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Suppose I have a probability measure $\pi$ and a Markov kernel $q$ which leaves $\pi$ invariant, in the sense that

\begin{align} \int_x \pi(dx) q(x \to dy) = \pi(dy) \end{align}

Then, a (the) time-reversal of $q$ is a (the) Markov kernel $r$ such that

\begin{align} \pi(dx) q(x \to dy) = \pi(dy) r(y \to dx) \end{align}

as measures on the product space.

My question is: what are some simple/standard conditions on $(\pi, q)$ which ensure that

  1. Such an $r$ exists, and
  2. It is essentially unique (i.e. up to changes of measure $0$ in a suitable sense).

My understanding is that this problem is equivalent to the existence of regular conditional probabilities for the joint measure

\begin{align} \mu(dx,dy) = \pi(dx) q(x \to dy). \end{align}

I can find some statements of theorems which give conditions for when a general measure admits regular conditional probabilities; I'd like a simpler (if possible) statement which gives the necessary conditions in terms of $(\pi,q)$.

References are gladly welcomed.

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    $\begingroup$ The conditions for existence of a regular conditional probability are not really about the measure but about whether the state space is not too horrible, as a measurable space. In particular, it's sufficient that your state space be standard Borel, which covers the vast majority of cases that people care about. $\endgroup$
    – Nate Eldredge
    Commented Sep 14, 2018 at 23:07
  • $\begingroup$ @NateEldredge Thank you - that's very useful. I'd seen a statement for the existence and was surprised that it depended so weakly on the measure, so wanted to do a sanity check and make sure I wasn't missing anything. So would it be true that if I'm working on a Polish space, and $(x,y)$ is jointly measurable w.r.t. the Borel $\sigma$-algebra, then $r$ should exist? $\endgroup$
    – πr8
    Commented Sep 14, 2018 at 23:21
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    $\begingroup$ Yes - this is true for the so-called Lebesgue (or, standard, or Lebesgue-Rokhlin) measure spaces. A Polish space endowed with a Borel probability measure is classically known to be of this kind. By the way, the right object to consider in this context is not the pair "initial distribution of $x$, transition probabilities $x\to y$", but the joint distribution of $x$ and $y$. $\endgroup$
    – R W
    Commented Sep 14, 2018 at 23:37
  • $\begingroup$ @RW Thanks - that's my question answered then! Do you happen to know references for i) the statement that this holds for Lebesgue-Rokhlin spaces and ii) the statement that [Polish $\cap$ Borel] $\subset$ Lebesgue-Rokhlin? Thanks also for the additional comment - it's all making a lot more sense now. $\endgroup$
    – πr8
    Commented Sep 14, 2018 at 23:47
  • $\begingroup$ The fact that standard Borel spaces admit regular conditional probabilities can be found in Durrett's Probability: Theory and Examples, Theorem 5.1.9 in the fourth edition. Note that his term "nice" simply means "standard Borel", which means, in effect, "a Polish space with its Borel $\sigma$-algebra". $\endgroup$
    – Nate Eldredge
    Commented Sep 15, 2018 at 0:37

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