Feature screening for ultrahigh-dimensional survival data when failure indicators are missing at random

Abstract

In modern statistical applications, the dimension of covariates can be much larger than the sample size, and extensive research has been done on screening methods which can effectively reduce the dimensionality. However, the existing feature screening procedure can not be used to handle the ultrahigh-dimensional survival data problems when failure indicators are missing at random. This motivates us to develop a feature screening procedure to handle this case. In this paper, we propose a feature screening procedure by sieved nonparametric maximum likelihood technique for ultrahigh-dimensional survival data with failure indicators missing at random. The proposed method has several desirable advantages. First, it does not rely on any model assumption and works well for nonlinear survival regression models. Second, it can be used to handle the incomplete survival data with failure indicators missing at random. Third, the proposed method is invariant under the monotone transformation of the response and satisfies the sure screening property. Simulation studies are conducted to examine the performance of our approach, and a real data example is also presented for illustration.

Appendix

Firstly, we introduce the following Lemma which is useful for proving Theorem 2.1.

Lemma 4.1

Under Assumptions 1–5, for any positive $\varepsilon$, we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\sup _{t\in (0,L]}P\left( \sqrt{n}|\{1-V(t)\} \{\tilde{G}(t)-G(t)\}|>\varepsilon \right) \le 2\exp \{-2\varepsilon ^{2}\}, \end{aligned}$$

(7)

and

$$\begin{aligned} \lim _{n\rightarrow \infty }\sup _{t\in (0,L]}P\left( \sqrt{n}|\{1-\tilde{V}(t)\} \{\tilde{G}(t)-G(t)\}|>\varepsilon \right) \le 2\exp \{-2\varepsilon ^{2}\}. \end{aligned}$$

(8)

Proof

Let $B^{0}$ denote a Brownian bridge. With the representation of that process from a Brownian bridge given by Bitouzé et al. (1999), we have

$$\begin{aligned} P\left( \sup _{t\in [0,L]}|B^{0}(t)|>\varepsilon \right) \le 2\exp \{-2\varepsilon ^{2}\}. \end{aligned}$$

(9)

Let

$$\begin{aligned} \varphi (t)=\int _{0}^{t}\frac{dG(s)}{\{1-G(s)\}^{2}\{1-V(s)\}}~~~\text {and}~~~K(t) =\frac{\varphi (t)}{1+\varphi (t)}. \end{aligned}$$

Based on Theorem 5.1 in van der Laan (1996), the estimator $\tilde{G}$ of G is efficient for the reduced data. Therefore, the result of Theorem 1.1 in Gill (1983) imply that

$$\begin{aligned} \sqrt{n}\frac{1-K}{1-G}(\tilde{G}-G){\mathop {\rightarrow }\limits ^{d}} B^{0}(K) ~~\text {in}~~ D[0,L] ~~\text {as}~~ n\rightarrow \infty , \end{aligned}$$

(10)

where ${\mathop {\rightarrow }\limits ^{d}}$ represents the convergence in distribution, and D[0, L] is the cadlag function space of real-valued functions on [0, L] endowed with the supremum norm. Hence it follows from (10) that

$$\begin{aligned}&\lim _{n\rightarrow \infty }\sup _{t\in [0,L]}P\left( \sqrt{n}\left| (1-V(t)) (\tilde{G}(t)-G(t))\right|>\varepsilon \right) \\&\quad =\lim _{n\rightarrow \infty }\sup _{t\in [0,L]}P\left( \left| \sqrt{n}\frac{1-K(t)}{1-G(t)} (\tilde{G}(t)-G(t))\right| \left| \frac{(1-G(t))(1-V(t))}{1-K(t)} \right|>\varepsilon \right) \\&\quad =\sup _{t\in [0,L]}P\left( \left| \frac{\{1-G(t)\}\{1-V(t)\}}{1-K(t)}\right| \left| B^{0}\{K(t)\}\right| >\varepsilon \right) . \end{aligned}$$

