# Doubt about Neyman-Pearson lemma and sufficiency (Fisher-Neyman factorization)

### 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 $$$$P_{T,\theta}(B)=\int_{B}\frac{dP_{\theta}\circ T^{-1}}{dx}dx=\int_{B}g_{\theta}\sum_{i=1}^{\infty}c_i\frac{dP_{\theta_i}\circ T^{-1}}{dx}dx\text{, }$$$$

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