# On hitting times of conditioned diffusions

I have a question of a conditioned diffusion processes. This question is somewhat related to an argument which appears in this article:

Let $$D=\{z=(x,y) \in \mathbb{R}^2 \mid |y|<1\}$$ and $$K=\{(x,y) \in D \mid x<1\}$$. We denote by $$X=(X_t,P_z)$$ the absorbing Brownian motion on $$D$$ conditioned to hit $$K$$. We let $$T_{K}=\inf\{t \ge 0 \mid X_t \in K\}$$.

My question

We set $$S=\inf\{t \ge 0 \mid \text{ the second coordinate of }X_{t}=-1/2\}$$.

Can we prove the following: \begin{align*} (1) \quad \lim_{x \to +\infty}\inf_{z=(x,y) \in D \\ \text{ with }-1/2

A claim similar to (1) should hold for more general conditioned diffusions on $$D$$. Can we prove (1) with some universal argument? I would like to know whether similar claims to (1) can be proved for more general conditioned diffusions.

Let $$h(x,y)$$ denote the probability that a Brownian motion started at $$(x,y)$$ hits $$K = \{x < 1, |y| < 1\}$$ before it leaves $$D = \{|y| < 1\}$$. That is, $$h = 1$$ on $$K$$, $$h = 0$$ on $$\partial D$$ and $$h$$ is harmonic in $$D \setminus K$$ (plus the usual continuity condition on regular boundary points of $$D \setminus K$$). The killed Brownian motion in $$D$$ conditioned to hit $$K$$ is the Doob $$h$$-transform of the usual killed Brownian motion in $$D \setminus K$$.

Now let $$L = \{y = \tfrac{1}{2}\}$$. The probability $$f(x, y)$$ that the conditioned process started at $$(x, y)$$ hits $$K$$ no later than it hits $$L$$ is an $$h$$-harmonic function in $$D \setminus (K \cup L)$$ which takes values $$1$$ in $$K$$ and $$0$$ in $$L \setminus K$$ and in $$\partial D$$ (plus the usual continuity condition). Therefore, $$g(x, y) = f(x, y) h(x, y)$$ is harmonic in $$D \setminus (K \cup L)$$, takes values $$1/h(x, y) = 1$$ on $$K$$ and $$0$$ on $$K \setminus L$$ and on $$\partial D$$.

Your question is: does $$f(x, y) = g(x, y) / h(x, y)$$ converge to zero as $$|(x, y)| \to \infty$$? Equivalently: does $$g(x, y)$$ converge to zero faster than $$h(x, y)$$ does?

It can be proved (using, for example, a boundary Harnack inequality argument) that $$h(x, y) \approx e^{-\pi x/2} \sin(\tfrac{1}{2} \pi (y + 1))$$ as $$x \to \infty$$, $$|y| < 1$$. Similarly, $$g(x, y) \approx e^{-2 \pi x} \sin(2 \pi (y + 1))$$ as $$x \to \infty$$, $$y \in (-1, -\tfrac{1}{2})$$, and $$g(x, y) \approx e^{-2 \pi x/3} \sin(\tfrac{2}{3} \pi (y + \tfrac{1}{2}))$$ as $$x \to \infty$$, $$y \in (-\tfrac{1}{2}, 1)$$. Thus, the answer is yes.

Of course, one can ask the same question for more general diffusions, as well as for more general $$K$$ and $$L$$. Whather there is a similar answer depends on what one knows about these objects: one needs some control over the behaviour at infinity of positive harmonic functions in $$D \setminus K$$ and in $$D \setminus (K \cup L)$$ at infinity.

• Thank you very much for your kind reply. Probably I understood. But I do not understand how to use a boundary Harnack inequality very well. Let $h_1=e^{-\pi x/2} \sin(\tfrac{1}{2} \pi (y + 1))$. $h_1$ is a bounded positive harmonic function on $D \setminus K$ vanishing at $\partial D$. Therefore, $h(z)/h(w) \le A h_1(z)/h_{1}(w)$ for $z,w \in (D \setminus K) \cap K'$. $K'$ is a compact subset. $A$ is a constant depending only on $D$, $K$ and $K'$. How do you prove from here? Jan 19, 2019 at 12:13
• @sharpe: Perhaps there is a shorter way, but here is what I had in mind: By BHI, $h$ and $h_1$ are comparable on $\{x=2,\,|y|<1\}$; say, $c h_1 \le h \le C h_1$. Thus $h - c h_1$ and $C h_1 - h$ are harmonic in $D':=\{x > 2,\,|y|<1\}$ and nonnegative on the boundary of $D'$. By the maximum principle, they are nonnegative in $D'$, and thus $c h_1 \le h \le C h_2$ in $D'$. Using more refined arguments one can in fact get $h \sim a h_1$ for some $a > 0$; actually, I am quite sure this follows from some general theorem on BHI at infinity, but I do not have a reference at hand. Jan 19, 2019 at 18:47
• Thank you very much for teaching me carefully. I learned a lot. Jan 20, 2019 at 8:49