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In $Y=X{\beta }^0+\epsilon$, put $\hat{\Sigma }=X^{\mathsf{T}}X/n$ and choose a matrix $M$ from $X$ alone. The corrected Lasso estimatehas the exact decompositionIf $\eta =\Vert I-M\hat{\Sigma }{\Vert }_{\max }$, every remainder coordinate is bounded by $\sqrt{n}\eta \Vert \hat{\beta }-{\beta }^0{\Vert }_1$. Thus a sufficiently accurate approximate inverse of a Gram matrix and a small coefficient L1 norm error make the remainder negligible.^[ref]

Given $X$, Gaussian noise $\epsilon \sim N(0,{\sigma }^{2}I)$ makes the first term exactly normal with covariance matrix ${\sigma }^{2}M\hat{\Sigma }{M}^{\mathsf{T}}$. For coordinate $j$, let $v_j=\sigma \sqrt{(M\hat{\Sigma }{M}^{\mathsf{T}}{)}_{jj}}>0$. If the remainder divided by $v_j$ tends to zero in probability, then $\sqrt{n}({\tilde{\beta }}_j-{\beta }_j^0)/v_j$ converges to a standard normal distribution. Replacing $\sigma$ requires a consistent estimator; non-Gaussian errors require a corresponding central limit theorem.

Put $\hat{\Sigma }=X^{T}X/n$. Direct substitution of $Y=X{\beta }^0+\epsilon$ into the Debiased Lasso givesConditionally on the deterministic design,The assumed approximate inverse of a Gram matrix property and Holder inequality implyThus $\rho (n,p)=\sqrt{\log p}$.