# Expected hitting time of Ornstein-Uhlenbeck process

If I recall correctly, it is known that for a standard brownian motion starting at $0$, that the expected time to hit some level $a>0$ is infinite. I'm curious if there's a proof of what the expected hitting time is for an Ornstein-Uhlenbeck process.

Let $\lambda >0$ and $X_t$ be the stochastic process which solves: $dX_t = -\lambda X_t dt + dW_t$. Let $\tau_a = \inf\left\lbrace t\geq 0: X_t = a >0\right\rbrace$. Assuming $X(0)=0$, is there a way to determine what the expected first hitting time should be? I would think it should also be infinite since the process $X_t$ drifts towards $0$ as it gets further away, so it should take longer to hit some level $a$, but I'm curious if there's a mathematical way to show this, or at least the idea of it.

Set $f(x,t)=xe^{\lambda t}$. By application of Ito's lemma, we have $$df(t,X_t)=\frac{\partial f}{\partial t}(t,X_t)dt+\frac{\partial f}{\partial x}(t,X_t)dX_t+\frac 12\frac{\partial^2 f}{\partial x^2}(t,X_t)d[X_t,X_t]$$ therefore $$df(X_te^{\lambda t})=\lambda X_te^{\lambda t}dt+e^{\lambda t}(-\lambda X_t dt+dW_t)$$ as a result
$$d(X_te^{\lambda t})=e^{\lambda t}dW_t$$ Since $X(0)=0$, thus $$X_te^{\lambda t}=\int_{0}^{t}e^{\lambda s}dW_s$$ or $$X_t=e^{-\lambda t}\int_{0}^{t}e^{\lambda s}dW_s$$ By the Dambis, Dubins-Schwartz theorem,there is a Brownian motion $B(t)$ such that $$\int_{0}^{t}e^{\lambda s}dW_s=B_{\tau(t)}$$ where $$\tau(t)=\frac{1}{2\lambda}(e^{2\lambda t}-1)$$ Hence, the representation $$X_t=e^{-\lambda t}B_{\tau(t)}$$ holds, which is known as the Doob's transformation.The conditional probability $$P(X_t\le x\,|\,X_0=0)=\frac{1}{\sqrt{2\pi \sigma(t)}}\int_{-\infty}^{x}\exp\left(-\frac{u^2}{2\sigma(t)}\right)du$$ where $$\sigma(t)=e^{-2\lambda t}\tau(t)$$ tends to the standard normal distribution, of course, as $t\to \infty$ (meaning that transients die out with time and don’t affect long-term behavior). Therefore I think (if I am right) that the solution of any problem involving the OU process would be similarly straightforward. Anyway , Thomas, Ricciardi and Sato showed that $$\mathbb{E}[\tau_a]=\sqrt{\frac{\pi}{2}}\int_{0}^{a}\left(1+\operatorname{erf}\left({\frac{t}{\sqrt{2}}}\right)\right)\exp\left(\frac{t^2}{2}\right)dt=\frac 12\sum_{n=1}^{\infty}\frac{\sqrt{2}^na^n}{n!}\Gamma\left(\frac{n}{2}\right)$$
• Hi, could you explain more about how to obtain $\tau(t)$? Many thank! Commented Dec 13, 2018 at 9:41