3
$\begingroup$

Let $\{X_t:t\geq 0\}$ be a Brownian motion with drift $\mu>0$ and define a stopping time $\tau$ by $$\tau=\inf\{t\geq 0:X_t=a\}.$$ Now I want to show that $$\mathbb{E}(e^{-\lambda\tau})=e^{(\mu-\sqrt{\mu^2+2\lambda})a}$$ for $\lambda>0$. Now as a hint I know that I need to use the martingale $M_t=e^{\alpha X_t-\alpha\mu t-\frac{1}{2}\alpha^2t}$. Obviously I need to use Doobs optional stopping theorem but I do not know how. Anyone has a suggestion?

$\endgroup$
2
$\begingroup$

Hints:

  1. Check that for fixed $\alpha>0$ the process $$M_t := \exp \left(\alpha X_t-\alpha \mu t- \frac{1}{2} \alpha^2 t \right)$$ is a martingale (with respect to the canonical filtration of the Brownian motion).
  2. By the optional stopping theorem, $$\mathbb{E}(M_{\tau \wedge t}) = \mathbb{E}(M_0) = 1, \qquad t \geq 0.$$
  3. Show that $|M_{t \wedge \tau}| \leq e^{\alpha a}$. Deduce from the dominated convergence theorem that $$\mathbb{E}(M_{\tau}) = 1.$$
  4. Since $(X_t)_{t \geq 0}$ has continuous sample paths, we have $X_{\tau}=a$. Hence, $$M_{\tau} = e^{\alpha a} \exp \left( - \left[ \mu \alpha + \frac{1}{2} \alpha^2 \right] \tau \right).$$
  5. It follows from step 3 and 4 that $$\mathbb{E} \exp\left( - \left[ \mu \alpha + \frac{1}{2} \alpha^2 \right] \tau \right) = e^{-\alpha a}.$$ Setting $\lambda := \mu \alpha + \frac{1}{2} \alpha^2$ proves the assertion.
$\endgroup$
  • $\begingroup$ Thanks for the help, it helped me a lot! I have a follow-up question though: I want to prove that with this drift $a>0$, that $\tau<\infty$ with probability one by taking the limit $\lambda \to 0$, but I can't figure out how $a>0$ plays a big role in proving this using equation from step 5? $\endgroup$ – higuys Jan 28 '17 at 15:26
  • 1
    $\begingroup$ @higuys I suppose you mean with drift $\mu>0$ (and not drift $a>0$)....? (Just as a side remark: If you find the answer useful, you can upvote it by clicking on the up arrow next to it.) $\endgroup$ – saz Jan 28 '17 at 15:34
  • 1
    $\begingroup$ @higuys I used in step 3 that $\tau<\infty$ almost surely. Note that if $\tau(\omega)=\infty$ for some $\omega$, then $$M_{t \wedge \tau}(\omega) = M_{t}(\omega) \xrightarrow[]{t \to \infty} 0,$$ this follows from the fact that $X_t(\omega) < a$ for all $t$ (as $\tau(\omega)=\infty$) and $\mu>0$. Hence, by the dominated convergence theorem, $$\mathbb{E}(M_{\tau} 1_{\{\tau<\infty\}}+ 0 \cdot 1_{\{\tau=\infty\}})=1.$$ Therefore, we get $$\mathbb{E}(e^{-\lambda \tau} 1_{\{\tau<\infty\}}) = e^{(\mu-\sqrt{\mu^2+2\lambda})a}.$$ Now you can let $\lambda \to 0$ to conclude $\mathbb{P}(\tau<\infty)=1$. $\endgroup$ – saz Jan 28 '17 at 17:55
  • 1
    $\begingroup$ @DaneelOlivaw 1.What do you mean by "couldn't we just state that [...]"? How do you justify that $\mathbb{E}(M_{t \wedge \tau}) = \mathbb{E}(M_{0 \wedge \tau})$? The optional sampling theorem shows that $(M_{t \wedge \tau})_t$ is a martingale and so the assertion follows; I fail to see how to prove this without the optional sampling theorem. 2. Typically you have to assume continuity of the sample paths. What exactly do you mean by "$X_{\tau}$ is not random variable"? $X_{\tau}=a$ is a random variable for sure (well, it's constant, but nevertheless it's a random variable). $\endgroup$ – saz Jun 30 '17 at 10:48
  • 1
    $\begingroup$ @DaneelOlivaw Well, yes, there are several names for this result (optional sampling/optional stopping/...). In my answer I use the fact that $(M_t)_t$ martingale implies that $(M_{t \wedge \tau})_t$ is a martingale, and that's it. $\endgroup$ – saz Jun 30 '17 at 11:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.