The Karlin–Rubin theorem for one-sided tests

By [F2], choose $c_{+},{\gamma }_{+}$ and $c_{-},{\gamma }_{-}$ so both threshold functions have expectation $\alpha$ under $P_{{\theta }_0}$. Since $T$ is measurable, their threshold events and functions ${\varphi }_{+},{\varphi }_{-}$ are measurable; disjointness of each pair of threshold events and ${\gamma }_{\pm }\in [0,1]$ give $0\le {\varphi }_{\pm }\le 1$. For each sign, define $K_{\pm }(x,B)={\varphi }_{\pm }(x){\mathbf{1}}_{\{1\in B\}}+(1-{\varphi }_{\pm }(x)){\mathbf{1}}_{\{0\in B\}}$ for $B\subseteq \{0,1\}$. The two weights are nonnegative and sum to one, and the formula is countably additive because each point belongs to at most one event in a disjoint sequence; thus $K_{\pm }(x,\cdot )$ is a probability measure. Each event evaluation is one of $0,1,{\varphi }_{\pm },1-{\varphi }_{\pm }$, hence measurable by [F9]. Thus [F10] gives probability kernels, which are randomized tests by [F4], with the stated functions. Select the finite-valued density representatives used below as in [F7] and [A1].

By [F3], $E_{\theta }{\varphi }_{+}\le E_{{\theta }_0}{\varphi }_{+}=\alpha$ for $\theta \le {\theta }_0$, and $E_{\theta }{\varphi }_{-}\le E_{{\theta }_0}{\varphi }_{-}=\alpha$ for $\theta \ge {\theta }_0$. Each null contains ${\theta }_0$, so its supremum size is at most $\alpha$; by [F8] and the empty-alternative convention in the statement, both candidates have level $\alpha$ for their stated nulls.

Fix ${\theta }_1>{\theta }_0$ with densities $p_0,p_1$, let $q$ be their Borel MLR function, and let $N$ be its pairwise $\mu$-null exceptional set from [F1]. By [F5], $P_{{\theta }_0}(N)=0$ and $P_{{\theta }_0}(p_0>0)=1$. Put $k=q(c_{+})$. The value $k$ is finite: if $P_{{\theta }_0}(T=c_{+})>0$, there is a point of this boundary outside $N$ with $p_0>0$ and finite $p_1$; otherwise $P_{{\theta }_0}(T>c_{+})=\alpha >0$ gives such a point above the threshold and monotonicity forces $q(c_{+})<\mathrm{\infty }$. On $\{p_0>0\}\setminus N$, $p_1-k{p}_0=p_0(q(T)-k)$ is nonnegative above $c_{+}$, nonpositive below it, and zero at $T=c_{+}$. On $\{p_0=0<p_1\}\setminus N$, $q(T)=+\mathrm{\infty }$; finite $k$ and monotonicity force $T>c_{+}$, where ${\varphi }_{+}=1$. Where both densities vanish, $p_1-k{p}_0=0$. By [F9], the products and differences are measurable, so every competing test $\psi$ satisfies $({\varphi }_{+}-\psi )(p_1-k{p}_0)\ge 0$ $\mu$-almost everywhere.

The measurable product in step 2.2 is integrable because $\mathrm{\mid }{\varphi }_{+}-\psi \mathrm{\mid }\le 1$ and $p_0,p_1$ each integrate to one. Apply [F5] to the positive and negative parts of ${\varphi }_{+}-\psi$; linearity [F6] then gives $E_{{\theta }_1}({\varphi }_{+}-\psi )-k{E}_{{\theta }_0}({\varphi }_{+}-\psi )=\int ({\varphi }_{+}-\psi )(p_1-k{p}_0)d\mu \ge 0$. For a level-$\alpha$ competitor, [F8] and step 1.1 give $E_{{\theta }_0}\psi \le \alpha =E_{{\theta }_0}{\varphi }_{+}$, while $k\ge 0$; thus $E_{{\theta }_1}{\varphi }_{+}\ge E_{{\theta }_1}\psi$. Since ${\theta }_1>{\theta }_0$ was arbitrary, ${\varphi }_{+}$ is UMP for the upper alternative.

Reparameterize by $\eta =-\theta$ and put $S=-T$. For ${\eta }_1<{\eta }_2$, let $q$ represent $p_{-{\eta }_1}\mathrm{/}{p}_{-{\eta }_2}$ as in [F1]. Define $h(u)=1\mathrm{/}u$ for $0<u<+\mathrm{\infty }$, $h(0)=+\mathrm{\infty }$, and $h(+\mathrm{\infty })=0$. The function $s\mapsto h(q(-s))$ is Borel and nondecreasing, since reflection and $h$ both reverse order. Outside the original pairwise exceptional set on the union of supports, it represents $p_{-{\eta }_2}\mathrm{/}{p}_{-{\eta }_1}$, including points where either density is zero; where both vanish no ratio is required. Thus the reflected family has MLR in $S$. Apply the upper-tail result of steps 1.1–3.1 at ${\eta }_0=-{\theta }_0$ with cutoff $-c_{-}$ and boundary probability ${\gamma }_{-}$, calibrated by step 1.1. The resulting UMP test is exactly ${\varphi }_{-}$ for the original lower alternative.