Conditional characteristic feature screening for massive imbalanced data

Abstract

Using conditional characteristic function as a screening index, a new model-free screening procedure is proposed to deal with variable screening problems in large-scale high-dimensional imbalanced data analysis. For binary response, our results show that the screening index under full data is proportional to the screening index under case–control sampling, an important sampling property for imbalanced data. This conclusion implies that we can apply this screening method to imbalanced data. Surely, the most appealing feature of the screening index is that it can be expressed as a simple linear combination of two first-order moments, so it is computationally simple. In addition, we successfully extend this method to multiple response. The theoretical properties are established under regularity conditions. To compare the performance of our method with its competitors, extensive simulations are conducted, which shows that the proposed procedure performs well in both the linear and nonlinear models. Finally, a real data analysis is investigated to further illustrate the effectiveness of the new method.

Appendix: Proof of main results

We first prove Lemma 2, which is the basis of our ideas.

Firstly, the proof of formula (a) of Lemma 2.

$$\begin{aligned}&\phi _{{X_k}|Y=1}(t)-\phi _{X_k}(t) \\&\quad =E(e^{itX_k}| Y=1)-E(E(e^{itX_k}|Y))\\&\quad =E(e^{itX_k}| Y=1)-(P_1E(e^{itX_k}|Y=1)+P_0E(e^{itX_k}| Y=0))\\&\quad =P_0(\phi _{{X_k}|Y=1}(t)-\phi _{{X_k}|Y=0}(t)). \end{aligned}$$

Similarly,

$$\begin{aligned} \phi _{{X_k}|Y=0}(t)-\phi _{X_k}(t)=P_1(\phi _{{X_k}|Y=0}(t)-\phi _{{X_k}|Y=1}(t)). \end{aligned}$$

Secondly, the proof of the formula (b) of Lemma 2.

$$\begin{aligned} \tau _k= & {} \sum _{y=0,1} p_y\int _{\mathbb {R}}\Vert \phi _{{X_k}|Y=y}(t)-\phi _{X_k}(t)\Vert ^2w(t) dt\\= & {} P_1P_0^2\int _{\mathbb {R}}\Vert \phi _{{X_k}|Y=1}(t)-\phi _{{X_k}|Y=0}(t)\Vert ^2w(t)dt\\&\quad +P_0P_1^2\int _{\mathbb {R}}\Vert \phi _{{X_k}|Y=0}(t)-\phi _{{X_k}|Y=0}(t)\Vert ^2w(t)dt\\= & {} P_0P_1\int _{\mathbb {R}}\Vert \phi _{{X_k}|Y=1}(t)-\phi _{{X_k}|Y=0}(t)\Vert ^2w(t)dt. \end{aligned}$$

$\square$

Proof of Theorem 1

$$\begin{aligned} \tau _k^*&=P_0^*P_1^*\int _{\mathbb {R}}\Vert \phi _{{X_k}^*|Y^*=1}(t)-\phi _{{X_k}^*|Y^*=0}(t))\Vert ^2w(t)dt \nonumber \\&=P_0^*P_1^*\int _{\mathbb {R}}\Vert \phi _{{X_k}|Y=1}(t)-\phi _{{X_k}|Y=0}(t))\Vert ^2w(t)dt \nonumber \\&=\frac{P_0^*P_1^*}{P_0P_1}P_0P_1\int _{\mathbb {R}}\Vert \phi _{{X_k}|Y=1}(t)-\phi _{{X_k}|Y=0}(t))\Vert ^2w(t)dt \nonumber \\&=a\tau _k ,\nonumber \end{aligned}$$

where $a=\frac{P_0^*P_1^*}{P_0P_1}$, $P_0^*$ and $P_1^*$ are the probabilities of response variable $Y^*$ respectively valued at $Y^*=0$ and $Y^*=1$ under case–control sampling setting. The second equality above holds due to the assumption of case–control sampling design. When the population and sampling are fixed, a is a constant. $\square$

Before proving Theorem 2, we introduce two important inequalities will be used repeatedly in the proof of Theorem 2.

