Low-dimensional confounder adjustment and high-dimensional penalized estimation for survival analysis

Abstract

High-throughput profiling is now common in biomedical research. In this paper we consider the layout of an etiology study composed of a failure time response, and gene expression measurements. In current practice, a widely adopted approach is to select genes according to a preliminary marginal screening and a follow-up penalized regression for model building. Confounders, including for example clinical risk factors and environmental exposures, usually exist and need to be properly accounted for. We propose covariate-adjusted screening and variable selection procedures under the accelerated failure time model. While penalizing the high-dimensional coefficients to achieve parsimonious model forms, our procedure also properly adjust the low-dimensional confounder effects to achieve more accurate estimation of regression coefficients. We establish the asymptotic properties of our proposed methods and carry out simulation studies to assess the finite sample performance. Our methods are illustrated with a real gene expression data analysis where proper adjustment of confounders produces more meaningful results.

Appendix: Conditions and proofs

1.1 Proposition 1

To obtain the screening consistency, we need the following technical assumptions.

• (S1) $p>n$ and $\log (p)=O(n^\xi )$ for some $\xi \in (0,1-2\kappa )$ where $\kappa$ is given in condition (S3).

• (S2) Denote $\varSigma =\text{ cov }(\mathbf{X})$. We assume $\varSigma ^{-1/2}{} \mathbf{X}$ has a spherically symmetric distribution and for any $n\times \tilde{p}$ submatrix ${\tilde{\mathbf{X}}}$ of $\varSigma ^{-1/2}{} \mathbf{X}$ with $cn<\tilde{p}\le p$,

  $$\begin{aligned} P(\lambda _{max}(\tilde{p}^{-1}\tilde{\mathbf{X}}\tilde{\mathbf{X}}^T)>c_1 \text{ or } \lambda _{min}(\tilde{p}^{-1}\tilde{\mathbf{X}}\tilde{\mathbf{X}}^T)<1/c_1) \le \exp (-C_1 n) \end{aligned}$$

  (15)

  for some $c_1>1$ and $C_1>0$, where $\lambda _{max}$ and $\lambda _{min}$ are the largest and smallest eigenvalues of a matrix.

• (S3) $\text{ var }[T-\sum _{j=1}^d g_j(U_j)]=O(1)$ and for some $\kappa \ge 0$ and $c_2,c_3>0$,

  $$\begin{aligned} \min _{j\in {\mathcal {M}}_*} |\beta _j|\ge \frac{c_2}{n^{\kappa }}\qquad \text{ and }\qquad \min _{j\in {\mathcal {M}}_*} \text{ cov }[\beta _j^{-1}\{T-\sum _{k=1}^d g_k(U_k)\},X_j]\ge c_3. \end{aligned}$$

  (16)

• (S4) There are some $\tau \ge 0$ and $c_4>0$ such that $\lambda _{max}(\varSigma )\le c_4 n^{\tau }$.

• (S5) The initial benchmark estimates satisfy $\sup _j \Vert g_j^*-g_j\Vert =O_p(\omega _n)$ and $\omega _n\rightarrow 0$ as $n\rightarrow \infty$.

Proof

(Sketch of Proof of Proposition 1) The proof can be completed in two-steps. In the first step, we define a submodel

$$\begin{aligned} {\mathcal {M}}_\delta =\{1\le j\le p: |\beta _j| \text{ is } \text{ among } \text{ the } \text{ first } [\delta p] \text{ largests }. \} \end{aligned}$$

(17)

and we may show that, if $\delta \rightarrow 0$ in such a way that $\delta n^{1-2\kappa -\tau }\rightarrow \infty$ as $n\rightarrow \infty$, we have for some $C>0$,

$$\begin{aligned} P({\mathcal {M}}_*\subset {{\mathcal {M}}_\delta })\ge 1- O\{\exp (-Cn^{1-2\kappa }/\log (n))\}. \end{aligned}$$

(18)

In the second step, we fix an $r\in (0,1)$ and choose a shrinking factor $\delta$ of the form $(n/p)^{1/(\kappa -r)}$. We successively carry out the argument in Step 1 to obtain submodels ${\mathcal {M}}_{\delta ^k}$, $k=1,\cdots ,r$. The final results may follow after some lengthy algebra. Similar derivation in Fan and Lv (2008) can be adopted. $\square$

