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Consider a particle starting at the origin in $\mathbb{R}^n$ and undergoing Brownian motion. Is there an expression known for the probability of the particle hitting the sphere $S^{n - 1}_r = \{x \in \mathbb{R}^n : \Vert x\Vert = r\}$ within time $t$ (obviously such an expression, if known, will be in terms of $r$ and $t$)? In other words, I am looking for a formula for $\mathbb{P}(\sup _{s \leq t}\Vert B(s)\Vert \geq r)$. Thanks!

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  • $\begingroup$ Do you mean rather $P(\sup_{s\leq t}\|B(s)\|\geq r)$? $\endgroup$ – Serguei Popov Nov 5 '15 at 12:24
  • $\begingroup$ @SergueiPopov Oops, you are right, fixed! $\endgroup$ – user82390 Nov 5 '15 at 12:29
  • $\begingroup$ @SergueiPopov Though, now that you mention it, isn't there some sort of a connection between $\mathbb{P}(\sup_{s \leq t} \Vert B(s)\Vert \geq r)$ and $\mathbb{P}(\Vert B(t)\Vert \geq r)$? I think in the one-dimensional case it is given by the reflection principle. $\endgroup$ – user82390 Nov 5 '15 at 12:33
  • $\begingroup$ Don't know if there is a easy connection between them for Bessel processes. Frankly, I doubt one can find it. $\endgroup$ – Serguei Popov Nov 5 '15 at 12:36
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The process $\|B(t)\|$ is called $n$-dimensional Bessel process (or Bessel process with parameter $\nu=\frac{n}{2}-1$). I think formula $\bf 4$.1.1.4 of Borodin-Salminen "Handbook of Brownian Motion -Facts and Formulae" is what you're looking for.

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    $\begingroup$ A more general case is discussed here: mathoverflow.net/a/96719/1847 $\endgroup$ – Steve Huntsman Nov 5 '15 at 13:31
  • $\begingroup$ @SergueiPopov Very nice! I am very interested in learning more. Could you shed a little light on how one could prove this formula? $\endgroup$ – HSM Dec 2 '15 at 20:37
  • $\begingroup$ I think it would be better to ask that to Borodin or Salminen :) $\endgroup$ – Serguei Popov Dec 2 '15 at 21:31
  • $\begingroup$ @SergueiPopov I will do that, thanks! One quick question, if I may: what does the $x$ in the expression $\mathbb{P}_x(\sup_{0 \leq s \leq t} R^{(n)}_s \geq y)$ denote? Does it mean that the particle starts at the point $x$? $\endgroup$ – HSM Dec 9 '15 at 11:29
  • $\begingroup$ Probably yes, should be the starting point. But I'm travelling now, have no access to the book. $\endgroup$ – Serguei Popov Dec 9 '15 at 16:38
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The distributions and properties of the maximum of a Bessel processes and Bessel bridges are discussed in details in

Jim Pitman: The law of of the maximum of a Bessel bridge

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