Subsampling estimates of the Lasso distribution.

4.2 Partial asymptotic normality 39
Finally assumption B.5 yields P (B n2 ) = o(1).
For B n3 , first note that
1
w nj
= | ˜β nj | ≤ M n2 + O P (1/r n ), j = k n + 1, . . . , p n
Furthermore, ‖(x j H n ) ′ ‖ ≤ √ n, so it follows that
⎛
⋃p n {
P (B n3 ) ≤ P ⎝ |x ′ jH n ε| ≥ (1 − κ − δ)λ n C −1 (M n2 + 1/r n ) −1}⎞ ⎠ + o P (1)
≤ m n q ∗ n
j=k n+1
((1 − κ − δ)λ n n −1/2 C −1 (M n2 + 1/r n ) −1)
for C large enough. Lemma 4.0.2.7 and assumption B.4 now imply that P (B n3 ) → 0.
Finally, for the set B n4 , recall that ‖x j ‖/n = 1, so Lemma 4.0.2.6 and assumption B.5
together imply
{∣ ∣∣x ′
max j X 1 C −1
k nk n
≤ τ −1/2
n1 o P (1) = o P (1).
∣
Now, by assumption B.3, it holds that ∣η nj x ′ j X 1C −1
0. Thich completes the proof.
∣
∣ /(nw nj ) − ∣η nj x ′ jX 1 C −1
) ∥ ′ ∥∥∥ ∥ }
∥∥| n1 ˜βnj˜s n1 − η nj s n1 | ∥
n1 s ∣
n1
}
n1 s n1∣
∣ ≤ κ, so we indeed obtain P (B n4 ) →
4.2 Partial asymptotic normality
The proof of the following result on partial asymptotic normality builds upon the fact that
estimates of relevant coefficients stay away from zero with high probability as n tends to
infinity. This is indeed a conclusion of Theorem 4.1.0.9 which asserts variable selection
consistency. Note however that this result is too strong for this purpose and a consistency
result for an l q -norm, q > 1 would actually be sufficient.
Theorem 4.2.0.10. (Huang et al., 2008, Theorem 2)(Asymptotic normality for nonzero
coefficients) Suppose that assumptions B.1 to B.6 are valid. For an arbitrary k n ×1-vector
α n with ‖α n ‖ ≤ 1, let
If M n1 λ n n −1/2 → 0, then
)
n 1/2 s −1
n α ′ n
(ˆβn − β 0 = n −1/2 s −1
n
s 2 n = σ 2 α ′ nC −1
n1 α′ n
n∑
ε i α ′ nC ′ n1u i + o P (1) N (0, 1).
i=1
where o P (1) is a term that converges to zero in probability uniformly with respect to α n .
Proof. Under assumptions B.1 to B.5, one has variable selection consistency according
to Theorem 4.1.0.9, in particular ˆβ n1 has no zero component on a set with probability
converging to one. On this set, one has ∂/∂β 1 L n ( ˆβ n1 , ˆβ n2 ) = 0, that is,
−2
n∑
i=1
(y i − u i ˆβn1 − z i ˆβn2
)
u i + 2λ n ψ n = 0 (4.2.0.8)