y ∣ z \sim N (z^{T} β, σ^{2}), β = Σ_{- 1, - 1}^{- 1} Σ_{- 1, 1}, σ^{2} = Σ_{11} - Σ_{1, - 1} Σ_{- 1, - 1}^{- 1} Σ_{- 1, 1} > 0.

For a jointly Gaussian vector, the residual is independent of the regressor, so

y = z^{T} β + σ e

with

e \sim N (0, 1)

independent of

z

\frac{1}{n} ∥ X - Zθ ∥_{2} + γ ∥ θ ∥_{1}, γ = \frac{2 l o g p}{n} .

At a nonzero residual

R

, its KKT condition is

\frac{Z ^{T} R}{n ∥ R ∥ _{2}} = γ u, ∥ u ∥_{\infty} \leq 1,

which proves the required inequality.

Y = f + σ e

, the reverse triangle inequality gives

σ - σ \frac{∥ e ∥ _{2}}{n} \leq \frac{∥ f - f ∥ _{2}}{n} = o_{P} (1) .

∥ e ∥_{2} / n \to_{P} 1

, hence

σ \to_{P} σ

Under

X ⊥ ⊥ Y ∣ Z

, Gaussianity makes

e

independent of

(X, Z)

and hence of

R

. Conditionally on

R

R^{T} e / ∥ R ∥_{2} \sim N (0, 1)

. The remaining numerator term obeys

\frac{∣ R ^{T} Z ( β - β ) ∣}{∥ R ∥ _{2}} \leq \frac{∥ Z ^{T} R ∥ _{\infty}}{∥ R ∥ _{2}} ∥ β - β ∥_{1} \leq 2 lo g p ∥ β - β ∥_{1} = o_{P} (1) .

Combining this with

σ \to_{P} σ

and Slutsky theorem proves the standard-normal limit.

Solved by gpt-5.6-sol high.

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