1
$\begingroup$

I've asked this question also on mathematics stackexchange, but despite nearly two dozen views, there isn't a single comment, nevermind an answer. Any help would be appreciated.

Update: See update 1 at bottom.


Let $(X,\mathcal A)$ and $(Y,\mathcal F)$ be measurable spaces.

Consider a probability kernel $\kappa : X \times \mathcal F \to [0,1]$.

I need to formalize the notion of a nonrandomized probability kernel. Two natural definitions are:

  1. for all $x \in X$, exists $y \in Y$, $\kappa(x,\{y\}) = 1$.
  2. for all $x \in X$, exists $y \in Y$, for all $A \in \mathcal F$, $(\kappa(x,A) = 1 \iff y \in A)$.

I believe the two definitions are equivalent if the singletons are measurable in $\mathcal F$. (Agree?)

In either case, consider the double integral: $$ \Phi = \int_X \Bigl \{ \int_Y f(x,y) \kappa(x,dy) \Bigr \} \mu(dx), $$ where $f$ is product measurable and $\mu$ is a probability measure on $(X,\mathcal A)$.

If $\kappa$ is nonrandomized (as in Defn 1 or 2 above), when can I assume that there exists a(n ostensibly measurable?) function $g : X \to Y$ such that $$ \Phi = \int_X f(x,g(x)) \mu(dx) $$ holds? (We can assume $f$ is integrable with respect to $\mu \otimes \kappa$, or alternative that $f$ is nonnegative (or nonpositive).)


Update 1

Commenters rightly pointed out that the question seems trivial. Indeed, the case where we assume definition 1 is straightforward and I followed the outline provided by Nate Eldredge to give sketched proofs.

The case where we assume merely definition 2 is still not clear to me. There may not be a unique $y$ for each $x$, and then we would need some sort of measurable selection, and I'm not versed in the requisite theorems. It would seem that I would need some structure on $Y$ beyond a $\sigma$-algebra. E.g., the Kuratowski and Ryll-Nardzewski measurable selection theorem would seem to require $(Y,\mathcal F)$ to be a Polish space with its Borel $\sigma$-algebra, but I believe that would imply that singletons are measurable, and so then the definitions collapse.

$\endgroup$
4
  • $\begingroup$ Maybe I am missing something, but this seems pretty trivial? You have assumed that for each $x$ there exists $y$ such that $\kappa(x,\{y\})=1$, and by definition of a probability kernel, such $y$ is unique. So just define $g(x)$ to be that unique $y$, and it's clear that $\int f(x,y) \kappa(x,dy) = f(x, g(x))$. Since probability kernels are measurable in the first variable, you will get that $g^{-1}(\{y\})$ is measurable for each singleton $\{y\}$, but $g$ need not be measurable. $\endgroup$ Commented Dec 28, 2016 at 17:02
  • $\begingroup$ Correction: $g$ is measurable, since for any measurable $A \subset Y$, we have $g^{-1}(A) = \kappa(\cdot, A)^{-1}(\{1\})$. $\endgroup$ Commented Dec 28, 2016 at 17:30
  • 3
    $\begingroup$ I suggest you work out the details and post your own answer. $\endgroup$ Commented Dec 28, 2016 at 17:30
  • $\begingroup$ I will do that. $\endgroup$ Commented Dec 28, 2016 at 18:42

1 Answer 1

-1
$\begingroup$

This is a partial answer. See updated question for seemingly trickier question.

Claim. Definitions 1 and 2 are equivalent if singletons are measurable.

Proof. To see this, consider the second definition, pick $x \in X$, and let $y \in Y$ be as in the definition. Since $y \in \{y\}$ and $\{y\}$ is measurable, $\kappa(x,\{y\}) = 1$. In the other direction, $\kappa(x,\{y\}) = 1$, which then implies, by the monotonicity of probability measures, that $\kappa(x,A) = 1$ if $y \in A$. If $y \not\in A$, then, again because $\{y\}$ is measurable and basic facts about probability distributions, $\kappa(x,A) = 1 - \kappa(x,Y \setminus A) \le 1 - \kappa(x,\{y\}) = 0$.

Note: I believe that the measurability of singletons is not necessary for the equivalence because it suffices that for all $x \in X$, there exists $y \in Y$, such that the singleton $\{y\}$ is measurable.

If we adopt definition 1, the comments by Nate Eldredge lead to the following proof.

Claim. Assume definition 1. Then such a $g$ exists and is measurable.

Proof. Define $g(x)$ to be the unique $y$ satisfying $\kappa(x,\{y\})=1$. To see that $g$ is measurable, note that, for all $A \in \mathcal F$, $\kappa(\cdot,A)$ is measurable and $$\begin{align}g^{-1}(A) &= \{ x \in X : (\exists y \in A)\, \kappa(x,\{y\}) = 1 \}\\ &= \{ x \in X : \kappa(x,A) = 1 \}\\ &= \kappa(\cdot,A)^{-1}(\{1\}).\end{align}$$ Then, for all $x \in X$, $$ \int_Y f(x,y) \kappa(x,dy) = f(x,g(x)), $$ and this quantity is $\mathcal A$-measurable because the l.h.s. was assumed to be.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.