1.2 Main theorems

We write ${\varvec{\theta }}=({\varvec{\beta }}^T,{\varvec{\gamma }}^T)^T$. Define $\tilde{\mathbf{Y}}=E(\mathbf{Y}|\mathbf{X}, \mathbf{U})$, $\tilde{\varvec{\theta }}=[\widetilde{\mathbf {X}}^T\mathbf { W} \widetilde{\mathbf {X}}]^{-1}\widetilde{\mathbf {X}}^T \mathbf {W} \tilde{\mathbf {Y}}=(\tilde{\varvec{\beta }}^T,\tilde{\varvec{\gamma }}^T)^T$, and $\tilde{g}_k(u)=\mathbf{B}(u)^T\tilde{\varvec{\gamma }}_k$. To prove the main theorems in this paper, we need the following assumptions.

• (C1) $E(\epsilon _i| \mathbf{X}_i, \mathbf{U}_i)=0$ and $E(T_i^2)$ is finite.

• (C2) $T_i$ and $C_i$ are independent and $P(Y_i\le C_i| \mathbf{X}_i, \mathbf{U}_i, Y_i)=P(Y_i\le C_i | Y_i)$.

• (C3) The eigenvalues of the matrix $E(\mathbf{X}_i\mathbf{X}_i^T)$ is bounded away from zero and finite.

• (C4) $\tau _T<\tau _C$ or $\tau _T=\tau _C=\infty$.

• (C5) The true parameter ${\varvec{\beta }}_0$ lives in a compact space $\varTheta$. Each true function $g_{k0}$ is from a second order Sobolev space.

• (C6) $E\left\{ \epsilon ^2\delta \mathbf{X}^{(1)}(\mathbf{X}^{(1)})^T \right\} <\infty$ and $E\left\{ |\epsilon \mathbf{X}^{(1)}| \sqrt{R(Y)}\right\} <\infty$ where $R(y)=\int _0^{y-}\{(1-H(w))(1-G(w))\}^{-1}G(dw)$ and G is the distribution function of the censoring time C.

The estimator $\hat{\varvec{\theta }}$ is of a form of weighted least squares estimator. However, the Kaplan-Meier weights $\{w_{ni}: i=1,\cdots ,n\}$ do not satisfy the assumptions which are usually required in weighted least squares estimation. We need Lemma 3 to ensure the validity of convergence argument used in the proof of Theorems 1 and 2. The proof of the lemma may follow Stute (1993).

Lemma 3

For an integrable function $\phi$, define a functional ${\mathcal {S}}_n\phi =\sum _{i=1}^n w_{ni} \phi (Y_i,\mathbf{X}_i,\mathbf{U}_i)$. Under (C1) and (C2), with probability one and in the mean we have

$$\begin{aligned} \lim _{n\rightarrow \infty }{\mathcal {S}}_n\phi =\int _{Y<\tau _H} \phi (Y,\mathbf{X},\mathbf{U}) dP + I(\tau _H\in A)\int _{Y=\tau _H}\phi (\tau _H,\mathbf{X},\mathbf{U})dP. \end{aligned}$$

(19)

We need the following lemma which summarizes necessary properties of the polynomial spline functions. The proof of the lemma may follow Lemma A.3 in Huang et al. (2004).

Lemma 4

Assume $\lim _{n\rightarrow \infty } n^{-1}M_n\log (M_n)=0$. Except on an event whose probability tends to zero, all the eigenvalues of $M_n/n \sum _{i=1}^n\mathbf{B}_i\mathbf{B}_i^T$ are bounded away from zero and infinity.

The next lemma establishes the consistency of the estimator.

Lemma 5

Assume the same conditions as Theorem 1. Then $\Vert \hat{\varvec{\theta }}-\tilde{\varvec{\theta }}\Vert =O_p(r_n +(\lambda _n\rho _n)^{1/2})$.

