Worst-law ellipsoid recovery has no training-size rate

The zero estimator gives risk at most $R^2$ for every allowed law and operator. If $\sigma <R$, put $t=(R\mathrm{/}\sigma {)}^{2\mathrm{/}(2s+2p+1)}>1$ and $k=\lfloor t\rfloor \ge 1$. Use the Borel finite-coordinate estimator ${\widehat{\theta }}_j=j^p{Y}_j$ for $j\le k$ and zero otherwise; it ignores the training data. Under any test truth its retained error is $P_{k}E\theta +(\sigma j^p{\xi }_j{)}_{j\le k}$ and its omitted error is $-(I-P_k)\theta$. These components are orthogonal, and the centered noise is independent of the truth. Hence its expected squared error is at most $R^2{\rho }_A^2+R^2(k+1{)}^{-2s}+{\sigma }^2{\sum }_{j=1}^k{j}^{2p}$. Because $k+1>t$ and ${\sum }_{j\le k}{j}^{2p}\le k^{2p+1}\le t^{2p+1}$, the last two terms are at most $2R^2{t}^{-2s}=2Q_{p,s}$. If $\sigma \ge R$, $Q_{p,s}=R^2$ and the zero estimator already gives the asserted upper bound. Taking the better of the zero and cutoff rules proves the displayed upper bound uniformly over both shift classes.

For the noise lower bound first suppose $\sigma \ge R$. Set $E=0$ and $\mu ={\mu }_0=({\delta }_{R{e}_1}+{\delta }_{-R{e}_1})\mathrm{/}2$. These laws have Wasserstein distance zero and are admissible at every ${\rho }_X$. The training sample is independent of the fresh test sign, and all other test coordinates are independent pure noise. Given those ancillary variables, testing the fresh sign reduces to [A1] with $\delta =R$. Thus any prediction of its first coordinate has squared error at least $R^2$ whenever its induced sign test is wrong, and its integrated risk is at least $R^2\mathrm{\Phi }(-R\mathrm{/}\sigma )\ge b{R}^2=b{Q}_{p,s}$. At a zero predicted coordinate, choose either sign; the squared error is still $R^2$.

Put $r={\rho }_A$. If $r>0$, define $a=R\mathrm{/}(1+r)$, $b_0=R$ and take the same source and test law ${\mu }_0=\mu =({\delta }_{a{e}_1}+{\delta }_{b_0{e}_1})\mathrm{/}2$. Compare the two allowed operators $E_1=0$ and $E_2=-r(1+r{)}^{-1}{P}_1$, whose norms are at most $r$. In the first world, the test truth $a{e}_1$ has observation mean $a{e}_1$; in the second, truth $b_0{e}_1$ has the same mean because $(I+E_2)b_0{e}_1=a{e}_1$. The joint law of the full training sample and this conditional test observation is identical in the two cases, including every noisy coordinate. Calling that common law $H$, the corresponding two full test risks satisfy $R_1\ge \frac{1}{2}{E}_H\mathrm{\parallel }G-a{e}_1{\mathrm{\parallel }}^2$ and $R_2\ge \frac{1}{2}{E}_H\mathrm{\parallel }G-b_0{e}_1{\mathrm{\parallel }}^2$. Therefore [A2] gives $\max (R_1,R_2)\ge (b_0-a{)}^2\mathrm{/}8=R^2{r}^2\mathrm{/}[8(1+r{)}^2]\ge R^2{r}^2\mathrm{/}18$. This holds for every Borel learner and all training sizes. If $r=0$, the asserted zero operator lower bound is immediate.

If $\sigma <R$, use $t$ from step 1.1 and $k=\lceil t\rceil \le 2t$. On the coordinate block $k\le j<2k$, put $a=2^{-s-1}R{k}^{-s-1\mathrm{/}2}$ and let ${\mu }_0=\mu$ be the uniform law on the $2^k$ vectors having independent coordinates ${\theta }_j=\pm a$ on this block and zero elsewhere. Each such vector lies in the ellipsoid because ${\sum }_j{j}^{2s}{\theta }_j^2\le k(2k{)}^{2s}{a}^2=R^2\mathrm{/}4$. Again choose $E=0$. For every block coordinate, its test sign is independent of the whole training sample and of all other test coordinates. Its signal amplitude is $j^{-p}a\le k^{-p}a\le \sigma$, by the definition of $t$ and $k\ge t$. Applying the density-crossing argument in [A1], conditional on all ancillary data, gives sign error at least $b$. Summing the coordinate squared errors gives risk at least $bk{a}^2=b{2}^{-2s-2}{R}^2{k}^{-2s}\ge b{2}^{-4s-2}{Q}_{p,s}$. Finite averaging over the sign prior and integration over the independent ancillary observations suffice; no choice of an unknown prior by the learner is being assumed.

Steps 1.2 and 2.1 give ${\mathfrak{R}}_N\ge c_0{Q}_{p,s}$ with $c_0=b\min \{1,2^{-4s-2}\}>0$, and step 1.3 gives ${\mathfrak{R}}_N\ge R^2{\rho }_A^2\mathrm{/}18$. The maximum of these bounds is at least $\frac{1}{2}\min \{c_0,1\mathrm{/}18\}(Q_{p,s}+R^2{\rho }_A^2)$, proving the lower bound with that explicit positive $c_{p,s}$. For fixed $p,s,R>0$, the upper bound tends to zero when $\sigma ,{\rho }_A\to 0$, and the lower bound forces both limits whenever the risk tends to zero. Both lower constructions have $\mu ={\mu }_0$, so they remain admissible even at zero truth-law shift; the upper construction ignores the training law entirely. This explains why $N$ and ${\rho }_X$ do not enter the matching-order rate for this worst-law formulation.