Seeking Guidance on Deriving the Passage Time Distribution for Geometric Brownian Motion

Question

I’m looking for some help from those with a solid understanding of stochastic processes. My classmate and I are studying how to derive the passage time distribution for a one-dimensional Geometric Brownian Motion (GBM) starting from a distribution . Specifically, we are interested in the passage time to reach a boundary.

We’ve been going through two different books for this topic. One is Brownian Motion and Stochastic Calculus by Karatzas and Shreve, and the other is Stochastic Processes for Physicists: Understanding Noisy Systems by Kurt Jacobs. However, we’ve encountered a few issues.

The problem is that these books approach the solution of the one-dimensional passage time problem in different ways:

Jacobs (2010) addresses this in Chapter 7.7 using the Fokker-Planck equation, but since this book is not a strictly mathematical text, some of the equation symbols are not well-defined. Karatzas & Shreve (1991) cover the topic in Chapter 2.8. According to the learning flowchart in the book, they seem not to use the Fokker-Planck equation, but the preceding material is quite dense, and there are many exercises. I’ve only completed the sections on Markov property and strong Markov property, but I’m still unsure of their relevance to the derivation. I’d appreciate it if someone familiar with this topic could guide us on which method is the correct approach for deriving the passage time distribution. Also, are there any more efficient learning strategies to approach this problem?

Thank you in advance! (I’ve attached the electronic versions of both books and my current study notes for reference.)

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I would like to express my gratitude to @​Kurt G. and ​@Snoop for their insightful suggestions, which greatly helped me approach this problem. With the assistance of a fellow student from my university, I was able to complete the derivation, and the detailed solution can be found in this GithubLink. This note is dedicated to deriving the first moment of the hitting time for a geometric Brownian motion with drift, defined by: \begin{equation*} dS_t=\mu S_tdt+\sigma S_tdW_t,\quad \text{where} \quad \mu \le 0, S_0 > a, \end{equation*} when it first hits a given lower boundary $a$. The derivation and an illustrative example are provided in the linked document.

Answer 1

Let $\tau_z:=\inf\{t:X_t=z\},z>0$ where $X$ is the GBM given by the unique strong solution of the SDE $dX_t=\mu X_t dt+\sigma X_t dW_t$. In addition to @KurtG.'s comment, a standard approach is to find the Laplace transform of $\tau_z$, which characterises the distribution. Suppose $X_0=x\in (0,z)$ and $\mu>\sigma^2/2$. In particular, the Laplace transform of $\tau_z$ given by $$E_x[e^{-\alpha \tau_z}]=\bigg(\frac{x}{z}\bigg)^{\sqrt{(\mu/\sigma^2-1/2)^2+2\alpha/\sigma^2)}-(\mu/\sigma^2-1/2)},\,\alpha \geq 0$$ and the density of $\tau_z$ you can find in (Borodin, Salminen 2.0.2 p. 622) with a change of parameters. To see this, first note that $\ln(z/x)>0$ and $$\begin{aligned} \tau_z=\inf\{t:(\mu-\sigma^2/2)t+\sigma W_t=\ln(z/X_0)\} \end{aligned}$$ so that $E_x[\tau_z]=\frac{\ln(z/x)}{\mu-\sigma^2/2}\in (0,\infty)$. The function $u(x)=E_x[e^{-\alpha \tau_z}]$ solves the ODE $\frac{\sigma^2x^2u''}{2}+\mu x u'=\alpha u$ on $(0,z)$ with boundary condition $u(z)=1$. You can find a general result along these lines in (Karatzas, Shreve Prop.7.2 p. 364). A formal argument would use Ito's formula on $(x,t)\mapsto u(x)e^{-\alpha t},\alpha>0$ for $u(x)\in C^2,u(z)=1$ which yields $$\begin{aligned} d(u(X_t)e^{-\alpha t})&=e^{-\alpha t}\bigg(\mu X_t u'(X_t)+\frac{\sigma^2}{2}X_t^2u''(X_t)-\alpha u(X_t)\bigg)dt\\ &+\sigma e^{-\alpha t}X_tu'(X_t)dW_t \end{aligned}$$ Under the assumption that the $dW_t$ part is a martingale and $u$ solves the ODE and is bounded, we get $$u(x)=E_x[u(X_{t\wedge \tau_z})e^{-\alpha {(t\wedge \tau_z)}}]\stackrel{t\to \infty}{\to}=E_x[e^{-\alpha {\tau_z}}]$$ since the paths of $X$ are continuous and $P(\tau_z<\infty)=1$ and $X_{\tau_z}=z$ $P$-a.s. Now, if $X_0\sim \mu,\mu((z,\infty))=0$ we have $E[e^{-\alpha \tau_z}]=\int_{(0,z]}E_x[e^{-\alpha \tau_z}]\mu(dx)$, so one approach would be to hopefully work out a density by inverting the Laplace transform (if feasible).