5
$\begingroup$

Assume that $X:(\mathbb R^+ \times \Omega) \to \mathbb{R}^n$ is a solution to an SDE of the form $dX = \mu(X,t) dt + \sigma(X,t)dW$ where $\mu, \sigma$ are continuous, Lipschitz continuous in the first variable (with a Lip constant idependent of $t$) and satisfy $||\mu(x,t)|| \le c||x|| +1 $ and $||\sigma(x,t)|| \le c||x|| +1 $ for some $c>0$ independent of $t$.


Q: Can we deduce that from here that, for almost every $\omega \in \Omega$ there exists $p(\omega)>0$ such that $X(t, \omega) \exp(-p(\omega)t) \to 0$ as $t \to \infty$?


In the deterministic case, this follows from the Gronwall lemma. I don't know if it is of any help, but it also resembles the exponential bound that one can obtain for semigroups of operators (via uniform boundedness), hence I assume that a proof could involve a Baire Category Theorem-like argument.

$\endgroup$
2
  • 1
    $\begingroup$ This is not what you are asking, but still it could help you. The following result appears in Theorem $V.2.4.$ of Ikeda & Watanabe's "Stochastic differential equations and diffusion processes". Assume $E(|X(0)|^2)<\infty$. Then : $$E(|X(t)|^2)\leq (1+E(|X(0)|^2)e^{ct}-1$$ for some constant $c>0$. Hence from here you could obtain the following $$E(|X(t)|^2)e^{-2ct}\leq (1+E(|X(0)|^2)e^{-ct}-e^{-2ct}$$ $$E(|X(t)e^{-ct}|^2)\leq (1+E(|X(0)|^2)e^{-ct}-e^{-2ct}$$ Letting $t\to \infty$ the RHS goes to $0$. Hence $X(t)e^{-ct} \to 0$ in quadratic mean. $\endgroup$
    – Chaos
    Apr 13, 2020 at 16:41
  • 2
    $\begingroup$ two things, I've realized that the theorem is actually in chapter IV, and the fact that since there's convergence in L2 there will be a subsequence converging a.s $\endgroup$
    – Chaos
    Apr 14, 2020 at 7:43

1 Answer 1

4
+50
$\begingroup$

The desired convergence holds even with a non-random constant in the exponent.

Indeed, the well-known estimates (see e.g. Theorem 2.2 in N. Touzi, A. Tourin Optimal Stochastic Control, Stochastic Target Problems, and Backward SDE) give $$ \mathrm E \biggl[\sup_{t\in [0,T]} ||X_t||^2\biggr]\le K e^{KT}. $$ Therefore, using Tonelli's theorem, $$ \mathrm E \biggl[\sum_{n=1}^\infty \sup_{t\in [0,n]} ||X_t||^2 e^{-2Kn}\biggr] = \sum_{n=1}^\infty \mathrm E \biggl[\sup_{t\in [0,n]} ||X_t||^2\biggr] e^{-2Kn}<\infty, $$ whence $$ \sum_{n=1}^\infty \sup_{t\in [0,n]} ||X_t||^2 e^{-2Kn} <\infty $$ almost surely. In particular, $\sup_{t\in [0,n]} ||X_t||^2 e^{-2Kn}\to 0, n\to\infty$, almost surely. Consequently, $$ ||X_t|| e^{-Kt} \to 0, t\to \infty, $$ almost surely.

$\endgroup$
2
  • $\begingroup$ Thank you very much for your answer. Could you please provide a reference for the first estimate? $\endgroup$ Apr 15, 2020 at 0:50
  • 1
    $\begingroup$ @MarkoKarbevski, see the updated answer. $\endgroup$
    – zhoraster
    Apr 15, 2020 at 6:48

You must log in to answer this question.