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Nawaf Bou-Rabee
Nawaf Bou-Rabee
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Suppose that $\mu=0$ --- for simplicity.

By Cauchy-Schwarz, \begin{align*} \mathbb{E} \left\{ \frac{|X_t|}{\int_0^t f(X_s) ds} \right\} \le \sqrt{\mathbb{E} \left\{ X_t^2 \right\} \mathbb{E} \left\{ \left( \frac{1}{\int_0^t f(X_s) ds}\right)^2 \right\} } \;. \tag{1} \end{align*} A direct calculation shows that, $$ \mathbb{E} \left\{ X_t^2 \right\} = e^{-2 t} X_0^2 +\frac{\sigma^2}{2} (1-e^{-2 t}) \;, \tag{2} $$\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*} and sinceSince $X$ is ergodic with non-normalized stationary density proportional to $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) and, (2) yields, $$ \mathbb{E} \left\{ \frac{|X_t|}{\int_0^t f(X_s) ds} \right\} \le \frac{\sqrt{ X_0^2 + \frac{\sigma^2}{2} }}{t} \sqrt{\mathbb{E} \left\{ \frac{1}{ \left( \frac{1}{t} \int_0^t f(X_s) ds \right)^2} \right\} } $$ Passing to the limit as $t \to \infty$, applyingand (3) and DCT, givesyields the desired result.

Suppose that $\mu=0$ --- for simplicity.

By Cauchy-Schwarz, \begin{align*} \mathbb{E} \left\{ \frac{|X_t|}{\int_0^t f(X_s) ds} \right\} \le \sqrt{\mathbb{E} \left\{ X_t^2 \right\} \mathbb{E} \left\{ \left( \frac{1}{\int_0^t f(X_s) ds}\right)^2 \right\} } \;. \tag{1} \end{align*} A direct calculation shows that, $$ \mathbb{E} \left\{ X_t^2 \right\} = e^{-2 t} X_0^2 +\frac{\sigma^2}{2} (1-e^{-2 t}) \;, \tag{2} $$ and since $X$ is ergodic with stationary density proportional to $e^{-\frac{x^2}{\sigma^2}} $ , $$ \lim_{t \to \infty} \frac{1}{t} \int_0^t f(X_s) ds = \frac{1}{2} \left( 1 + \operatorname{erf}\left( \frac{a}{\sigma} \right) \right) \;. \tag{3} $$ Combining (1) and (2) yields, $$ \mathbb{E} \left\{ \frac{|X_t|}{\int_0^t f(X_s) ds} \right\} \le \frac{\sqrt{ X_0^2 + \frac{\sigma^2}{2} }}{t} \sqrt{\mathbb{E} \left\{ \frac{1}{ \left( \frac{1}{t} \int_0^t f(X_s) ds \right)^2} \right\} } $$ Passing to the limit as $t \to \infty$, applying (3) and DCT, gives the desired result.

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.

Post Deleted by Nawaf Bou-Rabee
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Nawaf Bou-Rabee
Nawaf Bou-Rabee
  • 6k
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  • 40

Suppose that $\mu=0$ --- for simplicity.

By Cauchy-Schwarz, \begin{align*} \mathbb{E} \left\{ \frac{|X_t|}{\int_0^t f(X_s) ds} \right\} \le \sqrt{\mathbb{E} \left\{ X_t^2 \right\} \mathbb{E} \left\{ \left( \frac{1}{\int_0^t f(X_s) ds}\right)^2 \right\} } \;. \tag{1} \end{align*} A direct calculation shows that, $$ \mathbb{E} \left\{ X_t^2 \right\} = e^{-2 t} X_0^2 +\frac{\sigma^2}{2} (1-e^{-2 t}) \;, \tag{2} $$ and since $X$ is ergodic with stationary density proportional to $e^{-\frac{x^2}{\sigma^2}} $ , $$ \lim_{t \to \infty} \frac{1}{t} \int_0^t f(X_s) ds = \frac{1}{2} \left( 1 + \operatorname{erf}\left( \frac{a}{\sigma} \right) \right) \;. \tag{3} $$ Combining (1) and (2) yields, $$ \mathbb{E} \left\{ \frac{|X_t|}{\int_0^t f(X_s) ds} \right\} \le \frac{\sqrt{\mathbb{E} \left\{ X_0^2 + \frac{\sigma^2}{2} \right\}}}{t} \sqrt{\mathbb{E} \left\{ \frac{1}{ \left( \frac{1}{t} \int_0^t f(X_s) ds \right)^2} \right\} } $$$$ \mathbb{E} \left\{ \frac{|X_t|}{\int_0^t f(X_s) ds} \right\} \le \frac{\sqrt{ X_0^2 + \frac{\sigma^2}{2} }}{t} \sqrt{\mathbb{E} \left\{ \frac{1}{ \left( \frac{1}{t} \int_0^t f(X_s) ds \right)^2} \right\} } $$ Passing to the limit as $t \to \infty$, applying (3) and DCT, gives the desired result.

