A two-sided normal alternative need not have a UMP test

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.