First hitting time for a brownian motion with a exponential boundary Let $B_t$ be the standard Brownian Motion. Is the distribution/density of the first hitting time of $B_t$ for an exponential decaying boundary known?
Trying to be more formal, if
$$T=\inf\{t\geq0,B_t\geq e^{-\lambda t}\}$$ with $\lambda>0$, what is
$$E[T]$$
If it is known is it also known when, instead of a Brownian Motion, one has a simple Ornstein-Uhlenbeck Process with mean $0$?
Thank you very much
 A: Since $B_t \to \infty$ a.s. as $t \to \infty$, we conclude from the intermediate value theorem that $T<\infty$ a.s. Moreover, by the continuity of the sample paths,
$$\mathbb{E}B_T = \mathbb{E}(e^{-\lambda T}) \neq 0$$
and therefore Wald's identities imply that $T$ is not integrable (if $T$ would be integrable, then $\mathbb{E}B_T = 0$).
Concerning the Ornstein-Uhlenbeck process: Let 
$$X_t = \sigma \cdot e^{b \, t} \cdot \int_0^t e^{-b \, s} \, dB_s$$
where $\sigma>0$. Then
$$X_t \geq e^{-\lambda \, t} \Leftrightarrow M_t := \int_0^t e^{-b \, s} \, dB_s \geq \frac{1}{\sigma} e^{-(\lambda+b) \, t}$$
$M$ is a martingale, $M_0 =0$. A similar calculation as in the proof of Wald's identity shows that $\mathbb{E}M_T = 0$ for any integrable stopping time $T \in L^1$. With the same reasoning as above, we conclude $T \notin L^1$.
A: We know that for a stopping time
$T_a =\inf \{ t> 0 : B_t \geq a\}$ it follows that
$$\mathbb E \left\{\exp(-\frac{\lambda^2}{2}{T_a})\ |\ \mathcal F_s  \right\}= e^{-\lambda a}$$
This result is proven here. So, deviating both sides in $\lambda$ gives us
$$\mathbb E \left\{{T_a}\exp(-\frac{\lambda^2}{2}{T_a})\ |\ \mathcal F_s  \right\}= \frac{ae^{-\lambda a}}{\lambda}$$
then we make $\lambda \rightarrow 0$ and by the monotone convergence theorem we conclude that $T_a \notin L^1 $
