If we have that $B_t$ is a Brownian Motion process, and we define a hitting time as $\nu = \inf\{t: B_t = r\}$ where $r \in \mathbb{R}$, how can I show that:

$$ E(e^{-\alpha\nu}) = e^{-|r|\sqrt{2\alpha}} $$

for $\alpha >0$?

A hint that was given in my book is to consider another random variable,

$$ M_t = e^{\lambda B_t - \frac{\lambda^2t}{2}} $$

where $B_t$ is standard Brownian Motion. The idea here I think is that $M_t = e^{\lambda B_t - \frac{\lambda^2t}{2}}$ is a martingale, which I have already proved. I am able to get the above result, just without the absolute value signs around $r$, does anyone have any ideas?

EDIT: I included my result without the absolute value signs:

$1 = E(M_0)= E(M_{\nu \wedge t}) \to E(M_{\nu}) = 1$ by convergence and martingale properties. Then, $E(M_{\nu}) = 1 \implies E(e^{\lambda r}e^{-\lambda^2 \frac{\nu}{2}}) = 1 \implies E(e^{-\lambda^2 \frac{\nu}{2}}) = \frac{1}{e^{\lambda r}}$. Then we set $\alpha = \frac{\lambda^2}{2}$ to get $E(e^{-\alpha \nu}) = e^{-r \sqrt{2\alpha}}$.

  • $\begingroup$ How do you show the result "without the absolute value signs around $r$"? $\endgroup$ – Did Mar 7 '16 at 0:41
  • $\begingroup$ I added my proof $\endgroup$ – user136503 Mar 7 '16 at 0:52
  • $\begingroup$ "by convergence" is awfully vague. What makes that $E(M_{\nu \wedge t}) \to E(M_{\nu})$, already? $\endgroup$ – Did Mar 7 '16 at 6:10
  • $\begingroup$ @Did I used the Martingale Convergence Theorem for $E(M_{\nu \wedge t}) \rightarrow E(M_\nu)$, and the Optional Stopping Time Theorem for $E(M_{\nu \wedge t}) \rightarrow E(M_0) = 1$ because the stopping time $\nu \wedge t$ is bounded. $\endgroup$ – Broken_Window Dec 24 '16 at 2:02

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