UMP Tests

If $E_n\uparrow E$, write $D_0=E_0$ and $D_n=E_n\setminus E_{n-1}$ for $n\ge 1$; these sets are disjoint with union $E$, so [F1] gives $Q(E)={\sum }_{n}Q(D_n)={\lim }_{n}Q(E_n)$. If $E_n\downarrow E$ and $Q(E_0)<\mathrm{\infty }$, apply continuity from below to $E_0\setminus E_n\uparrow E_0\setminus E$ and subtract from $Q(E_0)$. Thus $Q$ is continuous from below and above.

Set $S(t)=Q((t,\mathrm{\infty }))$ and $C=\{t:S(t)\le \alpha \}$. By step 1.1, $S(n)\to 0$ and $S(-n)\to 1$, so $C$ is nonempty and bounded below; it is upward closed. Let $c_{+}=\inf C$ by [F3]. For $t>c_{+}$ there is $s\in C$ with $s<t$, hence $S(t)\le \alpha$; for $t<c_{+}$, $S(t)>\alpha$. Continuity from below and above gives $Q((c_{+},\mathrm{\infty }))\le \alpha \le Q([c_{+},\mathrm{\infty }))$. If $d=Q(\{c_{+}\})>0$, set ${\gamma }_{+}=(\alpha -Q((c_{+},\mathrm{\infty })))\mathrm{/}d\in [0,1]$; if $d=0$, the inequalities force $Q((c_{+},\mathrm{\infty }))=\alpha$, so take ${\gamma }_{+}=0$.

Define $Q^{-}(B)=Q(\{x:-x\in B\})$. Continuity of reflection and [F4] show every Borel $B$ has Borel preimage; [F1]–[F2] then make $Q^{-}$ a Borel probability. Apply step 2.1 to $Q^{-}$, obtaining $c^{\mathrm{\prime }}$ and ${\gamma }_{-}$. With $c_{-}=-c^{\mathrm{\prime }}$, $Q^{-}((c^{\mathrm{\prime }},\mathrm{\infty }))=Q((-\mathrm{\infty },c_{-}))$ and $Q^{-}(\{c^{\mathrm{\prime }}\})=Q(\{c_{-}\})$, which gives the lower-tail equality.

Choose simultaneous densities $p_1=p_{{\theta }_1}$ and $p_2=p_{{\theta }_2}$ using AC [A1], let $r$ be their nondecreasing Borel MLR function, and put $A=\{p_1>0\}$ and $S=\{p_1=0<p_2\}$ with $s=P_{{\theta }_2}(S)$. Let $N$ be the union of the $\mu$-null exceptional sets in [F1]. For $x\in S\setminus N$ and $y\in A\setminus N$, $r(T(x))=+\mathrm{\infty }$ while $r(T(y))=p_2(y)\mathrm{/}{p}_1(y)<+\mathrm{\infty }$; monotonicity therefore gives $T(y)<T(x)$. The two laws give $N$ measure zero, $P_{{\theta }_1}(A)=1$, and $P_{{\theta }_2}(S)=s$. Since ${\varphi }_{c,\gamma }^{+}$ is nondecreasing in $T$ and ${\varphi }_{c,\gamma }^{-}$ is nonincreasing in $T$, this support order gives $E_{{\theta }_2}[{\varphi }_{c,\gamma }^{+}{\mathbf{1}}_S]\ge s{E}_{{\theta }_1}{\varphi }_{c,\gamma }^{+},E_{{\theta }_2}[{\varphi }_{c,\gamma }^{-}{\mathbf{1}}_S]\le s{E}_{{\theta }_1}{\varphi }_{c,\gamma }^{-}\mathrm{.}$ By [F4], $r(T)$ is measurable.

Partition $\mathcal{X}$ into $B_{-}=\{T<c\}$, $B_0=\{T=c\}$, and $B_{+}=\{T>c\}$. Put $a_j=P_{{\theta }_1}(B_j)$ and $b_j=P_{{\theta }_2}(B_j\cap A)$. On $A$, [F1] gives $p_2=r(T)p_1$ $\mu$-almost everywhere, so [F2] yields $b_j={\int }_{B_j\cap A}{p}_{2}d\mu ={\int }_{B_j\cap A}r(T)p_{1}d\mu ={\int }_{B_j\cap A}r(T)d{P}_{{\theta }_1}$. The measurability of $r(T)$ was established in step 1.1. Also [F3] and step 1.1 give ${\sum }_j{b}_j=1-s$ and $P_{{\theta }_1}(A)=1$, hence $P_{{\theta }_1}(B_j\cap A)=a_j$. For each block with $a_j>0$, set $r_j=b_j\mathrm{/}{a}_j$. Monotonicity [F1] orders $r(T)$ values across every pair of occupied blocks, so their $P_{{\theta }_1}$-averages satisfy $r_i\le r_j$ whenever $i<j$.