Because $|[\{1-G(t)\}\{1-V(t)\}]/\{1-K(t)\}|\le 1$, combining (9), we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\sup _{t\in (0,L]}P\left( \sqrt{n}|\{1-V(t)\}\{\tilde{G}(t) -G(t)\}|>\varepsilon \right) \le 2\exp \{-2\varepsilon ^{2}\}. \end{aligned}$$

The first part of Lemma 4.1 is proved, and then we begin to prove the second part of it. By simple calculation, we have

$$\begin{aligned} \sqrt{n}\{1-\tilde{V}(t)\}\{\tilde{G}(t)-G(t)\}= & {} \sqrt{n}\{1-V(t)\} \{\tilde{G}(t)-G(t)\}+\sqrt{n}\{\tilde{V}(t)\nonumber \\&-V(t)\}\{\tilde{G}(t)-G(t)\}, \end{aligned}$$

(11)

and

$$\begin{aligned} \sqrt{n}\{\tilde{V}(t)-V(t)\}\{\tilde{G}(t)-G(t)\}=\frac{\tilde{G}(t)-G(t)}{1-G(t)} \sqrt{n}\{1-G(t)\}\{\tilde{V}(t)-V(t)\}. \end{aligned}$$

(12)

Because the $\tilde{V}(t)$ is obtained by switching the role of failure time and censoring time in (2), similar to the proof of (7), we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\sup _{t\in (0,L]}P\left( \sqrt{n}|\{1-G(t)\}\{\tilde{V}(t) -V(t)\}|>\varepsilon \right) \le 2\exp \{-2\varepsilon ^{2}\}. \end{aligned}$$

(13)

Moreover, we also have

$$\begin{aligned} \sup _{t\in (0,L]}\sqrt{n}\{\tilde{V}(t)-V(t)\}\{\tilde{G}(t)-G(t)\}\le & {} \sup _{t\in (0,L]}\sqrt{n}|\{1-G(t)\}\{\tilde{V}(t)\nonumber \\&-V(t)\}|\sup _{t\in (0,L]}\left| \frac{\tilde{G}(t)-G(t)}{1-G(t)}\right| . \end{aligned}$$

(14)

If we can show

$$\begin{aligned} \sup _{t\in (0,L]}\left| \tilde{G}(t)-G(t)\right| {\mathop {\rightarrow }\limits ^{p}} 0, \end{aligned}$$

(15)

where ${\mathop {\rightarrow }\limits ^{p}}$ represents the convergence in probability. Then, we can obtain, according to Lemma 2.6 in Gill (1983), that $\sup _{t\in (0,L]}|(\tilde{G}(t)-G(t))/(1-G(t))|$ is bounded in probability. Therefore, it follows by combining (7) and (11)–(14)

$$\begin{aligned} \lim _{n\rightarrow \infty }\sup _{t\in (0,L]}P\left( \sqrt{n}|\{1-\tilde{V}(t)\} \{\tilde{G}(t)-G(t)\}|>\varepsilon \right) \le 2\exp \{-2\varepsilon ^{2}\}. \end{aligned}$$

Next, we begin to prove (15). For the distribution function G(t), the cumulative hazard function H(t) can be defined as

$$\begin{aligned} H(t)=\int _{(0,t]}\frac{F(dy,1)}{1-F_{Y}(y)}, \end{aligned}$$

and its estimator $\hat{H}_{n}(t)$ is given by

$$\begin{aligned} \hat{H}_{n}(t)=\int _{(0,t]}\frac{F_{n}(dy,1)}{1-F_{Y,n}(y)}. \end{aligned}$$

