I have some question related to expectation and Brownian motion.
For $a,b > 0$ we define stopping time $T_{a,b}= \inf \{t \geq 0: B_t < bt-a \}. $
I have to show that for $b=1$: $E[e^{1/2 T_{a,1}}] = e^a.$
I am a little bit confused how to start this, because I usually had problems where $T$ is hitting time, but here $B_t < bt-a$, so I am not sure how to do this. Thanks for any help.
Elaboration: Abbreviate $T=T_{a,1}$ and $Z_t=\exp(B_t-t/2)$. There is no doubt that $P[T<\infty]=1$. Because $Z=(Z_t)$ is a martingale, so too is $Z$ stopped at time $T$; because $Z_0=1$, $$ 1=E[Z_{T\wedge t}]=E[e^{T/2-a};T\le t]+E[e^{B-t-t/2}; t<T]. $$ The first term on the right side of this display converges to $E[e^{T/2-a}]$ as $t$ increases to $+\infty$ by monotone convergence. It remains to show that the second term converges to $0$. By Girsanov's theorem, because $Z_t1_{\{t<T\}}$ is $\mathcal F_t$-measurable, $$ E[e^{B_t-t/2};t<T]=E[Z_t; t<T]=Q[t<T], $$ where $Q$ is the law of the Brownian motion with drift +1 (that is, the law of $B_t+t$). Thus $$ \lim_{t\to+\infty}E[e^{B_t-t/2};t<T]= Q[T=+\infty]. $$ Finally, $Q[T=+\infty]=0$, because the hitting timeof the boundary $t-a$ by a Brownian motion with unit positive drift has the same distribution as the hitting time of $-a$ by a standard Brownian motion, and the latter hitting time is finite, a.s.