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Let $\tau_a=\inf\{t: B_t=a\}$, the hitting time of the standard Brownian motion to reach the boundary $a$.

This is easily derived

$$E(e^{-\lambda \tau_a})=e^{-|a|\sqrt{2\lambda}}$$

But I am having a problem of using this formula to get the moments of $\tau_a$ by matching the coefficients of terms for each power of $\lambda$, specifically, the LHS, if expanded, has all integer powers of $\lambda$; however, the RHS has the various terms of $\sqrt{\lambda}$. How to reconcile the difference here? Could somebody please help? Thanks.

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2 Answers 2

up vote 6 down vote accepted

Well, in fact $E(\tau_a)$ is infinite...

A classic argument to see that $\tau_a$ does not have finite mean is to consider $\tau_{2a}$. On the one hand, to hit $2a$, one has to hit $a$ and then, starting from $a$, to hit $2a$. By the strong Markov property, this second duration is independent of $\tau_a$ and, by invariance by translations, distributed like $\tau_a$. On the other hand, by the scale invariance of Brownian motion, $(2B_t)$ is distributed like $(B_{4t})$, hence $\tau_{2a}$ is distributed like $4\tau_a$. Finally, $\tau_a$ solves a distribution equality: the distribution of $T+T'$ and the distribution of $4T$ coincide, where $T$ and $T'$ are independent copies of $\tau_a$. This is enough to prove that $\tau_a$ is not integrable (and almost enough to deduce its Laplace transform, for that matter).

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right!... forgot to see this. So higher moments will also be infinite then! there is no point to do the matching. correct? :) –  Qiang Li Feb 22 '11 at 21:41
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To add on Didier's answer, note that the PDF of $\tau_a$, for $a > 0$, is given by $$ f_{\tau _a } (t) = \frac{a}{{\sqrt {2\pi } }}t^{ - 3/2} e^{ - a^2 /(2t)} ,\;\; t > 0. $$ From this you immediately see that ${\rm E}(\tau_a) = \infty$.

EDIT: The distribution function of $\tau_a$ can be derived as follows (cf. p.1 here). First note that, for any $t > 0$, $$ {\rm P}(\tau _a < t) = {\rm P}(\tau _a < t,B_t < a) + {\rm P}(\tau _a < t,B_t > a). $$ The right-hand side is equal to $2{\rm P}(\tau _a < t,B_t > a)$, hence to $2{\rm P}(B_t > a)$. Thus, $$ {\rm P}(\tau _a < t) = 2 {\rm P}(B_t > a) = \frac{2}{{\sqrt {2\pi t} }}\int_a^\infty {e^{ - \frac{{y^2 }}{{2t}}} \,{\rm d}u} = \frac{2}{{\sqrt \pi }}\int_{\frac{a}{{\sqrt {2t} }}}^\infty {e^{ - y^2 } \,{\rm d}y}. $$ Differentiating with respect to $t$ gives the above formula for $f_{\tau _a } (t)$. (For arbitrary $a \in \mathbb{R}$, just put $|a|$ instead of $a$ in that formula.)

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