= Debiased-Lasso remainder bound
{title2=$\|\Delta\|_\infty\le\sqrt n\|\beta^0-\widehat\beta\|_1\max_j\lambda_j/\widehat\tau_j^2$}
Let $\widehat\Sigma=X^TX/n$ and use <Nodewise Lasso> with $\lambda_j>0$ and nonzero design columns. Form $c_j$ with entry $j$ equal to one and other entries $-\widehat\gamma^{(j)}$, and put row $j$ of $\widehat\Theta$ equal to $c_j^T/\widehat\tau_j^2$. The <nodewise-Lasso residual identity> and the <Karush-Kuhn-Tucker conditions> show that the diagonal of $\widehat\Theta\widehat\Sigma$ equals one and its row-$j$ off-diagonal entries have magnitude at most $\lambda_j/\widehat\tau_j^2$. For the <Debiased Lasso>,
$$
\sqrt n(\widehat b-\beta^0)=\widehat\Theta X^T\varepsilon/\sqrt n+\Delta,\qquad \Delta=\sqrt n(\widehat\Theta\widehat\Sigma-I)(\beta^0-\widehat\beta).
$$
The displayed bound follows row by row from <Holder inequality>. If the noise is conditionally <multivariate normal> with <covariance> $\sigma^2I$, and the nodewise fits and tuning parameters depend only on $X$, the first term is conditionally normal with <covariance> $\sigma^2\widehat\Theta\widehat\Sigma\widehat\Theta^T$.
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