A varying-coefficient partially linear transformation model for length-biased data with an application to HIV vaccine studies

Lemma 1

Assume that Conditions (C1)–(C6) hold, and h → 0 and nh → 0 as n → ∞. Also, assume that the matrix A _z (defined in the proof) is finite and non-degenerate for any z. Then the one-step estimators β ̃ ( z ) , α ̃ 1 ( z ) , and α ̃ 0 ( z ) are locally consistent.

Proof of Lemma 1

Let H ̃ z ( t ; β , α 0 , α 1 ) be the solution of (8) for fixed β , α ₀, and α ₁. Note that the left-hand side of (8) is monotone in H. By similar arguments to Chen et al. [19]; it can be shown that (8) has a unique solution H ̃ z ( t ; β , f 0 , f ̇ 0 ) , and for any t ∈ (0, τ],

where →_a.s. denotes the convergence almost surely. Let

Then H ̃ z ( t ; β , α 0 , α 1 ) may be considered as a random mapping from D _ϵ to another open connected set B _n in R p + 2 q .

We divide our proof into three steps. First, by the law of large numbers and Lemma 1 in Cai et al. [25]; we have

where g(z) is the density function of Z. Thus, B _n contains 0 with probability approaching 1.

Second, let

where k ₂ = ∫x ² k(x)dx. By techniques similar to those used above,

in probability. Using methods similar to Lu and Zhang [17] and the uniform law of large numbers, we can obtain,

as n → ∞, ϵ → ∞ and ϵ _z → ∞.

Third, as A _z is finite and non-degenerate by assumption, the map U ̃ z ( β , α 1 , α 0 ) is a homeomorphism from D _ϵ to B _n . However, U ̃ z ( β ̃ , α ̃ 1 , α ̃ 0 ) = 0 . Thus, by letting ϵ → 0, it can be shown that β ̃ ( z ) , α ̃ 1 ( z ) and α ̃ 0 ( z ) are locally consistent.

Proof of Theorem 1

We need to prove the local consistency of estimators β ̂ , H ̂ ( ⋅ ) , α ̂ 0 ( z ) and α ̂ 1 ( z ) . Given that β ̃ ( z ) , α ̃ 1 ( z ) and α ̃ 0 ( z ) are locally consistent, then so are β ̂ , α ̂ 0 ( z ) and α ̂ 1 ( z ) from results of Carroll et al. [24]. Thus, we only have to prove the consistency of H ̂ ( ⋅ ) . Let H ̆ be a limit of H ̂ ( t , β 0 , f 0 ) . By Eq. (5) and arguments similar to those used by Kim et al. [29]; we have

Differentiating (11) with respect to t, we have

As (12) is a Cauchy problem, it has a unique solution. As well, it is readily seen from Eq. (4) that H ₀, the true transformation function, satisfies (12). Hence, by Helly’s Lemma, H 0 = H ̆ and we have H ̂ ( t , β 0 , f 0 ) → a.s. H 0 . Note that ∂ H ̂ ( t , β , α 0 ) ∂ β and ∂ H ̂ ( t , β , α 0 ) ∂ α 0 are bounded on [0, τ] if β ∈ D ϵ 1 1 and α 0 ∈ D ϵ 2 2 for some ϵ ₁ and ϵ ₂. Then by using the Taylor-series expansion, we have H ̂ ( t ) → a.s. H 0 ( t ) .

Next, we prove the asymptotic normality of β ̂ . Our proof focuses on the following asymptotic representation of β ̂ :

If it can be proven that if (13) is true, then it is clear that Theorem 1 is also true by the martingale central limit theorem. In order to prove (13), we first give the representation of Λ ϵ * { H ̂ ( t ; β 0 , f 0 ) } − Λ ϵ * { H 0 ( t ) } and ∂ ∂ β H ̂ ( t , β 0 , f 0 ) that will be used to approximate the rate of Taylor-series expansion. Next, we give the representation of 1 n ∑ i = 1 n M i ( t ) and 1 n ∑ i = 1 n ∫ 0 τ X i d M i ( t ) . Finally, we combine the results in the first two steps to obtain (13).

From Eqs. (4) and (5), we have

By some tedious calculations and results of the empirical process theory for Z-estimator and the definition of λ ϵ * { H 0 ( t ) } , we obtain

Then it follows, for t ∈ (0, τ], that

Note that for any t ∈ (0, τ],

for the parameter β . Differentiating (16) with respect to β on both sides yields

Following arguments similar to Chen et al. [19] and Lu and Zhang [17]; it can be shown that

Next we give the representation of 1 n ∑ i = 1 n M i ( t ) and 1 n ∑ i = 1 n ∫ 0 τ X i d M i ( t ) .

Denote

Taking the derivative of U ₁( β , f ) with respect to β , setting β = β ₀ and f = f ₀, and using the law of large numbers and (17), we obtain

Let ( α ̂ 0 , α ̂ 1 ) be the solution of (7), and for t ∈ [0, τ], define H ̂ 0 ( t ; β ) = H ̂ ( t ; β , α ̂ 0 ) . For any z, let

Then we have

where ( α ̂ 0 , α ̂ 1 ) is the solution of (7) at convergence, and ( β ̂ , H ̂ 0 ( t ; β ̂ ) ) is the solution of (5) and (6) at convergence. By the Taylor-series expansion and the law of large numbers, we obtain

where

However, as H ̂ 0 ( t ; β ) = H ̂ ( t ; β , α ̂ 0 ) , we have

Taking derivative with respect to β on both sides of (20), applying the law of large numbers and using arguments similar to those in Step 2, we obtain

Multiplying λ ϵ * { H 0 ( s ) } to both sides of (21) and performing integration with respect to s over the range (0, t) yields

Substituting (22) into E ₁(z) leads to

where e 1 T ( z ) = g ( z ) E R ( t ) W λ ϵ H 0 ( t ) + β 0 T X + f 0 T ( z ) W ( X − a ( t ) ) T | Z = z . On the other hand,

where

Moreover,

Thus, if we define