A one-sided normal likelihood ratio at the boundary

Put $q_{\mu }(x)=(2\pi {\sigma }^2{)}^{-1\mathrm{/}4}{e}^{-(x-\mu {)}^2\mathrm{/}(4{\sigma }^2)}$. Direct differentiation gives $q_{\mu }^{\mathrm{\prime }}=(x-\mu )q_{\mu }\mathrm{/}(2{\sigma }^2)$ and $q_{\mu }^{\mathrm{\prime }\mathrm{\prime }}=((x-\mu {)}^2\mathrm{/}(4{\sigma }^4)-1\mathrm{/}(2{\sigma }^2))q_{\mu }$. Its square is a polynomial times $p_{\mu }=q_{\mu }^2$. By [F8] and [F9], $\int (x-\mu {)}^2{p}_{\mu }dx={\sigma }^2$ and $\int (x-\mu {)}^4{p}_{\mu }dx=3{\sigma }^4$, so $\mathrm{\parallel }{q}_{\mu }^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_2^2=3\mathrm{/}(16{\sigma }^4)<\mathrm{\infty }$. Also $s_{\mu }=2q_{\mu }^{\mathrm{\prime }}\mathrm{/}{q}_{\mu }=(x-\mu )\mathrm{/}{\sigma }^2$ has second moment $1\mathrm{/}{\sigma }^2$. The identity $q_a^{\mathrm{\prime }\mathrm{\prime }}(x)=q_0^{\mathrm{\prime }\mathrm{\prime }}(x-a)$ and [F12] give the same norm for every real shift $a$. With $\sigma =1$, Cauchy–Schwarz [F13] gives $E\mathrm{\mid }X-\mu \mathrm{\mid }\le 1$, so the even normal density [F1] gives $E(X-\mu )=0$ and $\mathrm{Var}(X)=1$. The two-variable functions $(x,v)\mapsto q_{\mu +v}^{\mathrm{\prime }\mathrm{\prime }}(x)$ and $(x,u)\mapsto q_{\mu +u}^{\mathrm{\prime }}(x)-q_{\mu }^{\mathrm{\prime }}(x)$ are continuous. To verify the needed product measurability locally, the rectangles $(a,b)\times (c,d)$ with rational endpoints form a countable base of ${\mathbb{R}}^2$: inside any open ball about a point, density of the rationals gives a rational rectangle containing that point and contained in the ball. Every open subset is therefore the union of the subfamily of these rectangles contained in it, hence a countable union of measurable product rectangles. Thus preimages of open subsets under these continuous functions belong to $\mathcal{B}(\mathbb{R})\otimes \mathcal{B}(\mathbb{R})$, and also to the product of the Lebesgue sigma-algebras. Their squares and restrictions to compact parameter intervals are consequently product measurable, as required by Tonelli below.

For fixed $x$, the functions $v\mapsto q_{\mu +v}^{\mathrm{\prime }\mathrm{\prime }}(x)$ and $u\mapsto q_{\mu +u}^{\mathrm{\prime }}(x)-q_{\mu }^{\mathrm{\prime }}(x)$ are continuous on each compact interval. By [F10] they are Riemann integrable, by [F15] their integrals are respectively $q_{\mu +u}^{\mathrm{\prime }}(x)-q_{\mu }^{\mathrm{\prime }}(x)$ and $q_{\mu +t}(x)-q_{\mu }(x)-t{q}_{\mu }^{\mathrm{\prime }}(x)$, and [F11] identifies these with their Lebesgue integrals under the assumed AC. Reversed endpoints give the same identities for negative $u,t$.

Cauchy–Schwarz [F13] on the interval between $0$ and $u$, then Tonelli [F14] and step 1.1, gives $\mathrm{\parallel }{q}_{\mu +u}^{\mathrm{\prime }}-q_{\mu }^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2\le u^2\mathrm{\parallel }{q}_0^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_2^2$. The second identity of step 2.1 and another Cauchy–Schwarz/Tonelli application give $\mathrm{\parallel }{q}_{\mu +t}-q_{\mu }-t{q}_{\mu }^{\mathrm{\prime }}{\mathrm{\parallel }}_2^2\le \mathrm{\mid }t\mathrm{\mid }{\int }_0^{\mathrm{\mid }t\mathrm{\mid }}{u}^2\mathrm{\parallel }{q}_0^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_2^{2}du=\mathrm{\mid }t{\mathrm{\mid }}^4\mathrm{\parallel }{q}_0^{\mathrm{\prime }\mathrm{\prime }}{\mathrm{\parallel }}_2^2\mathrm{/}3=o(t^2)$. For negative $t$ use $q_{\mu -u}^{\mathrm{\prime }}$ on the length-$\mathrm{\mid }t\mathrm{\mid }$ interval; the same first bound holds. This proves QMD with the explicitly supplied normal densities, score $s_{\mu }$, and positive information $1\mathrm{/}{\sigma }^2$ by [F7].

Under $\mu =0$, the local sets are $C_n=\mathbb{R}$ and $C_{0n}=(-\mathrm{\infty },0]$ for every $n$, so they are already their limiting closed convex cones. Direct expansion gives ${\mathrm{\ell }}_n(h\mathrm{/}\sqrt{n})-{\mathrm{\ell }}_n(0)=h{Z}_n-h^2\mathrm{/}2$ with $Z_n=\sqrt{n}{\bar{X}}_n$ and zero remainder. The full and null local maximizers are $Z_n$ and $\min (0,Z_n)$. By [F4] and the mean and variance from step 1.1, $Z_n\Rightarrow G\sim N(0,1)$. Choose a large $R$ with $P(\mathrm{\mid }G\mathrm{\mid }\ge R)$ small and apply [F5] to the closed tails to see that both maximizer sequences are tight. Thus all hypotheses of [F3] hold with $I=1$.

Completing the square in ${\mathrm{\ell }}_n(\mu )$ gives the full MLE ${\bar{X}}_n$ and the constrained MLE $\min (0,{\bar{X}}_n)$. Hence the finite-sample likelihood-ratio statistic is $T_n=2[{\mathrm{\ell }}_n({\bar{X}}_n)-{\mathrm{\ell }}_n(\min (0,{\bar{X}}_n))]=n({\bar{X}}_n\vee 0{)}^2=(Z_n\vee 0{)}^2$. By [F5], $T_n\Rightarrow (G\vee 0{)}^2$. Its CDF is $0$ for $t<0$ and $P(G\le \sqrt{t})=\mathrm{\Phi }(\sqrt{t})$ for $t\ge 0$, so it has mass $1\mathrm{/}2$ at zero. By symmetry, $P(G^2\le t)=2\mathrm{\Phi }(\sqrt{t})-1$ for $t\ge 0$, while [F1] gives $P(G=0)=0$; hence $(G\vee 0{)}^2$ has the mixture law $\frac{1}{2}{\delta }_0+\frac{1}{2}\mathcal{L}(G^2)=\frac{1}{2}{\chi }_0^2+\frac{1}{2}{\chi }_1^2$, not ${\chi }_1^2$.