For ordered block values $g_{-}\le g_0\le g_{+}$, omitting zero-mass blocks, expansion gives ${\sum }_j{a}_j{r}_j{g}_j-({\sum }_j{a}_j{r}_j)({\sum }_j{a}_j{g}_j)={\sum }_{i<j}{a}_i{a}_j(r_j-r_i)(g_j-g_i)\ge 0$. Step 2.1 gives ${\sum }_j{a}_j=1$ and ${\sum }_j{a}_j{r}_j=1-s$. For the upper threshold $g=(0,\gamma ,1)$, this identity makes $E_{{\theta }_2}[{\varphi }_{c,\gamma }^{+}{\mathbf{1}}_A]\ge (1-s)E_{{\theta }_1}{\varphi }_{c,\gamma }^{+}$. Step 1.1 gives $E_{{\theta }_2}[{\varphi }_{c,\gamma }^{+}{\mathbf{1}}_S]\ge s{E}_{{\theta }_1}{\varphi }_{c,\gamma }^{+}$, and $P_{{\theta }_2}(\mathcal{X}\setminus (A\cup S))=0$. Adding these bounds proves $E_{{\theta }_2}{\varphi }_{c,\gamma }^{+}\ge E_{{\theta }_1}{\varphi }_{c,\gamma }^{+}$.

The lower threshold $g=(1,\gamma ,0)$ is nonincreasing, so the covariance identity in step 3.1 is nonpositive and gives $E_{{\theta }_2}[{\varphi }_{c,\gamma }^{-}{\mathbf{1}}_A]\le (1-s)E_{{\theta }_1}{\varphi }_{c,\gamma }^{-}$. Step 1.1 gives $E_{{\theta }_2}[{\varphi }_{c,\gamma }^{-}{\mathbf{1}}_S]\le s{E}_{{\theta }_1}{\varphi }_{c,\gamma }^{-}$; the remaining set has $P_{{\theta }_2}$ measure zero. Adding these bounds proves $E_{{\theta }_2}{\varphi }_{c,\gamma }^{-}\le E_{{\theta }_1}{\varphi }_{c,\gamma }^{-}$.

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.

Write $\phi (x)=e^{-x^2\mathrm{/}2}\mathrm{/}\sqrt{2\pi }$ and $P_{\theta }=N(\theta ,1)$. The translation ${\tau }_{\theta }(x)=x+\theta$ is measurable and preserves Lebesgue measure by [F2]–[F3]; [F1] makes $P_{\theta }$ the push-forward of $\phi d\lambda$ under ${\tau }_{\theta }$. For each Borel $A$, $x\mapsto \phi (x-\theta )$ is continuous, so its product with the measurable indicator ${\mathbf{1}}_A$ is measurable by [F12]; integral invariance gives $P_{\theta }(A)={\int }_A\phi (x-\theta )d\lambda (x)$, so $p_{\theta }(x)=\phi (x-\theta )$ is a positive density for every $\theta$.

AC [A1] applied to any sequence of nonempty sets supplies a choice function on its values and hence a sequence choice, so [F5] shows every singleton is $\lambda$-null. By step 1.1 and [F4], $P_0(\{c\})={\int }_{\{c\}}{p}_{0}d\lambda =0$ for every real $c$. Applying [F9] under $P_0$ gives finite $c_{+},c_{-}$ with $P_0(X>c_{+})=P_0(X<c_{-})=\alpha$, since the boundary masses vanish. The threshold indicators are measurable by [F13]. For $B\subseteq \{0,1\}$, define $K_{\pm }(x,B)={\varphi }_{\pm }(x){\mathbf{1}}_{\{1\in B\}}+(1-{\varphi }_{\pm }(x)){\mathbf{1}}_{\{0\in B\}}$. 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; each event evaluation is measurable by [F12]–[F13]. Thus [F14] gives probability kernels with test functions ${\varphi }_{\pm }$, which are randomized tests by [F8] and have level $\alpha$ by [F10].

