Minimal sufficient statistic: a measurability issue in a well-known theorem

Question

Given a statistical model $\{\mathbb{P}_\theta\mid\theta\in\Theta\}$ on $(\Omega,\mathscr{F})$, and given a real-valued random variable $X$, we say a real-valued random variable $T$ is a sufficient statistic if $X_*\mathbb{P}_\theta(dx\mid T)$ (the conditional distribution of $X$ given $T$) is independent of $\theta$. We also say a sufficient statistic $S$ is minimal if for all other sufficient statistics $T$ there is a measurable $\varphi:(\mathbb{R},\operatorname{Bor}(\mathbb{R})) \to (\mathbb{R}, \operatorname{Bor}(\mathbb{R}))$ such that $S=\varphi(T)$ $\mathbb{P}_\theta$-a.e. for any $\theta$.

Suppose that all $X_*\mathbb{P}_\theta(dx)$ are absolutely continuous with respect to the Lebsegue measure with density $f(x\mid\theta)$, and that $X,S$ are real-valued random variables. A well-known theorem says that if "for all $x,y\in \mathbb{R}$ $\frac{f(x\mid\theta)}{f(y\mid\theta)}$ is independent of $\theta$ if and only if $S(x)=S(y)$" (let's call this condition $(\star)$), then $S$ is a minimal sufficient statistic.

A proof of this theorem can be found here (Thm 3.4): http://www.statslab.cam.ac.uk/Dept/People/djsteaching/S1B-15-03-sufficiency-4.pdf. This proof (which is the same as all other proofs that I saw) however only shows that, given another sufficient statistic $T$, there exists a function $\varphi:\mathbb{R}\to\mathbb{R}$ such that $S=\varphi(T)$ $\mathbb{P}_\theta$-a.e. for any $\theta$, but doesn't show it is measurable.

My question is simple: how does one show $\varphi$ is measurable?

Answer 1

The existence of a (measurable) minimal sufficient statistic for a family $(f_\theta)$ of densities with respect to the Lebesgue measure follows immediately from Theorem 6.3 by Lehmann and Scheffé (HS) (and the separability of $L^1(\mathbb R)$, as discussed on p. 335 of the HS paper) -- or, rather, concerning the measurability, from the proof of that theorem; cf. the bottom paragraph and the footnote on p. 309 of the HS paper.

Answer 2

I have done some more research about this question and here is my (partial) conclusion.

The theorem as I stated it in the body of the question is probably wrong (or still open) because it's too general. The only form of the theorem that seems to be known is essentially the original one in the paper of Lehmann and Scheffé linked by Iosif Pinelis (where they get a possibly non-measurable $\phi$). In facts, after consulting also some other references in the field (2, 3) I am now more or less convinced that the right way of thinking of minimal sufficient statistics in this theorem is in terms of set functions, not of measurable functions.

Of course one can obtain an obvious corollary for the measurable case from the non-measurable case, which I state below but which adds nothing to the original version.

If a set function $S:\mathcal{X}\to E$ satisfies condition $(\star)$, if $L^1(\mathcal{X})$ is separable, and if we endow $E$ with the final $\sigma$-algebra $\mathscr{E}$ ($A\in \mathscr{E}$ if and only if $S^{-1}(A)\in\mathscr{X}$), then $S:(\mathcal{X},\mathscr{X})\to (E,\mathscr{E})$ is a measurable sufficient statistic which is "measurably minimal" in the class of all measurable sufficient statistics $T$ with the final $\sigma$-algebra on the codomain, meaning that $S=\phi(T)$ for some measurable $\phi$.

I haven't found anything about the general case where the codomains are endowed with arbitrary $\sigma$-algebras.

I also mention that the concept of minimal sufficient $\sigma$-algebra, which seems to have become the standard approach to minimal sufficiency, is equivalent to the one of "measurably" minimal statistic on Polish sample spaces (see Theorem 4 in Rogge - and Landers and Rogge for a counterexample when the space is not Polish).

Answer 3

There is a theorem in descriptive set theory, which might be helpful to attack the measurability issue.

Theorem (Blackwell): Let $X$, $Y$ and $Z$ be measurable spaces, let $f:X\rightarrow Z$ and $g:X\rightarrow Y$ be measurable functions and let $h:Y \rightarrow Z$ be a (not necessarily measurable) function, sucht that $f = h\circ g$. Suppose that $X$ is a Blackwell space, $Y$ is a countably generated Hausdorff measurable space, and $Z$ is a Lusin space (also called a standard Borel space). Then one can replace $h$ with a measurable function. A proof is for example in Delacherie and Meyer, “Probabilities and Potential” on page 51.

I hope this helps.