CLT for integrated square error of density estimators with censoring indicators missing at random

Abstract

A popular stochastic measure of the distance between the density of the lifetimes and its estimator is the integrated square error (ISE) and Hellinger distance (HD). In this paper, we focus on the right-censored model when the censoring indicators are missing at random. Based on two density estimators defined by Wang et al.(J Multivar Anal 100:835–850, 2009), and another new kernel estimator of the density, we established the asymptotic normality of the ISE and HD for the proposed estimators. In addition, the uniformly strongly consistency of the new kernel estimator of the density is discussed. Also, a simulation study is conducted to compare finite-sample performance of the proposed estimators.

Appendix

In this section, we give some preliminary Lemmas, which have been used in Sect. 4.

Lemma 5.1

(Peter Hall (1984), Theorem1, p. 3) Let $X_1,\ldots ,X_n$ be i.i.d. random variables. Assume $\widetilde{H}_n(x,y)$ is symmetric, $E[\widetilde{H}_n(X_1,X_2)|X_1]=0$ almost surely and $E\widetilde{H}_n^2(X_1,X_2)<\infty$ for each n. Define $\widetilde{G}_n(x,y)=E[\widetilde{H}_n(X_1,x)\widetilde{H}_n(X_1,y)]$. If

$$\begin{aligned}{}[E\widetilde{G}_n^2(X_1,X_2)+n^{-1}E\widetilde{H}_n^2(X_1,X_2)]/[E\widetilde{H}_n^2(X_1,X_2)]^2\rightarrow 0, \end{aligned}$$

then $U_n=\sum _{1\le i<j \le n}\widetilde{H}_n(X_i,X_j)$ is asymptotically normally distributed with zero mean and variance given by $\frac{1}{2}n^2E\widetilde{H}_n^2(X_1,X_2)$.

Lemma 5.2

Let $V_n(X_i)=\frac{1}{nb_{n}}\sum _{j=1}^n\xi _j\Omega (\frac{X_i-X_j}{b_{n}})$ and $V(X_i)=\pi (X_i)h(X_i)$. If (A2), (A4), (A5) and (A7) hold, then

• (a) $\max _{1\le i\le n}|V_n(X_i)-V(X_i)|I(X_i\le \tau )=O(\gamma _{n})~a.s.$

• (b) $\max _{1\le i\le n}|\widehat{m}_n(X_i)-m(X_i)|I(X_i\le \tau )=O(\gamma _{n})~a.s.$

• (c) $\max _{1\le i\le n}|\widehat{G}_n(X_i-)-G(X_i-)|I(X_i\le \tau )=O(\gamma _{n})~a.s.$

• (d) $\max _{1\le i\le n}|\widehat{\pi }_n(X_i)-\pi (X_i)|I(X_i\le \tau )=O(\eta _{n})~a.s.$

Proof

Following the proof of Lemma A.3 in Zou and Liang (2017b), one can prove this Lemma. $\square$

Lemma 5.3

Suppose that (A2), (A4), (A5) and (A7) hold, then

$$\begin{aligned} \{\widehat{G}_n(X_i-)-G(X_i-)\}I(X_i\le \tau )= & {} \frac{1}{n}(1-G(X_i))\sum _{j=1}^n\zeta (X_j,\delta _j,\xi _j;X_i)I(X_i\le \tau )\\&+R_n(X_i)I(X_i\le \tau ), \end{aligned}$$

where $\max _{1\le i\le n}|R_{n}(X_i)I(X_i\le \tau )|=o_p(n^{-1/2})$ and

$$\begin{aligned} \zeta (X_j,\delta _j,\xi _j;X_i)=&\frac{(\xi _j-\pi (X_j))(m(X_j)-\delta _j)}{\pi (X_j)(1-H(X_j))}I(X_j\le X_i) +\int _0^{X_i\wedge X_j}\frac{d{H}_1(s)}{(1-H(s))^2}\\&+\frac{1-\delta _j}{1-H(X_j)}I(X_j\le X_i). \end{aligned}$$

