Appendix: Proofs
Before proceeding to the proofs, we list some notations. Put
$$\begin{aligned} \left\{ \begin{array}{ll} &{} \sigma _1^2 =\int _{0}^{\infty }\Big [\int _{{\mathscr {B}}}w(t)(x-\xi _t)I(x\le \xi _t)dt\Big ]^2 dF(x)-\big [\int _{{\mathscr {B}}}w(t)(\theta _1(t)-t\xi _t)dt\big ]^2,\\ &{} \sigma _2^2 =\int _{0}^{\infty }\Big [\int _{{\mathscr {B}}}w(t)(y-\zeta _t)I(y\le \zeta _t)dt\Big ]^2 dG(y)-\big [\int _{{\mathscr {B}}}w(t)(\theta _2(t)-t\zeta _t)dt\big ]^2,\\ &{} c_1=\int _{{\mathscr {B}}}w(t)\big [\xi _t(\theta _1(t)-t\mu _1)+\int _{0}^{\xi _t}x^2dF(x)-\mu _1\theta _1(t)\big ]dt,\\ &{} c_2=\int _{{\mathscr {B}}}w(t)\big [\zeta _t(\theta _2(t)-t\mu _2)+\int _{0}^{\zeta _t}y^2dG(y)-\mu _2\theta _2(t)\big ]dt,\\ &{} \sigma _{11}^2=\mu _2^2\sigma _1^2+\left[ \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\mu _2\Delta \right] ^2\text{ var }(X) -2\mu _2\left[ \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\mu _2\Delta \right] \cdot c_1,\\ &{} \sigma _{22}^2=\mu _1^2\sigma _2^2+\left[ \int _{{\mathscr {B}}}w(t)\theta _1(t)dt-\mu _1\Delta \right] ^2\text{ var }(Y) -2\mu _1\left[ \int _{{\mathscr {B}}}w(t)\theta _1(t)dt-\mu _1\Delta \right] \cdot c_2,\\ &{}\sigma ^2 =(1+c^{-1})\sigma _{11}^2 +(1+c)\sigma _{22}^2,~\mathrm {where~c~is~defined~in~condition~(C2)}, \\ &{} {\widehat{V}}^{(1)}_k =n {{\widehat{T}}}^{(1)}_n-(n-1){{\widehat{T}}}^{(1)}_{n-1,k} ,~k=1, 2, \ldots , n. \end{array}\right. \end{aligned}$$
Lemma 1
When conditions (C1), (C3) and (C4) hold, we have
$$\begin{aligned} \sqrt{n} \left[ {{\widehat{T}}}^{(1)}_n-\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\right] \xrightarrow {d} N(0,\sigma _1^2),\quad \text{ as } n \rightarrow \infty \text{. } \end{aligned}$$
(22)
Proof
First, from Eq. (10), we can divide Eq. (22) into three parts as follows
$$\begin{aligned}&\sqrt{n}\left[ {{\widehat{T}}}^{(1)}_n - \int _{{\mathscr {B}}}w(t)\theta _1(t)dt \right] \nonumber \\&\quad =\sqrt{n}\left[ \frac{1}{n}\sum _{i=1}^nX_i\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_n(X_i)}{h_1}\right) dt\nonumber \right. \\&\left. \qquad -\frac{1}{n}\sum _{i=1}^nX_i\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F(X_i)}{h_1}\right) dt\right] \nonumber \\&\qquad +\sqrt{n}\left[ \frac{1}{n}\sum _{i=1}^nX_i\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F(X_i)}{h_1}\right) dt-\int _{{\mathscr {B}}}w(t)\int _0^{\xi _t} xdF(x)dt\right] \nonumber \\&\quad = \int _{0}^{\infty }x\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(x)}{h_1}\right) -K_1\left( \frac{t-F(x)}{h_1}\right) \right] dtd[\sqrt{n}(F_n(x)-F(x))] \nonumber \\&\qquad +\sqrt{n}\int _{0}^{\infty }x\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(x)}{h_1}\right) -K_1\left( \frac{t-F(x)}{h_1}\right) \right] dtdF(x) \nonumber \\&\qquad +\sqrt{n}\left[ \frac{1}{n}\sum _{i=1}^nX_i\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F(X_i)}{h_1}\right) dt-\int _{{\mathscr {B}}}w(t)\int _0^{\xi _t} xdF(x)dt\right] \nonumber \\&\quad =: J_1+J_2+J_3. \end{aligned}$$
(23)
Note that \(\sqrt{n} [F_n(x)-F(x)] \xrightarrow d B(x)\), as \(n\rightarrow \infty \), where B(x) is a Gaussian process. By condition (C1), and the Taylor expansion, we obtain
$$\begin{aligned} J_2= & {} \sqrt{n}\int _{0}^{\infty } x\int _{{\mathscr {B}}}w(t)\left[ k_1\left( \frac{t-F(x)}{h_1}\right) dt\frac{F(x)-F_n(x)}{h_1} \right. \nonumber \\&\quad \left. + \frac{1}{2}k'_1\left( \frac{t-F(x)}{h_1}\right) \left( \frac{ F(x)-F_n(x)}{h_1}\right) ^2 \right] dtdF(x)+o_p(1)\nonumber \\= & {} -\int _{0}^{\infty } x\int _{{\mathscr {B}}}w(t) k_1\left( \frac{t-F(x)}{h_1}\right) \frac{\sqrt{n} (F_n(x)-F(x))}{h_1}dtdF(x) +O_p\left( \frac{1}{\sqrt{n}h_1^2}\right) \nonumber \\= & {} -\int _{{\mathscr {B}}}w(t)\int _{-1}^{1} F^{-1}(y) k_1\left( \frac{t-u}{h_1}\right) \frac{\sqrt{n} \left[ \frac{1}{n}\sum _{k=1}^nI(F(X_k)\le u)-u\right] }{h_1}dudt +o_p(1) \nonumber \\= & {} -\int _{{\mathscr {B}}}w(t)\int _{\frac{t-1}{h_1}}^{\frac{t+1}{h_1}} F^{-1}(t-h_1v) k_1(v) \sqrt{n}\nonumber \\&\quad \times \left[ \frac{1}{n}\sum _{k=1}^nI(F(X_k) \le t-h_1v)-(t-h_1v)\right] dvdt +o_p(1) \nonumber \\= & {} -\int _{{\mathscr {B}}}w(t)\xi _t\sqrt{n}(F_n(\xi _t)-t)dt+o_p(1). \end{aligned}$$
(24)
Similarly, we can obtain \(J_1=o_p(1)\).
It follows from the condition (C1) that
$$\begin{aligned} I(x \le \xi _t) - K_1\left( \frac{t-F(x)}{h_1}\right) = \left\{ \begin{matrix} 1 - K_1\left( \frac{t-F(x)}{h_1}\right) , &{} h_1 \ge t - F(x) > 0, \\ - K_1\left( \frac{t-F(x)}{h_1}\right) , &{} 0 \ge t - F(x) \ge -h_1, \\ 0, &{} \mathrm {otherwise}, \end{matrix}\right. \end{aligned}$$
and noting that \(k_1(x)\) and \(k_2(x)\) are even functions in the domain \([-1,1]\). So, for \(i=1,2\), we have
$$\begin{aligned} \int _{-1}^1K_i(v)dv= & {} \int _{-1}^1\int _{-1}^vk_i(x)dxdv=\int _{-1}^1k_i(x)\int _{-1}^xdvdx\\= & {} \int _{-1}^1xk_i(x)dx + \int _{-1}^1k_i(x)dx = 1. \end{aligned}$$
By condition (C3), it can be verified that
$$\begin{aligned}&\lim _{h_1\rightarrow 0}\sqrt{n} \Big | \int _0^{\xi _t} xdF(x) - E\left[ X K_1\left( \frac{t-F(X)}{h_1}\right) \right] \Big |\\&\quad =\lim _{h_1\rightarrow 0}\sqrt{n} \Big | \int _0^{\infty } xI(x\le \xi _t)dF(x) - \int _0^{\infty } xK_1\left( \frac{t-F(x)}{h_1}\right) dF(x) \Big |\\&\quad = \lim _{h_1\rightarrow 0}\sqrt{n}\Big |\int _{-h_1}^{h_1} F^{-1}(t - u) \left[ I(0\le u) - K_1\left( \frac{u}{h_1}\right) \right] du\Big |\\&\quad = \lim _{h_1\rightarrow 0}\sqrt{n}h_1\xi _t\Big |1- \int _{-1}^{1} K_1(v) dv\Big | = 0, \end{aligned}$$
which yields
$$\begin{aligned} \sqrt{n} \int _0^{\xi _t} xdF(x) =\sqrt{n} E\left[ X K_1\left( \frac{t-F(X)}{h_1}\right) \right] +o(1). \end{aligned}$$
Therefore,
$$\begin{aligned} J_3= & {} \sqrt{n}\left\{ \frac{1}{n}\sum _{i=1}^nX_i\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F(X_i)}{h_1}\right) dt\nonumber \right. \\&\quad \left. -E\left[ X\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F(X)}{h_1}\right) dt\right] \right\} +o(1). \end{aligned}$$
(25)
Furthermore, it follows from Eqs. (23), (24) and (25) that
$$\begin{aligned}&\sqrt{n}\left[ {{\widehat{T}}}^{(1)}_n - \theta _1(t) \right] \nonumber \\&\quad =\sqrt{n}\int _{{\mathscr {B}}}w(t)\left\{ \frac{1}{n}\sum _{i=1}^nX_i K_1\left( \frac{t-F(X_i)}{h_1}\right) -\xi _t\frac{1}{n}\sum _{i=1}^nI(X_i\le \xi _t) \right. \nonumber \\&\qquad \left. -E\left[ X K_1\left( \frac{t-F(X)}{h_1}\right) -\xi _t I(X \le \xi _t)\right] \right\} dt +o_p(1). \end{aligned}$$
(26)
For deriving the asymptotic distribution of \(\sqrt{n}\left[ {{\widehat{T}}}^{(1)}_n - \int _{{\mathscr {B}}}w(t)\theta _1(t)dt \right] \), write
$$\begin{aligned} \left\{ \begin{array}{ll} W_i:=\int _{{\mathscr {B}}}w(t)X_i K_1\left( \frac{t-F(X_i)}{h_1}\right) dt-\int _{{\mathscr {B}}}w(t)\xi _tI(X_i\le \xi _t)dt,\\ W:=\int _{{\mathscr {B}}}w(t)X K_1\left( \frac{t-F(X)}{h_1}\right) dt-\int _{{\mathscr {B}}}w(t)\xi _tI(X\le \xi _t)dt, \end{array}\right. \end{aligned}$$
we thus obtain
