Put $s=\lVert\widehat\beta\rVert_0+\lVert\beta^0\rVert_0$. Since $\delta$ has at most $s$ nonzero coordinates, $\lVert\delta\rVert_1^2\leq s\lVert\delta\rVert_2^2$. On $\Omega_2$,
$$
\delta^T\widehat\Sigma\delta
\geq\delta^T\Sigma^0\delta-
\lVert\widehat\Sigma-\Sigma^0\rVert_\infty\lVert\delta\rVert_1^2
\geq\frac\mu2\lVert\delta\rVert_2^2,
$$
where $\widehat\Sigma=X^TX/n$. Write $R=\delta^T\widehat\Sigma\delta$. On $\Omega_1$,
$$
R\leq2A\sigma v\sqrt{\frac{\log p}{n}}\lVert\delta\rVert_1
\leq2A\sigma v\sqrt{\frac{2s\log p}{\mu n}}\sqrt R.
$$
Squaring after division by $\sqrt R$ proves
$$
\frac1n\lVert X(\beta^0-\widehat\beta)\rVert_2^2
\leq8A^2\sigma^2v^2\frac{s\log p}{\mu n}.
$$


Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 205 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-205/2/c/solution