# A generalization of the conditional expectation to kernels

Let $\left(\Omega_1,\mathcal{A}_1, P\right)$ be a probability space, let $\left(\Omega_2,\mathcal{A}_2\right)$ be a measurable space and let $\kappa:\Omega_1\times\mathcal{A}_2\rightarrow\left[0,1\right]$ be a Markov kernel from $\mathcal{A}_1$ tp $\mathcal{A}_2$. Suppose $\mathcal{A}_3$ is a sub-$\sigma$-algebra of $\mathcal{A}_1$. Is there a Merkov kernel $\kappa'$ from $\mathcal{A}_3$ to $\mathcal{A}_2$ such that for all $B\in\mathcal{A}_2$,

$$\kappa\left(\omega,B\right)=\kappa'\left(\omega,B\right)\hspace{10mm}\left[\mathcal{A}_1,P\right]$$

Possible plan of attack: We could view $\kappa$ as a family of $\mathcal{A}_1$-measurable real functions indexed by $\mathcal{A}_2$ (viz. for every $B\in\mathcal{A}_2$, $\kappa_B$ is the function $\omega\mapsto\kappa\left(\omega,B\right)$) and define $\kappa'\left(\omega,B\right):=E\left(\kappa_B\mid\mathcal{A}_3\right)$. That would meet half the requirements for a Markov kernel, except $\kappa'$ is not guaranteed to be a probability in the second argument. In fact, for some combinations of $\omega\in\Omega$, $B\in\mathcal{A}_2$, $\kappa'\left(\omega,B\right)$ may not even be defined.

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The answer is yes, provided $\mathcal{A}_2$ is a Borel space.

This is an application of definition 5.1 and theorem 5.1 from Bahadur, R. R. Sufficiency and Statistical Decision Functions. Ann. Math. Statist. Volume 25, Number 3 (1954), 423-462 (p. 434).

Definition 5.1. A field $\mathbf{S}_0\subseteq\mathbf{S}$ is sufficient for the measures $P$ on $\mathbf{S}$ (briefly, $\mathbf{S}_0$ is sufficient for $P$) if corresponding to each $\mathbf{S}$-measurable set $A$, there exists and $\mathbf{S}_0$-measurable function $\varphi_A(x)$ such that $$\left(5.1\right)\hspace{10mm}\varphi_A(x)=E_p\left(\chi_A(x)\mid\mathbf{S}_0\right)\hspace{5mm}\left[\mathbf{S}_0,p\right]$$ for each $p$ in $P$.

Theorem 5.1. When $R$ is the real line, and $\mathbf{R}$ the class of Borel sets of $R$, the following statements are mutually equivalent.

(i) $\mathbf{S}_0$ is sufficient for $P$.

(ii) (...) [Omitted. --E.A.]

(iii) Corresponding to each function $\mu\left(B,x\right)$, defined for $B\in\mathbf{R}$ and $x\in X$ such that $\mu$ is $\mathbf{S}$-$P$ integrable for each $B$ and a measure on $\mathbf{R}$ for each $x$, there exists a function $\nu\left(B,x\right)$ such that $\nu$ is $\mathbf{S}_0$-$P$-integrable for each $B$ and a measure on $\mathbf{R}$ for each $x$, and such that $$\left(5.3\right)\hspace{10mm}\nu\left(B,x\right)=E_p\left(\mu\left(B,x\right)\mid\mathbf{S}_0\right)\hspace{5mm}\left[\mathbf{S}_0,p\right]$$ for each $B\in\mathbf{R}$ and $p\in P$.

Note that the order of parameters of a kernel in Bahadur's paper is non standard.

The proof of theorem 5.1 is rather involved, but apparently standard: Bahadur notes (p. 435) that it is a generalization of an argument used by Doob for the existence of a conditional probability measure on the real line.

Before applying these results to the problem at hand, it should be noted that theorem 5.1's proof, as given in the paper, can be readily modified to allow for the following enhancements:

1. $\left(R,\mathbf{R}\right)$ may be taken to be any Borel space (i.e. a measurable space isomorphic to a Borel subset of $\left[0,1\right]$; see Kallenberg's Fondations of Modern Probability (2002), p. 7).

2. If $\mu\left(B,x\right)$ is a probability measure in $B$ for any fixed $x$, $\nu$ can be chosen likewise.

Now all that is left is to notice that when $P$ consists of a single measure, every field $\mathbf{S}_0\subseteq\mathbf{S}$ is trivially sufficient.

As a corollary, since any $\sigma$-algebra admits of some probability measure, we get the following result.

If $\kappa$ is a Markov kernel from $\mathcal{A}_1$ to $\mathcal{A}_2$, with $\mathcal{A}_2$ being a Borel space, and if $\mathcal{A}_3\subseteq\mathcal{A}_1$ is a sub-$\sigma$-algebra, there exists some Markov kernel from $\mathcal{A}_3$ to $\mathcal{A}_2$.

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