Symmetric Gaussian product-mixture testing bound

For each sign vector $\tau \in \{-1,+1{\}}^J$, let $v_j^{\tau }={\tau }_j{v}_j$ on $J$ and zero elsewhere. Absolute values and squares are unchanged, so every $v^{\tau }$ belongs to ${\mathrm{\Theta }}_q(r)$.

Under the null, $X_j=Y_j\mathrm{/}\epsilon$ has density $\phi$. For a fixed sign and coordinate, the density ratio for mean ${\tau }_j{v}_j$ versus zero is $\exp ({\tau }_j{v}_j{Y}_j\mathrm{/}{\epsilon }^2-v_j^2\mathrm{/}(2{\epsilon }^2))$; completing the square using [F7] shows its null expectation is one.

Independence [F4] multiplies the one-coordinate ratios from step 1.2, giving the density ratio for the finite coordinate vector on each cylinder event. Its null expectation is the product of unit expectations, so it defines a probability measure on the full product space.

Cylinder events form a generating pi-system. The fixed-sign sampling law and the probability measure from step 2.1 agree on every cylinder, so [F5] gives equality on the product sigma-algebra and validates the full likelihood-ratio identity.

Averaging the fixed-sign likelihood ratios in step 3.1 over the finite sign set factors coordinatewise and gives $L_v={\prod }_{j\in J}{e}^{-{\delta }_j\mathrm{/}2}\cosh (\sqrt{{\delta }_j}{X}_j)$, where ${\delta }_j=v_j^2\mathrm{/}{\epsilon }^2$. This density is positive and has null expectation one.

For a standard normal $G$, [F7] and ${\cosh }^{2}t=(1+\cosh (2t))\mathrm{/}2$ give $e^{-\delta }\mathbb{E}{\cosh }^2(\sqrt{\delta }G)=e^{-\delta }(1+e^{2\delta })\mathrm{/}2=\cosh \delta$. Independence [F4] and step 4.1 therefore give ${\mathbb{E}}_0{L}_v^2={\prod }_{j\in J}\cosh {\delta }_j$.

The function $g(t)=t^2\mathrm{/}2-\log \cosh t$ is even, has $g(0)=0$, and for $t\ge 0$ has derivative $g^{\mathrm{\prime }}(t)=t-\tanh t\ge 0$, since $(\tanh t{)}^{\mathrm{\prime }}\le 1$ and $\tanh 0=0$. Thus $\cosh t\le e^{t^2\mathrm{/}2}$ for all real $t$. Since ${\sum }_j{\delta }_j^2=2u(v{)}^2$, step 5.1 yields ${\mathbb{E}}_0(L_v-1{)}^2={\prod }_j\cosh {\delta }_j-1\le e^{\frac{1}{2}{\sum }_j{\delta }_j^2}-1=e^{u(v{)}^2}-1$.

Since ${\mathbb{E}}_0(L_v-1)=0$, the positive and negative parts of $L_v-1$ have equal integrals; hence every event $E$ satisfies $\mathrm{\mid }{P}_{{\pi }_v}(E)-P_0(E)\mathrm{\mid }\le \frac{1}{2}{\mathbb{E}}_0\mathrm{\mid }{L}_v-1\mathrm{\mid }\le \frac{1}{2}\sqrt{{\mathbb{E}}_0(L_v-1{)}^2}$. Approximate a randomized test $\psi$ by $N^{-1}{\sum }_{k=1}^N{\mathbf{1}}_{\{\psi \ge k\mathrm{/}N\}}$; the event bound [F6] and uniform approximation error $N^{-1}$ give $\mathrm{\mid }{\mathbb{E}}_{{\pi }_v}\psi -{\mathbb{E}}_0\psi \mathrm{\mid }\le \mathrm{\parallel }{P}_{{\pi }_v}-P_0{\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}$.

The maximal type-II error is at least its mixture average because every sign alternative lies in ${\mathrm{\Theta }}_q(r)$. Therefore every test has total error at least $1-\mathrm{\parallel }{P}_{{\pi }_v}-P_0{\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}$, and any level-$\alpha$ test has maximal type-II error at least $1-\alpha -\mathrm{\parallel }{P}_{{\pi }_v}-P_0{\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}$. Taking the respective infima, then applying steps 6.1 and 7.1, proves both stated risk bounds.