$$\begin{aligned} \sqrt{n}\left[ {{\widehat{T}}}^{(1)}_n - \int _{{\mathscr {B}}}w(t)\theta _1(t)dt \right] =\frac{1}{\sqrt{n}}\sum _{i=1}^n\left[ W_i-EW_i\right] +o_p(1). \end{aligned}$$
(27)
Noticing that
$$\begin{aligned} \lim _{h_1\rightarrow 0}E [W]= & {} \lim _{h_1\rightarrow 0}E\int _{{\mathscr {B}}}w(t)X K_1\left( \frac{t-F(X)}{h_1}\right) dt-\xi _t E[I(X\le \xi _t)]dt \nonumber \\= & {} \lim _{h_1\rightarrow 0}\int _{0}^{\infty }x\int _{{\mathscr {B}}}w(t)\left[ \int _{-\infty }^{\frac{t-F(x)}{h_1}}k_1(u)du \right] f(x)dtdx\nonumber \\&\quad -\int _{{\mathscr {B}}}w(t)\xi _t\int _{-\infty }^{\xi _t}dF(x)dt\nonumber \\= & {} \int _{{\mathscr {B}}}w(t)\int _{0}^{\infty }x I(F(x)\le t)f(x)dxdt-\int _{{\mathscr {B}}}w(t)t \xi _tdt\nonumber \\= & {} \int _{{\mathscr {B}}}w(t)\theta _1(t)dt-\int _{{\mathscr {B}}}w(t)t\xi _tdt, \end{aligned}$$
(28)
and
$$\begin{aligned}&\lim _{h_1\rightarrow 0}E[W^2]=\lim _{h_1\rightarrow 0}E\left[ \int _{{\mathscr {B}}}w(t)X K_1\left( \frac{t-F(X)}{h_1}\right) dt-\int _{{\mathscr {B}}}w(t)\xi _tI(X\le \xi _t)dt\right] ^2 \nonumber \\&\quad =\lim _{h_1\rightarrow 0}E\left\{ \left[ \int _{{\mathscr {B}}}w(t)X K_1\left( \frac{t-F(X)}{h_1}\right) dt\right] ^2+\left[ \int _{{\mathscr {B}}}w(t)\xi _tI(X\le \xi _t)dt\right] ^2\right. \nonumber \\&\qquad \left. -2\int _{{\mathscr {B}}}w(t)X K_1\left( \frac{t-F(X)}{h_1}\right) dt\int _{{\mathscr {B}}}w(t)\xi _t I(X\le \xi _t)dt\right\} \nonumber \\&\quad =\lim _{h_1\rightarrow 0}\int _{0}^{\infty }x^2 \left[ \int _{{\mathscr {B}}}w(t)\int _{-\infty }^{\frac{t-F(x)}{h_1}}k_1(u)du dt\right] ^2f(x)dx\nonumber \\&\qquad +\int _{0}^{\infty }\left[ \int _{{\mathscr {B}}}w(t)\xi _tI(x\le \xi _t)dt\right] ^2f(x)dx\nonumber \\&\qquad -2\lim _{h_1\rightarrow 0}\int _{0}^{\infty }x \left[ \int _{{\mathscr {B}}}w(t)\int _{-\infty }^{\frac{t-F(x)}{h_1}}k_1(u)dudt \right] \left[ \int _{{\mathscr {B}}}w(t)\xi _tI(x\le \xi _t)dt\right] f(x)dx \nonumber \\&\quad =\int _{0}^{\infty }\left[ \int _{{\mathscr {B}}}w(t)(x-\xi _t)I(x\le \xi _t)dt\right] ^2 dF(x). \end{aligned}$$
(29)
Thus, we obtain
$$\begin{aligned} \text{ Var }[W]= & {} \int _{0}^{\infty }\left[ \int _{{\mathscr {B}}}w(t)(x-\xi _t)I(x\le \xi _t)dt\right] ^2 dF(x)\\&\quad -\left[ \int _{{\mathscr {B}}}w(t)(\theta _1(t)-t\xi _t)dt\right] ^2+o(1)=\sigma _1^2+o(1). \end{aligned}$$
Now, combining Eqs. (27)–(29) with the central limit theorem, leads to
$$\begin{aligned} \sqrt{n}\left[ {{\widehat{T}}}^{(1)}_n- \int _{{\mathscr {B}}}\theta _1(t)dt \right] \overset{d}{\rightarrow } N(0,\sigma _1^2),\quad \text{ as } n\rightarrow \infty . \end{aligned}$$
This completes the proof. \(\square \)
Lemma 2
Under the condition (C1), (C3) and (C4), we have
$$\begin{aligned} \sum _{k=1}^n\Big [{{\widehat{T}}}^{(1)}_n-{{\widehat{T}}}^{(1)}_{n-1,k}\Big ]= O_p\left( \frac{1}{nh_1^2}\right) ,\quad \text{ as } n\rightarrow \infty . \end{aligned}$$
Proof
From Eqs. (9) and (12), we have
$$\begin{aligned}&{{\widehat{T}}}^{(1)}_n-{{\widehat{T}}}^{(1)}_{n-1,k}\\&\quad =\frac{1}{n}\sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_n(X_i)}{h_1}\right) dt\\&\qquad - \frac{1}{n-1}\sum _{i\ne k}X_i\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) dt\\&\quad =\frac{1}{n}\sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_n(X_i)}{h_1}\right) dt\\&\qquad - \frac{1}{n-1}\sum _{i=1}^nX_i\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) dt\\&\qquad +\frac{1}{n-1}X_k\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) dt\\&\quad =\frac{1}{n-1}\sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_n(X_i)}{h_1}\right) dt\\&\qquad - \frac{1}{n-1}\sum _{i=1}^nX_i\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) dt\\&\qquad -\frac{1}{n(n-1)}\sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_n(X_i)}{h_1}\right) dt\\&\qquad +\frac{1}{n-1}X_k\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) dt\\&\quad =\frac{1}{n-1}\sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_i)}{h_1}\right) - K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) \right] dt\\&\qquad -\frac{1}{n(n-1)}\sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_n(X_i)}{h_1}\right) dt\\&\qquad +\frac{1}{n-1}X_k\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) dt, \end{aligned}$$
thus,
$$\begin{aligned}&\sum _{k=1}^n\Big [{{\widehat{T}}}^{(1)}_n-{{\widehat{T}}}^{(1)}_{n-1,k}\Big ]\\&\quad =\frac{1}{n-1}\sum _{k=1}^n\sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_i)}{h_1}\right) - K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) \right] dt\\&\qquad -\frac{1}{n-1}\sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_n(X_i)}{h_1}\right) dt\\&\qquad +\frac{1}{n-1}\sum _{k=1}^nX_k\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) dt\\&\quad =\frac{1}{n-1}\sum _{k=1}^n\sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_i)}{h_1}\right) - K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) \right] dt\\&\qquad -\frac{1}{n-1}\sum _{k=1}^n X_k\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_k)}{h_1}\right) -K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) \right] dt\\&\quad =:P_1 - P_2. \end{aligned}$$
Note that
$$\begin{aligned} F_{n,k}(X)-F_n(X)=\frac{1}{n-1}\{F_n(X)-I(X_k\le X)\}=O_{p}\left( \frac{1}{n}\right) , \end{aligned}$$
(30)
and \(\sum _{k=1}^n[F_n(X)-F_{n,k}(X)]=0\). By the Taylor expression, we have
$$\begin{aligned}&K_1\left( \frac{t-F_{n,k}(X)}{h_1}\right) -K_1\left( \frac{t-F_n(X)}{h_1}\right) \nonumber \\&\quad = k_1\left( \frac{t-F_n(X)}{h_1}\right) \frac{F_{n,k}(X)-F_n(X)}{h_1}+ \frac{1}{2}k^{\prime }_1\left( \frac{t-\rho _{n,k,i}}{h_1}\right) \left( \frac{F_{n,k}(X)-F_n(X)}{h_1} \right) ^2 \nonumber \\&\quad = k_1\left( \frac{t-F_n(X)}{h_1}\right) \frac{F_{n,k}(X)-F_n(X)}{h_1}+O_p\left( \frac{1}{n^2h_1^2}\right) \nonumber \\&\quad = O_{p}\left( \frac{1}{nh_1}\right) , \end{aligned}$$
(31)
where \(\rho _{n,k,i}\) is random variables between \(F_n(X)\) and \(F_{n,k}(X)\), the \(P_1\) thus can be written as
$$\begin{aligned} P_1&=\frac{1}{n-1}\sum _{k=1}^n\sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_i)}{h_1}\right) - K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) \right] dt\\&=\frac{1}{n-1}\sum _{k=1}^n\sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t)\left[ k_1\left( \frac{t-F_n(X_i)}{h_1}\right) \frac{F_n(X_i)-F_{n,k}(X_i)}{h_1}\right. \\&\left. \quad +O_p\left( \frac{1}{n^2h_1^2}\right) \right] dt\\&=\frac{1}{n-1}\sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t)k_1\left( \frac{t-F_n(X_i)}{h_1}\right) dt\left( \frac{\sum _{k=1}^n[F_n(X_i)-F_{n,k}(X_i)]}{h_1}\right) \\&\quad +O_p\left( \frac{1}{nh_1^2}\right) \\&=O_p\left( \frac{1}{nh_1^2}\right) . \end{aligned}$$
From Eq. (31), and similar to \(P_1\), we can get
$$\begin{aligned} P_2= & {} \frac{1}{n-1}\sum _{k=1}^n X_k\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_k)}{h_1}\right) -K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) \right] dt\\= & {} O_{p}\left( \frac{1}{nh_1}\right) . \end{aligned}$$
