# Laplace transform of Geometric Brownian Motion Hitting Time

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.

• Assuming $\mu>0$, the indicator takes care of itself because $e^{-\mu\tau}=0$ on $\{\tau=\infty\}$. – John Dawkins Jul 1 '17 at 16:27
• @JohnDawkins so are you saying I can simply state that: $\mathbb{E}[\mathbf{1}_{\{\tau \, < \, \infty\}}e^{-\mu \tau}] = \mathbb{E}[e^{-\mu \tau}]$? – Morris Fletcher Jul 1 '17 at 16:53
• @JohnDawkins There is something that bugs me, I have the impression that then I can both show that $\mathbb{E}[e^{-\mu \tau}] = x_0/x$ as you said, as well as $\mathbb{E}[\mathbf{1}_{\{\tau \, < \, \infty\}}] = x_0/x$ by letting $\mu \rightarrow \infty$ and invoking some theorem like dominated or monotone convergence. Where is the problem here? – Morris Fletcher Jul 1 '17 at 16:57
• Don't forget that $\tau$ itself depends on $\mu$! – John Dawkins Jul 1 '17 at 17:11
• Oh oh oh, true that! I hadn't realised it. – Morris Fletcher Jul 1 '17 at 17:13

## 1 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.