Proof

Following the proof line of Theorem 3.2 in Wang and Ng (2008), one can verify Lemma 5.5. $\square$

Lemma 5.4

If (A3), (A5) and (A7) hold, then we have $\widehat{f}_R(t):=\bar{f}_R(t)+\widetilde{f}_R(t)+r_{1n}(t)+r_{2n}(t),$ where $P_1(t)=[\pi (t)(1-G(t))]^{-1}$ and

$$\begin{aligned} \bar{f}_R(t)=&\frac{1}{nh_n}\sum _{i=1}^n\Big [\frac{\delta _i}{1-G(X_i)} +\frac{m(X_i)-\delta _i}{1-G(X_i)}+\frac{\xi _i(\delta _i-m(X_i))}{\pi (X_i)(1-G(X_i))}\Big ]K\Big (\frac{t-X_i}{h_{n}}\Big ),\\ \widetilde{f}_R(t)=&\frac{1}{nh_n}\sum _{i=1}^n\frac{ m(X_i)[\widehat{G}_n(X_i-)-G(X_i-)]}{(1- G(X_i))^2}K\Big (\frac{t-X_i}{h_{n}}\Big ),\\ r_{1n}(t)=&\frac{1}{nh_n}\sum _{j=1}^n\frac{\xi _j(\delta _j-m(X_j))}{\pi (X_j)[1-G(X_j)]}\int \Big [K\Big (\frac{t-X_j}{h_{n}}-\frac{b_n}{h_n}u\Big )- K\Big (\frac{t-X_j}{h_{n}}\Big )\Big ]\Omega (u)du\\&+\frac{1}{nh_n}\sum _{j=1}^n\xi _j(\delta _j-m(X_j))\int [P_1(X_j+b_nu)-P_1(X_j)]K\Big (\frac{t-X_j}{h_{n}}-\frac{b_n}{h_n}u\Big )\\&+\frac{1}{n^2h_nb_n}\sum _{i=1}^n\sum _{j=1}^n\xi _j(\delta _j-m(X_j))\Big \{ \frac{\Omega \big (\frac{X_i-X_j}{b_{n}}\big )}{V(X_i)(1-G(X_i))}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&-E_{X_j} \Big [ \frac{\Omega \big (\frac{X_i-X_j}{b_{n}}\big )}{V(X_i)(1-G(X_i))}K\Big (\frac{t-X_i}{h_{n}}\Big )\Big ]\Big \}\\&+\frac{1}{n^2h_nb_n}\sum _{i=1}^n\sum _{j=1}^n\frac{1}{V(X_i)(1-G(X_i))}K \Big (\frac{t-X_i}{h_{n}}\Big ) \xi _j(m(X_j)\\&-m(X_i))\Omega \Big (\frac{X_i-X_j}{b_{n}}\Big )\\&+\frac{1}{n^2h_nb_n}\sum _{i=1}^n\sum _{j=1}^n\frac{V(X_i)-V_n(X_i)}{V_n(X_i)V(X_i) (1-G(X_i))}K\Big (\frac{t-X_i}{h_{n}}\Big ) \xi _j(\delta _j\\&-m(X_i))\Omega \Big (\frac{X_i-X_j}{b_{n}}\Big ),\\ r_{2n}(t)=&\frac{1}{nh_n}\sum _{i=1}^n\frac{(\widehat{m}_n(X_i)-m(X_i))(\widehat{G}_n(X_i-)-G(X_i-))}{(1- G(X_i))(1-\widehat{G}_n(X_i-))}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&+\frac{1}{nh_n}\sum _{i=1}^n\frac{m(X_i)(\widehat{G}_n(X_i-)-G(X_i-))^2}{(1- G(X_i))^2(1-\widehat{G}_n(X_i-))}K\Big (\frac{t-X_i}{h_{n}}\Big ). \end{aligned}$$

