The lambda-zero sparse detection boundary

Substitution gives $\lambda =(1+1)\mathrm{/}2-1=0$ and $\mathrm{\Lambda }=(2\mathrm{/}(1+1){)}^{1\mathrm{/}4}=1$. To prove the rate order, first let $r\mathrm{/}{r}_{\epsilon }^{\ast }\to 0$, set $n=\lfloor 1\mathrm{/}(4r)\rfloor$, and for each $i\in \{n+1,\dots ,2n\}$ use the two alternatives with sole nonzero coordinate ${\eta }_i=\pm r\mathrm{/}i$. Each lies in ${\mathrm{\Theta }}_1(r)$ because $ri\le 1$ and $i^2{\eta }_i^2=r^2$. Put ${\mu }_i=r\mathrm{/}(\epsilon i)$; then their standardized means satisfy $\mathrm{\mid }{\eta }_i\mathrm{\mid }\mathrm{/}\epsilon ={\mu }_i\le 4r^2\mathrm{/}\epsilon$. If $s=r\mathrm{/}{r}_{\epsilon }^{\ast }\to 0$, then ${\max }_i{\mu }_i^2\le 16{\epsilon }^{-2}{r}^4=16s^4\log ({\epsilon }^{-1})=o(\log n)$, because $\log n\ge c\log ({\epsilon }^{-1})$ for some $c>0$ when $r\le r_{\epsilon }^{\ast }$ and $\epsilon$ is small. For each location $i$, the shifted scalar density divided by the null density is $e^{\pm {\mu }_i{G}_i-{\mu }_i^2\mathrm{/}2}$, where $G_i=Y_i\mathrm{/}\epsilon$ under the null; averaging the two signs gives $L_i=e^{-{\mu }_i^2\mathrm{/}2}\cosh ({\mu }_i{G}_i)$. Completing the square using [F3] and [F4] gives ${\mathbb{E}}_0{L}_i=1$ and ${\mathbb{E}}_0{L}_i^2=\cosh ({\mu }_i^2)$. The measure $L_{i}d{P}_0$ is therefore a probability measure. Finite-coordinate factorization in [F2] identifies its value on every finite-coordinate rectangular cylinder with the sign-mixture sampling law at location $i$: only the $i$th scalar density changes and all other coordinates retain their null law. These cylinders generate the sequence sigma-algebra, so finite-measure uniqueness in [F2] identifies the two measures on all measurable events. For distinct locations, [F2] also gives ${\mathbb{E}}_0{L}_i{L}_j=1$. Thus the uniform mixture $L=n^{-1}{\sum }_i{L}_i$ satisfies ${\mathbb{E}}_0(L-1{)}^2=n^{-2}{\sum }_i(\cosh ({\mu }_i^2)-1)\le n^{-1}\exp ({\max }_i{\mu }_i^2)=n^{-1}\exp (o(\log n))\to 0$. Since ${\mathbb{E}}_0(L-1)=0$, its positive and negative parts have equal integrals, so for every event $E$, $\mathrm{\mid }{P}_{\pi }(E)-P_0(E)\mathrm{\mid }\le \frac{1}{2}{\mathbb{E}}_0\mathrm{\mid }L-1\mathrm{\mid }\le \frac{1}{2}\sqrt{{\mathbb{E}}_0(L-1{)}^2}$. Approximation by averages of event indicators gives the same bound for randomized tests. Therefore the mixture total variation tends to zero, ${\gamma }_{\epsilon }({\mathrm{\Theta }}_1(r))\to 1$, and every level-$\alpha$ test has maximal type-II error at least $1-\alpha -o(1)$. The constant randomized test $\psi \equiv \alpha$ gives the matching upper bound $1-\alpha$.

