# Hitting time of a sphere by a Brownian particle

Consider a Brownian particle in $$\mathbb{R}^n$$, starting at the origin. Let us consider a sphere of radius $$r$$ in $$\mathbb{R}^n$$ centered at the origin. We know that the probability that the particle strikes the sphere within time $$r^2$$ is in fact a number independent of $$r$$.

My question is, is it known what the (approximate) numerical value of this probability is? In dimension $$n = 1$$, it is easy to calculate this directly from the standard normal distribution, but in higher dimensions, I do not know how to do this. Any suggestion will be highly appreciated.

• The euclidean norm of your Brownian motion forms a Bessel process (and the square of it a squared Bessel process.) As you already explained, you may just take $r=1$. So one could simply integrate the transition density $p(t,x,y)$ over time intervall $t\in [0,1]$, target $y\in [1,\infty)$ and starting point $x=0$. – maliesen Apr 10 at 14:22

According to https://fbe.unimelb.edu.au/__data/assets/pdf_file/0004/2592040/116.pdf page 6, the transition density function of the squared Bessel process with dimension $$\delta$$ (which is the dimension of you Brownian motion) is given by
$$f_t^\delta(0,y) = \frac{y^{\delta/2-1} }{(2t)^{\delta/2} \Gamma(\delta/2)} e^{-y/(2t)},$$ so if my suggestion is right, the answer to the question is the integral value
$$\int_0^1 \int_1^\infty f_t^\delta(0,y) \, dy\, dt.$$