Appendix
In this section, we collect and prove some lemmas, which have been used in Sect. 4.
Lemma 5.1
(Serfling 1980, Lemma A, page 95) Let \(\{Z_i\}\) be independent random variables satisfying \(P(|Z_i|\le M)=1, i\ge 1.\) Then, for any \(t>0,\)
$$\begin{aligned} P\Big (\Big |\sum _{i=1}^n(Z_i-EZ_i)\Big |>t\Big )\le 2\exp \Bigg \{-\frac{t^2}{2\left( \sum _{j=1}^n\mathrm{Var}Z_j+Mt/3\right) }\Bigg \}. \end{aligned}$$
Lemma 5.2
(Petrov 1995, Theorem 2.9, page 59) Let \(W_1,\ldots ,W_n\) be independent random variables with \(EW_k=0\)
\((k=1,2,\ldots ,n)\), and let \(p\ge 2\). Then
$$\begin{aligned} E\Big (\Big |\sum _{i=1}^nW_i\Big |^p\Big )\le C(p)\Big \{\sum ^n_{k=1}E(|W_k|^p)+\Big (\sum ^n_{k=1}EW_k^2\Big )^{p/2}\Big \}, \end{aligned}$$
where \(C(p)\) is a positive constant depending only on \(p\).
Lemma 5.3
Set \(V_n(\mathbf {X}_i)=\frac{1}{nh_{1n}^d}\sum _{i=1}^n\delta _jK_1(\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}})\) and \(V(\mathbf {X}_i)=g(\mathbf {X}_i)\Delta (\mathbf {X}_i).\) Suppose (A1)–(A3) hold. Then \(\max _{1\le i\le n}|V_n(\mathbf {X}_i)-V(\mathbf {X}_i)|=O(\gamma _{1n})\) a.s., where \(\gamma _{1n}=\big (\frac{\log n}{nh_{1n}^d}\big )^{1/2}+h_{1n}^r\rightarrow 0.\)
Proof
Let \(E_{\mathbf {X}_i}\) and \(P_{\mathbf {X}_i}\) denote conditional expectation and conditional probability given \(\mathbf {X}_i\), then
$$\begin{aligned} V_n(\mathbf {X}_i)-V(\mathbf {X}_i) =&\frac{1}{nh_{1n}^d}\sum _{j=1}^n\delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )-g(\mathbf {X}_i)\Delta (\mathbf {X}_i)\\ =&\frac{1}{nh_{1n}^d}\sum _{j=1}^n\delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big ) -E_{\mathbf {X}_i}\Big [\frac{1}{nh_{1n}^d}\sum _{j=1}^n\delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\Big ]\\&+E_{\mathbf {X}_i}\Big [\frac{1}{nh_{1n}^d}\sum _{j=1}^n\delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\Big ]-g(\mathbf {X}_i)\Delta (\mathbf {X}_i). \end{aligned}$$
Let \(\tilde{\gamma }_{1n}=(\log n/(nh_{1n}^d))^{1/2}\). For some large \(C_2>0\) we have
$$\begin{aligned}&P\Bigg (\max _{1\le i\le n}\Big |\frac{1}{nh_{1n}^d}\sum _{j=1}^n\delta _jK_1\Bigg (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Bigg ) -E_{\mathbf {X}_i}\Bigg [\frac{1}{nh_{1n}^d}\sum _{j=1}^n\delta _jK_1\Bigg (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Bigg )\Bigg ]\Big |>C_2\tilde{\gamma }_{1n}\Bigg )\nonumber \\&\le \sum _{i=1}^nE\Bigg \{P_{\mathbf {X}_i}\Bigg (\Big |\sum _{j=1}^n\Big \{\delta _jK_1\Bigg (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Bigg )-E_{\mathbf {X}_i}\Bigg [\delta _j K_1\Bigg (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Bigg )\Bigg ]\Bigg \}\Big | >C_2nh_{1n}^d\tilde{\gamma }_{1n}\Bigg )\Bigg \}. \end{aligned}$$
(5.1)
Applying Lemma 5.1, it follows that
$$\begin{aligned}&P_{\mathbf {X}_i}\Bigg (\Big |\sum _{j=1}^n\Bigg \{\delta _jK_1\Bigg (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Bigg )-E_{\mathbf {X}_i}\Bigg [\delta _j K_1\Bigg (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Bigg )\Bigg ]\Bigg \}\Big |>C_2nh_{1n}^d\tilde{\gamma }_{1n}\Bigg )\\&\quad \le 2\exp \Bigg \{-\frac{(C_2nh_{1n}^d\tilde{\gamma }_{1n})^2/2}{\sum _{j=1}^n E_{\mathbf {X}_i}\{\delta _jK_1(\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}})\}^2+MC_0nh_{1n}^d\tilde{\gamma }_{1n}/3}\Bigg \}\\&\quad =2\exp \Bigg \{-\frac{C_2^2nh_{1n}^d\tilde{\gamma }_{1n}^2/2}{\Delta (\mathbf {X}_i)g(\mathbf {X}_i)\int K_1^2(\mathbf {t})d\mathbf {t}+O(h_{1n})+MC_2\tilde{\gamma }_{1n}/3}\Bigg \}\\&\quad \le 2\exp \bigg \{-CC_2\log n\}\le Cn^{-3}, \end{aligned}$$
