Show that $\mathbb{P}(\tau_{0}>T)\approx\frac{1}{\sqrt{T}}$ where $\{ B(t) : t\geq 0\}$ is a linear brownian motion started at $B(0)=1$

Question

I'd appreciate if someone could provide me with a solution for the following problem:

Let $\left\{ B\left(t\right)\thinspace|\thinspace t\geq0\right\}$ be a linear brownian motion started at $B\left(0\right)=1$ and let $\tau_{x}$ be the hitting time of $x\in\mathbb{R}$ . Show that $\mathbb{P}\left(\tau_{0}>T\right)\approx\frac{1}{\sqrt{T}}$

Thanks in advance!!

Answer 1

Define $\tilde\tau_x=\inf\{t\ge 0:\tilde B(t)=x\}$ where $\tilde B(0)=0$. Using the translation invariance, symmetry, and the reflection principle for BM

$$P\{\tau_0>T\}=P\{\tilde\tau_1>T\}=\frac{2}{\sqrt{\pi}}\int_{0}^{\frac{1}{\sqrt{2T}}} e^{-u^2}du$$ $$=\frac{2}{\sqrt{\pi}}\left(\frac{1}{\sqrt{2T}}-\frac{1}{3}\left(\frac{1}{\sqrt{2T}}\right)^3+\cdots\right)\approx \frac{0.8}{\sqrt{T}}$$