Proof of Karlin-Rubin's theorem

Question

I asked this question on Math Exchange, but as I did not receive a successful answer, maybe you could help me.

Karlin-Rubin's theorem states conditions under which we can find a uniformly most powerful test (UMPT) for a statistical hypothesis:

Suppose a family of density or mass functions $\{f(\vec{x}|\theta):\,\theta\in\Theta\}$ and we want to test $$\begin{cases} H_0:\,\theta\leq\theta_0 \\ H_A:\,\theta>\theta_0.\end{cases}$$If the likelihood ratio is monotone on a statistic $T(\vec{x})$ (that is, for every fixed $\theta_1<\theta_2$ in $\Theta$, the ratio $\frac{f(\vec{x}|\theta_2)}{f(\vec{x}|\theta_1)}$ is nondecreasing on $\{\vec{x}:\,f(\vec{x}|\theta_2)>0\text{ or }f(\vec{x}|\theta_1)>0\}$ as a function of $T(\vec{x})$, interpreting $c/0=\infty$ if $c>0$), then the test of critical region $\text{CR}=\{\vec{x}:\,T(\vec{x})\geq k\}$, where $k$ is chosen so that $\alpha=P(\text{CR}|\theta=\theta_0)$, is the UMPT of size $\alpha$.

In all the proofs I have read (for instance, in page 22 here or in "Statistical inference" by Casella-Berger, 2n edition, page 391), it is (more or less) said: "we can find $k_1$ such that, if $T(\vec{x})\geq k$, then $\frac{f(\vec{x}|\theta_2)}{f(\vec{x}|\theta_1)}\geq k_1$, and if $T(\vec{x})<k$, then $\frac{f(\vec{x}|\theta_2)}{f(\vec{x}|\theta_1)}< k_1$". I would understand that statement if the likehood ratio were strictly increasing, but what about the case in which it is constant?

For example, if $X\sim U(0,\theta)$, the likelihood ratio is monotone on $T(\vec{x})=\max_{1\leq i\leq n}x_i$ ($n$ is the length of the sample $\vec{x}$), but not strictly increasing.

My questions are:

1. Is the assertion between quotation marks true for every density or mass function with (not strictly) monotone likelihood ratio on $T$?

2. And what about in the case of the uniform distribution?

Answer 1

$\newcommand\th\theta\newcommand\al\alpha$Even the following more general assertion is true:

Suppose that there is a statistic $T$ such that for all $\th_0$ and $\th_1$ such that $\th_0<\th_1$ there is a nondecreasing function $g_{\th_0,\th_1}$ such that for all $x$ we have $$r_{\th_0,\th_1}(x):=\frac{f_{\th_1}(x)}{f_{\th_0}(x)}=g_{\th_0,\th_1}(T(x)).$$ Suppose that a test $\phi$ is such that for some real $k$ and all $x$ $$\phi(x)=\begin{cases} 1&\text{ if }T(x)>k, \\ 0&\text{ if }T(x)<k \end{cases}$$ (the test $\phi$ may be randomized, taking any value $\phi(x)$ in the interval $[0,1]$ if $x$ is such that $T(x)=k$). Then the test $\phi$ is uniformly most powerful of level $\al:=E\phi(X)$ for the hypotheses as in the OP.

This follows because, for $c:=g_{\th_0,\th_1}(k)$, we have the implications $$r_{\th_0,\th_1}(x)>c\iff g_{\th_0,\th_1}(T(x))>c\implies T(x)>k\implies\phi(x)=1\tag{1}$$ (so that $H_0$ is rejected by the test $\phi$) and $$r_{\th_0,\th_1}(x)<c\iff g_{\th_0,\th_1}(T(x))<c\implies T(x)<k\implies\phi(x)=0\tag{2}$$ (so that $H_0$ is not rejected by the test $\phi$) -- which means that $\phi$ is a Neyman--Pearson testfor any $\th_0$ and $\th_1$ such that $\th_0<\th_1$.

That $T$ is not strictly increasing causes no problems whatsoever, because in (1) and (2) we only need the left-to-right implications.

Answer 2

If it were to be constant then the statistical model $\{f(\vec{x}|\theta):\,\theta\in\Theta\}$ would only contain copies of the same density, i.e. one single density. In this case, inference is quite useless, but nonetheless the test is UMPT (and unbiased) since power and size agree for every test.