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


where $f(x)=\mathbb{1}\{x\leq a\}$ for some $a>\mu$. Is there a simple way to prove this? Thanks!

  • 3
    $\begingroup$ 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

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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.