Proof

From the definition of $\widehat{f}_R(t)$ in Sect. 2, one can rewrite

$$\begin{aligned} \widehat{f}_R(t) =&\frac{1}{nh_n}\sum _{i=1}^n\frac{m(X_i)}{1-G(X_i)}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&+\frac{1}{nh_n}\sum _{i=1}^n\frac{ m(X_i)(\widehat{G}_n(X_i-)-G(X_i-))}{(1- G(X_i))^2}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&+\frac{1}{nh_n}\sum _{i=1}^n\frac{\widehat{m}_n(X_i)-m(X_i)}{1- G(X_i)}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&+\Big \{\frac{1}{nh_n}\sum _{i=1}^n\frac{(\widehat{m}_n(X_i)-m(X_i))(\widehat{G}_n(X_i-)-G(X_i-))}{(1- G(X_i))(1-\widehat{G}_n(X_i-))}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&+\frac{1}{nh_n}\sum _{i=1}^n\frac{m(X_i)(\widehat{G}_n(X_i-)-G(X_i-))^2}{(1- G(X_i))^2(1-\widehat{G}_n(X_i-))}K\Big (\frac{t-X_i}{h_{n}}\Big )\Big \}\\ =&\frac{1}{nh_n}\sum _{i=1}^n\frac{\delta _i}{1-G(X_i)}K\Big (\frac{t-X_i}{h_{n}}\Big ) +\frac{1}{nh_n}\sum _{i=1}^n\frac{m(X_i)-\delta _i}{1-G(X_i)}K \Big (\frac{t-X_i}{h_{n}}\Big )\\&+\widetilde{f}_R(t)+D_{1n}(t)+r_{2n}(t). \end{aligned}$$

From the definition of $\widehat{m}_n(\cdot )$ in Sect. 2, we have

$$\begin{aligned} D_{1n}(t)=&\frac{1}{n^2h_nb_n}\sum _{i=1}^n\sum _{j=1}^n\xi _j(\delta _j-m(X_j))E_{X_j}\Big [ \frac{\Omega \big (\frac{X_i-X_j}{b_{n}}\big )}{V(X_i)(1-G(X_i))}K\Big (\frac{t-X_i}{h_{n}}\Big )\Big ]\\&+\frac{1}{n^2h_nb_n}\sum _{i=1}^n\sum _{j=1}^n\xi _j(\delta _j-m(X_j))\Big \{ \frac{\Omega \big (\frac{X_i-X_j}{b_{n}}\big )}{V(X_i)(1-G(X_i))}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&-E_{X_j}\Big [ \frac{\Omega \big (\frac{X_i-X_j}{b_{n}}\big )}{V(X_i)(1-G(X_i))}K\Big (\frac{t-X_i}{h_{n}}\Big )\Big ]\Big \}\\&+\frac{1}{n^2h_nb_n}\sum _{i=1}^n\sum _{j=1}^n\frac{1}{V(X_i)(1-G(X_i))} K\Big (\frac{t-X_i}{h_{n}}\Big ) \\&\xi _j(m(X_j)-m(X_i))\Omega \Big (\frac{X_i-X_j}{b_{n}}\Big )\\&+\frac{1}{n^2h_nb_n}\sum _{i=1}^n\sum _{j=1}^n \frac{V(X_i)-V_n(X_i)}{V_n(X_i)V(X_i)(1-G(X_i))}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&\xi _j(\delta _j-m(X_i))\Omega \Big (\frac{X_i-X_j}{b_{n}}\Big ) \end{aligned}$$

