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This question is under the topic of Brownian Motion. The question: question enter image description here

I don't understand how the P{down 2 before up 1} can translate into 1/3. What's the logic behind that?

Thanks a lot!

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As was already mentioned to you, it is quite unfortunate that you say nothing of what you know about the topics your posts mention... Here is an explanation of the present one, using the symmetry of Brownian motion.

Call $u(x)$ the probability that the path hits $-1$ before $2$, starting from $x$ in $[-1,2]$. Then $u(-1)=1$ and $u(2)=0$. Furthermore, for every $x$ in $(-1,2)$, consider some positive $z$ such that $-1\lt x-z$ and $x+z\lt 2$. To hit $\{-1,2\}$ starting from $x$, one must hit $\{x-z,x+z\}$ first. By symmetry, one hits $x-z$ or $x+z$ first with equal probabilities. Once at $x\pm z$, one must hit $\{-1,2\}$, and the point where one hits first does not depend on the way the path reaches $x\pm z$ (this is not obvious, but is called the strong Markov property).

All this yields $u(x)=\frac12(u(x+z)+u(x-z))$ for every suitable $x$ and $z$. Thus, the function $x\mapsto u(x)$ is affine. Since one knows $u$ at the boundary points, this yields $u(x)=\frac13(2-x)$, in particular the quantity you are after is $u(1)=\frac13$.

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