Diagonal covariance lower bound with bounded-risk saturation

Put $s=\tau \mathrm{/}4$ and $\delta =\frac{1}{2}\min \{1,\sqrt{\log (p+1)\mathrm{/}n}\}$. Use $p+1$ Gaussian laws $P_0=N(0,s{I}_p)$ and $P_j=N(0,s(I_p+\delta e_j{e}_j^{\mathsf{T}}))$, $1\le j\le p$. Each covariance has operator norm at most $3\tau \mathrm{/}8<\tau$. If $Z$ is standard normal, $P(\mathrm{\mid }Z\mathrm{\mid }>t)=e^{-t^2\mathrm{/}2}\sqrt{2\mathrm{/}\pi }{\int }_0^{\mathrm{\infty }}{e}^{-w^2\mathrm{/}2-tw}dw\le e^{-t^2\mathrm{/}2}$. Hence every unit-direction projection of these laws obeys the required tail with parameter $\tau$. Their covariance matrices are diagonal, so [F2] puts every law in every class in the statement.

For $j\ge 1$, the Gaussian covariance KL formula with ${\mathrm{\Sigma }}_j=s(I_p+\delta e_j{e}_j^{\mathsf{T}})$ and ${\mathrm{\Sigma }}_0=s{I}_p$ gives $D(P_j\mathrm{\parallel }{P}_0)=\frac{1}{2}(\delta -\log (1+\delta ))\le {\delta }^2\mathrm{/}4,$ where $\delta -\log (1+\delta )={\int }_0^{\delta }t\mathrm{/}(1+t)dt\le {\delta }^2\mathrm{/}2$. By finite-product KL additivity, $D(P_j^{\otimes n}\mathrm{\parallel }{P}_0^{\otimes n})=nD(P_j\mathrm{\parallel }{P}_0)\le n{\delta }^2\mathrm{/}4\le \log (p+1)\mathrm{/}16.$ The $j=0$ divergence is zero, so the reference-law average KL has the same bound.

Use the full matrix action space $\mathcal{T}={\mathbb{R}}^{p\times p}$ and set $\rho (A,B)=\mathrm{\parallel }A-B{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$. For a unit vector $v$, rowwise Cauchy–Schwarz [F5] gives $\mathrm{\parallel }Av{\mathrm{\parallel }}_2^2\le {\sum }_{i,j}{a}_{ij}^2=\mathrm{\parallel }A{\mathrm{\parallel }}_F^2$, so [F4] gives $\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \mathrm{\parallel }A{\mathrm{\parallel }}_F$. Conversely each coordinate unit vector $e_j$ has norm one, so $\mathrm{\parallel }A{\mathrm{\parallel }}_F^2={\sum }_j\mathrm{\parallel }A{e}_j{\mathrm{\parallel }}_2^2\le p\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}^2$. Thus $\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \mathrm{\parallel }A{\mathrm{\parallel }}_F\le \sqrt{p}\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$. The maximum definition also gives homogeneity, symmetry under $A\mapsto -A$, and $\mathrm{\parallel }A+B{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}\le \mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}+\mathrm{\parallel }B{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}$ by the Euclidean triangle inequality on each unit vector. If $\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}=0$, all columns $A{e}_j$ vanish, so $A=0$. Hence $\rho$ is a finite metric by [F5]. The two norm inequalities imply each ball for either metric contains a ball for the other about the same center, so their open sets and therefore their Borel sigma-algebras coincide. Under the entry-list identification ${\mathbb{R}}^{p\times p}={\mathbb{R}}^{p^2}$ this is exactly the Euclidean Borel action space in the minimax-risk definition. In particular the diagonal embedding $a\mapsto \mathrm{diag}(a)$ is continuous: its Frobenius difference norm is $\mathrm{\parallel }a-b{\mathrm{\parallel }}_2$, and its operator difference norm is at most that. All targets in step 1.1 are thus legitimate Borel metric-space points, and every estimator in the minimax definition is measurable into this metric Borel space and induces an admissible deterministic kernel by [F6]. No new choice axiom is used.

For a diagonal matrix $D=\mathrm{diag}(a)$ and a unit vector $v$, $\mathrm{\parallel }Dv{\mathrm{\parallel }}_2^2={\sum }_j{a}_j^2{v}_j^2\le {\max }_j\mathrm{\mid }{a}_j{\mathrm{\mid }}^2$; equality holds at a coordinate unit vector attaining the finite maximum. Hence $\mathrm{\parallel }D{\mathrm{\parallel }}_{\mathrm{o}\mathrm{p}}={\max }_j\mathrm{\mid }{a}_j\mathrm{\mid }$. A target difference from the base has one diagonal entry $s\delta$; a difference between two nonbase targets has two entries $s\delta ,-s\delta$. Every distinct pair therefore has metric separation exactly $s\delta$.

Apply [F3] to these $p+1$ targets, with the base sample law as reference. Its bracket is at least $1-1\mathrm{/}16-\log 2\mathrm{/}\log 3>1\mathrm{/}4$. Thus the risk is at least $s^2{\delta }^2\mathrm{/}16\ge ({\tau }^2\mathrm{/}1024)\min \{1,\log p\mathrm{/}n\}$. These finite experiments are subsets of the full class, so the full minimax risk is no smaller.