Variable selection for bivariate interval-censored failure time data under linear transformation models

Condition 1

There exists a compact neighborhood B 0 of the true value β ₀ such that

where I β 0 is a positive definite matrix, n − 1 Ω n = ∇ 2 H β | Λ ̃ 01 , Λ ̃ 02 .

Condition 2

There exists a constant c > 1 such that c − 1 < eigen min n − 1 Ω n ≤ eigen max n − 1 Ω n < c for sufficiently large n, where eigen_min(⋅) and eigen_max(⋅) stand for the smallest and largest eigenvalues of the matrix. There exits positive constants a ₀ and a ₁ such that a 0 ≤ β 0 j ≤ a 1 , 1 ≤ j ≤ q n .

Condition 3

As n → ∞, p n q n / n → 0 , λ n / n → 0 , λ n q n / n → 0 and λ n 2 / ( p n n ) → ∞ .

Condition 4

The initial estimator β ̂ ( 0 ) satisfies β ̂ ( 0 ) − β 0 = O p ( p n / n ) .

Lemma 1

Let δ n be a sequence of positive real numbers satisfying δ _n → ∞ and δ n 2 p n / λ n → 0 . Define H n ≡ β = β 1 ′ , β 2 ′ ′ : β 1 1 / K 0 , K 0 q n , β 2 ≤ δ n p n / n , where K ₀ > 1 is a constant such that β 01 ∈ 1 / K 0 , K 0 q n . Then under the regular conditions (C1) − (C4) given above and with probability tending to 1, we have

1. sup β ∈ H n γ * ( β ) β 2 < 1 c 0 for some constant c ₀ > 1.

2. g(⋅) is a mapping from H _n to itself.

Proof

First from formula (3.11), it is easy to see that b ̂ is equal to g( β ) with λ _n = 0, which is equivalent to the maximizer of the log-likelihood function. By using arguments similar to those in Zhang et al. [29] and referencing from Zeng et al. [16], one can obtain that b ̂ − β 0 = O ( p n / n ) . Hence it follows from (A.2) that

By condition (C2) and the fact that

we can derive ‖ B ‖ ≤ 2 c . Furthermore, based on conditions (C2) and (C3) and note that β 1 ∈ 1 / K 0 , K 0 q n and α * ≤ ‖ g ( β ) ‖ ≤ ‖ b ̂ ‖ = O p ( p n ) , we have

Since λ _min( G ) > c ⁻¹, it follows from (A.2) that, with probability tending to 1

Let m γ * / β 2 = γ 1 * / β q n + 1 , γ 2 * / β q n + 2 , … , γ p n − q n * / β p n ′ . It then follows from the Cauchy–Schwarz inequality and the assumption β 2 ≤ δ n p n / n that

and

for all large n. Thus, from Eqs. (A.4) and (A.5), we have the following inequality:

Immediately from p n δ n 2 / λ n → 0 , we have that

with probability tending to one. Hence, it follows from Eqs. (A.5) and (A.6) that

which implies that conclusion (i) holds.

To prove (ii), we only need to verify that α * ∈ 1 / K 0 , K 0 q n with probability tending to 1 since (A.7) has showed that γ * ≤ δ n p n / n with probability tending to 1. Analogously, given condition ( C 2 ) , β 1 ∈ 1 / K 0 , K 0 q n and α * < O p ( p n ) , we have

Then from (A.2), we have

and according to (A.4) and (A.7), we have λ n n D 2 β 2 γ * ≤ 2 c δ n p n / n . Hence based on condition (C2), we know that as n → ∞ and with probability tending to one,

Therefore, from (A.8) and (A.9), we can get

with probability tending to one, which implies that for any ϵ > 0, P α * − β 01 ≤ ϵ → 1 . Thus, it follows from β 01 ∈ 1 / K 0 , K 0 q n that α * ∈ 1 / K 0 , K 0 q n holds for large n, which implies that conclusion (ii) holds. This completes the proof.