Suppose that $\mu=0$ --- for simplicity.

By Cauchy-Schwarz, \begin{align*} \mathbb{E} \left\{ \frac{|X_t|}{\int_0^t f(X_s) ds} \right\} \le \sqrt{\mathbb{E} \left\{ X_t^2 \right\} \mathbb{E} \left\{ \left( \frac{1}{\int_0^t f(X_s) ds}\right)^2 \right\} } \;. \tag{1} \end{align*} A direct calculation shows that, $$ \mathbb{E} \left\{ X_t^2 \right\} = e^{-2 t} X_0^2 +\frac{\sigma^2}{2} (1-e^{-2 t}) \;, \tag{2} $$ and since $X$ is ergodic with stationary density proportional to $e^{-\frac{x^2}{\sigma^2}} $ , $$ \lim_{t \to \infty} \frac{1}{t} \int_0^t f(X_s) ds = \frac{1}{2} \left( 1 + \operatorname{erf}\left( \frac{a}{\sigma} \right) \right) \;. \tag{3} $$ Combining (1) and (2) yields, $$ \mathbb{E} \left\{ \frac{|X_t|}{\int_0^t f(X_s) ds} \right\} \le \frac{\sqrt{\mathbb{E} \left\{ X_0^2 + \frac{\sigma^2}{2} \right\}}}{t} \sqrt{\mathbb{E} \left\{ \frac{1}{ \left( \frac{1}{t} \int_0^t f(X_s) ds \right)^2} \right\} } $$ Passing to the limit as $t \to \infty$, applying (3) and DCT, gives the desired result.

Suppose that $\mu=0$ --- for simplicity.

By Cauchy-Schwarz, \begin{align*} \mathbb{E} \left\{ \frac{|X_t|}{\int_0^t f(X_s) ds} \right\} \le \sqrt{\mathbb{E} \left\{ X_t^2 \right\} \mathbb{E} \left\{ \left( \frac{1}{\int_0^t f(X_s) ds}\right)^2 \right\} } \;. \tag{1} \end{align*} A direct calculation shows that, $$ \mathbb{E} \left\{ X_t^2 \right\} = e^{-2 t} X_0^2 +\frac{\sigma^2}{2} (1-e^{-2 t}) \;, \tag{2} $$ and since $X$ is ergodic with stationary density proportional to $e^{-\frac{x^2}{\sigma^2}} $ , $$ \lim_{t \to \infty} \frac{1}{t} \int_0^t f(X_s) ds = \frac{1}{2} \left( 1 + \operatorname{erf}\left( \frac{a}{\sigma} \right) \right) \;. \tag{3} $$ Combining (1) and (2) yields, $$ \mathbb{E} \left\{ \frac{|X_t|}{\int_0^t f(X_s) ds} \right\} \le \frac{\sqrt{ X_0^2 + \frac{\sigma^2}{2} }}{t} \sqrt{\mathbb{E} \left\{ \frac{1}{ \left( \frac{1}{t} \int_0^t f(X_s) ds \right)^2} \right\} } $$ Passing to the limit as $t \to \infty$, applying (3) and DCT, gives the desired result.

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Nawaf Bou-Rabee
Nawaf Bou-Rabee
  • 6k
  • 2
  • 20
  • 40

Suppose that $\mu=0$ --- for simplicity.

By Cauchy-Schwarz, \begin{align*} \mathbb{E} \left\{ \frac{|X_t|}{\int_0^t f(X_s) ds} \right\} \le \sqrt{\mathbb{E} \left\{ X_t^2 \right\} \mathbb{E} \left\{ \left( \frac{1}{\int_0^t f(X_s) ds}\right)^2 \right\} } \;. \tag{1} \end{align*} A direct calculation shows that, $$ \mathbb{E} \left\{ X_t^2 \right\} = e^{-2 t} X_0^2 +\frac{\sigma^2}{2} (1-e^{-2 t}) \;, \tag{2} $$ and since $X$ is ergodic with stationary density proportional to $e^{-\frac{x^2}{\sigma^2}} $ , $$ \lim_{t \to \infty} \frac{1}{t} \int_0^t f(X_s) ds = \frac{1}{2} \left( 1 + \operatorname{erf}\left( \frac{a}{\sigma} \right) \right) \;. \tag{3} $$ Combining (1) and (2) yields, $$ \mathbb{E} \left\{ \frac{|X_t|}{\int_0^t f(X_s) ds} \right\} \le \frac{\sqrt{\mathbb{E} \left\{ X_0^2 + \frac{\sigma^2}{2} \right\}}}{t} \sqrt{\mathbb{E} \left\{ \frac{1}{ \left( \frac{1}{t} \int_0^t f(X_s) ds \right)^2} \right\} } $$ Passing to the limit as $t \to \infty$, applying (3) and DCT, gives the desired result.