and denote the first term as $D_{2n}(t)$, then we have

$$\begin{aligned} D_{2n}(t)=&\frac{1}{nh_n}\sum _{j=1}^n\int \frac{\xi _j(\delta _j-m(X_j))}{\pi (X_j+b_nu)(1-G(X_j+b_{n}u))}K\Big (\frac{t-X_j}{h_{n}}-\frac{b_n}{h_n}u\Big ) \Omega (u)du\\ =&\frac{1}{nh_n}\sum _{j=1}^n\frac{\xi _j(\delta _j-m(X_j))}{\pi (X_j)(1-G(X_j))} K\Big (\frac{t-X_j}{h_{n}}\Big )\\&+\frac{1}{nh_n}\sum _{j=1}^n\frac{\xi _j(\delta _j-m(X_j))}{\pi (X_j)(1-G(X_j))} \int \Big [K\Big (\frac{t-X_j}{h_{n}}-\frac{b_n}{h_n}u\Big )\\&-K\Big (\frac{t-X_j}{h_{n}}\Big )\Big ]\Omega (u)du\\&+\frac{1}{nh_n}\sum _{j=1}^n\xi _j(\delta _j-m(X_j))\int [P_1(X_j+b_nu) -P_1(X_j)]\\&K\Big (\frac{t-X_j}{h_{n}}-\frac{b_n}{h_n}u\Big ) \Omega (u)du. \end{aligned}$$

Collecting the results above, this lemma is proved. $\square$

Lemma 5.5

Under Lemma 5.5, if (A0) and (A2)–(A8) hold, then $\int r_{1n}^2(t)w(t)dt=o_p(n^{-1/2}h_n^2).$

Proof

From the definition of $r_{1n}(t)$ in Lemma 5.4, let $r_{1n}(t):=\sum _{l=1}^5r_{1nl}(t)$. Due to the independence of $\{X_j,\delta _j,\xi _j\}$ for $j=1,\ldots ,n$, then, under MAR assumption, from (A5) to (A7) we have

$$\begin{aligned} \int E[r_{1n1}(t)]^2w(t)dt \le&\frac{C}{n^2h_n^2}\sum _{j=1}^n\int E\Big [\int \frac{\Omega (u)}{(r+1)!}\Big (-\frac{b_n}{h_n}u\Big )^r\int _0^1(1-\theta )^{r+1}\\&\times K^{(r)}\Big (\frac{t-X_j}{h_{n}}-\frac{b_n\theta }{h_n}u\Big )d\theta du\Big ]^2w(t)dt\\ =&\,O(n^{-1}b_n^{2r}h_n^{-(2r+1)}). \end{aligned}$$

Similarly, we have $\int E[r_{1n2}(t)]^2w(t)dt=O(n^{-1}b_n^{2r}h_n^{-(2r+1)})$.

In order to evaluate $r_{1n3}(t)$, let $U_j=\xi _j(\delta _j-m(X_j))$ and

$$\begin{aligned} Y_{ij}=\frac{\Omega \Big (\frac{X_i-X_j}{b_{n}}\Big )}{V(X_i)[1-G(X_i)]}K\Big (\frac{t-X_i}{h_{n}}\Big )-E_{X_j} \Big [\frac{\Omega \Big (\frac{X_i-X_j}{b_{n}}\Big )}{V(X_i)[1-G(X_i)]}K\Big (\frac{t-X_i}{h_{n}}\Big )\Big ], \end{aligned}$$

it is easy to verify $E[U_j^2 Y_{ij}^2]=O(h_nb_n)$, then

$$\begin{aligned} \int E[r_{1n3}(t)]^2w(t)dt =\frac{1}{n^4h_{n}^2b_n^2}\int \sum _{i=1}^n\sum _{j=1}^nE[U_j^2Y_{ij}^2]w(t)dt =O(n^{-2}h_n^{-1}b_n^{-1}). \end{aligned}$$

