Support recovery and irrepresentability
Use active and inactive optimality conditions to distinguish accurate estimation from exact variable selection.
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The question for today
What you’ll understand
- Construct an active-set candidate and its inactive dual scores.
- Interpret a signed irrepresentability condition with a strict margin.
- Check minimum signal strength separately from design incoherence.
An estimator can predict well while reporting the wrong variables. Exact support recovery asks for all true nonzero coordinates and no others, a substantially stronger demand than a norm bound. The Lasso’s subgradient equations explain why: an active coefficient must keep its sign after shrinkage and noise, while every inactive coordinate must satisfy a separate score constraint. Irrepresentability concerns that second requirement.
Build a candidate on the true support
Let , let , and assume the active principal block is invertible. If a candidate has the true signed support with sign vector , the active optimality equations determine its values explicitly. Set all inactive coordinates to zero and solve the displayed linear equation. This is a diagnostic construction using the unknown true support, not an implementable selection algorithm. It allows the proof to test whether a correctly signed solution can exist.
The candidate is valid only if every active entry has the proposed sign and every inactive residual score has absolute value at most the penalty. Strict inequality on inactive scores, together with invertibility of the active block, is sufficient for uniqueness. The sign requirement depends on the smallest true coefficient as well as the inverse active block and noise. Large signals can survive shrinkage; coefficients below the threshold scale can disappear even when the design is orthogonal and there is no observation noise.
Inactive correlations can import active signs
Substituting the active candidate into an inactive score gives the expression in the second display. In the noiseless case it reduces to the penalty times . A signed irrepresentability condition bounds the maximum absolute coordinate of this vector strictly below one, leaving a margin for noise. The stronger induced row-sum norm condition bounds it uniformly over all possible sign vectors. These signed and uniform conditions should not be treated as identical statements.
A positive minimum eigenvalue for the active block alone does not control the inactive vector. Several moderately correlated active variables can collectively represent an inactive variable well enough to push its score past the permitted threshold. Conversely, failure of a strict margin is not a blanket impossibility theorem for every target, noise realization, or estimator. A boundary equality can allow a Lasso solution but without the same robust strict-dual certificate. Exact claims must specify the sign pattern, signal size, and estimator under discussion.
Selection needs a signal condition as well
A sufficient active-sign condition is that the maximum absolute coordinate of be smaller than the minimum nonzero true coefficient magnitude. This is a finite deterministic comparison for the realized noise. High-probability selection theorems combine it with a noise bound and a design margin, then specify how the minimum signal scales with sample size, sparsity, and dimension. Removing the signal assumption from such a theorem changes its meaning: arbitrarily weak coefficients cannot generally be selected reliably.
Selection instability can coexist with stable prediction when correlated columns have similar effects on responses. Reporting one selected column as the unique scientific explanation requires more information than a penalized fit provides. After selection, ordinary fixed-model confidence intervals also require separate justification because the selected model depends on the same responses. This lesson’s KKT construction certifies a particular selection claim; it neither repairs post-selection coverage nor converts a predictive tuning criterion into a support-recovery theorem.
Work it through
Moderate correlations create a false inclusion
Take noiseless truth , penalty , and Gram matrix . Check whether Lasso can retain exactly the two positive active coordinates.
The active block is the identity, so the proposed signed-support candidate is . The inactive score is , exceeding the allowed . Hence that candidate cannot solve the convex problem.
The Gram matrix is positive definite: its eigenvalues are and . Thus ordinary nonsingularity is not the issue; the signed irrepresentability quantity is .
Solving the all-positive equations gives . Every entry is positive and the residual score equals the penalty in all coordinates, verifying the unique optimum and its false inclusion.
The unique Lasso optimum includes the inactive third variable despite a well-defined noiseless linear model and an invertible active block.
The living notebook
Explore the model
Interactive lab · vectors
A strict correlation margin
Pause/Reset, set angle 60 degrees and magnitude 1, and predict horizontal projection 0.5. For a fresh comparison Pause/Reset, set both angle 0 and magnitude 1, giving projection 1 and no angular separation. Reset restores both controls and the clock; changing an angle alone preserves time.
- x component
- 1.393
- y component
- 0.975
- Magnitude
- 1.7
Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.
Look forProjection measures how one direction represents another. This single-vector geometry illustrates a correlation margin; the full irrepresentability condition combines an active inverse block and multiple coordinates.
Pause & reason
Check your understanding
Your turn
Try it yourself
With an orthogonal normalized design, noiseless truth (0.05, 0.5), and penalty 0.1, compute the Lasso estimate. Explain why ideal inactive geometry does not save the smaller signal.
Take a moment to make a prediction or work it out on paper. Open a hint when you need one.
Give me a hint
Under orthogonality, apply the coordinate soft threshold. Compare each true magnitude directly with the penalty.
Compare with the solution
Soft thresholding gives . The first true coefficient is smaller than the penalty and disappears, whereas the second remains positive after shrinkage. There is no correlation-driven inactive violation in this example; the active minimum-signal condition fails instead. Exact support recovery therefore needs more than favorable design geometry. The resulting coefficient error has Euclidean norm , which can be small while the selected support is still wrong.
Keep the idea
Take it with you
- Active KKT equations produce a candidate, not yet a complete selection certificate.
- Irrepresentability controls inactive scores relative to the active signs.
- Exact support recovery also needs adequate signal magnitude and a quantified noise event.
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