Stationary Distribution of Langevin Dynamics driven by Lévy Process

Question

Let $f\geq 0$ be a Lipschitz function and let $(L_t)_{t\geq 0}$ be an $\alpha$-stable Lévy process ($0<\alpha<2$, possibly multivariate). Consider the process given by $$dX_t=-\nabla f(X_t)dt+\sigma dL_t$$ where $\sigma>0$ is a constant. This SDE is reminiscient of Langevin Dynamics, where we usually let the process be driven by Brownian Motion instead of a Lévy process.

I am interested in seeing which results from the "Brownian setting" extend to the setting with $\alpha$-stable Lévy processes as described above.

Most importantly, my main question is: Does $X_t$ admit a stationary distribution? When the process is driven by Brownian Motion, it is known that there exists a stationary distribution, for which we can also find a closed form expression for the density.

Example: Ornstein-Uhlenbeck Process

If we consider the case $f(x)=\frac12 |x|^2$ we get the Ornstein-Uhlenbeck process driven by the Lévy process $L$. There it is known, see e.g. Topics in Infinitely Divisible Distributions and Lévy Processes Theorem 2.17, that the process admits a stationary distribution.

Answer 1

There are several generalizations of the Brownian case results to general L'evy processes. For instance, under a classical log-concave assumption on $U$ we have a convergence to equilibrium in Wassertein distance to a unique stationary distribution. More precisely, consider the following SDE

$$ dX^x_t=-\nabla U(X^x_t) dt +dN_t, \quad X^x_0=x \in \mathbb R^n $$ where $N$ is any L'evy process such that $\mathbb{E} \left( \sup_{t\in [0,T]} | N_t |^p \right) <+\infty$ for some $p>1$. For stable processes one therefore asks that $\alpha>1$. Denote $P_t$ the semigroup of $X_t$.

Theorem: Assume that there exists $a>0$ such that $\nabla^2 U \ge a$ (uniformly in the sense of quadratic forms). Then, there exists a unique probability measure $\mu$ in the Wasserstein space $\mathcal{P}_p(\mathbb R^n)$ such that for every $t \ge 0$, $\mu P_t = \mu$. Moreover, for every $t \ge 0$, and $\nu \in \mathcal{P}_p(\mathbb R^n)$ one has, $$ W_p ( \nu P_t, \mu ) \le e^{-at} W_p ( \nu , \mu ). $$ Therefore $X_t$ converges exponentially fast to the invariant distribution in the Wasserstein distance $W_p$.

This can be proved using similar arguments as in Theorem 4.9 in the paper Transport inequalities for Markov kernels and their applications which treated the case $p=2$.

Here are the main steps.

Let $J_t=\frac{\partial X_t^x}{\partial x}$ be the first variation process. Let $f$ be a $C^1$ and bounded Lipschitz function. Since $P_tf(x)=\mathbb{E}( f(X_t^x))$, by the chain rule we have $$ \nabla P_t f (x)=\mathbb{E}\left( J_t^* \nabla f(X_t^x)\right). $$ Therefore, by Holder inequality, $$ | \nabla P_t f (x) | \le \mathbb{E}\left( | J_t^*|^p\right)^{1/p} \mathbb{E}\left( | \nabla f(X_t^x) |^q\right)^{1/q}. $$ where $q$ is the conjugate exponent of $p$. Observe that $$ dJ_t=-\nabla^2 U(X^x_t) J_t dt, \quad J_0=\mathbf{Id}_{\mathbb R^n}. $$ From the assumption $\nabla^2 U \ge a$ this yields $$ | J_t^*| \le e^{-at }. $$ One concludes $\mathbb{E}\left( | J_t^*|^p\right) \le e^{-pat }$ and therefore $$ | \nabla P_t f (x) | \le e^{-at} P_t (| \nabla f |^q)(x)^{1/q}. $$ By Kuwada duality, this yields that for every $\nu_0,\nu_1 \in \mathcal{P}_p(\mathbb R^n)$, $$ W_p ( \nu_0 P_t, \nu_1 P_t ) \le e^{-at} W_p ( \nu_0 , \nu_1 ). $$

A fixed point argument in the Wasserstein space allows then to conclude.