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I am struggling with the following problem:

Let $B$ be a one dimensional Brownian motion and $a,b>0$. Show that $$P[B_t=a + bt \text{ for some } t\geq 0] = e^{-2ab}.$$

The following hint is given: Consider the martingale $(X_t)_{t\geq 0} = (\exp(2bB_t -2b^2 t))_{t \geq 0}$.

I already showed that $(X_t)$ is a martingale but I do not have any idea how I can use this to prove the statement.

Could somebody help me? Thanks in advance!

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    $\begingroup$ You have to define a suitable stopping time and apply the optional stopping theorem. $\endgroup$
    – saz
    Dec 8, 2018 at 16:24
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    $\begingroup$ Sub-hint: $(B_t)$ hits the line $B_t=a+bt$ if and only if $(X_t)$ hits... $\endgroup$
    – Did
    Dec 8, 2018 at 16:25
  • $\begingroup$ Ok, so clearly $P[B_t = a+bt \text{ for some } t\geq0] = P[X_t = \exp(2ba) \text{ for some } t\geq0] = P[\tau < \infty]$ when I define $\tau$ to be the hitting time of $\exp(2ba)$. But how can I use the optional stopping theorem here? It says that $E(X_\tau) = E(X_0)$ if the necessary conditions are fulfilled, and then...? $\endgroup$
    – nabla
    Dec 9, 2018 at 17:21
  • $\begingroup$ See below for further hints. (I didn't get a notification about your comment because you didn't ping me... you have to write @xyz at the beginning of your comment in order to notify the person with username "xyz".) $\endgroup$
    – saz
    Dec 10, 2018 at 15:09

1 Answer 1

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Hints:

  1. Define a stopping time $\tau$ by $$\tau := \inf\{t \geq 0; X_t = e^{2ab}\}.$$ Show that $$\mathbb{P}(\tau<\infty) = \mathbb{P}(\exists t \geq 0: B_t = a+bt). \tag{1}$$
  2. Apply the optional stopping theorem to show that $$\mathbb{E}(X_{t \wedge \tau}) = \mathbb{E}(X_0)=1 \tag{2}$$ for all $t \geq 0$.
  3. Show that $$\lim_{t \to \infty} X_{t \wedge \tau}(\omega)=e^{2ab} \quad \text{for $\omega \in \{\tau<\infty\}$}$$ and $$\lim_{t \to \infty} X_{t \wedge \tau}(\omega)=0 \quad \text{for $\omega \in \{\tau=\infty\}$}$$ (use $\lim_{t \to \infty} B_t/t=0$ almost surely).
  4. By Step 3, we have $$\lim_{t \to \infty} X_{t \wedge \tau} = e^{2ab} \cdot 1_{\{\tau<\infty\}}.$$ Use the fact that $|X_{t \wedge \tau}| \leq e^{2ab}$ to conclude from $(2)$ and the dominated convergence theorem that $$\mathbb{P}(\tau<\infty) = e^{-2ab}.$$
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