Some context

This question came to my head when reading Corollary 8.3.13 from Casella-Berger's Statistical Inference book, whose proof relies completely on my question. Similar questions have been asked (check this) but none of them with a satisfactory answer.

Let $\mathcal{X}:=\mathbb{R}^n\supset\textrm{Im}(X_1,\dotsc,X_n)$ where $\{X_i\}_{i=1}^{n}$ are independent and identically distributed random variables. Also, let's say we know that the measure induced by $(X_1,\dots,X_n)$ over $\mathcal{X}$ is in the set $\mathcal{P}:=\{P_{\theta}:\ \theta\in\Theta\}$.

By the Neyman-Fisher factorization theorem, if $\mathcal{P}$ is a family dominated by $\mu$, then $T$ is a sufficient statistic if and only if $f_{\theta}(\mathbf{x}):=\frac{dP_{\theta}}{d\mu}=g_{\theta}(T(\mathbf{x}))h(\mathbf{x})$ for some non-negative measurable functions $g_{\theta}$ and $h$.

The question

If the measures of $\mathcal{P}$ are discrete, then it is easy to prove that $g_{\theta}$ if the p.m.f. of $T$. But what happens if $\mu$ is the Lebesgue measure? Is $g_{\theta}$ the p.d.f. of $T$? At least in some cases?

My try

I tried to re-read carefully the proof of the factorization theorem, but this is the best I could achieve \begin{equation} P_{T,\theta}(B)=\int_{B}\frac{dP_{\theta}\circ T^{-1}}{dx}dx=\int_{B}g_{\theta}\sum_{i=1}^{\infty}c_i\frac{dP_{\theta_i}\circ T^{-1}}{dx}dx\text{, } \end{equation}

  • $\begingroup$ The factorization $f=gh = (cg)(h/c)$ is not unique, so its hard to see how you could conclude $g$ integrates to $1$. $\endgroup$ Feb 16, 2021 at 11:01
  • 1
    $\begingroup$ So, is there a factorization such that $g$ is the p.d.f of $T$? $\endgroup$ Feb 16, 2021 at 11:07


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