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Let $$X_t = m \cdot t + W_t$$ where $W_t$ is a Brownian motion. Let $$Z = \sup \{ t\in [0,1] : X_t = 0\}.$$ It is known that if $m = 0$ then the distribution of $z$ is given by $$\mathbb{P}[Z \leq y ] = \frac{2}{\pi} \arcsin \left( \sqrt{ y } \right).$$

What is the distribution of $z$ for general values of $m$?

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Here is a simple description:

$$Z\stackrel{d}{=}A\cdot Y_m$$ i.e. $Z$ is distributed as the product of two random variables, where the the factors are independent, $A$ is $\arcsin$-distributed, and $Y_m$ is distributed as $\min\{1,\exp(m^2/2)\}$.
($\exp(\lambda)$ denotes an exponentially distributed rv. with exp. $1/\lambda$)

This can be proved using the random walk approximation of BM with drift which was used by Tak'acs (On a generalization of the arc-sine law, Ann. Appl. Prob., $\mathbf{3}$ (1996), 1035-1040.) to study the sojourn time. (In that paper he gave an expression for the distribution of the sojourn time of BM with drift, which was much simpler than the original solution by Akahori). In a similar way one can solve the corresponding random walk problem combinatorially and pass to the limit.

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  • $\begingroup$ Do you know of a reference for the product decomposition of $Z$? $\endgroup$ – HMPanzo Jun 10 '16 at 20:19
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    $\begingroup$ Please see arxiv.org/abs/1601.00609 $\endgroup$ – esg Jun 11 '16 at 15:47
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I don't know if there is a nice closed-form expression for this, but you can try to work it out writing $$ P[Z\leq y] = P[X_t\neq 0 \text{ for }x\in (y,1]] = \int_{-\infty}^{+\infty} f_{y}(v) P_v[X_t\neq 0 \text{ for }x\in (0,1-v]] \, dv, $$ where $f_y(\cdot)$ is the transition density of the drifted BM at time $y$ (see formula $\bf 2$.1.0.6 in Borodin-Salminen), and then use also formula $\bf 2$.1.1.4 for that probability in the integral.

Good luck :)

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To add some more detail to the excellent answer above. Let $\phi$ be the normal density. Using the reflection principle for the Brownian motion with drift yields

$$ \int_{-\infty}^\infty \phi(\frac{v-\mu\,t}{\sqrt{y}}) P_v[X_t \neq 0 \text{ for }t \in (0,1-y)] dv \\ = \int_{-\infty}^\infty \phi(\frac{v-\mu\,t}{\sqrt{y}}) \int_{0}^\infty \left( \phi(\frac{z-v-(1-y)\,t}{\sqrt{1-y}}) - \phi(\frac{-z-v-(1-y)\,t}{\sqrt{1-y}}) \right)\, dz\, dv $$

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