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I have a question on asymptotic behavior of distributions of Brownian hitting times.

Let $B_t$ and $W_t$ be independent one-dimensional Brownian motions starting at the origin. The joint law is denoted by $P$. For $x,y>0$, we set \begin{align*} \sigma_x&=\inf\{t>0 \mid B_t=x\} \\ \tau_y&=\inf\{t>0 \mid W_t \notin (-y,y)\}. \end{align*} Then, we can show that \begin{align*} (1)\quad P(\sigma_r>\tau_{r^{1/2}})=O(r^{1-\varepsilon})\quad \text{as}\quad r \to 0, \end{align*} for any $\varepsilon>1/2.$

For the proof of $(1)$, I used the fact that $P(\sigma_r>t) \lesssim r/\sqrt{t}$ and $P(\tau_{r^{1/2}} \le t) \lesssim \sqrt{t/r}\exp(-r/8t)$ valid for $r,t>0$.

Is my estimate optimistic? I think there is a sharper bound.

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  • $\begingroup$ Although I think you meant optimal, your estimate indeed seems optimistic, or even super-optimistic: unless I got something wrong, by scaling, we have $P(\sigma_r > \tau_{r^{1/2}}) = P(\sigma_{r^{1/2}} > \tau_1) \to 1$ as $r \to 0^+$. $\endgroup$
    – Mateusz Kwaśnicki
    Oct 29, 2019 at 9:08
  • $\begingroup$ Thank you for your comment. I may have misunderstood something. Why $\lim_{r \to 0}P(\sigma_{r^1/2}>\tau_1) =1$ ? There is the exact formula for the distribution of $\sigma_{r^1/2}$, which means that $P(\sigma_{r^{1/2}}>t) \lesssim (r/t)^{1/2}$. So, as $r \to 0$, $P(\sigma_{r^{1/2}}>t) \to 0$. $\endgroup$
    – sharpe
    Oct 29, 2019 at 10:39
  • $\begingroup$ Of course, you are right, sorry. For some reason I was thinking about the limit as $r \to \infty$. But then, it is relatively easy to see that $P(\sigma_{r^{1/2}} > \tau_1)$ is comparable with $r^{1/2}$: this is the value $u(0, r^{1/2})$ of a harmonic function $u$ in $(-1,1) \times (0, \infty)$, with boundary value $0$ on $(-1,1) \times \{0\}$ and $1$ on $\{-1,1\} \times (0, \infty)$. By the boundary Harnack inequality, the decay of $u$ is linear near $u$. In fact, no need to invoke BHI here: extend $u$ so that $u(x,-y)=-u(x,y)$, and observe that $u$ is harmonic in $(-1,1)\times \mathbb{R}$. $\endgroup$
    – Mateusz Kwaśnicki
    Oct 29, 2019 at 11:01
  • $\begingroup$ @MateuszKwaśnicki Thank you for your kind reply. I have a question on your comment. We denote by $P_{x,y}$ the joint distribution of independent one-dimensional Brownian motions $B$ and $W$ starting at $(x,y)$. I think $u(x,y)=P_{x,y}(\sigma_{r^{1/2}}>\tau_1)$. Then, $u$ is a harmonic function w.r.t. the two-dimensional Brownian motion $(B,W)$, right? How do you prove the decay of $u$ is linear near $(0,0)$? $\endgroup$
    – sharpe
    Oct 29, 2019 at 15:54
  • $\begingroup$ @MateuszKwaśnicki AH... I might understood. The identity $u(x,y)=-u(x,y)$ leads us to the lenearly decay. $\endgroup$
    – sharpe
    Oct 29, 2019 at 16:06

1 Answer 1

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OK, here is the extended version of my comment.

Let $u(x,y)$ be the (bounded) harmonic function in $D = (-1,1) \times (0, \infty)$, with boundary value $1$ along $\{-1,1\} \times (0, \infty)$ and $0$ along $(-1,1) \times \{0\}$. Consider a 2-D Brownian motion $(X_t, Y_t)$, started at $(X_0,Y_0) = (x,y)$, and denote the corresponding probability law by $P^{x,y}$. Either by Itô's lemma, or by potential-theoretic results, $u(X_t,Y_t)$ is a martingale up to $\tau_D$, the first time $(X_t,Y_t)$ hits the boundary of $D$. In particular, $$u(x,y) = E^{x,y} u(X_{\tau_D},Y_{\tau_D}).$$ At $t = \tau_D$, we either have $X_t = \pm 1$ or $Y_t = 0$. Let $\sigma$ be the first time $Y_t = 0$, and $\tau$ be the first time $X_t = \pm 1$. Clearly, $\tau_D = \min\{\sigma, \tau\}$. If $\sigma < \tau$, then $\tau_D = \sigma$ and $Y_{\tau_D} = 0$. Thus, $u(X_{\tau_D},Y_{\tau_D}) = 0$. On the other hand, if $\sigma > \tau$, then $X_{\tau_D} = \pm 1$, and hence $u(X_{\tau_D},Y_{\tau_D}) = 1$. It follows that $$u(x,y) = P^{x,y}(\sigma > \tau).$$ By symmetry and translation invariance, with the notation of the problem, $$u(0,y) = P(\sigma_y > \tau_1).$$ By self-similarity (a.k.a. scaling), $$u(0,\sqrt{r}) = P(\sigma_{\sqrt{r}} > \tau_1) = P(\sigma_r > \tau_{\sqrt{r}}).$$

By the boundary Harnack inequality, we know that a finite, positive limit $$ \lim_{y \to 0^+} \frac{u(0, y)}{y} $$ exists. This implies that $$P(\sigma_r > \tau_{\sqrt{r}}) \sim c \sqrt{r}$$ as $r \to 0^+$.

As a side remark: In fact, there is no need to employ BHI here, as soon as one observes that the formula $u(x,-y) = -u(x,y)$ extends $u$ to a harmonic function in $(-1, 1) \times \mathbb{R}$. It is then clear that $u(0,y)$ is a smooth function of $y$, and it remains to show that $\partial_y u(0, y) > 0$. This last stem follows for example from the explicit expression for the Poisson kernel of a strip; however, I think one can easily cook up a soft argument here.

By the way, Hopf's lemma provides a yet another proof of the asymptotic formula for $u(0, y)$.

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  • $\begingroup$ Thank you for your very kind reply. But I have one question. Sorry for my lack of study. In your answer, you first take the harmonic function $u$ on $D$ with the following boundary condition; $u=0$ on $(-1,1) \times \{0\}$ and $u=1$ on $\{-1,1\} \times (0,\infty)$. Why $u$ exists ? This may be a basic question. $\endgroup$
    – sharpe
    Oct 30, 2019 at 5:32
  • $\begingroup$ I can prove the uniqueness of $u$. This follows from the maximum principle. $\endgroup$
    – sharpe
    Oct 30, 2019 at 5:35
  • $\begingroup$ Well, that depends on your point of view. From my probabilistic perspective, you can simply define $u$ as $u(x,y) = P^{x,y}(\sigma > \tau)$. There are, however, alternative analytical arguments. For example, you can prove existence of a solution to the Dirichlet problem with continuous data and then extend it to piecewise continuous boundary values. Or, you can take any smooth function with these boundary values, and solve a Poisson problem instead. $\endgroup$
    – Mateusz Kwaśnicki
    Oct 30, 2019 at 6:04
  • $\begingroup$ Thank you for your reply. I understood. An answer of my question follows from some results of the Dirichlet problem., $\endgroup$
    – sharpe
    Oct 30, 2019 at 6:18

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