Expected time that a geometric brownian motion falls outside an interval.

Question

I was thinking about the use of geometric brownian motion for modeling stock prices after reading Chapter 21 of Achim Klenke's Probability Theory, and while playing around with the equations, I got stuck on the following problem about geometric brownian motion. Suppose that the stochastic process $(X_t)_{t \in [0,\infty)}$ follows geometric brownian motion. That is, for all $t \in [0,\infty)$, $$X_t = X_0e^{\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W_t},$$ where $W_t$ is the Wiener Process, $\mu$ and $\sigma$ are some constants, and $X_0 > 0$. Assume that $X_0 \in [a-b,a+b] \subset [0,\infty)$, and let $\tau$ be the first time $t$ that $X_t \notin [a-b,a+b]$. That is, $$\tau = \inf\{t > 0 : X_t \notin [a-b,a+b]\}.$$ Then, what is the expected value of $\tau$ (i.e., $E[\tau]$)? I thought the Reflection Principle for Wiener processes would be helpful, but the absolute value causes a lot of problems. My guess is that this is not going to have a closed form solution, but it would be awesome if this has some series that would allow for a close estimate computationally.

Here is my attempt to simplify $\tau$ based on @Kolakoski54 comment. $$\begin{align} \tau &= \inf\{t > 0 : X_t \notin [a-b,a+b]\} \\ \tau &= \inf\left\{t > 0 : X_0e^{\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W_t} \notin [a-b,a+b]\right\} \\ \tau &= \inf\left\{t > 0 : e^{\left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W_t} \notin \left[\frac{a-b}{X_0},\frac{a+b}{X_0}\right]\right\} \\ \tau &= \inf\left\{t > 0 : \left(\mu-\frac{\sigma^2}{2}\right)t+\sigma W_t \notin \left[\ln{\frac{a-b}{X_0}},\ln{\frac{a+b}{X_0}}\right]\right\} \\ \tau &= \inf\left\{t > 0 : \frac{1}{\sigma}\left(\mu-\frac{\sigma^2}{2}\right)t + W_t \notin \left[\frac{1}{\sigma}\ln{\frac{a-b}{X_0}},\frac{1}{\sigma}\ln{\frac{a+b}{X_0}}\right]\right\} \\ \tau &= \inf\left\{t > 0 : At + W_t \notin \left[B,C\right]\right\}, \end{align}$$ where $$\begin{align} A &:= \frac{1}{\sigma}\left(\mu-\frac{\sigma^2}{2}\right), \\ B &:= \frac{1}{\sigma}\ln{\frac{a-b}{X_0}}, \\ C &:= \frac{1}{\sigma}\ln{\frac{a+b}{X_0}}. \end{align}$$ But, I am not sure where to go from here.

Answer 1

Some thougts: First, to lessen to notation, I will write the interval $[A,B]$, instead of $[a-b,a+b]$ and assume $\sigma >0$. Then

$$ X_t \in [A,B] \iff \frac{1}{\sigma} \log\left( \frac{A}{X_0} \right) \leq W_t + \left( \frac{\mu}{\sigma} - \frac{\sigma}{2} \right)t \leq \frac{1}{\sigma} \log\left( \frac{B}{X_0} \right). $$

Let $C : = -\left( \frac{\mu}{\sigma} -\frac{\sigma}{2} \right) $, $D := \frac{1}{\sigma} \log\left( \frac{A}{X_0} \right) $, $E := \frac{1}{\sigma} \log\left( \frac{B}{X_0} \right)$, $\widetilde W := W-Ct $. Then, by Girsanov Theorem, $ (\widetilde W_t)_{0 \leq t \leq T}$ is a Brownian motion under the measure

$$ \forall \in \mathcal F, \quad \mathbb Q(O) := \mathbb E_{\mathbb P}\left[1_O\exp\left( CW_T - \frac{C^2}{2}T\right) \right] $$

Now we have

\begin{align} \mathbb P(\tau \leq T) &= \mathbb E_{\mathbb Q} \left[ 1_{\tau \leq T} \exp\left(- CW_T + \frac{C^2}{2}T\right) \right] \\ &= \mathbb E_{\mathbb Q} \left[ 1_{\tau \leq T} \exp\left(- C \widetilde W_T - \frac{C^2}{2}T\right) \right] \quad (*) \end{align}

Note that under $\mathbb Q$, $\tau$ is the first exit time of the Brownian motion $\widetilde W$ outside the interval $[D,E]$ and its distribution under $\mathbb Q$ is well-known. So we need to calculate the last term. When $\tau$ is just the hitting time of some level, say the first hitting time of level $E$, then this computation is know, check these answers.

It is tempting to imitate the answer of @user126540 there by reasoning

$$ \mathbb P(\tau \in dt) = \mathbb Q(\tau \in dt, \widetilde W_\tau = E)\exp\left(-CE-\frac{C^2}{2}T\right) + \mathbb Q(\tau \in dt, \widetilde W_\tau = D)\exp\left(-CD-\frac{C^2}{2}T\right)$$

But I'm not sure about the rigor and the joint density of $(\tau,W)$ is unknown is this case.

However, to calculate the probability $(*)$, we can use simulations to generate samples of Brownian motions on each interval $[0,T]$ for each T, and from this we can also approximate the expectation of $\tau$.

Another way to calculate $(*)$ could be try to compute for each $t>0$

$$ E[1_{\tau \leq t} \widetilde W_t^n], \quad n \in \mathbb N. $$

Again, this expectation can be estimated by simulations though.