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This question is a modified version of this unanswered question asked on MSE, which mainly concerns an Ornstein-Uhlenbeck process with discontinuous drift on $\mathbb R^n$(for simplicity let $n=2$ for now): \begin{equation} \label{OU}\tag{1} \mathrm d\mathbf X_t=-(1+f(\mathbf X_t))\mathbf X_t\mathrm dt+\sqrt{2}\mathrm d\mathbf W_t\end{equation} in which $\mathbf X=(X^1,X^2)\in\mathbb R^2,$ $\mathbf W_t$ is a standard Wiener process on $\mathbb R^2,$ and $f(\mathbf X)$ is a piecewise constant function in the form \begin{equation} f(\mathbf X)=\Theta(\mathbf a\cdot \mathbf X)+\Theta(\mathbf b\cdot \mathbf X) \end{equation} where $\mathbf a,\mathbf b\in\mathbb R^2$ are two noncollinear vectors, and $\Theta(x)=\left\{\begin{aligned}&1&x>0\\&0&x\leq 0\end{aligned}\right.$ is Heaviside function.

While solving the SDE \eqref{OU} may be difficult, I'm mainly interested in the evaluation of the moments $\mathbb E^{\mathbf X}\{\mathbf X_t\}$ and $\mathbb E^{\mathbf X}\{X^i_tX_t^j\}$ of the process \eqref{OU} as $t\to+\infty.$ Here for any function $g(\mathbf X_t)$ of $\mathbf X_t,$ \begin{equation}\mathbb E^{\mathbf X}\{g(\mathbf X_t)\}:=\mathbb E\{g(\mathbf X_t)|\mathbf X_0=\mathbf X\}\end{equation} is the (conditional) expectation of $g(\mathbf X_t)$ given initial value $\mathbf X_{t=0}=\mathbf X$. In particular, is it possible that the OU process \eqref{OU} has a non-zero equilibrium displacement, i.e. \begin{equation}\lim_{t\to+\infty}\mathbb E^{\mathbf X}\{\mathbf X_t\}\neq \mathbf X\ ?\end{equation} Thanks in advance. Recommendation of related literature is also welcomed.

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  • $\begingroup$ What is $\mathbb{E}^{\mathbf{X}}$, and what is $\mathbf{X}$ in your last equation? $\endgroup$
    – Kostya_I
    Commented May 22 at 8:25
  • $\begingroup$ $\mathbb E^{\mathbf X}\{f(\mathbf X)\}$ means the mathematical expectation of $f(\mathbf X)$ given initial value $X_{t=0}=\mathbf X.$ Thanks for pointing this out and I'll add it in the question @Kostya_I $\endgroup$
    – painday
    Commented May 22 at 13:07
  • $\begingroup$ But then, one shouldn't expect the limit in the left-hand side to depend on $\mathbf{X}$ at all, since the process converges to its stationary distribution at large $t$. $\endgroup$
    – Kostya_I
    Commented May 22 at 16:41
  • $\begingroup$ @Kostya_l Since the process in the question has singular(discontinuous)drift, I think this process may not have nice enough ergodic properties(uniqueness of invariant measure and convergence to a unique stationary measure regardless of initial distribution, etc.) In statistical physics nonergodic processes can lead to fascinating phase transition phenomena, which is also a motivation of this question. $\endgroup$
    – painday
    Commented May 22 at 17:08

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If you consider the process $Y_t=|X_t|,$ Ito's formula gives $$ dY_t=\sqrt{2}dB_t+\frac{dt}{Y_t}-l(\theta_t)Y_t\,dt, $$ where $\theta_t$ is an argument of $X_t$, $l(\theta_t)\in \{1,2,3\}$ is given by $l(\theta_t)=1+f(\mathbf{X_t}),$ and $B_t$ is the one-dimentional Brownian motion. You can now use comparison theorem for SDE to conclude that the solution to this equation is dominated by a solution to $$ d\tilde{Y}_t=d\tilde{B_t}+\frac{dt}{\tilde{Y}_t}-\tilde{Y}_t\,dt. $$ The latter process is just the absolute value of the usual two-dimensiona OU process, so is should be straightforward to see that it converges to a stationary distribution in any sense you want, in particular, in expectation. Using some theory of Markov processes, this should imply that the same is true for the original process (and so, in fact, the left-hand side of your last equation does not depend on $\mathbf{X}$), but what you immediately see is that you last equation cannot hold in general since $$|\mathbb{E}^{\mathbf{X}}\mathbf{X_t}|\leq \mathbb{E}^{\mathbf{X}}\mathbf{\tilde{Y}_t}\to\mathrm{const},$$where $\mathrm{const}$ is independent of $\mathbf{X}$.

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  • $\begingroup$ 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 :) $\endgroup$
    – painday
    Commented May 23 at 12:11
  • $\begingroup$ @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. $\endgroup$
    – Kostya_I
    Commented May 23 at 13:29
  • $\begingroup$ 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 ~ $\endgroup$
    – painday
    Commented May 23 at 15:20
  • $\begingroup$ @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. $\endgroup$
    – Kostya_I
    Commented May 23 at 18:25
  • $\begingroup$ 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 $0$. $\endgroup$
    – painday
    Commented May 23 at 23:29

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