On the same event, the <Lasso cone condition> and the coordinatewise bound from part a give
$$
\|\widehat\beta-\beta^0\|_1
=\|\delta_S\|_1+\|\delta_N\|_1
\leq4\|\delta_S\|_1
\leq4s\|\delta_S\|_\infty
\leq\frac{6s\lambda}{\psi}.
$$
With $\lambda=A\sqrt{\log p/n}$ this is exactly
$$
\|\widehat\beta-\beta^0\|_1
\leq\frac{6sA}{\psi}\sqrt{\frac{\log p}{n}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 205 / 6 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-205/6/b/solution