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Let $\{ X_t, t \geq 0 \}$ be a $\mathbb{R}^d$-valued stochastic process on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$.

Assumption

  • $X_t$ is a regenerative process in the sense of https://en.wikipedia.org/wiki/Regenerative_process

  • for all functions $f$ continuous and bounded on $\mathbb{R}^d$ $$ \lim_{t \to\infty} \frac{1}{t} \int_0^t \mathbb{E} f(X_s) \mbox{d} s = \nu(f) $$ with $s \to \mathbb{E}(f(X_s))$ is measurable and where $\nu$ is a probability measure on $\mathbb{R}^d$.

Question

How to get rid of the expectation? In the sense, $$ \lim_{t \to\infty} \frac{1}{t} \int_0^t f(X_s) \mbox{d} s = \nu(f), \: \mbox{a.s.} $$ I would like to avoid the renewal reward theorem.

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    $\begingroup$ What if $X_t$ is a process that starts at a random position, and never moves? $\endgroup$
    – Anthony Quas
    Commented Apr 23, 2018 at 3:37
  • $\begingroup$ Could you please specify your assumptions on the stochastic process $\{X_t, t \ge 0\}$? In the current form of your question I can't see why the function $s \mapsto \mathbb{E}f(X_s)$ should be measurable. $\endgroup$
    – Jochen Glueck
    Commented Apr 23, 2018 at 4:34
  • $\begingroup$ Thank you Anthony Quas and Jochen Glueck for your remarks. I am not sure what type of minimal assumption on $X$ would make $s \to \mathbb{E} f(X_s)$ measurable. I think we can assume it. Furthermore, I would like to add that $X_t$ is a regenerative process in the sense of en.wikipedia.org/wiki/Regenerative_process. $\endgroup$
    – megaproba
    Commented Apr 23, 2018 at 10:08

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