Inequality 1. Let $\mu =E(Y)$. If $P(a \le Y\le b)=1$, then

$$\begin{aligned} E[exp\{s(Y-\mu )\}] \le exp\{s^2(b-a)^2/8\} \quad \text {for any} \quad s>0. \end{aligned}$$

Inequality 2. Suppose that $h(Y_1,\cdots ,Y_m)$ is a kernel of the U statistics $U_n$, and let $\theta =E\{h(Y_1,\cdots ,Y_m)\}$. If $a\le h(Y_1,\cdots ,Y_m) \le b$, then, for any $t>0$ and $n\ge m$,

$$\begin{aligned} P(U_n-\theta \ge t) \le exp\{-2[n/m] t^2/(b-a)^2\}, \end{aligned}$$

where [n/m] denotes the integer part of n/m. Due to the symmetry of U statistics, we further have that

$$\begin{aligned} P(\left| U_n-\theta \ge t\right| ) \le exp\{-2[n/m] t^2/(b-a)^2\}, \end{aligned}$$

Proof of Theorem 2

The events satisfy $\{\left| {\tilde{\tau }}_{k}^*-\tau _{k}^*\right| >\varepsilon \}$ $\subseteq$ $\{\left| {\hat{\tau }}_{k}^*-\tau _{k}^*\right| >\varepsilon \}$. To prove sure screening property of ${\tilde{\tau }}_{k}^*$, we just need to prove that ${\hat{\tau }}_{k}^*$ also has the same properties. The ranking statistics ${\widehat{\tau }}_k^*$ can be written as

$$\begin{aligned} {\widehat{\tau }}_k^*={\hat{I}}_{k, 1}+{\hat{I}}_{k, 2}, \end{aligned}$$

where

$$\begin{aligned} \begin{array}{l} {\hat{I}}_{k, 1}=n^{-2} \sum _{i, j=1}^{n}\left| X_{ik}-X_{jk}\right| , \\ {\hat{I}}_{k, 2}=\sum _{y=1}^{R} {\hat{J}}_{y, k 2} ~ \text{ for } {\hat{J}}_{y, k 2}=\frac{n^{-2} \sum _{i, j=1}^{n}\left| X_{ik}-X_{jk}\right| {\mathbf {1}}\left( Y_{i}=y\right) {\mathbf {1}}\left( Y_{j}=y\right) }{n^{-1} \sum _{l=1}^{n} {\mathbf {1}}\left( Y_{l}=y\right) }. \end{array} \end{aligned}$$

Correspondingly, the population form $\tau _k^*$ can also be written as $\tau _{k}^*=I_{k, 1}+$ $I_{k, 2}$, where

$$\begin{aligned} \begin{array}{l} I_{k, 1}=E\left| X_{k}-X_{k}^{\prime }\right| , \\ I_{k, 2}=\sum _{\nu =1}^{R} J_{y, k 2} ~ \text{ for } J_{y, k 2}=\frac{E\left| X_{k}-X_{k}^{\prime }\right| {\mathbf {1}}(Y=y) {\mathbf {1}}\left( Y^{\prime }=y\right) }{E {\mathbf {1}}(Y=y)}. \end{array} \end{aligned}$$

We aim to show the uniform consistency of the ${\widehat{\tau }}_{k}^*$ under regularity conditions.

Firstly, we deal with the term ${\hat{I}}_{k, 1}$. Define

$$\begin{aligned} {\hat{I}}_{k 1}^{\star }=\frac{2}{n(n-1)} \sum _{i<j} h\left( X_{i k}, X_{j k}\right) , \end{aligned}$$

where $h\left( X_{i k},X_{j k}\right) =\frac{1}{2}\left| X_{i k}-X_{j k}\right| +\frac{1}{2}\left| X_{j k}-X_{i k}\right|$. Then it has that ${\hat{I}}_{k 1}^{\star }(n-1) / n={\hat{I}}_{k 1}$. ${\hat{I}}_{k 1}^{\star }$ is a usual U statistics with kernel h, thus we shall establish the uniform consistency of ${\hat{I}}_{k 1}^{\star }$ by invoking the theory of U statistics (Serfling, 1980, chap. 5). Condition (C1) states that $I_{k 1}$ is uniformly bounded in p, that is, $\sup _{p} \max _{1 \le k \le p} I_{k 1}<\infty .$ For any give $\varepsilon >0$, take n large enough such that $I_{k 1} / n<\varepsilon$, then it can be showed that

