first passage time, brownian motion

Hi,

If X(t) is Brownian motion in 2D, where X(0) = 0, then we can ask what is the expected time required to first hit a circle of radius R, centered at the origin. This is a First Passage Time problem. I believe that for Brownian motion this is a well understood subject. The First Passage Time, T, is usually expressed in the form of T as a function of R: T(R).

Now, please consider this related problem, What is the expected minimum radius R that fully contains the path of X(t) up to some time T? I'd like to write this as R(T).

Aren't the above two problems completely equivalent? For my work in physics, I'd like to formulate the problem in the second way. But it seems as though mathematicians never formulate it this way. They seem to always look at it as a First Passage Time problem. I would like to call the second formulation of the problem a "Boundary Expansion" problem. But I don't see any such phrase in the mathematics literature.

In summary, my question is, what do I call the function R(T)? It is fully equivalent to a First Passage Time problem (or so I think). But I can't really call it a First Passage Time, because it isn't a time, its a radius. Any suggestions welcome.

Thanks, Chris

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I think you are looking at the radial part of a 2D Brownian motion, which is well-known to be a Bessel process. So you are really asking for the expected value of the Bessel process at time t. This is also well-known. –  John Jiang Feb 15 '12 at 0:57
I interpreted the question as asking about the expected value of the maximum value of the radial part of Brownian motion for $t \leq T$. –  ShawnD Feb 15 '12 at 1:04
Oh sorry you are right. –  John Jiang Feb 15 '12 at 1:08
Yes, you could call in "the expected value of the maximum value of the radial part of the Brownian motion for t<T." Is there a succinct way of referring to this kind of problem? Would you agree this is equivalent to a first passage time problem. –  chris in physics Feb 15 '12 at 4:36
To use your notation, let $T(R)$ be the first time the radial part of the Brownian motion is equal to $R$ and $R(T)$ be the maximum value of the radial part of the Brownian motion for $t \leq T$. Then $\mathbf{P}\{R(T)>x\ }=\mathbf{P}\{T(x)<T \}$. So if you know the distribution of one, then you know the distribution of the other. –  ShawnD Feb 15 '12 at 5:44

While the expected value of first hitting time of circle of radius $r$ is certainly not directly related to the expected value of the maximum radius achieved up to time $t$, the following formula I found in Pitman and Yor's article does show the distributions of the two processes are related in a nice way: $$\displaystyle \mathcal{L}(M_{\delta*}^{-2}) = \mathcal{L}(\tau_\delta)$$ where $M_{\delta*} = M_{\delta*}(1)= \sup_{0 < t < 1} B_\delta(t)$ is the running maximum of the Bessel process of dimension $\delta$ up to time $1$, started at $0$, and $\tau_\delta = \inf\{t: B_\delta(t) =1\}$ is the first hitting time of $1$ of the same Bessel process. See this following linked article:

http://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.38.596

The formula can be seen by realizing that $P(\tau_\delta > t) = P(M_{\delta*(1)} > t) = P(M_{\delta*}(t) > t^{-1/2})$ by Brownian scaling: $c^{1/2} B_t$ has the same law as $B_{ct}$.

Now the distribution function of $\tau_\delta$ is given in the following article as an infinite series, coming from Laplace transforms.

http://arxiv.org/pdf/1106.6132v3.pdf

The expected value of $\tau_\delta$ has a closed formula in terms of Bessel functions (hence the name; see formula 2.1 in the linked article 2). I don't expect $\mathbb{E} M_{\delta*}$ to have a closed form formula since it amounts to $\mathbb{E} \tau_{\delta}^{-1/2}$, i.e., a negative fractional moment.

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