Timeline for If $X$ is a Markov process, can we find a mild assumption ensuring that $\frac1t\operatorname E_x\left[\int_0^tc(X_s)\:{\rm d}s\right]\to c(x)$?
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| Jul 19, 2022 at 6:17 | comment | added | 0xbadf00d | It seems like there is a counterexample: Take $E=\mathbb R$, $\mathcal E=\mathcal B(E)$, $c(x)=x$ and $X_s=s+tY$ for some $V\ge0$ with $\operatorname E_x[Y]=\infty$. If $U$ is an open neighborhood of $x$, there is a $\varepsilon>0$ with $N\subseteq U$ and $$\operatorname P_x[X_s\in U\text{ for all }s\in[0,t)]\ge\operatorname P_x[X_s\in B_\varepsilon(x)\text{ for all }s\in[0,t)]=\operatorname P_x[\|V\|_E<\frac\varepsilon s\text{ for all }s\in[0,t)]\xrightarrow{t\to0+}1.$$ On the other hand, $\operatorname E_x\left[\frac1tY_t\right]=x+\frac t2\operatorname E_x[V]=\infty$. | |
| Jul 18, 2022 at 19:26 | comment | added | 0xbadf00d | How would you conclude from $\mathbb P_x(X_s\in U \text{ for all }0\le s < t)\to 1$ for each open neighborhood $U$ of $x$? Please see math.stackexchange.com/q/4495504/47771. | |
| Jul 18, 2022 at 14:17 | comment | added | 0xbadf00d | Don't you think that we need to assume continuity of $c$? Actually, the only thing we need is continuity of $c\circ X$ at $0$, but I don't see any assumption other than continuity of $c$ which would ensure that. Morever, what do you think about the attempt I've described in the comments below the question? | |
| Jul 17, 2022 at 14:06 | comment | added | user1118 | True it doesn't take you all the way there, but I think it could be used to show that $\mathbb P_x(X_s\in U \text{ for all }0\le s < t)\to 1$ as $t\to0$ for all open $U$ containing $x$. | |
| Jul 16, 2022 at 11:05 | comment | added | 0xbadf00d | Please see mathoverflow.net/q/426699/91890. | |
| Jul 14, 2022 at 19:44 | comment | added | 0xbadf00d | Thank you for your input. I've checked Blumental and Getoor, but unless I'm missing something the bound for the set $A$ does depend on $\omega$. To be precise, if I understand the claim correctly, they say that for each $t\ge0$, there is a null set $N$ such that $\{X_s(\omega):s\in[0,t]\}$ is bounded for all $\omega\not\in N$. So, the bound should depend on $\omega$ and hence this is not enough to apply the DCT. | |
| S Jul 13, 2022 at 10:04 | review | First answers | |||
| Jul 13, 2022 at 11:35 | |||||
| S Jul 13, 2022 at 10:04 | history | answered | user1118 | CC BY-SA 4.0 |