= Gaussian identity design for debiased Lasso
{c}
In the <normal linear model>, assume $\varepsilon\sim N_n(0,\sigma^2I)$ is independent of $X$. Independent rows of $X$ with <multivariate normal distribution> $N_p(0,I_p)$ give a simple random design for <Debiased Lasso>. Let $p\to\infty$, $\log p=o(n)$ and $s\log p=o(\sqrt n)$, with fixed positive noise scale. Choose the <Lasso> penalty and all <Nodewise Lasso> penalties as sufficiently large constant multiples of $\sqrt{\log p/n}$, including the noise scale in the former. Uniform sample cross-correlations are bounded by this scale, so zero is the nodewise solution with <probability> tending to one. Column squared <norms> divided by $n$ converge uniformly to one. To check the <Compatibility condition for the Lasso> directly, put $\widehat\Sigma=X^TX/n$ and $\eta=\max_{j,k}|\widehat\Sigma_{jk}-\mathbf1_{\{j=k\}}|$. On the <Lasso cone condition>, $\|\delta\|_1\le4\|\delta_S\|_1$, and the <Cauchy-Schwarz inequality> gives
$$
\delta^T\widehat\Sigma\delta\ge\|\delta\|_2^2-\eta\|\delta\|_1^2\ge(1/s-16\eta)\|\delta_S\|_1^2.
$$
For $s\ge1$, <entrywise concentration of a Gaussian Gram matrix> gives $\eta\le D_0\sqrt{\log p/n}$ with <probability> tending to one for sufficiently large fixed $D_0$. The stated sparsity condition implies $s\eta\to0$ on this event, so one can take $\phi^2\ge1/2$ eventually. The conditional <Gaussian tail bound> for the noise score and the <Basic inequality for the Lasso> then give $\|\widehat\beta-\beta^0\|_1\le C s\sqrt{\log p/n}$ with <probability> tending to one. Consequently the <debiased-Lasso remainder bound> is at most $A s\log p/\sqrt n$ with <probability> tending to one, for a fixed constant $A$. These assumptions allow both $p$ and $s$ to grow, for example $p=n^2$, $s=\lfloor n^{1/4}\rfloor$.
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