By simple calculation, we have

$$\begin{aligned} \hat{H}_{n}(t)-H(t)&=\int _{(0,t]}\left( \frac{F_{n}(dy,1)}{1-F_{Y,n}(y)}- \frac{F(dy,1)}{1-F_{Y}(y)}\right) \\&=\int _{(0,t]}\frac{F_{n}(dy,1)(1-F_{Y}(y))-F(dy,1)(1-F_{Y,n}(y))}{(1-F_{Y}(y)) (1-F_{Y,n}(y))}\\&=\int _{(0,t]}\frac{F_{n}(dy,1)-F(dy,1)}{(1-F_{Y}(y))(1-F_{Y,n}(y))}\\&\quad +\int _{(0,t]} \frac{(F(dy,1)-F_{n}(dy,1))F_{Y,n}(y)}{(1-F_{Y}(y))(1-F_{Y,n}(y))}\\&\quad +\int _{(0,t]}\frac{F_{n}(dy,1)(F_{Y}(y)-F_{Y,n}(y))}{(1-F_{Y}(y)) (1-F_{Y,n}(y))}\\&=R_{1}+R_{2}+R_{3}. \end{aligned}$$

Based on Lemma 4.2 in van der Laan (1996) and Assumption 2, we can obtain

$$\begin{aligned} \sup _{t\in (0,L]}|F_{n}(dy,1)-F(dy,1)|{\mathop {\rightarrow }\limits ^{p}} 0. \end{aligned}$$

Moreover, similar to the proof of Theorem 1 in Wang (1987), we can show that

$$\begin{aligned} \sup _{t\in (0,L]}|F_{Y,n}(y)-F_{Y}(y)|{\mathop {\rightarrow }\limits ^{p}} 0. \end{aligned}$$

It follows form Assumptions 3 and 6 that $1/((1-F_{Y}(y))(1-F_{Y,n}(y)))$ is bounded. Therefore, we have

$$\begin{aligned} \sup _{t\in (0,L]}\left| R_{i}\right| {\mathop {\rightarrow }\limits ^{p}} 0,~~~~i=1,2,3, \end{aligned}$$

and

$$\begin{aligned} \sup _{t\in (0,L]}|\hat{H}_{n}(t)-H(t)|\le \left\{ \sup _{t\in (0,L]}\left| R_{1}\right| + \sup _{t\in (0,L]}\left| R_{2}\right| +\sup _{t\in (0,L]}\left| R_{3}\right| \right\} {\mathop {\rightarrow }\limits ^{p}}0. \end{aligned}$$

Now by using Lemma 2 in Gill (1981) we get

$$\begin{aligned} \sup _{t\in (0,L]}\left| \tilde{G}(t)-G(t)\right| {\mathop {\rightarrow }\limits ^{p}} 0. \end{aligned}$$

So (15) follows. The proof of Lemma 4.1 is completed. $\square$

Proof of Theorem 2.1

Let $\omega ^{*}_{k}=\{1/n\sum _{i=1}^{n}X_{ki}u(Y_{i})G_{k}(Y_{i})\}^{2}$, $k=1,\cdots ,p$. we have

$$\begin{aligned} |\hat{\omega }_{k}-\omega _{k}|&\le |\hat{\omega }_{k}-\tilde{\omega }_{k}|+ |\tilde{\omega }_{k}-\omega _{k}|\nonumber \\&\le |\hat{\omega }_{k}-\tilde{\omega }_{k}|+ |\tilde{\omega }_{k}-\omega ^{*}_{k}| +|\omega ^{*}_{k}-\omega _{k}|. \end{aligned}$$

(16)

Combining the strong law of large numbers and Assumption 1, we can obtain that

$$\begin{aligned} \frac{1}{n}\sum _{i=1}^{n}(X_{ki})^{2}{\mathop {\longrightarrow }\limits ^{a.s.}}O(1), ~~~k=1,\cdots ,p, \end{aligned}$$

where ${\mathop {\longrightarrow }\limits ^{a.s.}}$ represents almost sure convergence. Based on the Cauchy−Schwarz inequality and the boundedness of $\tilde{G}(t)$, G(t), $\tilde{u}(t)$ and u(t), combining Assumption 4, we can obtain that

$$\begin{aligned}&\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}\tilde{u}(Y_{i})\tilde{G}(Y_{i})\right|< +\infty ,~~~~~~\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}u(Y_{i})\tilde{G} (Y_{i})\right|<+\infty ,\\&\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}u(Y_{i})G(Y_{i})\right|<+\infty ~~~~ \text {and}~~~~E\left| X_{k}u(Y)G(Y)\right| <+\infty . \end{aligned}$$

