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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!

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    $\begingroup$ What is $X_{\infty}$? $\endgroup$ – S.Surace Jan 2 '18 at 10:44
  • $\begingroup$ @S.Surace $X_\infty$ is $\lim_{t\rightarrow\infty}X_t$. $\endgroup$ – Jackie Lu Jan 3 '18 at 3:47
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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.

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  • $\begingroup$ Sorry but can you explain the last step a bit more? (3) implies that $\frac{1}{\int_{0}^{\infty}f(X_s)ds}=0$ almost surely, and that does not necessarily mean $\mathbb{E}\left\{\left(\frac{1}{\int_{0}^{\infty}f(X_s)ds}\right)^2\right\}=0$, right? $\endgroup$ – Jackie Lu Jan 3 '18 at 3:44
  • $\begingroup$ Actually $\frac{1}{\int_{0}^{\infty}f(X_s)ds}=0$ almost surely does not imply $\lim_{t\rightarrow\infty}\mathbb{E}\left\{\left(\frac{1}{\int_{0}^{t}f(X_s)ds}\right)^2\right\}=0$, but what about $\mathbb{E}\left\{\left(\frac{1}{\int_{0}^{\infty}f(X_s)ds}\right)^2\right\}$? $\endgroup$ – Jackie Lu Jan 3 '18 at 4:11
  • $\begingroup$ Hinit: use limit rules. $\endgroup$ – Nawaf Bou-Rabee Jan 3 '18 at 11:51

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