Combining the \(P_1\) and \(P_2\), then
$$\begin{aligned} \sum _{k=1}^n\Big [{{\widehat{T}}}^{(1)}_n-{{\widehat{T}}}^{(1)}_{n-1,k}\Big ]=P_1-P_2= O_p\left( \frac{1}{nh_1^2}\right) . \end{aligned}$$
The proof is completed. \(\square \)
Lemma 3
Under the condition (C1), (C3) and (C4), we have
$$\begin{aligned} \sqrt{n} \left[ \frac{1}{n}\sum _{k=1}^n {\widehat{V}}^{(1)}_k-\int _{{\mathscr {B}}}\theta _1(t)dt \right] \xrightarrow {d} N(0,\sigma _1^2), \quad \text{ as } n \rightarrow \infty . \end{aligned}$$
(32)
Proof
From Lemmas 1 and 2, we have
$$\begin{aligned}&\sqrt{n} \left[ \frac{1}{n}\sum _{k=1}^n {\widehat{V}}^{(1)}_k-\int _{{\mathscr {B}}}\theta _1(t)dt \right] \\&\quad =\sqrt{n} \Big \{ \frac{1}{n}\sum _{k=1}^n\Big [n {{\widehat{T}}}^{(1)}_n-(n-1){{\widehat{T}}}^{(1)}_{n-1,k}\Big ]-\int _{{\mathscr {B}}}\theta _1(t)dt \Big \}\\&\quad =\sqrt{n}\left[ {{\widehat{T}}}^{(1)}_n- \int _{{\mathscr {B}}}\theta _1(t)dt \right] + \frac{n-1}{\sqrt{n}}\sum _{k=1}^n\Big [{{\widehat{T}}}^{(1)}_n-{{\widehat{T}}}^{(1)}_{n-1,k}\Big ]\\&\quad =\sqrt{n}\left[ {{\widehat{T}}}^{(1)}_n- \int _{{\mathscr {B}}}\theta _1(t)dt \right] + O_p\left( \frac{1}{\sqrt{n}h_1^2}\right) \\&\quad \xrightarrow {d} N(0,\sigma _1^2). \end{aligned}$$
This completes the proof. \(\square \)
Proof of Theorem 1
Using (8) and (9), we have
$$\begin{aligned}&\sqrt{m+n} {\widehat{D}}_{m,n}(\Delta ) \nonumber \\&\quad =\sqrt{m+n} \left[ {\bar{Y}} {{\widehat{T}}}^{(1)}_n - {\bar{X}}{{\widehat{T}}}^{(2)}_m -\Delta \cdot {\bar{X}}{\bar{Y}} \right] \nonumber \\&\quad =\sqrt{m+n} \left[ {\bar{Y}}{{\widehat{T}}}^{(1)}_n -\mu _2\int _{{\mathscr {B}}}w(t)\theta _1 (t)dt\right] \nonumber \\&\qquad -\sqrt{m+n}\left[ {\bar{X}}{{\widehat{T}}}^{(2)}_m - \mu _1\int _{{\mathscr {B}}}w(t)\theta _2 (t)dt \right] \nonumber \\&\qquad -\sqrt{m+n}\left[ \Delta \cdot {\bar{X}}{\bar{Y}}-\Delta \cdot \mu _1\mu _2 \right] \nonumber \\&\quad =\sqrt{m+n} \left[ {\bar{Y}}{{\widehat{T}}}^{(1)}_n -\mu _2\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\right] \nonumber \\&\qquad -\sqrt{m+n}\left[ {\bar{X}}{{\widehat{T}}}^{(2)}_m - \mu _1\int _{{\mathscr {B}}}w(t)\theta _2 (t)dt \right] \nonumber \\&\qquad -\sqrt{m+n}\Big \{ \Delta \cdot ({\bar{X}}-\mu _1 )\cdot \mu _2+\Delta \cdot ({\bar{Y}}-\mu _2)\cdot \mu _1 +\Delta \cdot ({\bar{X}}-\mu _1) \cdot ({\bar{Y}}- \mu _2)\Big \} \nonumber \\&\quad =\sqrt{m+n} \Big \{ {\bar{Y}}{{\widehat{T}}}^{(1)}_n -\mu _2\int _{{\mathscr {B}}}w(t)\theta _1(t)dt- \Delta \cdot ({\bar{Y}}-\mu _2)\cdot \mu _1 \Big \} \nonumber \\&\qquad -\sqrt{m+n}\Big \{{\bar{X}}{{\widehat{T}}}^{(2)}_m - \mu _1\int _{{\mathscr {B}}}w(t)\theta _2 (t)dt + \Delta \cdot ({\bar{X}}-\mu _1 )\cdot \mu _2\Big \}\nonumber \\&\qquad +\sqrt{m+n}\Delta \cdot ({\bar{X}}-\mu _1) \cdot ({\bar{Y}}- \mu _2) \nonumber \\&\quad =: I_1+I_2+I_3. \end{aligned}$$
(33)
Applying Lemma 1 yields that
$$\begin{aligned} {{\widehat{T}}}^{(1)}_n=\int _{{\mathscr {B}}}w(t)\theta _1(t)dt+O_{p}\left( \frac{1}{\sqrt{n}}\right) ,\quad {{\widehat{T}}}^{(2)}_m =\int _{{\mathscr {B}}}w(t)\theta _2(t)dt+O_{p}\left( \frac{1}{\sqrt{m}}\right) . \end{aligned}$$
Moreover, since
$$\begin{aligned}&I_1=\sqrt{m+n} \Big \{ {\bar{Y}}{{\widehat{T}}}^{(1)}_n -\mu _2\int _{{\mathscr {B}}}w(t)\theta _1(t)dt- \Delta \cdot ({\bar{Y}}-\mu _2)\cdot \mu _1 \Big \} \\&\quad =\sqrt{m+n} ({\bar{Y}}-\mu _2)\left[ {{\widehat{T}}}^{(1)}_n - \int _{{\mathscr {B}}}w(t)\theta _1(t)dt \right] \\&\qquad + \sqrt{m+n} \left[ {{\widehat{T}}}^{(1)}_n - \int _{{\mathscr {B}}}w(t)\theta _1(t)dt \right] \mu _2\\&\qquad +\sqrt{m+n}({\bar{Y}}-\mu _2)\int _{{\mathscr {B}}}w(t)\theta _1(t)dt -\sqrt{m+n}\Delta \cdot ({\bar{Y}}-\mu _2)\cdot \mu _1 \\&\quad =\sqrt{m+n} O_{p}\left( \frac{1}{\sqrt{m}}\right) O_{p}\left( \frac{1}{\sqrt{n}}\right) + \sqrt{m+n} \left[ {{\widehat{T}}}^{(1)}_n - \int _{{\mathscr {B}}}w(t)\theta _1(t)dt \right] \mu _2\\&\qquad +\sqrt{m+n}({\bar{Y}}-\mu _2) \left[ \int _{{\mathscr {B}}}w(t)\theta _1(t)dt-\Delta \cdot \mu _1 \right] \\&\quad = \sqrt{m+n} \left[ {{\widehat{T}}}^{(1)}_n - \int _{{\mathscr {B}}}w(t)\theta _1(t)dt \right] \mu _2\\&\qquad +\sqrt{m+n}({\bar{Y}}-\mu _2) \left[ \int _{{\mathscr {B}}}w(t)\theta _1(t)dt-\Delta \cdot \mu _1 \right] +o_{p}(1),\\&I_2=-\sqrt{m+n} \left[ {{\widehat{T}}}^{(2)}_m - \int _{{\mathscr {B}}}w(t)\theta _2(t)dt \right] \mu _1\\&\qquad -\sqrt{m+n}({\bar{X}}-\mu _1) \left[ \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\Delta \cdot \mu _2 \right] +o_{p}(1), \end{aligned}$$
\(I_3=o_{p}(1)\), which implies that
$$\begin{aligned}&\sqrt{m+n} {\widehat{D}}_{m,n}(\Delta ) \nonumber \\&\quad =\sqrt{m+n} \left[ {{\widehat{T}}}^{(1)}_n - \int _{{\mathscr {B}}}w(t)\theta _1(t)dt \right] \mu _2\nonumber \\&\qquad +\sqrt{m+n}({\bar{Y}}-\mu _2) \left[ \int _{{\mathscr {B}}}w(t)\theta _1(t)dt-\Delta \cdot \mu _1 \right] \nonumber \\&\qquad -\sqrt{m+n} \left[ {{\widehat{T}}}^{(2)}_m - \int _{{\mathscr {B}}}w(t)\theta _2(t)dt \right] \mu _1\nonumber \\&\qquad -\sqrt{m+n}({\bar{X}}-\mu _1) \left[ \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\Delta \cdot \mu _2 \right] +o_{p}(1)\nonumber \\&\quad =\sqrt{m+n} \Big \{\Big [ {{\widehat{T}}}^{(1)}_n \!-\! \int _{{\mathscr {B}}}w(t)\theta _1(t)dt \Big ]\mu _2\!-\!({\bar{X}}-\mu _1) \Big [ \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\Delta \cdot \mu _2 \Big ]\Big \}\nonumber \\&\qquad +\sqrt{m+n}\Big \{({\bar{Y}}-\mu _2) \Big [ \int _{{\mathscr {B}}}w(t)\theta _1(t)dt-\Delta \cdot \mu _1 \Big ]\nonumber \\&\qquad - \Big [ {{\widehat{T}}}^{(2)}_m - \int _{{\mathscr {B}}}w(t)\theta _2(t)dt \Big ]\mu _1\Big \}+o_{p}(1)\nonumber \\&\quad =: A_1+A_2+o_{p}(1). \end{aligned}$$
(34)
By (27), (28) and (29), and noting that \(A_1\) is independent of \(A_2\), we have that
$$\begin{aligned} A_1= & {} \sqrt{m+n} \Big \{\Big [ {{\widehat{T}}}^{(1)}_n \!-\! \int _{{\mathscr {B}}}w(t)\theta _1(t)dt \Big ]\mu _2\!-\!({\bar{X}}-\mu _1) \Big [ \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\Delta \cdot \mu _2 \Big ]\Big \}\\= & {} \frac{\sqrt{m+n}}{n}\left\{ \sum _{i=1}^n [W_i-EW_i]\mu _2\nonumber \right. \\&\quad \left. - \sum _{i=1}^n(X_i-\mu _1)\Big [ \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\Delta \cdot \mu _2 \Big ]\right\} +o_p(1). \end{aligned}$$
Again, write
$$\begin{aligned} \left\{ \begin{array}{ll} M_i=\mu _2W_i-X_i(\int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\Delta \cdot \mu _2 ),\\ M=\mu _2W-X( \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\Delta \cdot \mu _2), \end{array}\right. \end{aligned}$$
thus,
$$\begin{aligned} A_1 =\frac{\sqrt{m+n} }{\sqrt{n}} \frac{1}{\sqrt{n}}\sum _{i=1}^n [M_i-EM_i]+o_p(1) . \end{aligned}$$
(35)
By (28) and conditions (C1)–(C4), we can show that
$$\begin{aligned} \lim _{h_1\rightarrow 0}E[M]&=\lim _{h_1\rightarrow 0}E\left[ \mu _2W-X\left( \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\Delta \cdot \mu _2 \right) \right] \nonumber \\&= \mu _2\left[ \int _{{\mathscr {B}}}w(t)(\theta _1(t)-t\xi _t)dt\right] -\mu _1\left[ \left( \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\Delta \cdot \mu _2 \right) \right] \nonumber \\&= -\mu _2\left[ \int _{{\mathscr {B}}}w(t)t\xi _tdt\right] , \end{aligned}$$
