I am dealing with an Orstein-Uhlenbeck process $X_t$ with its stochastic differential equation being
$$dX_t=(\mu-X_t)dt+\sigma dW_t.$$
I want to show
$$\mathbb{E}\left[\frac{|X_\infty|}{\int_{0}^{\infty}f(X_s)ds}\right]=0,$$
where $f(x)=\mathbb{1}\{x\leq a\}$ for some $a>\mu$. Is there a simple way to prove this? Thanks!
Suppose that $\mu=0$ --- for simplicity.
By Cauchy-Schwarz, \begin{align*} \mathbb{E} \left\{ \frac{ |X_{\infty}|}{\int_0^{\infty} f(X_s) ds} \right\} \le \sqrt{\mathbb{E} \left\{ X_{\infty}^2 \right\} \mathbb{E} \left\{ \left( \frac{1}{\int_0^{\infty} f(X_s) ds}\right)^2 \right\} } \;. \tag{1} \end{align*} Since $X$ is ergodic with non-normalized stationary density $e^{-\frac{x^2}{\sigma^2}} $, $$ \mathbb{E} \left\{ X_{\infty}^2 \right\} = \frac{\sigma^2}{2} \;, \tag{2} $$ and , $$ \lim_{t \to \infty} \frac{1}{t} \int_0^t f(X_s) ds = \frac{1}{2} \left( 1 + \operatorname{erf}\left( \frac{a}{\sigma} \right) \right) \;. \tag{3} $$ Combining (1), (2) and (3) yields the desired result.
