Let us consider a process $(X_t)_{t \geq 0}$ which follows a Geometric Brownian Motion (GBM) $-$ assuming both $\mu>0$ and $\sigma>0$:
$$ \begin{align}
& dX_t = \mu X_tdt+\sigma X_tdW_t
\\[3pt]
& X_0 = x_0
\end{align}$$
Letting $w>0$, we define the hitting time of the Brownian motion $W_t$ as:
$$ \tau_W = \min\{t \geq 0: W_t=w\}$$
Letting $\lambda>0$, it is well known that the Laplace transform of the hitting time is given by:
$$ \mathbb{E}\left[e^{-\lambda\tau_W}\right] = e^{-w\sqrt{2\lambda}}$$
Now, letting $x>0$, we define the hitting time of the GBM $X_t$ as:
$$ \tau_X = \min\{t \geq 0: X_t=x\}$$
I am looking for a closed-form formula for the Laplace transform of $\tau_X$ with respect to the drift term:
$$ \mathcal{L}_{\mu}(\tau_X)=\mathbb{E}\left[e^{-\mu \tau_X}\right] $$
Building on the proof for the Laplace transform of the plain Brownian motion, up to now I have tried the following. Given:
$$ X_t = x_0e^{\left(\mu – \frac{\sigma^2}{2}\right)t+\sigma W_t}$$
I have defined the following drifted Brownian motion:
$$ B_t= W_t+\left(\mu – \frac{\sigma^2}{2}\right)\frac{t}{\sigma} $$
So that:
$$ \tau_X = \min\{t \geq 0: B_t=\frac{1}{\sigma}\log \frac{x}{x_0}\}$$
Letting $\tau = \tau_X$, I then have considered the auxiliary process $-$ which is a martingale:
$$ Y_t = e^{\sigma B_t – \mu t} = e^{\sigma W_t – \frac{\sigma^2}{2}t}$$
As well as the stopped process $Y_{t \wedge \tau}$. Given $Y_t$ is a martingale, so does the stopped process, from which we can conclude that:
$$ \mathbb{E}[Y_{t \wedge \tau}] = \mathbb{E}[Y_0] = 1$$
From here, analysing the behaviour of $Y_{t \wedge \tau}$ when $t \rightarrow \infty$ and using the dominated convergence theorem, I got:
$$ \lim_{t \rightarrow \infty}\mathbb{E}[Y_{t \wedge \tau}] = \mathbb{E}\left[\lim_{t \rightarrow \infty}Y_{t \wedge \tau}\right] = \mathbb{E}\left[\mathbf{1}_{\{\tau \, < \, \infty\}}\left(e^{\sigma B_{\tau} – \mu \tau}\right)\right] = 1$$
Thus by definition of $\tau$:
$$ \mathbb{E}\left[\mathbf{1}_{\{\tau \, < \, \infty\}}e^{- \mu \tau}\right] = \frac{x_0}{x} $$
However, I get stuck here as I cannot get rid of the indicator function.
Any hints/ideas on how to continue from here? Alternatively, is there any general closed-form formula for the Laplace transform of the GBM hitting time in all generality?
Any reference would be greatly appreciated.
Best Answer
Let $W$ be standard Brownian motion and $Y$ geometric Brownian motion, i.e., for $t\geq 0$ $$ Y_{t}=y_{0}\exp (\alpha t+\sigma W_{t}), $$ where $y_{0}\in \mathbb{R}_{++}$, $\alpha \in \mathbb{R}$, and $\sigma \in \mathbb{R}_{++}$.
The Handbook of Brownian Motion by Borodin and Salminen says that for any $y\geq y_{0}$ and $r>0$ $$ \Bbb{E}[e^{-r\tau _{y}}]=\left(\frac{y_{0}}{y}\right)^{\kappa}, $$ where $\tau_y$ is the time of the first transition to $y$ and $$ \kappa =\left(\sqrt{\alpha ^{2}+2r\sigma ^{2}}-\alpha \right)\sigma ^{-2} $$ is a strictly positive constant.
Which I think means you are close.