If $κ_i$ is a Markov kernel, is there an unique Markov kernel $κ$ on the product space with $κ((x_1,x_2),B_1\times B_2)=κ_1(x_1,B_1)κ_2(x_2,B_2)$?

Question

Let

• $(E_i,\mathcal E_i)$ be a measurable space
• $\kappa_i$ be a Markov kernel on $(E_i,\mathcal E_i)$

Is there an unique Markov kernel $(E_1\times E_2,\mathcal E_1\otimes\mathcal E_2)$ with $$\kappa((x_1,x_2),B_1\times B_2)=\kappa_1(x_1,B_1)\kappa_2(x_2,B_2)\tag1$$ for all $(x_1,x_2)\in E_1\times E_2$ and $B_1\in B_2\in\mathcal E_1\times\mathcal E_2$?

If $\kappa$ exists, it's at least clear to me that, for fixed $(x_1,x_2)$, $\kappa((x_1,x_2),\;\cdot\;)$ is uniquely determined by ist values on $\mathcal E_1\times\mathcal E_2$ (since $\mathcal E_1\times\mathcal E_2$ is a $\cap$-stable system containing $E_1\times E_2$ and generating $\mathcal E_1\otimes\mathcal E_2$).

Answer 1

Yes.

You know, surely, that product measures exist: given probability measures $\mu_1, \mu_2$ on $E_1, E_2$, there is a unique probability measure $\mu$ on $E_1 \times E_2$, denoted $\mu = \mu_1 \otimes \mu_2$, with $\mu(B_1 \times B_2) = \mu_1(B_1) \mu_2(B_2)$ for measurable sets $B_1, B_2$.

For each $(x_1, x_2) \in E_1 \times E_2$, set $\kappa((x_1, x_2), B) = (\kappa_1(x_1, \cdot) \otimes \kappa_2(x_2, \cdot))(B)$. As just mentioned, $\kappa((x_1, x_2), \cdot)$ is a probability measure that satisfies your desired condition. To show that $\kappa$ is a Markov kernel, it remains to show that for each $B$, the map $x \mapsto \kappa(x,B)$ is measurable. Let $\mathcal{L}$ be the collection of all sets $B \in \mathcal{E}_1 \otimes \mathcal{E}_2$ for which $x \mapsto \kappa(x,B)$ is measurable. For $B$ of the form $B = B_1 \times B_2$, we have $\kappa((x_1, x_2), B) = \kappa_1(x_1, B_1) \kappa_2(x_2, B_2)$, so $\kappa(\cdot, B)$ is a product of two measurable functions and hence is measurable. Hence if we let $\mathcal{P} = \{B_1 \times B_2 : B_i \in \mathcal{E}_i\}$, we have shown $\mathcal{P} \subset \mathcal{E}$ and $\mathcal{P}$ is a $\pi$-system that generates $\mathcal{E}_1 \otimes \mathcal{E}_2$.

It now suffices to show that $\mathcal{L}$ is a $\lambda$-system. Then the $\pi$-$\lambda$ lemma will imply that $\mathcal{L} = \sigma(\mathcal{P}) = \mathcal{E}_1 \otimes \mathcal{E}_2$, i.e. the map $\kappa(\cdot, B)$ is measurable for every $B \in \mathcal{E}_1 \otimes \mathcal{E}_2$.

If $A,B \in \mathcal{L}$ with $B \subset A$ then for every $x$, $\kappa(x, B \setminus A) = \kappa(x,B) - \kappa(x,A)$ since $\kappa(x,\cdot)$ is a probability measure. So $\kappa(\cdot, B \setminus A)$ is measurable because it is the difference of two measurable functions. Hence $B \setminus A \in \mathcal{L}$.

If $A_1 \subset A_2 \subset \dots \in \mathcal{L}$, and $A = \bigcup_n A_n$, then $\kappa(x,A) = \lim_{n \to \infty} \kappa(x, A_n)$. So $\kappa(\cdot, A)$ is measurable since it is the pointwise limit of a sequence of measurable functions. Hence $A \in \mathcal{L}$.

This completes the proof.