For $r_{1n4}(t)$, we write

$$\begin{aligned} r_{1n4}(t)=&\frac{1}{n^2h_nb_n}\sum _{i=1}^n\sum _{j=1}^n\frac{1}{V(X_i)(1-G(X_i))}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&\times E_{X_i} \Big [\xi _j(m(X_j)-m(X_i))\Omega \Big (\frac{X_i-X_j}{b_{n}}\Big )\Big ]\\&+\frac{1}{n^2h_nb_n}\sum _{i=1}^n\frac{1}{V(X_i)(1-G(X_i))} K\Big (\frac{t-X_i}{h_{n}}\Big )\\&\times \Big \{ \xi _j(m(X_j)-m(X_i)) \Omega \Big (\frac{X_i-X_j}{b_{n}}\Big )\\&-E_{X_i}\Big [\xi _j(m(X_j)-m(X_i))\Omega \Big (\frac{X_i-X_j}{b_{n}}\Big )\Big ]\Big \} \\ :=&r_{1n4}^{(1)}(t)+r_{1n4}^{(2)}(t). \end{aligned}$$

Using the Taylor expansion, it follows that $\int E[r_{1n4}^{(1)}(t)]^2w(t)dt=O(b_n^{2r})$.

Let $A_i=\frac{1}{V(X_i)[1-G(X_i)]}K(\frac{t-X_i}{h_{n}})$ and

$$\begin{aligned} B_{ij}=\xi _j(m(X_j)-m(X_i))\Omega \Big (\frac{X_i-X_j}{b_{n}}\Big ) -E_{X_i}\Big [\xi _j(m(X_j)\!-\!m(X_i))\Omega \Big (\frac{X_i-X_j}{b_{n}}\Big )\Big ]. \end{aligned}$$

It can be verified that $E[A_i^2B_{ij}^2]=O(h_nb_n^3)$ and $E[A_{i_1}A_{i_2}B_{i_1j}B_{i_2j}]=O(h_n^2b_n^4)$. Then

$$\begin{aligned} \int E[r_{1n4}^{(2)}(t)]^2w(t)dt&=\frac{1}{n^4h_n^2b_n^2}\int \Big \{\sum _{i=1}^n\sum _{j=1}^nE[A_i^2B_{ij}^2]\\&\quad +\sum _{i_1\ne i_2}^n\sum _{j=1}^nE[A_{i_1}A_{i_2}B_{i_1j}B_{i_2j}]\Big \}w(t)dt =O(n^{-1}b_n^2). \end{aligned}$$

For $r_{1n5}(t)$, we observe

$$\begin{aligned} r_{1n5}(t) =&\frac{1}{n^2h_nb_n}\sum _{i=1}^n\sum _{j=1}^n \frac{ V(X_i)-V_n(X_i)}{V^2(X_i)(1-G(X_i))} K\Big (\frac{t-X_i}{h_{n}}\Big )\xi _j(\delta _j-m(X_i)) \Omega \Big (\frac{X_i-X_j}{b_{n}}\Big )\\&+\frac{1}{n^2h_nb_n}\sum _{i=1}^n\sum _{j=1}^n \frac{(V(X_i)-V_n(X_i))^2}{V_n(X_i)V^2(X_i)(1-G(X_i))}\\&K\Big (\frac{t-X_i}{h_{n}}\Big )\xi _j(\delta _j-m(X_i)) \Omega \Big (\frac{X_i-X_j}{b_{n}}\Big )\\ :=&I_{1n5}^{(1)}(t)+r_{1n5}^{(2)}(t). \end{aligned}$$

