# Hitting probability of a transient diffusion process

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

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).

• 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. – sharpe Dec 21 '18 at 6:49
• @sharpe: I added some details, hope this is clearer now. – Mateusz Kwaśnicki Dec 21 '18 at 7:14
• 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 $R>R_1(\omega)$, $X_{T_{R}}(\omega) \in K$? – sharpe Dec 21 '18 at 8:06
• 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? – sharpe Dec 21 '18 at 8:27
• Hmm... Where do you use the transience of $X$? – sharpe Dec 21 '18 at 8:59