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I am considering a brownian motion with negative drift $X_t:=B_t-\mu t$, where $\mu>0$. I want to know what $\mathbb{E}(\tau_1)$ is, where $\tau_1$ is the first time that $X_t$ reaches $1$. Is $\tau_1<\infty$ a.s.? From this question, I suspect that $\tau_1$ is not integrable, since $e^{-t}$ grows slower than $t$.

I know that $X_t$ reaches any negative number in finite time a.s. and the hitting time is integrable since we can look at $-X_t$ and apply this result. A natural question to ask is what about positve numbers? This motivates the above question.

I try to argue as follows: assume $\mathbb{E}(\tau_1)<\infty$, then by Doob's optional sampling theorem, $0=\mathbb{E}(B_{\tau_1\wedge t}) = \mathbb{E}(X_{\tau_1\wedge t})+\mu\mathbb{E}(\tau_1\wedge t)\Rightarrow\mathbb{E}(\tau_1\wedge t)=-\mathbb{E}(X_{\tau_1\wedge t})/\mu$., which is non-negative iff $\mathbb{E}(X_{\tau_1\wedge t})\geq0$. Hence, it is not possible that $\tau_1$ is integrable. Is the argument flawed? If the argument is correct, then this would imply that $\tau_b,\forall b>0$ is not integrable. Strange...

XXX
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  • @KurtG. I have already looked at that question and I don't think it contains my case. And it is not really easy to integrate that density function... – XXX Nov 06 '22 at 20:17
  • That question gives the density of $\tau_1$ for any constant $\mu$. If that's not your case I must have missed something. – Kurt G. Nov 06 '22 at 20:23

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For BM with negative drift $-\mu$, the hitting time $\tau_1$ is infinite with positive probability. This probability is $1-e^{-2\mu}$, see, e.g., Proposition 1.4 page 10 in [1].

Note that for standard BM (without drift) $\tau_1$ is finite with probability 1, yet its expectation is still infinite, which one can deduce, e.g., by optional stopping.

[1] http://www.columbia.edu/~ww2040/4701Sum07/4701-06-Notes-BM.pdf

Yuval Peres
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