Proof

(Proof of Lemma 5) We note

$$\begin{aligned} Q(\hat{\varvec{\theta }})-Q(\tilde{\varvec{\theta }})= & {} \frac{1}{n}\left[ (\mathbf{Y}-\tilde{\mathbf{X}}\hat{\varvec{\theta }})^T\mathbf{W} (\mathbf{Y}-\tilde{\mathbf{X}}\hat{\varvec{\theta }}) - (\mathbf{Y}-\tilde{\mathbf{X}}\tilde{\varvec{\theta }})^T\mathbf{W} (\mathbf{Y}-\tilde{\mathbf{X}}\tilde{\varvec{\theta }})\right] \nonumber \\&+\sum _{k=1}^p\{p_{\lambda _n}(|\hat{\beta }_k|)-p_{\lambda _n}(|\tilde{\beta }_k|) \}\nonumber \\= & {} \frac{1}{n}\left[ -2(\mathbf{Y}-\tilde{\mathbf{X}}\tilde{\varvec{\theta }})^T\mathbf{W}\tilde{\mathbf{X}}(\hat{\varvec{\theta }}-\tilde{\varvec{\theta }}) +(\hat{\varvec{\theta }}-\tilde{\varvec{\theta }})^T\tilde{\mathbf{X}}^T\mathbf{W}\tilde{\mathbf{X}}(\hat{\varvec{\theta }}-\tilde{\varvec{\theta }})\right] \nonumber \\&+\sum _{k=1}^p\{p_{\lambda _n}(|\hat{\beta }_k|)-p_{\lambda _n}(|\tilde{\beta }_k|)\} \nonumber \\= & {} -2{\varvec{\epsilon }}^T\mathbf{W}\tilde{\mathbf{X}}\mathbf{v}M_n^{1/2}\delta _n/n+\delta _n^2/n M_n \mathbf{v}^T\tilde{\mathbf{X}}^T\mathbf{W}\tilde{\mathbf{X}} \mathbf{v}\nonumber \\&+\sum _{k=1}^p\{p_{\lambda _n}(|\hat{\beta }_k|)-p_{\lambda _n}(|\tilde{\beta }_k|) \}\le 0, \end{aligned}$$

(20)

where in the third equality we write $\hat{\varvec{\theta }}-\tilde{\varvec{\theta }}=\delta _nM_n^{1/2}\mathbf{v}$, with $\delta _n$ being a scalar and $\mathbf{v}$ being a vector satisfying $\Vert \mathbf{v}\Vert =1$, and use the fact that $\tilde{\mathbf{X}}^T\mathbf{W}(\mathbf{Y}-{\varvec{\epsilon }}-\tilde{\mathbf{X}}\tilde{\varvec{\theta }})=0$. The inequality follows from the definition of $\hat{\varvec{\theta }}$. We first show that $\delta _n=O_p(r_n+\lambda _n)$. To this end, we can show easily

$$\begin{aligned} \frac{M_n^{1/2}}{n}{\varvec{\epsilon }}^T\mathbf{W}\tilde{\mathbf{X}}\mathbf{v}=\frac{M_n^{1/2}}{n}\sum _{i=1}^n \epsilon _i w_{ni}\tilde{\mathbf{X}}_{(i)}{} \mathbf{v}=O_p(r_n). \end{aligned}$$

(21)

By assumption (C3), Lemma 4 and following Lemma A.2 of Liu et al. (2011), there exists a positive $c_1$ such that

$$\begin{aligned} \frac{M_n}{n} \mathbf{v}^T\tilde{\mathbf{X}}^T\mathbf{W}\tilde{\mathbf{X}} \mathbf{v}=\frac{M_n}{n} \sum _{i=1}^n w_{ni}{} \mathbf{v}^T \left( \begin{array}{cc} \mathbf{X}_{(i)}{} \mathbf{X}_{(i)}^T &{} \mathbf{X}_{(i)}{} \mathbf{B}_{(i)}^T \\ \mathbf{B}_{(i)}{} \mathbf{X}_{(i)}^T &{} \mathbf{B}_{(i)}{} \mathbf{B}_{(i)}^T \end{array} \right) \mathbf{v} \ge c_1, \end{aligned}$$

(22)

with probability approaching 1. Using inequality $| p_\lambda (a)-p_\lambda (b)|\le \lambda |a-b|$, we obtain

$$\begin{aligned} \sum _{k=1}^p\{p_{\lambda _n}(|\hat{\beta }_k|)-p_{\lambda _n}(|\tilde{\beta }_k|)\ge & {} \sum _{k=1}^p -\lambda _n |\hat{\beta }_k-\tilde{\beta }_k| \asymp -\lambda _n \delta _n. \end{aligned}$$

(23)

Therefore, $-O_p(r_n)\delta _n+c_1\delta _n^2-\lambda _n\delta _n\le 0$ with probability approaching 1, which implies that $\delta _n=O_p(r_n+\lambda _n)$.

