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If $X$ is a Markov process, can we find a mild assumption ensuring that $\frac1t\operatorname E_x\left[\int_0^tc(X_s)\:{\rm d}s\right]\to c(x)$?
Let
$(E,\mathcal E)$ be a measurable space with $\{x\}\in\mathcal E$ for all $x\in E$
$\mathcal E_b:=\{f:E\to\mathbb R\mid f\text{ is bounded and }\mathcal E\text{-measurable}\}$
$(\kappa_t)_{t\ge0}$ ...