Let $X$ and $Y$ be Polish spaces and $K$ a Markov kernel from $X$ to $Y$. That is, $K$ is a mapping $X \times \mathcal{B}_Y \rightarrow [0,1]$ (where $\mathcal{B}_Y$ is the $\sigma$-algebra of Borel sets on $Y$) s.t.

  • For every $A \in \mathcal{B}_Y$, the mapping $K^A: X \rightarrow [0,1]$ defined by $K^A(x):=K(x,A)$ is Borel measurable.

  • For every $x \in X$, the mapping $K_x: \mathcal{B}_Y \rightarrow [0,1]$ defined by $K_x(A):=K(x,A)$ is a probability measure on $Y$.

It seems natural to consider the additional condition on $K$ that $K_x$ depends continuously on $x$ in the sense of the given topology on $X$ and the weak topology on the space $\mathcal{P}(Y)$ of probability measures on $Y$. For example, if $K$ is deterministic (i.e. if for any $x \in X$, $K_x$ is a Dirac measure) this condition means that $K$ defines a continuous mapping from $X$ to $Y$.

What is the standard name for this condition?

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    $\begingroup$ Some kind of Feller continuity? $\endgroup$ – Mateusz Kwaśnicki Sep 25 '17 at 19:12
  • $\begingroup$ @MateuszKwaśnicki Thank you! This looks right, some writers even use the name "Feller Markov kernel" (arxiv.org/abs/1207.0086). Do you want to post this as an answer? $\endgroup$ – Vanessa Sep 27 '17 at 8:58
  • $\begingroup$ Is this motivated by the Robust filtering of Clarke in some way? (Out of curiosity) $\endgroup$ – BLBA Apr 29 '20 at 12:53
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    $\begingroup$ @AnnieTheKatsu Not really, it came up when I worked on this paper: arxiv.org/abs/1705.04630 $\endgroup$ – Vanessa Apr 30 '20 at 13:23

As stated in the comment, this seems to be some kind of Feller continuity.

Having said that, I should emphasize that there is some confusion about the name "Feller" with regard to the properties of a Markov process, Markov transition function or a Markov kernel, and link to Martin Hairer's comment about that.


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