2
$\begingroup$

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.

$\endgroup$

1 Answer 1

1
+100
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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
    Commented Jan 19, 2019 at 12:13
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Jan 19, 2019 at 18:47
  • $\begingroup$ Thank you very much for teaching me carefully. I learned a lot. $\endgroup$
    – sharpe
    Commented Jan 20, 2019 at 8:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.