What is the mathematical characterization of sufficient statistics of a given $\sigma$-dominated probability model?

Question

Given a probability model $\mathcal{P}=\{P_{\theta},\theta \in \Theta \}$ dominated by a $\sigma$-finite measure $\lambda$ (e.g. Lebesgue measure) on a locally compact space $\cal{X}$ along with $\sigma$-field $\cal{F}$. A sufficient statistic is defined to be an $\cal{F}$-measurable mapping $T:\cal{X}\rightarrow \mathbb{R}$ such that $P_{\theta}(X\mid T=t)$ does not depend on the $\theta\in\Theta$.

According to Neyman-Fisher factorization theorem, the sufficient and necessary condition that $T$ is a sufficient statistic for $\cal{P}$ is that $$p(X\mid \theta)=g(T,\theta)h(X),\forall\theta \in \Theta,X\in\cal X$$ where $p(X\mid \theta)=\frac{dP_{\theta}}{d\lambda}$ is the density/Radon-Nikodym derivative w.r.t. $\lambda$.

My questions:

(1) Since the conditional probability measure $P_{\theta}(X\mid T)$ is defined as a solution to the functional equation $\int_{B}P_{\theta}(A \mid \sigma(T))dP_{\theta}=P_{\theta}(A\cap B), \forall B\in \sigma(T)$, are there any characterization of the notion of sufficiency in terms of the $\sigma$-field $\sigma(T)$ generated by the $\cal{F}$-measurable mapping $T$ ,AND the $\sigma$-fields associated with $\cal{P}$?

(2) What is the implication of Neyman-Fisher Theorem in terms of the product measure $\lambda_\Theta\times\lambda$, assuming there is some measure $\lambda_\Theta$ given on the parameter space $\Theta$? It doe not seem like "independence" though.

I did not post this on stats.SE because I think it will receive better response here.

Answer 1

I figured it out by looking at the last few chapters in [Shiryayev] and some thoughts. The problem can be considered in following way with the aid of a formalized definition of conditional expectation.

Without loss of generality we assumed the sample random variable $X$ from the model $\cal{P}$ to be $L^1(P),\forall P\in \cal{P}$. This is easily verified to be true for any $P$ dominated by a $\sigma$-finite measure $\mu$ on the probability space $(\Omega,\mathcal{F},P)$. Also we discuss the situation where $X,T\geq 0,\forall\omega\in\Omega $. Here $$X:\Omega\rightarrow\mathbb{R},\omega\mapsto X(\omega)$$ and $$T:\Omega\rightarrow\mathbb{R},\omega\mapsto T(X(\omega))$$ are random variables. $X$ is the sample random variable; $T$ is the sufficient statistics/random variable.

Invoke Radon-Nikodym Theorem on the measure defined by $\nu(G):=\int_{G}XdP$ on subspace $(\Omega,\sigma(T),P)$ where $\sigma(T)\subset\cal{F}$ is the $\sigma$-field generated by sufficient statistic $T$, then the R-N derivative $\frac{d\nu}{d\mu}$ is the conditional expectation (random variable) $\boldsymbol{E}(X\mid T)(\omega)$. This can be directly checked: $$\int_{G}\boldsymbol{E}(X\mid T)(\omega)d\mu(\omega)$$ $$=\int_{G}\frac{d\nu}{d\mu}d\mu(\omega)=\int_{G}d\nu(\omega)=\nu(G)$$ for any $G\in \sigma(T)$. With this definition, we can directly assert that a conditional expectation is a simple random variable if $A_n$ is a partition of $\Omega$, i.e. $\cup_n A_n=\Omega,A_n\in \sigma(T)$ such that (since we want $1_{A_n}$ to be $\sigma(T)$-measurable functions.) $$\boldsymbol{E}(X\mid T)(\omega)=\frac{\nu(A_n)}{P(A_n)}1_{A_n}(\omega)$$

We know that if a sufficient statistics $T$ for model $\mathcal{P}=\{P\}$ , then from the expression above, the partition $\{A_n\}$ will satisfy that $\frac{\nu(A_n)}{P(A_n)}=\frac{\int_{A_n}XdP}{P(A_n)}=\frac{\int_{A_n}XdP}{\int_{A_n}1dP}$ is independent of the choice of $P\in\cal{P}$.

The level sets of $T$ form a partition as well as generating class(not necessarily atoms) of $\sigma(T)$. So a sufficient statistic will have its level sets partition the $\Omega$ in such a way that the conditional expectation is a constant random variable on each partition component independent of the choice of $P$. In other words, $T$ is a "piece-wise" constant set function, being constant on those sets $A_n\in\cal{F}$ such that $P\{X\in A_n\}\equiv C_n\in\mathbb{R},\forall P\in\cal{P}$

The minimal sufficient statistics $U=\phi(T)$ for some measurable function $\phi$ because $\sigma (U)$ is the coarest partition possible satisfying above requirements. i.e. $\sigma(U)\subset\sigma(T)$ so $U$ must be $\sigma(T)$-measurable, and the relation $U=\phi(T)$ follows from the fact that any $\sigma(T)$-measurable random variable $U$ can be written as $\phi(T)$ for some Borel function $\phi$.

A direct consequence is: there exists a Borel-measurable function $\phi$, $\boldsymbol{E}(U\mid T)=\phi(T)$