# 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 $\alpha^n$.

• Suggestion: Induction on $n$. Define $\alpha:=\sup_{a<x<b}P^x(\tau_a\wedge\tau_b>1)$. (1) Show that $\alpha\in(0,1)$. This covers the initial case $n=1$. (2) For the induction step use the simple Markov property of Brownian motion and the observation that $\{\tau_a\wedge\tau_b>n+1\}=\{\tau_a\wedge\tau_b>n\}\cap\{\tau^*_a\wedge\tau^*_b>1\}$, where $\tau_a^*$ is the hitting time of $a$ by the post$-n$ process $t\mapsto W_{t+n}$, etc. – John Dawkins Nov 11 '17 at 16:57
• Seems good to me, thank you. Just one question about the $\alpha$. It shouldn't matter if i take the maximum and include the bounds instead right? And does there exists an analytic solution for it? My guess would be that x=(b+a)/2 gives the sup. – Andrew Nov 11 '17 at 19:35

A number of ways of doing this:

1. 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}

1. 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).

2. 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.

• 1. A couple of additional things: $\Theta(t, B)$ is typically a "defective" probability measure, meaning that it integrates to less than 1, and here, it decays with $t$. You want the decay rate. 2. Have a look at optional stopping techniques, and how they are used to study properties of stopping times. The theory is quite beautiful. – zab Aug 12 '18 at 15:25
• 2. Brownian max and min before a fixed time $n$ are independent, but the times of their occurrence are not, so your factorization of $P(\tau_a \wedge \tau_b > n)$ isn't right. – zab Aug 13 '18 at 6:33

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.

• I think there is a problem in your proof: You are supposing that $P(\tau_a \wedge \tau_b>n )>\alpha^n$ for all $n \in \mathbb{N}$. Therefore you showed that there exists an $n \in \mathbb{N}$ such that the inequality holds and not that it holds for all $n \in \mathbb{N}$. – Andrew Nov 11 '17 at 21:33
• Yes, indeed, I didn‘t notice that. Well, hope these calculations could be useful in the future though. – Aleksandr Samarin Nov 12 '17 at 8:47