# The probability of a Brownian particle traveling a distance $L$ before returning to its point-of-origin

What's the probability that a Brownian particle diffusing along a one-dimensional interval returns to its point of origin before traveling a distance $L$?

We know that in the limit of a random walk on a discrete lattice of points, a random walker starting at some position $(X = 1)$ has a probability of reaching $L$ before reaching $0$ of $p(L)=\frac{1}{L}$ [Feller, W. An Introduction to Probability Theory and Its Applications (chapter 5). John Wiley and Sons (1958)]. What happens in the limit of a continuous Brownian motion process?

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You seem to be asking three different questions.

1. A Brownian motion returns to its starting position instantly, so the answer to the question in the title is "zero".

2. The question is reversed in the first paragraph of post, the answer now is "one".

3. Finally, if you start a Brownian motion at position 1, the chance that it will hit $L>1$ before hitting zero is $1/L$, just like for the random walk.

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Fix $n\in\mathbb N$ and consider the discrete walk given by the Brownian particle's passages at positions $kL/n$ for $-n\le k\le n$, where once the particle has reached a position $kL/n$ the next passage is recorded when it reaches either $(k-1)L/n$ or $(k+1)L/n$, but not when it reaches $kL/n$ again. This is a simple discrete one-dimensional random walk. After the first step away from the origin, the particle has probability $1/n$ of reaching $\pm L$ before returning to the origin in the discrete walk, and this is an upper bound for the probability of the Brownian motion reaching $\pm L$ before returning to the origin. Since we can choose $n$ freely, it follows that the latter probability is $0$.

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