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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<y<1}P_{z}(T_{K}<S)=0. \end{align*}

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.

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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.

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  • $\begingroup$ 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? $\endgroup$ – sharpe Jan 19 '19 at 12:13
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    $\begingroup$ @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. $\endgroup$ – Mateusz Kwaśnicki Jan 19 '19 at 18:47
  • $\begingroup$ Thank you very much for teaching me carefully. I learned a lot. $\endgroup$ – sharpe Jan 20 '19 at 8:49

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