(36)
and
$$\begin{aligned}&\lim _{h_1\rightarrow 0}E[X\cdot W]\nonumber \\&\quad =\lim _{h_1\rightarrow 0}E\left[ \int _{{\mathscr {B}}}w(t)X^2 K_1\left( \frac{t-F(X)}{h_1}\right) dt\right] -E\left[ X\int _{{\mathscr {B}}}w(t)\xi _tI(X\le \xi _t)dt\right] \nonumber \\&\quad =\int _{0}^{\infty }\int _{{\mathscr {B}}}w(t)x^2\Big [\lim _{h_1\rightarrow 0}\int _{-\infty }^{\frac{t-F(x)}{h_1}}k_1(z)dz\Big ]dtdF(x)-\int _{{\mathscr {B}}}w(t)\xi _t\theta _1(t)dt\nonumber \\&\quad =\int _{0}^{\infty }\int _{{\mathscr {B}}}w(t)x^2I(x\le \xi _t)dtdF(x)-\int _{{\mathscr {B}}}w(t)\xi _t\theta _1(t)dt\nonumber \\&\quad =\int _{{\mathscr {B}}}w(t)\Big [\int _{0}^{\xi _t}x^2dF(x)-\xi _t\theta _1(t)\Big ]dt. \end{aligned}$$
(37)
Likewise, using (29), (37) and condition (C4), we have
$$\begin{aligned}&\lim _{h_1\rightarrow 0}E[M^2] \nonumber \\&\quad =\mu _2^2\lim _{h_1\rightarrow 0}E[W^2]-\left( \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\Delta \cdot \mu _2 \right) ^2E[X^2]\nonumber \\&\qquad -2\mu _2\left( \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\Delta \cdot \mu _2\right) \lim _{h_1\rightarrow 0}E[X\cdot W] \nonumber \\&\quad =\mu _2^2\int _{0}^{\infty }\left[ \int _{{\mathscr {B}}}w(t)(x-\xi _t)I(x\le \xi _t)dt\right] ^2 dF(x)\nonumber \\&\qquad -\left( \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\Delta \cdot \mu _2 \right) ^2E[X^2]\nonumber \\&\qquad -2\mu _2\left( \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\Delta \cdot \mu _2\right) \int _{{\mathscr {B}}}w(t)\left[ \int _{0}^{\xi _t}x^2dF(x)-\xi _t\theta _1(t)\right] dt.\nonumber \\ \end{aligned}$$
(38)
It follows from (36) and (38) that
$$\begin{aligned} \text{ Var }[M]&=E[M^2] -(E[M])^2 \\&=\mu _2^2\int _{0}^{\infty }\left[ \int _{{\mathscr {B}}}w(t)(x-\xi _t)I(x\le \xi _t)dt\right] ^2 dF(x)\\&\quad -\left( \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\Delta \cdot \mu _2 \right) ^2E[X^2]\\&\quad -2\mu _2\left( \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\Delta \cdot \mu _2\right) \int _{{\mathscr {B}}}w(t)\left[ \int _{0}^{\xi _t}x^2dF(x)-\xi _t\theta _1(t)\right] dt\\&\quad -\mu _2^2\left[ \int _{{\mathscr {B}}}w(t)t\xi _tdt\right] ^2+o_p(1)\\&=\sigma _{11}^2+o_p(1). \end{aligned}$$
Now, by condition (C2) and Eq. (35), and the central limit theorem, we get
$$\begin{aligned} A_{1}=\frac{\sqrt{m+n} }{\sqrt{n}} \frac{1}{\sqrt{n}}\sum _{i=1}^n (M_i-EM_i) \xrightarrow {d}N\left( 0,\frac{c+1}{c}\sigma _{11}^2\right) . \end{aligned}$$
Similarly, we can show that
$$\begin{aligned} A_2 \xrightarrow {d}N(0,(c+1)\sigma _{22}^2), \quad \text{ as } n\rightarrow \infty . \end{aligned}$$
Based on these arguments, we then have, as \(n\rightarrow \infty \),
$$\begin{aligned} \sqrt{m+n} {\widehat{D}}_{m,n}(\Delta )=A_1+A_2+o_{p}(1) \xrightarrow {d} N(0,\sigma ^2), \end{aligned}$$
where \(\sigma ^2 =(1+c^{-1})\sigma _{11}^2 +(1+c)\sigma _{22}^2.\) \(\square \)
Lemma 4
Under the conditions (C1)–(C4), we have
$$\begin{aligned} \frac{1}{\sqrt{m+n}}\sum _{k=1}^{m+n}{\widehat{V}} _k(\Delta ) \xrightarrow {d} N(0,\sigma ^2 (t)),\quad \text{ as } n \rightarrow \infty . \end{aligned}$$
Proof
Since \(\sum _{k=1}^n({\bar{X}}-{\bar{X}}_k)=0\) and Eq. (14), for \(k=1, \ldots , n\), we have that
$$\begin{aligned} \sum _{k=1}^n{\widehat{V}} _k(\Delta )&= \sum _{k=1}^n\Big \{ (m+n)\big ({\bar{Y}}{{\widehat{T}}}^{(1)}_n-{\bar{X}}{{\widehat{T}}}^{(2)}_m-{\bar{X}}{\bar{Y}}\cdot \Delta \big )\nonumber \\&\quad -(m+n-1)\big ({\bar{Y}}{{\widehat{T}}}^{(1)}_{n-1,k}-{\bar{X}}_k{{\widehat{T}}}^{(2)}_m-{\bar{X}}_k{\bar{Y}}\cdot \Delta \big ) \Big \} \nonumber \\&= \sum _{k=1}^n \big ({\bar{Y}}{{\widehat{T}}}^{(1)}_n-{\bar{X}}{{\widehat{T}}}^{(2)}_m-{\bar{X}}{\bar{Y}}\cdot \Delta \big )\nonumber \\&\quad +(m+n-1)\sum _{k=1}^n \Big [{\bar{Y}}({{\widehat{T}}}^{(1)}_n-{{\widehat{T}}}^{(1)}_{n-1,k})-({\bar{X}}-{\bar{X}}_k){{\widehat{T}}}^{(2)}_m-({\bar{X}}-{\bar{X}}_k){\bar{Y}}\cdot \Delta \Big ] \nonumber \\&= n\cdot {\widehat{D}}_{m,n}(\Delta ) + (m+n-1)\sum _{k=1}^n \big [{\bar{Y}}({{\widehat{T}}}^{(1)}_n-{{\widehat{T}}}^{(1)}_{n-1,k})\big ] \nonumber \\&=: B_1+B_2. \end{aligned}$$
(39)
Based on Lemma 2, we have
$$\begin{aligned} B_2= & {} (m+n-1){\bar{Y}}\sum _{k=1}^n\left( {{\widehat{T}}}^{(1)}_n-{{\widehat{T}}}^{(1)}_{n-1,k}\right) \\= & {} (m+n-1) O_{p}(1) \cdot O_{p}\left( \frac{1}{nh_1^2 }\right) = O_{p}\left( \frac{1}{h_1^2 }\right) . \end{aligned}$$
Hence,
$$\begin{aligned} \sum _{k=1}^n{\widehat{V}}_k(\Delta )= B_1+B_2= n\cdot {{\widehat{D}}_{m,n}}(\Delta ) +O_{p}\left( \frac{1}{h_1^2 }\right) . \end{aligned}$$
Analogously, we can show that
$$\begin{aligned} \sum _{k=n+1}^{m+n}{\widehat{V}}_k(\Delta )= m\cdot {{\widehat{D}}_{m,n}}(\Delta ) +O_{p}\left( \frac{1}{h_2^2 }\right) . \end{aligned}$$
Now, it follows from conditions (C2), (C3) and Theorem 1 that
$$\begin{aligned} \frac{1}{\sqrt{m+n}}\sum _{k=1}^{m+n}{\widehat{V}} _k(\Delta )= & {} \frac{\sqrt{m+n}}{m+n} \left[ n\cdot {{\widehat{D}}_{m,n}}(\Delta )\right. \\&\quad \left. +O_{p}\left( \frac{1}{h_1^2 }\right) + m\cdot {{\widehat{D}}_{m,n}}(\Delta ) +O_{p}\left( \frac{1}{h_2^2 }\right) \right] \\= & {} \sqrt{m+n}{{\widehat{D}}_{m,n}}(\Delta ) + o_{p}(1) \\&\quad \xrightarrow {d} N(0,\sigma ^2). \end{aligned}$$
Thus, we reach the conclusion. \(\square \)
Lemma 5
Under the conditions (C1), (C3) and (C4), we have
$$\begin{aligned} \frac{1}{n}\sum _{k=1}^n (X_k-\mu _1)\big [{\widehat{V}}^{(1)}_k-\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\big ] =c_1 + o_{p}(1). \end{aligned}$$
Proof
First, we have the following decomposition
$$\begin{aligned}&\frac{1}{n}\sum _{k=1}^n (X_k-\mu _1)\big [{\widehat{V}}^{(1)}_k-\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\big ] \\&\quad =\frac{1}{n}\sum _{k=1}^n X_k\big [{\widehat{V}}^{(1)}_k-\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\big ]-\frac{1}{n}\sum _{k=1}^n \mu _1\big [{\widehat{V}}^{(1)}_k-\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\big ] \\&\quad =\frac{1}{n}\sum _{k=1}^n X_k{\widehat{V}}^{(1)}_k-\frac{1}{n}\sum _{k=1}^n X_k\int _{{\mathscr {B}}}w(t)\theta _1(t)dt+O_p\left( \frac{1}{\sqrt{n}}\right) \\&\quad =\frac{1}{n}\sum _{k=1}^n X_k{\widehat{V}}^{(1)}_k -\mu _1\int _{{\mathscr {B}}}w(t)\theta _1(t)dt+o_p(1). \end{aligned}$$
Note that
$$\begin{aligned} {\widehat{V}}^{(1)}_k= & {} \sum _{i\ne k} X_i\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_i)}{h_1}\right) -K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) \right] dt\\&\quad +X_k\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_n(X_k)}{h_1}\right) dt, \end{aligned}$$
then
