= Debiased-Lasso asymptotic normality
{c}
In $Y=X\beta^0+\varepsilon$, put $\widehat\Sigma=X^{\mathsf T}X/n$ and choose a <matrix> $M$ from $X$ alone. The corrected <Lasso> estimate
$$
\widetilde\beta=\widehat\beta+M X^{\mathsf T}(Y-X\widehat\beta)/n
$$
has the exact decomposition
$$
\sqrt n(\widetilde\beta-\beta^0)
=M X^{\mathsf T}\varepsilon/\sqrt n
+\sqrt n(I-M\widehat\Sigma)(\widehat\beta-\beta^0).
$$
If $\eta=\|I-M\widehat\Sigma\|_{\max}$, every remainder coordinate is bounded by $\sqrt n\eta\|\widehat\beta-\beta^0\|_1$. Thus a sufficiently accurate <approximate inverse of a Gram matrix> and a small coefficient <L1 norm> error make the remainder negligible.<Statistical Theory for High-Dimensional Models by Sara van de Geer>[Section 2.4, Theorems 2.4.1–2.4.2, printed pp. 18–19]{ref}
Given $X$, <Gaussian noise> $\varepsilon\sim N(0,\sigma^2I)$ makes the first term exactly <normal> with <covariance matrix> $\sigma^2M\widehat\Sigma M^{\mathsf T}$. For coordinate $j$, let $v_j=\sigma\sqrt{(M\widehat\Sigma M^{\mathsf T})_{jj}}>0$. If the remainder divided by $v_j$ tends to zero in <probability>, then $\sqrt n(\widetilde\beta_j-\beta_j^0)/v_j$ converges to a <standard normal distribution>. Replacing $\sigma$ requires a <consistent estimator>; non-Gaussian errors require a corresponding <central limit theorem>.
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