5
$\begingroup$

On a measurable space $(E,\mathcal E)$, a stochastic kernel is a function $p\colon E\times \mathcal E\to [0,1]$ such that:

  1. for each $x\in E$, the function $A\mapsto p(x,A)$ is a probability measure;
  2. for each $A\in\mathcal E$, the function $x\mapsto p(x,A)$ is measurable.

Such a kernel induces a linear operator $P$ on the space $\mathcal B(E)$ of bounded measurable functions $f\colon E\to\mathbb R$ as follows: $$\forall x\in E,\qquad Pf(x):=\int f(y)p(x,dy).$$ Clearly, the linear operator $P\colon \mathcal B(E)\to \mathcal B(E)$ preserves non-negativity and maps the constant function $1$ to itself.

My question is about the converse: is it true that any linear operator $P\colon \mathcal B(E)\to \mathcal B(E)$ preserving non-negativity and mapping $1$ to itself arises in this way ? Of course, the underlying stochastic kernel then has to be $$\forall(x,A)\in E\times\mathcal E,\qquad p(x,A) := P1_A(x),$$ but proving $\sigma-$additivity seems to require extra assumptions on the ambient space $(E,\mathcal E)$, and I was unable to find a reference.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

Not at all. This is a pretty common mistake, for instance reproduced in the erroneous Wikipedia definition of a Markov operator. For the simplest counterexample take just a countable state space $E$. Then the set of positive normalized functionals on $\ell^\infty(E)$ is not exhausted by probability measures on $E$, cf. the notion of Banach limit.

The proper definition of a Markov operator on $L^\infty$ (under which the transition probabilities are well-defined) in addition to positivity and normalization includes the order continuity condition: if $f_n \downarrow 0$, then $Pf_n \downarrow 0$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ The reason why the mistake is harmless though is that probabilists avoid the full axiom of choice like the plague, so effectively tend to work in ZF+DC which is consistent with the truth of that statement. $\endgroup$ Commented Nov 16 at 23:54
  • $\begingroup$ @Martin Hairer Not sure whether they do it fully consciously though :) $\endgroup$ Commented Nov 17 at 10:09
  • $\begingroup$ I see, thanks very much to both of you for the clarification... $\endgroup$ Commented Nov 17 at 11:53

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.