which, together with (5.1), yields
$$\begin{aligned} \max _{1\le i\le n}\Big |\sum _{j=1}^n\Bigg \{\delta _jK_1\Bigg (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Bigg )-E_{\mathbf {X}_i}\Bigg [\delta _jK_1\Bigg (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Bigg )\Bigg ]\Bigg \}\Big | =O(\tilde{\gamma }_{1n})~~a.s. \end{aligned}$$
By applying a Taylor expansion, we find
$$\begin{aligned}&\max _{1\le i\le n}\Big |E_{\mathbf {X}_i}\Bigg [\frac{1}{nh_{1n}^d}\sum _{j=1}^n\delta _jK_1\Bigg (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Bigg )\Bigg ]-g(\mathbf {X}_i)\Delta (\mathbf {X}_i)\Big |\\&\quad =\max _{1\le i\le n}\Big |\frac{1}{nh_{1n}^d}\sum _{j=1}^nE\Bigg \{K_1\Bigg (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Bigg )\Delta (\mathbf {X}_j)\Big |\mathbf {X}_i\Bigg \}-g(\mathbf {X}_i)\Delta (\mathbf {X}_i)\Big |\\&\quad =\max _{1\le i\le n}\Big |\int K_1(\mathbf {t})[\Delta (\mathbf {X}_i-h_{1n}\mathbf {t})g(\mathbf {X}_i-h_{1n}\mathbf {t})-g(\mathbf {X}_i)\Delta (\mathbf {X}_i)]d\mathbf {t}\Big |\\&\quad =O(h_{1n}^r)~~a.s. \end{aligned}$$
From the proof above, it follows that \(\max _{1\le i\le n}|V_n(\mathbf {X}_i)-V(\mathbf {X}_i)|=O(\gamma _{1n})~a.s.\)
\(\square \)
Lemma 5.4
Suppose (A1)–(A3) hold. Then \( \max _{1\le i\le n}|\Delta _n(\mathbf {X}_i)-\Delta (\mathbf {X}_i)|=O(\gamma _{2n})~a.s., \) where \(\gamma _{2n}=(\log n/(nh_{2n}^d))^{1/2}+h_{2n}^r\rightarrow 0.\)
Proof
Follows from the proof of Lemma 5.3 above. \(\square \)
Lemma 5.5
Let \(\widehat{b}_l\) and \(\widehat{b}_{kl}\) be defined in Section 2, and let \(\gamma _{1n}\) and \(\gamma _{2n}\) be given in Lemmas 5.3 and 5.4. Under assumptions (A1) and (A2), we have
$$\begin{aligned} \widehat{b}_l=&\frac{1}{n}\sum _{i=1}^n\frac{\delta _i\phi _l(Y_i)}{\Delta (\mathbf {X}_i)} +\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n\frac{\Delta (\mathbf {X}_i)-\delta _i}{\Delta (X_i)V(\mathbf {X}_i)} \delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\\&+O(\gamma _{2n})\frac{1}{n}\sum _{i=1}^n\frac{\delta _i|\phi _l(Y_i)|}{\Delta (\mathbf {X}_i)}\\&+O(\gamma _{1n})\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n\frac{|\Delta (\mathbf {X}_i)-\delta _i|}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} \delta _j\Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big |\\&+O(\gamma _{2n})\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n\frac{\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} \delta _j\Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big |\\&+O(\gamma _{1n})O(\gamma _{2n})\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n\frac{\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} \delta _j\Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big |\\ :=&\tilde{b}_l+B_{1l}+B_{2l}+B_{3l}+B_{4l}+B_{5l} ~~~ a.s.,\\ \widehat{b}_{kl}=&\frac{1}{n}\sum _{i=1}^n\frac{\delta _i\psi _{kl}(Y_i)}{\Delta (\mathbf {X}_i)} +\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n\frac{\Delta (X_i)-\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} \delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\psi _{kl}(Y_j)\\&+O(\gamma _{2n})\frac{1}{n}\sum _{i=1}^n\frac{\delta _i|\psi _{kl}(Y_i)|}{\Delta (\mathbf {X}_i)}\\&+O(\gamma _{1n})\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n\frac{|\Delta (\mathbf {X}_i)-\delta _i|}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} \delta _j\Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\psi _{kl}(Y_j)\Big |\\&+O(\gamma _{2n})\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n\frac{\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} \delta _j\Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\psi _{kl}(Y_j)\Big |\\&+O(\gamma _{1n})O(\gamma _{2n})\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n\frac{\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} \delta _j\Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\psi _{kl}(Y_j)\Big |\\ :=&\, \tilde{b}_{kl}+B_{1kl}+B_{2kl}+B_{3kl}+B_{3kl}+B_{4kl}+B_{5kl} ~~~ a.s. \end{aligned}$$
and \(E\tilde{b}_l=b_l\), \(E\tilde{b}_{kl}=b_{kl}.\)
Proof
We verify only the expression for \(\widehat{b}_l\), the proof for \(\widehat{b}_{kl}\) can be done similarly. Write
$$\begin{aligned} F_{n,W}(y) \!=\!\frac{1}{n}\sum _{i=1}^n\bigg \{\frac{\delta _i}{\Delta _n(\mathbf {X}_i)}I(Y_i\le y) \!+\!\bigg (1\!-\!\frac{\delta _i}{\Delta _n(\mathbf {X}_i)}\bigg )F_n(y|\mathbf {X}_i)\bigg \} :=\,F_{n1}(y)\!+\!F_{n2}(y). \end{aligned}$$
Then \( \widehat{b}_l=\int \phi _l(y)dF_{n,W}(y)=\int \phi _l(y)dF_{n1}(y)+\int \phi _l(y)dF_{n2}(y):=L_1+L_2. \) It is easy to see that \(L_1=\frac{1}{n}\sum _{i=1}^n\frac{\delta _i\phi _l(Y_i)}{\Delta _n(X_i)}.\)
From \(F_{n2}(y)=\frac{1}{n}\sum _{i=1}^n\big (1-\frac{\delta _i}{\Delta _n(\mathbf {X}_i)}\big )\frac{\sum _{j=1}^n\delta _jI(Y_j\le y)K_1(\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}})}{\sum _{k=1}^n\delta _kK_1(\frac{\mathbf {X}_i-\mathbf {X}_k}{h_{1n}})}\), we have
$$\begin{aligned} L_2=\frac{1}{n}\sum _{i=1}^n\Big (1-\frac{\delta _i}{\Delta _n(\mathbf {X}_i)}\Big ) \frac{\sum _{j=1}^n\delta _jK_1\big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\big )\phi _l(Y_j)}{\sum ^n_{k=1}\delta _kK_1\big (\frac{\mathbf {X}_i-\mathbf {X}_k}{h_{1n}}\big )}. \end{aligned}$$
Let the definition of \(V_n(\mathbf {x})\) and \(V(\mathbf {x})\) be same as in Lemma 5.3. Using Lemmas 5.3 and 5.4 we obtain
$$\begin{aligned} \widehat{b}_l =&\frac{1}{n}\sum _{i=1}^n\frac{\delta _i\phi _l(Y_i)}{\Delta _n(\mathbf {X}_i)} +\frac{1}{n}\sum _{i=1}^n\Big (1-\frac{\delta _i}{\Delta _n(\mathbf {X}_i)}\Big ) \frac{\frac{1}{nh_{1n}^d}\sum _{j=1}^n\delta _jK_1\big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\big )\phi _l(Y_j)}{\frac{1}{nh_{1n}^d}\sum _{j=1}^n\delta _jK_1\big (\frac{X_i-X_j}{h_{1n}}\big )}\\ =&\frac{1}{n}\sum _{i=1}^n\frac{\delta _i\phi _l(Y_i)}{\Delta (X_i)} +\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n\frac{\Delta (\mathbf {X}_i)-\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} \delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\\&+\frac{1}{n}\sum _{i=1}^n\frac{\delta _i\phi _l(Y_i)}{\Delta (\mathbf {X}_i)} \frac{\Delta (X_i)-\Delta _n(\mathbf {X}_i)}{\Delta _n(\mathbf {X}_i)}\\&+\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n\frac{\Delta (\mathbf {X}_i)-\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)}\frac{V(\mathbf {X}_i)-V_n(\mathbf {X}_i)}{V_n(\mathbf {X}_i)} \delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\\&+\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n\frac{\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)}\frac{\Delta _n(\mathbf {X}_i)-\Delta (\mathbf {X}_i)}{\Delta _n(\mathbf {X}_i)} \delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\\&+\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n\frac{\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)}\frac{\Delta _n(\mathbf {X}_i)-\Delta (\mathbf {X}_i)}{\Delta _n(\mathbf {X}_i)}\frac{V(\mathbf {X}_i)-V_n(\mathbf {X}_i)}{V_n(\mathbf {X}_i)} \delta _j\\&\quad K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j) :=\tilde{b}_l+B_{1l}+B_{2l}^{'}+B_{3l}^{'}+B_{4l}^{'}+B_{5l}^{'}, \end{aligned}$$
where
$$\begin{aligned} B_{2l}^{'} \le&\,\frac{\max _{1\le i\le n}|\Delta _n(\mathbf {X}_i)-\Delta (\mathbf {X}_i)|}{\inf _{\mathbf {x}\in D}\Delta (\mathbf {x}) -\max _{1\le i\le n}|\Delta _n(\mathbf {X}_i)-\Delta (\mathbf {X}_i)|}\cdot \frac{1}{n}\sum _{i=1}^n\frac{\delta _i|\phi _l(Y_i)|}{\Delta (X_i)}\\ =&\frac{O(\gamma _{2n})}{n}\sum _{i=1}^n\frac{\delta _i|\phi _l(Y_i)|}{\Delta (\mathbf {X}_i)}:=B_{2l};\\ B_{3l}^{'} \le&\,\frac{\max _{1\le i\le n}|V_n(\mathbf {X}_i)-V(\mathbf {X}_i)|}{\inf _{\mathbf {x}\in D}V(\mathbf {x})-\max _{1\le i\le n}|V_n(\mathbf {X}_i)-V(\mathbf {X}_i)|}\\&\cdot \frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n\frac{|\Delta (\mathbf {X}_i)-\delta _i|}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} \delta _j\Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big |\\ =&\frac{O(\gamma _{1n})}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n\frac{|\Delta (\mathbf {X}_i)-\delta _i|}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} \delta _j\Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big |:=B_{3l}. \end{aligned}$$
Similarly \(B_{4l}^{'} =\frac{O(\gamma _{2n})}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n\frac{\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} \delta _j\big |K_1\big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\big )\phi _l(Y_j)\big |:=B_{4l}\) and
$$\begin{aligned} B_{5l}^{'} =&\,O(\gamma _{1n})O(\gamma _{2n})\frac{1}{n^2a_d}\sum _{i=1}^n\sum _{j=1}^n\frac{\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} \delta _j\Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big |:=B_{5l}. \end{aligned}$$
In addition, under the MAR assumption, we have
$$\begin{aligned} E\tilde{b}_l =E\Bigg [\frac{1}{n}\sum _{i=1}^n\frac{\delta _i\phi _l(Y_i)}{\Delta (\mathbf {X}_i)}\Bigg ] =E\Bigg \{\frac{\phi _l(Y)}{\Delta (\mathbf {X})}E[\delta |\mathbf {X},Y]\Bigg \}=E\phi _l(Y)=b_l. \end{aligned}$$
\(\square \)
Lemma 5.6
Under the assumptions of Theorem 3.1, we have \(EB_{1l}^2=O(n^{-1}p^{-1})\) and \(EB_{1kl}^2=O(n^{-1}p_k^{-1})\), where \(B_{1l}\) and \(B_{1kl}\) are defined as in Lemma 5.5.
