Exit Time of an Interval Brownian Motion - Distribution

Question

Let $W_t$ be a Brownian motion, fix $a<0<b$ and let $\tau_x=\mathrm{inf}(t\ge0:W_t=x)$.

Show there is an $\alpha<1$: $P(\tau_a \wedge \tau_b>n )\le \alpha^n$ for all $n \in \mathbb{N}$.

Proof-Idea: Use the distribution of the min and max of the brownian motion and their independence, pray and find an estimate:

$$ \begin{eqnarray} P(\tau_a \wedge \tau_b>n ) &=& (1-P(\tau_a\le n ))(1-P(\tau_b\le n))\\ &=&(1-\Phi(\frac{-a}{\sqrt{n}})(1-\Phi(\frac{b}{\sqrt{n}})\\ \end{eqnarray} $$

But i am not able to find an estimation such that this expression is dominated be $\alpha^n$.

Answer 1

I will use three relations here, which are fairly easy to prove (or to find via googling):

• $\mathbb{E}[\tau_a \wedge \tau_b] = -ab$ (related question on MathExchange),
• $\sum_n x^n = \frac{1}{1-x}$ for $x < 1$,
• $\mathbb{E}[X] = \sum_{n=0}^{\infty} \mathbb{P}(X > n)$ for discrete random variable $X$, taking positive integer values.

Let $\alpha = \frac{ab}{ab-1}$, and suppose that $\mathbb{P}(\tau_a \wedge \tau_b > n) > \alpha^n$, then $$ \begin{aligned} -ab &= \mathbb{E}[\tau_a \wedge \tau_b] \geq \mathbb{E}[\lfloor \tau_a \wedge \tau_b\rfloor] = \sum_{n \in \mathbb{N_0}} \mathbb{P}(\lfloor \tau_a \wedge \tau_b\rfloor > n) \\ &= \sum_{n \in \mathbb{N}} \mathbb{P}( \tau_a \wedge \tau_b > n) > \sum_{n \in \mathbb{N}} \alpha^n = \frac{\alpha}{1-\alpha} = -ab, \end{aligned} $$ which is a contradiction.