Q) Let $B_t$ be a brownian motion and let $\tau = \text{inf}\{t:B_t=a+bt\}, a>0$. Use the martingale $\text{exp} (\theta B_t-\theta^2t/2)$ with $\theta = b+(b^2+2\lambda)^{1/2}$ to show
$$E_0\text{exp}(-\lambda\tau)=\text{exp}(-a[b+(b^2+2\lambda)]^{1/2})$$
Since $t\wedge \tau$ is a bounded stopping time and by martingale property $1= E_0[\text{exp}(\theta B_{t\wedge \tau}-\theta^2(t\wedge \tau))/2]$ and the answer follows if I send $t\rightarrow \infty$ and apply bounded convergence theorem but may I know why I can do that as $B_{t\wedge \tau} \leq a+b(t\wedge \tau)$ and for bounded convergence theorem, don't we need a constant bound? Thanks.
Best Answer
For $\lambda>0$, $$ \theta B_{t\wedge \tau}-\theta^2(t\wedge \tau)/2\le \theta a-\lambda(\tau \wedge t)\le \theta a. $$ Then, using the bounded convergence theorem, one gets $$ 1=\mathsf{E}_0\left[e^{\theta(a+b\tau)-\theta^2\tau/2}1\{\tau<\infty\}\right]. $$