Exit probability on a Brownian motion from an interval

Question

Suppose that $B_s, s \geq 0$ is a one dimensional Brownian motion starts at $0$.

Define the event $A$:$B_s$ exits the interval $(-a,a)$ before time $t$.

Prove that

$P(A)\leq \frac{2\sqrt{2t}}{a\sqrt{\pi}}*\exp(\frac{-a^2}{2t})$

Is there any hint how to do it?

Thanks a lot.

Answer 1

Let $T_a=\inf\{t:B_t=a\}$. Then by symmetry and the Reflection principle we get \begin{align} \mathsf{P}(A)&\le \mathsf{P}(T_a< t)+\mathsf{P}(T_{-a}< t)\\ &=4\mathsf{P}(B_t\ge a)=4\mathsf{P}(B_1\ge t^{-1/2}a)\le \frac{2\sqrt{2t}}{a\sqrt{\pi}}e^{-\frac{a^2}{2t}}. \end{align}