$$\begin{aligned} \frac{1}{n}\sum _{k=1}^n X_k{\widehat{V}}^{(1)}_k&=\frac{1}{n}\sum _{k=1}^n X_k\sum _{i\ne k} X_i\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_i)}{h_1}\right) \right. \\&\left. \quad -K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) \right] dt \\&\quad +\frac{1}{n}\sum _{k=1}^n X_k^2\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_n(X_k)}{h_1}\right) dt\\&=: I_1+I_2. \end{aligned}$$
From the Taylor expansion (31) of \(K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) -K_1\left( \frac{t-F_n(X_i)}{h_1}\right) \), then
$$\begin{aligned} I_1= & {} \frac{1}{n}\sum _{k=1}^n X_k\sum _{i\ne k} X_i\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_i)}{h_1}\right) -K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) \right] dt\nonumber \\= & {} \frac{1}{n}\sum _{k=1}^n X_k\sum _{i\ne k} X_i\int _{{\mathscr {B}}}w(t) k_1\left( \frac{t-F_n(X_i)}{h_1}\right) \frac{F_n(X_i)-F_{n,k}(X_i)}{h_1}dt+o_p(1)\nonumber \\= & {} \frac{1}{n}\sum _{k=1}^n X_k\sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t) k_1\left( \frac{t-F_n(X_i)}{h_1}\right) \frac{F_n(X_i)-F_{n,k}(X_i)}{h_1}dt+o_p(1)\nonumber \\= & {} \sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t) k_1\left( \frac{t-F_n(X_i)}{h_1}\right) \frac{1}{n}\sum _{k=1}^n X_k\frac{I(X_k\le X_i)-F_n(X_i)}{(n-1)h_1}dt+o_p(1)\nonumber \\= & {} \sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t) k_1\left( \frac{t-F_n(X_i)}{h_1}\right) \frac{\theta _1(F_n(X_i))-\mu _1F_n(X_i)+o_p(1)}{(n-1)h_1}dt+o_p(1)\nonumber \\= & {} \int _{0}^{\infty }x\int _{{\mathscr {B}}}w(t) k_1\left( \frac{t-F_n(x)}{h_1}\right) \frac{\theta _1(F_n(x))-\mu _1F_n(x)}{h_1}dtdF_n(x)+o_p(1)\nonumber \\= & {} \int _{{\mathscr {B}}}\int _{-1}^{1}F_n^{-1}(y)w(t) k_1\left( \frac{t-y}{h_1}\right) \frac{\theta _1(y)-\mu _1y}{h_1}dydt+o_p(1)\nonumber \\= & {} \int _{{\mathscr {B}}}\int _{\frac{t-1}{h_1} }^{\frac{t+1}{h_1} } F_n^{-1}(t-uh_1)w(t) k_1(u)[\theta _1(t-uh_1)-\mu _1(t-uh_1)]dudt+o_p(1)\nonumber \\= & {} \int _{{\mathscr {B}}}\int _{-1}^{1}\xi _t w(t) k_1(u)(\theta _1(t)-t\mu _1)dudt+o_p(1)\nonumber \\= & {} \int _{{\mathscr {B}}}w(t)\xi _t(\theta _1(t)-t\mu _1)dt+o_p(1). \end{aligned}$$
(40)
Another part \(I_2\) can be written as
$$\begin{aligned} I_2= & {} \frac{1}{n}\sum _{k=1}^n X_k^2\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_n(X_k)}{h_1}\right) dt\nonumber \\= & {} \int _{{\mathscr {B}}}w(t)\frac{1}{n}\sum _{k=1}^n X_k^2 [I(F_n(X_k)\le t)+o_p(1)]dt\nonumber \\= & {} \int _{{\mathscr {B}}}w(t)\int _{0}^{\xi _t}x^2dF(x)dt+o_p(1). \end{aligned}$$
(41)
Thus, from Eqs. (40) and (41), we have
$$\begin{aligned}&\frac{1}{n}\sum _{k=1}^n (X_k-\mu _1)\big [{\widehat{V}}^{(1)}_k-\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\big ]\\&\quad =\int _{{\mathscr {B}}}w(t)\xi _t(\theta _1(t)-t\mu _1)dt+\int _{{\mathscr {B}}}w(t)\int _{0}^{\xi _t}x^2dF(x)dt\\&\qquad -\mu _1\int _{{\mathscr {B}}}w(t)\theta _1(t)dt+o_p(1)\\&=\int _{{\mathscr {B}}}w(t)\Big [\xi _t(\theta _1(t)-t\mu _1)+\int _{0}^{\xi _t}x^2dF(x)-\mu _1\theta _1(t)\Big ]dt+o_p(1)\\&=c_1+o_p(1). \end{aligned}$$
\(\square \)
Lemma 6
Under the conditions (C1), (C3) and (C4), we have
$$\begin{aligned} \frac{1}{n}\sum _{k=1}^n \left[ {\widehat{V}}^{(1)}_k -\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\right] ^2 \xrightarrow {p} \sigma _1^2,\quad \text{ as } n \rightarrow \infty . \end{aligned}$$
Proof
Applying Lemma 3 yields that
$$\begin{aligned}&\frac{1}{n}\sum _{k=1}^n\left[ {\widehat{V}}^{(1)}_k-\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\right] ^2 \nonumber \\&\quad =\frac{1}{n}\sum _{k=1}^n [{\widehat{V}}^{(1)}_k]^2-2\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\cdot \frac{1}{n}\sum _{k=1}^n {\widehat{V}}^{(1)}_k+\left[ \int _{{\mathscr {B}}}w(t)\theta _1(t)dt\right] ^2\nonumber \\&\quad =\frac{1}{n}\sum _{k=1}^n [{\widehat{V}}^{(1)}_k]^2-\left[ \int _{{\mathscr {B}}}w(t)\theta _1(t)dt\right] ^2+ o_{p}(1), \end{aligned}$$
(42)
where
$$\begin{aligned} {\widehat{V}}^{(1)}_k= & {} \sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_i)}{h_1}\right) -K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) \right] dt\\&\quad +X_k\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) dt. \end{aligned}$$
We first deal with the term \(\frac{1}{n}\sum _{k=1}^n [{\widehat{V}}^{(1)}_k]^2\), that is,
$$\begin{aligned}&\frac{1}{n}\sum _{k=1}^n [{\widehat{V}}^{(1)}_k]^2\nonumber \\&\quad = \frac{1}{n}\sum _{k=1}^n\left\{ \sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_i)}{h_1}\right) -K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) \right] dt \right\} ^2 \nonumber \\&\qquad +\frac{1}{n}\sum _{k=1}^nX^2_k\left[ \int _{{\mathscr {B}}}w(t)K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) dt\right] ^2 \nonumber \\&\qquad +2\frac{1}{n}\sum _{k=1}^nX_k\left[ \int _{{\mathscr {B}}}w(t)K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) dt\right] \sum _{i=1}^n X_i \int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_i)}{h_1}\right) \nonumber \right. \\&\left. \qquad -K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) \right] dt \nonumber \\&\quad =: D_1 +D_2+D_3. \end{aligned}$$
(43)
By the Taylor expansion, condition (C3) and Eq. (30), we have
$$\begin{aligned}&D_1 =\frac{1}{n}\sum _{k=1}^n\left\{ \sum _{i=1}^n X_i \int _{{\mathscr {B}}}w(t)\left[ k_1\left( \frac{t-F_n(X_i)}{h_1}\right) \frac{F_{n,k}(X_i)-F_n(X_i)}{h_1} \right. \right. \\&\left. \left. \qquad - \frac{1}{2}k^{\prime }_1\left( \frac{t-\rho _{n,k,i}}{h_1}\right) \left( \frac{F_{n,k}(X_i)-F_n(X_i)}{h_1} \right) ^2 \right] \right\} ^2dt\\&\quad =\frac{1}{n}\sum _{k=1}^n\left\{ \sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t)\left[ k_1\left( \frac{t-F_n(X_i)}{h_1}\right) \frac{F_{n,k}(X_i)-F_n(X_i)}{h_1}\right. \right. \\&\left. \left. \qquad +O_p\left( \frac{1}{(nh_1)^2}\right) \right] \right\} ^2dt\\&\quad =\frac{1}{n}\sum _{k=1}^n \left\{ \sum _{i=1}^n X_i \int _{{\mathscr {B}}}w(t)k_1\left( \frac{1-F_n(X_i)}{h_1}\right) \frac{F_{n,k}(X_i)-F_n(X_i)}{h_1}dt\right\} ^2\\&\qquad +O_{p}\left( \frac{1}{nh^3_1}\right) +O_{p}\left( \frac{1}{nh^3_1}\right) +O_{p}\left( \frac{1}{n^2h^4_1 }\right) \\&\quad =\frac{1}{n}\sum _{k=1}^n \left\{ n\int _{0}^{\infty } x\int _{{\mathscr {B}}}w(t) k_1\left( \frac{t-F_n(x)}{h_1}\right) \frac{F_{n,k}(x)-F_n(x)}{h_1} dtdF_n(x)\right\} ^2+o_{p}(1), \end{aligned}$$
where \(\rho _{n,k,i}\) is the random variable between \(F_n(X_i)\) and \(F_{n,k}(X_i)\). Moreover, it can be checked that
$$\begin{aligned} D_1= & {} n\sum _{k=1}^n \left\{ \int _{{\mathscr {B}}}\int _{{\mathscr {B}}}\int _{0}^{\infty }\int _{0}^{\infty }w(t_1)w(t_2) x_1x_2 k_1\left( \frac{t_1-F_n(x_1)}{h_1}\right) k_1\left( \frac{t_2-F_n(x_2)}{h_1}\right) \right. \\&\quad \left. \times \frac{F_{n,k}(x_1)-F_n(x_1)}{h_1} \frac{F_{n,k}(x_2)-F_n(x_2)}{h_1} dF_n(x_1) dF_n(x_2)dt_1dt_2\right\} +o_{p}(1) \\= & {} \frac{n}{(n-1)^2 h_1^2 } \left\{ \int _{{\mathscr {B}}}\int _{{\mathscr {B}}}w(t_1)w(t_2)\int _{0}^{\infty }\int _{0}^{\infty } \sum _{k=1}^n[F_n(x_1)F_n(x_2)-I(X_k\le x_1)F_n(x_2)\right. \\&\quad -I(X_k\le x_2)F_n(x_1)+I(X_k\le x_1\wedge x_2)] \\&\quad \left. \times x_1x_2k_1\left( \frac{t_1-F_n(x_1)}{h_1}\right) k_1\left( \frac{t_2-F_n(x_2)}{h_1}\right) dF_n(x_1) dF_n(x_2)dt_1dt_2\right\} +o_{p}(1) \\= & {} \frac{n^2 }{(n-1)^2 h_1^2 } \left\{ \int _{{\mathscr {B}}}\int _{{\mathscr {B}}}w(t_1)w(t_2) \int _{0}^{\infty }\int _{0}^{\infty } [F_n(x_1\wedge x_2)-F_n(x_1)F_n(x_2)] \right. \\&\quad \left. \times x_1x_2k_1\left( \frac{t_1-F_n(x_1)}{h_1}\right) k_1\left( \frac{t_2-F_n(x_2)}{h_1}\right) dF_n(x_1)dF_n(x_2)dt_1dt_2\right\} +o_{p}(1). \end{aligned}$$
