Compact models admit tests at a unique KL projection under fractional-moment continuity

Fix ${\theta }_1\mathrm{\ne }{\theta }^{\ast }$ and put $r_1=p_{{\theta }_1}\mathrm{/}{p}_{{\theta }^{\ast }}$ on $\{p_{{\theta }^{\ast }}>0\}$, with value zero on its $P_0$-null complement. The finite likelihood-ratio mean makes $r_1$ integrable. For $r_1\ge 1$, [F11] gives $0\le \log r_1={\int }_1^{r_1}{u}^{-1}du\le r_1-1\le r_1$, so the positive part is integrable. The log ratio is measurable by [F14]. Also ${\mathbb{E}}_{P_0}\log r_1=-(K({\theta }_1)-K^{\ast })<0,$ by the extended relative-contrast hypothesis, with value $-\mathrm{\infty }$ allowed.

For $t\downarrow 0$, let $A(t)={\mathbb{E}}_{P_0}{r}_1^t$. Since $r_1^t\le 1+r_1$ for $0<t\le 1$, [F8] gives $A(t)\to P_0(r_1>0)\le 1$. If this limit is less than one, then $A(t)<1$ for all sufficiently small $t>0$. If the limit equals one, split into $\{r_1\ge 1\}$ and $\{0<r_1<1\}$. On the first set, $(r_1^t-1)\mathrm{/}t\to \log r_1$ and, for fixed $t_0<1$ and $0<t\le t_0$, the mean-value theorem bounds this quotient above by $r_1^{t_0}\log r_1\le C{r}_1$ for a finite constant $C$: the function $r^{t_0-1}\log r$ is continuous on $[1,\mathrm{\infty })$ and tends to zero at infinity. On the second set, write $y=-\log r_1>0$. Then $(1-r_1^t)\mathrm{/}t=y(1-e^{-yt})\mathrm{/}(yt)$. The function $u\mapsto (1-e^{-u})\mathrm{/}u$ decreases on $u>0$, since its derivative is $((u+1)e^{-u}-1)\mathrm{/}{u}^2\le 0$ by $e^u\ge 1+u$: for $u\ge 0$, $h(u)=\exp (u)-1-u$ has $h(0)=0$ by the exponential definition and $h^{\mathrm{\prime }}(u)=\exp (u)-1\ge 0$ by the derivative formula and exponential monotonicity, the mean value theorem then gives $h(u)\ge 0$. Hence the quotient increases to $y$ as $t\downarrow 0$. Dominated convergence on the first set and monotone convergence [F9] on the second show $\frac{A(t)-1}{t}⟶{\mathbb{E}}_{P_0}\log r_1<0$ in the extended sense. Thus $A(t)<1$ for all sufficiently small $t>0$.

Choose such a $t$ with $s:=1-t$ in the left-neighborhood from the fractional-moment hypothesis. Then $M_s({\theta }_1,{\theta }_1)=A(t)<1$. By continuity at ${\theta }_1$, there are an open neighborhood $G_{{\theta }_1}$ and $0<a_{{\theta }_1}<1$ such that ${\sup }_{\theta \in G_{{\theta }_1}}{M}_s({\theta }_1,\theta )\le a_{{\theta }_1}$.

Define ${\varphi }_{n,{\theta }_1}=\mathbf{1}\{{\prod }_{i=1}^n{r}_1(X_i)>1\}\mathrm{.}$ The score and test are measurable by [F10], [F14], and [F15]. On the rejection event the product raised to $t$ exceeds one. Repeated Tonelli integration under $P_0^n$ therefore gives $P_0^n{\varphi }_{n,{\theta }_1}\le {\left({\mathbb{E}}_{P_0}{r}_1^t\right)}^n=M_s({\theta }_1,{\theta }_1{)}^n\le a_{{\theta }_1}^n\mathrm{.}$ For the second error, let $g_{\theta ,{\theta }_1,s}$ be the integrand defining $M_s({\theta }_1,\theta )$, with the zero and infinite conventions in the Statement. On the $P_0$-full set where $p_{{\theta }^{\ast }}>0$, if $p_{\theta }>0$ and $p_{{\theta }_1}>0$ then algebra gives $g_{\theta ,{\theta }_1,s}=r_{\theta }{r}_1^{-s}$; the displayed conventions extend this identity to the boundary values. On nonrejection, ${\prod }_i{r}_1(X_i)\le 1$. If every factor is positive, this gives ${\prod }_i{r}_{\theta }(X_i)\le {\prod }_i(r_{\theta }(X_i)r_1(X_i{)}^{-s})={\prod }_i{g}_{\theta ,{\theta }_1,s}(X_i)\mathrm{.}$ If some $r_{\theta }(X_i)=0$, both the density product and the right-hand product are zero under $0\cdot \mathrm{\infty }=0$; if some $r_1(X_i)=0$ while every $r_{\theta }(X_i)>0$, the right-hand product is infinite and the inequality still holds. The product-density identity [F6] and repeated Tonelli [F7] now give, for $\theta \in G_{{\theta }_1}$, $Q_{\theta }^n(1-{\varphi }_{n,{\theta }_1})\le {({\mathbb{E}}_{P_0}{g}_{\theta ,{\theta }_1,s}(X))}^n=M_s({\theta }_1,\theta {)}^n\le a_{{\theta }_1}^n\mathrm{.}$ The moment hypothesis makes these integrals finite on the chosen neighborhood.

Put $C_{\eta }=\{\theta :\rho (\theta ,{\theta }^{\ast })\ge \eta \}$. If $\rho (\theta ,{\theta }^{\ast })<\eta$, then every ${\theta }^{\mathrm{\prime }}$ with $\rho ({\theta }^{\mathrm{\prime }},\theta )<\eta -\rho (\theta ,{\theta }^{\ast })$ satisfies $\rho ({\theta }^{\mathrm{\prime }},{\theta }^{\ast })<\eta$ by [F13]. Thus the complement of $C_{\eta }$ is open, so $C_{\eta }$ is closed by [F12]. By [F5], it is compact. For every point of $C_{\eta }$, step 3.1 supplies a neighborhood; [F4] gives a finite subcover $G_{{\theta }_1},\dots ,G_{{\theta }_k}$.

Set ${\varphi }_n={\max }_{1\le j\le k}{\varphi }_{n,{\theta }_j}$. The union bound and step 4.1 give $P_0^n{\varphi }_n\le {\sum }_{j=1}^k{a}_{{\theta }_j}^n,{\sup }_{\theta \in C_{\eta }}{Q}_{\theta }^n(1-{\varphi }_n)\le {\max }_{1\le j\le k}{a}_{{\theta }_j}^n\mathrm{.}$ Since the subcover is finite and each $a_{{\theta }_j}<1$, both bounds are at most $C_{\eta }{e}^{-c_{\eta }n}$ for suitable $c_{\eta },C_{\eta }>0$ and all sufficiently large $n$. The set $\{\rho (\theta ,{\theta }^{\ast })>\eta \}$ is contained in $C_{\eta }$, so the same type-II bound holds there. If $C_{\eta }=\mathrm{\varnothing }$, the zero test has both errors zero.