Expectation of $e^{-4B_\tau}$, where $\tau$ is an extended stopping time

Question

This is an specific example so with a bit of luck I can get some general methodology from your answers.

I have this stopping time: $$ \tau = \inf\{t \geq 0; B_t < t-2 \} $$

This is a clear example of the hitting time of the process $B_t$ to an open set, hence $\tau$ is an extended stopping time. I am now trying to calculate: $$ \mathbb Ee^{-4B_{\tau}} $$

My problem is that now I cannot substitute $B_{\tau}$ by any value at all like I usually do, because the stopping time is not of the form $B_t = x$. I was thinking of considering the event $B_t = t-3$ to try and compute this, but I am not sure if this is valid.

Any hints are more than welcome.

Answer 1

Hints

1. It follows from the continuity of the sample paths that $$B_{\tau} = \tau-2.$$ (Suppose that $B_{\tau}<\tau-2$. Then the continuous function $f(t) := t-2-B_{t}$ satisfies $f(\tau)>0$. Hence, there exists $s<\tau$ such that $f(s)>0$, i.e. $B_s<s-2$. This contradicts the definition of $\tau$.)
2. Using Itô's formula (or some straightforward calculations) it is not difficult to show that $$M_t := \exp\bigg(-2(B_t+t) \bigg)$$ is a martingale.
3. Apply the optional stopping theorem and the dominated convergence theorem in order to deduce $$\mathbb{E}M_{\tau} =1.$$ From 1. it follows that $$1 = \mathbb{E}M_{\tau} = \mathbb{E}\exp(-2((\tau-2)+\tau) = \exp(4) \cdot \mathbb{E}\exp(-4\tau)$$ i.e. $\mathbb{E}\exp(-4\tau) = e^{-4}$. Hence, $$\mathbb{E}e^{-4B_{\tau}} = \mathbb{E}e^{-4(\tau-2)} = e^{-4} e^{8} = e^4.$$