$$\begin{aligned} \begin{aligned}&\mathrm {P}\left( \left| {\hat{I}}_{k 1}-I_{k 1}\right| \ge 2 \varepsilon \right) \\&\quad \le ~ \mathrm {P}\left\{ \left| {\hat{I}}_{k 1}^{\star }-I_{k 1}\right| (n-1) / n \ge 2 \varepsilon -I_{k 1} / n\right\} \\&\quad \le ~ \mathrm {P}\left( \left| {\hat{I}}_{k 1}^{\star }-I_{k 1}\right| \ge \varepsilon \right) . \end{aligned} \end{aligned}$$

(a.1)

The U statistics ${\hat{I}}_{k 1}^{\star }$ can be written as

$$\begin{aligned} \begin{aligned} {\hat{I}}_{k 1}^{\star }=&~ \frac{1}{2} \frac{2}{n(n-1)} \sum _{i<j} h\left( X_{i k}, X_{j k}\right) {\mathbf {1}}\left\{ h\left( X_{i k},X_{j k}\right) \le M\right\} \\&+\frac{1}{2} \frac{2}{n(n-1)} \sum _{i<j} h\left( X_{i k},X_{j k}\right) {\mathbf {1}}\left\{ h\left( X_{i k},X_{j k}\right) >M\right\} \\ =&~ {\hat{I}}_{k 11}^{\star }+{\hat{I}}_{k 12}^{\star }, \end{aligned} \end{aligned}$$

where M will be specified later.

Accordingly, we decompose $I_{k 1}$ into two parts:

$$\begin{aligned} \begin{aligned} I_{k 1}=&~ E\left[ h\left( X_{i k},X_{j k}\right) {\mathbf {1}}\left\{ h\left( X_{i k},X_{j k}\right) \le M\right\} \right] \\&+E\left[ h\left( X_{i k},X_{j k}\right) {\mathbf {1}}\left\{ h\left( X_{i k},X_{j k}\right) >M\right\} \right] \\ =&~ I_{k 11}+I_{k 12}. \end{aligned} \end{aligned}$$

Obviously, ${\hat{I}}_{k 11}^{\star }$ and ${\hat{I}}_{k 12}^{\star }$ are unbiased estimators of $I_{k 11}$ and $I_{k 12}$. We deal with the consistency of ${\hat{I}}_{k 11}^{\star }$ first. With the Markov’s inequality,for any $t>0$, we obtain that

$$\begin{aligned} \begin{aligned}&\mathrm {P}\left( {\hat{I}}_{k 11}^{\star }-I_{k 11} \ge \varepsilon \right) \\&\quad = ~ \mathrm {P}\left( exp(t{\hat{I}}_{k 11}^{\star }) \ge exp(tI_{k 11}) exp(t\varepsilon )\right) \\&\quad \le ~ \exp (-t \varepsilon ) \exp \left( -t I_{k 11}\right) E\left[ \exp \left( t {\hat{I}}_{k 11}^{\star }\right) \right] . \end{aligned} \end{aligned}$$

Serfling (2009) showed that any U statistics can be represented as an average of of i.i.d random variables, i.e., ${\hat{I}}_{k 11}^{\star }=(n !)^{-1} \sum _{n !} \Omega _{1}\left( X_{1 k},\cdots ,X_{n k}\right)$ where $\sum _{n !}$ denotes the summation over all possible permutations of $(1, \cdots , n)$, and each $\Omega _{1}\left( X_{1 k},X_{n k} \right)$ is an average of $m=[n / 2]$ i.i.d. random variables (i.e., $\Omega _{1}=m^{-1} \sum _{l} h^{(l)} {\mathbf {1}}\left\{ h^{(l)} \le M\right\} .$ Since the exponential function is convex, it follows from the Jensen’s inequality that, for $0<t \le 2 s_{0}$,

