Timeline for Does the variance of a continuous time, time homogeneous, Markov process starting from one point necessarily not decrease?
Current License: CC BY-SA 3.0
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| Dec 18, 2013 at 17:05 | comment | added | Hans | I still do not understand your proof. There seems to be something wrong, if I may say. Consider $X$ of the kind where $|X_t|>at$, where $a$ is a positive constant. Since $Y(t)$ is to be chosen arbitrarily initially so long as it is independent of $X(t)$, consider $Y(t)\equiv 0$. Then the stopping time $\tau=\infty$. By construction, $0=|Y(t)|<|X(t)|, \forall t>0$. The laws of $X$ and $Y$ are different. Could you explain the contradiction, or tell me what I have missed? | |
| Dec 17, 2013 at 8:58 | comment | added | Martin Hairer | Sorry, I got myself confused with the indices... Now it should be correct. | |
| Dec 17, 2013 at 8:55 | history | edited | Martin Hairer | CC BY-SA 3.0 |
added 15 characters in body
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| Dec 16, 2013 at 16:20 | comment | added | Hans | I have some confusions with your proof. First you state "Let $\delta>0$ be fixed and construct a process $Y_t$ with the same law as $X_{t+\delta}$ as follows.", then you claim "$Y$ has the same law as $X$", which I interpret as $Y(t)$ has the same law as $X(t)$ which I suppose has a different law from that of $X(t+\delta)$. I am also having difficulty understanding how the laws of $X(t)$ and $Y(t)$ (or Y(t+\delta) ) are the same given $Y(t)<X(t+\delta),\,\forall t<\tau$ and $Y(t)$ is independent from $X(t)$ for $t<\tau$. Could you please explain? | |
| Dec 15, 2013 at 23:00 | comment | added | Hans | Excuse my obtuseness, but could you please explicate "by the strong Markov property, Y has the same law as X"? | |
| Dec 15, 2013 at 9:31 | history | answered | Martin Hairer | CC BY-SA 3.0 |