Let $(X_t)_{t\geq0}$ be Brownan motion on $\mathbb R$, and $\tau$ is a stopping time adapted with the natural filtration generated by the Brownian motion. If $X_0=0$, $E(e^{\tau/2})<+\infty$. Then $E(e^{X_{\tau}-\tau/2})=1$
I think because $e^{X_{t}-t/2}$ is martingale, then Doob's stopping time theorem can be used to get the result.
I am familier with the bounded stopping time version of Doob's stopping time theorem, but be confused with the version without the bounded condition on stopping time. As I have come across some versions of the theorem when the "bounded" dropped, but the conditions added to the martingale are not the same.
For the question above, which property of the martingale $X_{t}-t/2$ garante to use Doob's stopping time theory? Furthermore, can someone give me some advice to remember the Doob's stopping time theorem with unbounded stopping time?
Thank you very much!
Here is a sketch of the proof - a more detailed proof can be found for instance in "Brownian Motion - An Introduction to Stochastic Processes" - René L. Schilling/Lothar Partzsch (p.58).
Let $c \in (0,1)$ arbritary and $p>1$ such that $p < \frac{1}{c \cdot (2-c)}$. The process $$M_t := \exp \left( c \cdot B_{t \wedge \tau} - \frac{1}{2} c^2 \cdot (t \wedge \tau) \right) \qquad (t \geq 0)$$ is a martingale by optional stopping. Moreover, $$\begin{align} \mathbb{E}(M_t^p) &= \mathbb{E} \left[ \exp \left( pc \cdot (B_{t \wedge \tau} - \frac{1}{2} (t \wedge \tau)) \right) \cdot \exp \left( \frac{1}{2} pc \cdot (1-c) \cdot (t \wedge \tau) \right) \right] \\ &\leq \underbrace{(\mathbb{E}(M_t))^{pc}}_{1} \cdot \left(\mathbb{E} \left[ \exp \left( \frac{1}{2} \frac{pc \cdot (1-c)}{1-pc} \tau \right) \right] \right)^{1-pc} \leq (\mathbb{E}e^{\frac{\tau}{2}})^{1-pc} \end{align}$$ where we applied Hölder's inequality (to $\frac{1}{pc}$, $\frac{1}{1-pc}$) and used $$pc \cdot \frac{1-c}{1-pc} \leq pc \cdot \frac{1-c}{1-\frac{1}{2-c}} = pc \cdot (2-c) <1$$ in the last step. Note that $\frac{1}{1-pc}$ is well-defined since $p< \frac{1}{c \cdot (2-c)}<\frac{1}{c}$. By assumption $\mathbb{E}e^{\frac{\tau}{2}}<\infty$, so this calculation shows that $(M_t)_{t \geq 0}$ is uniformly integrable. Therefore, there exists $\lim_{t \to \infty} M_t$ in $L^1$ and a.s. and this shows $\mathbb{E}\exp \left(c \cdot B_{\tau}- \frac{1}{2} c^2 \cdot \tau \right)=1$ for $c \in (0,1)$. Now we apply again Hölder inequality (to $\frac{1}{c}$, $\frac{1}{1-c}$) to obtain
$$1 = \mathbb{E}\exp \left[c \cdot \left(B_{\tau}-\frac{1}{2} \tau \right) + \frac{1}{2} c \cdot (1-c) \cdot \tau \right] \\ \leq (\mathbb{E}e^{B_{\tau}-\frac{\tau}{2}})^c \cdot (\mathbb{E}e^{\frac{1}{2} c \cdot \tau})^{1-c} \to \mathbb{E}e^{B_{\tau}-\frac{\tau}{2}} \quad (c \to 1)$$ i.e. $\mathbb{E}e^{B_{\tau}-\frac{\tau}{2}} \geq 1$. On the other hand, by applying Fatous lemma one can easily prove $\mathbb{E}e^{B_{\tau}-\frac{\tau}{2}} \leq 1$: $$\mathbb{E}e^{B_{\tau}-\frac{\tau}{2}} \leq \liminf_{t \to \infty} \mathbb{E}e^{B_{t \wedge \tau}- \frac{1}{2} (t \wedge \tau)} = 1$$