Recalling the definition of $V_n(\cdot )$ in Lemma 5.2, it follows that

$$\begin{aligned} r_{1n5}^{(1)}(t)=&\frac{1}{n^2h_nb_n}\sum _{i=1}^n\sum _{j=1}^n\frac{ V(X_i) -\frac{1}{nb_n}\sum _{k\ne j}^n\xi _k\Omega \Big (\frac{X_i-X_k}{b_n}\Big )}{V^2(X_i)(1-G(X_i))}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&\times E_{X_i} \Big [\xi _j(\delta _j-m(X_i))\Omega \Big (\frac{X_i-X_j}{b_n}\Big )\Big ]\\&+\frac{1}{n^2h_nb_n}\sum _{i=1}^n\sum _{j=1}^n\frac{ V(X_i) -\frac{1}{nb_n}\sum _{k\ne j}^n\xi _k\Omega \Big (\frac{X_i-X_k}{b_n}\Big )}{V^2(X_i)(1-G(X_i))}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&\times \Big \{\xi _j (\delta _j-m(X_i))\Omega \Big (\frac{X_i-X_j}{b_n}\Big )\\&-E_{X_i}\Big [\xi _j(\delta _j-m(X_i))\Omega \Big (\frac{X_i-X_j}{b_n}\Big )\Big ]\Big \}\\&+\frac{1}{n^3h_nb_n^2}\sum _{i=1}^n\sum _{j=1}^n\frac{ 1}{V^2(X_i) (1-G(X_i))}K\Big (\frac{t-X_i}{h_{n}}\Big )\xi _j(\delta _j-m(X_i)) \Omega ^2\Big (\frac{X_i-X_j}{b_n}\Big )\\ :=&r_{1n5}^{(11)}(t)+r_{1n5}^{(12)}(t)+r_{1n5}^{(13)}(t). \end{aligned}$$

If $i=j$, it is easy to verify that $r_{1n5}^{(11)}(t)=0$, $\int E[r_{1n5}^{(12)}(t)]^2w(t)dt=O(\gamma _n^2n^{-3}b_n^{-2} h_n^{-1})$ and $\int E[r_{1n5}^{(13)}(t)]^2w(t)dt=O(n^{-5}b_n^{-4}h_n^{-1}).$

If $i\ne j$, we have $\int E[r_{1n5}^{(11)}(t)]^2w(t)dt=O(\gamma _n^2b_n^{2r}h_n^{-1})$. Let $S_i=\frac{ V(X_i)-\frac{1}{nb_n}\sum _{k\ne j}^n\xi _k\Omega (\frac{X_i-X_k}{b_n})}{V^2(X_i)[1-G(X_i)]}K\big (\frac{t-X_i}{h_{n}}\big )$,

$$\begin{aligned} Y_{ij}=\xi _j(\delta _j-m(X_i))\Omega \Big (\frac{X_i-X_j}{b_n}\Big ) -E_{X_i}\Big [\xi _j(\delta _j-m(X_i))\Omega \;\Big (\frac{X_i-X_j}{b_n}\Big )\Big ]. \end{aligned}$$

It can be verified that $E[S_i^2Y_{ij}^2]=O(\gamma _n^2h_nb_n)$ and $E[S_{i_1}S_{i_2}Y_{i_1j}Y_{i_2j}]=O(\gamma _n^2h_nb_n^2)$. Then

$$\begin{aligned} \int E[r_{1n5}^{(12)}(t)]^2w(t)dt=&\frac{1}{n^4h_n^2b_n^2}\int \Big \{\sum _{i=1}^n\sum _{j=1}^n E[S_i^2Y_{ij}^2]\\+&\sum _{i_1\ne i_2}\sum _{j=1}^nE[S_{i_1}S_{i_2}Y_{i_1j}Y_{i_2j}]\Big \}w(t)dt =O(\gamma _n^2n^{-1}h_n^{-1}). \end{aligned}$$

Similarly, $\int E[r_{1n5}^{(13)}(t)]^2w(t)dt=O(n^{-2}b_n^{-2}).$ From Lemma 5.2 (a), one can get $\int E[r_{1n5}^{(2)}(t)]^2w(t)dt=O(\gamma _n^4).$ Collecting the results above and using (A8), it yields that

$$\begin{aligned} \int r_{1n}^2(t)w(t)dt= & {} O_p(n^{-1}b_n^{2r}h_n^{-(2r+1)}+b_n^{2r}+n^{-1}b_n^2+(\log n/(nb_n))^2)\\ {}= & {} o_p(n^{-1/2}h_n^2). \square \end{aligned}$$