Now we notice for $1\le k \le p$, $|\hat{\beta }_k-\tilde{\beta }_k|=o_p(1)$. Next we can see

$$\begin{aligned} | \Vert \tilde{\gamma }_k\Vert -\Vert g_k\Vert _{L_2}|\le & {} | \Vert \tilde{g}_k\Vert _{L_2}-\Vert g_k\Vert _{L_2}|\end{aligned}$$

(24)

$$\begin{aligned}\le & {} \Vert \tilde{g}_k -g_k\Vert _{L_2} =O_p(\rho _n)=o_p(1). \end{aligned}$$

(25)

It then follows that $\hat{\beta }_k\rightarrow \beta _{k0}$, $\tilde{\beta }_k\rightarrow \beta _{k0}$, $\Vert \hat{g}_k\Vert _{L_2}\rightarrow \Vert g_{k0}\Vert _{L_2}$ and $\Vert \tilde{g}_k\Vert _{L_2}\rightarrow \Vert g_{k0}\Vert _{L_2}$ in probability. Because $|\beta _{k0}|>0$ for $1\le k\le s$ and $\lambda \rightarrow 0$, we have that, with probability approaching 1, $|\hat{\beta }_k|>a\lambda _n$ and $|\tilde{\beta }_k|>a\lambda _n$ for $1\le k\le s$. On the other hand, $\beta _{k0}=0$ for $s+1\le k \le p$, so the previous results imply $\tilde{\beta }_k=O_p(\rho _n)$. Since $\lambda _n/\rho _n \rightarrow \infty$, we have $|\tilde{\beta }_k| < \lambda _n$, $s+1\le k \le p$. Consequently by the definition of $p_\lambda (\cdot )$, we have $P(p_{\lambda _n}(|\tilde{\beta }_k|)=p_{\lambda _n}(|\hat{\beta }_k|))\rightarrow 1$ when $1\le k\le s$; and $P(p_{\lambda _n}(|\tilde{\beta }_k|)=\lambda _n |\tilde{\beta }_k|) \rightarrow 1$ when $s+1\le k \le p$. Therefore

$$\begin{aligned} \sum _{k=1}^p\{p_{\lambda _n}(|\hat{\beta }_k|)-p_{\lambda _n}(|\tilde{\beta }_k|)\}=\lambda _n \sum _{k=s+1}^p |\tilde{\beta }_k| \ge -O_p(\lambda _n\rho _n). \end{aligned}$$

(26)

Combining with previous results, we have

$$\begin{aligned} Q(\hat{\varvec{\theta }})-Q(\tilde{\varvec{\theta }}) \ge -O_p(r_n)\delta _n+c_1\delta _n^2-O(\lambda _n\rho _n), \end{aligned}$$

(27)

which implies that $\delta _n=O_p(r_n+(\lambda _n\rho _n)^{1/2})$. $\square$

Proof

(Proof of Theorem 1) We prove part a by contradiction. Suppose that for a sufficiently large n there exists a constant $\eta >0$ such that with probability at least $\eta$ there exists a $k^*>s$ such that $\hat{\beta }_{k^*}\not =0$. Let $\hat{\varvec{\theta }}^*$ be a vector constructed by replacing $\hat{\beta }_{k^*}$ with 0 in $\hat{\varvec{\theta }}$. Then

$$\begin{aligned} Q(\hat{\varvec{\theta }})-Q(\hat{\varvec{\theta }}^*)= & {} \frac{1}{n}\left[ (\mathbf{Y}-\tilde{\mathbf{X}}\hat{\varvec{\theta }})^T\mathbf{W} (\mathbf{Y}-\tilde{\mathbf{X}}\hat{\varvec{\theta }}) - (\mathbf{Y}-\tilde{\mathbf{X}}\hat{\varvec{\theta }}^*)^T\mathbf{W} (\mathbf{Y}-\tilde{\mathbf{X}}\hat{\varvec{\theta }}^*)\right] \nonumber \\&+\,p_{\lambda _n}(|\hat{\beta }_{k^*}|). \end{aligned}$$

(28)

