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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?

saz
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higuys
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1 Answers1

6

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.
saz
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  • 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? – higuys Jan 28 '17 at 15:26
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    @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.) – saz Jan 28 '17 at 15:34
  • sorry my bad, the drift is $\mu>0$ indeed. – higuys Jan 28 '17 at 16:32
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    @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 $\mathbb{P}(\tau<\infty)=1$. – saz Jan 28 '17 at 17:55
  • Your answer is very useful @saz. I have a couple of questions if I may: first, I do not understand why the optional sampling theorem is needed; given the stopped process is a martingale and that $t, \tau >0$, couldn't we just state that: $\mathbb{E}[M_{t \wedge \tau}] = \mathbb{E}[M_{0 \wedge \tau}] = \mathbb{E}[M_0] = 1$? My second question refers to the conditions under which we can state that a stopped process $X_{\tau}$ takes the value defining the stopping time, $a$ in this case. Indeed, the way I see it $X_{\tau}$ is not a random variable. Would you have any reference? – Morris Fletcher Jun 30 '17 at 10:19
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    @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). – saz Jun 30 '17 at 10:48
  • @saz My understanding is that the optional sampling theorem (OST) $-$ at least as stated in Wikipedia $-$ is about the process $M_{\tau}$, not $M_{t \wedge \tau}$, and that the claim that $M_{t \wedge \tau}$ is a martingale is a more general result independent of the OST $-$ I interpret $M_{t \wedge \tau}$ as a stochastic process indexed on $t$ so that $M_{0 \wedge \tau}$ makes sense. Isn't that right? For "$X_{\tau}$ is not a random variables" I meant that, that it is constant. Thanks for the sample paths continuity condition. – Morris Fletcher Jun 30 '17 at 11:04
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    @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. – saz Jun 30 '17 at 11:24
  • If we take a Brownian martingale obtained from a Hermite polynomial as in your other solutions, can we get the same result for the entry time of a closed set ${a}$ @saz –  Apr 11 '20 at 19:27
  • @saz Would anything change in our thinking process if we would assume our drift parameter is of negative value , for example equal to -1 ? I don't see where are we using positivity of the parameter. – DGF Jun 15 '20 at 20:00
  • @MorrisFletcher Yes, it does change because then it is no longer clear that $|M_{t \wedge \tau}| \leq e^{\alpha a}$. In the current setting, I use $e^{-\alpha \mu t} \leq 1$, which is of course no longer true if $\mu <0$. – saz Jun 16 '20 at 04:37
  • Does this means the expected stopping time is $a/\mu$, regardless of the volatility? – Lei Hao Nov 08 '21 at 11:22