$$\begin{aligned} \begin{aligned} E\left\{ \exp \left( t {\hat{I}}_{k 11}^{\star }\right) \right\}&=E\left[ \exp \left\{ t(n !)^{-1} \sum _{n !} \Omega _{1}\left( X_{1 k},\cdots ,X_{n k}\right) \right\} \right] \\&\le (n !)^{-1} \sum _{n !} E\left[ \exp \left\{ t \Omega _{1}\left( X_{1 k},\cdots ,X_{n k}\right) \right\} \right] \\&=E^{m}\left\{ \exp \left( m^{-1} t h^{(l)} 1\left\{ h^{(l)} \le M\right\} \right) \right\} , \end{aligned} \end{aligned}$$

which together with Inequality 2, entails immediately that

$$\begin{aligned} \begin{aligned}&\mathrm {P}\left( {\hat{I}}_{k 11}^{\star }-I_{k 11} \ge \varepsilon \right) \\&\quad \le \exp (-t \varepsilon ) E^{m}\left\{ \exp \left( m^{-1} t\left[ h^{(l)} {\mathbf {1}}\left\{ h^{(l)} \le M\right\} -I_{k 11}\right] \right) \right\} \\&\quad \le \exp \left\{ -t \varepsilon +M^{2} t^{2} /(8 m)\right\} . \end{aligned} \end{aligned}$$

By choosing $t=4 \varepsilon m / M^{2}$, we have $\mathrm {P}\left( {\hat{I}}_{k 11}^{\star }-I_{k 11} \ge \varepsilon \right) \le \exp \left( -2 \varepsilon ^{2} m / M^{2}\right)$. Therefore, by the symmetry of U statistics, we can obtain easily that

$$\begin{aligned} \mathrm {P}\left( \left| {\hat{I}}_{k 11}^{\star }-I_{k 11}\right| \ge \varepsilon \right) \le 2 \exp \left( -2 \varepsilon ^{2} m / M^{2}\right) . \end{aligned}$$

(a.2)

Next, we show the consistency of ${\hat{I}}_{k 12}^{\star }.$ With Cauchy-Schwartz and Markov’s inequalities, for any $s^{\prime }>0$,

$$\begin{aligned} \begin{aligned} I_{k 12}^{2}&\le E\left\{ h^{2}\left( X_{i k},X_{j k}\right) \right\} \mathrm {P}\left\{ h\left( X_{i k},X_{j k}\right) >M\right\} \\&\le E\left\{ h^{2}\left( X_{i k},X_{j k}\right) \right\} E\left\{ \exp \left[ s^{\prime } h\left( X_{i k},X_{j k}\right) \right] \right\} \exp \left( -s^{\prime } M\right) . \end{aligned} \end{aligned}$$

We have

$$\begin{aligned} \begin{aligned} h\left( X_{i k},X_{j k}\right) \le \left| X_{i k}\right| +\left| X_{j k}\right| , \end{aligned} \end{aligned}$$

together with the Cauchy-Schwartz inequality, yields that

$$\begin{aligned} E\left\{ \exp \left[ sh\left( X_{i k},X_{j k}\right) \right] \right\} \le E\left\{ \exp \left[ s\left( \left| X_{i k}\right| +\left| X_{j k}\right| \right) \right] \right\} . \end{aligned}$$

If we choose $M=c n^{\gamma }$ for $0<\gamma <1 / 2-\kappa$, then by condition $(\mathrm {C} 1), I_{k 12} \le \varepsilon / 2$ when n is sufficiently large. Consequently,

$$\begin{aligned} \mathrm {P}\left( \left| {\hat{I}}_{k 12}^{\star }-I_{k 12}\right| \ge \varepsilon \right) \le \mathrm {P}\left( \left| {\hat{I}}_{k 12}^{\star }\right| >\varepsilon / 2\right) . \end{aligned}$$

(a.3)