By Lemma 3 and the fact that $\beta _{k^*0}=0$, $\hat{\beta }_{k^*}=O_p(r_n+(\lambda _n\rho _n)^{1/2})$. Because $\lambda _n/\max (r_n,\rho _n)\rightarrow \infty$, we have $|\hat{\beta }_{k^*}| < \lambda _n$ and thus $p_{\lambda _n}(|\hat{\beta }_{k^*}|)=\lambda _n |\hat{\beta }_{k^*}|$ with probability approaching 1. For the first term in (28), simple algebra leads to

$$\begin{aligned}&(\mathbf{Y}-\tilde{\mathbf{X}}\hat{\varvec{\theta }})^T\mathbf{W} (\mathbf{Y}-\tilde{\mathbf{X}}\hat{\varvec{\theta }}) - (\mathbf{Y}-\tilde{\mathbf{X}}\hat{\varvec{\theta }}^*)^T\mathbf{W} (\mathbf{Y}-\tilde{\mathbf{X}}\hat{\varvec{\theta }}^*) \nonumber \\&\quad \ge - (\mathbf{Y}-\tilde{\mathbf{X}}\hat{\varvec{\theta }})^T\mathbf{W}\tilde{\mathbf{X}}(\hat{\varvec{\theta }}-\hat{\varvec{\theta }}^*) \nonumber \\&\quad = -2 (\mathbf{Y}-\tilde{\mathbf{X}}\tilde{\varvec{\theta }})^T\mathbf{W}\tilde{\mathbf{X}}(\hat{\varvec{\theta }}-\hat{\varvec{\theta }}^*) \nonumber \\&\qquad -2 (\tilde{\varvec{\theta }}-\hat{\varvec{\theta }}^*)\tilde{\mathbf{X}}^T\mathbf{W}\tilde{\mathbf{X}}(\hat{\varvec{\theta }}-\hat{\varvec{\theta }}^*). \end{aligned}$$

(29)

By the Cauchy–Schwartz inequality,

$$\begin{aligned}&(\tilde{\varvec{\theta }}-\hat{\varvec{\theta }}^*)\tilde{\mathbf{X}}^T\mathbf{W}\tilde{\mathbf{X}}(\hat{\varvec{\theta }}-\hat{\varvec{\theta }}^*)\\&\quad \le \{(\tilde{\varvec{\theta }}-\hat{\varvec{\theta }}^*)\tilde{\mathbf{X}}^T\mathbf{W}\tilde{\mathbf{X}}(\tilde{\varvec{\theta }}-\hat{\varvec{\theta }}^*)\}^{1/2}\\&\qquad \times \{(\hat{\varvec{\theta }}-\hat{\varvec{\theta }}^*)\tilde{\mathbf{X}}^T\mathbf{W}\tilde{\mathbf{X}}(\hat{\varvec{\theta }}-\hat{\varvec{\theta }}^*)\}^{1/2}\\&\quad \le c_2 \frac{n}{M_n}\Vert \tilde{\varvec{\theta }}-\hat{\varvec{\theta }}^*\Vert \Vert \hat{\varvec{\theta }}-\hat{\varvec{\theta }}^*\Vert . \end{aligned}$$

From the triangle inequality and Lemma 3, it follows that

$$\begin{aligned} \Vert \tilde{\varvec{\theta }}-\hat{\varvec{\theta }}^*\Vert\le & {} \Vert \tilde{\varvec{\theta }}-\hat{\varvec{\theta }}\Vert +|\tilde{\beta }_{k^*}|\\= & {} O_p(M_n^{1/2}\{r_n+(\lambda _n\rho _n)^{1/2}+\rho _n \}), \end{aligned}$$

thus

$$\begin{aligned} \frac{1}{n}(\tilde{\varvec{\theta }}-\hat{\varvec{\theta }}^*)\tilde{\mathbf{X}}^T\mathbf{W}\tilde{\mathbf{X}}(\hat{\varvec{\theta }}-\hat{\varvec{\theta }}^*)=O_p(M_n^{-1/2} (r_n+(\lambda _n\rho _n)^{1/2}+\rho _n))|\hat{\beta }_{k^*}|. \end{aligned}$$

(30)

We can also show that

$$\begin{aligned} |(\mathbf{Y}-\tilde{\mathbf{X}}\tilde{\varvec{\theta }})^T\mathbf{W}\tilde{\mathbf{X}}(\hat{\varvec{\theta }}-\hat{\varvec{\theta }}^*)|=|{\varvec{\epsilon }}^T \mathbf{W}\tilde{\mathbf{X}}(\hat{\varvec{\theta }}-\hat{\varvec{\theta }}^*)| =O_p(\frac{nr_n}{M_n^{1/2}})|\hat{\beta }_{k^*}|. \end{aligned}$$

