Fixed-scale matrix-truncated covariance deviation under finite matrix variance

Intersect the two oracle events [F1], at cost $h_p{e}^{-\beta }$. Put $R=1\mathrm{/}\theta$ and $w_u(z)=\min \{1,R\mathrm{/}\mathrm{\parallel }z-u{\mathrm{\parallel }}^2\}$, interpreted as one at $z=u$. For $\mathrm{\parallel }u\mathrm{\parallel }\le B$, the ratio $h_u(z)=w_u(z)\mathrm{/}w(z)=\max (R,\mathrm{\parallel }z{\mathrm{\parallel }}^2)\mathrm{/}\max (R,\mathrm{\parallel }z-u{\mathrm{\parallel }}^2)$ lies in $[(1+B\mathrm{/}\sqrt{R}{)}^{-2},(1+B\mathrm{/}\sqrt{R}{)}^2]$. Indeed $\max (\sqrt{R},\mathrm{\parallel }z\mathrm{\parallel })\le \max (\sqrt{R},\mathrm{\parallel }z-u\mathrm{\parallel })+B$, and the reverse follows by swapping the two vectors. Thus $\mathrm{\mid }{h}_u-1\mathrm{\mid }\le q=2B\sqrt{\theta }+B^2\theta$.

Expand $S_u=m^{-1}{\sum }_i{w}_u(Z_i)(Z_i-u)(Z_i-u{)}^{\mathsf{T}}$. The perturbation of its $Z_i{Z}_i^{\mathsf{T}}$ part has norm at most $q\mathrm{\parallel }S\mathrm{\parallel }$, because its scalar coefficients have absolute value at most $q$ and the summands are positive semidefinite. For the cross vector, scalar Cauchy–Schwarz and $w^2\le w$ give $\mathrm{\parallel }{m}^{-1}{\sum }_i(h_u-1)w(Z_i)Z_i\mathrm{\parallel }\le q\sqrt{\mathrm{\parallel }S\mathrm{\parallel }}$. Hence $\mathrm{\parallel }{S}_u-\mathrm{\Sigma }\mathrm{\parallel }\le \mathrm{\parallel }S-\mathrm{\Sigma }\mathrm{\parallel }+q\mathrm{\parallel }S\mathrm{\parallel }+2B(\mathrm{\parallel }V\mathrm{\parallel }+q\sqrt{\mathrm{\parallel }S\mathrm{\parallel }})+B^2$, since $0\le w_u\le 1$. This bound is pointwise for every $u$ in the ball, so it applies to the random same-sample $u=\widehat{\mu }-\mu$ on $E$, without conditioning on it. The first conclusion follows using $\mathrm{\parallel }S\mathrm{\parallel }\le M$ and [F1].

For the prescribed pilot take $B=11\sqrt{2Tt}<16\sqrt{Tt}$. The condition $m\ge {10}^{16}D\beta$ and [F3] imply $k\le 4.5\beta \le m\mathrm{/}2$, so [F2] applies. By homogeneity, replace $Z$ by $Z\mathrm{/}\sqrt{L}$ and $\sigma$ by $\sigma \mathrm{/}L$: then $T,{\sigma }_0,\sigma$ scale by $1\mathrm{/}L$, $\theta$ by $L$, $S,\mathrm{\Sigma }$ by $1\mathrm{/}L$, and $D$ is unchanged. It therefore suffices to normalize by $L$ so $L=1$, $T=\rho$, ${\sigma }_0=\sqrt{d}$ and $\sigma \ge \sqrt{d}\ge 1$. Put $\eta =Dt\le {10}^{-16}$ and $H=\sigma \sqrt{t}$. Using $D\ge d$, ${\rho }^2\mathrm{/}d\le D$ and $t\le \eta \le 1$, direct substitution gives $B^2\mathrm{/}H\le 256\sqrt{\eta }$, $B\mathrm{/}\sqrt{H}\le 16{\eta }^{1\mathrm{/}4}$ and $q\le 288{\eta }^{1\mathrm{/}4}\sqrt{t}$. Therefore $qM\mathrm{/}H\le 1066{\eta }^{1\mathrm{/}4}$ and $2Bq\sqrt{M}\mathrm{/}H\le 18432\sqrt{\eta }$: indeed $q\sqrt{M}\mathrm{/}\sqrt{H}\le 288{\eta }^{1\mathrm{/}4}\sqrt{H(1+\frac{27}{10}H)}\mathrm{/}\sigma$, and $H\le \sigma$, $\sigma \ge 1$ give $H(1+\frac{27}{10}H)\mathrm{/}{\sigma }^2\le \sigma (1+\frac{27}{10}\sigma )\mathrm{/}{\sigma }^2\le \frac{37}{10}<4$, hence the square-root factor is at most $\sqrt{37\mathrm{/}10}<2$. The three terms in $2BW\mathrm{/}H$ are at most $32{\eta }^{1\mathrm{/}4}$, $64\sqrt{\eta }$ and $86{\eta }^{1\mathrm{/}4}$, respectively: use $B{\sigma }_0\sqrt{\theta }\mathrm{/}H\le 16{\eta }^{1\mathrm{/}4}$, $\rho \sqrt{t}\mathrm{/}\sigma \le \sqrt{\eta }$ and $Bt\mathrm{/}(\sqrt{\theta }H)\le 16{\eta }^{1\mathrm{/}4}$. In total the perturbation beyond $\frac{27}{10}H$ is at most $1184{\eta }^{1\mathrm{/}4}+18752\sqrt{\eta }\le \frac{118588}{{10}^6}<\frac{3}{10}$. Step 2.1 is thus strictly below $3H$. Undo the normalization and add the single pilot failure to obtain $(h_p+1)e^{-\beta }\le 5p{e}^{-\beta }$. For known mean, $u=B=0$ and only the oracle matrix event is needed, proving the last assertion.