First suppose $r\mathrm{/}{r}_{\epsilon }^{\ast }\to \mathrm{\infty }$ and $r\to 0$. Put $N=\lceil 2\mathrm{/}r\rceil$, $t_N=4\sqrt{\log N}$, and reject if ${\max }_{i\le N}\mathrm{\mid }{Y}_i\mathrm{\mid }\mathrm{/}\epsilon >t_N$. Under the null, [F5] and a union bound give type-I error at most $2N{e}^{-t_N^2\mathrm{/}2}=2N^{-7}\to 0$. For any alternative, $\mathrm{\mid }{\eta }_i\mathrm{\mid }\le i^{-2}$ follows from [F1], so the omitted weighted energy obeys ${\sum }_{i>N}{i}^2{\eta }_i^2\le N^{-2}{\sum }_{i>N}{i}^2\mathrm{\mid }{\eta }_i\mathrm{\mid }\le N^{-2}\le r^2\mathrm{/}4$. Hence ${\sum }_{i\le N}{i}^2{\eta }_i^2\ge 3r^2\mathrm{/}4$. On the event that all $N$ null-noise coordinates satisfy $\mathrm{\mid }{\xi }_i\mathrm{\mid }\le t_N\mathrm{/}2$, nonrejection would imply $\mathrm{\mid }{\eta }_i\mathrm{\mid }\le (3\mathrm{/}2)\epsilon t_N$ for every $i\le N$, and therefore ${\sum }_{i\le N}{i}^2{\eta }_i^2\le (3\mathrm{/}2)\epsilon t_N{\sum }_{i\le N}{i}^2\mathrm{\mid }{\eta }_i\mathrm{\mid }\le (3\mathrm{/}2)\epsilon t_N$. But $r\mathrm{/}{r}_{\epsilon }^{\ast }\to \mathrm{\infty }$ makes $r^2\mathrm{/}(\epsilon \sqrt{\log N})\to \mathrm{\infty }$, a contradiction for small $\epsilon$. The probability that the stated noise event fails is at most $2N{e}^{-(t_N\mathrm{/}2{)}^2\mathrm{/}2}=2\mathrm{/}N\to 0$, so the maximal type-II error tends to zero.

For an arbitrary positive radius sequence with $r_{\epsilon }\mathrm{/}{r}_{\epsilon }^{\ast }\to \mathrm{\infty }$, use the positive representative $r_{\epsilon }^{\ast }={\epsilon }^{1\mathrm{/}2}(\log {\epsilon }^{-1}{)}^{1\mathrm{/}4}$ and put ${\bar{r}}_{\epsilon }=\min \{r_{\epsilon },\sqrt{r_{\epsilon }^{\ast }}\}$. Then ${\bar{r}}_{\epsilon }\to 0$, ${\bar{r}}_{\epsilon }\le r_{\epsilon }$, and ${\bar{r}}_{\epsilon }\mathrm{/}{r}_{\epsilon }^{\ast }=\min \{r_{\epsilon }\mathrm{/}{r}_{\epsilon }^{\ast },1\mathrm{/}\sqrt{r_{\epsilon }^{\ast }}\}\to \mathrm{\infty }$. Apply step 2.1 with ${\bar{r}}_{\epsilon }$ in place of $r_{\epsilon }$. Its cutoff $N=\lceil 2\mathrm{/}{\bar{r}}_{\epsilon }\rceil$ now tends to infinity, so both its null-size bound $2N^{-7}$ and its noise-event bound $2\mathrm{/}N$ vanish, and the energy-tail and signal-to-threshold estimates in that step apply at ${\bar{r}}_{\epsilon }$. Since ${\mathrm{\Theta }}_1(r_{\epsilon })\subseteq {\mathrm{\Theta }}_1({\bar{r}}_{\epsilon })$, the same test has unchanged null size and no larger maximal type-II error on the requested alternative. Its total error therefore tends to zero, and its null size is eventually below any fixed significance level, giving both consistency limits in [F1]. For an empty requested alternative both minimax risks are zero by that definition, with the never-reject test attaining zero total error. This covers the nonvanishing-radius case without using a bounded or nondiverging cutoff.

The impossibility implication in step 1.1 and the all-positive-radius consistency implication in step 3.1 establish the separation-rate order ${\epsilon }^{1\mathrm{/}2}(\log {\epsilon }^{-1}{)}^{1\mathrm{/}4}$. The parameter and constant were computed in step 1.1, proving the example.