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The fast-rate Lasso prediction bound under compatibility implies $\Vert {\hat{\beta }}_S-{\beta }_S^0{\Vert }_1\le (1+c)\lambda s/{\varphi }^2$. Each true coefficient larger in absolute value than this bound has the correct estimated sign. This conclusion alone does not ensure absence of false positives.

Fix the normalizationThe Lasso penalty uses the L1 norm. All arguments below apply to every minimizer; uniqueness of the coefficient vector is not assumed.

The Karush-Kuhn-Tucker conditions are the subgradient optimality conditionThese conditions are necessary and sufficient because the objective is a convex function. They use the subgradient of the absolute value, coordinate by coordinate.

Put $\delta =\hat{\beta }-{\beta }^0$. Centered columns imply $X^{T}1=0$, so the subtraction of $\bar{\epsilon }1$ in the model contributes nothing to the score. The Karush-Kuhn-Tucker conditions becomeMultiply by ${\delta }^T$. Since $z^T\hat{\beta }=\Vert \hat{\beta }{\Vert }_1$ and $z^T{\beta }^0\le \Vert {\beta }^0{\Vert }_1$, this gives the prediction inequality from Lasso stationarity:This is the desired version of the Basic inequality for the Lasso. Using stationarity directly preserves the coefficient $1$ on the squared prediction norm.

For the Gaussian score event for the Lasso, write $X_j$ for column $j$. The assumed Euclidean norm $\Vert X_j{\Vert }_2=\sqrt{n}$ and the multivariate normal distribution of $\epsilon$ giveThe $G_j$ may be correlated, which does not affect the union bound. Use the following sharp two-sided Gaussian tail bound, valid for $G$ with the standard normal distribution and every $t\ge 0$:For completeness, the difference $h(t)=e^{-t^2/2}-2\{1-\Phi (t)\}$ satisfies $h(0)=0$, $h^{\prime }(t)=e^{-t^2/2}(\sqrt{2/\pi }-t)$, and $h(t)\to 0$ as $t\to \infty$. It first increases and then decreases to zero, so it is nonnegative. This bound avoids introducing an unnecessary factor two.

Here $\Phi$ is the standard normal distribution function. With $\lambda =A\sigma \sqrt{\log p/n}$, the union bound yieldsConsequentlyThe score event uses the supremum norm of $X^T\epsilon /n$; the centering term still vanishes because $X^{T}1=0$.

We next work deterministically on $\Omega$. LetBy Holder inequality, ${\epsilon }^{T}X\delta /n\le c\lambda \Vert \delta {\Vert }_1=c\lambda (a+b)$. Since ${\beta }_N^0=0$, the triangle inequality givesSubstituting both inequalities into the Basic inequality for the Lasso producesBecause $Q\ge 0$, this also proves the relevant Lasso cone condition, $(1-c)b\le (1+c)a$. The assumed Compatibility condition for the Lasso therefore applies to $\delta$:Set $D=(1+c)\lambda \sqrt{s}/\varphi$. Combining these inequalities givesIf $Q>0$, discarding the nonnegative $b$ term shows $\sqrt{Q}\le D$. If $Q=0$, compatibility gives $a=0$, and the preceding inequality then gives $b=0$. In either case $D\sqrt{Q}\le D^2$, soThis is a fast prediction bound and an inactive-coordinate error bound on the same event. It is the fast-rate Lasso prediction bound under compatibility.

Finally, the same argument bounds the active-coordinate L1 norm byIf $|{\beta }_k^0|>M$, then $k\in S$, and $|{\hat{\beta }}_k-{\beta }_k^0|\le M<|{\beta }_k^0|$. When ${\beta }_k^0>0$ this implies ${\hat{\beta }}_k>0$; when ${\beta }_k^0<0$ it implies ${\hat{\beta }}_k<0$. Thus the sign function obeysThis is Lasso sign recovery under compatibility. It concerns the sufficiently large nonzero coefficients; it does not assert that every estimated inactive coefficient is zero.