(31)

Combining (28) to (31), we arrive at

$$\begin{aligned} Q(\hat{\varvec{\theta }})-Q(\hat{\varvec{\theta }}^*)\ge & {} \frac{\lambda _n}{M_n^{1/2}}|\hat{\beta }_{k^*}| -O_p(\frac{r_n}{M_n^{1/2}})|\hat{\beta }_{k^*}| \nonumber \\&-O_p(\frac{r_n+(\lambda _n\rho _n)^{1/2}+\rho _n}{M_n^{1/2}})| \hat{\beta }_{k^*}|. \end{aligned}$$

(32)

We note that the first term on the right hand side of (32) dominates the other two terms since $\lambda _n/\max (\rho _n,r_n)\rightarrow \infty$. This contradicts the fact that $Q(\hat{\varvec{\theta }})-Q(\hat{\varvec{\theta }}^*)\le 0$. Hence the proof of part a is completed.

To prove parts b and c, we define the oracle version of $\tilde{\varvec{\theta }}$,

$$\begin{aligned} \tilde{\varvec{\theta }}_{\varOmega }=\text{ arg }\min _{{\varvec{\theta }} =({\varvec{\beta }}_1^T,\mathbf{0}^T,{\varvec{\gamma }}^T)}\frac{1}{n}(\mathbf{Y}-\tilde{\mathbf{X}}{\varvec{\theta }})^T\mathbf{W} (\mathbf{Y}-\tilde{\mathbf{X}}{\varvec{\theta }}) \end{aligned}$$

(33)

which is obtained as if the information of the nonzero components were given. By the construction and Lemma 4, we have $\Vert \tilde{\varvec{\theta }}_{\varOmega }-{\varvec{\theta }}_0\Vert =O_p(\rho _n)$ and $\Vert \hat{\varvec{\theta }}-{\varvec{\theta }}_0\Vert =o_p(1)$. Thus, with probability approaching 1, $\tilde{\beta }_{k,\varOmega }\rightarrow \beta _{k0}$, $\hat{\beta }_{k}\rightarrow \beta _{k0}$ ($1\le k\le s$), $\Vert \tilde{g}_{k,\varOmega }\Vert \rightarrow \Vert g_{k0}\Vert$, and $\Vert \hat{g}_{k}\Vert \rightarrow \Vert g_{k0}\Vert$ ($1\le k\le d$). On the other hand, for $s+1\le k\le p$, by the definition $\tilde{\beta }_{k,\varOmega }=0$, and by part a, with probability approaching 1, $\hat{\beta }_{k}=0$. Consequently, we have

$$\begin{aligned} \sum _{k=1}^p p_{\lambda _n}(|\tilde{\beta }_{k,\varOmega }|)=\sum _{k=1}^p p_{\lambda _n}(|\hat{\beta }_k|) \end{aligned}$$

(34)

with probability approaching 1. Now write $\hat{\varvec{\theta }}-\tilde{\varvec{\theta }}_{\varOmega }=\delta _nM_n^{1/2}\mathbf{v}$, with $\Vert \mathbf{v}\Vert =1$. By (20) and (34),

$$\begin{aligned} 0\ge & {} Q(\hat{\varvec{\theta }})-Q(\tilde{\varvec{\theta }}_{\varOmega })\\= & {} -2{\varvec{\epsilon }}{} \mathbf{W}\tilde{\mathbf{X}}\mathbf{v}M_n^{1/2}\delta _n/n+\delta ^2/n M_n \mathbf{v}^T\tilde{\mathbf{X}}^T\mathbf{W}\tilde{\mathbf{X}} \mathbf{v}\\\ge & {} -O_p(r_n)\delta _n+c_1 \delta _n^2. \end{aligned}$$

Thus $\Vert \hat{\varvec{\theta }}-\tilde{\varvec{\theta }}_{\varOmega }\Vert \asymp \delta _n=O_p(r_n)$, which, together with $\Vert \tilde{\varvec{\theta }}_{\varOmega }-{\varvec{\beta }}_0\Vert =O_p(\rho _n)$, implies that $\Vert \hat{\varvec{\theta }}-\tilde{\varvec{\theta }}\Vert =O_p(\rho _n+r_n)$. Hence the claims in parts b and c follow. $\square$

