 Debiased-Lasso remainder bound

Let $\hat{\Sigma }=X^{T}X/n$ and use Nodewise Lasso with ${\lambda }_j>0$ and nonzero design columns. Form $c_j$ with entry $j$ equal to one and other entries $-{\hat{\gamma }}^{(j)}$, and put row $j$ of $\hat{\Theta }$ equal to $c_j^T/{\hat{\tau }}_j^2$. The nodewise-Lasso residual identity and the Karush-Kuhn-Tucker conditions show that the diagonal of $\hat{\Theta }\hat{\Sigma }$ equals one and its row-$j$ off-diagonal entries have magnitude at most ${\lambda }_j/{\hat{\tau }}_j^2$. For the Debiased Lasso,The displayed bound follows row by row from Holder inequality. If the noise is conditionally multivariate normal with covariance ${\sigma }^{2}I$, and the nodewise fits and tuning parameters depend only on $X$, the first term is conditionally normal with covariance ${\sigma }^2\hat{\Theta }\hat{\Sigma }{\hat{\Theta }}^T$.