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
9 events
| when toggle format | what | by | license | comment | |
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| Dec 14, 2013 at 13:15 | comment | added | Hans | Would you mind writing out both your proof for the strong Markov case and a counter example for the weak Markov case? Thank you, Martin. | |
| Dec 14, 2013 at 8:48 | comment | added | Martin Hairer | I think that a variant of Yuri's proof works in $1D$ for any strong Markov process with continuous sample paths that is symmetric under $x \mapsto -x$. Strong Markov (rather than just Markov) is essential since one can build a counterexample that first runs away from the origin on dyadics and then comes back on irrationals. Not sure whether the symmetry assumption is also needed... | |
| Dec 14, 2013 at 1:14 | comment | added | Hans | Do you have any idea on how to proceed to prove it? Yuri Bakhtin had a special case of diffusion with symmetry proved as shown in his answer. | |
| Dec 12, 2013 at 14:38 | comment | added | Martin Hairer | My hunch would be that if you insist on continuous paths, mean 0, and dimension 1, then the statement might be correct... | |
| Dec 11, 2013 at 19:52 | comment | added | Hans | I did up-vote your solution though. | |
| Dec 11, 2013 at 19:02 | comment | added | Hans | Tricky! Putting the value of the process in 2 dimension avoids having to revisit the same point before coming back to the origin. The injective map is nice, too, to keep the transformed process Markovian. It cleverly circumvents the difficult point. I would very much like to accept this marvelous solution, but since only one solution can be "accepted", I would want to wait for the one dimensional case (or cylindrical symmetric case for higher dimension) which tackles the difficulty of revisiting-same-points head on. Thank you, very much, Martin. | |
| Dec 11, 2013 at 18:49 | vote | accept | Hans | ||
| Dec 11, 2013 at 18:50 | |||||
| Dec 11, 2013 at 18:49 | vote | accept | Hans | ||
| Dec 11, 2013 at 18:49 | |||||
| Dec 11, 2013 at 10:35 | history | answered | Martin Hairer | CC BY-SA 3.0 |