Alphabeta Math
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Lasso on an orthonormal design is coordinatewise soft-thresholding

Statement

Let n,p≥1, X∈Rn×p satisfy XTX=Ip, and let y∈Rn and λ>0. The objective 12n∥y−Xβ∥22+λ∥β∥1 has the unique minimizer β^j=Snλ((XTy)j), where Sa(t)=sgn⁡(t)(∣t∣−a)+, (u)+=max⁡{u,0}, and sgn⁡(0)=0. Thus the threshold is nλ under this loss normalization.

Facts & Assumptions

Given: n,p≥1, XTX=Ip, y∈Rn, and λ>0.

[F1]

The fixed-design empirical squared loss is ∥y−Xβ∥22/(2n) and the Lasso penalty is λ∥β∥1 (Fixed-design least-squares, ridge, lasso, group, and fused objectives).

Proof

1.1F1givenalgebra

Put z=XTy. Expanding the squared norm and using XTX=Ip gives 12n∥y−Xβ∥22+λ∥β∥1=∥y∥22−∥z∥222n+∑j=1p((βj−zj)22n+λ∣βj∣). The first term is constant in β, so the minimization separates into scalar strictly convex objectives.

2.1step 1.1algebra

For a scalar z, the derivative of (t−z)2/(2n)+λ∣t∣ on t>0 is (t−z)/n+λ, and on t<0 it is (t−z)/n−λ. The positive stationary point is t=z−nλ exactly when z>nλ; the negative stationary point is t=z+nλ exactly when z<−nλ. When ∣z∣≤nλ, the one-sided derivatives at zero satisfy (−z/n−λ)≤0≤(−z/n+λ), so zero minimizes. Strict convexity makes each minimizer unique, and these cases are precisely Snλ(z).

3.1step 1.1step 2.1∎

Applying step 2.1 to every coordinate yields the stated vector formula and uniqueness.

Depends on

Used by

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Sources