Let $(Ω, A,P)$ be a statistical model, $H_{0} = \{P_{0}\}\subseteq P$ a simple null hypothesis, and $H_{1} = \{P_{1}\} ⊂ P$ a different simple alternative, so that $P_{1}$ with respect to $P_{0}$ has a density $L = dP_{1}/dP_{0}$. We assume the distribution $\mathcal{L}_{P_{0}}(L)$ of the Likelihood-Quotient $L$ under $P_{0}$ has a density $f$ w.r.t. the Lebesgue-measure. Let $p : Ω → \mathbb R$ be the $p-$value of the Likelihood-Quotient-Test (Neyman-Pearson-Test) for both Hypotheses. Which of the following statements $A-E$ is definitely true, and why? Only one is true:
$A$ The distribution $L_{P_{0}}(p)$ of $p$-value w.r.t. the null hypothesis has the density $f$ w.r.t the Lebesgue Measure.
$B$ The distribution $L_{P_{1}}(p)$ of $p$-value w.r.t. the alternative hypothesis has the density $f$ w.r.t the Lebesgue Measure.
$C$ The $p-$value is uniformly distributed w.r.t the null hypothesis on the unit interval
$D$ The $p-$value is uniformly distributed w.r.t the alternative hypothesis on the unit interval
$E$ The statements $A-D$ are incorrect.
I am new to statistics. But we do not have any critical region $V$ that is given. I know that the Likelihood Quotient Test is given by the critical region $V$, where $\{ \frac{dP_{1}}{dP_{0}}>c\}\subseteq V \subseteq \{ \frac{dP_{1}}{dP_{0}}\geq c\}$. Furthermore, the $p-$value is based on the infimum of significance levels $a_{V}$. I do not see how any of this may help me prove that statements $A-E$. Any ideas?