To prove Theorem 2, we need the following lemma which gives the asymptotic behavior of the AFT estimator under Kaplan-Meier weights. Denote the right hand side of (19) to be ${\mathcal {S}}\phi$. Introduce the following sub-distribution functions:

$$\begin{aligned} \tilde{H}_{1}(\mathbf{x},\mathbf{u},y)= & {} P(\mathbf{X}\le \mathbf{x}, \mathbf{U}\le \mathbf{u}, Y\le y, \delta =1)\\ \tilde{H}_0(y)= & {} P(Y\le y, \delta =1). \end{aligned}$$

Put

$$\begin{aligned} \xi _0(y)= & {} \exp \left\{ \int _0^{y-} \frac{\tilde{H}_0(dz)}{1-H(z)} \right\} \\ \xi _1^\phi (y)= & {} \frac{1}{1-H(y)}\int _{w>y}\phi (\mathbf{x},\mathbf{u},w)\xi _0(w)\tilde{H}_1(d\mathbf{x}, d\mathbf{u}, dw)\\ \xi _2^\phi (y)= & {} \int \int \frac{I(v<y,v<w)\phi (\mathbf{x},\mathbf{u},w)\xi _0(w)}{(1-H(v))^2}\tilde{H}_0(dv)\tilde{H}_1(d\mathbf{x},d\mathbf{u}, dw). \end{aligned}$$

Let $\{\phi _1,\cdots ,\phi _J\}$ be a set of measurable functions. Write

$$\begin{aligned} \underline{\mathcal {S}}_n=({\mathcal {S}}_n\phi _1,\cdots ,{\mathcal {S}}_n\phi _J)^T \end{aligned}$$

and

$$\begin{aligned} \underline{\mathcal {S}}=({\mathcal {S}}\phi _1,\cdots ,{\mathcal {S}}\phi _J)^T. \end{aligned}$$

Lemma 6

Assume that (C1) and (C2) hold. In addition, assume the following two integrability conditions hold for all $\phi _j$, $1\le j\le J$,

$$\begin{aligned} \int \phi _j(\mathbf{X},\mathbf{U},W)\xi _0(W)\delta ^2 d\mathbf{P}_{\mathbf{X},\mathbf{U},Y}< & {} \infty \end{aligned}$$

(35)

$$\begin{aligned} \int \phi _j(\mathbf{X},\mathbf{U},W)\sqrt{R(W)}d\mathbf{P}_{\mathbf{X},\mathbf{U},Y}< & {} \infty . \end{aligned}$$

(36)

Then in distribution

$$\begin{aligned} \sqrt{n}(\underline{\mathcal {S}}_n - \underline{\mathcal {S}}) \rightarrow N(0,\varSigma ), \end{aligned}$$

(37)

where $\varSigma =(\sigma _{jj'})$, $\sigma _{jj'}=\text{ cov }(\psi _j,\psi _{j'})$ and $\psi _j=\phi _j(\mathbf{X},\mathbf{U},Y)\xi _0(Y)\delta + \xi _1^{\phi _j}(Y)(1-\delta )-\xi _2^{\phi _j}(Y)$.

The proof of this lemma may follow Stute (1996). We now proceed to prove Theorem 2.

Proof

(Proof of Theorem 2) According to the proof of Lemma 5, with probability approaching 1, $|\tilde{\beta }_k|>a\lambda _n$, $|\hat{\beta }_k|>a\lambda _n$ and thus $p_{\lambda _n}(|\tilde{\beta }_k|)=p_{\lambda _n}(|\hat{\beta }_k|)$ for $1\le k\le s$. By Theorem 1, with probability approaching 1, $\hat{\varvec{\theta }}=(\hat{\varvec{\beta }}_{1}^T, \mathbf{0}^T,\hat{\varvec{\gamma }}^T)^T$ is a local minimizer of $Q({\varvec{\theta }})$. We may note that $Q({\varvec{\theta }})$ is quadratic in $({\varvec{\beta }}_1^T,{\varvec{\gamma }}^T)^T$ when $|\beta _k|>a\lambda _n$ for $1\le k\le s$. Therefore $\partial Q({\varvec{\theta }})/\partial {\varvec{\theta }}|_{{\varvec{\beta }}_1=\hat{\varvec{\beta }}_1,{\varvec{\beta }}_2=\mathbf{0}, {\varvec{\gamma }}=\hat{\varvec{\gamma }}}=\mathbf{0}$, which implies that

