Ornstein Uhlenbeck process hits zero with probability 1 in finite time

Question

I am looking for a reference or a proof which shows that $P(\tau_0^Y<\infty)=1$ for an ornstein uhlenbeck process $Y$ given by $$ dY(t)=-\frac{1}{2}\alpha Y(t)dt+ \frac{1}{2} \sigma dW(t),Y(0)=y>0 $$ where $\sigma,\alpha>0$ and $\tau_0^Y=\inf\{t \ge 0:Y(t)=0\}$

I tried to explicitly solve for $Y(t)$ using Ito's lemma from which I could conclude that $Y(t)$ is normally distributed(I explicitly computed the mean and the variance). But I couldn't use it to prove that the process hits zero in finite time with probability one. Any hints on how could I show this?

Answer 1

Since you're asking for a reference: Example 9.12' (a) in the book "Asymptotic Statistics with a View to Stochastic Processes" by R. Höpfner shows that $Y$ is positive Harris-recurrent and that its unique invariant law is Gaussian (in particular its support is the entire space $\mathbb R$). Together with the continuity of its trajectories, applying the definition of Harris recurrence implies that for any starting point $Y(0)=y\in\mathbb R$ and any $x\in\mathbb R$ with $x\ge y$ we have $$ P(\tau_x^Y<\infty)\ge P\left(\int_0^\infty 1_{[x,\infty)}(Y(t))dt=\infty\right) =1. $$ The case $x\le y$ is dealt with in the same way.

The ninth chapter of the mentioned book is a nice survey of the basics about Harris recurrence and how to check it for one-dimensional diffusions.