Denote \(y_1=F_n(x_1)\), \(y_2=F_n(x_2)\), \(u_1=\frac{t_1-y_1}{h_1}\) and \(u_2=\frac{t_2-y_2}{h_1}\). From conditions (C1)–(C4), we can get
$$\begin{aligned} D_1= & {} \frac{n^2 }{(n-1)^2 h_1^2 } \left\{ \int _{{\mathscr {B}}}\int _{{\mathscr {B}}}w(t_1)w(t_2)\right. \\&\quad \left. \int _{-1}^{1}\int _{-1}^{1}F_n^{-1}(y_1)F_n^{-1}(y_2)k_1\left( \frac{t_1-y_1}{h_1}\right) k_1\left( \frac{t_2-y_2}{h_1}\right) \right. \\&\quad \left. \times (y_1\wedge y_2-y_1y_2) dy_1dy_2dt_1dt_2\right\} +o_{p}(1)\\= & {} \frac{n^2 }{(n-1)^2 } \left\{ \int _{{\mathscr {B}}}\int _{{\mathscr {B}}}w(t_1)w(t_2)\right. \\&\quad \left. \int _{0}^{\infty }\int _{0}^{\infty }F_n^{-1}(t_1-u_1h_1)F_n^{-1}(t_2-u_2h_1)k_1(u_1) k_1(u_2)\right. \\&\quad \left. \times [(t_1-u_1h_1)\wedge (t_2-u_2h_1)-(t_1-u_1h_1)(t_2-u_2h_1)] du_1du_2dt_1dt_2\right\} +o_{p}(1)\\= & {} \int _{{\mathscr {B}}}\int _{{\mathscr {B}}}w(t_1)w(t_2)\int _{-1}^{1}\int _{-1}^{1}\xi _{t_1}\xi _{t_2}k_1(u_1) k_1(u_2)\\&\quad \cdot ( t_1\wedge t_2-t_1\cdot t_2) du_1du_2dt_1dt_2 +o_{p}(1)\\= & {} \int _{{\mathscr {B}}}\int _{{\mathscr {B}}}w(t_1)w(t_2)\xi _{t_1}\xi _{t_2}\cdot \left[ \int _{0}^{\infty }I(F(x)\le t_1)I(F(x)\le t_2)dF(x)\right] dt_1dt_2\\&\quad -\Big [\int _{{\mathscr {B}}}w(t_1)\xi _{t_1}t_1 dt_1\Big ]\cdot \Big [\int _{{\mathscr {B}}}w(t_2)\xi _{t_2}t_2dt_2\Big ] +o_{p}(1)\\= & {} \int _{0}^{\infty }\Big [\int _{{\mathscr {B}}}w(t)\xi _{t}I(x\le \xi _t)dt\Big ]^2dF(x)-\Big [\int _{{\mathscr {B}}}w(t)t\xi _t dt\Big ]^2 +o_{p}(1). \end{aligned}$$
As \(n\rightarrow \infty \), \(\sqrt{n}[ F_n(x)-F(x) ] \xrightarrow {d} B(x)\), where B(x) is a Gaussian process, applying the Taylor expansion again, we have
$$\begin{aligned}&K_1\left( \frac{t-F(X_k)}{h_1}\right) -K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) \\&\quad = k_1\left( \frac{t-F(X_k)}{h_1}\right) \frac{F_{n,k}(X_k)-F(X_k)}{h_1}\\&\qquad - \frac{1}{2}w^{\prime }_1\left( \frac{t-\rho '_{n,k}}{h_1}\right) \left\{ \frac{F_{n,k}(X_k)-F(X_k)}{h_1}\right\} ^2 \\&\quad = O_{p}\left( \frac{1}{\sqrt{n}h_1}\right) , \end{aligned}$$
where \(\rho '_{n,k}\) is a random variable between \(F(X_k)\) and \(F_{n,k}(X_k)\). Recalling (37) in Theorem 1, we obtain
$$\begin{aligned} D_2= & {} \frac{1}{n}\sum _{k=1}^nX^2_k\left[ \int _{{\mathscr {B}}}w(t)K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) dt\right] ^2 \\= & {} \frac{1}{n}\sum _{k=1}^nX^2_k\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) dt\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) \right. \\&\quad \left. -K_1\left( \frac{t-F(X_k)}{h_1}\right) \right] dt \\&\quad +\frac{1}{n}\sum _{k=1}^nX^2_k\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) \right. \\&\quad \left. -K_1\left( \frac{t-F(X_k)}{h_1}\right) \right] dt\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F(X_k)}{h_1}\right) dt \\&\quad + \frac{1}{n}\sum _{k=1}^nX^2_k \left[ \int _{{\mathscr {B}}}w(t)K_1\left( \frac{t-F(X_k)}{h_1}\right) dt\right] ^2 \\= & {} \frac{1}{n}\sum _{k=1}^nX^2_k \left[ \int _{{\mathscr {B}}}w(t)K_1\left( \frac{t-F(X_k)}{h_1}\right) dt\right] ^2 + o_{p}(1) \\= & {} \int _{0}^{\infty }\left[ x\int _{{\mathscr {B}}}w(t)I(x\le \xi _t)dt\right] ^2dF(x) + o_{p}(1). \end{aligned}$$
Thirdly,
$$\begin{aligned} D_3= & {} \frac{2}{n}\sum _{k=1}^nX_k\left[ \int _{{\mathscr {B}}}w(t)K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) dt\right] \sum _{i=1}^n X_i \int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_i)}{h_1}\right) \nonumber \right. \\&\quad \left. -K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) \right] dt \nonumber \\= & {} \frac{2}{n}\sum _{k=1}^nX_k\left[ \int _{{\mathscr {B}}}w(t)K_1\left( \frac{t-F_n(X_k)}{h_1}\right) dt\right] \sum _{i=1}^n X_i \int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_i)}{h_1}\right) \nonumber \right. \\&\quad \left. -K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) \right] dt \nonumber \\&\quad +\frac{2}{n}\sum _{k=1}^nX_k\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) -K_1\left( \frac{t-F_n(X_k)}{h_1}\right) \right] dt \nonumber \\&\quad \times \sum _{i=1}^n X_i \int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_i)}{h_1}\right) -K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) \right] dt \nonumber \\=: & {} D_{31}+D_{32}. \end{aligned}$$
(44)
By Eq. (30) and the Taylor expansion, we have
$$\begin{aligned} D_{31}&=\frac{2}{n}\sum _{k=1}^nX_k\left[ \int _{{\mathscr {B}}}w(t)K_1\left( \frac{t-F_n(X_k)}{h_1}\right) dt\right] \sum _{i=1}^n X_i \int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_i)}{h_1}\right) \nonumber \right. \\&\left. \quad -K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) \right] dt \nonumber \\&=\frac{2}{n}\sum _{k=1}^nX_k\int _{{\mathscr {B}}}w(t) K_1\left( \frac{t-F_n(X_k)}{h_1}\right) dt\sum _{i=1}^n X_i\nonumber \\&\quad \times \int _{{\mathscr {B}}}w(t)\left\{ k_1\left( \frac{t-F_n(X_i)}{h_1}\right) \frac{F_{n,k}(X_i)-F_n(X_i)}{h_1} \right. \nonumber \\&\left. \quad - \frac{1}{2}k^{\prime }_1\left( \frac{t-\rho _{n,k,i}}{h_1}\right) \left( \frac{F_{n,k}(X_i)-F_n(X_i)}{h_1} \right) ^2 \right\} dt\nonumber \\&=\frac{2}{n}\sum _{k=1}^nX_k\int _{{\mathscr {B}}}w(t)K_1\left( \frac{t-F_n(X_k)}{h_1}\right) dt \sum _{i=1}^n X_i\nonumber \\&\quad \times \int _{{\mathscr {B}}}w(t) k_1\left( \frac{t-F_n(X_i)}{h_1}\right) \frac{F_n(X_i)-I (X_k\le X_i )}{(n-1)h_1}dt +O_{p}\left( \frac{1}{n h_1^2 }\right) \nonumber \\&= \frac{2}{n}\sum _{k=1}^nX_k\int _{{\mathscr {B}}}w(t)K_1\left( \frac{t-F_n(X_k)}{h_1}\right) dt \sum _{i=1}^n X_i\nonumber \\&\quad \times \int _{{\mathscr {B}}}w(t) k_1\left( \frac{t-F_n(X_i)}{h_1}\right) \frac{F_n(X_i)}{(n-1)h_1}dt\nonumber \\&\quad -\frac{2}{n}\sum _{k=1}^nX_k\int _{{\mathscr {B}}}w(t)K_1\left( \frac{t-F_n(X_k)}{h_1}\right) dt \sum _{i=1}^n X_i\nonumber \\&\quad \times \int _{{\mathscr {B}}}w(t) k_1\left( \frac{t-F_n(X_i)}{h_1}\right) \frac{I(X_k\le X_i )}{(n-1)h_1}dt+O_{p}\left( \frac{1}{nh_1^2 }\right) \nonumber \\&=: E_1+E_2+O_{p}\left( \frac{1}{nh_1^2 }\right) , \end{aligned}$$
(45)
where \(\rho _{n,k,i}\) is a random variable between \(F_n(X_i)\) and \(F_{n,k}(X_i)\). By conditions (C1), (C3) and Lemma 1, it can be verified that
