# Sufficient statistic and hypothesis testing

Suppose I have a family of (continuous) distributions $\mathcal{P}=\{P_\theta(x),\theta\in\mathbb{R}^+\}$. I also have a statistic $T(x)$ that is sufficient for $\theta$.

The value of the parameter $\theta$ is unknown and my interest in it is limited to testing $H_0:\theta=0$ vs. $H_1:\theta>0$.

Clearly, the alternate hypothesis $H_1$ is composite, thus Neyman-Pearson lemma doesn't apply. Also, unfortunately, the "easy" uniformly most powerful (UMP) test formulation doesn't apply as I don't think that the family of densities $p_\theta(x)$ has a monotone likelihood ratio (see Lehmann and Romano p. 65). As the matter of fact, while, for any $\theta <\theta'$, the distributions $P_\theta$ and $P_{\theta'}$ are distinct, the ratio $p_{\theta'}(x)/p_\theta(x)$ is a non-increasing (as opposed to non-decreasing) function of $T(x)$.

I understand that the maximum likelihood estimator (MLE) $\hat{\theta}_{MLE}$ of $\theta$ is a function of $T(x)$. But what is $T(x)$'s relationship to hypothesis testing?

I ask because a trivial test that solves my problem can be constructed as follows: first, pick threshold $\eta$, then compute estimate $\hat{\theta}_{MLE}$, and accept $H_0$ if $\hat{\theta}_{MLE}<\eta$, reject otherwise. The error probabilities then depend on the choice of $\eta$ and the value of $\theta$. However, can anything be said about optimality (in some way, shape or form) of the said test? Are there any conditions on $\mathcal{P}$ other than monotone likelihood function described above that lead to optimality (maybe something weaker than UMP)? The test using $\hat{\theta}_{MLE}$ seems like the best I can do, however, is there a way to prove this statement?

First, based on the Fisher–Neyman Factorization, if $T(\mathbf{x})$ is a sufficient statistic, then the likelihood function factorizes to the product of (1) a function that does not involve $\theta$; times (2) a function that depends on the sample only through the sufficient statistic $T(\mathbf{x})$. So the first function out of the factorization cancels when one looks at the likelihood ratio. In other words, if there is a sufficient statistic $T(\mathbf{x})$, the likelihood ratio's dependency on the sample is only through $T(\mathbf{x})$. Then, assessing the plausibility of a bigger $\theta$ versus a smaller $\theta$ (i.e., whether or not to reject the null hypothesis) based on the sample $\mathbf{x}$ has to be tied to $T(\mathbf{x})$.
Second, if the ratio is "non-increasing (as opposed to non-decreasing)" function of $T(\mathbf{x})$, wouldn't the ratio automatically be a non-decreasing function of $-T(\mathbf{x})$? Doesn't the theorem about UMP test then apply, now using the "different" sufficient statistic $-T(\mathbf{x})$?