# Do Lyapunov functions imply exponential integrability of hitting times?

I have a question of some integrability of hitting times.

Let $$X=(\{X_t\}_{t \ge0},\{P_x\}_{x \in E})$$ be a diffusion process on a locally compact separable metric space $$E$$.

We assume that there exist Borel measurable functions $$f\colon E \to [1,\infty)$$ and $$g \colon E \to \mathbb{R}$$ such that $$\left\{f(X_t)-f(x)-\int_{0}^{t}g(X_s)\,ds\right\}$$ is a local martingale in $$P_x$$ for any $$x \in E$$. By convention, we will write $$\mathcal{L}f$$ for $$g$$. We further assume that $$f$$ is bounded, and that there exist $$\alpha,\beta \in (0,\infty)$$ and compact subset $$K \subset E$$ such that \begin{align*} \mathcal{L}f(x) \le -\alpha f(x) +\beta\textbf{1}_{K}(x),\quad x \in E. \end{align*}

Under the conditions stated above, we can prove that \begin{align*} \sup_{x \in E}E_x[\sigma_K]<\infty, \end{align*} where $$\sigma_K$$ is the first hitting time of $$K$$. Indeed, fix $$x \in E \setminus K$$ and a localizing sequence $$\{\tau_l\}_{l \ge 1}$$ . Then, for any $$l \ge 1$$, $$\left\{ f(X_{t\wedge \sigma_K \wedge \tau_l})-f(x)-\int_{0}^{t\wedge \sigma_K \wedge \tau_l}\mathcal{L}f(X_s)\,ds \right\}_{t \ge 0}$$ is a $$P_x$$-martingale. Therefore, we obtain that for any $$t>0$$ and $$l \ge 1$$, \begin{align*} -E_{x} \left[\int_{0}^{t \wedge \tau_l \wedge \sigma_K} \mathcal{L}f(X_s)\,ds\right] &= f(x)-E_{x}[f(X_{t \wedge \tau_l \wedge \sigma_K})] \le f(x). \end{align*} Because $$\mathcal{L}f \le -\alpha f$$ on $$E \setminus K$$, it follows that \begin{align*} E_{x}[t \wedge \sigma_K \wedge \tau_l] \le f(x)/\alpha. \end{align*} Because $$f$$ is bounded, Fatou's lemma shows that $$\sup_{x \in E \setminus K}E_x[\sigma_K]<\infty$$. It clearly holds that $$\sup_{x \in K}E_x[\sigma_K]=0.$$

My question

We can take increasing compact subsets $$\{K_n\}_{n=1}^{\infty}$$ of $$E$$ such that $$E=\bigcup_{n=1}^{\infty}K_n$$. In this situation, I would want to expect that \begin{align*} (1)\quad \lim_{n \to \infty}\sup_{x \in E}E_{x}[\sigma_{K_n}]=0. \end{align*} Then, there exists $$N \in \mathbb{N}$$ such that for any $$n \ge N$$, $$\sup_{x \in E}E_x[e^{\sigma_{K_n}}]<\infty$$.

Can we prove $$(1)$$? If not, please tell me a counterexample.

• Do the $\{ K_n \}$ in (1) have any relation to $K$ in the infinitesimal drift condition? Also, can you expand a bit more on the display before (1) that shows that $E_x( \sigma_K ) \le f(x) / \alpha$? – Nawaf Bou-Rabee May 29 at 7:14
• @NawafBou-Rabee Thank you for your comment. There exists $N \in \mathbb{N}$ such that $K \subset K_N$. Therefore, we can prove that $\mathcal{L}f \le -\alpha f$ on $E \setminus K_n$ for any $n \ge N$ and $\sup_{n \in \mathbb{N}}\sup_{x \in E}E_{x}[\sigma_{K_n}]<\infty$. – sharpe May 29 at 9:16
• Thanks; an expectation seems to be missing in the LHS of the inequality with the upper bound $f(x)$. How is the limit in (1) related to integrability of the exponential moment of $\sigma_{K_n}$? – Nawaf Bou-Rabee May 29 at 9:30
• @NawafBou-Rabee The exponential moment of $\sigma_{K_n}$ is proved by the Khasminskii's lemma. This allows us to conclude that $\sup_{x \in E}E_{x}[e^{\sigma_L}]<\infty$ if $\sup_{x \in E}E_{x}[\sigma_L]<1$, where $L$ is a compact subset of $E$. – sharpe May 29 at 10:44

It may depend on what exactly you mean by a diffusion on a general metric space.

Here is a counterexample for a discontinuous process. Take $$E = \mathbb{R}$$, and let $$(X_t)$$ be the process started at $$x \in \mathbb{R}$$ and jumping to the origin after a unit exponential time $$\tau$$. In other words, $$(X_t)$$ jumps to the origin at rate $$1$$, and then stays there forever. Take $$f(x) = I \{x \ne 0 \}$$. Then $$g(x) = -f$$. The conditions on $$f$$ are satisfied with $$K = \{ 0\}$$.

In this situation (1) fails because $$\sup_{x \in E}E_{x}[\sigma_{K_n}] = \sup_{x \in E}E_{x}[\sigma_{K}] = E_{x} \tau = 1$$.

EDIT. Here is a set-up that should give a counterexample with a continuous process. Take $$E = [0, \infty) \times [0,1]$$, and consider the system $$dX _t = h(X_t, Y_t), \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \$$ $$dY _t = g(X_t, Y_t), \ \ \ \ \ \ \ (X_0, Y_0) \in E$$ with $$h$$ and $$g$$ satisfying the following conditions:

1. $$h([0, \infty) \times (0, 1]) = \{0\}$$.
2. $$g([0, \infty) \times (0, 1] ) = \{-1\}$$
3. $$h(x, 0) = - \varphi (x)$$, where $$\varphi > 0$$ is a function such that the ODE $$z' = \varphi (z), \ z(0) = 0$$ escapes to infinity in a finite time (for example, $$\varphi (x) = 1 + x ^2$$) .
4. $$g(x, 0 ) = 0$$.

For $$(x,y) \in E$$, take now $$f(x,y) = (y + t_x) \vee 1$$, where $$t_x$$ is the time when the solution to the ode $$z' = \varphi (z), \ z(0) = 0$$ reaches $$x$$. It holds that $$t_x\leq t_{expl}$$, where $$t_{expl}$$ is the explosion time for $$z$$. Hence $$f$$ is bounded. Furthermore, $$f(X_t, Y_t) = (f(X_0, Y_0) - t) \vee 1$$, hence $$f$$ satisfy the inequality \begin{align*} \mathcal{L}f \le -\alpha f +\beta\textbf{1}_{K}, \end{align*}

with $$\alpha = \frac{1}{\|f \|_{\infty}}$$ and $$K = \{ (0,0)\}$$.

However, (1) is not satisfied because any compact $$\mathcal{K} \subset E$$ is bounded, and for $$x \in (0,\infty)$$ such that $$\{x\} \times [0,1] \cap \mathcal{K} = \varnothing$$ we have $$E_{(x,1)}[\sigma_{\mathcal{K}}] \geq 1$$ because $$\sigma_{\mathcal{K}} \geq 1$$ under $$P_{(x,1)}$$.

Remark. The example can be modified without much trouble to make $$h$$ and $$g$$ continuous. I have a feeling that (1) might actually be true if $$E = \mathbb{R}$$.

• Thank you for your reply. This is a counterexample for a discontinuous process. – sharpe May 31 at 13:01
• Thank you for your very kind reply. – sharpe Jun 2 at 12:12