It remains to bound the probability $\mathrm {P}\left( \left| {\hat{I}}_{k 12}^{\star }\right| >\varepsilon / 2\right)$. Observe that

$$\begin{aligned} \left\{ \left| {\hat{I}}_{k 12}^{\star }\right|>\varepsilon / 2\right\} \subseteq \left\{ \left| X_{i k}\right| >M / 2, \text{ for } \text{ some } 1 \le i \le n\right\} . \end{aligned}$$

(a.4)

By invoking condition (C1) and Markov’s inequality for $s>0$, there must be some constant C such that

$$\begin{aligned} \mathrm {P}\left( \left| X_{i k}\right| >M / 2\right) \le 2 C \exp (-s M / 2). \end{aligned}$$

Consequently,

$$\begin{aligned} \max _{1 \le k \le p} \mathrm {P}\left( \left| {\hat{I}}_{k 12}^{\star }\right|>\varepsilon / 2\right) \le n \max _{1 \le k \le p} \mathrm {P}\left( \left| X_{i k}\right| >M / 2\right) \le 2 n C \exp (-s M / 2). \end{aligned}$$

(a.5)

Combing the results (a.1), (a.2), (a.3) and (a.5) and $M=c n^{\gamma }$, it is easily obtained that

$$\begin{aligned} \max _{1 \le k \le p} \mathrm {P}\left( \left| {\hat{I}}_{k 1}^{\star }-I_{k 1}\right| >4 \varepsilon \right) \le 2 \exp \left( -\varepsilon ^{2} n^{1-2 \gamma }\right) +2 n C \exp \left( -s n^{\gamma } / 2\right) . \end{aligned}$$

(a.6)

Thus, we have

$$\begin{aligned} \mathrm {P}\left( \left| {\hat{I}}_{k 1}-I_{k 1}\right| >4 \varepsilon \right) \le 2 \exp \left( -\varepsilon ^{2} n^{1-2 \gamma }\right) +2 n C \exp \left( -s n^{\gamma } / 2\right) . \end{aligned}$$

(a.7)

It remains to prove the uniform consistency of ${\hat{I}}_{k, 2} .$ We still first deal with the term $J_{y, k 2}=E\left| X_{k}-X_{k}^{\prime }\right| {\mathbf {1}}(Y=y) {\mathbf {1}}\left( Y^{\prime }=y\right) / p_{y} .$ Let ${\hat{J}}_{y, k 2}=\hat{{\bar{J}}}_{y, k 2} / {\hat{p}}_{y}$, where

$$\begin{aligned} \hat{{\bar{J}}}_{y, k 2}=n^{-2} \sum _{i \ne j}\left| X_{i k}-X_{j k}\right| {\mathbf {1}}\left( Y_{i}=y\right) {\mathbf {1}}\left( Y_{j}=y\right) \text{ and } {\hat{p}}_{y}=n^{-1} \sum _{i=1}^{n} {\mathbf {1}}\left( Y_{i}=y\right) . \end{aligned}$$

Accordingly, $J_{y, k 2}$ can be decomposed into $J_{y, k 2}={\bar{J}}_{y, k 2} / p_{y}$ where ${\bar{J}}_{y, k 2}=$ $E\left| X_{k}-X_{k}^{\prime }\right| {\mathbf {1}}(Y=y) {\mathbf {1}}\left( Y^{\prime }=y\right)$. Following the argument for proving (a.7), we get that

$$\begin{aligned} \mathrm {P}\left( \left| \hat{{\bar{J}}}_{y, k 2}-{\bar{J}}_{y, k 2}\right| >4 \varepsilon \right) \le 2 \exp \left( -\varepsilon ^{2} n^{1-2 \gamma }\right) +2 n C \exp \left( -s n^{\gamma } / 2\right) . \end{aligned}$$

(a.8)

