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.



Asymptotics of distributions of hitting times
OK, here is the extended version of my comment.




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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),
$$u(0,\sqrt{r}) = P(\sigma\_{\sqrt{r}} > \tau\_1) = P(\sigma\_r > \tau\_{\sqrt{r}}).$$




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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 $u(0, y)$.