There exists positive constants $c_{1}$, $c_{2}$ and $c_{3}$ such that

$$\begin{aligned} |\hat{\omega }_{k}-\tilde{\omega }_{k}|&=\left| \left\{ \frac{1}{n}\sum _{i=1}^{n}X_{ki} \tilde{u}(Y_{i})\tilde{G}(Y_{i})\right\} ^{2}-\left\{ \frac{1}{n} \sum _{i=1}^{n}X_{ki}u(Y_{i})\tilde{G}(Y_{i})\right\} ^{2}\right| \nonumber \\&=\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}\tilde{u}(Y_{i})\tilde{G}(Y_{i})+\frac{1}{n} \sum _{i=1}^{n}X_{ki}u(Y_{i})\tilde{G}(Y_{i})\right| \nonumber \\&\quad \left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}\tilde{G}(Y_{i})\{\tilde{u}(Y_{i})-u(Y_{i})\} \right| \nonumber \\&\le c_{1}\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}\tilde{G}(Y_{i})\{\tilde{u}(Y_{i}) -u(Y_{i})\}\right| \nonumber \\&\le c_{1}\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}\{1-\tilde{G}(Y_{i})\}\{\tilde{u} (Y_{i})-u(Y_{i})\}\right| \nonumber \\&\quad +c_{1}\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}\{\tilde{u}(Y_{i})-u(Y_{i})\}\right| , \end{aligned}$$

(17)

$$\begin{aligned} |\tilde{\omega }_{k}-\omega ^{*}_{k}|&=\left| \left\{ \frac{1}{n}\sum _{i=1}^{n} X_{ki}u(Y_{i})\tilde{G}(Y_{i})\right\} ^{2}-\left\{ \frac{1}{n} \sum _{i=1}^{n}X_{ki}u(Y_{i})G(Y_{i})\right\} ^{2}\right| \nonumber \\&\le c_{2}\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}u(Y_{i})\{\tilde{G}(Y_{i}) -G(Y_{i})\}\right| \end{aligned}$$

(18)

and

$$\begin{aligned} |\omega ^{*}_{k}-\omega _{k}|&=\left| \left\{ \frac{1}{n}\sum _{i=1}^{n}X_{ki}u(Y_{i}) G(Y_{i})\right\} ^{2}-\left[ E\left\{ X_{k}u(Y)G(Y)\right\} \right] ^{2}\right| \nonumber \\&\le c_{3}\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}u(Y_{i})G(Y_{i})- E\left\{ X_{k}u(Y)G(Y)\right\} \right| . \end{aligned}$$

(19)

Based on $V(t)=1-u(t)$ and $\tilde{V}(t)=1-\tilde{u}(t)$, we have

$$\begin{aligned}&\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}\{1-\tilde{G}(Y_{i})\}\{\tilde{u}(Y_{i}) -u(Y_{i})\}\right| \\&\quad =\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}\{1-\tilde{G}(Y_{i})\}\{\tilde{V}(Y_{i}) -V(Y_{i})\}\right| \nonumber \\&\quad \le c_{4}\left( \frac{1}{n}\sum _{i=1}^{n}\left[ \{1-\tilde{G}(Y_{i})\}\{\tilde{V} (Y_{i})-V(Y_{i})\}\right] ^{2}\right) ^{1/2}\nonumber \\&\quad \le c_{4}\sup _{0\le t\le L}\left| \{1-\tilde{G}(t)\}\{\tilde{V}(t)-V(t)\} \right| \end{aligned}$$

