Exit Time of an Interval Brownian Motion - Distribution 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 $\alpha^n$.
 A: A number of ways of doing this:


*

*Let $\tau = \tau_a \wedge \tau_b$. Then 
\begin{align}
P[\tau > n]
&= \int^b_aP[\tau > n | \tau > n-1, W_{n-1} = x] P[\tau > n-1, W_{n-1} \in dx] \\ 
&= \int^b_aP[\tau > n | W_{n-1} = x] P[\tau > n-1, W_{n-1} \in dx] \quad(\text{Markov property}) \\
&= \int^b_aP[\tau > 1 | W_0 = x] P[\tau > n-1, W_{n-1} \in dx] \quad(\text{Stationarity of BM segments}) \\ 
&\leq \max_{x}P[\tau > 1 | W_0 = x] \int^b_a P[\tau > n-1, W_{n-1} \in dx] \\
&= \max_x P[\tau > 1 | W_0 = x] P[\tau > n-1].
\end{align}
Writing $\alpha :=  \max_x P[\tau > 1 | W_0 = x]$, we have 
$
P[\tau > n] \leq \alpha^n.
$


Now to show that $\alpha < 1$: Exact calculation of the distribution of $\tau$ requires some heavy machinery (optional stopping & Laplace transforms), but here's an easy but crude upper bound: 
Let $x^* := \text{argmax}_x P[\tau > 1 | W_0 = x]$. We know that $x^*$ cannot be $a$ or $b$, and in fact $a < x^* < b$ (strict ineqs.) so that $b - x^* > 0$.
\begin{align}
\alpha &= P[\tau > 1 | W_0 = x^*] \\
&\leq P[\tau_b > 1 | W_0 = x^*] \quad (\tau_b \text{ never comes sooner than }\tau) \\
&= 1 - 2P[W_1 > b | W_0 = x^*] \\
&= 1 - 2 \bar{\Phi}(b - x^*) < 1 \quad \text{(since $b - x^* > 0$ so that $\bar{\Phi}(b - x^*) < 1/2$}).
\end{align}


*Let $B$ be a Borel subset of the interval $[a,b]$. Then $\Theta(t,B):= P[\tau > t, W_t \in B]$ is a (family of) probability measure(s) (indexed by $t$) on $[a, b]$ with Lebesgue density $\theta(t,x)dx:= P[\tau > t, W_t \in dx]$. If you know your PDE theory, $\theta(t,x)$ solves the heat equation in the strip $(t,x) \in R^+ \times [a,b]$ with initial heat atom at $(0,0)$ and boundary condition $\theta(t,a) = \theta(t,b) \equiv 0.$ The probability you are interested in is simply $P[\tau > n] = \int^b_a \theta(n, x)dx$. You can then invoke results from PDE theory about the decay rate of temperatures (parabolic functions) in such problems (Widder etc).

*If you know some optional stopping techniques, you can calculate the Laplace transform of the distribution of $\tau$, and use a Tauberian theorem to find the exact decay rate of $P[\tau > n]$ in terms of the radius of convergence of the Laplace transform, like here:Burq & Jones, Exact Simulation of Brownian Motion at First Passage Times.
A: 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.