$$\begin{aligned} (\hat{\varvec{\beta }}_{1}^T, \hat{\varvec{\gamma }}^T)^T=\left( \sum _{i=1}^nw_{ni}\left[ \begin{array}{c@{\quad }c} \mathbf{X}_i^{(1)}(\mathbf{X}_i^{(1)})^T&{}\mathbf{X}_i^{(1)}{} \mathbf{B}_i^T\\ \mathbf{B}_i(\mathbf{X}_i^{(1)})^T&{}\mathbf{B}_i\mathbf{B}_i^T] \end{array}\right] \right) ^{-1} \left( \sum _{i=1}^nw_{ni}\left[ \begin{array}{c} \mathbf{X}_i^{(1)}\\ \mathbf{B}_i \end{array}\right] {Y}_i\right) . \end{aligned}$$

Next we put

$$\begin{aligned} (\tilde{\varvec{\beta }}_{1}^T, \tilde{\varvec{\gamma }}^T)^T= & {} \left( \sum _{i=1}^nw_{ni}\left[ \begin{array}{c@{\quad }c} \mathbf{X}_i^{(1)}(\mathbf{X}_i^{(1)})^T&{}\mathbf{X}_i^{(1)}{} \mathbf{B}_i^T\\ \mathbf{B}_i(\mathbf{X}_i^{(1)})^T&{}\mathbf{B}_i\mathbf{B}_i^T \end{array}\right] \right) ^{-1}\nonumber \\&\times \left( \sum _{i=1}^n\left[ \begin{array}{c} (\mathbf{X}_i^{(1)})\\ \mathbf{B}_i \end{array}\right] E\{w_{ni}{Y}_i|\mathbf{X}_i,\mathbf{U}_i\}\right) . \end{aligned}$$

Invoking Lemmas 3 and 6 while applying a version of Lindeberge central limit theorem (cf. Petrov 1975), we obtain that for any vector $\mathbf{c}_n$ with dimension $s+dM_n$ and components not all 0,

$$\begin{aligned} \{ \mathbf{c}_n^T\varUpsilon \mathbf{c}_n\}^{-1/2}{} \mathbf{c}_n^T\left( \left[ \begin{array}{c} \hat{\varvec{\beta }}_1\\ \hat{\varvec{\gamma }} \end{array}\right] -\left[ \begin{array}{c} \tilde{\varvec{\beta }}_1\\ \tilde{\varvec{\gamma }}\end{array}\right] \right) \rightarrow _d N(0,1), \end{aligned}$$

(38)

where $\varUpsilon = \mathbf{H}^{-1} \varSigma ^* \mathbf{H}^{-1}$, $\mathbf{H}=E\left[ \begin{array}{cc} (\mathbf{X}_i^{(1)})(\mathbf{X}_i^{(1)})^T&{}(\mathbf{X}_i^{(1)})\mathbf{B}_i^T\\ \mathbf{B}_i(\mathbf{X}_i^{(1)})^T&{}\mathbf{B}_i\mathbf{B}_i^T \end{array}\right]$ and $\varSigma ^*=\text{ Var }[ \delta _i \xi _0^*(Y_i)(Y_i-E(Y_i|\mathbf{X}_i^{(1)}, \mathbf{U}_i))((\mathbf{X}_i^{(1)})^T, \mathbf{B}_i^T)^T+(1-\delta _i)\xi _1^*(Y_i)-\xi _2^*(Y_i)]$. Part a of Theorem 2 follows from (38) immediately. Further, if we choose $\mathbf{c}_n=(\mathbf{0}^T,\mathbf{B}(\mathbf{u})^T\mathbf{a}_n)$ such that not all elements of $\mathbf{a}_n$ are 0, we obtain

$$\begin{aligned} \{ \mathbf{a}_n^T\varGamma (\mathbf{u}) \mathbf{a}_n\}^{-1/2}{} \mathbf{a}_n^T\left\{ \left[ \begin{array}{c} \hat{g}_1(u_1)\\ \cdots \\ \hat{g}_d(u_d) \end{array}\right] -\left[ \begin{array}{c} \tilde{g}_1(u_1)\\ \cdots \\ \tilde{g}_d(u_d) \end{array}\right] \right\} \rightarrow _d N(0,1) \end{aligned}$$

which leads to part b. $\square$