Timeline for Does the variance of a continuous time, time homogeneous, Markov process starting from one point necessarily not decrease?
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| Dec 14, 2013 at 21:59 | comment | added | Yuri Bakhtin | @Hansen: Sure. These continuous Markov processes with indeterminacy only at the starting point are not diffusions. As for a reference, it should be in any book on diffusions. My favorite stochastic processes book is the textbook by Wentzell, but it is not easy to find a copy. Another useful book on Markov processes is Ethier&Kurtz. A sufficient condition is essentially a combination of stochastic continuity and nice behavior of (truncated) mean and variance of increments of the process. | |
| Dec 13, 2013 at 18:04 | comment | added | Hans | @YuriBakhtin: By "the example in my first update", are you referring to the stochastic process in 2 dimension constructed along $\gamma$? I may well have missed something, but it does not seems that address the question of whether any continuous path stochastic process can be represented by a diffusion. Can you give the regularity condition or cite a, perhaps linked, reference? | |
| Dec 13, 2013 at 17:54 | comment | added | Yuri Bakhtin | @Hansen: No, not every process, as the example in my first update shows. You have to impose some regularity conditions, roughly speaking, on mean and variance of local increments of the process. | |
| Dec 13, 2013 at 17:31 | comment | added | Hans | @YuriBakhtin: Also, I want to ask my previous second question again: Can every continuous path Markovian (whether time homogeneous or not) process be represented as an Ito or a diffusion process? | |
| Dec 13, 2013 at 17:29 | comment | added | Hans | @YuriBakhtin: I see. This statement using comparison principle says each path a.s. keeps moving in one direction all the time and seems much stronger than the variance increase claim for diffusion process. Is it stemming from the Markovian property of $X^2$? Can we get the same result dealing with just $X$ and therefore not imposing the symmetry condition on $b$ and $\sigma$? By the way, I up-voted your ingenious setup. Thank you, Yuri! | |
| Dec 13, 2013 at 14:05 | comment | added | Yuri Bakhtin | @Hansen: This setup is needed so $X^2$ is also a Markov diffusion. Probably there is some symmetrization trick to treat the general diffusion situation. | |
| Dec 13, 2013 at 4:30 | comment | added | Hans | @YuriBakhtin: Ingenious setup and proof for a Wiener process! I have two questions. 1. Why do you need a(X) and b(X) to be odd and even, respectively? I understand that it guarantees the mean zero. Perhaps, it is also necessary for the mean to be zero at very small time when the path starts deterministically from $0$? It think it is plausible but can you supply or reference a proof? 2. Can every continuous path Markovian (whether time homogeneous or not) process be represented as a Wiener process? What can we do about the general continuous path Markovian process if it can not be? | |
| Dec 12, 2013 at 14:50 | history | edited | Yuri Bakhtin | CC BY-SA 3.0 |
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| Dec 11, 2013 at 21:20 | history | edited | Yuri Bakhtin | CC BY-SA 3.0 |
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| Dec 11, 2013 at 21:13 | history | edited | Yuri Bakhtin | CC BY-SA 3.0 |
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| Dec 11, 2013 at 20:13 | comment | added | Hans | @YuriBakhtin: I agree with your edit regarding the 2d construction inspired by Martin Hairer's answer. Please see my comment below his answer and my modification of question in case 3). I would like to tackle the difficulty of the process forced to revisit the same points. So I restrict the process to 1 dimension especially for case 3) of continuous path. Any ideas? | |
| Dec 11, 2013 at 20:12 | history | edited | Yuri Bakhtin | CC BY-SA 3.0 |
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| Dec 11, 2013 at 20:01 | history | edited | Yuri Bakhtin | CC BY-SA 3.0 |
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| Dec 11, 2013 at 19:08 | comment | added | Hans | @YuriBakhtin: I agree with ofer zeritouni that the fixing of $0$ starting point by adding a jump at $t=0$, would make the process time inhomogeneous. I thought of that jump in my attempt for case 2) but had to make the jump time homogenous which destroyed the variance decreasing property. Any other ideas? | |
| Dec 11, 2013 at 18:49 | vote | accept | Hans | ||
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| Dec 11, 2013 at 7:30 | comment | added | Yuri Bakhtin | Oh, I missed the mean zero requirement. Of course, Nate and Ofer, you are right. | |
| Dec 11, 2013 at 7:25 | comment | added | ofer zeitouni | Hi Yuri, there is no way by a simple shift to bring it back to mean $0$, as the mean of your process is time dependent. But you can easily fix it by starting a process at $0$, making a jump to $\pm 1$, and then start your geometric BM. This is the continuous version of my example, and it is time inhomogeneous (since you will never revisit $0$). However, this still has discontinuous paths (at $x=0$). | |
| Dec 11, 2013 at 7:19 | comment | added | Nate Eldredge | The first sentence asks for the process to have zero mean. | |
| Dec 11, 2013 at 7:04 | history | answered | Yuri Bakhtin | CC BY-SA 3.0 |