$$\begin{aligned} E_1= & {} \frac{2}{n}\sum _{k=1}^nX_k\int _{{\mathscr {B}}}w(t)K_1\left( \frac{t-F_n(X_k)}{h_1}\right) dt \sum _{i=1}^n X_i\nonumber \\&\quad \times \int _{{\mathscr {B}}}w(t) k_1\left( \frac{t-F_n(X_i)}{h_1}\right) \frac{F_n(X_i)}{(n-1)h_1}dt \nonumber \\= & {} \frac{2}{n}\sum _{k=1}^nX_k\int _{{\mathscr {B}}}w(t)K_1\left( \frac{t-F_n(X_k)}{h_1}\right) dt \left\{ n\int _{-\infty }^{\infty }\nonumber \right. \\&\quad \left. \times \int _{{\mathscr {B}}}w(t)xk_1\left( \frac{t-F_n(x)}{h_1}\right) \frac{F_n(x)}{(n-1)h_1}dt dF_n(x) \right\} \nonumber \\= & {} \frac{2}{n-1}\sum _{k=1}^nX_k\int _{{\mathscr {B}}}w(t)K_1\left( \frac{1-F_n(X_k)}{h_1}\right) dt\nonumber \\&\quad \times \int _{{\mathscr {B}}}w(t)\int _{-1}^{1} F^{-1}_n(y)k_1\left( \frac{t-y}{h_1}\right) \frac{y}{h_1} dydt \nonumber \\= & {} 2\left\{ \int _{{\mathscr {B}}}w(t)\theta _1(t)dt+O_{p}\left( \frac{1}{\sqrt{n}}\right) \right\} \nonumber \\&\quad \times \left\{ \int _{{\mathscr {B}}}w(t) \int _{\frac{t-1}{h_1} }^{\frac{t+1}{h_1} } F^{-1}_n(t-h_1z)k_1(z)(t-h_1z) dz dt\right\} \nonumber \\= & {} 2\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\int _{{\mathscr {B}}}w(t) t\xi _t dt +O_{p}\left( \frac{1}{\sqrt{n}}\right) , \end{aligned}$$
(46)
and
$$\begin{aligned} E_2&= -\frac{2}{n}\sum _{k=1}^nX_k\int _{{\mathscr {B}}}w(t_1)K_1\left( \frac{t_1-F_n(X_k)}{h_1}\right) dt_1 \sum _{i=1}^n X_i\nonumber \\&\quad \times \int _{{\mathscr {B}}}w(t_2) k_1\left( \frac{t_2-F_n(X_i)}{h_1}\right) \frac{I(X_k\le X_i )}{(n-1)h_1}dt_2\nonumber \\&=-\frac{2}{n-1}\sum _{k=1}^nX_k\int _{{\mathscr {B}}}w(t_1) K_1\left( \frac{t_1-F_n(X_k)}{h_1}\right) dt_1 \int _{-\infty }^{\infty } x\nonumber \\&\quad \times \int _{{\mathscr {B}}}w(t_2)k_1\left( \frac{t_2-F_n(x)}{h_1}\right) \frac{I(X_k\le x )}{h_1} dt_2dF_n(x) \nonumber \\&=-\frac{2}{n-1}\sum _{k=1}^nX_k\int _{{\mathscr {B}}}w(t_1)I(X_k\le \xi _t)dt_1\nonumber \\&\quad \times \int _{{\mathscr {B}}}w(t_2)\int _{\frac{t_2-1}{h_1}}^{\frac{t_2+1}{h_1}} F_n^{-1}(t_2-uh_1)k_1(u)I(X_k\le t_2-uh_1 ) dudt_2 +o_p(1)\nonumber \\&=-2\int _{0}^{\infty }x\left[ \int _{{\mathscr {B}}}w(t_1)I(x\le \xi _t)dt_1\right] \left[ \int _{{\mathscr {B}}}w(t_2)\xi _{t_2}I(x\le t_2 )dt_2\right] dF(x)+o_p(1). \end{aligned}$$
(47)
Recalling (30), we also have that
$$\begin{aligned}&D_{32}=\frac{2}{n}\sum _{k=1}^nX_k\int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_{n,k}(X_k)}{h_1}\right) -K_1\left( \frac{t-F_n(X_k)}{h_1}\right) \right] dt \nonumber \\&\qquad \times \sum _{i=1}^n X_i \int _{{\mathscr {B}}}w(t)\left[ K_1\left( \frac{t-F_n(X_i)}{h_1}\right) -K_1\left( \frac{t-F_{n,k}(X_i)}{h_1}\right) \right] dt \nonumber \\&\quad =-\frac{2}{n}\sum _{k=1}^nX_k\int _{{\mathscr {B}}}w(t) \left\{ k_1\left( \frac{t-F_n(X_k)}{h_1}\right) \frac{F_{n,k}(X_k)-F_n(X_k)}{h_1}\nonumber \right. \\&\left. \qquad - \frac{1}{2}k^{\prime }_1\left( \frac{t-\rho _{n,k}}{h_1}\right) \left( \frac{F_{n,k}(X_k)-F_n(X_k)}{h_1} \right) ^2 \right\} dt \nonumber \\&\qquad \times \sum _{i=1}^n X_i\int _{{\mathscr {B}}}w(t) \left\{ k_1\left( \frac{t-F_n(X_i)}{h_1}\right) \frac{F_{n,k}(X_i)-F_n(X_i)}{h_1}\nonumber \right. \\&\left. \qquad - \frac{1}{2}k^{\prime }_1\left( \frac{t-\rho _{n,k,i}}{h_1}\right) \left( \frac{F_{n,k}(X_i)-F_n(X_i)}{h_1} \right) ^2 \right\} dt\nonumber \\&\quad = \frac{2}{n} nO_{p}\left( \frac{1}{nh_1}\right) nO_{p}\left( \frac{1}{nh_1}\right) = O_{p} \left( \frac{1}{nh_1^2 } \right) , \end{aligned}$$
(48)
where \(\rho _{n,k}\) is a random variable between \(F_n(X_k)\) and \(F_{n,k}(X_k)\). Now using condition (C3), together with Eqs. (44)–(48) yields that
$$\begin{aligned} D_3= & {} 2\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\int _{{\mathscr {B}}}w(t) t\xi _t dt \\&\quad -2\int _{0}^{\infty }x\left[ \int _{{\mathscr {B}}}w(t_1)I(x\le \xi _t)dt_1\right] \left[ \int _{{\mathscr {B}}}w(t_2)\xi _{t_2}I(x\le t_2 )dt_2\right] dF(x)+o_p(1). \end{aligned}$$
Finally, we combine the above component-wise results to achieve the following asymptotic result,
$$\begin{aligned}&\frac{1}{n}\sum _{k=1}^n\left[ {\widehat{V}}^{(1)}_k-\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\right] ^2 \\&\quad = \sum _{i=1}^3 D_i-\left[ \int _{{\mathscr {B}}}w(t)\theta _1(t)dt\right] ^2+o_{p}(1)\\&\quad = \int _{0}^{\infty }\Bigg [\int _{{\mathscr {B}}}w(t)\xi _{t}I(x\le \xi _t)dt\Bigg ]^2dF(x)-\Bigg [\int _{{\mathscr {B}}}w(t)t\xi _t dt\Bigg ]^2+o_{p}(1) \\&\qquad +\int _{0}^{\infty }\Bigg [x\int _{{\mathscr {B}}}w(t)I(x\le \xi _t)dt\Bigg ]^2dF(x) + o_{p}(1)\\&\qquad +2\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\int _{{\mathscr {B}}}w(t) t\xi _t dt \\&\qquad -2\int _{0}^{\infty }x\Bigg [\int _{{\mathscr {B}}}w(t_1)I(x\le \xi _t)dt_1\Bigg ]\Bigg [\int _{{\mathscr {B}}}w(t_2)\xi _{t_2}I(x\le t_2 )dt_2\Bigg ]dF(x)+o_p(1)\\&\qquad -\left[ \int _{{\mathscr {B}}}w(t)\theta _1(t)dt\right] ^2+o_{p}(1)\\&\quad =\int _{0}^{\infty }\Bigg [\int _{{\mathscr {B}}}w(t)(x-\xi _t)I(x\le \xi _t)dt\Bigg ]^2 dF(x)\\&\qquad -\Bigg [\int _{{\mathscr {B}}}w(t)(\theta _1(t)-t\xi _t)dt\Bigg ]^2+o_p(1)\\&\quad = \sigma _1^2+o_p(1). \end{aligned}$$
This completes the proof of Lemma 6. \(\square \)
Lemma 7
Under the conditions (C1)–(C4), we have
$$\begin{aligned} \frac{1}{m+n}\sum _{k=1}^{m+n} \left[ {\widehat{V}} _k(\Delta ) \right] ^2 \xrightarrow {p} \sigma ^2,\quad \text{ as } n\rightarrow \infty . \end{aligned}$$
Proof
Noting that the symmetries of \({\widehat{V}}_k(\Delta )\)’s by Eq. (13), for \(1\le k \le n\), we can write
$$\begin{aligned}&{\widehat{V}} _k(\Delta )= (m+n)\big ({\bar{Y}}{{\widehat{T}}}^{(1)}_n-{\bar{X}}{{\widehat{T}}}^{(2)}_m-{\bar{X}}{\bar{Y}}\cdot \Delta \big )\nonumber \\&\qquad -(m+n-1)\big ({\bar{Y}}{{\widehat{T}}}^{(1)}_{n-1,k}-{\bar{X}}_k{{\widehat{T}}}^{(2)}_m-{\bar{X}}_k{\bar{Y}}\cdot \Delta \big ) \nonumber \\&\quad =\left( \frac{n(m+n-1)}{n-1}-\frac{m}{n-1}\right) \big ({\bar{Y}}{{\widehat{T}}}^{(1)}_n-{\bar{X}}{{\widehat{T}}}^{(2)}_m-{\bar{X}}{\bar{Y}}\cdot \Delta \big )\nonumber \\&\qquad -(m+n-1)\big ({\bar{Y}}{{\widehat{T}}}^{(1)}_{n-1,k}-{\bar{X}}_k{{\widehat{T}}}^{(2)}_m-{\bar{X}}_k{\bar{Y}}\cdot \Delta \big ) \nonumber \\&\quad =\frac{m+n-1}{n-1}\Big [n\big ({\bar{Y}}{{\widehat{T}}}^{(1)}_n-{\bar{X}}{{\widehat{T}}}^{(2)}_m-{\bar{X}}{\bar{Y}}\cdot \Delta \big )\nonumber \\&\qquad -(n-1)\big ({\bar{Y}}{{\widehat{T}}}^{(1)}_{n-1,k}-{\bar{X}}_k{{\widehat{T}}}^{(2)}_m-{\bar{X}}_k{\bar{Y}}\cdot \Delta \big )\Big ] \nonumber \\&\qquad -\frac{m}{n-1}\big ({\bar{Y}}{{\widehat{T}}}^{(1)}_n-{\bar{X}}{{\widehat{T}}}^{(2)}_m-{\bar{X}}{\bar{Y}}\cdot \Delta \big ) \nonumber \\&\quad = \frac{m+n-1}{n-1}\left\{ {\bar{Y}}{\widehat{V}}^{(1)}_k -X_k({{\widehat{T}}}^{(2)}_m+{\bar{Y}}\Delta )\right\} -\frac{m}{n-1}{\widehat{D}}_{m,n}(\Delta ) \nonumber \\&\quad = \frac{m+n-1}{n-1}\Big \{{\bar{Y}}\big [{\widehat{V}}^{(1)}_k-\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\big ] -(X_k-\mu _1)({{\widehat{T}}}^{(2)}_m+{\bar{Y}}\Delta )\Big \} \nonumber \\&\qquad +\frac{m+n-1}{n-1}\Big \{{\bar{Y}}\int _{{\mathscr {B}}}w(t)\theta _1(t)dt -\mu _1({{\widehat{T}}}^{(2)}_m+{\bar{Y}}\Delta )\Big \} -\frac{m}{n-1}{\widehat{D}}_{m,n}(\Delta ) \nonumber \\&\quad =: H_1+H_2. \end{aligned}$$
(49)
Thus,