As a result, it has that

$$\begin{aligned} \begin{aligned}&\mathrm {P}\left( \left| {\hat{J}}_{y, k 2}-J_{y, k 2}\right| \ge 2 \varepsilon \right) \\&\quad \le ~ \mathrm {P}\left( \left| \frac{\hat{{\bar{J}}}_{y, k 2}-{\bar{J}}_{y, k 2}}{{\hat{p}}_{y}}-\frac{{\bar{J}}_{y, k 2}\left( p_{y}-{\hat{p}}_{y}\right) }{{\hat{p}}_{y} p_{y}}\right| \ge 2 \varepsilon \right) \\&\quad \le ~ \mathrm {P}\left( \left| \frac{\hat{{\bar{J}}}_{y, k 2}-{\bar{J}}_{y, k 2}}{{\hat{p}}_{y}}\right| \ge \varepsilon \right) +\mathrm {P}\left( \left| \frac{{\bar{J}}_{y, k 2}\left( p_{y}-{\hat{p}}_{y}\right) }{{\hat{p}}_{y} p_{y}}\right| \ge \varepsilon \right) . \end{aligned} \end{aligned}$$

(a.9)

We first deal with the first term in (a.9). Under condition (C2), we have

$$\begin{aligned} \begin{aligned}&\mathrm {P}\left( \left| \frac{\hat{{\bar{J}}}_{y, k 2}-{\bar{J}}_{y, k 2}}{{\hat{p}}_{y}}\right| \ge \varepsilon \right) \\&\quad \le ~ \mathrm {P}\left( \left| \frac{\hat{{\bar{J}}}_{y, k 2}-{\bar{J}}_{y, k 2}}{{\hat{p}}_{y}}\right| \ge \varepsilon , {\hat{p}}_{y} \ge \frac{c_{1}}{2 R}\right) +\mathrm {P}\left( {\hat{p}}_{y}<\frac{c_{1}}{2 R}\right) \\&\quad \le ~ \mathrm {P}\left( \left| \hat{{\bar{J}}}_{y, k 2}-{\bar{J}}_{y, k 2}\right| \ge \frac{c_{1} \varepsilon }{2 R}\right) +\mathrm {P}\left( \left| {\hat{p}}_{y}-p_{y}\right| \ge \frac{c_{1} \varepsilon }{2 R}\right) \\&\quad \le ~ 2 \exp \left( -\frac{c_{1}^{2} \varepsilon ^{2}}{32 R^{2}} n^{1-2 \gamma }\right) +2 n C \exp \left( -\frac{s n^{\gamma }}{2}\right) +2 \exp \left( -2 n \frac{c_{1}^{2} \varepsilon ^{2}}{4 R^{2}}\right) . \end{aligned} \end{aligned}$$

(a.10)

The last inequality holds because of (a.8) and Inequality 1. We next deal with the second term in (a.9),

$$\begin{aligned} \begin{aligned}&\mathrm {P}\left( \left| \frac{{\bar{J}}_{y, k 2}\left( p_{y}-{\hat{p}}_{y}\right) }{{\hat{p}}_{y} p_{y}}\right| \ge \varepsilon \right) \\&\quad \le ~\mathrm {P}\left( \left| \frac{{\bar{J}}_{y, k 2}\left( p_{y}-{\hat{p}}_{y}\right) }{{\hat{p}}_{y} p_{y}}\right| \ge \varepsilon , {\hat{p}}_{y} \ge \frac{c_{1}}{2 R}\right) +\mathrm {P}\left( {\hat{p}}_{y}<\frac{c_{1}}{2 R}\right) \\&\quad \le ~ \mathrm {P}\left( \left| {\bar{J}}_{y, k 2}\left( p_{y}-{\hat{p}}_{y}\right) \right| \ge \frac{c_{1}^{2} \varepsilon }{2 R^{2}}\right) +\mathrm {P}\left( \left| {\hat{p}}_{y}-p_{y}\right| \ge \frac{c_{1} \varepsilon }{2 R}\right) \\&\quad \le ~ \mathrm {P}\left( 2 E\left| X_{k}\right| \cdot \left| p_{y}-{\hat{p}}_{y}\right| \ge \frac{c_{1}^{2} \varepsilon }{2 R^{2}}\right) +\mathrm {P}\left( \left| {\hat{p}}_{y}-p_{y}\right| \ge \frac{c_{1} \varepsilon }{2 R}\right) \\&\quad \le ~ \mathrm {P}\left( \left| p_{y}-{\hat{p}}_{y}\right| \ge \frac{c_{1}^{2} \varepsilon }{4 C_{0} R^{2}}\right) +2 \exp \left( -2 n \frac{c_{1}^{2} \varepsilon ^{2}}{4 R^{2}}\right) \\&\quad \le ~ 2 \exp \left( -2 n \frac{c_{1}^{4} \varepsilon ^{2}}{16 C_{0}^{2} R^{4}}\right) +2 \exp \left( -2 n \frac{c_{1}^{2} \varepsilon ^{2}}{4 R^{2}}\right) \end{aligned}. \end{aligned}$$

