Timeline for Does the entropy of a SDE with nondegenerate noise always increase?
Current License: CC BY-SA 4.0
18 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 28, 2023 at 21:31 | vote | accept | Nate River | ||
| May 28, 2023 at 16:59 | answer | added | Iosif Pinelis | timeline score: 2 | |
| May 28, 2023 at 14:41 | comment | added | Nate River | Hmm you are right… | |
| May 28, 2023 at 14:29 | comment | added | Iosif Pinelis | @NateRiver : I don't think so. If you look back at my two-state Markov chain example and modify it slightly by making $2$ only 'almost absorbing" (with some small enough transition probability $p_{21}$), then, by continuity, the entropy can still decrease after one step. | |
| May 28, 2023 at 14:18 | comment | added | Nate River | @Iosif Pinelis Ah, I was trying to say that the analogue of the condition “$\sigma$ bounded away from $0$” in the case of a countable state Markov chain is that the transition probabilities are all uniformly bounded away from $0$. It’s plausible that the entropy inequality holds for the Markov chain under these conditions. | |
| May 28, 2023 at 14:15 | comment | added | Iosif Pinelis | @NateRiver : Sorry, I don't understand your "@IosifPinelis" comment. Can you rephrase it, without using terms like "this"? | |
| May 27, 2023 at 5:58 | comment | added | mike | No, you are right, I bungled that, fwiw, I had in mind problem 3.14 in Karlin and Taylor vol 1 | |
| May 27, 2023 at 5:49 | comment | added | Nate River | An elementary idea would be to compare with the trivial case $dY_t = c \, dW_t$, where $c > 0$ is the uniform bound on $\sigma$. The inequality clearly holds in this case, but exactly how to compare with this is unclear. The result here may be helpful somehow. | |
| May 27, 2023 at 2:32 | comment | added | Nate River | @IosifPinelis I suppose the analogue of non degenerate noise in this case would be nonzero probability of reaching any state from any other state. | |
| May 26, 2023 at 17:39 | comment | added | Iosif Pinelis | @mike : Markov chains? What about a chain on $\{1,2\}$, with $2$ absorbing and $1$ not absorbing? Then the entropy will be $0$ after one step, but not necessarily intially. | |
| May 26, 2023 at 6:08 | comment | added | Nate River | @mike By Markov chains, you mean in discrete time? A reference would be great! | |
| May 26, 2023 at 6:04 | comment | added | mike | you know that this is known for markov chains, I think it is an exercise in karlin & taylor. It seems to me the same argument works here, i.e., that the markov property is crucial, but I don't know what technicalities may arise. | |
| May 26, 2023 at 0:22 | comment | added | Nate River | Thanks for your ideas, they sound promising indeed. Will definitely give this a go and let you know how it goes. | |
| May 26, 2023 at 0:14 | comment | added | Thomas Kojar | another thought, I had was the situation with super-harmonic functions and supermartingales. math.stackexchange.com/questions/304050/…. The log function is harmonic and then the density satisfies the elliptic equation given by Fokker-Plank. So then I was thinking to then possibly use the mean value property for elliptic equations to get that supermartingale property. | |
| May 26, 2023 at 0:13 | comment | added | Thomas Kojar | yes and I was thinking that then maybe the Fokker-Plank equation for densities might then come into use to simplify that Ito expression. | |
| May 25, 2023 at 23:59 | comment | added | Nate River | Right i guess assuming smoothness of the densities this is a pretty good idea! | |
| May 25, 2023 at 16:47 | comment | added | Thomas Kojar | Since this is a supermartingale type question, one good start is computing the Ito for that log expression to see exactly what integrals are involved. | |
| May 25, 2023 at 13:51 | history | asked | Nate River | CC BY-SA 4.0 |