$$\begin{aligned} \frac{1}{m+n}\sum _{k=1}^n \left[ {\widehat{V}} _k(\Delta ) \right] ^2 =\frac{1}{m+n}\sum _{k=1}^n ( H_1+H_2 )^2. \end{aligned}$$
Recalling Eq. (8), it holds that
$$\begin{aligned} H_2= & {} \frac{m+n-1}{n-1}\Big \{{\bar{Y}}\int _{{\mathscr {B}}}w(t)\theta _1(t)dt -\mu _1({{\widehat{T}}}^{(2)}_m+{\bar{Y}}\Delta )\Big \} -\frac{m}{n-1}{\widehat{D}}_{m,n}(\Delta )\\= & {} \frac{m+n-1}{n-1}\Big \{({\bar{Y}}-\mu _2)\left[ \int _{{\mathscr {B}}}w(t)\theta _1(t)dt-\mu _1\Delta \right] \\&\quad -\mu _1\left[ {{\widehat{T}}}^{(2)}_m-\int _{{\mathscr {B}}}w(t)\theta _2(t)dt\right] \Big \} \\&\quad +\frac{m+n-1}{n-1}\Big \{\mu _2\left[ \int _{{\mathscr {B}}}w(t)\theta _1(t)dt-\mu _1\Delta \right] -\mu _1\int _{{\mathscr {B}}}w(t)\theta _2(t)dt\Big \}\\&\quad -\frac{m}{n-1}{\widehat{D}}_{m,n}(\Delta )\\= & {} \frac{m+n-1}{n-1}\left\{ ({\bar{Y}}-\mu _2)\left[ \int _{{\mathscr {B}}}w(t)\theta _1(t)dt-\mu _1\Delta \right] \right. \\&\quad \left. -\mu _1\left[ {{\widehat{T}}}^{(2)}_m-\int _{{\mathscr {B}}}w(t)\theta _2(t)dt\right] \right\} \\&\quad -\frac{m}{n-1}{\widehat{D}}_{m,n}(\Delta ). \end{aligned}$$
Now, Lemma 1 in conjunction with Theorem 1 imply that
$$\begin{aligned} {{\widehat{T}}}^{(2)}_m-\int _{{\mathscr {B}}}w(t)\theta _2(t)dt=O_{p}\left( \frac{1}{{\sqrt{m}}}\right) ,\quad {\widehat{D}}_{m,n}(\Delta )=O_{p}\left( \frac{1}{m+n}\right) . \end{aligned}$$
By Eq. (49), we can easily find \(H_2\) is not involved with k. Then
$$\begin{aligned} H_2=O_{p}\left( \frac{1}{\sqrt{m+n}}\right) ,\quad \frac{1}{m+n}\sum _{k=1}^n H_2^2 =\frac{n}{m+n} H_2^2 =O_{p}\left( \frac{1}{m+n}\right) =o_{p}(1). \end{aligned}$$
Applying Lemma 3 yields that \(\frac{1}{n}\sum _{k=1}^n{\widehat{V}}^{(1)}_k-\int _{{\mathscr {B}}}w(t)\theta _1(t)dt =O_{p}\left( \frac{1}{{\sqrt{n}}}\right) \), which in turn implies that \(\frac{1}{m+n}\sum _{k=1}^n H_1=O_{p}\left( \frac{1}{{\sqrt{m+n}}}\right) \). Notice that \(H_2\) is not involved with k, we then have
$$\begin{aligned} \frac{1}{m+n}\sum _{k=1}^n H_1H_2=H_2\left( \frac{1}{m+n}\sum _{k=1}^n H_1 \right) =O_{p}\left( \frac{1}{m+n}\right) =o_{p}(1). \end{aligned}$$
Applying Lemmas 5 and 6, we have that
$$\begin{aligned}&\frac{1}{m+n}\sum _{k=1}^n H_1^2 \\&\quad =\frac{(m+n-1)^2 }{(n-1)^2(n+m) }\sum _{k=1}^n \left\{ {\bar{Y}}\left[ {\widehat{V}}^{(1)}_k-\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\right] -(X_k-\mu _1)({{\widehat{T}}}^{(2)}_m+{\bar{Y}}\Delta ) \right\} ^2 \\&\quad =\frac{n(m+n-1)^2 }{(n-1)^2 (m+n)}{\bar{Y}}^2 \frac{1}{n}\sum _{k=1}^n \left[ {\widehat{V}}^{(1)}_k-\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\right] ^2 \\&\qquad -2\frac{n(m+n-1)^2 }{(n-1)^2 (m+n)}{\bar{Y}}({{\widehat{T}}}^{(2)}_m+{\bar{Y}}\Delta )\frac{1}{n}\sum _{k=1}^n (X_k-\mu _1)\left[ {\widehat{V}}^{(1)}_k-\int _{{\mathscr {B}}}w(t)\theta _1(t)dt\right] \\&\qquad + \frac{n(m+n-1)^2 }{(n-1)^2 (m+n)} ({{\widehat{T}}}^{(2)}_m+{\bar{Y}}\Delta )^2 \cdot \frac{1}{n}\sum _{k=1}^n(X_k-\mu _1)^2 \\&\quad =(1+c^{-1})\mu _2^2\sigma _1^2+(1+c^{-1})\left( \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\mu _2\Delta \right) ^2\cdot \text{ var }(X) \\&\qquad -2(1+c^{-1})\mu _2\left( \int _{{\mathscr {B}}}w(t)\theta _2(t)dt+\mu _2\Delta \right) c_1 +o_{p}(1)\\&\quad =(1+c^{-1})\sigma _{11}^2+o_p(1). \end{aligned}$$
Then,
$$\begin{aligned} \frac{1}{m+n}\sum _{k=1}^n \left[ {\widehat{V}} _k(\Delta ) \right] ^2= & {} \frac{1}{m+n}\sum _{k=1}^n H_1^2 +2\frac{1}{m+n}\sum _{k=1}^n H_1H_2+\frac{1}{m+n}\sum _{k=1}^n H_2^2 \\= & {} (1+c^{-1})\sigma _{11}^2+o_p(1). \end{aligned}$$
Similarly, for \(k \in \{n+1,\,n+2,\ldots ,m+n \} \), we can easily obtain
$$\begin{aligned} \frac{1}{m+n}\sum _{k=n+1}^{m+n} \left[ {\widehat{V}} _k(\Delta ) \right] ^2=(c+1)\sigma _{22}^2 +o_{p}(1). \end{aligned}$$
Therefore, by the above arguments, we have
$$\begin{aligned} \frac{1}{m+n}\sum _{k=1}^{m+n} \left[ {\widehat{V}} _k(\Delta ) \right] ^2 =(1+c^{-1})\sigma _{11}^2+(1+c)\sigma _{22}^2+o_p(1) \xrightarrow {p} \sigma ^2. \end{aligned}$$
This completes the proof. \(\square \)
Proof of Theorem 2
The proof follows directly from Lemmas 4 and 7. \(\square \)
Proof of Theorem 3
Define
$$\begin{aligned} g(\lambda )= \frac{1}{m+n} \sum _{i=1}^{m+n}\frac{{\widehat{V}}_i(\Delta ) }{ 1+\lambda {\widehat{V}}_i(\Delta ) }, \end{aligned}$$
it is easy to check that
$$\begin{aligned} 0=|g(\lambda )|= & {} \frac{1}{m+n}\left| \sum _{i=1}^{m+n} {\widehat{V}}_i(\Delta ) -\lambda \sum _{i=1}^{m+n}\frac{{\widehat{V}}_i^2(\Delta ) }{ 1+\lambda {\widehat{V}}_i(\Delta ) }\right| \\\ge & {} \frac{1}{m+n}\left| \lambda \sum _{i=1}^{m+n}\frac{{\widehat{V}}_i^2(\Delta ) }{ 1+\lambda {\widehat{V}}_i(\Delta ) }\right| -\frac{1}{m+n}\left| \sum _{i=1}^{m+n} {\widehat{V}}_i(\Delta ) \right| \\\ge & {} \frac{|\lambda |S_{m+n}}{1+|\lambda |Z_{m+n}}-\frac{1}{m+n}\left| \sum _{i=1}^{m+n}{\widehat{V}}_i(\Delta ) \right| , \end{aligned}$$
where \(S_{m+n}=\frac{1}{m+n} \sum _{i=1}^{m+n} \left[ {\widehat{V}}_i(\Delta ) \right] ^2 \), \(Z_{m+n}=\underset{1 \le i \le m+n}{\max }| {\widehat{V}}_i(\Delta )| \). By Lemmas 3 and 6, we obtain \(|\lambda |=O_{p}\left( \frac{1}{\sqrt{m+n}}\right) \). Denote \(\gamma _i=\lambda {\widehat{V}}_i(\Delta ) \), then \(\underset{1 \le i \le m+n}{\max } |\gamma _i|= o_{p}(1)\). Using the above equations and the Taylor expansion, we have
$$\begin{aligned} 0=g(\lambda )= & {} \frac{1}{m+n} \sum _{i=1}^{m+n} {\widehat{V}}_i(\Delta ) \left( 1-\gamma _i+\frac{\gamma _i^2 }{1+\gamma _i}\right) \\= & {} \frac{1}{m+n} \sum _{i=1}^{m+n} {\widehat{V}}_i(\Delta ) -S_{m+n}\lambda + \frac{1}{m+n} \sum _{i=1}^{m+n} \frac{{\widehat{V}}_i(\Delta ) \gamma _i^2 }{1+\gamma _i}\\= & {} \frac{1}{m+n} \sum _{i=1}^{m+n} {\widehat{V}}_i(\Delta ) -S_{m+n}\lambda + o_{p}\left( \frac{1}{\sqrt{m+n}}\right) , \end{aligned}$$
which implies that
$$\begin{aligned} \lambda = S_{m+n}^{-1}\frac{1}{m+n} \sum _{i=1}^{m+n} {\widehat{V}}_i(\Delta ) + \beta _n, \end{aligned}$$
where \(\beta _n=o_{p}\left( \frac{1}{\sqrt{m+n}}\right) \), hence
$$\begin{aligned} l_{m,n}(\Delta )= & {} 2\sum _{i=1}^{m+n} \log \left\{ 1+\lambda {\widehat{V}}_i(\Delta ) \right\} \\= & {} 2\sum _{i=1}^{m+n} \gamma _i-\sum _{i=1}^{m+n} \gamma _i^2 +O_{p}(\gamma _i^3) \\= & {} 2 (m+n)\lambda \frac{1}{m+n} \sum _{i=1}^{m+n} {\widehat{V}}_i(\Delta )-(m+n)S_{m+n}\lambda ^2 +o_{p}(1)\\= & {} 2 (m+n)\left[ S_{m+n}^{-1}\frac{1}{m+n} \sum _{i=1}^{m+n} {\widehat{V}}_i(\Delta ) + \beta _n\right] \frac{1}{m+n} \sum _{i=1}^{m+n} {\widehat{V}}_i(\Delta )\\&\quad -(m+n)S_{m+n}\left[ S_{m+n}^{-1}\frac{1}{m+n} \sum _{i=1}^{m+n} {\widehat{V}}_i(\Delta ) + \beta _n\right] ^2 +o_{p}(1)\\= & {} \frac{(m+n)\left\{ \frac{1}{m+n} \sum _{i=1}^{m+n}{\widehat{V}}_i(\Delta ) \right\} ^2 }{S_{m+n}}-(m+n)S_{m+n}\beta _n^2 +o_{p}(1)\\= & {} \frac{(m+n)\left\{ \frac{1}{m+n} \sum _{i=1}^{m+n}{\widehat{V}}_i(\Delta ) \right\} ^2 }{S_{m+n}}+o_{p}(1)\\&\quad \xrightarrow {d} \chi ^2_1. \end{aligned}$$
The proof of Theorem 3 is completed. \(\square \)