Proof
We evaluate only \(EB_{1l}^2,\) the evaluation for \(EB_{1kl}^2\) is similar. From Lemma 5.5 we write
$$\begin{aligned} B_{1l}=&\frac{1}{n}\sum _{i=1}^n\frac{\Delta (\mathbf {X}_i)-\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)}\frac{1}{nh_{1n}^d}\sum _{j=1}^n \Big \{\delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\\&-E\Big [\delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big |\mathbf {X}_i\Big ]\Big \}\\&+\frac{1}{n}\sum _{i=1}^n\frac{\Delta (\mathbf {X}_i)-\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)}\frac{1}{nh_{1n}^d}\sum _{j=1}^n E\Big [\delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big |\mathbf {X}_i\Big ]\\ :=&\,T_1+T_2. \end{aligned}$$
Let \(A_i=\frac{\Delta (\mathbf {X}_i)-\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)}\) and \(B_{ij}=\{\delta _jK_1(\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}})\phi _l(Y_j) -E[\delta _jK_1(\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}})\phi _l(Y_j)|\mathbf {X}_i]\}\). Then using independence of \((\mathbf {X}_i, Y_i, \delta _i)\) for \(i=1,\ldots ,n\), it is easy to see that
$$\begin{aligned} ET_1^2&=\frac{1}{n^4h_{1n}^{2d}}E\Big (\sum _{i=1}^n\sum _{j=1}^nA_iB_{ij}\Big )^2\\&=\frac{1}{n^4h_{1n}^{2d}}\Big (\sum _{i=1}^nE(A_i^2B_{ii}^2)+\sum _{i=1}^n\sum _{j=1}^nE(A_i^2B_{ij}^2)\Big ):=T_{11}+T_{12}. \end{aligned}$$
Under conditions (A1)–(A3), we have
$$\begin{aligned} T_{11}&=\frac{1}{n^4h_{1n}^{2d}}\sum _{i=1}^n E\Big \{\Big (\frac{\Delta (\mathbf {X}_i)-\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)}\Big )^2[\delta _{i}K_1(0)\phi _l(Y_{i}) -E(\delta _{i}K_1(0)\phi _l(Y_{i})|\mathbf {X}_i)]^2\Big \}\\&\le Cn^{-3}h_n^{-2d}. \end{aligned}$$
Note that
$$\begin{aligned} T_{12} =&\frac{1}{n^4h_{1n}^{2d}}\sum _{i=1}^n\sum _{j=1}^nE\Big \{\Big (\frac{\Delta (\mathbf {X}_i)-\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)}\Big )^2 \Big [\delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\\&{\quad }-E\Big (\delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big |\mathbf {X}_i\Big )\Big ]^2\Big \}\\ \le&\frac{1}{n^4h_{1n}^{2d}}\sum _{i=1}^n\sum _{j=1}^nE\Big \{\delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j) -E\Big [\delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big |\mathbf {X}_i\Big ]\Big \}^2\\ \le&\,\frac{2}{n^4h_{1n}^{2d}}\sum _{i=1}^n\sum _{j=1}^n\Big \{E\Big [\delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big ]^2\\&{\quad } +E\Big [E\Big (\delta _jK_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big |\mathbf {X}_i\Big )\Big ]^2\Big \}\\ :=&\,T_{121}+T_{122}. \end{aligned}$$
Since \(\phi \) has a compact support, and under conditions (A2), (A3) and (A5), we have