Lemma 2

Under the regular conditions (C1)–(C4) given above and with probability tending to 1, the equation α = Ω n ( 1 ) + λ n D 1 ( α ) − 1 v n ( 1 ) has a unique fixed-point α ̂ * in the domain 1 / K 0 , K 0 q n

Proof

Define

where α = α 1 , … , α q n ′ . By multiplying Ω n ( 1 ) − 1 Ω n ( 1 ) + λ n D 1 ( α ) and then minus β ₀₁ on both sides of (A.10), we have

where X ₁ is the first q _n columns of X ̂ , and ϵ = y − X ̂ β . Therefore,

Similar to (A.3), it can be shown that

Thus

which implies that f ( α ) ∈ 1 / K 0 , K 0 q n with probability tending to one. That is, f( α ) is a mapping from 1 / K 0 , K 0 q n to itself.

Also by multiplying Ω n ( 1 ) + λ n D 1 ( α ) and taking derivative with respect to α on both sides of (A.10), we have

where f ̇ ( α ) = ∂ f ( α ) / ∂ α ′ . Then

According to condition (C3) and the fact that α ∈ 1 / K 0 , K 0 q n , we can derive

Thus, we have that sup α ∈ 1 / K 0 , K 0 q n ‖ f ̇ ( α ) ‖ → 0 , which implies that f(⋅) is a contraction mapping from 1 / K 0 , K 0 q n to itself with probability tending to one. Hence, according to the contraction mapping theorem, there exists one unique fixed-point α ̂ * ∈ 1 / K 0 , K 0 q n such that

Proof of Theorem 1

First according to the definition of β ̂ * and β ̂ 2 ( k ) , it follows from (A.7) that

with the probability tending to 1. That is, the conclusion (1) holds. Now we will show that P β ̂ 1 * = α ̂ * → 1 . For this, consider (A.2) and define γ * = 0 if β ₂ = 0. Note that for any fixed large n, from (A.2), we have

Furthermore, by multiplying Ω n + λ n D ( β ) on both sides of (A.1), we can get

By combining Eqs. (A.12) and (A.13), it follows that

since f(⋅) is a contract mapping. Hence (A.11) yields

Let h k = β ̂ 1 ( k ) − α ̂ * ⋅ It then follows from (A.14) and (A.15) that

From (A.14), for any ϵ ≥ 0, there exists N > 0 such that when k > N, η k < ϵ . Employing some recursive calculation, we have h _k → 0 as k → ∞. Hence, with probability tending to one, we have

since β ̂ 1 ≡ lim k → ∞ β ̂ 1 ( k ) , and it follows from the uniqueness of the fixed-point that

Finally, based on (A.11), we have n α ̂ * − β 01 = Π 1 + Π 2 , where

and

It follows from the first-order resolvent expansion formula that

This yields that

By the assumption (C2) and (C3), we have

Furthermore, it follows from (A.16) and the assumption λ n / n → 0 that

We denote −n*H( β |Λ ₀) as l _n ( β |Λ ₀) and Λ ₀ as Λ₀₁, Λ₀₂. Then we have that n − 1 / 2 v n ( 1 ) − Ω n ( 1 ) β 01 = n − 1 / 2 l ̇ n ( 1 ) β ̂ | Λ ̂ 0 + o p ( 1 ) with l ̇ n ( 1 ) β ̂ | Λ ̂ 0 denoting the first q _n components of l ̇ n β ̂ | Λ ̂ 0 . Let I ( β ) = E − l ̈ n β | Λ ̂ 0 be the Fisher information matrix, where l ̈ n ( β | Λ 0 ) is the partial Hessian matrix about β . Since n − 1 / 2 l n β ̂ | Λ ̂ 0 → N 0 , n − 1 I β 0 , we have n α ̂ * − β 01 → N q n ( 0 , Σ ) with Σ = n Ω n ( 1 ) β 0 − 1 I ( 1 ) β 0 Ω n ( 1 ) β 0 − 1 , where I ( 1 ) β 0 is the leading q _n × q _n submatrix of I β 0 . This completes the proof.