Exponential or sub-exponential ergodicity?

Question

Consider the one-dimensional stochastic differential equation $$d X(t) = -sgn(X(t))dt + dW(t),$$ where $W$ is a standard Brownian motion, and $sgn(x) = 1$ if $x > 0$ and $-1$ if $x\le 0$. It can be shown that this process possesses an invariant distribution $\pi$ with density $m(x) = C e^{-|x|}$ for some normalizing constant $C> 0$. Denote by $P_t(x,\cdot)$ the transition probabilities of the process.

My question is: does $P_t(x,\cdot)$ converge to $\pi$ in total variation norm at exponential rate or strictly sub-exponential rate? Thanks.

Answer 1

Answer given is already nice; alternatively, exponential convergence in the total variation distance can be verified by an application of Harris's Theorem with Foster-Lyapunov function $\mathcal{V}(x)=\exp(\varphi_{\epsilon}(|x|^2))$ where $\varphi_{\epsilon}(\cdot)$ is a $C^2$ approximation to the square root, $$ \varphi_{\epsilon}(x) = \begin{cases} \sqrt{x} & x \ge \epsilon^2 \;, \\ -\frac{1}{8 \epsilon^3} x^2 + \frac{3}{4 \epsilon} x + \frac{3 \epsilon}{8} & x \le \epsilon^2 \;. \end{cases} $$ For a $C^2$ function $f$, let $$\mathscr{L} f(x) = -sgn(x) \frac{\partial f}{\partial x}(x) + \frac{1}{2} \frac{\partial^2 f}{\partial x^2}(x)$$ denote the action of the infinitesimal generator of the process $X$ on the function $f$. Then for $x \ge \epsilon^2$ $$ \mathscr{L} \mathcal{V}(x) = - \frac{1}{2} \mathcal{V}(x) $$ and similarly for $x \le -\epsilon^2$. Thus, globally, $$ \mathscr{L} \mathcal{V}(x) \le - \frac{1}{2} \mathcal{V}(x) + K \;, \quad \text{where $K = \sup\left\{ \mathscr{L} \mathcal{V}(x) + \frac{1}{2} \mathcal{V}(x) : |x| \le \epsilon^2 \right\}$}\;. $$ Combining this with a minorization condition (which follows from ellipticity of the SDE), Harris's theorem can now be invoked to obtain the required exponential convergence in the total variation distance. For a reference see, e.g., Theorem 2.5 of the following paper.

Mattingly, J. C.; Stuart, A. M.; Higham, D. J., Ergodicity for SDEs and approximations: locally Lipschitz vector fields and degenerate noise., Stochastic Processes Appl. 101, No. 2, 185-232 (2002). ZBL1075.60072.

Here are some additional references.

Meyn, Sean; Tweedie, Richard L., Markov chains and stochastic stability. Prologue by Peter W. Glynn., Cambridge Mathematical Library. Cambridge: Cambridge University Press (ISBN 978-0-521-73182-9/pbk; 978-0-511-62663-0/ebook). xviii, 594 p. (2009). ZBL1165.60001.

Hairer, Martin; Mattingly, Jonathan C., Yet another look at Harris’ ergodic theorem for Markov chains, Dalang, Robert C. (ed.) et al., Seminar on stochastic analysis, random fields and applications VI. Centro Stefano Franscini, Ascona, Italy, May 19–23, 2008. Basel: Birkhäuser (ISBN 978-3-0348-0020-4/pbk; 978-3-0348-0021-1/ebook). Progress in Probability 63, 109-117 (2011). ZBL1248.60082.

Answer 2

The rate is exponential. First, let me show that $|X(t)|$ (which is a reflected BM with constant drift towards the origin) has exponential rate of convergence. I do it by a coupling argument: you run two independent copies of the process, one started from $x$ and another from the stationary distribution, until they first collide, and therefrom run them together. The TV distance between their distributions at time $t$ is bounded by twice the probability that they do not collide by that time. But, if by time $t$ each process has hit the origin, they must have collided. Hence, it suffices to show that for each process, the probability that it does not hit the origin before time $t$ decays exponentially. This is simply a large deviation estimate for BM.

It's not hard to modify this argument so that it works for $X(t)$. For example, I can run the two processes independently until their absolute values coincide, then run them as mirror reflections of each other until they hit the origin, and therefrom run them together.

Alternatively, you could probably compute the transition density for $|X(t)|$ explicitly, from Kolmogorov's equation. The spectrum of the operator in the RHS (once you divide $P_t(x,y)$ by $e^{-y}$ to bring it into symmetric form) has eigenvalue $0$ and a continuous part $(-\infty,-1]$, hence there's a spectral gap. But the reflecting boundary conditions are somewhat subtle, so I am not sure how to cleanly write the relevant self-adjoint extension...