feb 22, 2011 at 12:34 | comment | added | Did | @vinay For $x\ne0$ and $t>0$, let $p_t(x,\cdot)$ denote the Gaussian distribution with mean $x$ and variance $t$ and $p_t(0,\cdot)$ denote the Dirac measure at $0$. For every $x$, let $p_0(x,\cdot)$ denote the Dirac measure at $x$. Then, for every bounded measurable $\varphi$, initial distribution $\nu$ and times $0=t_0\le t_1\le \cdots\le t_n$, $E_\nu[\varphi(X(t_0),X(t_1),\ldots,X(t_n))]$ is the integral you know, involving $\varphi$, $\nu$ and the semi-group $(p_t)_{t\ge0}$. QED. In fact, a good way to understand this example is to try to prove that $(p_t)_{t\ge0}$ is indeed a semi-group. | |
| Feb 22, 2011 at 12:14 | comment | added | user13166 | I did not quite get the first answer(the one using Brownian Motion). If the process starts at x(not equal to 0), the distribution of X(0) is delta(x) and transition kernels are that of brownian motion and if x = 0 then distribution of x(0) is delta(0) and transition kernels according as a constant stochastic process. How do we mix the 2 processes? Sorry if I am missing something silly. | |
| Oct 28, 2010 at 11:46 | vote | accept | Simon Lyons | ||
| Oct 27, 2010 at 17:24 | answer | added | George Lowther | timeline score: 18 | |
| Oct 27, 2010 at 17:20 | answer | added | Andrey Rekalo | timeline score: 12 | |
| Oct 27, 2010 at 17:17 | answer | added | user6096 | timeline score: 18 | |
| Oct 27, 2010 at 17:00 | history | asked | Simon Lyons | CC BY-SA 2.5 |