Lemma 5.6

Under (A3), (A5) and (A7), we have $\widehat{f}_I(t):=\bar{f}_I(t)+\widetilde{f}_I(t)+s_{1n}(t)+s_{2n}(t),$ where

$$\begin{aligned} \bar{f}_I(t)=&\frac{1}{nh_n}\sum _{i=1}^n\Big [\frac{\delta _i}{1-G(X_i)} +\frac{(1-\xi _i)(m(X_i)-\delta _i)}{1-G(X_i)}\\&+\frac{\xi _i(\delta _i-m(X_i))(1-\pi (X_i))}{\pi (X_i)(1-G(X_i))}\Big ] K\Big (\frac{t-X_i}{h_{n}}\Big ),\\ \widetilde{f}_I(t)=&\frac{1}{nh_n}\sum _{i=1}^n\frac{ (\xi _i\delta _i +(1-\xi _i)m(X_i))(\widehat{G}_n(X_i-)-G(X_i-))}{(1- G(X_i))^2}K \Big (\frac{t-X_i}{h_{n}}\Big ),\\ s_{1n}(t)=&\frac{1}{nh_n}\sum _{j=1}^n\frac{\xi _j(\delta _j-m(X_j)) (1-\pi (X_j))}{\pi (X_j)(1-G(X_j))}\int \Big [K\Big (\frac{t-X_j}{h_{n}} -\frac{b_n}{h_n}u\Big )\\&-K\Big (\frac{t-X_j}{h_{n}}\Big )\Big ]\Omega (u)du\\&+\frac{1}{nh_n}\sum _{j=1}^n\xi _j(\delta _j-m(X_j))(1-\pi (X_j))\\&\times \int [P_1(X_j+b_nu)-P_1(X_j)]K\Big (\frac{t-X_j}{h_{n}}-\frac{b_n}{h_n}u\Big )\\&+\frac{1}{n^2h_nb_n}\sum _{i=1}^n\sum _{j=1}^n\xi _j(\delta _j-m(X_j)) \Big \{ \frac{(1-\xi _i)\Omega \big (\frac{X_i-X_j}{b_{n}}\big )}{V(X_i) (1-G(X_i))}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&-E_{X_j} \Big [ \frac{(1-\xi _i)\Omega \big (\frac{X_i-X_j}{b_{n}}\big )}{V(X_i)(1-G(X_i))}K\Big (\frac{t-X_i}{h_{n}}\Big )\Big ]\Big \}\\&+\frac{1}{n^2h_nb_n}\sum _{i=1}^n\sum _{j=1}^n\frac{1-\xi _i}{V(X_i) (1-G(X_i))}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&\times \xi _j(m(X_j)-m(X_i)) \Omega \Big (\frac{X_i-X_j}{b_{n}}\Big )\\&+\frac{1}{n^2h_nb_n}\sum _{i=1}^n\sum _{j=1}^n\frac{(1-\xi _i)(V(X_i) -V_n(X_i))}{V_n(X_i)V(X_i)(1-G(X_i))}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&\times \xi _j(\delta _j-m(X_i))\Omega \Big (\frac{X_i-X_j}{b_{n}}\Big ),\\ s_{2n}(t)=&\frac{1}{nh_n}\sum _{i=1}^n\frac{(1-\xi _i) (\widehat{m}_n(X_i)-m(X_i))(\widehat{G}_n(X_i-)-G(X_i-))}{(1- G(X_i))(1-\widehat{G}_n(X_i-))}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&+\frac{1}{nh_n}\sum _{i=1}^n\frac{(\xi _i\delta _i+(1-\xi _i)m(X_i)) (\widehat{G}_n(X_i-)-G(X_i-))^2}{(1- G(X_i))^2(1-\widehat{G}_n(X_i-))} K\Big (\frac{t-X_i}{h_{n}}\Big ). \end{aligned}$$