This, together with (a.10), we have that

$$\begin{aligned} \begin{aligned} \mathrm {P}\left( \left| {\hat{I}}_{k 2}-I_{k 2}\right| >4 \varepsilon \right) \le&~R\left\{ \exp \left( -\frac{c_{1}^{2} \varepsilon ^{2}}{22 R^{2}} n^{1-2 \gamma }\right) +2 n C \exp \left( -\frac{s n^{\gamma }}{2}\right) \right. \\&\left. +4 \exp \left( -2 n \frac{c_{1}^{2} \varepsilon ^{2}}{4 R^{2}}\right) +2 \exp \left( -2 n \frac{c_{1}^{4} \varepsilon ^{2}}{16 C_{0}^{2} R^{4}}\right) \right\} . \end{aligned} \end{aligned}$$

(a.11)

Combining (a.7), (a.11) and the Bonferroni’s inequality, we can easily get that

$$\begin{aligned} \begin{aligned}&\mathrm {P}\left( |({\hat{I}}_{k, 1}+{\hat{I}}_{k, 2})-I_{k, 1}-I_{k, 2}| \ge \varepsilon \right) \\&\quad \le ~ \mathrm {P}\left( \left| {\hat{I}}_{k, 1}-I_{k, 1}\right| \ge \varepsilon / 2\right) +\mathrm {P}\left( \left| {\hat{I}}_{k, 2}-I_{k, 2}\right| \ge \varepsilon / 2\right) \\&\quad = ~ O\left\{ n^{\kappa } \exp \left( -C_{1} \varepsilon ^{2} n^{1-2 \gamma -\kappa }\right) +n^{1+\kappa } \exp \left( -C_{2} n^{\gamma }\right) \right\} , \end{aligned} \end{aligned}$$

for some positive constants $C_{1}$ and $C_{2}.$ The convergence rate of ${\widehat{\tau }}_{k}^*$ is now achieved by

$$\begin{aligned} \begin{aligned}&\mathrm {P}\left\{ \max _{1 \le k \le p}\left| {\widehat{\tau }}_{k}^*-\tau _{k}\right| \ge \varepsilon \right\} \\&\quad \le ~ p \mathrm {P}\left\{ \left| {\widehat{\tau }}_{k}^*-\tau _{k}\right| \ge \varepsilon \right\} \\&\quad = ~ p O\left\{ n^{\kappa } \exp \left( -c_{1} \varepsilon ^{2} n^{1-2 \gamma -2 \kappa } / 2\right) +n^{1+\kappa } \exp \left( -C_{2} n^{\gamma }\right) \right\} . \end{aligned} \end{aligned}$$

Let $\varepsilon =c n^{-\varsigma }$, where $\varsigma$ satisfies that $0 \le \varsigma <1 / 2$. We thus have

$$\begin{aligned} \mathrm {P}\left\{ \max _{1 \le k \le p}\left| {\widehat{\tau }}_{k}^*-\tau _{k}\right| \ge c n^{-\varsigma }\right\} =p O\left\{ n^{\kappa } \exp \left( -C_{1} n^{1-2 \gamma -2 \kappa -2 \varsigma }\right) +n^{1+\kappa } \exp \left( -C_{2} n^{\gamma }\right) \right\} . \end{aligned}$$

