# Ornstein Uhlenbeck process hits zero with probability 1 in finite time

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?

• Can't you use a similar reasoning as in this answer, i.e. write $(Y_t)_{t \geq 0}$ as a time-changed Brownian motion and then use that a Brownian motion hits any point $x \in \mathbb{R}$ with probability $1$ (infinitely often)?
– saz
Commented Feb 5, 2019 at 17:49
• @saz I am going to try to adapt this proof . Thank you Commented Feb 5, 2019 at 18:02

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.
• We have $\{\tau_x^Y=\infty\}\subset\{1_{[x,\infty)}(Y(t))=0 \text{ for all$t\in[0,\infty)$}\}\subset\{\int_0^\infty1_{[x,\infty)}(Y(t))dt<\infty\}$. Now, take the complement. Commented Feb 28, 2019 at 12:12