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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

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2 Answers 2

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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 $T \notin L^1$.

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    $\begingroup$ +1. Similar to your argument in the first case: with positive probability $B_1\leqslant-1$ without hitting the curve before time $1$. After that, one needs at least to hit $0$. The mean time this takes is infinite. Hence $E[T]$ is indeed infinite. $\endgroup$
    – Did
    Commented Sep 12, 2013 at 7:12
  • $\begingroup$ @saz: Nice answer $\endgroup$
    – Paul
    Commented Oct 5, 2013 at 16:54
  • $\begingroup$ So basically by saying that $T \notin L^1$ you are saying that it is Infinite? This is not intuitive for me as an Ornstein-Uhlenbeck process with mean $0$ wanders around $0$ and the boundary is approaching $0$. Did I interpret your answer correctly? $\endgroup$
    – gota
    Commented Oct 7, 2013 at 9:20
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    $\begingroup$ @NunoCalaim No. The very first sentence of my answer states that $T$ is almost surely finite. But $T$ is not integrable! $\endgroup$
    – saz
    Commented Oct 7, 2013 at 9:54
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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 $

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