Exponential decay of tail probability for Brownian stopping time

brownian motionmarkov-processstopping-times

A previous question, Exit Time of an Interval Brownian Motion – Distribution, asked about the tail probabilities for exit times from a region $(-a,b)$. In particular, the question was about exponential decay of the tail probability:

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 $n \in \mathbb{N}$.

The accepted answer showed that this was indeed the case, and the resulting $\alpha$ depended implicitly on $a$ and $b$. However, I would like a more explicit form for $\alpha$'s dependence on $a$ and $b$. In particular, my question is:

Does $P(\tau_{-a} \wedge \tau_b>t )\le e^{-ct/(a+b)}$ for some constant $c$? If no, is there another way to express a simple bound showing both exponential decay in $t$ and some dependence on $a$ and $b$?

Best Answer

You can actually compute the moment generating function of that random variable with some cleverness using an appropriate martingale. This computation is one I saw in "Continuous Exponential Martingales and BMO" by Kazamaki. It is Lemma 1.3 in Chapter 1.2 and is currently available on Springer's website. The idea is recorded there as being due to Leipngle. I'll just comment that the relevant local martingale is $$ M_t^{\theta} = e^{\frac{1}{2}\theta^2t}\cos\left(\theta B_t - \frac{\theta(b-a)}{2}\right). $$ One has to do a little work to show that optional stopping can be applied, but I'll skip that and refer you to the book for the relevant estimates if you need them. The end result is that for $0 \leq \theta < \frac{\pi}{a+b}$, $$ \mathbb{E}[e^{\frac{1}{2}\theta^2 \tau_{-a}\wedge \tau_b}] = \frac{\cos\left(\frac{a-b}{2} \theta \right)}{\cos\left(\frac{a+b}{2} \theta\right)} $$ Now we can use Markov's inequality: $$ \mathbb{P}(\tau_{-a}\wedge \tau_b > t) \leq \frac{\mathbb{E}\left[e^{\theta^2 \tau_{-a} \wedge \tau_b}\right]}{e^{\frac{1}{2}\theta^2 t}} = \frac{\cos\left(\frac{a-b}{2} \theta \right)}{\cos\left(\frac{a+b}{2} \theta\right)} e^{-\frac{1}{2}\theta^2 t} $$ Choosing for example $\theta = \frac{\pi}{2(a+b)}$, and noticing that the numerator is non-negative and less than or equal to $1$ we obtain $$ \mathbb{P}(\tau_{-a}\wedge \tau_b > t) \leq \sqrt{2}e^{-\frac{\pi^2 t}{8(a+b)^2}}. $$

Related Solutions

Brownian Motion – Exit Time of an Interval Distribution

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.

Random time vs stopping time & expectation

The following R-code serves to compute the mean of $B_{\tau/2}$ conditionally on $B_\tau = 1$. Paths of Brownian motions on $[0,1]$ are simulated via independent Gaussian increments. For the subsample where $\tau \leq 1$ and $B_{\tau} = 1$, the values $B_{\tau/2}$ are computed, if necessary via linear interpolation. The mean is then approximated by taking the average of these values.

(Intuitively, conditioning on $\tau \leq 1$, shouldn't be too problematic; but a rigorous argument escapes me.)

I find that $\mathbb{E}[B_{\tau/2} \, | \, B_\tau = 1] \approx 0.26$. Replacing $\tau$ by $\bar{\tau} := \inf\{t > 0 : B_t = 1\}$, one also finds $\approx 0.26$. I find these numerical results to be somewhat consistent with an earlier comment of mine.

set.seed(2)

N <- 5000
n <- 5000

h <- 1/n

X <- apply(rbind(rep(0,N),matrix(rnorm(n*N, mean = 0, sd = sqrt(h)), nrow = n, ncol = N)), 2, cumsum)
    
tau <- apply(X, 2, function(y){Position(function(x){x>=1 | x<=-1}, y)})

B_tau <- vector()
B_tau_half <- vector()
for (i in 1:N){
  if(!is.na(tau[i]) & X[tau[i],i] >= 1){
    B_tau <- c(B_tau,X[tau[i],i])
    if(tau[i] %% 2 == 0){
      B_tau_half <- c(B_tau_half,X[tau[i]/2,i])
    }
    else{
      B_tau_half <- c(B_tau_half,X[(tau[i]-1)/2,i]/2 + X[(tau[i]+1)/2,i]/2)
    }
  }
}

mean(B_tau_half)

Best Answer

Related Solutions

Brownian Motion – Exit Time of an Interval Distribution

Random time vs stopping time & expectation

Related Question