Because the estimator $\tilde{V}(t)$ is obtained by switching the role of failure time and censoring time and combining (2), we can show that, by using (8) in Lemma 4.1 and switching the role of $\tilde{V}(t)$ and $\tilde{G}(t)$,

$$\begin{aligned}&\lim _{n\rightarrow \infty }P\left( c_{1}\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}\{1- \tilde{G}(Y_{i})\}\{\tilde{u}(Y_{i})-u(Y_{i})\}\right| \ge Mn^{-\alpha }\right) \\&\quad \le \lim _{n\rightarrow \infty }P\left( c_{1}c_{4}\sup _{0\le t\le L}\left| \{1 -\tilde{G}(t)\}\{\tilde{V}(t)-V(t)\}\right| \ge Mn^{-\alpha }\right) \\&\quad =\lim _{n\rightarrow \infty }P\left( \sup _{0\le t\le L}\sqrt{n}\left| \{1 -\tilde{G}(t)\}\{\tilde{V}(t)-V(t)\}\right| \ge (c_{1}c_{4})^{-1}Mn^{1/2- \alpha }\right) \\&\quad \le 2\exp \{-2(c_{1}c_{4})^{-2}M^{2}n^{1-2\alpha }\}. \end{aligned}$$

Without loss of generality, when n is large enough, we can show that

$$\begin{aligned}&P\left( c_{1}\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}\{1-\tilde{G}(Y_{i})\} \{\tilde{u}(Y_{i})-u(Y_{i})\}\right| \ge Mn^{-\alpha }\right) \nonumber \\&\quad \le 2\exp \{-2(c_{1}c_{4})^{-2}M^{2}n^{1-2\alpha }\}. \end{aligned}$$

(20)

Combining Assumption 6 and the Cauchy–Schwarz inequality, we have

$$\begin{aligned} \left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}\{\tilde{u}(Y_{i})-u(Y_{i})\}\right|&\le \frac{1}{\rho _{1}}\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}\{1-G(Y_{i})\}\{\tilde{u} (Y_{i})-u(Y_{i})\}\right| \\&=\frac{1}{\rho _{1}}\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}\{1-G(Y_{i})\} \{\tilde{V}(Y_{i})-V(Y_{i})\}\right| \\&\le \left\{ \frac{1}{n}\sum _{i=1}^{n}X_{ki}^{2}\frac{1}{n}\sum _{i=1}^{n} \{1-G(Y_{i})\}\{\tilde{V}(Y_{i})-V(Y_{i})\}^{2}\right\} ^{1/2}\\&\le c_{5}\left\{ \frac{1}{n}\sum _{i=1}^{n}\{1-G(Y_{i})\}\{\tilde{V}(Y_{i})- V(Y_{i})\}^{2}\right\} ^{1/2}\\&\le c_{5}\sup _{0\le t\le L}\left| \{1-G(t)\}\{\tilde{V}(t)-V(t)\}\right| , \end{aligned}$$

where $c_{5}$ is a positive constant. Therefore, based on (7) in Lemma 4.1, we can obtain that

$$\begin{aligned} P\left( c_{1}\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}\{\tilde{u}(Y_{i})-u(Y_{i})\} \right| \ge Mn^{-\alpha }\right) \le 2\exp \{-2(c_{1}c_{5})^{-2}M^{2}n^{1-2\alpha }\}. \end{aligned}$$

(21)

Similarly, there exists a positive constant $c_{6}$ such that

$$\begin{aligned} P\left( c_{2}\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}u(Y_{i})\{\tilde{G}(Y_{i})- G(Y_{i})\}\right| \ge Mn^{-\alpha }\right) \le 2\exp \{-2(c_{2}c_{6})^{-2}M^{2}n^{1-2\alpha }\}. \end{aligned}$$

(22)