Proof

Following the proof line as for Lemma 5.4, one can verify Lemma 5.6. $\square$

Lemma 5.7

Following simple calculation, we have $\widehat{f}_W(t):=\bar{f}_W(t)+\widetilde{f}_W(t)+v_{1n}(t)+v_{2n}(t),$ where

$$\begin{aligned} \bar{f}_W(t)=&\frac{1}{nh_n}\sum _{i=1}^n\Big [\frac{\delta _i}{1-G(X_i)} +\frac{\Big (1-\frac{\xi _i}{\pi (X_i)}\Big )(m(X_i)-\delta _i)}{1-G(X_i)}\Big ]K\Big (\frac{t-X_j}{h_{n}}\Big ),\\ \widetilde{f}_W(t)=&\frac{1}{nh_n}\sum _{i=1}^n \frac{ \Big (\frac{\xi _i\delta _i}{\pi (X_i)}+\Big (1-\frac{\xi _i}{\pi (X_i)}\Big )m(X_i)\Big )}{(1- G(X_i))^2}(\widehat{G}_n(X_i-)-G(X_i-))K\Big (\frac{t-X_i}{h_{n}}\Big ),\\ v_{1n}(t)=&\frac{1}{nh_n}\sum _{i=1}^n\frac{\Big (1-\frac{\xi _i}{\pi (X_i)}\Big )(\widehat{m}_n(X_i)-m(X_i))}{1-G(X_i)}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&+\frac{1}{nh_n}\sum _{i=1}^n\frac{\xi _i( m(X_i)-\delta _i)(\widehat{\pi }_n(X_i)-\pi (X_i))}{\pi ^2(X_i)(1-G(X_i))}K\Big (\frac{t-X_i}{h_{n}}\Big ),\\ v_{2n}(t)=&\frac{1}{nh_n}\sum _{i=1}^n\frac{\Big (\frac{\xi _i\delta _i}{\pi (X_i)}+\Big (1-\frac{\xi _i}{\pi (X_i)}\Big )m(X_i)\Big )(\widehat{G}_n(X_i-)-G(X_i-))^2}{(1- G(X_i))(1-\widehat{G}_n(X_i-))}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&+\frac{1}{nh_n}\sum _{i=1}^n\frac{\Big (1-\frac{\xi _i}{\pi (X_i)}\Big )(\widehat{m}_n(X_i)-m(X_i))(\widehat{G}_n(X_i-)-G(X_i-))}{(1- G(X_i))(1-\widehat{G}_n(X_i-))}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&+\frac{1}{nh_n}\sum _{i=1}^n\frac{\xi _i(\delta _i-m(X_i))(\widehat{\pi }_n(X_i)-\pi (X_i))^2}{\pi ^2(X_i)\widehat{\pi }_n(X_i)(1-\widehat{G}_n(X_i-))}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&+\frac{1}{nh_n}\sum _{i=1}^n\frac{\xi _i(m(X_i)-\delta _i)(\widehat{\pi }_n(X_i)-\pi (X_i))(\widehat{G}_n(X_i-)-G(X_i-))}{\pi ^2(X_i)(1-\widehat{G}_n(X_i-))(1- G(X_i)}K\Big (\frac{t-X_i}{h_{n}}\Big )\\&+\frac{1}{nh_n}\sum _{i=1}^n\frac{\xi _i(\widehat{\pi }_n(X_i)-\pi (X_i))(\widehat{m}_n(X_i)-m(X_i))}{\pi (X_i)\widehat{\pi }_n(X_i)(1-\widehat{G}_n(X_i-))}K\Big (\frac{t-X_i}{h_{n}}\Big ). \end{aligned}$$

Proof

Following simple computation, one can get Lemma 5.7. Here, we omit the proof line. $\square$