# Proof of Karlin-Rubin's theorem

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?

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.
• But I mean constant around $k$, not everywhere (as in the case of $U(0,\theta)$, which is piecewise constant). The thing is that, if the likelihood ratio is constant around $k$, it is not true (I think) that "we can find $k_1$ such that, if $T(\vec{x})\geq k$, then $f(\vec{x}|\theta_2)/f(\vec{x}|\theta_1)\geq k_1$, and if $T(\vec{x})< k$, then $f(\vec{x}|\theta_2)/f(\vec{x}|\theta_1)< k_1$". So, why is that assertion written in all proofs? – user39756 Jan 15 '17 at 14:15
• The thing is that if for $\theta_1<\theta_2$ we have a constant likelihood ratio around $k$ then one density is $k$ times the other. Since they both integrate to $1$, then necessarily $k=1$ and the densities are equal. – mbe Jan 15 '17 at 14:23
• I see this: Let $\theta_1<\theta_2$. We have $$\frac{f(\vec{x}|\theta_2)}{f(\vec{x}|\theta_1)}=g_{\theta_1,\theta_2}(T(\vec{x})),$$ where $g=g_{\theta_1,\theta_2}$ is monotone and constant around the point $k\in \mathbb{R}$. That is, there is an $r>0$ such that $g|_{]k-r,k+r[}=C=\text{constant}$. Then $f(\vec{x}|\theta_2)=C\,f(\vec{x}|\theta_1)$ for all $\vec{x}\in T^{-1}(]k-r,k+r[)$. The set of equality is not the whole $\mathbb{R}$ so I cannot integrate. And $C$ is not $k$. – user39756 Jan 15 '17 at 14:33