Note that $\{X_{ki}u(Y_{i})G(Y_{i}):~i=1,\cdots ,n\}$ are independent identically sample form $X_{k}u(Y)G(Y)$, then, based on Hoeffding’s inequality, we have

$$\begin{aligned}&P\left( c_{3}\left| \frac{1}{n}\sum _{i=1}^{n}X_{ki}u(Y_{i})G(Y_{i})-E\left\{ X_{k}u (Y)G(Y)\right\} \right| \ge Mn^{-\alpha }\right) \nonumber \\&\quad \le 2\exp \left\{ -\frac{2(c_{3})^{-2}M^{2}n^{2-2\alpha }}{nc_{7}^{2}}\right\} =2\exp \{-2(c_{3}c_{7})^{-2}M^{2}n^{1-2\alpha }\}. \end{aligned}$$

(23)

According to (16), it is easy to show that

$$\begin{aligned}&P(|\hat{\omega }_{k}-\omega _{k}|\ge 3Mn^{-\alpha })\le P(|\hat{\omega }_{k}- \tilde{\omega }_{k}|+|\tilde{\omega }_{k}-\omega _{k}|\ge 3Mn^{-\alpha })\nonumber \\&\quad =P(|\hat{\omega }_{k}-\tilde{\omega }_{k}|+|\tilde{\omega }_{k}-\omega _{k}| \ge 3Mn^{-\alpha },|\hat{\omega }_{k}-\tilde{\omega }_{k}|\ge Mn^{-\alpha })\nonumber \\&\qquad +P(|\hat{\omega }_{k}-\tilde{\omega }_{k}|+|\tilde{\omega }_{k}-\omega _{k}| \ge 3Mn^{-\alpha }, |\hat{\omega }_{k}-\tilde{\omega }_{k}|< Mn^{-\alpha })\nonumber \\&\quad \le P(|\hat{\omega }_{k}-\tilde{\omega }_{k}|\ge Mn^{-\alpha })+P(| \tilde{\omega }_{k}-\omega _{k}|\ge 2Mn^{-\alpha })\nonumber \\&\quad \le P(|\hat{\omega }_{k}-\tilde{\omega }_{k}|\ge Mn^{-\alpha })+P(| \tilde{\omega }_{k}-\omega ^{*}_{k}|\ge Mn^{-\alpha })+P(|\omega ^{*}_{k}- \omega _{k}|\ge Mn^{-\alpha })\nonumber \\ \end{aligned}$$

(24)

Therefore, from (17)–(24), there exists a positive constant $\gamma$ such that

$$\begin{aligned} P(|\hat{\omega }_{k}-\omega _{k}|\ge 3Mn^{-\alpha })&\le 2\exp \{-2(c_{1}c_{4})^{-2} M^{2}n^{1-2\alpha }\}\nonumber \\&\quad +2\exp \{-2(c_{1}c_{5})^{-2}M^{2}n^{1-2\alpha }\}\nonumber \\&\quad +2\exp \{-2(c_{2}c_{6})^{-2}M^{2}n^{1-2\alpha }\}\nonumber \\&\quad +2\exp \{-2(c_{3}c_{7})^{-2}M^{2}n^{1-2\alpha }\}\nonumber \\&\le 8\exp \{-2\gamma ^{-2}M^{2}n^{1-2\alpha }\}. \end{aligned}$$

(25)

Immediately, we have

$$\begin{aligned} P(\max _{1\le k\le p}|\hat{\omega }_{k}-\omega _{k}|>Mn^{-\alpha })\le 8p\exp \left\{ \frac{-2\gamma ^{-2}M^{2}n^{1-2\alpha }}{9}\right\} . \end{aligned}$$

(26)

On the other hand, similar to the proof of the second part of Theorem 2.1 in Zhang et al. (2017), we can prove that

$$\begin{aligned} P(\mathscr {A}\subseteq \hat{\mathscr {A}})\ge 1-O\left( d\exp \left\{ \frac{-2\gamma ^{-2}M^{2}n^{1-2\alpha }}{9}\right\} \right) , \end{aligned}$$

where $d=|\mathscr {A}|$ is the cardinality of $\mathscr {A}$. The proof of Theorem 2.1 is completed. $\square$