Now, we deal with the second part of Theorem 2. If ${\mathcal {A}} \subsetneq \hat{{\mathcal {A}}}$, then there must be some $k \in {\mathcal {A}}$ such that ${\widehat{\tau }}_{k}^*<c n^{-\varsigma }$. It follows from Condition (C3) that $\left| {\widehat{\tau }}_{k}^*-\tau _{k}\right| >c n^{-\varsigma }$, for some $k \in {\mathcal {A}}$, indicating that the events satisfy $\{{\mathcal {A}} \subsetneq \hat{{\mathcal {A}}}\} \subseteq \left\{ \left| {\widehat{\tau }}_{k}^*-\tau _{k}\right| >c n^{-\varsigma }\right.$, for some $\left. k \in {\mathcal {A}}\right\} .$ Hence $D_{n}=\left\{ \max _{k \in {\mathcal {A}}} \mid {\widehat{\tau }}_{k}^*-\right.$ $\left. \tau _{k} \mid \le c n^{-\varsigma }\right\} \subset \{{\mathcal {A}} \subset \hat{{\mathcal {A}}}\}.$ Consequently,

$$\begin{aligned} \begin{aligned} \mathrm {P}\{{\mathcal {A}} \subset \hat{{\mathcal {A}}}\}&=\mathrm {P}\left\{ D_{n}\right\} =1-\mathrm {P}\left\{ D_{n}^{c}\right\} =1-\mathrm {P}\left\{ \min _{k \in {\mathcal {A}}}\left| {\hat{\tau }}_{k}^*-\tau _{k}\right| \ge c n^{\varsigma }\right\} \\&=1-s_{n} \mathrm {P}\left\{ \left| {\widehat{\tau }}_{k}^*-\tau _{k}\right| \ge c n^{-\varsigma }\right\} \\&\ge 1-s_{n} O\left\{ n^{\kappa } \exp \left( -C_{1} n^{1-2 \gamma -2 \kappa -2 \varsigma }\right) +n^{1+\kappa } \exp \left( -C_{2} n^{\gamma }\right) \right\} . \end{aligned} \end{aligned}$$

where $s_{n}$ is the cardinality of ${\mathcal {A}}$. This proves the second result in Theorem 2. $\square$

Proof of Theorem 3

Define

$$\begin{aligned} \delta =\min _{j \in {\mathcal {A}}} \tau _{k}^*-\max _{j \in {\mathcal {I}}} \tau _{k}^*. \end{aligned}$$

Then, we have

$$\begin{aligned} \begin{aligned}&\mathrm {P}\left( \min _{k \in {\mathcal {A}}} {\tilde{\tau }}_{k}^* \le \max _{k \in {\mathcal {I}}} {\tilde{\tau }}_{k}^*\right) \\&\quad = ~\mathrm {P}\left( \min _{k \in {\mathcal {A}}} {\tilde{\tau }}_{k}^*-\min _{k \in {\mathcal {A}}} \tau _{k}^*+\delta \le \max _{k \in {\mathcal {I}}} {\tilde{\tau }}_{k}^*-\max _{j \in {\mathcal {I}}} \tau _{k}^*\right) \\&\quad \le ~ \mathrm {P}\left( \max _{j \in {\mathcal {I}}}\left| {\tilde{\tau }}_{k}^*-\tau _{k}^*\right| +\max _{k \in {\mathcal {A}}}\left| {\tilde{\tau }}_{k}^*-\tau _{k}^*\right| \ge \delta \right) \\&\quad \le ~ \mathrm {P}\left( \max _{1 \le k \le p}\left| {\tilde{\tau }}_{k}^*-\tau _{k}^*\right| \ge \delta / 2\right) \le \sum _{k=1}^{p} \mathrm {P}\left( \left| {\tilde{\tau }}_{k}^*-\tau _{k}^*\right| \ge \delta / 2\right) \\&\quad \le ~ p O\left\{ n^{\kappa } \exp \left( -c_{1} \delta ^{2} n^{1-2 \gamma -2 \kappa } / 8\right) +n^{1+\kappa } \exp \left( -C_{2} n^{\gamma }\right) \right\} . \end{aligned} \end{aligned}$$

The last term goes to 0 under condition (C2) as $n \rightarrow \infty$. $\square$