# Asymptotics of distributions of hitting times

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.

• 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 $r \to 0^+$. – Mateusz Kwaśnicki Oct 29 '19 at 9:08
• 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 $r \to 0$, $P(\sigma_{r^{1/2}}>t) \to 0$. – sharpe Oct 29 '19 at 10:39
• 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 $(-1,1)\times \mathbb{R}$. – Mateusz Kwaśnicki Oct 29 '19 at 11:01
• @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 $(0,0)$? – sharpe Oct 29 '19 at 15:54
• @MateuszKwaśnicki AH... I might understood. The identity $u(x,y)=-u(x,y)$ leads us to the lenearly decay. – sharpe Oct 29 '19 at 16:06

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 $$u(0, y)$$.

• 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. – sharpe Oct 30 '19 at 5:32
• I can prove the uniqueness of $u$. This follows from the maximum principle. – sharpe Oct 30 '19 at 5:35
• 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. – Mateusz Kwaśnicki Oct 30 '19 at 6:04
• Thank you for your reply. I understood. An answer of my question follows from some results of the Dirichlet problem., – sharpe Oct 30 '19 at 6:18