# Expected value of exponential of hitting time of GBM

We have a stopping time $$\tau=\inf\{t\geq 0: S_0e^{\sigma B_t+(r-\sigma^2/2)t}=S^* \}$$ where $$S_0,\sigma,r,S^*$$ are constants and $$S^*, and $$B_t$$ is a brownian motion. I wish to compute the expected value $$\mathbb{E}[1_{\tau \leq \infty }e^{-r\tau}].$$ I realize that we can see that $$\tau$$ can be rewritten as $$\tau=\inf\{t\geq 0: B_t+\frac{(r-\sigma^2/2)}{\sigma}t=\frac{1}{\sigma}\log{\frac{S^*}{S_0}} \}$$ and that it from here would be possible to see that if we do an appropriate measure change, $$B_t+\xi t$$ will be a brownian motion under the new measure. What I tried to do was to set the RN-derivative $$e^{\sigma B_{\tau}-\sigma^2/2\tau }$$

But I never get anywhere further than this -I try to perform the meausre change in order to end up only with the expectance that $$\tau$$ is finite, but I never manage to do that.

## 1 Answer

Let $$W_t= -B_t$$. Moreover, let $$a= - \frac{r-\frac{1}{2}\sigma^2}{\sigma}$$ and $$b= -\frac{1}{\sigma}\ln \frac{S^*}{S_0}$$. Then, as in this question, \begin{align*} \mathbb{P}\left(\tau \ge T \mid W_T\right)\pmb{1}_{\{W_T \le b-aT\}} &= \mathbb{P}\left(W_t+at \le b, t\in[0, T] \mid W_T\right)\pmb{1}_{\{W_T \le b-aT\}}\\ &=\Big[1-\exp\Big(-\frac{2}{T}b\big(b-W_T-aT\big)\Big)\Big] \pmb{1}_{\{W_T \le b-aT\}}, \end{align*} and, consequently, \begin{align*} \mathbb{P}(\tau \ge t) &= \Phi\left(\frac{b-at}{\sqrt{t}}\right) - e^{2ab}\Phi\left(\frac{-b-at}{\sqrt{t}}\right), \end{align*} where $$\Phi$$ is the cumulative distribution function of a standard normal random variable. The density function is then given by \begin{align*} \frac{b}{\sqrt{2\pi t^3}}e^{ab}e^{-\frac{1}{2}\left(\frac{b^2}{t}+a^2t \right)}\pmb{1}_{t> 0} = \frac{b}{\sqrt{2\pi t^3}} e^{-\frac{1}{2}\left(\frac{b-at}{\sqrt{t}}\right)^2}\pmb{1}_{t> 0} \end{align*} Therefore, \begin{align*} E\left(\pmb{1}_{\tau \leq \infty }e^{-r\tau}\right) &= \int_0^{\infty} e^{-rt}\frac{b}{\sqrt{2\pi t^3}} e^{-\frac{1}{2}\left(\frac{b-at}{\sqrt{t}}\right)^2}dt \\ &=e^{ab - b \sqrt{2r+a^2}}\int_0^{\infty} \frac{b}{\sqrt{2\pi t^3}} e^{-\frac{1}{2}\left(\frac{b-\sqrt{2r+a^2}t}{\sqrt{t}}\right)^2}dt \\ &=e^{ab - b \sqrt{2r+a^2}}\left[\Phi\left(\frac{b-\sqrt{2r+a^2}t}{\sqrt{t}}\right) - e^{2b\sqrt{2r+a^2}}\Phi\left(\frac{-b-\sqrt{2r+a^2}t}{\sqrt{t}}\right) \right]_{\infty}^0\\ &=e^{ab - b \sqrt{2r+a^2}}\\ &=\left(\frac{S^*}{S_0}\right)^{\frac{2r}{\sigma^2}}. \end{align*}

Alternative Solution

We define the probability measure $$\tilde{P}$$ such that \begin{align*} \frac{d\tilde{P}}{dP}\big|_t = e^{-\frac{1}{2}a^2 t - aW_t}, \end{align*} where $$P$$ is the original probability measure. Then $$\tilde{W}_t = W_t + at$$ is a standard Brownian motion under $$\tilde{P}$$. Let $$E$$ and $$\tilde{E}$$ be expectations with respect to measures $$P$$ and $$\tilde{P}$$. Then, \begin{align*} E\left(\pmb{1}_{\tau \leq \infty }e^{-r\tau}\right) &= \tilde{E}\left(\frac{dP}{d\tilde{P}}\big|_{\tau} \pmb{1}_{\tau \leq \infty }e^{-r\tau}\right)\\ &= \tilde{E}\left(\left(\frac{d\tilde{P}}{dP}\big|_{\tau}\right)^{-1} \pmb{1}_{\tau \leq \infty }e^{-r\tau}\right)\\ &= \tilde{E}\left(e^{\frac{1}{2}a^2 \tau + aW_{\tau}} \pmb{1}_{\tau \leq \infty }e^{-r\tau}\right)\\ &= \tilde{E}\left(e^{-\frac{1}{2}a^2 \tau + a\tilde{W}_{\tau}} \pmb{1}_{\tau \leq \infty }e^{-r\tau}\right)\\ &= \tilde{E}\left(e^{-\frac{1}{2}\left(2r+a^2\right) \tau + \sqrt{2r+a^2} \tilde{W}_{\tau} +(a-\sqrt{2r+a^2}) \tilde{W}_{\tau} } \pmb{1}_{\tau \leq \infty }\right)\\ &= \tilde{E}\left(e^{-\frac{1}{2}\left(2r+a^2\right) \tau + \sqrt{2r+a^2} \tilde{W}_{\tau} +(a-\sqrt{2r+a^2}) b } \right)\\ &= e^{(a-\sqrt{2r+a^2}) b}, \end{align*} by the optional sampling theorem.

• Hi, are you sure? Your answer is not in line with what I would expect, I am trying to understand what is going on in for instance: stat.uchicago.edu/~lalley/Courses/391/Lecture15.pdf section 4, lemma 1... – edo May 18 '19 at 9:28
• You need one more step algebraic manipulation; see the edit. – Gordon May 18 '19 at 11:25