1
$\begingroup$

Trying to obtain some exchangeability-related results, I ended up with the following questions, which I couldn't answer (at least, not in the negative); this is also related to this MO thread (edit: actually, not really)

Given a measurable space $(S,\mathscr{S})$, let $\pi$ be a probability measure on the product space $(S^2,\mathscr{S}^2)$. Also, endow the set $\mathcal{P}(S)$ of probability measures over $(S,\mathscr{S})$ with the $\sigma$-algebra generated by the measurable maps $\nu \mapsto \nu(F)$, with $F \in \mathscr{S}$ (and similarly for $\mathcal{P}(S)^2$).

Question. Is it true that we can find always a probability measure $\mu$ on $\mathcal{P}(S)^2$ such that $$ \pi(A)=\int_{\mathcal{P}(S)^2} (\alpha \otimes \beta)(A) \,\mu(\mathrm{d}(\alpha\otimes \beta))\,\,\,\,\,? $$ for all $A \in \mathscr{S}^2$.

[Here, given probability measures $\alpha,\beta$ on $(S,\mathscr{S})$, we let $\alpha \otimes \beta$ denote their product measure]

Improve this question
$\endgroup$
8
  • $\begingroup$ You're trying to integrate a function whose codomain is $\mathcal{P}(S^2)$ - can you explain how this is to be interepreted? Most "vector-valued" integrals (Bochner, Pettis, etc) tend to need a codomain which is a Banach space, or at least some sort of "complete" topological vector space. Here we do not have a vector space (though it does have a convex structure), and you have not specified a topology. $\endgroup$
    – Nate Eldredge
    Commented Mar 9, 2016 at 21:14
  • $\begingroup$ Or do you simply want $\pi(A) = \int_{\mathcal{P}(S^2)} (\alpha \otimes \beta)(A) \mu(d\alpha, d\beta)$ for all measurable sets $A$? $\endgroup$
    – Nate Eldredge
    Commented Mar 9, 2016 at 21:15
  • $\begingroup$ I mean the second one, for all $A \in \mathscr{S}^2$; sorry for having been not clear, I add it in the text.. $\endgroup$
    – Paolo Leonetti
    Commented Mar 9, 2016 at 21:16
  • $\begingroup$ And you meant $\mu$ to be a probability measure on $\mathcal{P}(S^2)$? $\endgroup$
    – Nate Eldredge
    Commented Mar 9, 2016 at 21:19
  • $\begingroup$ So it seems an obvious thing to do is to consider the map $S^2 \to \mathcal{P}(S^2)$ that sends $(x,y)$ to the Dirac mass $\delta_{(x,y)}$, and try letting $\mu$ be the pushforward of $\pi$ under this map. There are some measurability things to check, though. $\endgroup$
    – Nate Eldredge
    Commented Mar 9, 2016 at 21:21

1 Answer 1

Reset to default
1
$\begingroup$

It's true; we can use the following rather trivial construction.

Notation: for measurable $A \subset S^2$, let $F_A : \mathcal{P}(S)^2 \to [0,1]$ be the "evaluation map" $F_A(\alpha, \beta) = (\alpha \otimes \beta)(A)$.

Consider the map $T : S^2 \to \mathcal{P}(S)^2$ defined by $T(x,y) = (\delta_x, \delta_y)$. This is the natural way to embed a space into its space of probability measures: map a point to a point mass. We note that $T$ is measurable, since for any $F_A$ we have $F_A(T(x,y)) = (\delta_x \otimes \delta_y)(A) = 1_A(x,y)$. So $F_A \circ T = 1_A$ is a measurable map from $S^2$ to $[0,1]$. Since the $F_A$ generate the $\sigma$-algebra on $\mathcal{P}(S^2)$, the measurability of $T$ follows.

Now for a given measure $\pi$, simply let $\mu = \pi \circ T^{-1}$ be the pushforward of $\pi$ under $T$. Then by change of variables $$\begin{align*}\int_{\mathcal{P}(S)^2} F_A(\alpha, \beta) \,\mu(d\alpha, d\beta) &= \int_{S^2} F_A(T(x,y))\,\pi(dx, dy) \\ &= \int_{S^2} 1_A(x,y)\,\pi(dx,dy) \\ &= \pi(A).\end{align*}$$

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .