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Jul 30, 2024 at 13:30 comment added Ziv Oh of course, my example admits both bad and good kernels, which is fine—never mind :).
Jul 30, 2024 at 12:38 comment added No-one Yes my concern is if it is always possible to find some kernels such that the composition holds everywhere. In other words, I'm asking if it is always possible to define the kernels in the unconstrained parts in a coherent way. In your specific example it surely is, but I would guess that in general it is not because one would have "uncountably many coherent choices" to make.
Jul 30, 2024 at 12:13 comment added Ziv Ahh, sorry, I missed the everywhere (not merely almost everywhere) subtlety. But I realize now there is an easy counterexample via the "what if you can't reach this part of the state space" idea. Take $E = \mathbb{R}$, set $X_0 = 0$, then have $X_t$ always moving at rate $\pm 1$, switching directions at the increments of a Poisson process. Then the effective state space for time $t$ is $[-t, +t]$, so $P_{t, u}(x, \cdot)$ is unconstrained for $|x| > t$. But maybe there's a way to exclude this concern?
Jul 30, 2024 at 12:08 history edited Ziv CC BY-SA 4.0
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Jul 30, 2024 at 9:31 comment added No-one Thank you, but I am aware of the fact that for all $A\in Bor(E)$ and all $t\leq u\leq v$ one has $(P_{u,v}\circ P_{t,u})(x,A)=P_{t,v}(x,A)$ for $\mathbb{P}\circ X_t^{-1}$-almost every $x\in E$. What I'm really wondering is if the kernels can be chosen so that the composition holds everywhere, or at least, if for all $t$ it holds on a set $\Omega_t$ of full measure which is the same for all $A$ and $u,v$. Since $A\in Bor(E)$, which is countably generated if $E$ is Polish, the fact that one can take $\Omega_t$ independent of $A$ should follow. The difficulties lie in the independence of $u,v$.
Jul 30, 2024 at 3:27 history answered Ziv CC BY-SA 4.0