Exit Time of an Interval Brownian Motion - Distribution Let $W_t$ be a Brownian motion, fix $a&lt;0&lt;b$ and let $\tau_x=\mathrm{inf}(t\ge0:W_t=x)$. 

Show there is an $\alpha&lt;1$: $P(\tau_a \wedge \tau_b&gt;n )\le \alpha^n$ 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:

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

But i am not able to find an estimation such that this expression is dominated be $\alpha^n$.
A number of ways of doing this:


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


Now to show that $\alpha &lt; 1$: Exact calculation of the distribution of $\tau$ requires some heavy machinery (optional stopping &amp; Laplace transforms), but here's an easy but crude upper bound: 

Let $x^* := \text{argmax}_x P[\tau &gt; 1 | W_0 = x]$. We know that $x^*$ cannot be $a$ or $b$, and in fact $a &lt; x^* &lt; b$ (strict ineqs.) so that $b - x^* &gt; 0$.
\begin{align}
\alpha &amp;= P[\tau &gt; 1 | W_0 = x^*] \\
&amp;\leq P[\tau_b &gt; 1 | W_0 = x^*] \quad (\tau_b \text{ never comes sooner than }\tau) \\
&amp;= 1 - 2P[W_1 &gt; b | W_0 = x^*] \\
&amp;= 1 - 2 \bar{\Phi}(b - x^*) &lt; 1 \quad \text{(since $b - x^* &gt; 0$ so that $\bar{\Phi}(b - x^*) &lt; 1/2$}).
\end{align}


Let $B$ be a Borel subset of the interval $[a,b]$. Then $\Theta(t,B):= P[\tau &gt; t, W_t \in B]$ is a (family of) probability measure(s) (indexed by $t$) on $[a, b]$ with Lebesgue density $\theta(t,x)dx:= P[\tau &gt; t, W_t \in dx]$. 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 $P[\tau &gt; n] = \int^b_a \theta(n, x)dx$. 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 $P[\tau &gt; n]$ in terms of the radius of convergence of the Laplace transform, like here:Burq &amp; Jones, Exact Simulation of Brownian Motion at First Passage Times.
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 $x &lt; 1$,
$\mathbb{E}[X] = \sum_{n=0}^{\infty} \mathbb{P}(X &gt; n)$ for discrete random variable $X$, taking positive integer values.


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