# Stochastic kernel as linear operator

Let $K$ be a stochastic kernel for a set $S$ equipped with a countably generated $\sigma$-Algebra $B(S)$, i.e.

$K:S\times B(S)\rightarrow [0,1]$

such that $K(\cdot,A)$ is a measurable function for all $A\in B(S)$ and $K(x,\cdot)$ is a probability measure for all $x\in S$.

Now $K$ operates linearly on $M_{1}(S)$, the space of probability measures on the measurable space $(S,B(S))$, by setting for all probability measures $\mu$

$K\mu := \int_S K(y,\cdot)\mu(dy)$.

Question: Is the converse also true, that each linear operator on $M_{1}(S)$, the space of probability measures on $(S,B(S))$, gives us a stochastic kernel again?

Each such linear operator O can be understood as mapping

$O:S \times B(S)\rightarrow[0,1]$

$\hphantom{O:}(x,A)\mapsto (O\delta_x)(A)$,

where $\delta_x$ is the Dirac-measure. So it basically boils down to the question if this last mapping is measurable in x, or if there are linear Operators on $M_{1}(S)$, so that the above mapping is not measurable in x.

I try it like this: Every probability measure $\mu$ on $(S,B(S))$ can be written as:

$\mu = \int_S \delta_y \mu(dy)$ (I hope this is right, but I cannot remember having seen such an argument before), then we have

$O\mu = O\int_S \delta_y \mu(dy)=\int_S O\delta_y \mu(dy)$ (last equality holds for finite sums because of the linearity of O, but I don't know why that is applicable to integrals).

This cannot be defined, if $O\delta_y$ is not measurable as a function of $y$, it then follows $O\mu$ is no probability measure. Contradiction.

Clarification: By linear operator $O$ I mean that for all $\mu,\nu\in M_{1}(S)$ and all $0\leq \alpha \leq 1$ we have $O(\alpha \mu + (1-\alpha)\nu)=\alpha O\mu + (1-\alpha)O\nu$ (thanks @ByronSchmuland for pointing this out).

• I think MathOverflow would be better place for your question. – Mohamed Mostafa Jul 23 '15 at 15:10
• $M_1(S)$ is not a linear space, so the term "linear operator" doesn't quite fit. I guess that you mean $O(\alpha\mu+(1-\alpha)\nu)=\alpha O\mu+(1-\alpha)O\nu$ for $0\leq \alpha\leq 1$ and $\mu,\nu\in M_1(S).$ Could you confirm this? – user940 Jul 23 '15 at 15:48
• @ByronSchmuland Yes, that is what I mean. Not sure how to call that attribute. – Paul Jul 23 '15 at 15:54
• Careful here: "... last equality holds for countable sums". Linearity only helps with finite sums, not countable sums. – user940 Jul 23 '15 at 18:43

The result is false in this generality.

Fix some $z\in S$ and define $O\mu=\mu_c+\mu_d(S)\delta_z$, where $\mu_d$ and $\mu_c=\mu-\mu_d$ are the discrete and continuous parts of $\mu$, respectively. Then $O$ satisfies your conditions, but cannot possibly be given by a kernel, if there is a non-discrete probability measure $\nu$ on $S$.

Why not?

If $O$ is given by the kernel $K$ then $O\mu=\int \mu(dx) K(x,\cdot)$. Setting $\mu=\delta_x$ shows $K(x,\cdot)=O\delta_x$.

If $\nu$ has no discrete part, then $O\nu=\nu$ but $\int\nu(dx) K(x,\cdot)=\delta_z\neq \nu.$

In order for $O$ to be represented by a kernel you need two things:

1. $x\mapsto O\delta_x$ should be measurable from $S$ to $M_1(S)$.
2. $O\mu=\int\mu(dx) O\delta_x$ for every $\mu\in M_1(S)$.

My counterexample above satisfies 1, but not 2.

• Ok, I see it now. I don't know what the right way for follow-up questions is, so I ask it here: Can you think of a counterexample that does not satisfy your condition 1? Because it is of some importance for me, if my (ill-named) linearity condition implies condition 1. – Paul Jul 23 '15 at 20:54
• I highly doubt it, but let me think about a counterexample. – user940 Jul 23 '15 at 21:14
• On a positive note, if $S$ is a decent topological space and you are willing to assume that $\mu\mapsto O\mu$ is continuous, I think you might get somewhere. – user940 Jul 23 '15 at 21:23
• I would need a topology on the measure space for that, and I think there is no canonical way to choose one depending on the topology on $S$. I usually have $S$ being a Polish space (separable completely metrizable topological space) if that makes anything easier – Paul Jul 23 '15 at 21:32
• By "topology on the measure space" do you mean "topology on $M_1(S)$"? You could use the topology of weak convergence in that case. – user940 Jul 23 '15 at 22:21