Let $D$ be the block-diagonal part of $\widehat\Sigma=n^{-1}X^TX$. The within-block eigenvalue assumption gives $\delta^TD\delta\geq\eta\lVert\delta\rVert_2^2$. On the compatibility cone with $\lVert\delta_S\rVert_1=1$,
$$
\lVert\delta\rVert_1\leq4,
\qquad
\lVert\delta\rVert_2^2\geq\frac1s.
$$
The off-block assumption therefore gives
$$
\left|\delta^T(\widehat\Sigma-D)\delta\right|
\leq\frac{\eta}{32p}\lVert\delta\rVert_1^2
\leq\frac{\eta}{2p}
\leq\frac{\eta}{2s}.
$$
It follows that $\delta^T\widehat\Sigma\delta\geq\eta/(2s)$, and hence
$$
\boxed{\phi_{\widehat\Sigma}^2(S)\geq\eta/2}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 205 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-205/2/b/solution