3
$\begingroup$

We define a Brownian motion $W$, and two stopping times as follow :

$$\tau_a=\inf(t \ge 0 | W_t>a)$$ $$\tau_b=\inf(t \ge 0 | W_t<-b)$$

where $a,b >0$

We can define another stopping time as follow $$\tau=\min(\tau_a,\tau_b)$$

While the density functions of $\tau_a$ and $\tau_b$ are known (by using the Brownian motion reflection principal), how about $\tau$ ?

$\endgroup$
1
+50
$\begingroup$

This is best done by first solving for $u(x,t)=\mathbb{P}_x[\tau>t]$ where $\mathbb{P}_x$ means the probability measure if start at $x$ at time $0$. One can show (or read in a book) that $u$ solves the PDE $\frac{1}{2}\frac{d^2}{dx^2}u=\frac{d}{dt}u$ (in general $Lu=\frac{d}{dt}u$ for a process with generator $L$). The boundary conditions are easier to see, $u(0,x)=1$ for all $x\in(a,b)$ and $u(a,t)=u(b,t)=0$ for $t>0$. The solution can be given as the Fourier series $u(x,t)=\sum_{n=1}^\infty\frac{2}{n\pi}(1-(-1)^n)\sin\left(\frac{n\pi(x-a)}{b-a}\right)e^{-\frac{n^2\pi^2t}{2(b-a)^2}}$. Then your density is $-\frac{d}{dt}u(x,t)=\sum_{n=1}^\infty\frac{n\pi}{(b-a)^2}(1-(-1)^n)\sin\left(\frac{n\pi(x-a)}{b-a}\right)e^{-\frac{n^2\pi^2t}{2(b-a)^2}}$. You can't do much better than this but it could be useful for simulations.

$\endgroup$
0
$\begingroup$

Let $f_{\tau_a}(t)$ and $f_{\tau_b}(t)$ be the density functions of $\tau_a$ and $\tau_b$. Let $F_{\tau_a}(t)$ and $F_{\tau_b}(t)$ be the corresponding CDFs, then, assuming that $\tau_a$ and $\tau_b$ are independent, we get,

$$P(\tau > t) = P(\tau_a > t, \tau_b > t) = P(\tau_a>t)P(\tau_b>t) = (1-F_{\tau_a}(t))(1-F_{\tau_b}(t))$$

$$\implies F_{\tau}(t) = 1-(1-F_{\tau_a}(t))(1-F_{\tau_b}(t))$$

$$\implies f_{\tau}(t) = f_{\tau_a}(t)(1-F_{\tau_b}(t)) +f_{\tau_b}(t)(1-F_{\tau_a}(t))$$

The $3$rd equation follows by taking derivative of $2$nd equation with respect to $t$.

$\endgroup$
  • $\begingroup$ With the independence assumption , of course it makes things easier $\endgroup$ – Canardini Jun 15 '17 at 13:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.