Put $\widehat\Sigma=X^TX/n$. Direct substitution of $Y=X\beta^0+\varepsilon$ into the <Debiased Lasso> gives
$$
\sqrt n(\widehat b-\beta^0)
=\underbrace{\frac1{\sqrt n}\widehat\Theta^TX^T\varepsilon}_{W}
+\underbrace{\sqrt n(I-\widehat\Theta^T\widehat\Sigma)(\widehat\beta-\beta^0)}_{\Delta}.
$$
Conditionally on the deterministic design,
$$
\boxed{W\sim N_p(0,\widehat\Theta^T\widehat\Sigma\widehat\Theta)}.
$$
The assumed <approximate inverse of a Gram matrix> property and <Holder inequality> imply
$$
\lVert\Delta\rVert_\infty
\leq\sqrt n\lVert I-\widehat\Theta^T\widehat\Sigma\rVert_{\max}
\lVert\widehat\beta-\beta^0\rVert_1
\leq\sqrt{\log p}\,\lVert\widehat\beta-\beta^0\rVert_1.
$$
Thus $\rho(n,p)=\sqrt{\log p}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 205 / 6 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-205/6/b/solution