$$\begin{aligned} T_{121} =&\,\frac{2}{n^4h_{1n}^{2d}}\sum _{i=1}^n\sum _{j=1}^nE\Big \{E\Big [\delta _jK_1^2\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l^2(Y_j)\Big |\mathbf {X}_i,\mathbf {X}_j,Y_j\Big ]\Big \}\\ =&\,\frac{2}{n^4h_{1n}^{2d}}\sum _{i=1}^n\sum _{j=1}^nE\Big \{K_1^2\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l^2(Y_j)\Delta (\mathbf {X}_j)\Big \}\\ =&\,\frac{2}{n^4h_{1n}^{2d}}\sum _{i=1}^n\sum _{j=1}^nE\Big \{E\Big [K_1^2\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l^2(Y_j)\Delta (\mathbf {X}_j)\Big |\mathbf {X}_i\Big ]\Big \}\\ =&\,\frac{2}{n^3h_{1n}^{2d}}\sum _{i=1}^nE\Big \{\int \int K_1^2\Big (\frac{\mathbf {X}_i-\mathbf {x}}{h_{1n}}\Big )p\phi ^2(py-l)\Delta (\mathbf {x})\xi (\mathbf {x},y)d\mathbf {x}dy\Big \}\\ =&\,\frac{2}{n^2h_{1n}^d}E\Big \{\int \int K_1^2(\mathbf {t})\phi ^2(u)\Delta (\mathbf {X}_i-h_{1n}\mathbf {t})\xi \Big (\mathbf {X}_i-h_{1n}\mathbf {t},\frac{u+l}{p}\Big )d\mathbf {t}du\Big \}\\ \le&\, Cn^{-2}h_{1n}^{-d}. \end{aligned}$$
Similarly \(T_{122}\le Cn^{-2}p^{-1}.\)
For \(T_2\) we have
$$\begin{aligned} T_2=&\,\frac{1}{n}\sum _{i=1}^n\frac{\Delta (\mathbf {X}_i)-\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} h_{1n}^{-d}E\Big \{E\Big [\delta K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}}{h_{1n}}\Big )\phi _l(Y)\Big |\mathbf {X}_i,\mathbf {X},Y\Big ]\Big |\mathbf {X}_i\Big \}\\ =&\,\frac{1}{n}\sum _{i=1}^n\frac{\Delta (\mathbf {X}_i)-\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} h_{1n}^{-d}E\Big \{K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}}{h_{1n}}\Big )\phi _l(Y)\Delta (\mathbf {X})\Big |\mathbf {X}_i\Big \}\\ =&\,\frac{1}{np^{1/2}}\sum _{i=1}^n\frac{\Delta (\mathbf {X}_i)-\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} \!\int \int K_1(\mathbf {t})\phi (u)\Delta (\mathbf {X}_i\!-\!h_{1n}\mathbf {t})\xi \Big (\mathbf {X}_i-h_{1n}\mathbf {t},\frac{u\!+\!l}{p}\Big )d\mathbf {t}du. \end{aligned}$$
It is easy to see that under the MAR assumption
$$\begin{aligned} E\Bigg \{\frac{\Delta (\mathbf {X}_i)-\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} \int \int K_1(\mathbf {t})\phi (u)\Delta (\mathbf {X}_i-h_{1n}\mathbf {t})\xi \Big (\mathbf {X}_i-h_{1n}\mathbf {t},\frac{u+l}{p}\Big )d\mathbf {t}du\Bigg \}=0. \end{aligned}$$
Note that \(\frac{\Delta (\mathbf {X}_i)-\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} \int \int K_1(\mathbf {t})\phi (u)\Delta (\mathbf {X}_i-h_{1n}\mathbf {t})\xi (\mathbf {X}_i-h_{1n}\mathbf {t},\frac{u+l}{p})d\mathbf {t}du\)
\((0\le i\le n)\) are independent and bounded, so from conditions (A2), (A3) and (A5), we have
$$\begin{aligned} ET_2^2&=\frac{1}{n^2p}\sum _{i=1}^nE\Big \{\frac{\Delta (\mathbf {X}_i)-\delta _i}{\Delta (\mathbf {X}_i)V(\mathbf {X}_i)} \int \int K_1(\mathbf {t})\phi (u)\Delta (\mathbf {X}_i-h_{1n}\mathbf {t})\\&\qquad \xi \Big (\mathbf {X}_i-h_{1n}\mathbf {t},\frac{u+l}{p}\Big )d\mathbf {t}du\Big \}^2 \le Cn^{-1}p^{-1}. \end{aligned}$$
Since \(nh_{1n}^dp^{-1}\rightarrow \infty ,\) we collect the above calculation and obtain \(EB_{1l}^2=O(n^{-1}p^{-1})\). \(\square \)
Lemma 5.7
Under the assumptions of Theorem 3.1, we have
