Timeline for Markov process on a torus with prescribed invariant distribution
Current License: CC BY-SA 4.0
17 events
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| Jun 1 at 19:23 | comment | added | 0xbadf00d | Thank you for your edit. Meanwhile I was able to prove that the toroidal wrapping of any Lévy process is a Markov process and I could identify its transition semigroup in terms of the transition semigroup of the original process. However, the open question is whether the same can be shown for the toroidal wrapping of a general SDE (under the condition of a periodic drift and a periodic diffusion coefficient); at least for the one with Langevin drift and constant diffusion coefficient. | |
| Jun 1 at 18:35 | history | edited | Nawaf Bou-Rabee | CC BY-SA 4.0 |
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| Jun 1 at 18:26 | history | edited | Nawaf Bou-Rabee | CC BY-SA 4.0 |
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| May 31 at 17:39 | comment | added | 0xbadf00d | @MartinHairer I think I start to see where my confusion came from. They define a toroidal diffusion to be induced by a SDE with periodic coefficients, which is why I didn't get why you said Prop 1 wouldn't apply. But the subtlety is that the "wrapped process" considered there is NOT a toroidal diffusion in the sense of Definition 1, right? The missing part is the periodicity assumption on the coefficients. (And maybe it would even not possible for a toroidal diffusion to be ergodic on all of $\mathbb R^d$, but I'm not sure about that) | |
| May 31 at 16:01 | comment | added | Martin Hairer | @0xbadf00d Prop 1-iv of that paper doesn't apply here since it assumes the diffusion is ergodic on $\mathbb{R}^d$ which isn't the case for the diffusion with periodic coefficients (which is Markov when projected down to the torus). | |
| May 31 at 13:19 | comment | added | 0xbadf00d | I was too hasty and I think we are both missing something. It seems like the wrapped diffusion will not be Markov in general. The measurable transform of a Markov process is not a Markov process, in general. But there are conditions on the transform which ensure that; see my latest edit. | |
| May 30 at 21:30 | comment | added | 0xbadf00d | I start to think that Proposition 1-iv is really wrong. If I'm not mistaken, the Markov property should be preserved under measurable transformations and $\mathbb R^d\to[0,1)^d,x\mapsto x-\lfloor x\rfloor$ is clearly (Borel) measurable. | |
| May 30 at 13:42 | comment | added | 0xbadf00d | I got all of that, but do you disagree with Proposition 1-iv or not? Isn't the wrapped diffusion defined there precisely the process you are saying that it is invariant with respect to my $p$ on $[0,1)^d$? I need it to be Markov, which is why Proposition 1-iv worries me. | |
| May 30 at 13:39 | comment | added | Nawaf Bou-Rabee | Yes, that is how BM on $\mathbb{T}$ identified with $[0,1)$ can be simulated. And the corresponding generator acts on $C^1(\mathbb{T}^d)$ functions that are $1$-periodic. This is again a special case of BM on a general manifold, since $\mathbb{T}^d$ is also a boundaryless manifold and $[0,1)^d$ is just a coordinate chart, see mathoverflow.net/a/358491/64449. (BTW, I would email the authors of that paper directly for clarification on "wrapped" diffusion processes.) | |
| May 30 at 13:29 | comment | added | 0xbadf00d | First of all, I am saying nothing. It is Proposition 1-iv which says that the wrapped diffusion process is not Markov anymore. I took that for granted and didn't validate it yet. Do you say the authors are wrong? (And as a side note: Interesting that you say that uniform distribution (on $[0,1)^d$, I guess) is the invariant measure for BM on $[0,1)^d$. How do we see that? Do you got a reference? And how is it defined at all? Simply by $W_t=B_t-\lfloor B_t\rfloor$, where $B_t$ is a standard BM?) | |
| May 30 at 13:25 | comment | added | Nawaf Bou-Rabee | Are you saying that BM on $\mathbb{T}^d$ which leaves invariant the uniform distribution is not Markovian? Isn't that statement self-contradicting? | |
| May 30 at 13:18 | history | edited | Nawaf Bou-Rabee | CC BY-SA 4.0 |
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| May 30 at 13:16 | comment | added | 0xbadf00d | You got me wrong. There is no issue with the generator. My problem is that I need to numerically simulate the process and for that I need to know how the process looks like? Intuitively I would think it is precisely the process defined in Definition 1 here with $b$ as in $(2)$ of my question. However, by Proposition 1-iv this process is not even a Markov process anymore. | |
| May 30 at 13:05 | comment | added | Nawaf Bou-Rabee | The generator is characterized by (local) differential operators, so I don't see what the issue is, and in the special case where $\Sigma(x)$ is the identity map, $p(x)$ is uniform and $b(x) \equiv 0$, then the corresponding process is indeed BM on $\mathbb{T}^d$. Writing down the adjoint generator is discussed in Lemma 2.2 of Barp et al. (2021). | |
| May 30 at 12:50 | comment | added | 0xbadf00d | Having said that, since the generator is the same, all of my concerns are still present. It's promising for me to hear that you seem to say, that what I'm trying to achieve actually works, but how does the corresponding $[0,1)^d$-valued(!) Markov process look like? Without that knowledge, I cannot simulate it in practice. (And as a minor thing; I also need to identify the $\mathcal L^\ast$, where it hear should be understand wrt the bracket $\langle f,g\rangle:=\int_{[0,\;1)^d}fg$ and hence should differ from the adjoint of $\mathcal L$ wrt $\langle f,g\rangle:=\int_{\mathbb R^d}fg$) | |
| May 30 at 12:48 | comment | added | 0xbadf00d | Thank you for your effort. This is the same generator as the diffusion on $\mathbb R^d$ has (though your formula differs by a factor of $\frac12$ and I guess you got $p=e^{-\tilde p}$ in mind and intended to write $\nabla\tilde p$, cause otherwise it should be $\nabla\ln p$ in the formula instead. And you also modeled the nonreversible part by a function $b$, whereas I used the matrix $U$ in the question.) | |
| May 30 at 12:17 | history | answered | Nawaf Bou-Rabee | CC BY-SA 4.0 |