Step 1.1 gives $p_1(x)\mathrm{/}{p}_0(x)=e^{x-1\mathrm{/}2}$, strictly increasing, and $p_{-1}(x)\mathrm{/}{p}_0(x)=e^{-x-1\mathrm{/}2}$, strictly decreasing. Put $k_{+}=e^{c_{+}-1\mathrm{/}2}>0$. For any level-$\alpha$ test $\psi$, $({\varphi }_{+}-\psi )(p_1-k_{+}{p}_0)\ge 0$: above $c_{+}$ both factors are nonnegative, below it both are nonpositive, and at $c_{+}$ the second vanishes. [F12] makes this product measurable; since $\mathrm{\mid }{\varphi }_{+}-\psi \mathrm{\mid }\le 1$ and each density integrates to one, [F7] makes it integrable. Apply [F4] to the positive and negative parts of ${\varphi }_{+}-\psi$, then [F7] gives $E_1({\varphi }_{+}-\psi )-k_{+}{E}_0({\varphi }_{+}-\psi )=\int ({\varphi }_{+}-\psi )(p_1-k_{+}{p}_0)d\lambda \ge 0$. By [F10] and step 2.1, $E_0{\varphi }_{+}=\alpha \ge E_0\psi$, so $E_1{\varphi }_{+}\ge E_1\psi$. If equality holds, the two nonnegative terms in $E_1({\varphi }_{+}-\psi )=\int ({\varphi }_{+}-\psi )(p_1-k_{+}{p}_0)d\lambda +k_{+}{E}_0({\varphi }_{+}-\psi )$ vanish. [F11] makes the integrand zero $\lambda$-almost everywhere; strict monotonicity then forces $\psi ={\varphi }_{+}$ off $c_{+}$ almost everywhere, and step 2.1 gives $P_0(\{c_{+}\})=0$. Since $P_0$ has density $p_0$, equality forces $\psi ={\varphi }_{+}$ $P_0$-almost surely, so ${\varphi }_{+}$ is most powerful against $\theta =1$, uniquely up to $P_0$-almost-sure equality of rejection functions.

Put $k_{-}=e^{-c_{-}-1\mathrm{/}2}>0$. For $x<c_{-}$, ${\varphi }_{-}(x)-\psi (x)\ge 0$ and $p_{-1}(x)-k_{-}{p}_0(x)>0$; for $x>c_{-}$ both are nonpositive, and at $c_{-}$ the density difference is zero. By [F12] the product is measurable; it is nonnegative and integrable by [F7]. Apply [F4] to the positive and negative parts of ${\varphi }_{-}-\psi$, then [F7] gives $E_{-1}({\varphi }_{-}-\psi )-k_{-}{E}_0({\varphi }_{-}-\psi )=\int ({\varphi }_{-}-\psi )(p_{-1}-k_{-}{p}_0)d\lambda \ge 0$. By [F10] and step 2.1, $E_0{\varphi }_{-}=\alpha \ge E_0\psi$, so $E_{-1}{\varphi }_{-}\ge E_{-1}\psi$. If equality holds, both nonnegative terms in the sum vanish; [F11] makes the integrand zero $\lambda$-almost everywhere, and strict decrease of $p_{-1}\mathrm{/}{p}_0$ forces $\psi ={\varphi }_{-}$ off $c_{-}$ almost everywhere. Step 2.1 gives $P_0(\{c_{-}\})=0$, and its density then gives equality $P_0$-almost surely, proving uniqueness up to $P_0$-almost-sure equality of rejection functions against $\theta =-1$.

If a UMP test ${\psi }^{\ast }$ existed, its optimality at $\theta =1$ and step 3.1 would force ${\psi }^{\ast }={\varphi }_{+}$ $P_0$-almost surely, while its optimality at $\theta =-1$ and step 3.2 would force ${\psi }^{\ast }={\varphi }_{-}$ $P_0$-almost surely. Choose $x_0>\max (c_{+},c_{-})$; positivity and continuity of $p_0$ give a nondegenerate interval above both cutoffs on which $p_0$ is bounded below by a positive constant. By [F6] and the countable choice in step 2.1, that interval has positive Lebesgue and hence positive $P_0$ measure by [F4] and [F7], but ${\varphi }_{+}=1$ and ${\varphi }_{-}=0$ there, contradicting the two almost-sure equalities. Therefore no UMP level-$\alpha$ test exists.