Does the pushforward of a sufficient statistic induce unique probability measures?

Question

Consider a collection of probability measures $\{P_\theta | \theta \in \Theta\}$ and a sufficient statistic $T$, that is for all $A \in \Sigma$ (the $\sigma$-algbera): $\mathbb{E}_\theta(1_A|T)$ is $\theta$-free. We assume that for all $\theta \in \Theta$, $P_\theta$ is unique. Are, then, the pushforwards $P_\theta T^{-1}$ also unique for all $\theta \in \Theta?$ I have tried finding a simple counterexample, but I did not succeed.

Answer 1

Yes. Suppose $\mathbb P_\theta T^{-1} = \mathbb P_{\theta'}T^{-1}$. Then for any $\sigma(T)$-measurable non-negative r.v. $X$, we have $\mathbb E_\theta X = \mathbb E_{\theta'} X$; if you want to be rigorous, you can prove this e.g. by taking limits of simple functions. We deduce for any measurable $A$,

$$\mathbb P_\theta(A) = \mathbb E_\theta[\mathbb E[1_A \mid T]] = \mathbb E_{\theta'}[\mathbb E[1_A \mid T]] = \mathbb P_{\theta'}(A),$$

so $\mathbb P_\theta = \mathbb P_ {\theta'}$, and $\theta = \theta'$.