$$\begin{aligned} E\left\{ \frac{1}{n^2h_{1n}^d}\sum _{i=1}^n \sum _{j=1}^n\delta _j\left| K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\right| \right\} ^2=O(p^{-1}) \end{aligned}$$
and \(E\Big \{\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n \sum _{j=1}^n\delta _j\big |K_1\big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\big )\psi _{kl}(Y_j)\big |\Big \}^2=O(p_k^{-1}).\)
Proof
Write
$$\begin{aligned}&\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n \sum _{j=1}^n\delta _j\Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big |\\ =&\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n \Bigg \{\delta _j\Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big | -E\Big [\delta _j\Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big |\Big |\mathbf {X}_i\Big ]\Bigg \}\\&+\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^nE\Big [\delta _j\Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big |\Big |\mathbf {X}_i\Big ]\\ :=&P_1+P_2. \end{aligned}$$
Following the arguments as for \(T_1\) in the proof of Lemma 5.6, we have \(EP_1^2=O(n^{-2}h_{1n}^{-d}).\)
As for \(P_2\), and under conditions (A1)–(A4), we have
$$\begin{aligned} P_2=&\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n E\Big [\delta _j\Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y)\Big |\Big |\mathbf {X}_i\Big ]\\ =&\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n E\Big \{\Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big |E[\delta _j|\mathbf {X}_i,\mathbf {X}_j,Y_j]|\mathbf {X}_i\Big \}\\ =&\frac{1}{n^2h_{1n}^d}\sum _{i=1}^n\sum _{j=1}^n E\Big \{\Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {X}_j}{h_{1n}}\Big )\phi _l(Y_j)\Big |\Delta (\mathbf {X}_j)\Big |\mathbf {X}_i\Big \}\\ =&\frac{1}{nh_{1n}^d}\sum _{i=1}^n \int \int \Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {x}}{h_{1n}}\Big )\phi _l(y)\Big |\Delta (\mathbf {x})\xi (\mathbf {x},y)d\mathbf {x}dy. \end{aligned}$$
Therefore, we can get
$$\begin{aligned} EP_2^2=&E\Big \{\frac{1}{nh_{1n}^d}\sum _{i=1}^n\int \int \Big |K_1\Big (\frac{\mathbf {X}_i-\mathbf {x}}{h_{1n}}\Big )\phi _l(y)\Big |\Delta (\mathbf {x})\xi (\mathbf {x},y)d\mathbf {x}dy\Big \}^2\\ =&E\Big \{\frac{1}{n}\sum _{i=1}^np^{-1/2}\int \int |K_1(\mathbf {t})\phi (u)|\Delta (\mathbf {X}_i-h_{1n}\mathbf {t})\xi \Big (\mathbf {X}_i-h_{1n}\mathbf {t},\frac{u+l}{p}\Big )d\mathbf {t}dy\Big \}^2\\ \le&\,\frac{n}{n^2}\sum _{i=1}^np^{-1}E\Big \{\int \int |K_1(\mathbf {t})\phi (u)|\Delta (\mathbf {X}_i-h_{1n}\mathbf {t})\xi \Big (\mathbf {X}_i-h_{1n}\mathbf {t},\frac{u+l}{p}\Big )d\mathbf {t}dy\Big \}^2\\ =&\,p^{-1}\int \Big \{\int \int |K_1(\mathbf {t})\phi (u)|\Delta (\mathbf {x}-h_{1n}\mathbf {t})\xi \Big (\mathbf {x}-h_{1n}\mathbf {t},\frac{u+l}{p}\Big )d\mathbf {t}dy\Big \}^2g(\mathbf {x})d\mathbf {x}\\ =&\,O(p^{-1}). \end{aligned}$$
Since \(nh_{1n}^dp^{-1}\rightarrow \infty \), we get the result. \(\square \)
