I have a question about properties of transient diffusion process.

In the case of $d$-dimensional Brownian motion $B=(B_t,P_x)$ ($d \ge 3$), we can prove that \begin{align} (1)&\quad 0<P_{x}(\sigma_{K}<\infty)\quad \text{ for any }x \in \mathbb{R}^d,\\ (2)&\quad \lim_{|x| \to \infty,\ x \in \mathbb{R}^d}P_{x}(\sigma_{K}<\infty)=0. \end{align} Here, $K$ is a compact subset of $\mathbb{R}^d$ with positive Lebesgue measure and $\sigma_{K}=\inf\{t \ge0 \mid X_t \in K\}$. $|\cdot|$ is the Eucledean metric. Note that (2) follows from the heat kernel estimate of the Brownian motion and the strong Markov property.

My question

We consider the next diffusion $X=(X_t,P_x)$ on $\mathbb{R}^d$: \begin{equation*} X_t=x+\int_{0}^{t}a(X_s)\,dB_s+\int_{0}^{t}b(X_s)\,ds, \end{equation*} where $a$ and $b$ are bounded continuous functions on $\mathbb{R}^d$ and $B$ is the $d$-dim Brownian motion starting from the origin. We assume that $X$ is transient.

Is there conditions on $a$ and $b$ such that $\inf_{x \in \mathbb{R}^d}P_{x}(T_{K}<\infty)>0?$

Here, $T_K=\inf\{t\ge0 \mid X_t \in K\}$.


1 Answer 1


I believe that the infimum is always zero. This follows by the following argument, valid for any standard Markov process with infinite lifetime.

Let $T$ be the hitting time of $K$, and write $f(x) = P_x(T_K < \infty)$. Suppose, contrary to our claim, that $f(x) \ge c > 0$ for all $x$, and define $g(x) = (f(x) - c) / (1 - c)$.

Fix $x \in \mathbb{R}^d \setminus K$ and $R > |x|$. Let $T_R$ be the hitting time of $K \cup (\mathbb{R}^d \setminus B(0, R))$ (the first time the process either hits $K$ or exits the centred ball with radius $R$). Then:

  1. $T_R$ is a family of stopping times, increasing in $R$.
  2. $T_R$ converges to $T$, and $T < \infty$ if and only if $T = T_R$ for $R$ large enough. Indeed, denote by $S$ the limit of $T_R$. Then $S \le T$, and hence if $S = \infty$, then $T = \infty$. Let us see what happens if $S < \infty$. By quasi-left continuity, the finite limit $\lim_{R \to \infty} X_{T_R}$ exists and it is equal to $X_S$. Since $X_{T_R}$ either belongs to $K$ or to $\mathbb{R}^d \setminus B(0, R)$, convergence of $X_{T_R}$ to a finite limit implies that $X_{T_R} \in K$ for all $R$ large enough. In particular, $T = T_R$ for large enough, and consequently $S = T = T_R$ and $X_T = X_{T_R} \in K$ for $R$ large enough.
  3. By the strong Markov property, $$f(x) = P_x(T < \infty) = E_x(f(X_{T_R})) , $$ so similarly $g(x) = E_x(g(X_{T_R}))$ (here we use infinite lifetime: constants are harmonic).

Since $g$ is non-negative and equal to $1$ on $K$, we have $$ P_x(T = T_R) = P_x(X_{T_R} \in K) \le E_x(g(X_{T_R})) = g(x) $$ for $R$ large enough. Passing to the limit as $R \to \infty$, we find that $$ f(x) = P_x(T < \infty) = \lim_{R \to \infty} P_x(T = T_R) \le g(x) , $$ contrary to the definition of $g$.

(I may be getting something slightly wrong due to the usual problem whether the hitting time is defined with $t > 0$ or $t \ge 0$ in the infimum, but this should be minor).

  • $\begingroup$ Thank you very much for your very kind reply. However, I couldn't understand Step2 in your argument. You asserts that $T=T_R$ for large $R$ enough. Why this holds? By your argument and the definition of $T$, it follows that $T=S$. But $T_R \nearrow S\le T$ in general. $\endgroup$
    – sharpe
    Commented Dec 21, 2018 at 6:49
  • $\begingroup$ @sharpe: I added some details, hope this is clearer now. $\endgroup$ Commented Dec 21, 2018 at 7:14
  • $\begingroup$ Thank very much for your reply. The quasi left continuity of $X$ yields $\lim_{R \to \infty}X_{T_R}(\omega)=X_{S}(\omega) \in K$, $P_x$-a.s. $\omega$. You mean there exists $R_{1}(\omega)$ and for all $R>R_1(\omega)$, $X_{T_{R}}(\omega) \in K$? $\endgroup$
    – sharpe
    Commented Dec 21, 2018 at 8:06
  • $\begingroup$ Oh I mostly understood. $\{T <\infty\} \subset \{\lim_{R \to \infty} T_R=T\}$ and this implies $P_x(T<\infty) \le \lim_{R \to \infty}P_x(T_R=T) \le g(x)$, right? $\endgroup$
    – sharpe
    Commented Dec 21, 2018 at 8:27
  • $\begingroup$ Hmm... Where do you use the transience of $X$? $\endgroup$
    – sharpe
    Commented Dec 21, 2018 at 8:59

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