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Let $B_t$ be the standard Brownian motion process, $a > 0$, and let $H_a = \inf \{ t : B_t > a \}$ be a stopping time. I want to show that the Laplace transform of $H_a$ is $$\mathbb{E}[\exp(-\lambda H_a)] = \exp (-\sqrt{2\lambda} H_a)$$ by considering the martingale $$M_t = \exp \left(\theta B_t -\frac{1}{2}\theta^2 t\right)$$

There's an obvious argument to follow here: assuming the optional stopping theorem applies, we have $$1 = \mathbb{E}[M_{H_a}] = \mathbb{E} \left[ \exp \left(\theta a - \frac{1}{2}\theta^2 H_a\right) \right] = \exp(\sqrt{2\lambda} a) \mathbb{E} \left[ \exp(-\lambda H_a) \right]$$ where $\theta = \sqrt{2\lambda}$. This is exactly what we wished to show. However, as far as I can tell, the hypotheses of the optional stopping theorem are not satisfied here. Here is the statement I have:

If $(X_n)$ is a martingale and $T$ is an a.s. bounded stopping time, then $\mathbb{E}[X_T] = \mathbb{E}[X_0]$.

I think not all is lost yet. $M_t > 0$ for all $t$, so the martingale convergence theorem applies, and $M_t \to M_\infty$ a.s. for some integrable random variable $M_\infty$. For each $t$, $H_a \wedge t = \min \{ H_a, t \}$ is a bounded stopping time, so certainly $\mathbb{E}[M_{H_a \wedge t}] = \mathbb{E}[M_0]$. But, $$\mathbb{E}[M_{H_a \wedge t}] = \mathbb{E}[M_{H_a} \mathbf{1}_{\{H_a \le t\}}] + \mathbb{E}[M_t \mathbf{1}_{\{H_a > t\}}]$$ and clearly what one wants to do is to take $t \to \infty$ on both sides. But here's where I get stuck: I'm sure I need a convergence theorem here in order to conclude that the equation remains valid in the limit.

Now, $0 < M_{H_a} = \exp \left(\theta a - \frac{1}{2} \theta^2 H_a \right) \le \exp(\theta a)$, so the dominated convergence theorem applies, and so $$\lim_{t \to \infty} \mathbb{E}[M_{H_a} \mathbf{1}_{\{H_a \le t\}}] = \mathbb{E}[M_{H_a} \mathbf{1}_{\{H_a < \infty\}}]$$ and I believe Fatou's lemma gives me that $$\liminf_{t \to \infty} \mathbb{E}[M_t \mathbf{1}_{\{H_a > t\}}] \ge \mathbb{E}[M_{\infty} \mathbf{1}_{\{H_a = \infty\}}]$$ but I think what I need is the equality $$\lim_{t \to \infty} \mathbb{E}[M_t \mathbf{1}_{\{H_a > t\}}] = \mathbb{E}[M_\infty \mathbf{1}_{\{H_a = \infty\}}]$$ and as far as I can tell neither the monotone convergence theorem nor the dominated convergence theorem applies here. Is there anything I can do to rescue this line of thought?

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2  
Use $0\le M_t\mathbf{1}_{\{H_a>t\}}\le\exp(\theta a-\frac12\theta^2t)$. –  Did Mar 10 '11 at 21:05
    
@Didier: Thanks! Not sure why I keep missing the obvious. –  Zhen Lin Mar 10 '11 at 21:24
3  
+1 Way to show your work! –  Byron Schmuland Mar 10 '11 at 23:34
    
This is to second Byron's comment. In fact, I wanted to mention this in my first comment but I forgot. –  Did Mar 11 '11 at 6:12
2  
@Didier: If you post your comment as an answer, we can vote it up and put this question to bed. –  Nate Eldredge Jun 9 '11 at 19:24

3 Answers 3

up vote 7 down vote accepted

Use the fact that $0\leqslant M_t\mathbf{1}_{\{H_a>t\}}\leqslant\exp\left(\theta a−\frac12\theta^2t\right)$.

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I don't think you need to worry about the case $H_a = \infty$ - for the simple reason that this is an event of probability zero.

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I m not sure that I understand properly the question.
We have $M_{H_a\wedge t}\to M_{H_a}$ almost surely, and
$M_{H_a\wedge t}<e^{\theta.a}$ The right hand side is constant so integrable, so doesn't the dominated convergence readily applies for $t \to +\infty$ ? Am i missing something here ?

Regards

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No, I'm afraid uniform integrability is something I never really understood well and have completely forgotten. –  Zhen Lin Mar 11 '11 at 7:27
    
It is not true that a positive martingale converges in $L^1$. It converges a.s. and is bounded in $L^1$, but may not be uniformly integrable. The critical binary branching process is an example. –  Byron Schmuland Mar 11 '11 at 16:06
    
In fact, the OP's positive martingale does converge to zero almost surely and therefore it cannot be UI. –  Byron Schmuland Mar 11 '11 at 16:10
    
@Byron Schmuland : You are right I edited my post consequently, sorry about that Zhen Lin. –  TheBridge Mar 11 '11 at 16:53

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