Proof of Theorem 2.2

Let $\eta =\min \limits _{k\in \mathscr {A}}\{\omega _{k}\}-\max \limits _{k\not \in \mathscr {A}}\{\omega _{k}\}$. It follows from Assumption 5 that $\eta >0$. Similar to the proof of (24) and (25) in Theorem 2.1, we can show that

$$\begin{aligned}&P\left( \min _{k\in \mathscr {A}}\hat{\omega }_{k}<\max _{k\not \in \mathscr {A}} \hat{\omega }_{k}\right) \nonumber \\&\quad =P\left( (\max _{k\not \in \mathscr {A}}\hat{\omega }_{k} -\max _{k\not \in \mathscr {A}}\omega _{k})-(\min _{k\in \mathscr {A}}\hat{\omega }_{k}- \min _{k\in \mathscr {A}}\omega _{k})>\eta \right) \nonumber \\&\quad \le P\left( \max _{k\not \in \mathscr {A}}|\hat{\omega }_{k}-\omega _{k}|+\min _{k\in \mathscr {A}}|\hat{\omega }_{k}-\omega _{k}|>\eta \right) \nonumber \\&\quad =P\left( \max _{k\not \in \mathscr {A}}|\hat{\omega }_{k}-\omega _{k}|+\min _{k\in \mathscr {A}}|\hat{\omega }_{k}-\omega _{k}|>\eta , \max _{k\not \in \mathscr {A}}| \hat{\omega }^{*}_{k}-\omega _{k}|\ge \frac{\eta }{2}\right) \nonumber \\&\qquad +P\left( \max _{k\not \in \mathscr {A}}|\hat{\omega }_{k}-\omega _{k}|+\min _{k\in \mathscr {A}}|\hat{\omega }_{k}-\omega _{k}|>\eta , \max _{k\not \in \mathscr {A}}| \hat{\omega }^{*}_{k}-\omega _{k}|<\frac{\eta }{2}\right) \nonumber \\&\quad \le P\left( \max _{k\not \in \mathscr {A}}|\hat{\omega }_{k}-\omega _{k}|\ge \frac{\eta }{2}\right) +P\left( \min _{k\in \mathscr {A}}|\hat{\omega }_{k}- \omega _{k}|>\frac{\eta }{2}\right) \end{aligned}$$

(27)

Therefore, combining (26) and (27), when n is large enough, we can obtain that

$$\begin{aligned} P\left( \min _{k\in \mathscr {A}}\hat{\omega }_{k}<\max _{k\not \in \mathscr {A}} \hat{\omega }_{k}\right)&\le P\left( \max _{k\not \in \mathscr {A}}|\hat{\omega }_{k}-\omega _{k}|\ge Mn^{-\alpha }\right) \\&\quad +P\left( \min _{k\in \mathscr {A}}|\hat{\omega }_{k}- \omega _{k}|>Mn^{-\alpha }\right) \\&\le 2P(\max _{1\le k\le p}|\hat{\omega }_{k}-\omega _{k}|\ge Mn^{-\alpha })\\&\le O\left( p\exp \left\{ \frac{-2\gamma ^{-2}M^{2}n^{1-2\alpha }}{9}\right\} \right) , \end{aligned}$$

and

$$\begin{aligned} P(\max _{k\not \in \mathcal {A}}\hat{\omega }^{*}_{k}<\min _{k\in \mathcal {A}} \hat{\omega }^{*}_{k})\ge 1-O\left( p\exp \left\{ \frac{-2\gamma ^{-2}M^{2}n^{1-2\alpha }}{9}\right\} \right) . \end{aligned}$$

Thus, the proof of Theorem 2.2 is completed $\square$