Timeline for Ornstein Uhlenbeck process with discontinuous drift
Current License: CC BY-SA 4.0
8 events
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| May 23 at 23:35 | comment | added | painday | And in some cases, e.g. the (piece-wise)asymmetric linear damping force is given by $\mathbf F=-\begin{bmatrix}X_1\\\Theta(X_1)X_2\end{bmatrix},$ the drift $\mathbf F$ is indeed discontinuous at $\mathbf X=0.$ And of course If we allow non-linear drift then the drift can be chosen to be continuous and even $C^\infty$. | |
| May 23 at 23:29 | comment | added | painday | That's true, but for physical picture of the Ornstein-Uhlenbeck process, if one let $\mathbf X$ denote the velocity(or momentum) of a Brownian particle diffusion in a medium, then the most physically relevant drift describing the damping force experienced by the particle is the linear drift $-\alpha \mathbf X,$ hence in this case the asymmetry of the drift typically implies that the coefficient $\alpha$ is only piece-wise constant, and hence $\alpha(\mathbf X)$ is discontinuous at $\mathbf X=0.$ Of course in this very situation $\alpha\cdot\mathbf X$ is still continuous at $0$. | |
| May 23 at 18:25 | comment | added | Kostya_I | @painday, I don't think this has anything to do with discontinuity though - a process with a drift non-symmetric about the origin has no particular reason to have zero equilibrium expectation. | |
| May 23 at 15:20 | comment | added | painday | Nice! The fact that the expectation $\mathbb E^{\mathbf X}\{\mathbf X_t\}$ can be nonzero would have many surprising physical consequences(e.g. noise-induced transitions)! The Ornstein-Uhlenbeck process with discontinuous drift is so interesting :) Thanks again ~ | |
| May 23 at 13:29 | comment | added | Kostya_I | @painday, I don't think there's any reason to expect it to be zero. Intuitively, the process will spend less time in a quadrant where drift towards the origin is stronger. | |
| May 23 at 12:12 | vote | accept | painday | ||
| May 23 at 12:11 | comment | added | painday | Excellent answer! Now I see why $\mathbb E^{\mathbf X}\{\mathbf X_t\}$ is a constant. BTW I wonder whether this constant can be nonzero? Thanks :) | |
| May 23 at 10:38 | history | answered | Kostya_I | CC BY-SA 4.0 |