5

Problem

Let $a>0$ and $B$ be a standard $\mathbb{R}$-valued Brownian motion. Define the stopping time $S_a:=\inf\{t\geq 0\ \vert \left\lvert B_t\right\rvert = a\}$. Compute $\mathbb{E}\left[e^{-\frac{\lambda^2}{2}S_a}\right]$ for $\lambda\in\mathbb{R}$.

My (edited) attempt

I noticed that the expression I needed to compute looks similiar to the martingale $Z=(e^{\lambda B_t - \frac{1}{2}\lambda^2t})_{t\geq 0}$. Consider the bounded stopping time $S_a \land n\leq n$ for some $n\in\mathbb{N}$. Doob’s optional sampling theorem yields:

\begin{align*} \mathbb{E}\left[Z_{S_a\land n}\right] = \mathbb{E}\left[Z_0\right] = \mathbb{E}\left[e^{\lambda B_0 - \frac{1}{2}\lambda^20}\right] = 1\ \text{(a.s.)} \end{align*}

We now have $\lim_{n\to\infty} Z_{S_a\land n} = Z_{S_a}$ (a.s.) and: $$\left\lvert Z_{S_a\land n}\right\rvert = e^{\lambda B_{S_a\land n} - \frac{1}{2}\lambda^2\left(S_a\land n\right)} \leq e^{\lambda B_{S_a\land n}} \leq e^{\left\lvert \lambda a\right\rvert}\ \text{(a.s.)}$$

Lebesgue’s dominated convergence theorem yields: $$\mathbb{E}\left[Z_{S_a}\right] = \mathbb{E}\left[\lim_{n\to\infty} Z_{S_a\land n}\right] = \lim_{n\to\infty} \mathbb{E}\left[Z_{S_a\land n}\right] = \lim_{n\to\infty} 1 = 1\ \text{(a.s.)}$$

Because of symmetry we have: $$x:=\mathbb{E}\left[e^{-\frac{\lambda^2}{2}S_a}1_{\left\{B_{S_a}=a\right\}}\right]=\mathbb{E}\left[e^{-\frac{\lambda^2}{2}S_a}1_{\left\{B_{S_a}=-a\right\}}\right]\quad \text{(a.s.)}$$ $$\mathbb{E}\left[e^{-\frac{\lambda^2}{2}S_a}\right] = \mathbb{E}\left[e^{-\frac{\lambda^2}{2}S_a}1_{\left\{B_{S_a}=a\right\}}\right]+\mathbb{E}\left[e^{-\frac{\lambda^2}{2}S_a}1_{\left\{B_{S_a}=-a\right\}}\right] = 2x\quad \text{(a.s.)}$$ $x$ can now be computed through the first equation: \begin{align*} 1 =\ &\mathbb{E}\left[e^{\lambda B_{S_a} - \frac{1}{2}\lambda^2S_a}\right]\\ =\ &\mathbb{E}\left[e^{\lambda B_{S_a} - \frac{1}{2}\lambda^2S_a} 1_{\left\{B_{S_a}=a\right\}}\right] + \mathbb{E}\left[e^{\lambda B_{S_a} - \frac{1}{2}\lambda^2S_a} 1_{\left\{B_{S_a}=-a\right\}}\right]\\ =\ &\mathbb{E}\left[e^{\lambda a - \frac{1}{2}\lambda^2S_a} 1_{\left\{B_{S_a}=a\right\}}\right] + \mathbb{E}\left[e^{-\lambda a - \frac{1}{2}\lambda^2S_a} 1_{\left\{B_{S_a}=-a\right\}}\right]\\ =\ &e^{\lambda a}\mathbb{E}\left[e^{-\frac{1}{2}\lambda^2S_a} 1_{\left\{B_{S_a}=a\right\}}\right] + e^{-\lambda a}\mathbb{E}\left[e^{- \frac{1}{2}\lambda^2S_a} 1_{\left\{B_{S_a}=-a\right\}}\right]\\ =\ &e^{\lambda a}x + e^{-\lambda a} x\\ =\ &\left(e^{\lambda a} + e^{-\lambda a}\right) x\quad \text{(a.s.)} \end{align*} The result is therefore: $$\mathbb{E}\left[e^{-\frac{\lambda^2}{2}S_a}\right] = \frac{2}{e^{\lambda a} + e^{-\lambda a}}\quad \text{(a.s.)} $$

Questions

  1. I am not very comfortable with how the result looks. Is my proof correct?
  2. I feel like I used a trick to obtain the result. Is there a more general way to make this computation?
Wilfred Montoya
  • 651

1 Answers1

1

All good. Perhaps, you can add more details on the "by symmetry" Independence of $T$ and $B_T$ i.e. we have $B_t\stackrel{d}{=}-B_{t}$ and $S_{a}$ is only a function of $|B_{t}|$, which is symmetric already.

One can also use pde methods to find the law for the exit time double barrier stopping time density function.

Here is also the more general formula for exit time from (a,b) Distribution of first exit time of Brownian motion, where symmetry doesn't apply and instead one has to use the strong Markov property.

Thomas Kojar
  • 7,349