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Consider an Ornstein-Uhlenbeck position process:

$dV_t=dB_t-\lambda V_tdt$

$dX_t=V_tdt$

where $B_t,V_t,X_t$ are all in $R^d$ with $d\geq 3$. Let $X_0\neq0$, $V_0=0$ .

Let $r>0$ and $S_r$ be the sphere with centre $0$ and radius $r$. Let $H(r)$ be the probability that $X_t$ ever hits $H(r)$. The problem is to find $\lim_{r\rightarrow0} H(r)r^{1-d}$.

I am thinking about the following, I want to prove that probability that $X_t$ hits the boundary of the ball more than twice can be ignored. And then can integrate the probability density that $X_t$ is on the boundary of $S_r$ at time t and integrate over t. However, I don't know how to prove that.

Another way is to write an PDE of the hitting probability P(X_t,V_t) as function of current state, it satisfies

$V_t\cdot \frac{\partial P}{\partial X}-\lambda V\cdot\frac{\partial P}{\partial V}+\frac{\partial^2 P}{\partial V^2}=0$.

Then, maybe we can do some asymptotics about the solution. But again, I have no idea how to do it.

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  • $\begingroup$ Since $V$ can stay an arbitrarily long time in any ball of R^d isn't it obvious that H(r)=1 if H(r) is the probability that X ever hit S_r. $\endgroup$ – Alekk Oct 25 '12 at 16:28
  • $\begingroup$ How is that obvious? I don't see any direct link between the recurrence of V and that of X. $\endgroup$ – Guolong Li Oct 29 '12 at 12:18

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